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

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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

Output

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

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 __int64
char 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;
}

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

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原文地址：http://blog.csdn.net/winddreams/article/details/43021933