Day 5 E. Arranging The Sheep

Problem:
You are playing the game “Arranging The Sheep”. The goal of this game is to make the sheep line up. The level in the game is described by a string of length n, consisting of the characters ‘.’ (empty space) and ‘*’ (sheep). In one move, you can move any sheep one square to the left or one square to the right, if the corresponding square exists and is empty. The game ends as soon as the sheep are lined up, that is, there should be no empty cells between any sheep.

For example, if n=6 and the level is described by the string " * * . * . . ",then the following game scenario is possible:

the sheep at the 4 position moves to the right, the state of the level: " * * . .* . ";
**the sheep at the 2 position moves to the right, the state of the level: " * . * . * . " ;
the sheep at the 1 position moves to the right, the state of the level: " . * * . * . ";
the sheep at the 3 position moves to the right, the state of the level: " . * . * * . ";
the sheep at the 2 position moves to the right, the state of the level: " . . * * * . ";
the sheep are lined up and the game ends.
For a given level, determine the minimum number of moves you need to make to complete the level.

Input
The first line contains one integer t (1≤t≤10^4). Then t test cases follow.
The first line of each test case contains one integer n (1≤n≤10^6).
The second line of each test case contains a string of length n, consisting of the characters ‘.’ (empty space) and ‘*’ (sheep) — the description of the level.
It is guaranteed that the sum of n over all test cases does not exceed 10^6.

Output
For each test case output the minimum number of moves you need to make to complete the level.

Example
input
5

6
* * . * . .

5
* * * * *

3
. * .

3
. . .

10
* . * . . . * . * *

output
1

题目大致意思为：给你一串长为n 仅由 *和.组成的字符串求至少经过多少次变换 *号能排成一行每次变换只能左右移动一个 *的位置这里我用到了中位数定理…
中位数定理
背景:如果有一个数轴数轴上有若干个点现要在数轴上找一点使得它到各个点的距离之和最短。
结论: 中位数就是最优解
中位数有这样的性质: 所有数与中位数的绝对差之和最小
中位数是数列中间的那个数或者是中间的那两个数之一.
在这道题目中中位数就是中间 *的位数

代码如下

#include <iostream>
#define ll long long
using namespace std;
string s;
ll f(int x)//计算最少变换次数
{ll cnt = 0;ll mid = 0;for (int i = 0; i < s.length(); i++)//获取中间*的位数 赋值给mid{if (s[i] == '*'){cnt++;}if (cnt == x){mid = i;break;}}ll ans = 0;ll cur = 0;for (int i = mid - 1; i >= 0; i--)//在中间*的前面的*所需要变换次数{if (s[i] == '.'){cur++;}if (s[i] == '*'){ans += cur;}}cur = 0;for (int i = mid + 1; i < s.length(); i++)//在中间*的后面的*所需要变换次数{if (s[i] == '.'){cur++;}if (s[i] == '*'){ans += cur;}}return ans;
}
int main()
{int t;cin >> t;while (t--){int n;cin >> n;cin >> s;ll ans = 0;ll cnt = 0;for (int i = 0; i < n; i++){if (s[i] == '*'){cnt++;}}if (cnt % 2 == 1)//找到中间*{ans = f(cnt / 2 + 1);}else{ans = min(f(cnt / 2), f(cnt / 2 + 1));}cout << ans << endl;}return 0;
}

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