南阳题目216-A problem is easy

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

213

样例输出

01

上传者

这是一个数学题目，找出规律就可以了

我们看题目要求，N= i * j + i + j (0 < i <= j) 

我们让等式两边同时加以1               i*(j+1)+j+1=N+1

所以     (i+1)*(j+1)=N+1

所以只要是N+1可以整除j即可

j的取值范围就是1-sqrt（N）-1

所以计算个数就行了

#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;int main(){int a,b,c,d,i,j,m,n,sum;scanf("%d",&m);while(m--){sum=0;scanf("%d",&n);for(i=1;i<=sqrt(n)-1;i++){if((n+1)%(i+1)==0)sum++;}printf("%d\n",sum);}return 0;}