Codeforces 626B

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。
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
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");
	}
}


内容概要:本文系统介绍了算术优化算法(AOA)的基本原理、核心思想及Python实现方法,并通过图像分割的实际案例展示了其应用价值。AOA是一种基于种群的元启发式算法,其核心思想来源于四则运算,利用乘除运算进行全局勘探,加减运算进行局部开发,通过数学优化器加速函数(MOA)和数学优化概率(MOP)动态控制搜索过程,在全局探索与局部开发之间实现平衡。文章详细解析了算法的初始化、勘探与开发阶段的更新策略,并提供了完整的Python代码实现,结合Rastrigin函数进行测试验证。进一步地,以Flask框架搭建前后端分离系统,将AOA应用于图像分割任务,展示了其在实际工程中的可行性与高效性。最后,通过收敛速度、寻优精度等指标评估算法性能,并提出自适应参数调整、模型优化和并行计算等改进策略。; 适合人群:具备一定Python编程基础和优化算法基础知识的高校学生、科研人员及工程技术人员,尤其适合从事人工智能、图像处理、智能优化等领域的从业者;; 使用场景及目标:①理解元启发式算法的设计思想与实现机制;②掌握AOA在函数优化、图像分割等实际问题中的建模与求解方法;③学习如何将优化算法集成到Web系统中实现工程化应用;④为算法性能评估与改进提供实践参考; 阅读建议:建议读者结合代码逐行调试,深入理解算法流程中MOA与MOP的作用机制,尝试在不同测试函数上运行算法以观察性能差异,并可进一步扩展图像分割模块,引入更复杂的预处理或后处理技术以提升分割效果。
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