ZOJ

https://vjudge.net/problem/ZOJ-3981

题意：
第一行三个数字n, m, q表示有m个座位围成一个环，n个队伍，q次A题
接下来n个数表示n个队伍所在位置(1<=ai<=m)
再接下来q行，每行a, b表示第a个队伍在第b秒A了一道题
有一个只会每一秒顺时针移动一个位置的发气球机器人
只要当前队伍有题目已经A了就会给他对应数量的气球（当然每道题最多1个气球）
如果a队伍在b时刻A了一道题，并在c时刻才拿到气球，那么这个队伍就会积累c-b点不开心值
求一个机器人起始位置（一开始是第0秒）使得所有队伍最终不开心值之和最小

做法：

假设机器人位置在1位置上，对于每个a b，那么每次产生的不开心值就是s[a]-1-b，移动到a位置所需时间-ac时间，当然这个值

有正有负，那么我们从for(1-q)所有的不开心值加起来，肯定是最后的总不开心值，无论这个是负还是正。对于负值，我们对这

个值%m取其正，那么排序一遍

然后枚举每一道题作为起点时候的情况。
对于第i道题目,因为位置往后走,后面的题目的时间就会减少 tid[i],前面的题目的时间就会增加 m - tid[i]。
因此递推式是sum - (p - i) * tid[i] + i * (m - tid[i]) = sum + m * (i - 1) - tid[i] * p

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