UVa 10879 Code Refactoring (因数分解)-优快云博客

本文链接：https://blog.youkuaiyun.com/synapse7/article/details/12219853

本文介绍了一个基于整数分解难题的加密方案挑战，通过给出具体的算法实现来解决特定输入数据下的因数分解问题。该算法能够在限定时间内找到两个不同的整数对，使得它们的乘积等于给定的密钥。

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10879 - Code Refactoring

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=100&page=show_problem&problem=1820

"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

water.

完整代码：

/*0.038s*/

#include <cstdio>
#include <cmath>

int main()
{
	int n, k, cnt, i, j;
	scanf("%d", &n);
	for (j = 1; j <= n; ++j)
	{
		scanf("%d", &k);
		printf("Case #%d: %d", j, k);
		cnt = 0;
		for (i = 2;; ++i)
		{
			if (k % i == 0)
			{
				printf(" = %d * %d", i, k / i);
				if (++cnt == 2) break;
			}
		}
		putchar('\n');
	}
	return 0;
}