C. Mahmoud and a Message
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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： 给定一个长为n的字符串，字符的组成全是小写英文字母，然后给出26个英文字母的所在分割字符串的长度范围，将字符串分割，问有多少种分割方法，分割后最长字符串的长度，最少的分割个数。
题解：用到dp了，转移方程dp[i] = dp[i] + dp[i-j]；i是第i个字符的位置，j是i的前缀范围，那么i-j就是第i-j个字符和i个字符分割在一起了，在加上一个对是否能将第i-j个字符和第i个字符分割到一起，以此类推就可以得到dp[n]；分割的最长字符串就是j了，求j的最大值；每次分割的分割数都加1。
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
const int mod=1e9+7;
const int inf=0x3f3f3f3f;
int a[30];
char str[1010];
int dp[1010];
int judge(int i, int j)
{
for(int k = i-j+1; k <= i; k++)
{
if(a[str[k]-'a'] < j)
return 0;
}
return 1;
}
int main()
{
int n, i, j;
int minn[1010];
scanf("%d", &n);
scanf("%s", str+1);
for(i = 0; i < 26; i++)
{
scanf("%d", &a[i]);
}
memset(dp, 0, sizeof(dp));
dp[0] = 1;
minn[0] = 0;
int maxx = 0;
for(i = 1; i <= n; i++)
{
minn[i] = inf;
for(j = 1; j <= i; j++)
{
if(judge(i, j))
{
if(dp[i-j])
maxx = max(j, maxx);
dp[i] = (dp[i] + dp[i-j])%mod;
minn[i] = min(minn[i], minn[i-j]+1);
}
}
}
printf("%d\n%d\n%d\n", dp[n], maxx, minn[n]);
return 0;
}