Little Gyro and Sets（第二届中国计量大学ACM程序设计竞赛个人赛）

链接：https://ac.nowcoder.com/acm/contest/3190/B
来源：牛客网

示例1
输入
3
3 5
2 10
3 16

输出
9
-5
46

说明
For the first sample, after these operations, A = < 3 >, B = < 1, 2, 4, 5 >, and the answer is 9.
For the second sample, after these operations, A = < 2, 4, 6, 8, 10 >, B = < 1, 3, 5, 7, 9 >, and the answer is -5.

本题是个数学题。B数组是以n为公差的等差数列，我们只需要求改等差数列的前m/n项和，再用首项加末项乘以项数除以二减去B数组和，即得到A数组的和，最后再用B数组和减去A数组和即可得到答案。

#include <iostream>
#include <cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std
int main()
{
    long long int T
    scanf("%lld", &T)
    while (T--)
    {
        long long int a, b
        long long int suma, sumb
        suma = sumb = 0
        scanf("%lld %lld", &a, &b)
        long long int q = b / a
        suma = q * a + (q * (q - 1) * a) / 2
        sumb = (1 + b) * b / 2 - suma
        printf("%lldn", sumb-suma)
    }
    return 0
}