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.
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; } } }