The Embarrassed Cryptographer Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 10450 Accepted: 2752 Description The young and very promising cryptographer Odd Even has implemented the security module of a large system with thousands of users, which is now in use in his company. The cryptographic keys are created from the product of two primes, and are believed to be secure because there is no known method for factoring such a product effectively. 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 這題大意啊 ，確定思路後不斷的Wa，對著數據無奈的發現我數組開小了了，總的思路來說還是枚舉 接近700ms過了 [cpp] #include <stdio.h> #include <string.h> #include <math.h> int a[1000001]; char s1[101]; __int64 b[30]; int main() { __int64 i,j,n,m,t,l,weishu; __int64 s; memset(a,0,sizeof(a)); for(i=2;i<=1000000;i++) { if(a[i]==0) { for(j=2;i*j<=1000000;j++) { a[i*j]=1; } } } while(scanf("%s %I64d",s1,&m)!=EOF) { l=strlen(s1); for(i=0;i<=l-1;i++) { if(s1[i]!='0') { break; } } if(i==l&&m==0) { break; } memset(b,0,sizeof(b)); for(i=0,j=0,s=0;i<=l-1;i++) { b[j]=b[j]*10+(s1[i]-'0'); s++; if((i+1)%9==0||i==l-1) { j++; weishu=s; s=0; } } n=j; for(i=2;i<=m-1;i++) { if(a[i]==0) { for(j=0,s=0;j<=n-1;j++) { if(j!=n-1) { s=s*1000000000+b[j]; }else { for(t=1;t<=weishu;t++) { s=s*10; } s+=b[j]; } s=s%i; } if(s==0) { break; } } } if(i==m) { www.2cto.com printf("GOOD\n"); }else { printf("BAD %I64d\n",i); } } return 0; }
