zoukankan      html  css  js  c++  java
  • Codeforces--626B--Cards(模拟)

    

    Cards

    Time Limit: 2000MS

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

    Status

    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
    2
    RB
    
    Output
    G
    
    Input
    3
    GRG
    
    Output
    BR
    
    Input
    5
    BBBBB
    
    Output
    B
    

    Sample Output

    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.

    Source


    如果只有一种颜色,那么只可能生成一种颜色,如果有两种就要分情况了,如果两种颜色各一个,输出的也只能有一种颜色,,,,,,,,

    #include<cstdio>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    char str[10010];
    int main()
    {
    	int n;
    	cin>>n;
    	cin>>str;
    	int R,G,B;
    	R=B=G=0;
    	for(int i=0;i<n;i++)
    	{
    		if(str[i]=='R') R++;
    		if(str[i]=='G') G++;
    		if(str[i]=='B') B++;
    	}
    	if(R&&G==0&&B==0) cout<<"R"<<endl;
    	else if(R==0&&G&&B==0) cout<<"G"<<endl;
    	else if(R==0&&G==0&&B) cout<<"B"<<endl;
    	else if(R==1&&G==1&&B==0) cout<<"B"<<endl;
    	else if(R==1&&G==0&&B==1) cout<<"G"<<endl;
    	else if(R==0&&G==1&&B==1) cout<<"R"<<endl;
    	
    	else if(R==1&&G==0&&B>1) cout<<"GR"<<endl;
    	else if(R==1&&G>1&&B==0) cout<<"BR"<<endl;
    	else if(R==0&&G==1&&B>1) cout<<"GR"<<endl;
    	else if(R==0&&G>1&&B==1) cout<<"BR"<<endl;
    	else if(R>1&&G==1&&B==0) cout<<"BG"<<endl;
    	else if(R>1&&G==0&&B==1) cout<<"BG"<<endl;
    	
    	else cout<<"BGR"<<endl;
    	return 0;
    }

  • 相关阅读:
    fedora 20 安裝 及 配置桌面環境
    2014上半年—Linux操作系统—嵌入式开发—中考
    【hdu 1864】最大报销额
    【hdu 1067】Gap
    【t055】成绩统计
    【b094&&z14】靶形数独
    【z06】观光公交
    【u213&&t037】修剪花卉
    【t062】最厉害的机器人
    【t075】郁闷的记者
  • 原文地址:https://www.cnblogs.com/playboy307/p/5273424.html
Copyright © 2011-2022 走看看