题目链接：http://codeforces.com/contest/766/problem/C
题目：
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 thanai. For example, ifa1 = 2 he can't write character 'a' on this paper in a string of length3 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 ben and they shouldn't overlap. For example, ifa1 = 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 use3 magical papers. He can't split it into "aa" and "aa" because the sum of their lengths is greater thann. 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 strings, 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 isn and they don't overlap? Compute the answer modulo109 + 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 lengthn 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 modulo109 + 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 length2.
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 length3, while a1 = 2.
题目大意：
给你一个长度为n的串s，给出每个字母（a~z）能够最长存在长度的字串，有3个问题：
1.有多少种分配方案
2.这些分配方案中最长的一段字串的长度
3.这些分配方案中最少可以把串s分成几份
题目分析;
显然要用动态规划来求解这题
对于问题1，用dp1[k]来表示前k个字符最多分成多少份，如果和前一个字符可以组成一个字符段有状态转移方程，dp1[i]+=dp1[j-1];由于这里是要满足该一小段里面所有字符的约束条件，所以要控制ml=min(ml,v[s[j]-'a']);
对于问题2，只需要在每个符合条件的子段中max即可
对于问题3，用dp2[k]代表前k份能分成的最少份数，初始化dp2[i]=dp2[i-1]+1;（当i-1和i不在同一段），如果前些字符可以与之处于同一字符小段内，则有状态转移方程dp2[i]=min(dp2[i],dp2[j-1]+1);
AC代码：
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
#define LL long long
const int maxn=1e3+5;
const LL mod=1e9+7;
int n;
char s[maxn];
int v[26];
int dp1[maxn];//dp1[k],前k个最多分成多少份
int dp2[maxn];//dp2[k]代表前k份能分成的最少份数
int mx,ml;
int main()
{
while(~scanf("%d",&n))
{
scanf("%s",s+1);
for(int i=0;i<26;i++)
scanf("%d",&v[i]);
memset(dp1,0,sizeof(dp1));
memset(dp2,0,sizeof(dp2));
dp1[0]=1;
mx=1;
for(int i=1;i<=n;i++)
{
dp1[i]=dp1[i-1];
dp1[i]%=mod;
dp2[i]=dp2[i-1]+1;
ml=v[s[i]-'a'];
for(int j=i-1;j>=1;j--)
{
ml=min(ml,v[s[j]-'a']);
if(i-j+1<=ml)
{
dp1[i]+=dp1[j-1];
dp1[i]%=mod;
mx=max(i-j+1,mx);
dp2[i]=min(dp2[i],dp2[j-1]+1);
}
else
break;
}
}
cout<<dp1[n]<<endl;
cout<<mx<<endl;
cout<<dp2[n]<<endl;
}
return 0;
}