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.
C：
#include<stdio.h>
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
printf("%.f\n\n",(1+n)/2.0*n);
}
return 0;
}
C++:
#include<iostream>
using namespace std;
int main()
{
int n;
int sum = 0;
while(cin >> n)
{
if(n&1) //奇数
sum = (n+1)/2*n;
else //偶数
sum = n/2*(n+1);
cout << sum << endl <<endl;
}
return 0;
}
Java:
import java.util.*;
import java.math.*;
public class Main
{
public static void main(String []args)
{
BigInteger [] b=new BigInteger [2];
b[0]=BigInteger.ONE;
b[1]=b[0].add(b[0]);
Scanner cin=new Scanner(System.in);
while(cin.hasNext())
{
BigInteger a=cin.nextBigInteger();
a=a.multiply(a.add(b[0]));
a=a.divide(b[1]);
System.out.println(a);
System.out.println();
}
}
}
