POJ 2635 The Embarrassed Cryptographer（高精度取模 + 同余模定理）

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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

    题意：给定一个大数str，str是两个大素数的乘积的值。再给定一个int内的数m问这两个大素数中最小的一个是否小于m，如果小于则输出这个素数。

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>

using namespace std;

char str[100010];
int a[100100];
int m;
const int MAXN = 20000000;
bool flag[MAXN];
int primes[MAXN / 3], pi;

void GetPrime()
{
    int i, j;
    pi = 0;
    memset(flag, false, sizeof(flag));
    for (i = 2; i < MAXN; i++)
    {
        if (!flag[i])
        {
            primes[pi++] = i;
        }
        for (j = 0; (j < pi)  && (i * primes[j] < MAXN); j++)
        {
            flag[i * primes[j]] = true;
            if (i % primes[j] == 0)
            {
                break;
            }
        }
    }
}

int main()
{
    GetPrime();
    while(scanf("%s%d",str,&m)!=EOF)
    {
        if(m == 0 && strcmp(str,"0") == 0)
        {
            break;
        }
        int len = strlen(str);
        int ll = (len+2)/3;
        int ans = len%3;
        memset(a,0,sizeof(a));
        for(int i=0; i<len; i++)   ///将10进制数字转变成1000制的数字，
        {
            int pk = i/3;
            a[pk] = a[pk]*10 + (str[i] - '0');
        }
        int p = 0;
        int flag = 0;
        int num;
        while(primes[p]<m)   
        {
            int sum = 0;
            for(int i=0;i<ll;i++)  /// 同余模定理
            {
                if(i == ll-1)
                {
                    if(ans == 0)
                    {
                        sum = (sum*1000 + a[i])%primes[p];
                    }
                    else if(ans == 1)
                    {
                        sum = (sum*10 + a[i])%primes[p];
                    }
                    else
                    {
                        sum = (sum*100 + a[i])%primes[p];
                    }
                }
                else
                {
                    sum = (sum*1000 + a[i])%primes[p];
                }

            }
            if(sum == 0)
            {
                num = primes[p];
                flag = 1;
                break;
            }
            p++;
        }
        if(flag == 1)
        {
            printf("BAD %d\n",num);
        }
        else
        {
            printf("GOOD\n");
        }
    }
}

POJ 2635 The Embarrassed Cryptographer（高精度取模 + 同余模定理）

标签：

原文地址：http://blog.csdn.net/yeguxin/article/details/47046037