ACM 数论 A problem is easy

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

0
1

题意简单，就数n有多少种可能摆成i*j+i+j的形势。

 重点：！！！学会数学推导变换

n=i*j+i+j

n+1=i*j+i+j+1

n+1=（i+1）*（j+1）

由于i，j均为整数，所以n+1也是整数

所以可以将n+1分解成两个整数的乘积。

     从1到根下n暴力求解。

#include<stdio.h>

#include<math.h>

intmain()

{

                            int i,j,k,l,m,n;

                            scanf("%d",&m);

                            while(m--)

                            {

                                scanf("%d",&n);

                                n=n+1;

                                l=sqrt(double(n));

                                for(i=2,k=0;i<=l;i++)

                                {

                                   if(n%i==0)

                                       k++;

                                }

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

                            }

}