Problem--266B--Codeforces--Queue at the School

B. Queue at the School
time limit per test:2 seconds
memory limit per test:256 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

#include<stdio.h>
int main(){
    int n,h;
    scanf("%d%d",&n,&h);
    char str[n+1];
    scanf("%s",str);
    for(int i=0;i<h;i++){
        for(int i=0;str[i]!='\0';i++){
            if(str[i]<str[i+1]){
                char temp=str[i];
                str[i]=str[i+1];
                str[i+1]=temp;
                i++;
            }
        }
     }
    printf("%s",str);
}

感覺有些投機取巧了，既然只有男（B），女（G）兩種字符，那乾脆比較大小好啦，事實上應該判斷是否==’G’，是否==’B’的，哈哈。