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    B. DZY Loves Strings
    time limit per test1 second
    memory limit per test256 megabytes
    inputstandard input
    outputstandard output
    DZY loves collecting special strings which only contain lowercase letters. For each lowercase letter c DZY knows its value wc. For each special string s = s1s2... s|s| (|s| is the length of the string) he represents its value with a function f(s), where


    Now DZY has a string s. He wants to insert k lowercase letters into this string in order to get the largest possible value of the resulting string. Can you help him calculate the largest possible value he could get?

    Input
    The first line contains a single string s (1 ≤ |s| ≤ 103).

    The second line contains a single integer k (0 ≤ k ≤ 103).

    The third line contains twenty-six integers from wa to wz. Each such number is non-negative and doesn't exceed 1000.

    Output
    Print a single integer — the largest possible value of the resulting string DZY could get.

    Sample test(s)
    input
    abc
    3
    1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
    output
    41
    Note
    In the test sample DZY can obtain "abcbbc", value = 1·1 + 2·2 + 3·2 + 4·2 + 5·2 + 6·2 = 41.

    简单贪心。

    因为填的字母没有次数限制,所以最优策略很容易想到,就是在最后面填最大的。

    不用实际去填,算出ans就可以。

    #include <iostream>
    #include <cmath>
    #include <cstring>
    #include <algorithm>
    
    using namespace std;
    
    int main()
    {
        int k,len;
        char st[2000];
        int a[2000];
        memset(a,0,sizeof(a));
        int w[50];
        cin>>st>>k;
        int m=-1;
        for (int i=1;i<=26;i++)
        {
    
            cin>>w[i];
            if (w[i]>m)
                m=w[i];
        }
        len=strlen(st);
        for (int i=0;i<len;i++)
            a[i]=(int)(st[i]-96);
            long long ans=0;
            for (int i=0;i<len;i++)
            {
                ans=ans+w[a[i]]*(i+1);
               //    cout<<ans<<endl;
            }
            for (int i=len;i<len+k;i++)
                ans=ans+m*(i+1);
            cout<<ans<<endl;
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/111qqz/p/4378701.html
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