A problem is easy时间限制:1000 ms | 内存限制:65535 KB难度:3描述
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 ?
输入
The first line contain a T(T <= 2000) . followed by T lines ,each line contain an integer N (0<=N <= 10^11).
输出
For each case, output the number of ways in one line
样例输入
2
1
3
样例输出
0
1
一下子没想到!
看下吧! 代码ac
#include<stdio.h>
int main()
{long long n,m;scanf("%lld",&m);while(m--){scanf("%lld",&n);int i,s=0;for(i=1;(i+1)*(i+1)<=n+1;i++){if((n+1)%(i+1)==0)s++;}printf("%d\n",s);}return 0;
}