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.
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.
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.
5 1 3 2
GBGBG
7 2 2 5
BBGBGBB
4 3 4 0
NO
Source
Codeforces Round #386 (Div. 2)
My Solution
題意：喝掉n袋茶，其中a袋綠茶b袋紅茶，連續喝相同的茶最多k次，如果可以全喝完則輸出喝茶的序列，如果不能則輸出NO
數論+貪心
char a為個數多的那個茶的字母的代表，同理b為少的那個字母，aa為a的個數，bb為b的個數
首先int realk = aa / (bb + 1); if(realk * (bb + 1) < aa) realk++; 如果 realk > k 則 ans 為 NO;
否則把aa 分成 bb + 1組，int ok = aa % (bb + 1);if(ok == 0){ok = bb + 1;}
從而有ok組是realk個a，剩余的是realk⑴個a，每組間用b隔開。
復雜度 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
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