hdoj 1017 （注意细节）

最新推荐文章于 2019-05-11 17:25:25 发布

原创最新推荐文章于 2019-05-11 17:25:25 发布 · 546 阅读

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本文深入探讨了C++编程的高级技巧与实用实例，涵盖了从基础语法到复杂概念的全面指南，帮助开发者提高编程效率与解决问题的能力。

这道题的英文太多，尤其是一开始我都不知道第一个字母是用来干啥的！

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.

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

Sample Output

  
   Case 1: 2
Case 2: 4
Case 3: 5

Problem Description

Input

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

Sample Output

  
   Case 1: 2
Case 2: 4
Case 3: 5

先输入一个数N然后会分N块输入，每块每次输入2个数，n,m，n=m=0时结束，当a和b满足0<a<b<n且使(a^2+b^2 +m)/(ab) 的值为整数时，那么这对a和b就是一组，输出这样的组数，一行输入，跟着一样输出!

AC码如下

<span style="font-size:24px;">#include<cstdio>
//#include<math.h>//以后要学会调用函数，用这个头文件！！
int main()
{	  int r;
     scanf("%d",&r);
     while(r--)
     {	int n,m,s,a,b,p,flag=0;//</span><span style="font-size:18px;">注意这个案例计数器的位置，当r减少时，计数器也要重新从1开始</span><span style="font-size:24px;">
     	while(scanf("%d%d",&n,&m))//第一次就定义到外边了，wa！
	{
		if(n==m&&n==0)
		break;
		for(p=0,a=1;a<n-1;a++)
		{
			for(b=a+1;b<n;b++)
			{
				s=(a*a+b*b+m)%(a*b);//切记不能换成pow！！！
				if(s==0)
				p++;
			}
		}
		flag++;
		printf("Case %d: %d\n",flag,p);
	}</span>

c++的pow( )函数调用的问题

2013-11-23 21:44plkm3 分类：C/C++ | 浏览 612 次

编程语言

我在程序中添加了头文件#include <cmath>；

但调用函数pow（）后，程序报错：error: call of overloaded `pow(int, int)' is ambiguous；
就算是最简单的调用都报错；
答

cmath里面对于Pow函数只有

这些重载。

你得把第一个参数转换成以上所需要的类型。

Problem Description

Input

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