codeforces 1520E. Arranging The Sheep(1400)

题目：codeforces 1520E. Arranging The Sheep(1400)
链接：https://codeforces.ml/problemset/problem/1520/E

题目大意：给你一个只有.和*的字符串，用最少的步数把*都凑在一起；

题解：有个东西叫中位数定理（假设有一个数轴，数轴上的点到中间点的距离之和最小……大概）

那这题只需要先找到在所有 * 处于中间位置的 * ，然后遍历计算；

PS：因为*移动会受到其他*的限制，所以距离还要减去目标和终点之间*的个数；

代码：

#include<bits/stdc++.h>using namespace std;
typedef long long ll;int a[1000005];int main()
{int t;cin >> t;while (t--) {int n;cin >> n;string s;cin >> s;int cnt = 0;for (int i = 0; i < n; i++) {if (s[i] == '*')a[cnt++] = i;}int mid = cnt / 2;ll sum = 0;for (int i = 0; i < cnt; i++){sum += abs(a[mid] - a[i]) - abs(mid - i);}cout << sum << endl;}}

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