Codeforces Round #FF (Div. 2):B. DZY Loves Strings

B. DZY Loves Strings

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

DZY loves collecting special strings which only contain lowercase letters. For each lowercase letter
c DZY knows its value
w_c. 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 w_a to
w_z. 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

题意：输入子串， 大小相应以下你输入的26个字母的大小， 后面再输入一个数字K，即你加入的过少个字母，要想结果最大。。就要找出你输入的26个字母相应数值大小最大的字母。。

就像案列中的B和C是最大的， 所以就加入其。

。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<sstream>
#include<cmath>

using namespace std;

#define f1(i, n) for(int i=0; i<n; i++)
#define f2(i, m) for(int i=1; i<=m; i++)
#define f3(i, n) for(int i=n; i>=0; i--)
#define M 1005

const int INF = 0x3f3f3f3f;

int main()
{
    char s[1005];
    int a, b, i, t=-1;
    __int64 d=0;
    int eng[27], c[1005];
    scanf("%s", s);
    b=strlen(s);
    scanf("%d", &a);
    for(i=1; i<=26; i++)
    {
        scanf("%d",&eng[i]);
        if( t<eng[i] )
            t = eng[i];
    }
    for(i=0; i<b; i++)
    {
        c[i] = s[i]-'a'+1;
        d = d + eng[c[i]]*(i+1);
    }
    for(i=1; i<=a; i++)
        d = d + t*(b+i);
    printf("%I64d\n",d);
    return 0;
}