A+B probiem

最新推荐文章于 2024-08-03 14:51:53 发布

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Sum Problem

Time Limit: 1000/500 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 167462 Accepted Submission(s): 39789

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

我的最先想到的挫逼代码，真的是wa啊：

#include<stdio.h>
int main()
{
	int i,j,n,a,sum=0;
	while(scanf("%d%d",&a,&n)!=EOF)
	{
	   //sum=n*(n+a)/2;
             sum=(int)(n*(n+1.0)/2.0);
	    printf("1\n\n");
	    //printf("%d\n",sum);
	   
		printf("%d\n",sum); 
	}
	
	return 0;
	
}

比较好的代码：

#include <stdio.h>
int main()
{
    int n, sum;
    while (scanf("%d", &n) !=EOF)
    {
        sum = 0;
        while (n != 0)
	   {
            sum += n--;
            }
        printf("%d\n\n", sum);
    }
    return 0;
}

其实我在想它为什么没有超时。。。。。。

最不能让人忍受的是像下面这种代码也能AC

#include <stdio.h>
int sum(int);

int main(){
    int input,output;
    while (scanf("%d",&input)==1)
    {
        output = sum(input);
        printf("%d\n\n",output);
    }
    return 0;
}
int sum(int n){
    int sum,i;
    sum=0;
    for (i=1;i<=n;i++)
    {
        sum+=i;
    }
    return sum;
}