CodeForces - 266B

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概述

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".

    Examples

Input

5 1
BGGBG

Output

GBGGB

Input

5 2
BGGBG

Output

GGBGB

Input

4 1
GGGB

Output

GGGB

题目简述：一队列有n个学生，为体现男生的绅士风度，站在队列靠前的男生与紧邻靠后的女生交换一次位置，每秒交换一次，经过t秒后，打印队列的分布状况
题目思路：两层循坏，外层为时间t的循环，内层为遍历队列，满足条件的进行交换
程序说明：在此题中，每秒交换的可能不止一对。比如：
BGBG这一队列，经过1秒后的队列分布情况为：GBGB

AC代码如下：

#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
 int n, t;
 scanf("%d %d", &n, &t);
 char a[55];
 scanf("%s", a, 55);
 while (t--)
 {
  for (int i = 0; i < n; i++)
  {
   if (a[i] == 'B'&&a[i + 1] == 'G')
   {
    int j = a[i];
    a[i] = a[i + 1];
    a[i + 1] = j;
    i += 1;
   }
  }
 }
 printf("%s", a);
 return 0;
}