题目描述

Description

Farmer John has purchased a subscription to Good Hooveskeeping magazine for his cows, so they have plenty of material to read while waiting around in the barn during milking sessions. Unfortunately, the latest issue contains a rather inappropriate article on how to cook the perfect steak, which FJ would rather his cows not see (clearly, the magazine is in need of better editorial oversight).
FJ has taken all of the text from the magazine to create the string S of length at most 10^6 characters. From this, he would like to remove occurrences of a substring T to censor the inappropriate content. To do this, Farmer John finds the first occurrence of T in S and deletes it. He then repeats the process again, deleting the first occurrence of T again, continuing until there are no more occurrences of T in S. Note that the deletion of one occurrence might create a new occurrence of T that didn't exist before.
Please help FJ determine the final contents of S after censoring is complete
有一个S串和一个T串，长度均小于1,000,000，设当前串为U串，然后从前往后枚举S串一个字符一个字符往U串里添加，若U串后缀为T，则去掉这个后缀继续流程。

Input

The first line will contain S. The second line will contain T. The length of T will be at most that of S, and all characters of S and T will be lower-case alphabet characters (in the range a..z).

Output

The string S after all deletions are complete. It is guaranteed that S will not become empty during the deletion process.

Sample Input

whatthemomooofun
moo

Sample Output

whatthefun

题目分析

我们维护一个栈 先把长度和T串相等的S串踹进栈里 顺便维护一下从第一个字符到每个字母之间的Hash值 再预处理出T串的Hash值
之后枚举剩下的T串字符 每次向栈内加入一个字符 就判断下以栈顶为结尾的T个连续字母的Hash值是否与T串相同 如果相同就踢掉这T个字符
最后输出栈内字符即可
这里试了一发自然溢出

#include <cstdio>
#include <cstring>
#include <set>
#include <map>
#include <vector>
#include <cmath>
#include <queue>
#include <algorithm>
using namespace std;
unsigned long long seed=131;
char s[1000100],t[1000100];
int lens,lent;
unsigned long long po[1000100],hsh[1000100],st;
void init()
{
    po[0]=1;
    for(int i=1;i<=lens;i++) po[i]=po[i-1]*seed;
    for(int i=0;i<lent;i++)
        hsh[i+1]=hsh[i]*seed+(s[i]-'a');
    for(int i=0;i<lent;i++)
        st=st*seed+(t[i]-'a');
}
int top;
char sta[1000100];
int main()
{
    scanf("%s%s",&s[0],&t[0]);
    lens=strlen(s),lent=strlen(t);
    init();
    for(int i=0;i<lent;i++)
        sta[++top]=s[i];
    for(int i=lent;i<lens;i++)
    {
        while(top>=lent&&hsh[top]-hsh[top-lent]*po[lent]==st)
            for(int j=0;j<lent;j++)
                sta[top--]='\0';
        sta[++top]=s[i];
        hsh[top]=hsh[top-1]*seed+(s[i]-'a');
    }
    while(top>=lent&&hsh[top]-hsh[top-lent]*po[lent]==st)
        for(int j=0;j<lent;j++)
            sta[top--]='\0';
    for(int i=1;i<=top;i++) printf("%c",sta[i]);
    return 0;
}