Alternating Sums

FJNU.1281

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
udel - fall 2002

My Program 

#include<iostream>
using namespace std;

int main()
...{
    int i,s,m,n;
    while(scanf("%d%d",&m,&n)!=EOF)
    ...{
        s=m;
        for(i=m+1;i<=n;i++)
            if((i-m)%2)
                s+=i;
            else
                s-=i;
        cout<<s<<endl;
    }
    return 0;
}

YOYO’s Note:
秒杀的简单计算题……为什么大二才写呢？
因为一直不知道没有结束条件的输入怎么写……后来才知道了直接while(cin>>xx)……