An easy problem

时间：2016-11-15 10:13:28 阅读：128 评论：0 收藏：0 [点我收藏+]

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An easy problem

Time Limit:3000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

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..
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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 <= 10 ¹⁰).

Output

For each case, output the number of ways in one line.

Sample Input

2
1
3

Sample Output

0
1

#include<iostream>
#include<stdio.h>
#include<math.h>
using namespace std;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        long long n;
        scanf("%I64d",&n);
        n++;
        int ans=0;
        long long m=sqrt(n);
        for(long long i=2;i<=m;i++)
        {
            if(n%i==0) ans++;
        }
        printf("%d\n",ans);
    }
}

View Code

n=i*j+i+j

==>n+1=(i+1)*(j+1)

An easy problem

标签：who close blog lines title div scanf lap blocks

原文地址：http://www.cnblogs.com/superxuezhazha/p/6064322.html

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