HDU——1001

最新推荐文章于 2024-08-22 20:10:34 发布

yeah_i_am_bruce

最新推荐文章于 2024-08-22 20:10:34 发布

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本文介绍了一个经典的HDOJ题目——求和问题。该问题要求计算从1加到n的总和，并通过等差数列求和公式进行解答。文章提供了C++代码示例，强调了使用long long数据类型的重要性，以避免整数溢出。

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                     ** Sum Problem**

Time Limit: 1000/500 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 469958 Accepted Submission(s): 118457

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

Author
DOOM III

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分析：此题一般会想到一个循环，其实数学家高斯的方法也可以等差数列求和的求出，这题有个地方需要注意
用等差数列求和需要使用longlong数据类型，应该是相乘时临时变量时int ，超出了int的表示范围，导致报错

考点：数学公式的灵活运用，ACM就是用编程语言代替人工计算解决数学问题吧

#include<bits/stdc++.h>
using namespace std;
int main()
{
 long long n;
 while(~scanf("%d",&n))
 {
  cout<<(n+1)*n/2<<endl;
  cout<<endl;
 }

}