A problem is easy

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2017/02/09 17:27
阅读数 15

 

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A problem is easy

http://acm.nyist.net/JudgeOnline/problem.php?pid=216

时间限制:1000 ms  |  内存限制:65535 KB

难度:3

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

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 ?

 

我的代码,根据已知条件,计算i的上限,化简式子

#include <stdio.h>
#include <math.h>
int main(int argc, char *argv[])
{
	int T,count;
	long long int n,i,j;
	scanf("%d",&T);
	while(T--){
		count=0;
		scanf("%lld",&n);
		j=(int)sqrt(n+1)-1;			//计算出i的上限,不能在for循环中重复计算,否则超时 
		for(i=1;i<=j ;i++){			
			if( (n+1) % (i+1) == 0)
				count++;
		}
		printf("%d\n",count);
	}
	return 0;
}

 

题目推荐

 
#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\n",cnt);
	}
	
}
        

 

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