NYOJ 46 sum of all integer numbers

sum of all integer numbers

时间限制：1000 ms  |  内存限制：65535 KB

难度：0

描述

Your task is to find the sum of all integer numbers lying between 1 and N inclusive.

输入

There are multiple test cases.
The input consists of a single integer N that is not greater than 10000 by it's absolute value.

输出

Write a single integer number that is the sum of all integer numbers lying between 1 and N inclusive.

样例输入

3

样例输出

6

大意是：输入一个N，（可正可负），输出1-N的和 

需要注意的是：如果N是负数就输出N-1的和

#include<stdio.h>#include<stdlib.h> int main(){  int n;  while(scanf("%d", &n) != EOF){    if(n < 1)      printf("%d\n", (1+n)*(abs(n)+2)/2);//负数情况，项数得加上2     else      printf("%d\n", (1+n)*n/2);  }}