CodeForces 626B CodeForces 626B【暴力】

B. Cards

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 626B

Description

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.

Sample Input

Input

2RB

Output

G

Input

3GRG

Output

BR

Input

5BBBBB

Output

B

Hint

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.

题目：

给出若干个三种颜色的卡片，有两种操作

1，两张相同颜色的能合并成一张

2，两张不同颜色的能合并成另外一种颜色

经过多次操作使得只剩下一种颜色，按字典序输出所有可能出现的结果

找规律的题目，确实伤不起....

规律如下：

1，如果只有一种颜色，不用说了，就是这一种

2，如果有一种颜色有0个，一种颜色有1个，分情况考虑：

 如果第三种颜色只有1 个，最终结果只可能是颜色为0的那种

 如果第三种颜色有多个，最终结果可以是前两种的任意一个（可以自己动手试试）

3，否则的话，三种颜色都可以组成

规律是推导出来了，只是代码写起来实在麻烦，想了好久，才想出来优化的办法......貌似是简化了一点...

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;char table[]={"BGR"};void slove(char s[],int n){sort(s,s+n);int x[3]={0},vis[3]={1,1,1};// 1 为输出，0，为不输出 for(int i=0;i<3;++i){x[i]=upper_bound(s,s+n,table[i])-lower_bound(s,s+n,table[i]);}for(int i=0;i<3;++i)//双循环枚举所有情况 {for(int j=0;j<3;++j){if(i==j) {continue;}if(x[i]==0&&x[j]==0)//其中任意两个为0 的只输出第三个 {vis[i]=vis[j]=0; break;}else if(x[i]==0&&x[j]==1)//一个0，一个1的 {if(x[3-i-j]==1)//另外一个只有1个，那么输出为零的那种 {vis[j]=vis[3-i-j]=0; }else if(x[3-i-j]>1)//有多个{vis[3-i-j]=0; }}}}for(int i=0;i<3;++i){if(vis[i]){printf("%c",table[i]);}}printf("\n");}int main(){//freopen("shuju.txt","r",stdin);int n;char s[205];while(~scanf("%d",&n)){scanf("%s",s);slove(s,n);}return 0;}