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 ai. 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:

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?≤?ax?≤?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
40143
竟然想成了一個(gè)區(qū)間dp，直接混了，從兩邊dp去了。其實(shí)就是一個(gè)很基礎(chǔ)的dp。
對(duì)于新來(lái)的一個(gè)字符，新生成的子串自然就是從后面開始長(zhǎng)度分別為1.2.3...的一些子串。
所以dp[i]表示長(zhǎng)度為i的字符串的總的分隔方法，然后用f[i]表示分隔出的最小的子串?dāng)?shù)。
j為帶有后面新加字符的子串
從i-j+1到i的字符串合法，那么有
1.dp[i]+=dp[i-j];
2.f[i]=min(f[i],f[i-j]+1);
3.如果1~i-j的字符串也合法，那么字符串長(zhǎng)度的最大值為max(ans,j);其實(shí)前面的必定合法。
因?yàn)殚L(zhǎng)度為1的必定合法。
#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <vector>#include <queue>using namespace std;int n,m;const int MAXN=1e3+7;const int mod=1e9+7;int limit[30];long long dp[MAXN];int f[MAXN];char s[MAXN];int check(int i,int j){    int l=j-i+1;    for(int k=i;k<=j;++k)    {        if(limit[s[k]-'a']<l)return 0;    }    return 1;}int main(){    int i,j;    scanf("%d",&n);    scanf("%s",s+1);    for(i=0;i<26;++i)scanf("%d",&limit[i]);    int MAX=0;    dp[0]=1;    for(i=1;i<=n;++i)//長(zhǎng)度為i    {        f[i]=1e9;        for(j=1;j<=i;++j)//長(zhǎng)度        {            if(check(i-j+1,i))//后面的當(dāng)前序列合法            {                dp[i]=(dp[i]+dp[i-j])%mod;                f[i]=min(f[i],f[i-j]+1);                MAX=max(MAX,j);            }        }    }    printf("%I64d/n%d/n%d/n",dp[n],MAX,f[n]);}