基础算法递推法Recursive algorithmWord文件下载.docx

基础算法递推法Recursive algorithmWord文件下载.docx

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{proceedfromtheboundaryconditionF1{}}

TheresultsofFiwhiledoFnnonfinalresultsbyFi=g（Fi-1）CISpushback;

TheoutputoftheFnresultsandthepushingprocess;

End;

{else}

I.backwardpushingmethod

Thepushdownmethod,isnotintheinitialvalueofthecase,bysomerecursiverelationsandinformedthesolutionoftheproblemorgoal,andthenpushdownover,inferitsinitialconditions.Becausetheoperationsofsuchproblemsaremappedonebyone,therecurrenceformulacanbeanalyzed.Then,fromthissolutionorgoal,takethepushbacksection,stepbystepintotheinitialstatementoftheproblem.

Herearesomeexamples.

[1]oilstoragepoint

Aheavytruckwantstocross1000kilometersofdesert,andthetruckuses1liters/kmoffuel,andthetotalcapacityofthetruckis500liters.Obviously,atrucknevergetsthroughthedesertatonetime.Asaresult,driversmusttrytosetupseveralstoragepointsalongthewaysothattruckscancrossthedesertsmoothly.Howcandriversbuildtheseoilstoragepoints?

Howmuchoilshouldbestoredateachpointtomakethetruckpassthroughthedesertattheexpenseofleastoil?

Algorithmanalysis:

Oilstoragepointnumbercalculationprogramandprinttheestablishment,theoilstoragefromtheedgeofthedesertofthedistanceandthestorageofoil.

No.Distance（K.M.）oil（litre）

1,X,X,X,X

2,X,X,X,X

3,X,X,X,X

........

Setdis[i]asthedistancebetweentheIoilstoragepointandtheendpoint（i=0）;

Oil[i]isthestoragequantityoftheIoilstoragepoints;

Wecansolvethisproblembythebackpushingmethod.Fromtheendtothebeginningofthepushdown,onebyonetofindthelocationofeachoilstoragepointandtheamountofoilstorage.

Butwhenthereturnpointareshown:

ThestrategyofpushingbackfromtheoilstoragepointItotheoilstoragepointi+1isthatthetruckmovesbackandforthbetweenpointIandpointi+1.Eachtimethetruckreturnstoi+1,itrunsoutof500litersofgasoline,andeachtimeitleavesthei+1,ithastohold500litersofgasoline.ThedistancebetweenthetwopointsmustsatisfytheIpointi*500gasolinefuelstorageatleasttherequirements（0<

=i<

=n-1）.Specifically,thefirstoilstoragepointi=1shouldbe500kmfromtheterminali=0andstore500litresofgasolineattheendofthetanksothatthetruckcanreachtheterminali=0ati=1.Thatistosay

Dis[1]=500oil[1]=500;

Inordertoi=1storageof500litersofgasoline,cartoi=1truckatleastfromi=2opentwofulloil.Sothei=2storesatleast2*500litersofgasoline,oroil[2]=500*2=1000.Inaddition,combinedwithareturnfromi=1toi=2atano-load,thetotalround-trip3times.Thethreeroundtripusesonly500litresoffuelatthemostprovinciallevel.Thatis,d12=500/3km

Dis[2]=dis[1]+d12=dis[1]+500/3

Inordertostore1000litersofgasolineati=2,thetruckwilldriveatleastthreefullloadvehiclesfromi=3toi=2.

Withi=3,atleast3*500litersofgasoline,oroil[3]=500*3=1500.Plusi=2toi=3atthetworeturnemptycar,total5times.Thefuelmileageshouldalsobe500liters,ord23=500/5,

Dis[3]=dis[2]+d23=dis[2]+500/5;

Andsoon,inordertostorek*500litersofgasolineatthei=k,thetruckwilldriveatleastkfullloadvehiclesfromi=k+1toi=k,i.e.

Oil[k+1]=[k+1]*500=oil[k]+500,plusthek-1returnspacefromi=ktoi=k+1,total2k-1times.This2k-1timesthetotalfuelconsumption,accordingtotheprovincialminimumrequirementof500liters,thatis

DK,k+1=500/（2k-1）

Dis[k+1]=dis[k]+dk,k+1

=dis[k]+500/（2k-1）;

Finally,thedistancefromi=ntothestartingpointis1000-dis[n],oil[n]=500*n.Inordertoobtainn*500litersofpetroltruckati=n,atleastfromthestartingpointn+1fullcartoi=n,andreturnedfromthei=npointofNtimesthetotalreturnempty,2n+1times,2n+1timesthetotalfuelconsumptionshouldbejustfor（1000-dis[n]）*（2n+1）,isthestartingpointforoilreservoir（oil[n]+1000-dis[n]）*（2n+1）.

Here'

stheprogramcode:

Programoil_lib;

Var

K:

integer;

{oilpointpositionserialnumber}

D,{thedistancefromthecumulativefinishtothecurrentpointofoilstorage

D1:

real;

{i=ntostartpointdistance}

Oil,dis:

array[1..10],of,real;

I:

{auxiliaryvariable}

Begin

Writeln（'

NO.'

'

distance（K.M）:

30,'

oil（1.）:

80）;

=1;

D:

=500;

{startati=1,pushbackwardtothestartpoint}

Dis[1]:

=500;

Oil[1]:

Repeat

=k+1;

=d+500/（2*k-1）;

Dis[k]:

=d;

Oil[k]:

=oil[k-1]+500;

Untild>

=1000;

=1000;

{startingpointtoenddistancevalue}

=1000-dis[k-1]={i=ntostartpoint}

=d1*（2*k+1）+oil[k-1];

{startpointoilstorage}

Fori:

=0tokdo{startfromthestartpoint,printthestartingpointtothedistancefromthecurrentoilstoragepointandtheamountofoilstorage}

Writeln（I,1000-dis[k-i]:

30,oil[k-i]:

End.{main}

ConvertstotheClanguageprogramasfollows:

#include<

stdio.h>

Void,main（）

{

Intk;

/**/oilstoragelocationnumber

FloatD,D1/*d:

;

thecumulativeendpointtothecurrentoilstoragepointdistance,d1:

i=ndistancetothestartingpoint.

Float,oil[10],dis[10];

Inti;

Printf（"

NO.distance（K.M.）\toil（L.）\n"

）;

K=1;

D=500;

/**/startingpointbegantopushdownfromi=1

Dis[1]=500;

Oil[1]=500;

Do{

K=k+1;

D=d+500/（2*k-1）;

Dis[k]=d;

Oil[k]=oil[k-1]+500;

}while（（d>

=1000））;

Dis[k]=1000;

/*thestartpointtotheendpointofthedistance.

D1=1000-dis[k-1];

/*fori=ntothestartingpointofthedistance.

Oil[k]=d1*（2*k+1）+oil[k-1];

/*forstartingpointofoilreservoir.

For（i=0;

i<

k;

i++）/*startingfromthestartpointtothestartingpointtoprintthecurrentoilstoragepointdistanceandtheamountofoilreservoir.

%d\t%f\t%f\t\n"

"

I"

1000-dis[k-i]"

oil[k-i]"

}

Practicalalgorithm（basicalgorithm-recursivemethod-02）

Dateofissue:

April10,2003provenance:

analysisandprogrammingofpracticalalgorithms.Author:

Clanguagehomecollation,1317readershavereadthisarticle

Forwardpushingmethod

Thereverseprocessofthepushdownmethodispushedalong,startingfromtheboundaryconditions,therecursiverelationslaunchedbythelattervalue,valueaccordingtotherecursiveformulaofrelaunch,followedbyconsequentvalue......recursive,untiltheproblemfromtheinitialstatementtomoveforwardtothesolutionofthisproblem.

Realnumbersequence:

arealnumberserieswithNitemsknown

Ai=（ai-1-ai+1）/2+d,（1<

i<

N）（N<

60）

ThekeyboardinputsareN,D,A1,an,m,andoutputam

Noerrorisrequiredforinputdata.

Analysisoftheproblem,theformula:

Ai=（Ai-1-Ai+1）/2+d（1<

N）（n<

Makeathoroughstudyanddiscussitslawofdigitaltransformation.Otherwise,therewillbenowaytostart.

MakeX=A2s2[i]=（PI,Qi）,

Ri）representsAi=PiX+QiD+RiA1

Wecanaccordingto

Ai=Ai-2-2Ai-1+2D

=PiX+QiD+RiA1

Formularelease

PiX+QiD+RiA1=（Pi-2-2Pi-1）,X+（Qi-2-2Qi-1+2）,D+（Ri-2-2Ri-1）,A1

ComparingthecoefficientsofX,DandA1attheendsofanequalsign

Pi=Pi-2-2Pi-1

Qi=Qi-2-2Qi-1+2

Ri=Ri-2-2Ri-1

Withtwoboundaryconditions

P1=0Q1=0R1=1（A1=A1）

P2=1Q2=0R2=0（A2=A2）

AccordingtotherecursionformulaofPi,QiandRi,itcanbecalculated

S2[1]=（0,0,1）;

S2[3]=（-2,2,1）;

S2[4]=（5,-2,-2）;

S2[5]=（-12,8,5）;

...

S2[i]=（Pi,Qi,Ri）;

S2[N]=（PN,QN,RN）;

Withthesefundamentals,AMisnothardtoget.Therearetwoways:

1,becauseAN,A1andPN,QN,RNknown,soyoucanfirstaccordingtoformula:

A2=AN-QND-RNA1/PN

FindA2.ThenA2issubstitutedintotheformula

A3=A1-2A2+2D

FindA3.ThenA3issubstitutedintotheformula

A4=A2-2A3+2D

FindA4.ThenA4issubstitutedintotheformula

And...

FindAi-1.ThenAi-1issubstitutedintotheformula

FindAi.Andsoon,untilthedeliveryuntilAM.

ThedefectisduetotheA2algorithmistheresultofthedivision,andthedivisorPNincrements,sotheerrorcanhardlybeavoided,recursiveprocesswillcontinuetoexpandaftertheerror,evenwhenMismorethan40AMwascalculatedfromthecorrectvaluebias.Obviously,thismethodissimplebutunreliable.

2,wemakeA2=A2,A3=X,representedbyS3[i]=（Pi,Qi,Ri）,andAi=PiX+QiD+RiA2（i>

=2）canbecalculated:

S3[2]=（0,0,1）=S2[1];

S3[3]=（1,0,0）=S2[2];

S3[4]=（-2,2,1）=S2[3];

S3[5]=（5,-2-2）=S2[4];

S3[i]=（....）=S2[i-1];

................