ACM之杭电——1001_Sum Problem

最新推荐文章于 2020-07-07 17:49:14 发布

cmolong

最新推荐文章于 2020-07-07 17:49:14 发布

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分类专栏： ACM

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本文介绍了一个简单的编程挑战，即计算从1到n的所有整数之和，并提供了使用C语言、C++及Java实现的示例代码。

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http://acm.hdu.edu.cn/showproblem.php?pid=1001

Sum Problem

Time Limit: 1000/500 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 342740 Accepted Submission(s): 86458

Problem Description

Hey, welcome to HDOJ(Hangzhou Dianzi University Online Judge).

In this problem, your task is to calculate SUM(n) = 1 + 2 + 3 + ... + n.

Input

The input will consist of a series of integers n, one integer per line.

Output

For each case, output SUM(n) in one line, followed by a blank line. You may assume the result will be in the range of 32-bit signed integer.

Sample Input

  
   1
100

Sample Output

  
   1

5050
  
  
   

  
  
   解答：
  
  
   

  
  
   C语言：


#include <stdio.h>


int main(){
	int n;
	while(scanf("%d", &n) != EOF){
		int sum = 0;
		for(int i = 0; i <=n; i++ )
		{
			sum += i;
		}
		printf("%d\n\n", sum);
	}


	return 0;
}




C++语言：


#include <iostream>


using namespace std;


int main()
{
   int n;


   while(cin>>n)
   {
       int sum=0;
       for(int i=0;i<=n;++i)
       {
           sum=sum+i;
       }
       cout<<sum<<endl;
       cout<<endl;
   }


}


JAVA语言：


import java.util.Scanner;


public class Main{


	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		while (sc.hasNext()) {
			int sum = 0;
			int n = sc.nextInt();
			for (int i = 1; i <= n; i++) {
				sum += i;
			}
			System.out.println(sum);
			System.out.println();
			sum = 0;
		}


	}


}