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?=?2and he wants to send string "aaa", he can split it into "a" and "aa" and use 2magical 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:

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".

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.

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.

3aab2 3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1Output
322Input
10abcdeabcde5 5 5 5 4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1Output
40143Note
In the first example the three ways to split the message are:
a|a|baa|ba|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.
題目大意：字符串全部都是小寫字母，每一個小寫字母都會被賦值，這個值就是這個字母在一個字符串里最多出現的次數。對這個字符串進行拆分成一個個子串，問你有幾種情況拆分，然后這些情況中，一個子串里最多有多少個字符，這個字符串最少被拆成幾個字符子串。
題目分析：雖說是基礎dp，我還是鼓搗了老半天，最后還因為那個數組開小了檢查了好久才檢查出來，坑爹cf，居然不報內存的錯，只說是wrong answer。
   因為每個字母都有只能出現幾次的要求，所以要開數組存儲下來。
   然后對于這個字符數組遍歷，定義len為當前這個字母所能出現的最大次數。然后再循環，用j往后遍歷，同時更新len，只要滿足i-j+1<=len,那么從i-j+1到i就能劃分成一組，所以此時
    dp1[i]=(dp1[i]+dp1[j-1])%mod;dp1數組表示從0到i所能劃分的最大組數。
  然后那個一個子串里有多少個字符這個比較簡單，直接比較即可。
   最后一個問題則用dp2表示
   dp2[i]=min(dp2[i],dp2[j-1]+1);
   dp1和dp2都需要一定的初始化。
   做完之后感覺這個dp的確比較簡單，可惜就是沒做出來。。。
#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define mod 1000000007using namespace std;const int maxn = 1005;int a[maxn];int dp1[maxn];int dp2[maxn];char s[maxn];int main(){	int n;	while((scanf("%d",&n))!=EOF){		scanf("%s",s+1);		for(int i=0;i<26;i++)		   scanf("%d",&a[i]);		memset(dp1,0,sizeof(dp1));		memset(dp2,1,sizeof(dp2));		dp1[0]=1;		dp2[0]=0;		int lmax=0;		for(int i=1;i<=n;i++){			int len=a[s[i]-'a'];			for(int j=i;j>0;j--){				len=min(len,a[s[j]-'a']);				if(i-j+1>len)				  break;				lmax=max(i-j+1,lmax);				dp1[i]=(dp1[i]+dp1[j-1])%mod;				dp2[i]=min(dp2[i],dp2[j-1]+1);			}		}		printf("%d/n%d/n%d/n",dp1[n],lmax,dp2[n]);	}	return 0;}