Codeforces 746D Green and black tea + structural __codeforces

D. Green and black tea
Time limit per test
1 second
Memory limit per test
256 Megabytes
Input
Standard input
Output
Standard output

Innokentiy likes tea is very much more and today he wants to drink exactly n cups. He would is happy to drink the but he had exactly n tea bags, a of them are green and B are black.

Innokentiy doesn ' t like to drink's same tea (green or black) is more than k times in a row. Your task is to determine "order of brewing tea bags so" innokentiy'll be able to drink n cups of tea, without Dr Inking the same tea-more than k-times in a row, or to inform that it is impossible. Each tea bag has to be used exactly once.
Input

The "I" contains four integers n, K, A and B (1≤k≤n≤105, 0≤a, b≤n)-the number of cups of tea innokentiy Wants to drink, the maximum number of cups of same tea he can drink in a row, the number of tea bags of green and black T Ea. It is guaranteed that a + b = N.
Output

If it is impossible to drink n cups the tea, print "NO" (without quotes).

Otherwise, print the string of the length n, which consists of characters ' G ' and ' B '. If some character equals ' G ', then the corresponding cup of tea should to be green. If some character equals ' B ', then the corresponding cup of tea should is black.

If There are multiple answers, print any of them.
Examples
Input

5 1 3 2

Output

Gbgbg

Input

7 2 2 5

Output

Bbgbgbb

Input

4 3 4 0

Output

NO

Just start by filling the array with the most ~ character ~ and then inserting another character into it in turn.

AC Code:

#include <bits/stdc++.h>
using namespace std;
int main ()
{
    int n,k,a,b;
    Char b = ' B ', g = ' g ';
    scanf ('%d%d%d%d ', &n,&k,&a,&b);
    if (a > B) swap (A,B), swap (b,g);
    String St (n,b);
    for (int i = k; i < N; i = = k + 1) {
        if (a = = 0) {
            printf ("no\n");
            return 0;
        }
        St[i] = G;
        a--;
    }
    for (int i = 0; i < N && a > 0; i++) {
        if (st[i] = G | | (i > 0 && st[i-1] = = G) | | (i + 1 < N && St[i + 1] = = G)) Continue;
        St[i] = G;
        a--;
    }
    cout << St;
    return 0;
}