2 seconds
256 megabytes
standard input
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".
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.
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
3
2
2
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
401
4
3
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.
——————————————————————————————————————
题目的意思是给出一个字符串，然后给出26个字母各自最大的字符串长度限制，求分割方案数，分割的最长长度限制，分割的最少段数。
思路：DP，DP[i]表示到第i位的方案数，他的值为第i位的字母能往前最大的位数到他前一位的和。
求最长长度是开个mx记录最长
求最小段数是开个Min[i]，表示到第i位的最小段数，他的值为第i位的字母能往前最大的位数到他前一位的最小值+1；
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <string>
#include <set>
#include <map>
using namespace std;
const int MAXN=100005;
const int inf=0x3f3f3f3f;
const int mod=1e9+7;
int dp[MAXN],Min[MAXN],a[MAXN];
char s[MAXN];
int n; int main()
{
while(~scanf("%d",&n))
{
scanf("%s",s);
for(int i=0; i<26; i++)
scanf("%d",&a[i]);
memset(dp,0,sizeof dp);
memset(Min,inf,sizeof Min);
dp[0]=1;
Min[0]=0;
int mx=0;
for(int i=1; i<=n; i++)
{
int mn=MAXN;
for(int j=1; j<=a[s[i-1]-'a']; j++)
{
if(i-j<0)
break;
mn=min(mn,a[s[i-j]-'a']);
if(j>mn)
break;
mx=max(mx,j);
dp[i]+=dp[i-j];
dp[i]%=mod;
Min[i]=min(Min[i],Min[i-j]);
}
Min[i]++;
}
printf("%d\n",dp[n]);
printf("%d\n",mx);
printf("%d\n",Min[n]);
}
return 0;
}