nyoj_216_A problem is easy_201312051117-白红宇

nyoj_216_A problem is easy_201312051117

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发布时间：2019-06-09

本文共 1839 字，大约阅读时间需要 6 分钟。

A problem is easy

时间限制：

1000 ms | 内存限制：

65535 KB

难度：

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

上传者

1 #include 
       
         2 #include 
        
          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\n",num);20     }21     return 0;22 }

优秀代码：

#include
       
        #include
        
         #include
         #include
          
           #include
           
            #include
            
             #include
             
              #include
              
               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\n",cnt);} }

简单数学题

转载于:https://www.cnblogs.com/xl1027515989/p/3459189.html

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