NYOJ-sum of all integer numbers

最新推荐文章于 2017-10-06 13:24:17 发布

凌摩绛霄

最新推荐文章于 2017-10-06 13:24:17 发布

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本文链接：https://blog.youkuaiyun.com/qq_18062811/article/details/38929229

本文介绍了一个简单的算法题目，任务是计算从1到N（包含N）的所有整数的总和。该算法适用于绝对值不超过10,000的整数N，并提供了C语言实现代码。

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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

样例输出

代码:

#include<stdio.h>
#include<stdlib.h>
int main()
{
	int n;
	while(~scanf("%d",&n))
	{
		if(n==0)
		{
			printf("1\n");
			continue;
		}
		int k=1,t=0;  
		if(n<0)
		k=-1,t=1;
		n=abs(n);
		if(n&1)
		printf("%d\n",((n+1)>>1)*n*k+t);
		else
		printf("%d\n",(n>>1)*(n+1)*k+t);
	   
	}
	return 0;
}