• CF447B DZY Loves Strings 贪心


    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.

    Examples
    Input
    Copy
    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
    Copy
    41
    Note

    In the test sample DZY can obtain "abcbbc", value = 1·1 + 2·2 + 3·2 + 4·2 + 5·2 + 6·2 = 41.

    贪心地选取 maxx 加在最后面即可;

    #include<iostream>
    #include<cstdio>
    #include<algorithm>
    #include<cstdlib>
    #include<cstring>
    #include<string>
    #include<cmath>
    #include<map>
    #include<set>
    #include<vector>
    #include<queue>
    #include<bitset>
    #include<ctime>
    #include<deque>
    #include<stack>
    #include<functional>
    #include<sstream>
    //#include<cctype>
    //#pragma GCC optimize(2)
    using namespace std;
    #define maxn 20005
    #define inf 0x7fffffff
    //#define INF 1e18
    #define rdint(x) scanf("%d",&x)
    #define rdllt(x) scanf("%lld",&x)
    #define rdult(x) scanf("%lu",&x)
    #define rdlf(x) scanf("%lf",&x)
    #define rdstr(x) scanf("%s",x)
    typedef long long  ll;
    typedef unsigned long long ull;
    typedef unsigned int U;
    #define ms(x) memset((x),0,sizeof(x))
    const long long int mod = 1e9 + 7;
    #define Mod 1000000000
    #define sq(x) (x)*(x)
    #define eps 1e-3
    typedef pair<int, int> pii;
    #define pi acos(-1.0)
    //const int N = 1005;
    #define REP(i,n) for(int i=0;i<(n);i++)
    typedef pair<int, int> pii;
    inline ll rd() {
        ll x = 0;
        char c = getchar();
        bool f = false;
        while (!isdigit(c)) {
            if (c == '-') f = true;
            c = getchar();
        }
        while (isdigit(c)) {
            x = (x << 1) + (x << 3) + (c ^ 48);
            c = getchar();
        }
        return f ? -x : x;
    }
    
    ll gcd(ll a, ll b) {
        return b == 0 ? a : gcd(b, a%b);
    }
    ll sqr(ll x) { return x * x; }
    
    /*ll ans;
    ll exgcd(ll a, ll b, ll &x, ll &y) {
        if (!b) {
            x = 1; y = 0; return a;
        }
        ans = exgcd(b, a%b, x, y);
        ll t = x; x = y; y = t - a / b * y;
        return ans;
    }
    */
    
    
    int val[30];
    bool cmp(int x, int y) {
        return x > y;
    }
    
    int main() {
        //ios::sync_with_stdio(0);
        string s;
        cin >> s;
        int n; cin >> n; int maxx = -1;
        for (int i = 1; i <= 26; i++) {
            rdint(val[i]), maxx = max(maxx, val[i]);
        }
    
        ll sum = 0; int i;
        int len = s.length();
        for (i = 0; i < len; i++)sum += ((ll)val[s[i] - 'a' + 1] * (i + 1));
        for (int j = 0; j < n; j++, i++) {
            sum += ((i + 1)*maxx);
        }
        cout << sum << endl;
        return 0;
    }
    
    EPFL - Fighting
  • 相关阅读:
    java数据类型
    索引的种类和优缺点
    IntelliJ IDEA 自动导入快捷键
    KTV点歌系统------LinkedList
    KTV 点歌系统------ArrayList
    超市购物程序
    awk 入门教程
    Git 分支开发规范
    私有镜像仓库Harbor设置https、http访问
    私有镜像仓库Harbor部署
  • 原文地址:https://www.cnblogs.com/zxyqzy/p/10215148.html
Copyright © 2020-2023  润新知