HDU 2601 An easy problem （数学）

最新推荐文章于 2022-11-11 17:52:42 发布

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最新推荐文章于 2022-11-11 17:52:42 发布

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分类专栏： HDUOJ ACM_物理/化学/数学公式文章标签： HDU 2601 An easy problem 数学

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本文介绍了一个数学问题的解决方法，该问题是求解特定形式下整数N的组合方式数量。通过将问题转化为寻找N+1的有效因数对来简化计算过程，并提供了一段使用C++实现的AC代码。

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An easy problem

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9133 Accepted Submission(s): 2243

Problem Description

When Teddy was a child , he was always thinking about some simple math problems ,such as “What it’s 1 cup of water plus 1 pile of dough ..” , “100 yuan buy 100 pig” .etc..

One day Teddy met a old man in his dream , in that dream the man whose name was“RuLai” gave Teddy a problem :
Given an N , can you calculate how many ways to write N as i * j + i + j (0 < i <= j) ?
Teddy found the answer when N was less than 10…but if N get bigger , he found it was too difficult for him to solve.
Well , you clever ACMers ,could you help little Teddy to solve this problem and let him have a good dream ?

Input

The first line contain a T(T <= 2000) . followed by T lines ,each line contain an integer N (0<=N <= 10¹⁰).

Output

For each case, output the number of ways in one line.

Sample Input

Sample Output

0
1

Author

Teddy

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

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题解：

因为n=i*j+i+j+1-1=(i+1)(j+1)-1;
n+1=(i+1)(j+1);
所以sqrt(n+1)>=(i+1)。

AC代码：

#include <iostream>
#include<stdio.h>
using namespace std;
#include<cmath>
//因为n=i*j+i+j+1-1=(i+1)(j+1)-1;
//n+1=(i+1)(j+1); 
//所以sqrt(n+1)>=(i+1) 
int main()
{
    ios::sync_with_stdio(false);
    long long n;
    int t;
    cin>>t;
    while(t--)
    {

    	cin>>n;
    	int num;
    	int sum=0;
    	num=sqrt(n+1);
    	for(int i=1;i<num;i++)
    	{
    		if((n+1)%(i+1)==0)  //所以(j+1)为整数 
    		sum++;
		}
	cout<<sum<<endl;
	}
    return 0;
}