【21.58%】【codeforces 746D】Green and Black Tea

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

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

Innokentiy doesn’t like to drink the same tea (green or black) more than k times in a row. Your task is to determine the order of brewing tea bags so that Innokentiy will be able to drink n cups of tea, without drinking 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 first line 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 tea. It is guaranteed that a + b = n.

Output

If it is impossible to drink n cups of 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 be green. If some character equals ‘B’, then the corresponding cup of tea should be 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

【题目链接】:http://codeforces.com/contest/746/problem/D

【题解】



最后必然是分成a/k 个绿茶块,b/k个黑茶块;

如果a/k==b/k

显然可以交替出现;

如果a/k>b/k

则先放k个a，然后放一个b;

这样a-=k,b-=1;

则a/k会越来越逼近b/k;

最后交替出现就可以了。

balabala



【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

//const int MAXN = x;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);

int n,k,a,b;
char t[2];
string ans = "";

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    rei(n);rei(k);rei(a);rei(b);
    t[0]='G',t[1] = 'B';
    if (a < b)
        swap(a,b),swap(t[0],t[1]);
    bool ok = false;
    while ( (a/k)>(b/k))
    {
        if (a<k || b<1)
            break;
        rep1(i,1,k)
            ans+=t[0];
        ans+=t[1];
        a-=k;
        b--;
    }
    while (a>0 && b>0)
    {
        int ma = min(a,k);
        rep1(i,1,ma)
            ans+=t[0];
        a-=ma;
        ma = min(b,k);
        rep1(i,1,ma)
            ans+=t[1];
        b-=ma;
    }
    if (a>k)
    {
        puts("NO");
        return 0;
    }
    rep1(i,1,a)
        ans+=t[0];
    if (b>k)
    {
        puts("NO");
        return 0;
    }
    rep1(i,1,b)
        ans+=t[1];
    cout << ans << endl;
    return 0;
}