POJ 2635.The Embarrassed Cryptographer

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题目:http://poj.org/problem?id=2635

AC代码(C++):

#include <iostream>#include <algorithm>#include <stdio.h>#include <vector>#include <queue>#include <math.h>#include <string>#include <string.h>#include <bitset>#define INF 0xfffffff#define MAXN 100105using namespace std;int primes[1000000];int k[50];int cnt;void table(){int pNum=0;      primes[pNum++]=2;        for(int i=3;i<=1000010;i+=2){          bool flag=true;          for(int j=0;primes[j]*primes[j]<=i;j++)            if(!(i%primes[j]))              {                  flag=false;                  break;              }          if(flag)              primes[pNum++]=i;      }      return;}int main(){char str[105];int l,len,maxcnt;table();while(cin>>str>>l){if(str[0]=='0'&&l==0)break;len = strlen(str);maxcnt = len/3;if(len%3==0)maxcnt--;for(cnt = maxcnt; cnt >= 0; cnt--){k[cnt] = 0;for(int j = 3; j >= 1; j--){if(len-(maxcnt-cnt)*3-j < 0)continue;k[cnt]*=10;k[cnt]+=str[len-(maxcnt-cnt)*3-j]-'0';}}int tmp;bool flag = true;for(int i = 0; primes[i] < l; i++){tmp = k[0]%primes[i];for(int j = 1; j <= maxcnt; j++){tmp = (tmp*1000+k[j])%primes[i];}if(tmp!=0)continue;cout<<"BAD "<<primes[i]<<endl;flag = false;break;}if(flag)cout<<"GOOD\n";}}

总结: 题意很简单, 就是大数有点难处理. 看了dalao解题报告用了千进制来储存k, 然后用同余定理做.

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