Codeforces Round #396 (Div. 2) C题Mahmoud and a Message（dp）解题报告

Posted 2020-08-30 惜取少年时

Mahmoud wrote a message s of length n. He wants to send it as a birthday present to his friend Moaz who likes strings. He wrote it on a magical paper but he was surprised because some characters disappeared while writing the string. That‘s because this magical paper doesn‘t allow character number i in the English alphabet to be written on it in a string of length more than a_i. For example, if a₁?=?2 he can‘t write character ‘a‘ on this paper in a string of length 3 or more. String "aa" is allowed while string "aaa" is not.

Mahmoud decided to split the message into some non-empty substrings so that he can write every substring on an independent magical paper and fulfill the condition. The sum of their lengths should be n and they shouldn‘t overlap. For example, if a₁?=?2 and he wants to send string "aaa", he can split it into "a" and "aa" and use 2 magical papers, or into "a", "a" and "a" and use 3 magical papers. He can‘t split it into "aa" and "aa" because the sum of their lengths is greater than n. He can split the message into single string if it fulfills the conditions.

A substring of string s is a string that consists of some consecutive characters from string s, strings "ab", "abc" and "b" are substrings of string "abc", while strings "acb" and "ac" are not. Any string is a substring of itself.

While Mahmoud was thinking of how to split the message, Ehab told him that there are many ways to split it. After that Mahmoud asked you three questions:

• How many ways are there to split the string into substrings such that every substring fulfills the condition of the magical paper, the sum of their lengths is n and they don‘t overlap? Compute the answer modulo 10⁹?+?7.
• What is the maximum length of a substring that can appear in some valid splitting?
• What is the minimum number of substrings the message can be spit in?

Two ways are considered different, if the sets of split positions differ. For example, splitting "aa|a" and "a|aa" are considered different splittings of message "aaa".

Input

The first line contains an integer n (1?≤?n?≤?10³) denoting the length of the message.

The second line contains the message s of length n that consists of lowercase English letters.

The third line contains 26 integers a₁,?a₂,?...,?a₂₆ (1?≤?a_x?≤?10³) — the maximum lengths of substring each letter can appear in.

Output

Print three lines.

In the first line print the number of ways to split the message into substrings and fulfill the conditions mentioned in the problem modulo 10⁹??+??7.

In the second line print the length of the longest substring over all the ways.

In the third line print the minimum number of substrings over all the ways.

Example

Input

3
aab
2 3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Output

3
2
2

Input

10
abcdeabcde
5 5 5 5 4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Output

401
4
3

Note

In the first example the three ways to split the message are:

• a|a|b
• aa|b
• a|ab

The longest substrings are "aa" and "ab" of length 2.

The minimum number of substrings is 2 in "a|ab" or "aa|b".

Notice that "aab" is not a possible splitting because the letter ‘a‘ appears in a substring of length 3, while a₁?=?2.

 

解题思路：

建立数组记录字符串以第i个字母结束为分割时情况数，进行动态规划（递推），最长序列在此过程中时刻刷新，最少分割次数建立另一数组记录。

参考代码：

 1 #include<bits/stdc++.h>
 2 #include <iostream>
 3 using namespace std;
 4 typedef long long ll;
 5 typedef unsigned long long ull;
 6 int n;
 7 const int MOD=1e9+7;
 8 char a[1005];
 9 int b[1005],c[1005];//b数组记录以第i个字母结束有多少种情况，c数组记录最少分次
10 int len_1=0;//时刻刷新最长序列长度
11 int dis[1005];//记录每个字母最多在多长的序列中
12 int x;
13 int main()
14 {
15     scanf("%d",&n);
16     scanf("%s",a);
17     int i;
18     for(i=0;i<=n;i++)
19     {
20         c[i]=10000;
21     }
22     for(i=0;i<26;i++)
23     {
24         scanf("%d",&dis[i]);
25     }
26     int j;
27     b[0]=1;
28     c[0]=0;
29     for(i=1;i<=n;i++)
30     {
31         x=10000;
32         for(j=1;j<=i&&j<=dis[a[i-1]-‘a‘]&&j<=x;j++)//动态规划的条件循环
33         {
34             x=min(x,dis[a[i-j-1]-‘a‘]);
35             b[i]+=b[i-j];
36             b[i]%=MOD;
37             len_1=max(len_1,j);
38             c[i]=min(c[i-j]+1,c[i]);
39         }
40     }
41     printf("%d\n%d\n%d\n",b[n],len_1,c[n]);
42     return 0;
43 }