Codeforces Round #555 (Div. 3) B. Long Number 【仔细读题】

B. Long Number

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a long decimal number aa consisting of nn digits from 11 to 99. You also have a function ff that maps every digit from 11 to 99 to some (possibly the same) digit from 11 to 99.

You can perform the following operation no more than once: choose a non-empty contiguous subsegment of digits in aa, and replace each digit xx from this segment with f(x)f(x). For example, if a=1337a=1337, f(1)=1f(1)=1, f(3)=5f(3)=5, f(7)=3f(7)=3, and you choose the segment consisting of three rightmost digits, you get 15531553 as the result.

What is the maximum possible number you can obtain applying this operation no more than once?

Input

The first line contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the number of digits in aa.

The second line contains a string of nn characters, denoting the number aa. Each character is a decimal digit from 11 to 99.

The third line contains exactly 99 integers f(1)f(1), f(2)f(2), ..., f(9)f(9) (1≤f(i)≤91≤f(i)≤9).

Output

Print the maximum number you can get after applying the operation described in the statement no more than once.

Examples

input

4
1337
1 2 5 4 6 6 3 1 9

output

1557

input

5
11111
9 8 7 6 5 4 3 2 1

output

99999

input

2
33
1 1 1 1 1 1 1 1 1

output

33
题意：在字符串中任意选择一个子串，将该子串中的数字按照题目的替换规则进行替换（无论替换之后，新数字比旧数字大还是小，只要在字串范围之内都要替换），找出替换后的最大数字；
这个数字替换是在字符串的某一段区间的，在这段区间中所有的就数字都要被新数字替换，所以我们最先找到的子串（其中替换后的新数字都比就数字大，遇到新数字比旧数字小，就停止），就可以的得到最大数字
错因：一开始以为替换区间是整个字符串，然后按照新数字比就数字大就替换，新数字比就数字小就不替换，然后 WA6了。

 #include<stdio.h>
 #include<string>
 #include<string.h>
 using namespace std;

 int n,num[],judge[];
 char m[];
 int main(){
     scanf("%d",&n);getchar();
     for(int i =  ; i <= n ; i++){
         scanf("%c",&m[i]);
     }
     for(int i =  ; i <= n ; i++){
         num[i] = m[i] - ;
     }
     int cnt = n;
     for(int i =  ; i <=  ; i++){
         scanf("%d",&judge[i]);
     }
     int ret = ;
     for(int i =  ; i <= n ; i++){
         if(ret ==  && judge[num[i]] < num[i]){
             ret++;
         }
         if(judge[num[i]] > num[i] && ret == ){
             num[i] = judge[num[i]];
             ret = ;
         }else if(judge[num[i]] > num[i] && ret == ){
             num[i] = judge[num[i]];
         }

         printf("%d",num[i]);
     }
     printf("\n");
     return ;
 }

AC代码

一个从很久以前就开始做的梦。