计算概论poj部分水题代码四

authorKINGRAIN@EECSPKUtimeApril4,2010#include<iostream>#include<cstring>usingnamespacestd;constintLENGTH=charintintint{cin>>t;for(i=0;i<t;{cin>>num1>>num2;length1=strlen(num1);length2=strlen(num2);_strrev(num2), 序delta=length1-length2;if(delta<0){}for(j=0;{sum[j]='0'+Divide(delta--}sum[j]=0;for(j=0;ptr[j]=='0'; cout<<ptr+j<<}return}intDivide(int {inti=0,N=0;for(i=length2+n-1;i>=n;i--)temp[i]=num2[i-for(;i>=0;i--temp[i]='0';{if{}

charstrcpy(N1,num1),_strrev(N1),ifreturnfor(i=0;i<length2+n;{num1[i]=num1[i]-temp[i]+'0';if(num1[i]<'0'){num1[i]+=10;}for(i=length1-1;((num1[i]=='0')&&(i>=0));i--,length1--);num1[i+1]=0;}return}authorKINGRAIN@EECSPKUtimeMar.20,2010#include<iostream>#include<cstring>usingnamespacestd;intmain(){charadd1[200]={0},add2[200]={0};//add1&add2charinteger1[100]={0},integer2[100]={0};n=0,i=0,j=0,k=0,l=0,m=0,length1=0,length2=0,pointpos1=0,pointpos2=0;pointpos记录小数点的位置length记录加数的长度int cin>>for(i=0;i<n;{cin>>add1>>add2;length1=strlen(add1);length2=for(j=0;add1[j]!='.'; pointpos1=for(j=0;add2[j]!='.'; pointpos2=for(m=1,j=pointpos1+1,k=pointpos2+1;add1[j]||add2[k]; //小数部分相加{ifderesult[m]+=add1[j]-'0';if(k<length2)deresult[m]+=add2[k]-}//此时m-1for(j=m-1;j>0;j--if{deresult[j]-=10;}for(l=0,j=pointpos1-1,k=pointpos2-1;j>=0||k>=0;j--,k--{ifinresult[l]+=add1[j]-'0';if(k>=0)inresult[l]+=add2[k]-}if for(j=0;j<l;j++)if{inresult[j+1]+=inresult[j]-=}while l= lfor(j=l-1;j>=0;j--for(j=m;deresult[j]==0;j--); 0for(j=1;j<=m;}return}authorKINGRAIN@EECSPKUtimeMay17,2010#include<iostream>#include<cstring>#include<memory>usingnamespaceconstintcharboolCOMPARE(int[],int[]);voidint{cin>>n;for(i=0;i<n;{cin>>A>>B;length1=strlen(A);for(k=length1-1;k>=0;k--a[length1-k-1]=A[k]-'0';length2=strlen(B);for(k=length2-1;k>=0;k--b[length2-k-1]=B[k]-'0';length=length1;for(j=0;j<21;{if{cout<<j<<endl;}}}return}void{intfor(inti=0;i<length;for(intj=0;j<length1;temp[i+j]+=result[i]*intfor(;i<length;{if{temp[i+1]+=temp[i]/10;temp[i]%=10;}}while{if{}

temp[i+1]+=temp[i]/10;temp[i]%=10;for(i=MAX-1;temp[i]==0;i--);length=i+1;}boolCOMPARE(intr[],int{for(inti=MAX-1;i>=0;i--ifreturn1;elseifreturn}authorKINGRAIN@EECSPKUtimeApril17,2010#include<iostream>usingnamespacestruct{intMONKEY*int{intwhile(cin>>n>>{

ifif m==1{}MONKEYfor(inti=0;i<n; {if{}{}

current=newMONKEY;current->num=1;head=current;current->nextcurrent;current->next=newMONKEY;current=current->next;current->num=i+1;current->next }current=head;MONKEY*temp=0;inti=1;while(current-{

for(inti=1;i!=m-1 current=current->next;temp=current->next;current->next=current->next->next; currentcurrent }

delete}return}#include<cstring>usingnamespacestd;intmain(){cin>>t;for(i=0;i<t;{cin>>n;max=0;for(j=0;j<=n;for(k=0;k<=n;{iffor(l=0;l<=n;l++){ifmax=l+j+k;}}return}

cout<<max<<#include<cstring>usingnamespaceintlength=0,k=0,n=0;voidOut_put(intorder,int pare(constvoid*elem1,constvoidint{intcin>>n=strlen(string_string); return} pare(constvoid*elem1,constvoid{char*p1=0,*p2=0;p1=(char*)elem1;p2=(char*)elem2;return(p1-p2);}voidOut_put(intorder,int{boolflag=for(inti=0;i<n;{temp[num]=string_string[i];for(intj=0;j<num;j++)if{}if

flag=1;{

if(num==n-cout<<temp<<}}}

authorKINGRAIN@EECSPKUtimeDec.3,2009addresscomputerroomusingnamespacevoidTRY(int, //voidtry()tryunsignedshortintnum=1; 整数本身与1相乘就是一种所以初始值为1int{unsignedshortintn,i;intcin>>for(i=1;i<=n;{cin>> cout<<num<<endl;num=1;}return}voidTRY(intfactor,int{for(;factor*factor<=INTEGER;{数能否被分解。同时这也是程序的出口，如果一直不满足该条件，那么就直接退出if(INTEGER%factor=={}

}}authorKINGRAIN@EECSPKUtimeFeb.8,2010usingnamespacestd;intmain(){cin>>n;for(i=0;i<n;{cin>>a;elseif(a%4==0){}{}}

results[i][0]=a/4;results[i][1]=results[i][0]=a/4+1;results[i][1]=a/2;for(i=0;i<n;cout<<results[i][0]<<''<<results[i][1]<<return}authorKINGRAIN@EECSPKUtimeMar.22,2010#include<iostream>#include<cstring>usingnamespace pareInc(constvoid*elem1,constvoid*elem2); pareDec(constvoid*elem1,constvoid*elem2); pareNcinc(constvoid*elem1,constvoid*elem2); pareNcdec(constvoid*elem1,constvoid*int{intn=0,i=0;cin>>kind;cin>>n;cin.getline(str[0],3);//cin回for(i=0;i<n;if(strcmp(kind,"inc")==0) elseif(strcmp(kind,"dec")==0) elseif(strcmp(kind,"ncinc")==0) elseif(strcmp(kind,"ncdec")==0) for(i=0;i<n;i++)return0;} pareInc(constvoid*elem1,constvoid*{charp1[200],p2[200];intj=0,k=0,l=0;strcpy(p1,(constchar*)elem1);strcpy(p2,(constchar*)elem2);l=strlen(p1);for(j=0;p1[j]=='';j++);//空循环找前导空格iffor(k=0;k<l-j;p1[k]=l=for(j=l-1;p1[j]=='';j p1[j+1]=l=for(j=0;p2[j]==' iffor(k=0;k<l-j;p2[k]=l=for(j=l-1;p2[j]=='';j p2[j+1]=return} pareDec(constvoid*elem1,constvoid*{charp1[200],p2[200];intj=0,k=0,l=0;strcpy(p1,(constchar*)elem1);strcpy(p2,(constchar*)elem2);l=strlen(p1);for(j=0;p1[j]=='';j++);//空循环找前导空格iffor(k=0;k<l-j;p1[k]=l=for(j=l-1;p1[j]=='';j p1[j+1]=l=for(j=0;p2[j]==' iffor(k=0;k<l-j;p2[k]=l=for(j=l-1;p2[j]=='';j p2[j+1]=return} pareNcinc(constvoid*elem1,constvoid*{charp1[200],p2[200];intj=0,k=0,l=0;strcpy(p1,(constchar*)elem1);strcpy(p2,(constchar*)elem2);l=strlen(p1);for(j=0;p1[j]=='';j++);//空循环找前导空格iffor(k=0;k<l-j;p1[k]=l=for(j=l-1;p1[j]=='';j p1[j+1]=l=for(j=0;p2[j]==' iffor(k=0;k<l-j;p2[k]=l=for(j=l-1;p2[j]=='';j p2[j+1]=return} pareNcdec(constvoid*elem1,constvoid*{charp1[200],p2[200];intj=0,k=0,l=0;strcpy(p1,(constchar*)elem1);strcpy(p2,(constchar*)elem2);l=strlen(p1);for(j=0;p1[j]=='';j++);//空循环找前导空格iffor(k=0;k<l-j;p1[k]=l=for(j=l-1;p1[j]=='';j p1[j+1]=l=for(j=0;p2[j]==' iffor(k=0;k<l-j;p2[k]=l=for(j=l-1;p2[j]=='';j p2[j+1]=intn=_stricmp(p1,p2);return-n;}usingnamespacestd;intmain(){longinta,b,c,d,m,n,i;cin>>n;//输入n{ {b=c;c=d;}}return}authorKINGRAIN@EECSPKUtimeApril3,2010#include<iostream>usingnamespacestd;voidSet_position(intn);intqueen[8]={0},order=0;charint{intn=0,i=0,t=0;cin>>n;for(i=0;i<n;{cin>>cout<<solution[t-1]<<}return}vo