NYOJ A problem is easy(水题）_this problem is easy. you just need to output an i-优快云博客

本文链接：https://blog.youkuaiyun.com/chat_c/article/details/50478858

本文探讨了如何通过计算特定数学表达式的组合方式来解决复杂的数学问题，详细介绍了问题背景、求解思路及算法实现，并提供了实例输入输出解析。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Time Limit:3000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

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

题意：给一个N，求有多少个0<i<=j,满足i*j+i+j=N?

分析：求i*j+i+j=N 就是求（i+1)*(j+1)=N+1 的个数，所以求N+1的所有小于等于sqrt(N+1)的因数（不含1）个数即可

#include<stdio.h>
#include<math.h>
#define ll long long
int main()
{
    ll N;
    int t,ans=0;
    scanf("%d",&t);
    while(t--)
    {
        ans=0;
        scanf("%I64d",&N);
        N++;
        ll m=(int)sqrt(N)+1;
        for(int i=2;i<m;i++)if(N%i==0)ans++;
        printf("%d\n",ans);
    }
}