fjnu 1281 Alternating Sums

最新推荐文章于 2024-08-20 10:12:01 发布

核动力蜗牛Killua

最新推荐文章于 2024-08-20 10:12:01 发布

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分类专栏： ☆ACM 解题报告☆ 文章标签： input math.h output each integer bi

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本文介绍了一种计算从a到b范围内整数的交替求和的算法实现，即首两个数相加，随后的数依次进行加减运算。通过一个具体的C++程序示例展示了如何根据输入的两个整数计算这一特定序列的和，并提供了样本输入输出以验证算法的正确性。

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Description

Alternating sums happen when you add one number in the sequence and then subtract off the following number from the sum. Repeat. Weird things happen when dealing with infinite alternating sums as some can made to converge to different values by rearranging terms, etc. (This applies to conditionally convergent series.) For this problem we'll be concerned with finite integer sums. The first two terms are added, the third subtracted, the fourth added, etc. For example, the alternating sum from 2 to 4 is 2 + 3 - 4 = 1. The alternating sum from 4 to 7 is 4 + 5 - 6 + 7 = 10. The sum from 4 to 4 is boring as it only includes one term, 4.

Input

Each line of input has two positive integers, a and b, each smaller than 1000. You may assume that a <= b.

Output

For each line of input, print out the alternating sum from a to b, inclusive.

Sample Input

2 4
4 4 
10 20

Sample Output

1 
4 
5

 Source:#include<iostream.h>
#include<math.h>

int sum(int a,int b)
...{
    if(a==b) return a;
    int s=a;
    int k=0;
    for(int i=a+1;i<=b;i++)
    ...{
        s+=(int)pow(-1,k++)*i;
    }
    return s;
}

int main()
...{
    int a,b;
    while(cin>>a>>b)
    ...{
        cout<<sum(a,b)<<endl;
    }
    return 0;
}