Codeforces 746D Green and Black tea (construction) __ Construction

Green and black tea time limit/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

The point: There are n cups of tea, divided into two kinds of g,b, there are a,b cups respectively. It is not required to drink a tea in a row for more than k times. Output the order of tea

The puzzle: The construction is good, the code writes the good setback Ah Σ (° °| | |) ︴

The code is as follows:

#include <cstdio> #include <cstring> #include <algorithm> using namespace std;
const int MAXN = 1E5+10;
Char STR[MAXN];
	int main () {int n,k,a,b; while (scanf ("%d%d%d%d", &n,&k,&a,&b)!=eof) {if (a>=b) {int cnt=a/k;//a is divided into several double num=a*1.
			0/k;
			if (num==cnt) cnt--;
			if (cnt==0) cnt=1;
			int count=cnt;
			cnt=b/count;//b each block must fill in at least several if (cnt>k) cnt=k; 
			if (cnt==0) cnt=1;
			int temp=0;
				while (a>0&&b>0) {cnt=a/k;
				num=a*1.0/k;
				if (num==cnt) cnt--;
				if (cnt==0) cnt=1;
				count=cnt;
				cnt=b/count;//each time to update the CNT if (cnt>k) cnt=k; 
				if (cnt==0) cnt=1;
					if (a>=k) {for (int i=0;i<k;++i) str[temp++]= ' G ';
				A-=k;
					else {for (int i=0;i<a;++i) str[temp++]= ' G ';
				A-=a;
				} if (a==0) cnt=k;
					if (b>=cnt) {for (int i=0;i<cnt;++i) str[temp++]= ' B ';
				b-=cnt;
		} else {			for (int i=0;i<b;++i) str[temp++]= ' B ';
				B-=b;
				} if (A>=k) {for (int i=0;i<k;++i) str[temp++]= ' G ';
			A-=k;
				else {for (int i=0;i<a;++i) str[temp++]= ' G ';
			A-=a;
			} if (A!=0 | | b!=0) printf ("no\n");
				else{str[temp]= ';
			printf ("%s\n", str);
			} else {int cnt=b/k;
			Double num=b*1.0/k;
			if (num==cnt) cnt--;
			if (cnt==0) cnt=1;
			int count=cnt;
			Cnt=a/count;
			if (cnt>k) cnt=k;
			if (cnt==0) cnt=1;
			int temp=0;
				while (a>0&&b>0) {cnt=b/k;
				num=b*1.0/k;
				if (num==cnt) cnt--;
				if (cnt==0) cnt=1;
				count=cnt;
				Cnt=a/count;
				if (cnt>k) cnt=k;
				if (cnt==0) cnt=1;
					if (b>=k) {for (int i=0;i<k;++i) str[temp++]= ' B ';
				B-=k;
					else {for (int i=0;i<b;++i) str[temp++]= ' B ';
				B-=b;
				} if (b==0) cnt=k; if (a>=cnt) {for (int i=0;i<cnt;++i) str[temp++]= ' G ';
				a-=cnt;
					else {for (int i=0;i<a;++i) str[temp++]= ' G ';
				A-=a;
				} if (B>=k) {for (int i=0;i<k;++i) str[temp++]= ' B ';
			B-=k;
				else {for (int i=0;i<b;++i) str[temp++]= ' B ';
			B-=b;
			} if (A!=0 | | b!=0) printf ("no\n");
				else{str[temp]= ';
			printf ("%s\n", str);
}} return 0; }/* 10 4 9 1 10 5 5 5 10 3 5 5 16 3 8 8 16 2 4 12 15 3 5 10 15 3 4 11 20 2/12 * *