字符串 hash - 不停地删掉字母

Censor

frog is now a editor to censor so-called sensitive words (敏感词).

She has a long text p

. Her job is relatively simple -- just to find the first occurence of sensitive word w

and remove it.

frog repeats over and over again. Help her do the tedious work.

Input

The input consists of multiple tests. For each test:

The first line contains 1

string w. The second line contains 1 string p

.

(1≤length of w,p≤5⋅106

, w,p

consists of only lowercase letter)

Output

For each test, write 1

string which denotes the censored text.

Sample Input

    abc
    aaabcbc
    b
    bbb
    abc
    ab

Sample Output

    a

    ab

题意： 给你两个字符串，要求每次可执行的操作是再下面的串里删去第一个串，直到不能删为止
思路分析 ： 先对模式串进行一次 hash ,然后再对主串进行依次 hash ，然后依次检索每个位置，当二者的 hash 值相同时即代表是相同的字符串，在这里我们只是 hash 了一次，并且默认了不同的字符串hash值一定是不同的
代码示例：

using namespace std;
#define ll unsigned long long
const ll maxn = 5e6+5;

char a[maxn], b[maxn];
ll len1, len2;
ll hash_num;
ll p = 19873;
ll pp[maxn];

void init(){
    pp[0] = 1;
    for(ll i = 1; i <= 5000000; i++){
        pp[i] = pp[i-1]*p;
    }
}

void cal(){
    hash_num = 0;

    for(ll i = 1; i <= len1; i++){
        hash_num = hash_num*p+(a[i]-'a');
    }
}

ll sta[maxn];
ll ha[maxn];

bool check(ll pos){

    if (pos >= len1 && ha[pos]-ha[pos-len1]*pp[len1] == hash_num){
        return true;
    }
    return false;
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    init();

    while(~scanf("%s%s", a+1, b+1)){
        len1 = strlen(a+1);
        len2 = strlen(b+1);

        cal();
        //prllf("+++  %llu \n", hash);    

        ll top = 1;
        ha[0] = 0;
        for(ll i = 1; i <= len2; i++){
            sta[top] = b[i];
            ha[top] = ha[top-1]*p+(b[i]-'a');
            if (check(top)) top -= len1;
            top++;
        }
        for(ll i = 1; i < top; i++){
            printf("%c", sta[i]);
        }
        printf("\n");
    }
    return 0;
}