您的位置:首页

CF 447B(DZY Loves Strings-贪心)

2017-06-23 21:05 369 查看
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 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.

贪心,不难想到取最大的w扔在末尾

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (1000+10)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
char s[MAXN];
int k,w[MAXN];
int main()
{
//	freopen("Strings.in","r",stdin);
//	freopen(".out","w",stdout);

scanf("%s\n%d",s+1,&k);

int len=strlen(s+1);

int p=0;
Fork(i,'a','z') cin>>w[i],p=max(p,w[i]);

ll ans=0;
For(i,len) ans+=w[s[i]]*i;

if (k) ans+=(len+1+len+k)*k/2*p;

cout<<ans<<endl;

return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: