HDU - Mr. Frog’s Problem(数学判断)

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5924
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

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  and 

Input

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

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

Problem solving report:

Description: 给你两个正整数A和B，要求找出所有的整数对(C,D)满足：A≤C≤B,A≤D≤B且A/B+B/A≤C/D+D/C。
Problem solving: 因为C/D+D/C是对称的，所以我们不妨假设D≥C，故可令D=C+e，(e>=0)
为了简化运算,我们令D=C+k(k≥0)，即C=D-k，

因由上式可知，当k越大时，C/D+D/C越大，而k又是D与C的差值，故当C=min(C)=A，D=max(D)=B时，k才是最大的，此时C/D+D/C达到最大，等于A/B+B/A，即C/D+D/C的最大值为A/B+B/A，故满足A/B+B/A≤C/D+D/C的情况只有A/B+B/A=C/D+D/C这种情况，故满足A/B+B/A≤C/D+D/C的解仅有A==C&&B==D||A==D&&B==C，而当A==B时,解唯一,即A==B==C==D。

Accepted Code:

/* 
 * @Author: lzyws739307453 
 * @Language: C++ 
 */
#include <bits/stdc++.h>
using namespace std;
int main() {
    int t, kase = 0;
    long long A, B;
    scanf("%d", &t);
    while (t--) {
        scanf("%lld%lld", &A, &B);
        printf("Case #%d:\n", ++kase);
        if (A != B) {
            puts("2");
            if (A > B)
                swap(A, B);
            printf("%lld %lld\n", A, B);
            printf("%lld %lld\n", B, A);
        }
        else printf("1\n%lld %lld\n", A, B);
    }
    return 0;
}