zoukankan      html  css  js  c++  java
  • CF w3d2 A. Tritonic Iridescence

    Overlooking the captivating blend of myriads of vernal hues, Arkady the painter lays out a long, long canvas.

    Arkady has a sufficiently large amount of paint of three colours: cyan, magenta, and yellow. On the one-dimensional canvas split into n consecutive segments, each segment needs to be painted in one of the colours.

    Arkady has already painted some (possibly none or all) segments and passes the paintbrush to you. You are to determine whether there are at least two ways of colouring all the unpainted segments so that no two adjacent segments are of the same colour. Two ways are considered different if and only if a segment is painted in different colours in them.

    Input

    The first line contains a single positive integer n (1 ≤ n ≤ 100) — the length of the canvas.

    The second line contains a string s of n characters, the i-th of which is either 'C' (denoting a segment painted in cyan), 'M' (denoting one painted in magenta), 'Y' (one painted in yellow), or '?' (an unpainted one).

    Output

    If there are at least two different ways of painting, output "Yes"; otherwise output "No" (both without quotes).

    You can print each character in any case (upper or lower).

    Examples

    inputCopy
    5
    CY??Y
    outputCopy
    Yes
    inputCopy
    5
    C?C?Y
    outputCopy
    Yes
    inputCopy
    5
    ?CYC?
    outputCopy
    Yes
    inputCopy
    5
    C??MM
    outputCopy
    No
    inputCopy
    3
    MMY
    outputCopy
    No

    Note

    For the first example, there are exactly two different ways of colouring: CYCMY and CYMCY.

    For the second example, there are also exactly two different ways of colouring: CMCMY and CYCMY.

    For the third example, there are four ways of colouring: MCYCM, MCYCY, YCYCM, and YCYCY.

    For the fourth example, no matter how the unpainted segments are coloured, the existing magenta segments will prevent the painting from satisfying the requirements. The similar is true for the fifth example.

    #include<bits/stdc++.h>
    using namespace std;
    int main()
    {
    	string s;
    	int n;
    	bool ans=false;
    	cin>>n>>s;
    	for(int i=0;i<n;i++){
    		if(s[i]=='?'){
    			if(i==0||i==n-1)ans=true;
    			if(s[i-1]==s[i+1]||s[i+1]=='?'||s[i-1]=='?')ans=true;
    		}
    		else {
    			if(s[i]==s[i-1]||s[i]==s[i+1]){
    				ans=false;
    				break;
    			}
    		}
    	}
    	if(ans)cout<<"YES";
    	else cout<<"NO";
    	return 0;
    }
    
  • 相关阅读:
    自动同步日期dos命令 | DOS命令自动同步时间
    Mysql字符串截取,去掉时间,匹配日期等于今日
    HTML指定页面编码
    Mysql连接字符,字段函数concat()
    功能强大的截图工具snipaste
    当页面提交时,执行相关JS函数检查输入是否合法
    DOM和BOM
    JS内建對象(Math,Number,String,Date)
    JS数组基础01(数组的创建,push,pop,unshift,shift,concat,join,splice,slice,sort.reverse,indexOf,三种排序)
    总结01(对象引用的赋值与对象的复制,函数作为对象及回调递归,区分数组与对象)
  • 原文地址:https://www.cnblogs.com/LiangYC1021/p/12796462.html
Copyright © 2011-2022 走看看