HDU_1017_AMathematicalCuriosity
来源:互联网 发布:20岁的眼泪 知乎 编辑:程序博客网 时间:2024/06/06 01:03
A Mathematical Curiosity
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 31813 Accepted Submission(s): 10197
Problem Description
Given two integers n and m, count the number of pairs of integers (a,b) such that 0 < a < b < n and (a^2+b^2 +m)/(ab) is an integer.
This problem contains multiple test cases!
The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.
The output format consists of N output blocks. There is a blank line between output blocks.
This problem contains multiple test cases!
The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.
The output format consists of N output blocks. There is a blank line between output blocks.
Input
You will be given a number of cases in the input. Each case is specified by a line containing the integers n and m. The end of input is indicated by a case in which n = m = 0. You may assume that 0 < n <= 100.
Output
For each case, print the case number as well as the number of pairs (a,b) satisfying the given property. Print the output for each case on one line in the format as shown below.
Sample Input
110 120 330 40 0
Sample Output
Case 1: 2Case 2: 4Case 3: 5
Source
East Central North America 1999, Practice
Recommend
JGShining
简单模拟题目条件,穷举就可以了。
注意题目中说的两块之间有空行,因此最后一块完了没有
#include <iostream>#include <stdio.h>using namespace std;int main(){ int t; int n,m; int ca; int co; scanf("%d",&t); while(t--) { ca=1; while(1) { co=0; scanf("%d%d",&n,&m); if(!n&&!m) break; for(int i=1;i<n;i++) for(int j=i+1;j<n;j++) if((i*i+j*j+m)%(i*j)==0) co++; printf("Case %d: %d\n",ca,co); ca++; } if(t) //这里的条件要注意 printf("\n"); } return 0;}
0 0
- HDU_1017_AMathematicalCuriosity
- uC/OS-II 函数之OSInit()
- 手动安装fastboot驱动
- Letter Combinations of a Phone Number
- thinkpad T430s 在win8上安装ubuntu双系统(UEFI开启)
- RHEL7破解root密码
- HDU_1017_AMathematicalCuriosity
- leetcode-190-Reverse Bits
- 转吧总结,一些零零碎碎的东西
- python encode和decode函数说明
- datatable
- leetcode--Maximum Depth of Binary Tree
- 手机APP自动化持续集成方案
- java trim
- netfilter 讲解 ,讲的很好