1688网站上自己做模版,网页设计模板套用步骤,wordpress 文艺主题,房地产如何做网站推广文章目录 CF778A String Game 题解题面翻译Input DataOutput DataInput Sample 1Output Sample 1题目描述输入格式输出格式样例 #1样例输入 #1样例输出 #1 样例 #2样例输入 #2样例输出 #2 提示算法#xff1a;二分代码#xff1a; CF778A String Game 题解
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题面翻译
… 文章目录 CF778A String Game 题解题面翻译Input DataOutput DataInput Sample 1Output Sample 1题目描述输入格式输出格式样例 #1样例输入 #1样例输出 #1 样例 #2样例输入 #2样例输出 #2 提示算法二分代码 CF778A String Game 题解
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题面翻译
给定两个由小写字母构成的字符串p和t同时给定一个由数字 1 , 2 , 3... ∣ P ∣ 1,2,3...∣P∣ 1,2,3...∣P∣ 组成的排列。其中 ∣ p ∣ ∣p∣ ∣p∣ 表示字符串p的长度按该排列顺序依次删除字符串 p p p 相应位置上的字母删除过程中约定各个字符的位置不变。请计算最多可以删除几次字符串 p p p 中仍然包含字符串 t t t。即字符串 t t t 仍然是字符串 p p p 的子序列
数据保证有解。
Input Data
第一行一个字符串 p p p 1 ≤ ∣ p ∣ ∣ t ∣ ≤ 200 , 0000 1≤∣p∣∣t∣≤200,0000 1≤∣p∣∣t∣≤200,0000
第二行一个字符串 t t t
第三行数字 1 1 1 到 ∣ p ∣ ∣p∣ ∣p∣ 组成的一个排列。
Output Data
一行一个整数表示最多删除的次数。
Input Sample 1
ababcbaabb5 3 4 1 7 6 2Output Sample 1
3题目描述
Little Nastya has a hobby, she likes to remove some letters from word, to obtain another word. But it turns out to be pretty hard for her, because she is too young. Therefore, her brother Sergey always helps her.
Sergey gives Nastya the word t t t and wants to get the word p p p out of it. Nastya removes letters in a certain order (one after another, in this order strictly), which is specified by permutation of letters’ indices of the word t t t : a 1 . . . a ∣ t ∣ a_{1}...\ a_{|t|} a1... a∣t∣ . We denote the length of word x x x as ∣ x ∣ |x| ∣x∣ . Note that after removing one letter, the indices of other letters don’t change. For example, if t t t “nastya” and a [ 4 , 1 , 5 , 3 , 2 , 6 ] a[4,1,5,3,2,6] a[4,1,5,3,2,6] then removals make the following sequence of words “nastya” “nastya” “nastya” “nastya” “nastya” “nastya” “nastya”.
Sergey knows this permutation. His goal is to stop his sister at some point and continue removing by himself to get the word p p p . Since Nastya likes this activity, Sergey wants to stop her as late as possible. Your task is to determine, how many letters Nastya can remove before she will be stopped by Sergey.
It is guaranteed that the word p p p can be obtained by removing the letters from word t t t .
输入格式
The first and second lines of the input contain the words t t t and p p p , respectively. Words are composed of lowercase letters of the Latin alphabet ( $ |p||t|200000$ ). It is guaranteed that the word $ p $ can be obtained by removing the letters from word t t t .
Next line contains a permutation a 1 , a 2 , . . . , a ∣ t ∣ a_{1},a_{2},...,a_{|t|} a1,a2,...,a∣t∣ of letter indices that specifies the order in which Nastya removes letters of t t t ( 1 a i ∣ t ∣ 1a_{i}|t| 1ai∣t∣ , all a i a_{i} ai are distinct).
输出格式
Print a single integer number, the maximum number of letters that Nastya can remove.
样例 #1
样例输入 #1
ababcba
abb
5 3 4 1 7 6 2样例输出 #1
3样例 #2
样例输入 #2
bbbabb
bb
1 6 3 4 2 5样例输出 #2
4提示
In the first sample test sequence of removing made by Nastya looks like this:
“ababcba” “ababcba” “ababcba” “ababcba”
Nastya can not continue, because it is impossible to get word “abb” from word “ababcba”.
So, Nastya will remove only three letters.
算法二分 二分枚举什么我们可以枚举删除的元素个数。 可行性假设删去 x x x 个元素可行那么删去 x − 1 x - 1 x−1 个元素也肯定可行。因此二分的序列有单调性该二分成立。 check 函数怎么写判断删掉 x x x 个元素后是否包含序列 t t t 即可。
代码
#includebits/stdc.h
using namespace std;
#define ll long long
const ll N2e610;
ll n,nt,a[N],l,r,mid,ans;
string p,t;
bool check(ll x){string kp;ll ct0;for(int i1;ix;i) k[a[i]-1] ;for(int i0;in;i){if(k[i]t[ct]) ct;if(ctnt) return 1; } return 0;
}
int main(){ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);cinpt;np.size(),ntt.size();for(int i1;in;i) cina[i];rn;while(lr){midlr1;if(check(mid)) ansmid,lmid1;else rmid-1;}coutans;return 0;
}
感谢大家的支持~
