Codeforces 626B

最新推荐文章于 2017-02-02 02:18:43 发布

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探讨了在一个包含红、绿、蓝三种颜色卡片的牌组中，通过特定规则交换卡片颜色直至只剩一张卡片时，该卡片可能的颜色。分析了不同初始状态下的卡片组合如何影响最终结果。

Catherine has a deck of n cards, each of which is either red, green, or blue. As long as there are at least two cards left, she can do one of two actions:

take any two (not necessarily adjacent) cards with different colors and exchange them for a new card of the third color;
take any two (not necessarily adjacent) cards with the same color and exchange them for a new card with that color.

She repeats this process until there is only one card left. What are the possible colors for the final card?

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 200) — the total number of cards.

The next line contains a string s of length n — the colors of the cards. s contains only the characters 'B', 'G', and 'R', representing blue, green, and red, respectively.

Output

Print a single string of up to three characters — the possible colors of the final card (using the same symbols as the input) in alphabetical order.

Examples

input
2
RB

output
G

input
3
GRG

output
BR

input
5
BBBBB

output
B

Note

In the first sample, Catherine has one red card and one blue card, which she must exchange for a green card.

In the second sample, Catherine has two green cards and one red card. She has two options: she can exchange the two green cards for a green card, then exchange the new green card and the red card for a blue card. Alternatively, she can exchange a green and a red card for a blue card, then exchange the blue card and remaining green card for a red card.

In the third sample, Catherine only has blue cards, so she can only exchange them for more blue cards.

刚开始看到这个题时，以为会有什么规律，想了半天没想到，看了师傅的代码才知道是把情况一一列举出来，坑，，，，，并且情况比较多，不注意就会漏掉一些

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char s[1100];
int i,j,k,l,m,n,p;
int main()
{
	while(scanf("%d",&n)!=EOF)
	{
		scanf("%s",s);
		int B,R,G;//这个最好用与蛋糕颜色相同的字母 
		B=R=G=0;
		for(i=0;i<n;i++)
		{
			if(s[i]=='B')
			B++;
			if(s[i]=='G')
			G++;
			if(s[i]=='R')
			R++;
		} 
		if(B>0&&G>0&&R>0)
		{
			printf("BGR\n");
			continue;
		}
		
		if(B==1&&G==1&&R==0)
		printf("R\n");
		if(B==1&&G==0&&R==1)
		printf("G\n");  
		if(B==0&&G==1&&R==1)
		printf("B\n"); 
		if(B>=2&&G>=2)
		{
			printf("BGR\n");
			continue;
		}
		
		if(B>=2&&R>=2)
		{
			printf("BGR\n");
			continue;
		}
		
		if(G>=2&&R>=2)
		{
			printf("BGR\n");
			continue;
		}
		if(G==0&&B>=2&&R==1)
		printf("GR\n");
		if(G==0&&R>=2&&B==1)
		printf("BG\n");
		if(G==1&&B>=2&&R==0)
		printf("GR\n");
		if(G==1&&B==0&&R>=2)
		printf("BG\n");
		if(G>=2&&B==1&&R==0)
		printf("BR\n");
		if(G>=2&&B==0&&R==1)
		printf("BR\n");
		if(B==0&&G==0&&R!=0)
		printf("R\n");
		if(B==0&&G!=0&&R==0)
		printf("G\n");
		if(B!=0&&G==0&&R==0)
		printf("B\n");
	}
}