Lambda Polarization in Polarized Proton-Proton Collisions at RHIC
a r X i v :h e p -p h /0002081v 1 8 F e
b 2000
ADP-99-53/T389IU/NTC 00-02
Lambda Polarization in Polarized Proton-Proton Collisions at
RHIC
C.Boros,J.T.Londergan 1and A.W.Thomas
Department of Physics and Mathematical Physics,and Special Research Center for the Subatomic
Structure of Matter,University of Adelaide,Adelaide 5005,Australia
1Department of Physics and Nuclear Theory Center,Indiana University,Bloomington,IN
47408,USA (February 1,2008)
Abstract
We discuss Lambda polarization in semi-inclusive proton-proton collisions,with one of the protons longitudinally polarized.The hyper?ne interaction responsible for the ?-N and Σ-Λmass splittings gives rise to ?avor asym-metric fragmentation functions and to sizable polarized non-strange fragmen-tation functions.We predict large positive Lambda polarization in polarized proton-proton collisions at large rapidities of the produced Lambda,while other models,based on SU (3)?avor symmetric fragmentation functions,pre-dict zero or negative Lambda polarization.The e?ect of Σ0and Σ?decays is also discussed.Forthcoming experiments at RHIC will be able to di?erentiate between these predictions.
I.INTRODUCTION
Measurements of the polarization dependent structure function,g1,in deep inelastic scattering[1]have inspired considerable experimental and theoretical e?ort to understand the spin structure of baryons.While most of these studies concern the spin structure of the nucleons,it has become clear that similar measurements involving other baryons would provide helpful,complementary information[2–9].The Lambda baryon plays a special role in this respect.It is an ideal testing ground for spin studies since it has a rather simple spin structure in the naive quark parton model.Furthermore,its self-analyzing decay makes polarization measurements experimentally feasible.
Forthcoming experiments at RHIC could measure the polarization of Lambda hyperons produced in proton-proton collisions with one of the protons longitudinally polarized,p↑p→Λ↑X.The polarization dependent fragmentation function of quarks and gluons into Lambda hyperons can be extracted from such experiments.These fragmentation functions contain information on how the spin of the polarized quarks and gluons is transferred to the?nal state Lambda.The advantage of proton proton collisions,as opposed to e+e?annihilation, whereΛproduction and polarization is dominated by strange quark fragmentation,is that Lambdas at large positive rapidity are mainly fragmentation products of up and down valence quarks of the polarized projectile.Thus,the important question,intimately related to our understanding of the spin structure of baryons,of whether polarized up and down quarks can transfer polarization to the Lambda can be tested at RHIC[5].
In a previous publication,we have shown that the hyper?ne interaction,responsible for the?-N andΣ0-Λmass splittings leads to non-zero polarized non-strange quark fragmen-tation functions[14].These non-zero polarized up and down quark fragmentation functions give rise to sizeable positiveΛpolarization in experiments where the strange quark fragmen-tation is suppressed.On the other hand,predictions based either on the naive quark model or on SU(3)?avor symmetry predict zero or negative Lambda polarization[2].
In section II,we brie?y discuss fragmentation functions and show how the hyper?ne interaction leads to polarized non-strange fragmentation functions.We?x the parameters of the model by?tting the data onΛproduction in e+e?annihilation.In section III,we discussΛproduction in pp collisons at RHIC energies.We point out that the production ofΛ’s at high rapidities is dominated by the fragmentation of valence up and down quarks of the polarized projectile,and is ideally suited to test whether non-strange quarks transfer their polarization to the?nal stateΛ.We predict signi?cant positiveΛ-polarization at large rapidities of the producedΛ.
II.FRAGMENTATION FUNCTIONS
Fragmentation functions can be de?ned as light-cone Fourier transforms of matrix ele-ments of quark operators[12,13]
1
4 n dξ?ψ(ξ?)|0 },(1)
where,Γis the appropriate Dirac matrix;P and p n refer to the momentum of the produced Λand of the intermediate system n;S is the spin of the Lambda and the plus projections
of the momenta are de?ned by P+≡12(P0+P3).z is the plus momentum fraction of the quark carried by the producedΛ.
Translating the matrix elements,using the integral representation of the delta function and projecting out the light-cone plus and helicity±components we obtain
1
2
√
2
γ?γ+1√
z D±ˉqΛ(z)=
1
2
n
δ[(1/z?1)P+?p+n]| 0|ψ?±+(0)|Λ(P S );n(p n) |2.(3)
D±qΛcan be interpreted as the probability that a quark with positive/negative helicity frag-ments into aΛwith positive helicity and similarly for antiquarks.
The operatorψ+(ψ?+)either destroys a quark(an antiquark)or it creates an antiquark (quark)when acting on theΛon the right hand side in the matrix elements.Thus,whereas, in the case of quark fragmentation,the intermediate state can be either an anti-diquark state,ˉqˉq,or a four-quark-antiquark state,qˉqˉqˉq,in the case of antiquark fragmentation,only four-antiquark states,ˉqˉqˉqˉq,are possible assuming that there are no antiquarks in theΛ. (Production ofΛ’s through coupling to higher Fock states of theΛis more complicated and involves higher number of quarks in the intermediate states.As a result it would lead to contributions at lower z values.)Thus,we have
?(1a)q→qqq+ˉqˉq=Λ+ˉqˉq
?(1b)q→qqq+qˉqˉqˉq=Λ+qˉqˉqˉq,
for the quark fragmentation and
?(2)ˉq→qqq+ˉqˉqˉqˉq=Λ+ˉqˉqˉqˉq,
for the antiquark fragmentation.
While,in case(1a),the initial fragmenting quark is contained in the produced Lambda, in case(1b)and(2),the Lambda is mainly produced by quarks created in the fragmentation process.Therefore,we not only expect that Lambdas produced through(1a)usually have larger momenta than those produced through(1b)or(2)but also that Lambdas produced through(1a)are much more sensitive to the?avor spin quantum numbers of the fragment-ing quark than those produced through(1b)and(2).In the following we assume that(1b) and(2)lead to approximately the same fragmentation functions.In this case,the di?er-ence,D qΛ?DˉqΛ,responsible for leading particle production,is given by the fragmentation functions associated with process(1a).
Similar observations also follow from energy-momentum conservation built in Eqs.(2) and(3).The delta function implies that the function,D q(z)/z,peaks at[14]
z max≈M
Here,M and M n are the mass of the produced particle and the produced system,n,and we
work in the rest frame of the produced particle.We see that the location of the maxima of the fragmentation function depends on the mass of the system n.While the high z region is
dominated by the fragmentation of a quark into the?nal particle and a small mass system, large mass systems contribute to the fragmentation at lower z values.The maxima of the
fragmentation functions from process(1a)are given by the mass of the intermediate diquark
state and that of the the fragmentation functions from the processes(1b)and(2b)by the masses of intermediate four quark states.Thus,the contribution from process(1a)is harder
than those from(1b)and(2).
Energy-momentum conservation also requires that the fragmentation functions are not ?avor symmetric.While the assertion of isospin symmetry,D uΛ=D dΛ,is well justi?ed,
SU(3)?avor symmetry is broken not only by the strange quark mass but also by the hyper?ne interaction.Let us discuss the fragmentation of a u(or d)quark and that of an s quark into
a Lambda through process(1a).While the intermediate diquark state is always a scalar
in the strange quark fragmentation,it can be either a vector or a scalar diquark in the fragmentation of the non-strange quarks.The masses of the scalar and vector non-strange
diquarks follow from the mass di?erence between the nucleon and the Delta[10],while those of the scalar and vector diquark containing a strange quark can be deduced from the mass
di?erence betweenΣandΛ[11].They are roughly m s≈650MeV and m v≈850MeV for the scalar and vector non-strange diquarks,and m′s≈890MeV and m′v≈1010MeV for scalar and vector diquarks containing strange quarks,respectively[11,14].According to
Eq.(4),these numbers lead to soft up and down quark fragmentation functions and to hard
strange quark fragmentation functions.
Energy-momentum conservation,together with the splitting of vector and scalar diquark
masses,has the further important consequence that polarized non-strange quarks can trans-fer polarization to the?nal state Lambda.To see this we note that the probabilities for the intermediate state to be a scalar or vector diquark state in the fragmentation of an up or down quark with parallel or anti-parallel spin to the spin of the Lambda can be obtained from the SU(6)wave function of theΛ
Λ↑=1
3
[2s↑(ud)0,0+
√
2u↓(ds)1,1+u↑(ds)1,0?u↑(ds)0,0].(5)
While the u or d quarks with spin anti-parallel to the spin of theΛare always associated with a vector diquark,u and d quarks with parallel spin have equal probabilities to be accompanied by a vector or scalar diquark.The fragmentation functions of non-strange quarks with spin parallel to theΛspin are harder than the corresponding fragmentation functions with anti-parallel spins.Thus,?D uΛis positive for large z values and negative for small z.Their total contribution to polarized Lambda production might be zero or very small.Nevertheless,?D uΛand?D dΛcan be sizable for large z values,since both D uΛand ?D uΛare dominated by the spin-zero component in the large z limit.Furthermore,they will dominate polarized Lambda production whenever the production from strange quarks is suppressed.
The matrix elements can be calculated using model wave functions at the scale relevant to the speci?c model and the resulting fragmentation functions can be evolved to a higher scale
to compare them to experiments.In a previous paper[14],we calculated the fragmentation functions in the MIT bag model and showed that the resulting fragmentation functions
give a very reasonable description of the data in e+e?annihilation.Since the mass of the intermediate states containing more than two quarks are not known we only calculate the
contributions of the diquark intermediate states in the bag model.The other contributions have been determined by performing a global?t to the e+e?data.For this,we used the simple functional form
DˉqΛ(z)=Nˉq zα(1?z)β(6) to parameterize DˉqΛ=DˉuΛ=DˉdΛ=...DˉbΛand also set D gΛ=0at the initial scale,μ=0.25GeV.
The fragmentation functions have to be evolved to the scale of the experiment,μ.The evolution of the non-singlet fragmentation functions in LO is given by[15,16]
d
z′P qq(
z
d lnμ2 q D qΛ(z,μ2)= 1
z
dz′
z′
) q D qΛ(z,μ2)+2n f P gq(z
d lnμ2
D gΛ(z,μ2)= 1z dz′z′) q D qΛ(z,μ2)+P gg(z
so that one has to select certain kinematic regions to suppress the unwanted contributions.In particular,in order to test whether polarized up and down quarks do fragment into polarized Lambdas the rapidity of the produced Lambda has to be large,since at high rapidity,Λ’s are mainly produced through valence up and down quarks.(We count positive rapidity in the direction of the polarized proton beam.)
The di?erence of the cross sections to produce a Lambda with positive helicity through the scattering of a proton with positive/negative helicity on an unpolarized proton is given
in leading order perturbative QCD(LO pQCD)by
1
E C ?dσ
d3p C
(A↑B→C↑+X)?E C
dσ
πz2c
?dσ
se?y/z c,?u=?x b p⊥
√
s is the total center of mass energy.The summation in Eq.(10)runs over all possible parton-parton combinations,qq′→qq′,qg→qg,qˉq→qˉq...The elementary unpolarized and polarized cross sections can be found in Refs.[24,25].Performing the integration in Eq.
(10)over z c one obtains
E C
?dσ
πz c
?dσ
2x b e?y+
x⊥
2x a?x⊥e y,x amin=
x⊥e y
1Since the relevant spin dependent cross sections on the parton level are only known in LO we perform a LO calculation here.
√
where x⊥=2p⊥/
2While there is only one integration variable,x a,in inclusive jet production,once p⊥and y are ?xed,both x a and x b have to be integrated over the allowed kinematic region in inclusive Lambda production,since the producedΛcarries only a fraction of the parton’s momentum.
predicted Lambda polarization is shown in Fig5a.It is positive at large rapidities where the contributions of polarized up and down quarks dominates the production process.At smaller rapidities,where x a is small,strange quarks also contribute.However,since the ratios of the polarized to the unpolarized parton distributions are small at small x a the Lambda polarization is suppressed.The result also depends on the parameterization of the polarized quark distributions.In particular,the polarized gluon distribution is not well con-strained.However,it is clear from the kinematics that the ambiguity associated with the polarized gluon distributions only e?ects the results at lower rapidities.This can be seen in Fig.5b where we plot the contribution from gluons,up plus down quarks and strange quarks to the Lambda polarization.
Next,we contrast our prediction with the predictions of various SU(3)?avor symmetric models which use
D uΛ=D dΛ=D sΛ.(14) We?tted the cross sections in e+e?annihilation using Eq.(14)and the functional form given in Eq.(6).For the polarized fragmentation functions,we discuss two di?erent scenarios: The model,SU(3)A(c.f.Fig.5a),corresponds to the expectations of the naive quark model that only polarized strange quarks can fragment into polarized Lambdas
?D uΛ=?D dΛ=0?D sΛ=D sΛ.(15) It gives essentially zero polarization because the strange quarks contribute at low rapidities where the polarization is suppressed.Model,SU(3)B(c.f.Fig.5a),which was proposed in Ref.[2],is based on DIS data,and sets
?D uΛ=?D dΛ=?0.20D uΛ?D sΛ=0.60D sΛ.(16) This model predicts negative Lambda polarization.
Finally,we address the problem of Lambdas produced through the decay of other hyper-ons,such asΣ0andΣ?.In order to estimate the contribution of hyperon decays we assume, in the following,that
(1)theΛ’s produced through hyperon decay inherit the momentum of the parent hyperon
(2)and that the total probability to produceΛ,Σ0orΣ?from a certain uds state is given by the SU(6)wave function and is independent of the mass of the produced hyperon.
Further,in order to estimate the polarization transfer in the decay process we use the constituent quark model.The polarization can be obtained by noting that the boson emitted in both theΣ0→Λγand theΣ?→Λπdecay changes the angular momentum of the nonstrange diquark from J=1to J=0,while the polarization of the spectator strange quark is unchanged.Then,the polarization of theΛis determined by the polarization of the strange quark in the parent hyperon,since the polarization of theΛis exclusively carried by the strange quark in the naive quark model.
First,let us discuss the case when the parent hyperon is produced by a strange quark. Since the strange quark is always accompanied by a vector ud diquark,in bothΣ0andΣ?the fragmenation functions of strange quarks into these hyperons are much softer than the corresponding fragmentation function into aΛ.Thus,in the high z limit,the contributions from the processes,s→Σ0→Λand s→Σ?→Λ,are negligible compared to the direct
production,s→Λ.Furthermore,both channels,s→Σ0→Λand s→Σ?→Λ,enhance the already positive polarization from the direct channel,s→Λ.
This is di?erent in the case when the parent hyperon is produced by an up or down quark. BothΛandΣ0can be produced by an up(down)quark and a scalar ds(us)diquark—a process which dominates in the large z limit.(The component with a vector diquark can be neglected in this limit).Furthermore,the up and down fragmentation function of theΣ?are as important as those of theΛandΣ0in the large z limit.This is because the u fragmentation function ofΣ?peaks at about1385/(1010+1385)≈0.58which is almost the same as the peak of the scalar components of theΛandΣ,which are1115/(890+1115)≈0.57and 1190/(890+1190)≈0.57,respectively.Thus,for the up and down quark fragmentation,it is important to include theΛ’s from these decay processes.
The relevant probabilities to produce aΛwith positive and negative polarization from a fragmenting up quark with positive polarization and an ds diquark are shown in Table III. We assumed that all spin states of the ds diquark are produced with equal probabilities. The?nal weights which are relevant in the large z limit are set in bold.We?nd that if we include all channels,which survive in the large z limit,the polarization of theΛis reduced by a factor of10/27compared to the case where only the directly producedΛ’s are included. Since theΣ?decay is a strong decay it is sometimes included in the fragmentation function of theΛ.Including onlyΣ?,the suppression factor we obtain is49/81.(Note that our model predicts that u↑(ds)0,0→Λ,u↑(ds)0,0→Σ0and u↑(ds)1,0→Σ?have approximately the same z dependence and are approximately equal(up to the Clebsch-Gordon factors)since the ratios,M/(M+M n),have roughly the same numerical values.Thus,the e?ect of the Σ0andΣ?decays can be taken into account by a multiplicative factor.) In order to illustrate the e?ect of these decays on the?nalΛpolarization,we multiplied our results with these factors.The results are shown in Fig.6b as dotted lines.We note that our implementation of this correction relies on the assumptions that the producedΛcarries all the momentum of the parent hyperon and that all states are produced with equal probabilities.Since neither of these assumptions is strictly valid,we tend to overestimate the importance of hyperon decays.Note also that the inclusion ofΣ0decay in the SU(3) symmetric models makes the resulting polarization more negative.As a result,even if e?ects ofΣdecays are included,large discrepancies still persist between our predictions and those of SU(3)symmetric models.
IV.CONCLUSIONS
Measurements of the Lambda polarization at RHIC would provide a clear answer to the question of whether polarized up and down quarks can transfer polarization to the?nal state Λ.We predict positive Lambda polarization at high rapidities,in contrast with models based on SU(3)?avor symmetry and DIS which predict zero or negative Lambda polarization.Our prediction is based on the same physics which led to harder up than down quark distributions in the proton and to the?-N andΣ-Λmass splittings.We also estimated the importance ofΣ0andΣ?decays which tend to reduce the predictedΛpolarization.
ACKNOWLEDGMENTS
This work was partly supported by the Australian Research Council.One of the authors [JTL]was supported in part by National Science Foundation research contract PHY-9722706. One author[CB]wishes to thank the Indiana University Nuclear Theory Center for its hospitality during the time part of this work was carried out.
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TABLES
TABLE I.Fit parameters obtained by?tting the e+e?data.We also parametrized D qΛ?DˉqΛand?D qΛ??DˉqΛ,calculated in the bag.
D sΛ?DˉsΛDˉqΛ?D uΛ??DˉuΛ
5.81×10999.76?
6.25×1010
21.551.2532.48
13.6011.6027.72
——0.52
TABLE II.Fit parameters obtained by?tting the e+e?data and asumming that the fragmen-tation fucntions are?avor symmetric.
D qΛ?DˉqΛ
N99.76
7.47
β11.60
TABLE III.Di?erent channels for the production ofΛhyperons from a positively polarized up quark and a ds diquark.It is assumed that all spin states of the ds diquark are produced with the same probabilities.Σ?↑andΣ??stand for the1/2and3/2spin component of theΣ?.See text for further details.
u(ds)states u↑(ds)0,0u↑(ds)1,1u↑(ds)1,0u↑(ds)1,?1
41
4
1
productsΛ↑Σ0↑Σ?0↑Σ?0?Λ↑Σ0↑Σ?0↑Λ↓Σ0↓Σ?0↓
43
4
1
3
1
6
1
decay productsΛ↑Λ↑Λ↓Λ↑Λ↓Λ↑Λ↓Λ↑Λ↓
3301012
33
33
12
?nal weights1
1680101011
918
8
11
3618
FIGURES
(1/σh ) (d σ/d x E )
1010.10.01
1010.11010.11010.11010.1x E
FIG.1.Inclusive Lambda production in e +e ?annihilation.The solid lines are the result of the global ?t.They contain two parts,the ?xed contributions from D q Λ?D ˉq Λcalculated in the bag (dashed line only shown for the Aleph data)and D ˉq Λobtained from the ?t (dash-dotted line).x E
is de?ned as x E =2E Λ/
√
s is the total center of mass enery.
z FIG.2.Fragmentation functions.The solid and dashed lines stand for the calculated fragmen-
tation functions of up and strange quarks into Lambda baryons through production of a Lambda
and an anti-diquark and correspond to D uΛ?DˉuΛand D sΛ?DˉsΛ,respectively.The dash-dotted line represents the contributions from higher intermediate states,and is obtained by?tting the
e+e?data and corresponds to DˉqΛ.The short dashed line is the gluon fragmentation function. The light and heavy lines are the fragmentation functions at the scales Q2=μ2and Q2=M2Z,
respectively.Note that D gΛ=0at Q2=μ2.
FIG.3.z c as a function of x a and y for two di?erent transverse momenta,p⊥=10GeV(left) and p⊥=30GeV(right)and for two di?erent values of x b,x b=x bmin+0.01(top)and and x b=x bmin+0.1(bottom).
10
101010101010E d 3
σ/d p 3
(m b /G e V 2
)
(a)
→ qq’→
qq –
→ qq –
→ qg –
→ gg
→ qq –
→ gg 10
10101010101010
E d 3
σ/d p 3
(m b /G e V 2
)
(b)
FIG.4.Contributions from the various channels (a)to the inclusive Lambda production cross section (pp →Λ+X )and (b)to the inclusive jet production cross section (pp →jet +X )at
p ⊥=10GeV (left)and p ⊥=30GeV (right)at
√
y https://www.docsj.com/doc/1310660765.html,mbda polarization at RHIC.(a)The solid line represents our prediction.The pre-dictions of SU(3)symmetric fragmentation models are shown for comparision.The model labeled as SU(3)A is based on the quark model expectation that only the polarized strange quark may fragment into polarized Lambdas,while SU(3)B,is based on DIS data.(b)Contributions of dif-ferent?avors to theΛ-polarization.The light dashed,dash-dotted and heavy dashed lines stand for the contributions from up plus down,from strange and from gluon fragmentation,respectively, as calculated here.The estimated polarization including bothΣ0andΣ?(lower dotted line)and
onlyΣ?(upper dotted line)decays are also shown.See text for further details.
化学平衡常数表达式的书写
化学平衡常数表达式的书写 1、写出铁与水蒸汽反应的化学方程式，如果它是一可逆反应，请写出其平衡常数表达式 2、写出工业上制水煤气的反应方程式，如它是一可逆反应，请写出其平衡常数表达式； 3、写出工业上合成氨的反应方程式，如它是一可逆反应，请写出其平衡常数表达式； 4、写出氨催化氧化成一氧化氮的反应方程式，如它是一可逆反，请写出其平衡常数表达式； 5、HAC + H2O H3O+ + AC—这是醋酸的电离方程式，请写出其电离平衡常数表达式 6、写出碳酸根离子水解的离子方程式，并写出其水解平衡常数的表达式； 7、2CrO42—+2 H+Cr2O72—+ H2O，写出其平衡常数表达式；8、写出乙酸与乙醇的酯化反应方程式，并写出其平衡常数表达式；9、写出乙酸乙酯在酸性环境下水解的反应方程式，并写出其平衡常数表达式；10、如果在常温下的饱和氯化钠溶液中，通入大量的氯化氢气体，有什么现象？你能用平衡 移动原理来解释这个现象吗？请写出其平衡的方程式，并写出其常数表达式。 11、写出氢氧化铝沉淀与水的混和体系中的各种平衡的方程式；并写出其对应的平衡常数表达 式； 化学平衡常数的计算 1、298K时，K sp[Ce(OH)4]＝1×10—29。Ce(OH)4的溶度积表达式为K sp＝ ____________ 。 为了使溶液中Ce4＋沉淀完全，即残留在溶液中的c(Ce4+)小于1×10—5mol·L－1，需调节pH为 ______ 以上。
2、某温度下，将2.0 mol CO和6.0 molH2充入2 L的密闭容器中，CO(g)+2H2(g) CH3OH(g) 充分反应后，达到平衡时测得c(CO)=0.25 mol/L，则CO的转化率=__ ___，此温度下的平衡常数K=___ __(请写出计算过程，保留二位有效数字)。 3、PCl5分解成PCl3和Cl2的反应是可逆反应。T℃时，向2.0 L恒容密闭容器中充入1.0 mol PCl5，经过250 s达到平衡。反应过程中测定的部分数据见下表： t / s050150250350 n(PCl3) / mol00. 160. 190. 200. 20 3 ②试计算该温度下反应的平衡常数(写出计算过程，保留 2 位有效数字) 4、不同温度下，向装有足量I2O5固体的2 L 恒容密闭容器中通入2molCO，5CO(g)＋I2O5 (s) 5CO2(g)＋I2(s)测得CO2的体积分数φ(CO2) 随时间t 变化曲线如右图。请回答： ①从反应开始至 a 点时的反应速率为v(CO)＝，b 点 时化学平衡常数K b＝。 5、对反应CO(g) + H2O(g) CO2 (g)+ H2(g) ΔH 2 = -41 kJ/mol，起始时在密闭容器中充 入 1.00 molCO 和 1.00 molH2O ，分别进行以下实验，探究影响平衡的因素(其它条件相同且不考 实验①中c(CO2)随时间变化的关系见下图，实验编号容器体积/L温度/°C 在与实验①相同的条件下，起始时充入① 2.01200 容器的物质的量：n(CO)=n(H2O)=n(CO2)② 2.01300 =n( H2)=1.00mol 。③ 1.01200 通过计算，判断出反应进行的方向。(写出计算过程。)
(完整版)数学表达式计算(c语言实现)
一、设计思想 计算算术表达式可以用两种方法实现： 1.中缀转后缀算法 此算法分两步实现：先将算术表达式转换为后缀表达式，然后对后缀表达式进行计算。具体实现方法如下： （1）中缀转后缀 需要建一个操作符栈op和一个字符数组exp，op栈存放操作符，字符数组用来存放转换以后的后缀表达式。首先，得到用户输入的中缀表达式，将其存入str数组中。 对str数组逐个扫描，如果是数字或小数点，则直接存入exp数组中，当扫描完数值后，在后面加一个#作为分隔符。 如果是操作符，并且栈为空直接入栈，如果栈不为空，与栈顶操作符比较优先等级，若比栈顶优先级高，入栈；如果比栈顶优先级低或相等，出栈将其操作符存到exp数组中，直到栈顶元素优先等级低于扫描的操作符，则此操作符入栈；如果是左括号，直接入栈，如果是右括号，出栈存入exp数组，直到遇到左括号，左括号丢掉。然后继续扫描下一个字符，直到遇到str中的结束符号\0，扫描结束。结束后看op栈是否为空，若不为空，继续出栈存入exp数组中，直到栈为空。到此在exp数组最后加结束字符\0。 我们就得到了后缀表达式。 （2）后缀表达式计算 此时需要一个数值栈od来存放数值。对exp数组进行逐个扫描，当遇到数字或小数点时，截取数值子串将其转换成double类型的小数，存入od栈中。当遇到操作符，从栈中取出两个数，进行计算后再放入栈中。继续扫描，知道扫描结束，此时值栈中的数值就是计算的结果，取出返回计算结果。 2.两个栈实现算法 此算法需要两个栈，一个值栈od，一个操作符栈op。将用户输入的数学表达式存入str数组中，对其数组进行逐个扫描。 当遇到数字或小数点，截取数值子串，将其转换成double类型的数值存入od栈中； 当遇到左括号，直接入op栈；遇到右括号，op栈出栈，再从值栈od中取出两个数值，计算将其结果存入值栈中，一直进行此操作，直到操作符栈栈顶为左括号，将左括号丢掉。 如果遇到操作符，若op栈为空，直接入栈；若栈不为空，与栈顶元素比较优先等级，若比栈顶操作符优先等级高，直接入op栈，如果低于或等于栈顶优先等级，op栈出栈，再从值栈中取出两个数值，计算将其结果存入值栈中，一直进行此操作，直到栈顶优先等级低于扫描的操作符等级，将此操作符入op栈。继续扫描直到遇到str中的结束字符\0，扫描结束。此时看操作符栈是否为空，若不为空，出栈，再从值栈中取出两个数值进行计算，将其结果存入值栈，一直进行此操作，直到操作符栈为空。此时把值栈中的数值取出，即为所得的最终计算结果。 二、算法流程图 第一种算法：中缀转后缀算法
名词性从句用法详解
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交互式多模型算法仿真与分析
硕037班 刘文3110038020 2011/4/20交互式多模型仿真与分析IMM算法与GBP算法的比较，算法实现和运动目标跟踪仿真，IMM算法的特性分析 多源信息融合实验报告
交互式多模型仿真与分析 一、 算法综述 由于混合系统的结构是未知的或者随机突变的，在估计系统状态参数的同时还需要对系统的运动模式进行辨识，所以不可能通过建立起一个固定的模型对系统状态进行效果较好的估计。针对这一问题，多模型的估计方法提出通过一个模型集{}(),1,2,,j M m j r == 中不同模型的切换来匹配不同目标的运动或者同一目标不同阶段的运动，达到运动模式的实时辨识的效果。 目前主要的多模型方法包括一阶广义贝叶斯方法（BGP1），二阶广义贝叶斯方法（GPB2）以及交互式多模型方法等（IMM ）。这些多模型方法的共同点是基于马尔科夫链对各自的模型集进行切换或者融合，他们的主要设计流程如下图： M={m1,m2,...mk} K 时刻输入 值的形式 图一 多模型设计方法 其中，滤波器的重初始化方式是区分不同多模型算法的主要标准。由于多模型方法都是基于一个马尔科夫链来切换与模型的，对于元素为r 的模型集{}(),1,2,,j M m j r == ，从0时刻到k 时刻，其可能的模型切换轨迹为 120,12{,,}k i i i k trace k M m m m = ，其中k i k m 表示K-1到K 时刻，模型切换到第k i 个， k i 可取1,2,,r ，即0,k trace M 总共有k r 种可能。再令1 2 1 ,,,,k k i i i i μ+ 为K+1时刻经由轨迹0,k trace M 输入到第1k i +个模型滤波器的加权系数，则输入可以表示为 0,11 2 1 12|,,,,|,,,???k k trace k k k i M k k i i i i k k i i i x x μ++=?∑ 可见轨迹0,k trace M 的复杂度直接影响到算法计算量和存储量。虽然全轨迹的
数据结构课程设计报告(精选.)
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报告： 报告内容：□详细□完整□基本完整□不完整设计方案：□非常合理□合理□基本合理□较差 算法实现：□全部实现□基本实现□部分实现□实现较差测试样例：□完备□比较完备□基本完备□不完备文档格式：□规范□比较规范□基本规范□不规范 答辩： □理解题目透彻，问题回答流利 □理解题目较透彻，回答问题基本正确 □部分理解题目，部分问题回答正确 □未能完全理解题目，答辩情况较差 总评成绩： □优□良□中□及格□不及格
运行环境：CodeBlocks 完成的题目：表达式类型的实现（难度系数：1.2） 选做的内容：(4)在表达式内增加对三角函数等初等函数的操作。 一、需求分析【课程设计要求】 【问题的描述】 一个表达式和一棵二叉树之间，存在着自然的对应关系。写一个程序，实现基于二叉树表示的算术表达式Expression的操作。 【基本要求】 【一】【必做部分】 假设算术表达式Expression内可以含有变量（a-z）,常量（0-9）和二元运算符（+，-，*，/，^（乘幂））。实现以下操作： （1）ReadExpr(E)――以字符序列的形式输入语法正确的前缀表达式并构造表达式E。 （2）WriteExpr(E)――用带括号的中缀表达式输出表达式E。 （3）Assign(V，c)――实现对变量V的赋值（V=c），变量的初值为0。（4） Value(E)――对算术表达式E求值。 （5）CompoundExpr(p,E1,E2)――构造一个新的复合表达式（E1）p（E2）。 【二】【选做部分】 （1）以表达式的原书写形式输入，支持大于0的正整数常量； （2）增加常数合并操作MergeConst(E)——合并表达式E中所有常数运算。例如， 对表达式E=(2+3-a)*(b+3*4)进行合并常数的操作后，求得E=(5-a)*(b+12) (3)增加对求偏导数的运算Diff（E，V）——求表达式E对V的导数 (4)在表达式内增加对三角函数等初等函数的操作。 【测试数据】 (1)分别输入0；a;-91;+a*bc;+*5x2*8x;+++*3^*2^x2x6并输出。 （2）每当输入一个表达式后，对其中的变量赋值，然后对表达式求值。 二、【概要设计】 1、数据类型的声明： 在这个课程设计中，采用了链表二叉树的存储结构，以及两个顺序栈的辅助存储结构
五种大数据压缩算法
?哈弗曼编码 A method for the construction of minimum-re-dundancy codes, 耿国华1数据结构1北京:高等教育出版社,2005:182—190 严蔚敏,吴伟民.数据结构(C语言版)[M].北京:清华大学出版社,1997. 冯桂,林其伟,陈东华.信息论与编码技术[M].北京:清华大学出版社,2007. 刘大有,唐海鹰,孙舒杨,等.数据结构[M].北京:高等教育出版社,2001 ?压缩实现 速度要求 为了让它(huffman.cpp)快速运行，同时不使用任何动态库，比如STL或者MFC。它压缩1M数据少于100ms（P3处理器，主频1G）。 压缩过程 压缩代码非常简单，首先用ASCII值初始化511个哈夫曼节点： CHuffmanNode nodes[511]; for(int nCount = 0; nCount < 256; nCount++) nodes[nCount].byAscii = nCount; 其次，计算在输入缓冲区数据中，每个ASCII码出现的频率： for(nCount = 0; nCount < nSrcLen; nCount++) nodes[pSrc[nCount]].nFrequency++; 然后，根据频率进行排序： qsort(nodes, 256, sizeof(CHuffmanNode), frequencyCompare); 哈夫曼树，获取每个ASCII码对应的位序列： int nNodeCount = GetHuffmanTree(nodes); 构造哈夫曼树 构造哈夫曼树非常简单，将所有的节点放到一个队列中，用一个节点替换两个频率最低的节点，新节点的频率就是这两个节点的频率之和。这样，新节点就是两个被替换节点的父
Book6 Unit1 词汇讲解 答案讲解学习
B o o k6U n i t1词汇 讲解答案
Book6 Unit1 Art 词汇学案 1．faith n. 信任；信心；信念 E.g Faith can move mountains. 精诚所至，金石为开。 【归纳】have faith in sb./sth. 对某人/某事有信心 lose faith in 不再信任；对……失去信心完成句子 (1) I ________________ (相信) his ability to succeed. (2) Though it’s only a small trouble, it made Rose （对……失去信心）success. 拓展 faithful adj. 忠实的 faithfully adv. 忠实地 2 . aim n. 目标；目的vi. & vt. 瞄准;(向某方向)努力 Eg: What is your aim in life? 你生活的目的是什么? 【归纳】 n. achieve one’s aim 达到目标 v. aim at瞄准；以…为目的 aim to do sth./ aim at doing sth. 力求/旨在做某事 运用完成句子 （1）He the gun the enemy officer. 他用枪瞄准了敌教官。 （2）He swim a mile. 他力求游一英里。 3.typical adj. 典型的；有代表性的 Eg: typical examples 典型事例 typical work 代表作 【归纳】 be typical of...是…的代表； It is typical of sb.to do sth.某人做某事是他的特点【拓展】 typically adv. 典型地；具有代表性地 （1）她这个人就是爱迟到。 （2）下雪是北方的特征。_______________________in the north. 4.adopt vt.采用，采纳，收养 5.e.g （1）Miss Gao has adopted a new teaching method. 高老师采用了一种新的教学法。（2）As they had no children of their own, they adopted an orphan. 翻译 玛丽采纳了我的意见去收养一个小孩。 ______________________________________________________ 【拓展】 adoption n.采用;收养 adopted adj. 收养的；领养的 5. possess vt. 拥有；具有；支配 1）He possesses two cars. 2) They are still in possession of the house, so we have to wait for another time. 他们依然拥有房子的产权，所以我们还得等待一段时间。 3) The factory is in the possession of one of my friend, who moved to Australia several years ago.这家工厂是属于我一个朋友的，几年前他搬到澳洲去了。 【用法】 sb be in possession of sth. 占有某物 sth be in the possession of sb. 为某人所有 【拓展】possession c n.财产(常用复数) un. 拥有，占有 收集于网络，如有侵权请联系管理员删除
动名词的用法详解
动名词的用法详解 今天给大家带来动名词的用法详解，我们一起来学习吧，下面就和大家分享，来欣赏一下吧。 英语语法：动名词的用法详解 动名词因同时拥有动词和名词两者的特点而拥有及其丰富 的用法，熟练的掌握这些用法不仅可以使口语表达更地道生动，也能在写作中增分添彩。 动名词主要有四种用法，做主语，作宾语，作表语，作定语，每种用法下又分小类别，是一个非常复杂庞大的系统，学习者们往往会理不清脉络，今天就为大家带来动名词的用法讲解。 一.作主语 1.直接位于句首 eg.Swimming is a good sport in summer. 2.用it作形式主语，把动名词(真实主语)置于句尾作后置主语。 eg.It is no use telling him not to worry.
.mportant，essential，necessary等形容词不能用于上述结构。 3.用于“There be”结构中 eg.There is no saying when hell come. 4.动名词的复合结构作主语: 当动名词有自己的逻辑主语时，常可以在前面加上一个名词或代词的所有格，构成动名词的复合结构,动名词疑问句通常使用这种结构做主语 eg.Their coming to help was a great encouragement to us. Does your saying that mean anything to him? 二.作宾语 1.作动词的宾语 某些动词后出现非限定性动词时只能用动名词作宾语，不能用不定式。不定式通常指某种特定的动作，但动名词表示泛指，常见的此类动词有： admit,appreciate,excuse,stand,advise,allow,permit,avoid,consider,e njoy,finish,give up,cannot help,imagine,include,keep,understand,keepon,mind,report,risk,mis s,put off,delay,practise,resist,suggest,depend on,think about,set about,succeed in,worry about,burst out,insist on,feel like,be used
五种类型的逻辑函数
五种类型的逻辑函数 逻辑函数式有五种表达式：与或、或与、与非与非、或非或非、与或非。例如 C A AB F += 与或型 C A AB F ?= 与非与非型 ))((C A B A F ++= 或与型 C A B A F +++= 或非或非型 C A B A F += 与或非型 它们的逻辑关系都相等，这很容易用真值表加以证明，也可以将它们的与或标准型写出，它们的最小项都相同。它们的最小项如下 ∑=+++=+=) 7,6,3,1(m BC A C B A ABC C AB C A AB F ∑=+=?=)7,6,3,1(m C A AB C A AB F ∑=+++=++=) 7,6,3,1())((m BC C A AB A A C A B A F ∑=++=+++=)7,6,3,1())((m C A B A C A B A F ∑=++=?=+=)7,6,3,1())((m C A B A C A B A C A B A F 这些逻辑表达式都可以用相应的与门、或门、与非门、或非门以及与或非门来实现，其电路见图17-7-1所示。 F (a) 与或型 (b) 与非与非型 C B A A F (c) 或与型 (d) 或非或非型
(e) 与或非型 图 17-7-1 同一逻辑关系的五种逻辑表达式 与或型转换为与非与非型 逻辑电路用与或式实现时，需要两种类型的逻辑门，与门和或门。用小规模集成电路实现时，要用一片四2输入与门，例如CT74LS08；一片四2输入或门CT74LS32。门的利用率很低，CT74LS08中有四个2输入的与门，只用了二个；CT74LS32中有四个或门，只用了一个。如果变换为与非与非型，需要2输入的与非门三个，这样用一片CT74LS00就可以了。74LS00中有四个2输入与非门，用去三个，只剩一个。 下面就以C A AB F +=为例说明逻辑式的变换问题。 将与或逻辑式转换为与非与非型，方法是对与或式二次求反。 C A A B C A AB C A AB F ?=+=+= 变换中主要利用了摩根定理，具体用与非门实现的电路见图17-7-1(b)。 与或型转换为或与型 将与或式转换为或与型的基本方法是：利用对偶规则求出与或式的对偶式，将对偶式展开，化简；最后将对偶式进行对偶变换，即可得到或与型逻辑式。这里请注意，与或式进行对偶变换，得到或与式，展开就得到与或式，再一次对偶就得到或与式。 将与或式C A A B F +=转化为最简的或与表达式。 B A AC BC B A A C C A B A F +=++=++=))((' ))(()'(B A C A F F ++='= 用或门和与门实现的电路见图17-7-1(c)。 与或型转换为或非或非型 基本方法是，将与或式先变换为最简或与式，对或与式进行二次求反，即
LZ77压缩算法实验报告
LZ77压缩算法实验报告 一、实验内容 使用C++编程实现LZ77压缩算法的实现。 二、实验目的 用LZ77实现文件的压缩。 三、实验环境 1、软件环境：Visual C++ 6.0 2、编程语言：C++ 四、实验原理 LZ77 算法在某种意义上又可以称为“滑动窗口压缩”，这是由于该算法将一个虚拟的，可以跟随压缩进程滑动的窗口作为术语字典，要压缩的字符串如果在该窗口中出现，则输出其出现位置和长度。使用固定大小窗口进行术语匹配，而不是在所有已经编码的信息中匹配，是因为匹配算法的时间消耗往往很多，必须限制字典的大小才能保证算法的效率；随着压缩的进程滑动字典窗口，使其中总包含最近编码过的信息，是因为对大多数信息而言，要编码的字符串往往在最近的上下文中更容易找到匹配串。 五、LZ77算法的基本流程 1、从当前压缩位置开始，考察未编码的数据，并试图在滑动窗口中找出最长的匹 配字符串，如果找到，则进行步骤2，否则进行步骤3。 2、输出三元符号组( off, len, c )。其中off 为窗口中匹
配字符串相对窗口边 界的偏移，len 为可匹配的长度，c 为下一个字符。然后将窗口向后滑动len + 1 个字符，继续步骤1。 3、输出三元符号组( 0, 0, c )。其中c 为下一个字符。然后将窗口向后滑动 len + 1 个字符，继续步骤1。 六、源程序 /********************************************************************* * * Project description: * Lz77 compression/decompression algorithm. * *********************************************************************/ #include 名词所有格的用法 1. 名词的格的种类 英语名词有三个格，即主格、宾格和所有格。名词的主格和宾格形式相同，所以它们又统称作通格。当名词用作主语、宾语、表语时，用通格。英语名词的所有格表示所属关系，它分-?s 所有格和of 所有格两种形式。 Tom loves Mary. (Tom 为主格，Mary 为宾格，均为通格形式) Tom?s best friend is Mary.(Tom?s 是所有格，Mary 为通格) The title of the book is interesting. (of the book 为所有格) 2. -?s所有格的构成方法 (1) 一般情况(包括单数名词和不带词尾s的复数名词)加-?s： children?s books 儿童图书 today?s paper 今天的报纸 (2) 带词尾s的复数名词只加省字撇(…)： girls? school 女子学校 the Smiths? car 史密斯家的小汽车 注：带词尾s的单数名词，通常仍加?s： the boss?s plan 老板的计划 the hostess?s worry 女主人的担心 (3) 带词尾s的人名，可加?s 或只加省字撇(…)： Dickens? novels 狄更斯的小说 Charles?s job 查理斯的工作 不带词尾-s却以咝音结尾者，一律加?s： Marx?s works 马克思的著作 George?s room 乔治的房间 (4) 用and连接的并列连词的所有格要分两种情况，即表示各自的所有关系时，要分别在并列连词后加-?s，表示共同的所有关系时，只在最后一个名词后加-?s： Tom?s and Jim?s rooms 汤姆和吉姆(各自)的房间 Tom and Jim?s rooms 汤姆和吉姆(共同)的房间 3. -?s所有格的用法 -?s 所有格主要用于有生命的东西，但有时也可用于无生命的东西，这主要见于： (1) 用于表时间的名词后： tomorrow?s weather 明天的天气 two days? journey 两天的旅程 比较：ten minutes? break = a ten-minute break 10分钟的休息 (2) 用于表国家、城市的名词后： America?s policy 美国的政策 the city?s population 这个城市的人口 (3) 用于某些集合名词后： the majority?s view 多数人的观点 the government?s policy 政府的政策 (4) 用于组织机构后： the station?s waiting-room 车站候车室 the newspaper?s editorial policy 这家报纸的编辑方针 (5) 用于度量衡及价值名词后： 英语单词惯用法集锦 习惯接动词不定式的动词（V to inf) adore(vi极喜欢) dread （vt.不愿做，厌恶）plan 计划 afford（+to,vt有条件，能承担）endeavour （vt，竭力做到，试图或力图）prefer（vt.宁可；宁愿(选择)；更喜欢）agree 同意endure(忍受.cannot ~ to) prepare准备 aim (vi[口语]打算：) engage (vi.保证，担保；) presume(vt.冒昧；敢于[用于第一人称时为客套话]：) appear (vi.似乎；显得) essay(vt.尝试，试图) pretend(vt.自命；自称；敢于；妄为) apply （申请）expect（期望，希望）proceed（开始，着手，）arrange （vi.做安排，(事先)筹划）fail （vt.未做…；疏忽)promise(许诺，保证做 ask （要求）forget （vt. 忘记）purpose （vt.决心，打算） beg （vt.正式场合的礼貌用语]请(原谅)，请(允许)：I beg to differ.恕我不能赞同）guarantee（保证，担保）refuse（拒绝）bear 承受，忍受hate（[口语]不喜欢；不愿意；）regret （vt. 抱歉；遗憾）begin help （有助于，促进）remember（记住） bother （vi.通常用于否定句]麻烦，费心）hesitate（vi.犹豫；有疑虑，不愿）scheme（策划做）care （vt.想要；希望；欲望[后接不定式，常用于否定、疑问及条件句中]）hope （vt.希望，盼望，期待）seek（vt.谋求，图谋[后接不定式]） cease （停止; 不再(做某事)[正式] intend （打算；想要）seem（似乎，好像[后接不定式或从句]；觉得像是，以为[ choose （意愿；选定；决定）itch start开始claim （vt. 主张；断言；宣称） continue （继续）like 喜欢swear（vt.起誓保证；立誓要做(或遵守） dare （vt.敢，敢于，勇于，胆敢）long（vi.渴望；热望；极想） decline（vt.拒绝，拒不(做、进入、考虑等) manage（设法完成某事）threaten（vt.威胁，恐吓，恫吓）deign （屈尊做）mean（有意[不用进行时）trouble（vi.费心，费神；麻烦）demand（vi.要求，请求：）need （需要）try（设法做） deserve (应得) neglect (疏忽) undertake(承诺,答应，保证) desire （希望渴望）offer（表示愿意(做某事)，自愿；）venture（冒险(做某事））determine（vi.决心，决意，决定，）omit （疏忽，忘记）want 想要 die （誓死做）pine （渴望）wish （希望） 习惯接“疑问词+动词不定式”的动词（有时也包括VN wh-+to do) advise 建议explain 解释perceive 觉察，发觉 answer 答复find 得知，察觉persuade 说服，劝说；使某人相信 ask 询问，问forget 忘记phone 打电话 assure 保证guess 臆测，猜度pray 祈祷 beg 请求，恳求hear 小心聆听（法庭案件）promise 允诺 conceive 想象，设想imagine 以为，假象remember记得 consider 考虑，思考indicate 暗示remind 提醒，使想起 convince 使相信inform告知通知instruct告知，教导 see 看看，考虑，注意decide 解决，决定know 学得，得知 show 给人解释；示范；叙述；discover发现；知道learn 得知，获悉 signal以信号表示doubt 怀疑，不相信look 察看；检查；探明 strike 使想起；使突然想到；使认为suggest 提议，建议tell 显示，表明；看出，晓得；warn 警告，告诫think 想出；记忆，回忆；想出，明白wonder 纳闷，想知道 wire 打电报telegraph 打电报 习惯接动名词的动词（包括v+one’s/one+v+ing) acknowledge 认知，承认…之事实escape免除，避免omit疏忽，忽略 admit 承认，供认excuse 原谅overlook 放任，宽容，忽视adore （非正式）极为喜欢fancy 构想，幻想，想想postpone 延期，搁置 advise 劝告，建议finish完成prefer较喜欢 appreciate 为…表示感激(或感谢)forbid 不许，禁止prevent预防 avoid 逃避forget 忘记prohibit 禁止，妨碍 第二章数据类型、运算符与表达式 2.1 选择题 **2.1C语言中(以16位PC机为例)，各数据类型的存储空间长度的排列顺序为。 A)char 实验名称：LZSS压缩算法实验报告 一、实验内容 使用Visual 6..0 C++编程实现LZ77压缩算法。 二、实验目的 用LZSS实现文件的压缩。 三、实验原理 LZSS压缩算法是词典编码无损压缩技术的一种。LZSS压缩算法的字典模型使用了自适应的方式，也就是说，将已经编码过的信息作为字典， 四、实验环境 1、软件环境：Visual C++ 6.0 2、编程语言：C++ 五、实验代码 #include -- need byte-swap function for big endian CPU */ #define READ_LE32(X) *(uint32_t *)(X) #define WRITE_LE32(X,Y) *(uint32_t *)(X) = (Y) /* this assumes sizeof(long)==4 */ typedef unsigned long uint32_t; /* text (input) size counter, code (output) size counter, and counter for reporting progress every 1K bytes */ static unsigned long g_text_size, g_code_size, g_print_count; /* ring buffer of size N, with extra F-1 bytes to facilitate string comparison */ static unsigned char g_ring_buffer[N + F - 1]; /* position and length of longest match; set by insert_node() */ static unsigned g_match_pos, g_match_len; /* left & right children & parent -- these constitute binary search tree */ static unsigned g_left_child[N + 1], g_right_child[N + 257], g_parent[N + 1]; /* input & output files */ static FILE *g_infile, *g_outfile; /***************************************************************************** initialize trees *****************************************************************************/ static void init_tree(void) { unsigned i; /* For i = 0 to N - 1, g_right_child[i] and g_left_child[i] will be the right and left children of node i. These nodes need not be initialized. Also, g_parent[i] is the parent of node i. These are initialized to NIL (= N), which stands for 'not used.' For i = 0 to 255, g_right_child[N + i + 1] is the root of the tree for strings that begin with character i. These are initialized to NIL. Note there are 256 trees. */ for(i = N + 1; i <= N + 256; i++) g_right_child[i] = NIL; for(i = 0; i < N; i++) g_parent[i] = NIL; } /***************************************************************************** Inserts string of length F, g_ring_buffer[r..r+F-1], into one of the trees (g_ring_buffer[r]'th tree) and returns the longest-match position and length via the global variables g_match_pos and g_match_len. If g_match_len = F, then removes the old node in favor of the new one, because the old one will be deleted sooner.英语名词所有格的用法讲解
英语单词惯用法集锦解析
C习题一表达式
LZSS压缩算法实验报告