HDU-5924:Mr. Frog’s Problem（不等式变换）

Mr. Frog’s Problem

                                                                        Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                                                                               Total Submission(s): 1506    Accepted Submission(s): 871

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

210 109 27

 

Sample Output

Case #1:110 10Case #2:29 2727 9

思路：把上面那个等式化简去掉分母后得（B*C-A*D）*（A*C-B*D）>=0。即B/A>=D/C且A/B>=D/C。

#include<bits/stdc++.h>using namespace std;int main(){    long long A,B;    int T,cas=1;cin>>T;    while(T--)    {        scanf("%lld%lld",&A,&B);        printf("Case #%d:\n",cas++);        if(A==B)printf("1\n%lld %lld\n",A,B);        else printf("2\n%lld %lld\n%lld %lld\n",A,B,B,A);    }    return 0;}

BA