hdu5924Mr. Frog’s Problem

Mr. Frog’s Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1337    Accepted Submission(s): 765

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

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

题目大意：A≤C≤B,A≤D≤B and A/B+B/A≤C/D+D/C,z找到满足关系式的c，d对数 
解题思路：其实经过一些实验你会发现，如果A，B相等，那满足式子的只有一对，即c,d,与a,,b相等，如果两个数不相等，则只有两对，c=a,d=b或c=b,d=a,这样这道题就变得相当简单了

#include <iostream>
#include<stdio.h>
using namespace std;

typedef long long ll;

int main()
{
    int t;
    scanf("%d",&t);
    ll path=1;
    while(t--)
    {
        ll a,b;
        scanf("%lld%lld",&a,&b);
        if(a==b)
        {

            printf("Case #%lld:\n",path++);
            printf("1\n");
            printf("%lld %lld\n",a,b);
        }
        else if(a!=b)
        {
             printf("Case #%lld:\n",path++);
            printf("2\n");
            if(a<b)
            {

            printf("%lld %lld\n",a,b);
            printf("%lld %lld\n",b,a);
            }
            else if(a>b)
            {
                printf("%lld %lld\n",b,a);
            printf("%lld %lld\n",a,b);
            }
        }
    }
    return 0;
}