Codeforces 266b solution: Queue at the School

This problem’s main point is each second a boy will let every girl swap positions if a girl is behind the boy. This happens for all the girls and boys who meet the same condition.

Sample Input:

5 1

BGBGB

Here, if you look at sample input you will see a single string containing only ‘B’ or ‘G’ letters denoting boy and girl respectively, where the sequence of the whole string represents the line containing boys and girls int the line/queue at a certain time.

The first integer represents the number of children ( in this case string size ) and the 2nd integer represents seconds that will pass.

Steps to solve:

• First, we will take two integers as asked in the problem. (number of children and passable time).
•  After inputting them we will declare a string whose length will be the 1st inputted integer to create the line of students as a string.
• Then, we will call a loop until we pass all iterations of the second Input integer so that we can do operations of swapping boys and girls for the given seconds. (line 8 of code )
• Now we will loop through all the index of the input string ( boys, girls queue representation ) to check if current_index = ‘B’ and next_index = ‘G’.
• If the previous comparison is correct we will swap current_index and next_index values. ( Boys exchanging position with girls if the girl is behind ). (line 10 – 12 of code)
• Then increment the loop iterator by 2 because we had done work with two indexes and we don,t need to do anything u til next second. (line 13 of code)
• Else if the comparison didn’t match we will increment the loop iterator only by one so we can check the next index.
• After the looping process is done just print the string to show the changed output.

Solutions of Codeforces 226b in C/C++

---

#include<bits/stdc++.h>
int main() {
int x,y;
scanf("%d%d",&x,&y);
char str[x],tmp;
scanf(" %s",str);
int len = strlen(str);
while(y--) {
   for(int i = 0; i<len-1;) {
      if(str[i]=='B' && str[i+1]=='G') {
         tmp = str[i];
         str[i] = str[i+1];
         str[i+1] = tmp;
         i+=2;
      }else{
         i++;
          }
       }
    }
printf("%s\n",str);
return 0;
}