Uva 10879 Code Refactoring

最新推荐文章于 2019-02-21 13:33:47 发布

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本文介绍了一个基于质因数分解的加密方案挑战，通过寻找特定数值的两组不同因子来破解加密消息。文章提供了一段C语言代码示例，该程序能够接受输入并输出所需的因子组合。

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

题意：

找出一个数的两个不同的因子，直接暴力求解。

#include <stdio.h>

int main() {
	int t;
	int n;
	int cut;
	while(scanf("%d",&t) != EOF) {

		for(int j=1;j<=t;j++) {
			scanf("%d",&n);
			cut = 0;

			int ans1,ans2,ans3,ans4;

			for(int i=2 ;i<=1000000;i++) {
				if(n % i == 0) {
					cut++;
				if(cut == 1) {
					ans1 = i;
					ans2 = n/i;
				}
				else
					if(cut == 2) {
						ans3 = i;
						ans4 = n/i;
					}
					else
						break;
				}
			}
			printf("Case #%d: %d = %d * %d = %d * %d\n",j,n,ans1,ans2,ans3,ans4);
		}	
	}
	return 0;
}