CodeForces - 266B - B. Queue at the School

怼烎@ 2022-06-12 07:18 130阅读 0赞

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题目连接：http://codeforces.com/problemset/problem/266/B

## 题目描述 ##

### Description ###

During the break the schoolchildren, boys and girls, formed a queue of n people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.

Let’s describe the process more precisely. Let’s say that the positions in the queue are sequentially numbered by integers from 1 to n, at that the person in the position number 1 is served first. Then, if at time x a boy stands on the i-th position and a girl stands on the (i + 1)-th position, then at time x + 1 the i-th position will have a girl and the (i + 1)-th position will have a boy. The time is given in seconds.

You’ve got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after t seconds.

### Input ###

The first line contains two integers n and t (1 ≤ n, t ≤ 50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.

The next line contains string s, which represents the schoolchildren’s initial arrangement. If the i-th position in the queue contains a boy, then the i-th character of string s equals “B”, otherwise the i-th character equals “G”.

### Output ###

Print string a, which describes the arrangement after t seconds. If the i-th position has a boy after the needed time, then the i-th character a must equal “B”, otherwise it must equal “G”.

### Sample Input ###

5 1  
BGGBG

### Sample Output ###

GBGGB

### Sample Input ###

5 2  
BGGBG

### Sample Output ###

GGBGB

### Sample Input ###

4 1  
GGGB

### Sample Output ###

GGGB

## 解题思路 ##

题意大致为男女生排队，G为女生，B为男生，交换的要求是：男生在女生前面就交换位置，其他条件不变，如下图 （n为人的个数，t为交换次数）

*  GG → GG
 *  GB → GB
 *  BG → GB
 *  BB → BB

思路：从头遍历，分条件控制 **i** 的位置，如果匹配成功的话往后移两位，如果没有成功移一位。

## AC代码 ##

#include<iostream>
        using namespace std;
        int main() { 
        	int n, t;
        	cin >> n >> t;
        	char str[60];
        	scanf("%s", str);
        	while(t--) { 
        		for(int i = 0; i <= n-1;) { 
        			if(str[i] == 'B' && str[i+1] == 'G') { 
        				swap(str[i], str[i+1]); //交换 
        				i += 2;
        			}
        			else  i++;
        		}
        	}
        	puts(str);
        	return 0;
        }