JHDU 2601 An easy problem （数学）

title: An easy problem 数学杭电2601

tags: [数学]

题目链接

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

分析：

注意到 ( i +1 ) * ( j + 1 ) = i * j + i + j + 1= n + 1；而且这里的 n 比较大，只能跑一层循环

代码：

#include<cstdio>

#include<cstring>

#include<cmath>

#define LL long long

using namespace std;

LL n;

int main()

{

    int t;

    scanf("%d",&t);

    while(t--)

    {

        scanf("%lld",&n);

        int ans=0;

        for(LL i=2; i*i<=n+1; i++)

        {

            if((n+1)%i==0)

                ans++;

        }

        printf("%d\n",ans);

    }

    return 0;

}

巴特西

JHDU 2601 An easy problem （数学）

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巴特西

JHDU 2601 An easy problem （数学 ）

最新文章

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JHDU 2601 An easy problem （数学）