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  • Codeforces Round #354 (Div. 2) C. Vasya and String 二分

    C. Vasya and String

    题目连接:

    http://www.codeforces.com/contest/676/problem/C

    Description

    High school student Vasya got a string of length n as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty 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.

    Sample Input

    4 2
    abba

    Sample Output

    4

    Hint

    题意

    有一个长度为n的字符串,你可以改变最多k次,问你最长的全是一样字符的串是多长

    这个字符串只含有a和b字符

    题解:

    枚举结尾,然后暴力二分就好了

    用前缀和预处理一下,那么前缀和之差,就是你需要改变的个数……

    代码

    #include<bits/stdc++.h>
    using namespace std;
    
    const int maxn = 1e5+7;
    int n,k,sum[maxn];
    char s[maxn];
    
    int main()
    {
        scanf("%d%d",&n,&k);
        scanf("%s",s+1);
        for(int i=1;i<=n;i++)
        {
            sum[i]=sum[i-1];
            if(s[i]=='a')sum[i]++;
        }
        int ans = 0;
        for(int i=1;i<=n;i++)
        {
            int l = 1,r = i,Ans = i;
            while(l<=r)
            {
                int mid = (l+r)/2;
                if(sum[i]-sum[mid-1]<=k)r=mid-1,Ans=mid;
                else l=mid+1;
            }
            if(sum[i]-sum[Ans-1]<=k)
                ans=max(ans,i-Ans+1);
        }
        for(int i=1;i<=n;i++)
        {
            sum[i]=sum[i-1];
            if(s[i]=='b')sum[i]++;
        }
        for(int i=1;i<=n;i++)
        {
            int l = 1,r = i,Ans = i;
            while(l<=r)
            {
                int mid = (l+r)/2;
                if(sum[i]-sum[mid-1]<=k)r=mid-1,Ans=mid;
                else l=mid+1;
            }
            if(sum[i]-sum[Ans-1]<=k)
                ans=max(ans,i-Ans+1);
        }
        cout<<ans<<endl;
    }
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  • 原文地址:https://www.cnblogs.com/qscqesze/p/5531376.html
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