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:
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 ≤ ax ≤ 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.
Examples 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.
题意：给出n作为字符串长度，再给出字符串，再给出26个字母每个字母的长度，规定每个字母所在子字符串不能超过这个长度，允许把原字符串进行分割。输出有多少种分组方案。分组方案中子字符串最长是多少。最少的分组方案是多少。涉及到每个字母的决策。所以是dp，三个dp的问题合并，对于像我这样的新手绝对是好题。 问题一：多少种分组方案。 状态转移方程 dp[0]=1 分组方案至少一个 ,从后往前，如果j字母所能在的长度符合i-j的长度 for i 1~n for j i-1~0 只要len>=i-j; 因为 len起码为1 所以 i起码比j大于1，才能不能更新。（关于基底的技巧）因为len>i-j 所以j可以和i在同一个子字符串，那么状态转移方程： dp[i]=dp[i]+dp[j] 因为dp[j]里面的所有方案没有第j个字母也是可以单独成立。然后一步步 j–，累加i-j的方案。知道len
#include <bits/stdc++.h>using namespace std;const int inf=(int)1e9;const int mod=inf+7;int a[30];int dp[3][2000];int main(){ int n; string s; while(cin>>n) { cin>>s; for(int i=0;i<26;i++) cin>>a[i]; dp[0][0]=1; for(int i=1;i<=n;i++) { int len=inf; dp[1][i]=-inf; dp[2][i]=inf; for(int j=i-1;j>=0;--j) { len=min(len,a[s[j]-'a']); if(len<i-j) break; dp[0][i]=(dp[0][i]+dp[0][j])%mod; dp[1][i]=max(dp[1][i],max(i-j,dp[1][j])); dp[2][i]=min(dp[2][i],dp[2][j]+1); } } cout<<dp[0][n]<<endl; cout<<dp[1][n]<<endl; cout<<dp[2][n]<<endl;}}新闻热点
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