文章目录
一、Queue at the School
总结
一、Queue at the School
本题链接：Queue at the School
题目：
B. Queue at the School
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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
本博客给出本题截图：
题意：每次交换一个'B'和'G'的位置，求交换t次后的结果
AC代码
#include <iostream> #include <string> #include <algorithm> using namespace std; int main() { int n, t; cin >> n >> t; string a; cin >> a; for (int i = 0; i < t; i ++ ) for (int j = 0; j < n - 1; j ++ ) if (a[j] == 'B' && a[j + 1] == 'G') { swap(a[j], a[j + 1]); j ++; } cout << a; return 0; }
总结
水题，不解释
