time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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.
【题目链接】:http://codeforces.com/contest/766/problem/C
【题意】
给你一个长度为n的字符串;
这个字符串只包含小写字母;
然后让你把这个字符串进行分割;形成若干个小的字符串;
但是不是任意分割的;
每个小写字母都有一个数字ma[i];表示这个字母能够存在于长度不超过ma[i]的字符串内;
在这个条件下分割;
【题解】
设f[i]表示从i开始进行分割的方案数;
f[i] += ∑f[j];这里min(ma[s[i]..s[j]])>=j-i+1,且j>=i;
ma[x]是x这个字母能够待在的最长的字符串的长度;
然后每次都用j-i+1尝试更新“段”的最大值;
用一个num[i]表示以i作为分割的起点需要分成几段；
num[i]=min(num[i],num[j]+1);
边界:
f[n+1]=1,num[n+1]=0;
逆序更新;
最后输出f[1]就好;
【完整代码】
#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int MAXN = 1100;
const int MOD = 1e9+7;
int n,ma[MAXN],a[MAXN],mal = 0,mi,num[MAXN];
LL ans = 0,f[MAXN];
char s[MAXN];
void solve(int x)
{
int change = 1e8;
rep1(i,x,n)
{
change = min(ma[a[i]],change);
int ll = i-x+1;
if (ll>change)
break;
if (f[i+1]!=-1)
{
if (f[x]==-1) f[x] = 0;
f[x] = (f[x]+f[i+1])%MOD;
num[x] = min(num[i+1]+1,num[x]);
mal = max(mal,ll);
}
}
}
int main()
{
//freopen("F:\rush.txt","r",stdin);
memset(f,255,sizeof f);
memset(num,0x3f3f3f3f,sizeof num);
rei(n);
scanf("%s",s+1);
rep1(i,1,n)
a[i] = s[i]-'a'+1;
rep1(i,1,26)
rei(ma[i]);
f[n+1] = 1,num[n+1] = 0;
rep2(i,n,1)
solve(i);
mi = n;
cout << f[1] << endl;
printf("%d
",mal);
printf("%d
",num[1]);
return 0;
}