04 2012 档案
摘要：#include<stdio.h>#include<string.h>char a[30][30];int d(int h,int l,int n,int m){if(h<0||l<0||h>n-1||l>m-1)return 0;else if(a[h][l]=='#')return 0;else if(a[h][l]=='.'){a[h][l]='#';return1+d(h,l-1,n,m)+d(h,l+1,n,m)+d(h-1,l,n,m)+d(h+1,l,n,m);}}int main()
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摘要：问题描述 人自出生起就有体力，情感和智力三个生理周期，分别为23，28和33天。一个周期内有一天为峰值，在这一天，人在对应的方面（体力，情感或智力）表现最好。通常这三个周期的峰值不会是同一天。现在给出三个日期，分别对应于体力，情感，智力出现峰值的日期。然后再给出一个起始日期，要求从这一天开始，算出最少再过多少天后三个峰值同时出现。问题分析 首先我们要知道，任意两个峰值之间一定相距整数倍的周期。假设一年的第N天达到峰值，则下次达到峰值的时间为N+Tk(T是周期，k是任意正整数)。所以，三个峰值同时出现的那一天(S)应满足 S = N1 + T1*k1 = N2 + T2*k2 = N3 + T3
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摘要：#include<stdio.h>#include<string.h>int main(){ int a; scanf("%d",&a); int a10=0; int mult=1; while(a){ a10+=mult*(a%10); mult*=8; a/=10; } printf("%d\n",a10); return 0;}
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摘要：#include<iostream>#include<stdio.h>#include<string.h>#define N 80using namespace std;int main(){ char test[N]; int counta=0,counte=0,counti=0,counto=0,countu=0; gets (test); int i; for(i=0;i<N && test[i]!='\0';i++){ if(test[i]=='a') counta++; else if(test
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摘要：#include <iostream>#include <cmath>using namespace std;int find(int x){ int i; for(i = 1; ;i++){ if(pow((double)2, i-1) <= x && x < pow((double)2, i)) return i; }}int main(){int n, high, m, low;int nCount, i;while (cin >> m >> n && m && n){nCount
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摘要：#include <string.h>#include <iostream>using namespace std;#define MAX 200unsigned int num[MAX+10];unsigned int num1[MAX+10];int main(){ string str1,str2; cin>>str1>>str2; memset(num,0,sizeof(num)); memset(num1,0,sizeof(num1)); int len = str1.length();int index = 0; for (int i
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摘要：#include<string>#include<iostream>usingnamespacestd;#defineMAX200intnum1[MAX+10];intnum2[MAX+10];unsignedinttotal[MAX+10];intSub(inta,intb){if(b>=a){if(a==b)return0;elsereturn-1;}inti=0;for(;a>b;i++){b++;}returni;}intmain(){intn;cin>>n;while(n--){memset(num1,0,sizeof(num1));m
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摘要：#include <iostream>#include <stdio.h>using namespace std;int main(){int num;cin>>num;int graphic[100][100];int i=0,j=0,rightEst,bottumEst;int width=0,lengh=0;for(i=0;i<num;i++){for(j=0;j<num;j++){scanf("%d",&graphic[i][j]);if(graphic[i][j]==0){rightEst=i;bottumE
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摘要：#include<iostream>#defineN100usingnamespacestd;intmain(){intcount,total;cin>>total;inta[N];for(count=0;count<total;count++)cin>>a[count];for(count=total-1;count>=0;count--)cout<<a[count]<<"";return0;}
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摘要：#include<stdio.h>#include<string.h>#include<iostream>using namespace std;#define N 110char str1[N],str2[N];int a[N],b[N],c[N];int main(){ int n,i,j; int len1,len2; cin>>n; while(n--) { memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); memset(c,0,sizeof(c)); cin>>str1>>
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摘要：#include <iostream>using namespace std;int common(int x,int y){ if (x==y) return x; if (x>y) return common(x/2,y); else return common(x,y/2);}int main(){ int x,y; cin>>x>>y; cout<<common(x,y)<<endl; return 0;}
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摘要：#include<iostream>using namespace std;int main(){int arr[100],i,x=0,y=0,z=0,n;cin>>n;for(i=0;i<n;i++)cin>>arr[i];for(i=0;i<n;i++){if(arr[i]==1)x++;else if(arr[i]==5)y++;else if(arr[i]==10)z++;}cout<<x<<endl;cout<<y<<endl;cout<<z<<endl;retur
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摘要：#include<iostream>usingnamespacestd;intmain(){intL,i,j,n;booltrees[10001];for(i=0;i<10001;i++)trees[i]=true;cin>>L>>n;for(i=0;i<n;i++){intbegin,end;cin>>begin>>end;for(j=begin;j<=end;j++)trees[j]=false;}intcount=0;for(i=0;i<=L;i++)if(trees[i])count++;cout<
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摘要：利用归并求逆序对DescriptionIn this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. For the input sequence9 1 0 5 4 ,Ultra-QuickSort produces the
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摘要：#include <iostream>using namespace std;int jz(int n,int y){if(n/8==0) return (n+y);elsey=y+n/8*10;y=jz(n%8,y);return y;}int main(){int x,y;cin>>x;y=jz(x,0);cout<<y<<endl;return 0;}
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摘要：#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int N=10;char a[N];int main(){int t;scanf("%d",&t);while(t--){int z=0;scanf("%s",a);int len=strlen(a);for(int i=0;i<len;i++){z*=16;if(a[i]>='A')z+=a[i]-'A'+
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摘要：#include<iostream>#defineN100usingnamespacestd;structInfo{ charname[80]; intlasttest; intdiscuss; charmana; charwest; inttext; intsum;};intmain(){ Infoa[N];intmax,i;cin>>max;for(i=0;i<max;i++){ cin>>a[i].name;cin>>a[i].lasttest;cin>>a[i].discuss;cin>>a[i].mana;
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摘要：1.#include<iostream>2.usingnamespacestd;3.intmain()4.{5. intn;6. cin>>n;7. inti,j,sum=0;8. for(i=1;i<=n;i++)9. if(i%7!=0&&i/10!=7&&i%10!=7)10. sum+=i*i;11.cout<<sum<<endl;12.return0;13.}
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摘要：#include<iostream>usingnamespacestd;intcount(intx,inty){if(y==1||x==0)return1;if(x<y)returncount(x,x);returncount(x,y-1)+count(x-y,y);}intmain(){intt,m,n;cin>>t;for(inti=0;i<t;i++){cin>>m>>n;cout<<count(m,n)<<endl;}while(1);return0;}
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摘要：#include<stack>#include<cmath>#include<iostream>usingnamespacestd;voidconversion(intdata){stack<int>s;if(data==0)cout<<"0";elseif(data<0){intdatai;datai=abs(data);while(datai!=0){s.push(datai%6);datai=datai/6;}cout<<"-";while(s.empty()!=true
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摘要：#include<string>#include<iostream>usingnamespacestd;structInfo{stringsex;floata,b,c,d,e;};intmain(){Infoinfo[101];intn,i,k=0;cin>>n;while(k<n){intcountN=0,countI=0;cin>>info[i].sex>>info[i].a>>info[i].b>>info[i].c>>info[i].d>>info[i].e;if(info
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摘要：#include<iostream>using namespace std;int a[100000]={0};int main(){int n;int i,len=1,j,temp=0;cin>>n;a[0]=1;for(i=2;i<=n;i++){for(j=0;j<len;j++){a[j]=i*a[j]+temp;temp=a[j]/10;a[j]%=10;}while(temp!=0){a[j++]=temp%10;temp/=10;}len=j;}len+=50-len%50;for(i=len-1;i>=0;i--){if((i+1)%5
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摘要：#include <stdio.h>#include <stdlib.h>int main(){ int i,n,a,sum; sum=0; scanf("%d%d",&a,&n); for(i=0;i<n;i++) { sum+=a; a=a%10+10*a; } printf("%d\n",sum); return 0;}
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摘要：#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int N=1000;char a[N];int b[3*N];int main(){ while(scanf("%s",a)!=EOF) { memset(b,0,sizeof(b)); int len=strlen(a); int iKeep,iTez,iToz=0; for(int i=len-1;i>=2;i--) { iTez=0; iKeep=0; int iNu
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摘要：POJ Biorhythms解题一、题目Description人生来就有三个生理周期，分别为体力、感情和智力周期，它们的周期长度为23天、28天和33天。每一个周期中有一天是高峰。在高峰这天，人会在相应的方面表现出色。例如，智力周期的高峰，人会思维敏捷，精力容易高度集中。因为三个周期的周长不同，所以通常三个周期的高峰不会落在同一天。对于每个人，我们想知道何时三个高峰落在同一天。对于每个周期，我们会给出从当前年份的第一天开始，到出现高峰的天数（不一定是第一次高峰出现的时间）。你的任务是给定一个从当年第一天开始数的天数，输出从给定时间开始（不包括给定时间）下一次三个高峰落在同一天的时间（距给定时间
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摘要：#include<iostream>using namespace std;int main(){int i,j=0,k=1;cin>>i;while(i){j=k*(i%10);k*=8;i=i/10;}cout<<j<<endl;return 0;}
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摘要：#include <iostream>#include<string.h>using namespace std;int qnum[93];void Display(bool flag[][8]){ static int num = 0; num++; int temp = 0; for (int i=0;i<8;i++) { for (int j=0;j<8;j++) { if (flag[i][j]) temp = temp*10+j+1; } } qnum[num] = temp;}bool Search(bool flag[][8],int m,in
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摘要：#include<stdio.h> int main() { int nCase,nFeet,i;scanf("%d",&nCase);for(i=1;i<=nCase;i++){ scanf("%d",&nFeet); if(nFeet%2 != 0)printf("0 0\n"); else if(nFeet%4 != 0) printf("%d %d\n",nFeet/4+1,nFeet/2); else printf("%d %d\n",nFeet/4,
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摘要：#include <iostream>#include <string>#include <vector>using namespace std;char sixteen[]={'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};
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