NYOJ -216 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

2
1
3

样例输出

0
1

虽然是英文题，但不是难理解，但凡对数字有些敏感的，应该都能看懂吧。。。

很显然，找出满足n=i * j + i + j (0 < i <= j) 的个数。。

但很多人（也包括我T T）第一想法都是用两个for循环，但无论怎样改都是超时ヽ(≧□≦)ノ。因为N (0<=N <= 10^11)，，这说明什么呢，反正我以后要好好看题了。

超时代码：

#include<stdio.h>
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        int sum=0;
        for(int j=1; j<=n; j++)
        {
            for(int i=1; i<=j; i++)
            {
                if(n==(i*j+i+j))
                    sum++;
            }
        }
        printf("%d\n",sum);
    }
    return 0;
}

其实这道题用了一个技巧，我看别人都是这样写的，才发现数学很重要T T.

n=i*j+i+j转化为n+1=i*j+i+j+1即（n+1）=(i+1)*(j+1),

代码如下：

#include<stdio.h>
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        int sum=0;
        for(int i=2; i*i<=(n+1); i++)
        {
            if((n+1)%i==0)sum++;
        }
        printf("%d\n",sum);
    }
    return 0;
}

本人菜鸟一个，，多多见谅。