nyoj_216_A problem is easy_201312051117

A  problem is easy

时间限制：1000 ms  |           内存限制：65535 KB

难度：3

 

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

上传者

苗栋栋

 #include <stdio.h>
 #include <math.h>

 int main()
 {
     int T;
     scanf("%d",&T);
     while(T--)
     {
         int i,j,t,n;
         int num=;
         scanf("%d",&n);
         for(i=;i<=sqrt((double)n)+;i++)
         {
             if((n-i)%(i+)==)
             if((n-i)/(i+)>=i)
             num++;
         }
         printf("%d\n",num);
     }
     return ;
 }

优秀代码：

#include<cstring>

#include<cstdio>

#include<map>

#include<string>

#include<algorithm>

#include<vector>

#include<iostream>

#include<cmath>

using namespace std;

#define CLR(arr,val) memset(arr,val,sizeof(arr))

int main()

{

int t,n,cnt=;

//long long num;

int num;

scanf("%d",&t);

while(t--)

{

//  scanf("%lld",&num);

scanf("%d",&num);

int nn=(int)(sqrt(num+1.0)+0.5);

num++;

cnt=;

for(int i=;i<=nn;i++)

if(num%i==) cnt++;

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

}

}

简单数学题