hdu 2601

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An easy problem

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5693 Accepted Submission(s): 1391

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

Sample Output

Author

Teddy

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

其实这个问题还是比较水的，思路其实不难，但是想要将其做出来并不是件容易的事情，i+j+i*j=n，(i+1)*(j+1)=n+1，求出有多少对i，j,其实就是再n+1求有多少个解罢了，所以，直接挨个判断就是了

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>using namespace std;int main(){    int i,j,k;    __int64 n;    int g;    int t;    scanf("%d",&t);    while(t--)    {        scanf("%I64d",&n);        n++;        //printf("%d\n",n);        g=(int)sqrt((double)n);        int sum=0;        for(i=2;i<=g;i++)        {            if(n%i==0)            {                sum++;            }        }        printf("%d\n",sum);    }    return 0;}

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