Arranging The Sheep

本文介绍了游戏'安排羊群'的玩法,目标是通过最少的移动次数使羊群整齐排列。文章给出了一个示例,展示了如何在特定关卡中通过策略完成任务,并提供了一个C++程序来计算完成关卡所需的最小移动次数。程序读取关卡描述,计算并输出结果。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

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:

1.the sheep at the 4 position moves to the right, the state of the level: “**…*.”;

2.the sheep at the 2 position moves to the right, the state of the level: “..*.”;

3.the sheep at the 1 position moves to the right, the state of the level: “.**.*.”;

4.the sheep at the 3 position moves to the right, the state of the level: “.*.**.”;

5.the sheep at the 2 position moves to the right, the state of the level: “…***.”;

6.the sheep are lined up and the game ends.

输入

The first line contains one integer t(1≤t≤104). Then tt test cases follow.

The first line of each test case contains one integer n(1≤n≤106).

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 nn over all test cases does not exceed 106.

输出

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

样例输入

5
6
**.*…
5


3
..
3

10
.
.**

样例输出

1
0
0
0
9

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e6+5;
int n;
int a[N],b[N];
char s[N];
int main(){
	cin>>n;
    while(n--)
    {
    	int k=0;
    	int t;
    	cin>>t;
       scanf("%s",s);
     for(int i=0;i<t;i++)
     {
     	if(s[i]=='*')
     	{
     		a[k]=i+1;
			 b[k]=k; 
			 k++;
		 }
	 }
	 ll res=0;
	 ll sum=0;
	 for(int i = 0; i<k; i ++) 
	 {
	 res += abs(a[i]-a[k/2]);
	 sum+=abs(b[i]-b[k/2]);
     }  
    printf("%lld\n",res-sum);
}
    return 0;
}

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值