大数取MOD。。。
The Embarrassed Cryptographer
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 11359 | Accepted: 3026 |
Description

What Odd Even did not think of, was that both factors in a key should be large, not just their product. It is now possible that some of the users of the system have weak keys. In a desperate attempt not to be fired, Odd Even secretly goes through all the users keys, to check if they are strong enough. He uses his very poweful Atari, and is especially careful when checking his boss' key.
Input
The input consists of no more than 20 test cases. Each test case is a line with the integers 4 <= K <= 10100 and 2 <= L <= 106. K is the key itself, a product of two primes. L is the wanted minimum size of the factors in the key. The input set is terminated by a case where K = 0 and L = 0.
Output
For each number K, if one of its factors are strictly less than the required L, your program should output "BAD p", where p is the smallest factor in K. Otherwise, it should output "GOOD". Cases should be separated by a line-break.
Sample Input
143 10
143 20
667 20
667 30
2573 30
2573 40
0 0
Sample Output
GOOD
BAD 11
GOOD
BAD 23
GOOD
BAD 31
Source
Nordic 2005#include <iostream> #include <cstdio> #include <cstring> using namespace std; int prim[80010],pn=0; int p[1000010]; int big[100],cnt;/// wan jing zhi char str[5000]; void getPRIM() { for(int i=2;i<=1000000;i++) p for(int i=2;i*i<=1000000;i++) { if(p for(int j=2;j*i<=1000000;j++) { p[j*i]=0; } } for(int i=0;i<=1000000;i++) { if(p { prim[pn++]=i; } } } void change2big(char str[500]) { int len=strlen(str),e=0,sum=0;cnt=0; for(int i=0;i<len;i++) { sum=sum*10+str e++; if(e>=4) { big[cnt++]=sum; sum=0;e=0; } } if(sum!=0) { big[cnt++]=sum; } } bool letsMOD(int mod) { unsigned long long int ans=0; for(int i=0;i<cnt;i++) { ans=((ans*10000)+(big // printf("%d%%%d--->%d ",big } if((int)ans==0) return true; else return false; } int main() { getPRIM();int L; while(scanf("%s%d",str,&L)!=EOF) { if(L==0&&strcmp(str,"0")==0) break; memset(big,0,sizeof(big));cnt=0; change2big(str); int pos=-1; for(int i=0;prim { //cout<<prim if(letsMOD(prim { pos=prim break; } } if(pos!=-1) printf("BAD %d ",pos); else printf("GOOD "); } return 0; } |