暴力 hdu2601 An easy problem

最新推荐文章于 2020-07-21 21:21:14 发布

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最新推荐文章于 2020-07-21 21:21:14 发布

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本文探讨了儿童时期的数学奇思妙想，通过描述一个梦境中的数学难题，即给定一个数值N，计算有多少种方式可以将其表示为i*j+i+j的形式（其中i和j都是非负整数）。文章提供了解决问题的方法，并通过实例展示如何应用该方法，旨在激发读者对数学的兴趣并提升解决实际问题的能力。

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An easy problem

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7448 Accepted Submission(s): 1768

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

2
1
3

Sample Output

0
1

Author

Teddy

求一个N=i*j+i+j的i j的个数

N+1=(i+1)*(j+1)；所以转换为 N+1的因子数

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
    long long int a;
    int ncase,sum;
    scanf("%d",&ncase);
    while(ncase--)
    {
        sum=0;
        scanf("%lld",&a);
        a++;
        long long k=sqrt(a);
        for(int i=2;i<=k;i++)
            if(a%i==0)
                sum++;
            printf("%d\n",sum);
    }
    return 0;
}