UVa 10879 - Code Refactoring

最新推荐文章于 2017-09-06 09:23:10 发布

原创最新推荐文章于 2017-09-06 09:23:10 发布 · 654 阅读

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#refactoring #encryption #cryptography #scheme #output #algorithm

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本文介绍了一种基于质因数分解的加密算法挑战，讨论了通过找到特定数值的不同质因数对来破解这一加密方案的方法，并提供了一个简单的C语言实现。

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Problem B
Code Refactoring
Time Limit: 2 seconds

"Harry, my dream is a code waiting to be
broken. Break the code, solve the crime."

Agent Cooper

Several algorithms in modern cryptography are based on the fact that factoring large numbers is difficult. Alicia and Bobby know this, so they have decided to design their own encryption scheme based on factoring. Their algorithm depends on a secret code, K, that Alicia sends to Bobby before sending him an encrypted message. After listening carefully to Alicia's description, Yvette says, "But if I can intercept K and factor it into two positive integers, A and B, I would break your encryption scheme! And the K values you use are at most 10,000,000. Hey, this is so easy; I can even factor it twice, into two different pairs of integers!"

Input
The first line of input gives the number of cases, N (at most 25000). N test cases follow. Each one contains the code, K, on a line by itself.

Output
For each test case, output one line containing "Case #x: K = A * B = C * D", where A, B, C and D are different positive integers larger than 1. A solution will always exist.

Sample Input	Sample Output
3 120 210 10000000	Case #1: 120 = 12 * 10 = 6 * 20 Case #2: 210 = 7 * 30 = 70 * 3 Case #3: 10000000 = 10 * 1000000 = 100 * 100000

Problemsetter: Igor Naverniouk

对于这道题目其实非常简单，但是看样例输入和样例输出却很容易让人糊涂，其实只要是找出a*b=k的即可，这题的结果并不唯一。

#include <stdio.h>
#include <string.h>
#include <math.h>
int main()
{
    int i,j,n,m,s,t,t1;
    scanf("%d",&t);
    t1=1;
    while(t--)
    {
        scanf("%d",&n);
        i=2;
        j=0;
        while(1)
        {
            if(n%i==0)
            {
                if(j==0)
                {
                    printf("Case #%d: %d = %d * %d",t1,n,i,n/i);
                    j=1;
                }else
                {
                    printf(" = %d * %d\n",i,n/i);
                    break;
                }
            }
            i++;
        }
        t1+=1;
    }
    return 0;
}