POJ2635----The Embarrassed Cryptographer

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The Embarrassed Cryptographer

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 13032 Accepted: 3515

Description

题意：给定一个数<10的100次幂，求在n范围内是否存在该数的质因子；

思路：同余模的高进制；

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 10143 20667 20667 302573 302573 400 0

Sample Output

GOODBAD 11GOODBAD 23GOODBAD 31

Source

Nordic 2005

10000进制版；

Source CodeProblem: 2635User: 14110103069Memory: 2268KTime: 860MSLanguage: G++Result: Accepted    Source Code    #include <iostream>    #include<cstdio>    #include<cstring>    #include<algorithm>    #include<cmath>    #include<queue>    #include<stack>    #include<set>    #include<map>    #include<cctype>    #define LL long long    #define INF 0x3f3f3f3f    using namespace std;    const int N = 1e6+1e4;    LL prime[N>>2];    bool flag[N];    int k;    char s[110];    LL a[110];    void getprime()    {        k=0;        memset(flag,0,sizeof(flag));        int i,j;        for(i=2;i<N;i++)        {            if(!flag[i])            {                prime[++k]=i;            }            for(j=1;j<=k&&prime[j]*i<N;j++)            {                flag[prime[j]*i]=1;                if(i%prime[j]==0)                    break;            }        }    }    int main()    {         getprime();         int n;         while(~scanf("%s%d",s,&n))         {             if(s[0]=='0'&&n==0)                break;             int len=strlen(s);             int i=0;             int l=0,j;             LL tmp;    bool first=1;             while(i<len)             {                 j=4;                 tmp=0;                 if(first)                 {                     j=len%4;                 }                 if(j==0)                    j=4;                     while(i<len&&j>0)                     {                         tmp=tmp*10+s[i]-'0';                         i++;j--;                     }                     first=0;                 a[++l]=tmp;             }             first =1;            for( i=1;prime[i]<n;i++)            {                tmp=0;                for( j=1;j<=l;j++)                {                    tmp=(tmp*10000+a[j])%prime[i];                }                if(tmp==0)                {                    first=0;                    break;                }            }            if(first)                printf("GOOD\n");            else                printf("BAD %lld\n",prime[i]);         }        return 0;    }

100000版

Source CodeProblem: 2635User: 14110103069Memory: 2268KTime: 688MSLanguage: G++Result: Accepted    Source Code    #include <iostream>    #include<cstdio>    #include<cstring>    #include<algorithm>    #include<cmath>    #include<queue>    #include<stack>    #include<set>    #include<map>    #include<cctype>    #define LL long long    #define INF 0x3f3f3f3f    using namespace std;    const int N = 1e6+1e4;    LL prime[N>>2];    bool flag[N];    int k;    char s[110];    LL a[110];    void getprime()    {        k=0;        memset(flag,0,sizeof(flag));        int i,j;        for(i=2;i<N;i++)        {            if(!flag[i])            {                prime[++k]=i;            }            for(j=1;j<=k&&prime[j]*i<N;j++)            {                flag[prime[j]*i]=1;                if(i%prime[j]==0)                    break;            }        }    }    int main()    {         getprime();         int n;         while(~scanf("%s%d",s,&n))         {             if(s[0]=='0'&&n==0)                break;             int len=strlen(s);             int i=0;             int l=0,j;             LL tmp;            bool first=1;             while(i<len)             {                 j=5;                 tmp=0;                 if(first)                 {                     j=len%5;                     first=0;                 }                 if(j==0)                    j=5;                     while(i<len&&j>0)                     {                         tmp=tmp*10+s[i]-'0';                         i++;j--;                     }                 a[++l]=tmp;             }             first =1;            for( i=1;prime[i]<n;i++)            {                tmp=0;                for( j=1;j<=l;j++)                {                    tmp=(tmp*100000+a[j])%prime[i];                }                if(tmp==0)                {                    first=0;                    break;                }            }            if(first)                printf("GOOD\n");            else                printf("BAD %lld\n",prime[i]);         }        return 0;    }

1000000版

Source CodeProblem: 2635User: 14110103069Memory: 2268KTime: 610MSLanguage: G++Result: Accepted    Source Code    #include <iostream>    #include<cstdio>    #include<cstring>    #include<algorithm>    #include<cmath>    #include<queue>    #include<stack>    #include<set>    #include<map>    #include<cctype>    #define LL long long    #define INF 0x3f3f3f3f    using namespace std;    const int N = 1e6+1e4;    LL prime[N>>2];    bool flag[N];    int k;    char s[110];    LL a[110];    void getprime()    {        k=0;        memset(flag,0,sizeof(flag));        int i,j;        for(i=2;i<N;i++)        {            if(!flag[i])            {                prime[++k]=i;            }            for(j=1;j<=k&&prime[j]*i<N;j++)            {                flag[prime[j]*i]=1;                if(i%prime[j]==0)                    break;            }        }    }    int main()    {         getprime();         int n;         while(~scanf("%s%d",s,&n))         {             if(s[0]=='0'&&n==0)                break;             int len=strlen(s);             int i=0;             int l=0,j;             LL tmp;            bool first=1;             while(i<len)             {                 j=6;                 tmp=0;                 if(first)                 {                     j=len%6;                     first=0;                 }                 if(j==0)                    j=6;                     while(i<len&&j>0)                     {                         tmp=tmp*10+s[i]-'0';                         i++;j--;                     }                 a[++l]=tmp;             }             first =1;            for( i=1;prime[i]<n;i++)            {                tmp=0;                for( j=1;j<=l;j++)                {                    tmp=(tmp*1000000+a[j])%prime[i];                }                if(tmp==0)                {                    first=0;                    break;                }            }            if(first)                printf("GOOD\n");            else                printf("BAD %lld\n",prime[i]);         }        return 0;    }

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