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?
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.
Print a single integer — the largest possible value of the resulting string DZY could get.
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
41
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; }