Mr. Frog’s Problem 水题，打表

最新推荐文章于 2019-04-17 10:48:26 发布

原创最新推荐文章于 2019-04-17 10:48:26 发布 · 706 阅读

0 ·

CC 4.0 BY-SA版权

在一场梦中，主人公遇到了Mr.Frog，并被赋予了一个有趣的数学挑战：寻找符合条件的整数对(C,D)，使得给定的两个正整数A和B与这对整数满足特定的数学不等式。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Mr. Frog’s Problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 18 Accepted Submission(s): 14

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

AB+BA≤CD+DC

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

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

找规律，可以打表，发现只有输入的这两个点符合题意

部分打表结果：

AC代码：很丑~~

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;


const int maxm = 100002;
const int maxn = 100002;
const int inf = 0x3f3f3f3f;

int main()
{
	int t;
	scanf("%d", &t);
	int cnt = 0;
	while(t--) {
		int a, b;
		scanf("%d%d", &a, &b);
		printf("Case #%d:\n", ++cnt);
		if(a == b)
			printf("1\n%d %d\n", a, b);
		else {
			printf("2\n");
			if(a < b) {
				printf("%d %d\n", a, b);
				printf("%d %d\n", b, a);
			}
			else {
				printf("%d %d\n", b, a);
				printf("%d %d\n", a, b);
			}
		}
	}
}