CodeForces 626B Cards（数学思维）

http://codeforces.com/problemset/problem/626/B

B. Cards

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

2RB

output

G

input

3GRG

output

BR

input

5BBBBB

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.

题意：

有 n 张只是 R G B 颜色的卡片，有两种变换合成形式：
1.两个卡片颜色相同，合成第三种颜色；
2.两个卡片颜色相同，合成同种颜色的一张卡片；

思路：

我们可以考虑假设最后合成是 R 颜色的条件：
G 和 B 都有或者都没有，或者在有 R 的条件下，R 或者 G 存在一个至少有两个的条件下，
最终就能够合成颜色 R 。
同理 G B 一样。 

参考Code:

#include<stdio.h>#include<cstring>#include<algorithm>#define AC main()using namespace std;const int MYDD = 1103;bool Judge(int Get, int One, int Two) {if((!One && !Two)||(One && Two)||(One > 1 && Get)||(Two > 1&& Get))return true;return false;}int AC {int n, sumR = 0, sumG = 0, sumB = 0;char dd[MYDD];scanf("%d", &n);scanf("%s", dd);for(int j = 0; j < n; j++) {if(dd[j] == 'R')    sumR++;if(dd[j] == 'G')    sumG++;if(dd[j] == 'B')    sumB++;}if(Judge(sumB, sumG, sumR)) printf("B");/*注意输出顺序*/if(Judge(sumG, sumR, sumB)) printf("G");if(Judge(sumR, sumG, sumB)) printf("R");return 0;}