HDU 2601 An easy problem(数学转换)

An easy problem

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9162    Accepted Submission(s): 2254

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

213

 

Sample Output

01

 

题如其名，真的是一道easy problem。

要注意的是N的数据范围很大，因此需要用long long这种类型定义N；

下面说数学转换：

N=i*j+i+j=（i+1）*（j+1）-1     

让i=i+1；j=j+1；

那么N+1=i*j（i<=j<=2）;

根据这个算是可以得到

i<=sqrt(N+1);

那么题目就可以得到解决了，

下面的代码采用cin，运行时间2.9s，题目要求时间是3s，所以水水的过了。

用scanf的话会稍微快一点。

下面是AC代码：

#include<iostream>
#include<string.h>
#include<string>
#include<stack>
#include<cmath>
#include<queue>
#include<vector>
#include<algorithm>


using namespace std;


const int inf=0x7FFFFFFF;
const int M=1000+10;


int dp[M];


int v,t,val[M],vo[M];
long long n;
int main()
{
    cin>>t;
    while(t--)
    {
        cin>>n;
        long long cnt=0;
        n++;
        for(int i=2;i<=sqrt(n);i++)
        {
            if(n%i==0) cnt++;
        }
        cout<<cnt<<endl;
    }
    return 0;
}