国产探花免费观看_亚洲丰满少妇自慰呻吟_97日韩有码在线_资源在线日韩欧美_一区二区精品毛片,辰东完美世界有声小说,欢乐颂第一季,yy玄幻小说排行榜完本

首頁 > 學院 > 開發設計 > 正文

【codeforces 766C】Mahmoud and a Message

2019-11-10 22:38:24
字體:
來源:轉載
供稿:網友

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.

【題目鏈接】:http://codeforces.com/contest/766/problem/C

【題意】 給你一個長度為n的字符串; 這個字符串只包含小寫字母; 然后讓你把這個字符串進行分割;形成若干個小的字符串; 但是不是任意分割的; 每個小寫字母都有一個數字ma[i];表示這個字母能夠存在于長度不超過ma[i]的字符串內; 在這個條件下分割;

【題解】 設f[i]表示從i開始進行分割的方案數; f[i] += ∑f[j];這里min(ma[s[i]..s[j]])>=j-i+1,且j>=i; ma[x]是x這個字母能夠待在的最長的字符串的長度; 然后每次都用j-i+1嘗試更新“段”的最大值; 用一個num[i]表示以i作為分割的起點需要分成幾段; num[i]=min(num[i],num[j]+1); 邊界: f[n+1]=1,num[n+1]=0; 逆序更新; 最后輸出f[1]就好; 【完整代碼】

#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/n",mal); printf("%d/n",num[1]); return 0;}
上一篇:文章標題

下一篇:新學內容

發表評論 共有條評論
用戶名: 密碼:
驗證碼: 匿名發表
主站蜘蛛池模板: 汾阳市| 清丰县| 彰化县| 凤冈县| 陆丰市| 诸暨市| 阳东县| 晋中市| 富阳市| 通许县| 涟源市| 无极县| 桦川县| 克东县| 辛集市| 寻甸| 白山市| 临江市| 潮安县| 万载县| 根河市| 洛浦县| 阳东县| 甘泉县| 盘锦市| 那曲县| 开鲁县| 永吉县| 建德市| 左云县| 丹东市| 武安市| 清苑县| 泾川县| 拜泉县| 纳雍县| 奉贤区| 固阳县| 休宁县| 定南县| 凌云县|