博客
关于我
强烈建议你试试无所不能的chatGPT,快点击我
Uva 10879 - Code Refactoring
阅读量:7065 次
发布时间:2019-06-28

本文共 1746 字,大约阅读时间需要 5 分钟。

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 #xK = A * B = C * D", where ABC and D are different positive integers larger than 1. A solution will always exist.

 

Sample Input Sample Output
312021010000000
Case #1: 120 = 12 * 10 = 6 * 20Case #2: 210 = 7 * 30 = 70 * 3Case #3: 10000000 = 10 * 1000000 = 100 * 100000

 


Problemsetter: Igor Naverniouk

 

#include
#include
#define MAXN 10000int prime[MAXN];int flag[MAXN];is_prime(){ int i, j, n = 0; memset(flag, 0, sizeof(flag)); for(i=2; i

解题思路:

这题比较简单,因为是特判题(Special judge)题目的要求很宽,题目的背景设得太渣了,什么  After listening carefully to Alicia's description 然后就没了逻辑性了,还是直接看Sample好,

思路跟上面几题一样,还是找因子,然后累乘找到能整除n的两个不同的数a, c ( a*c != n),然后按题目要求输出就行了

转载于:https://www.cnblogs.com/liaoguifa/archive/2013/03/10/2952308.html

你可能感兴趣的文章
RabbitMQ 基础概念介绍
查看>>
1117bootstrap
查看>>
centos6.5上卸载和安装JDK7
查看>>
从文件加载至NSData
查看>>
分享一下自己用c++写的小地图
查看>>
马尔可夫模型
查看>>
bzoj 1697: [Usaco2007 Feb]Cow Sorting牛排序
查看>>
js面向对象编程
查看>>
Tensorflow serving的编译
查看>>
JAVA API----Math类和Random类
查看>>
155. Min Stack
查看>>
Android深度探索(卷1)HAL与驱动开发学习笔记(5)
查看>>
JavaScript高级
查看>>
静态成员函数访问构造函数
查看>>
scla-基础-函数-元组(0)
查看>>
How to Convert a Single-Instance ASM to Cluster ASM [ID 452758.1]
查看>>
触发a标签
查看>>
所有Windows7下游戏的全屏问题
查看>>
UIImage转换成UIView
查看>>
一种专门用于前后端分离的web服务器(JerryServer)
查看>>