nyoj_216_A problem is easy

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

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

巴特西

nyoj_216_A problem is easy_201312051117

A problem is easy

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