Codeforces Round #386 (Div. 2) D. Green and Black Tea 数论+贪心

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 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 thata + 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

Source

Codeforces Round #386 (Div. 2)

My Solution

题意：喝掉n袋茶，其中a袋绿茶b袋红茶，连续喝相同的茶最多k次，如果可以全喝完则输出喝茶的序列，如果不能则输出NO

数论+贪心

char a为个数多的那个茶的字母的代表，同理b为少的那个字母，aa为a的个数，bb为b的个数

首先int realk = aa / (bb + 1); if(realk * (bb + 1) < aa) realk++; 如果 realk > k 则 ans 为 NO;

否则把aa 分成 bb + 1组，int ok = aa % (bb + 1);if(ok == 0){ok = bb + 1;}

从而有ok组是realk个a，剩余的是realk-1个a，每组间用b隔开。

复杂度 O(n)

#include <iostream>
#include <cstdio>
#include <string>
using namespace std;
typedef long long LL;
const int maxn = 1e6 + 8;

string ans;

int main()
{
#ifdef LOCAL
freopen("d.txt", "r", stdin);
//freopen("d.out", "w", stdout);
int T = 6;
while(T--){
#endif // LOCAL
ios::sync_with_stdio(false); cin.tie(0);

LL n, k, a, b;
cin >> n >> k >> a >> b;
int aa = max(a, b), bb = min(a, b);
char x, y;
if(aa == a){
x = 'G';
y = 'B';
}
else{
x = 'B';
y = 'G';
}

int realk = aa / (bb + 1);
if(realk * (bb + 1) < aa) realk++;

if(realk <= k){

//cout << realk << endl;
int ok = aa % (bb + 1);
if(ok == 0){
ok = bb + 1;
}
int cnt = 0, oki = 0;
ans.clear();
for(int i = 0; i < n; i++){
if(cnt == realk){
ans += y;
cnt = 0;
oki++;
if(oki == ok){
realk--;
}
}
else{
cnt++;
ans += x;
}
}
cout << ans << endl;
}
else cout << "NO" << endl;

#ifdef LOCAL
cout << endl;
}
#endif // LOCAL
return 0;
}



  Thank you!

                                                                                                           
                                   ------from ProLights