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  • Codeforces Round #354 (Div. 2)——C. Vasya and String(尺取)

    C. Vasya and String
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    High school student Vasya got a string of length n as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotesbeauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters.

    Vasya can change no more than k characters of the original string. What is the maximum beauty of the string he can achieve?

    Input

    The first line of the input contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ n) — the length of the string and the maximum number of characters to change.

    The second line contains the string, consisting of letters 'a' and 'b' only.

    Output

    Print the only integer — the maximum beauty of the string Vasya can achieve by changing no more than k characters.

    Examples
    input
    4 2
    abba
    
    output
    4
    
    input
    8 1
    aabaabaa
    
    output
    5
    
    Note

    In the first sample, Vasya can obtain both strings "aaaa" and "bbbb".

    In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa".

    一开始的渣代码运气好到test65才跪……稍微改下形式就行

    代码:

    #include<iostream>
    #include<algorithm>
    #include<cstdlib>
    #include<sstream>
    #include<cstring>
    #include<cstdio>
    #include<string>
    #include<deque>
    #include<stack>
    #include<cmath>
    #include<queue>
    #include<set>
    #include<map>
    using namespace std;
    #define INF 0x3f3f3f3f
    #define MM(x,y) memset(x,y,sizeof(x))
    typedef pair<int,int> pii;
    typedef long long LL;
    const double PI=acos(-1.0);
    const int N=100010;
    char s[N];
    int main(void)
    {
    	int n,i,j,cnt,k;
    	while (~scanf("%d%d",&n,&k))
    	{
    		scanf("%s",s);
    		int L,R,ma=0,mb=0,c;
    		L=R=c=0;
    		while (L<n)
    		{
    			while (R<n&&c<=k)
    			{
    				if(s[R]=='b')
    				{
    					if(c==k)
    						break;
    					c++;
    				}	
    				R++;
    			}
    			ma=max(ma,R-L);
    			L++;
    			c=c-(s[L-1]=='b');
    		}
    		L=R=c=0;
    		while (L<n)
    		{
    			while (R<n&&c<=k)
    			{
    				if(s[R]=='a')
    				{
    					if(c==k)
    						break;
    					c++;
    				}	
    				R++;
    			}
    			mb=max(mb,R-L);
    			L++;
    			c=c-(s[L-1]=='a');
    		}
    		printf("%d
    ",max(ma,mb));
    	}
    	return 0;
    }
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  • 原文地址:https://www.cnblogs.com/Blackops/p/5766298.html
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