（HDU 5924）Mr. Frog’s Problem 思维水题 2016CCPC东北地区大学生程序设计竞赛 - 重现赛

本文主要是介绍（HDU 5924）Mr. Frog’s Problem 思维水题 2016CCPC东北地区大学生程序设计竞赛 - 重现赛，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=5924

Mr. Frog’s Problem
Problem Description
One day, you, a clever boy, feel bored in your math class, and then fall asleep without your control. In your dream, you meet Mr. Frog, an elder man. He has a problem for you.

He gives you two positive integers A and B, and your task is to find all pairs of integers (C, D), such that A≤C≤B,A≤D≤B and A/B+B/A≤C/D+D/C

Input
first line contains only one integer T (T≤125), which indicates the number of test cases. Each test case contains two integers A and B (1≤A≤B≤1018).

Output
For each test case, first output one line “Case #x:”, where x is the case number (starting from 1).

Then in a new line, print an integer s indicating the number of pairs you find.

In each of the following s lines, print a pair of integers C and D. pairs should be sorted by C, and then by D in ascending order.

Sample Input
2
10 10
9 27

Sample Output
Case #1:
1
10 10
Case #2:
2
9 27
27 9

Source
2016CCPC东北地区大学生程序设计竞赛 - 重现赛

分析：
水题。由于A,B是区间的两个端点，所以A/B + B/A 已经就是最大值了。所以当A==B 时答案就是A，B；当不相等时，答案为A,B和B,A

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;int main()
{int t,kase=1;char a[30],b[30];scanf("%d",&t);while(t--){cin>>a>>b;if(strcmp(a,b)==0){printf("Case #%d:\n1\n",kase++);cout<<a<<" "<<b<<endl;}else{printf("Case #%d:\n2\n",kase++);cout<<a<<" "<<b<<endl;cout<<b<<" "<<a<<endl;}}return 0;
}

这篇关于（HDU 5924）Mr. Frog’s Problem 思维水题 2016CCPC东北地区大学生程序设计竞赛 - 重现赛的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！

---