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

#include <bits/stdc++.h> using namespace std; typedef long long ll; using namespace std;int main(){int t;long long n,i,j,ans;scanf ("%d",&t);while (t--){ans=0;scanf ("%lld",&n);n++;for (i=2;i*i<=n;i++){if (n%i==0)ans++;}printf ("%lld\n",ans);}return 0; }

總結

以上是生活随笔為你收集整理的数学--数论-- HDU 2601 An easy problem（约束和）的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。