zoukankan      html  css  js  c++  java
  • CodeForces 729A Interview with Oleg (模拟)

    Polycarp has interviewed Oleg and has written the interview down without punctuation marks and spaces to save time. Thus, the interview is now a string s consisting of n lowercase English letters.

    There is a filler word ogo in Oleg's speech. All words that can be obtained from ogo by adding go several times to the end of it are also considered to be fillers. For example, the words ogo, ogogo, ogogogo are fillers, but the words go, og, ogog, ogogog and oggo are not fillers.

    The fillers have maximal size, for example, for ogogoo speech we can't consider ogo a filler and goo as a normal phrase. We should consider ogogo as a filler here.

    To print the interview, Polycarp has to replace each of the fillers with three asterisks. Note that a filler word is replaced with exactly three asterisks regardless of its length.

    Polycarp has dealt with this problem in no time. Can you do the same? The clock is ticking!

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

    The second line contains the string s of length n, consisting of lowercase English letters.

    Output
    Print the interview text after the replacement of each of the fillers with "". It is allowed for the substring "" to have several consecutive occurences.

    Example
    Input
    7
    aogogob
    Output
    a***b
    Input
    13
    ogogmgogogogo
    Output
    gmg
    Input
    9
    ogoogoogo
    Output


    Note
    The first sample contains one filler word ogogo, so the interview for printing is "a***b".

    The second sample contains two fillers ogo and ogogogo. Thus, the interview is transformed to "gmg".

    题意:

    应该很好理解。就是ogo替换成,ogogogo同样视为ogo,只用一个替代。

    题解:

    模拟即可。

    #include<iostream>
    #include<cstring>
    using namespace std;
    int main()
    {
    	int n;
    	char s[105];
    	while(cin>>n>>s)
    	{
    		for(int i=0;i<n;i++)
    		{
    			if(i+2<n&&s[i]=='o'&&s[i+1]=='g'&&s[i+2]=='o')
    			{
    				cout<<"***";
    				i+=2;
    				while(s[i+1]=='g'&&s[i+2]=='o')
    					i+=2;
    			}
    			else
    			{
    				cout<<s[i];
    			}
    		}
    		cout<<endl;
    	}
    	return 0;
    }
  • 相关阅读:
    Sql例子Sp_ExecuteSql 带参数
    Flex显示麦克风当前音量
    无法将 flash.display::Sprite@156b7b1 转换为 mx.core.IUIComponent
    FMS (端口问题)如何穿透防火墙
    19:A*B问题
    6264:走出迷宫
    2753:走迷宫
    1792:迷宫
    换钱问题(经典枚举样例)
    1943(2.1)
  • 原文地址:https://www.cnblogs.com/orion7/p/7703598.html
Copyright © 2011-2022 走看看