HDU 1017 A Mathematical Curiosity (输出格式,穷举)

#include<stdio.h>
int main()
{
    int N;
    int n,m;
    int a,b;
    int cas;
    scanf("%d",&N);
    while(N--)
    {
        cas=1;//必须在这儿初始化cas,坑
        while(scanf("%d%d",&n,&m),n||m)
        {
            int count=0;
            for(a = 1; a < n; a++)//穷举法
            {
            	for(b = a + 1; b < n; b++)
            	{
                	if((a*a + b*b + m) % (a * b) == 0)
                		count++;
            	}
        	}
    		printf("Case %d: %d\n",cas++,count);
     	}
        if(N)
        	printf("\n");//注意输出换行的位置
    }
    return 0;
}

  

时间: 2024-12-25 10:05:14

HDU 1017 A Mathematical Curiosity (输出格式,穷举)的相关文章

HDU 1017 A Mathematical Curiosity【看懂题意+穷举法】

//2014.10.17    01:19 //题意: //先输入一个数N,然后分块输入,每块输入每次2个数,n,m,直到n,m同一时候为零时 //结束,当a和b满足题目要求时那么这对a和b就是一组 //注意: //每一块的输出中间有一个回车 A Mathematical Curiosity Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s)

Hdu 1017 A Mathematical Curiosity

 A Mathematical Curiosity Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 41387    Accepted Submission(s): 13311 Problem Description Given two integers n and m, count the number of pairs of inte

HDU 1017 A Mathematical Curiosity (数学)

A Mathematical Curiosity Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 25871    Accepted Submission(s): 8174 Problem Description Given two integers n and m, count the number of pairs of integ

HDU 1017 A Mathematical Curiosity (枚举水题)

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

(HDU)1017 --A Mathematical Curiosity(数学好奇心)

描述 给定两个整数n和m,计数整数对(a,b)的数目,使得0 <a <b <n,并且(a ^ 2 + b ^ 2 + m)/(ab)是一个整数. 这个问题包含多个测试用例! 输入的第一行是整数N,然后是空白行,后跟N个输入块. 每个输入块采用问题说明中指示的格式. 输入块之间有空行. 输出格式由N个输出块组成. 输出块之间有一个空行. 输入 您将在输入中获得多个样例. 每个情况由包含整数n和m的行指定. 输入结束由n = m = 0的情况表示.您可以假设0 <n <= 100

Hdu oj 1017 A Mathematical Curiosity

题目:点击打开链接 #include<stdio.h> int main() { int t; scanf("%d",&t); while(t--) { int a=0; int m,n; while(scanf("%d%d",&n,&m)) { if(m==0&&n==0) break; a++; int i,j,k,z=0; for(i=1;i<n;i++) for(j=i+1;j<n;j++)//

HDU ACM 1017 A Mathematical Curiosity 水题

分析:水题,但要注意格式. #include<iostream> using namespace std; int core(int n,int m) { int i,j,ans=0; for(i=1;i<n;i++) for(j=i+1;j<n;j++) if((i*i+j*j+m)%(i*j)==0) ans++; return ans; } int main() { int T,t,n,m; bool fg=false; scanf("%d",&T

HDU 1017A Mathematical Curiosity (暴力统计特殊要求个数)

传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1017 A Mathematical Curiosity Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 47658    Accepted Submission(s): 15285 Problem Description Given two

HDU 5339 Untitled (递归穷举)

题意:给定一个序列,要求从这个序列中挑出k个数字,使得n%a1%a2%a3....=0(顺序随你意).求k的最小值. 思路:排个序,从大的数开始模起,这是因为小的模完还能模大的么? 每个元素可以选,也可以不选,两种情况.递归穷举每个可能性,O(2n). 1 //#include <bits/stdc++.h> 2 #include <cstdio> 3 #include <cstring> 4 #include <map> 5 #include <al