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(HDU5924)Mr.Frog’sProblem思维水题

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题目连接:http:acm.hdu.edu.cnshowproblem.php?pid5924Mr.Frog’sProblemProblemDescriptionOn 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
题目连接:http:acm.hdu.edu.cnshowproblem.php?pid5924Mr.Frog’sProblemProblemDescriptionOn

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 #include #include #include #include using namespace std;int main(){ int t,kase=1; char a[30],b[30]; scanf("%d", while(t--) { cin>>a>>b; if(strcmp(a,b)==0) { printf("Case #%d:\n1\n",kase++); cout<" "<
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