#include <bits/stdc++.h>
using namespace std;
const int maxn = 100010;
#define ull unsigned long long 
int c,t,len,state;
string s;
int n[maxn];
long long sm[maxn][10],tag[10],cnt[10],vlu[10],e[10],f[10],deta[10];
long long ans = LLONG_MAX,nans;
long long mb[10] = {0,1,11,111,1111,11111,111111},me[10] = {0,9,99,999,9999,99999,999999},jc[10] = {0,1,10,100,1000,10000,100000,1000000};
void pre(){
	ans = 0;
	cin>>s;
	memset(n,0,sizeof(n));
	memset(cnt,0,sizeof(cnt));
	memset(tag,0,sizeof(tag));
	memset(e,0,sizeof(e));
	len = s.length();
	for(int i=1;i<=len;i++){
		n[i] = s[i-1] - '0';
		cnt[n[i]]++;
		if(!tag[n[i]]) tag[n[i]] = i;
		for(int x=1;x<=9;x++) sm[i][x] = sm[i-1][x];
		sm[i][n[i]]++;
	}
	for(int i=1;i<=9;i++) cin>>vlu[i];
	return;
}
inline bool check(int mj){//(l,r) check the under
	int l=tag[e[1]],r=len+1,mid=(l+r)>>1;
	int now = l;
	if(len-l+1 < mj) return 0;
	for(int i=2;i<=mj;i++){
		l = now,r = len+1;
		while(l<r-1){
			mid = (l+r)>>1;
			if(sm[mid][e[i]] - sm[now][e[i]] > 0) r = mid;
			else l = mid;
		}
		now = r;
		if(now == len+1) return 0;
	}
	return 1;
}
void slove(){
	ans = LLONG_MAX;
	bool pd = 0;
	long long opmin = LLONG_MAX;
	for(int mj=1;mj<=min(6,len);mj++){
		for(int now=mb[mj];now<=me[mj];now++){
			memset(deta,0,sizeof(deta));
			nans = 0;
			pd = 0;
			int temp = now,pll = 0;
			while(temp){
				f[++pll] = temp%10;
				temp/=10;
			}
			for(int i=1;i<=mj;i++){
				e[i] = f[mj-i+1];
				deta[e[i]]++;
				if(!e[i]){pd = 1;break;}
				if(deta[e[i]] > cnt[e[i]]){pd = 1;break;}
			}
			if(pd) continue;
			if(!check(mj)) continue;
			for(int i=1;i<=9;i++) nans += (cnt[i] - deta[i]) * vlu[i];
			if(nans >= ans) continue;
			if(len > 10){
				ans = min(nans+now,ans);
				continue;
			}
			opmin = LLONG_MAX;
			for(int state=0;state<(1<<mj+1);state++){
				long long op = 0,res = 0;
				for(int i=0;i<=mj-1;i++){
					if((state>>i)&1){
						int ch = e[i+1];
						op += vlu[ch];
					}
					else res = res * 10 + e[i+1];
				}
				op += res;
				opmin = min(op,opmin);
			}
			nans += opmin;
			ans = min(nans,ans);
		}
	}
	return;
}
int main(){
//	freopen("bargain4.in","r",stdin);
	cin>>c>>t;
	while(t){
		pre();
		slove();
		cout<<ans;
		if(t != 1) cout<<endl;
		t--;
	}
	return 0;
}

思路：根据vlu的上界得到6位以上删数更优，枚举所有可能的数，二分判断是否可取，然后枚举小于等于六位里面可不可取（这里直接换成了ans = min(nans+now,ans)，now是枚举的六位数，nans是删去数的总和）

为什么直接ans = min(nans+now,ans);是可行的？（带if是因为如果去掉后面枚举会WA#1#2）

测评记录：here

AC