nyoj_216_A problem is easy

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

 1 #include <stdio.h>
 2 #include <math.h>
 3 
 4 int main()
 5 {
 6     int T;
 7     scanf("%d",&T);
 8     while(T--)
 9     {
10         int i,j,t,n;
11         int num=0;
12         scanf("%d",&n);
13         for(i=1;i<=sqrt((double)n)+1;i++)
14         {
15             if((n-i)%(i+1)==0)
16             if((n-i)/(i+1)>=i)
17             num++;
18         }
19         printf("%d
",num);
20     }
21     return 0;
22 }

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=0;

//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=0;

for(int i=2;i<=nn;i++)

if(num%i==0) cnt++;

printf("%d
",cnt);

}

 

}

简单数学题