poj2635--The Embarrassed Cryptographer(数论篇1，大数取模)

The Embarrassed Cryptographer

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 12496 Accepted: 3330

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 10143 20667 20667 302573 302573 400 0

Sample Output

GOODBAD 11GOODBAD 23GOODBAD 31

Source

Nordic 2005

 

 

题目要求：给出两个数k和l，k是由两个素数相乘得到，如果素数小于l，输出BAD 和那个素数，否则输出GOOD。

先打出素数表，其中一定要有大于100万的素数，把在l内的素数与k去模，其中k是10^100用到大数取余，将k分为三位数的段a[0] a[1]，在计算

temp = a[0]%phi[i]

temp = (a[1]*1000+temp)%phi[i]

 

#include <cstdio>#include <cstring>#include <algorithm>using namespace std ;#define LL __int64char str[200] ;int n , a[100] , num ;int vis[2100000] , phi[1100000] , cnt ;void init(){    int i , j ;    cnt = 0 ;    memset(vis,0,sizeof(vis)) ;    for(i = 2 ; i <= 2000000 ; i++)    {        if( !vis[i] ) phi[cnt++] = i ;        for(j = 0 ; j < cnt ; j++)        {            if( i*phi[j] >= 2000000 )                break ;            vis[i*phi[j]] = 1 ;            if( i%phi[j] == 0 )                break ;        }    }}int f(int k){    LL sum = 0 , i , l ;    for(i = num-1 ; i >= 0 ; i--)    {        sum = sum*1000 + a[i] ;        sum %= k ;    }    if( sum == 0 )        return 1 ;    return 0;}int main(){    int i , j , l , flag , s ;    init() ;    while( scanf("%s %d", str, &n) != EOF )    {        if( strlen(str) == 1 && str[0] == '0' && n == 0 ) break ;        l = strlen(str) ;        memset(a,0,sizeof(a)) ;        num = 0 ;        flag = 0 ;        for(i = l-1 ; i >= 0 ; i--)        {            flag++ ;            if( flag == 3 )            {                a[num] = (str[i]-'0')*100 + (str[i+1]-'0')*10 + ( str[i+2] -'0' ) ;                num++ ;                flag = 0 ;            }        }        if( flag == 1 )            a[num++] = str[0] - '0' ;        else if( flag == 2 )            a[num++] = (str[0] - '0') * 10 + str[1] - '0' ;        for(i = 0 ; i < cnt ; i++)        {            if( phi[i] >= n ) break ;            if( f(phi[i]) )                break ;        }        if( phi[i] >= n || i == cnt )            printf("GOOD\n") ;        else            printf("BAD %d\n", phi[i]);    }    return 0;}