ACM之杭电——1001_Sum Problem

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; } } }

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