CF1056E Check Transcription 字符串哈希

Posted 2021-02-07 itst

传送门

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暴力枚举(0)的长度，如果对应的(1)的长度也是一个整数就去check是否合法。check使用字符串哈希。

复杂度看起来是(O(st))的，但是因为(01)两个数中数量较多的至少有(frac{|s|}{2})个，那么最多有(frac{2|t|}{|s|})个可能的答案，而每一次check是(O(|s|))的，所以总复杂度是(O(|t|))的

#include<bits/stdc++.h>
#define ll long long
#define PLL pair < ll , ll > 
//This code is written by Itst
using namespace std;

inline int read(){
    int a = 0;
    char c = getchar();
    bool f = 0;
    while(!isdigit(c) && c != EOF){
        if(c == ‘-‘)
            f = 1;
        c = getchar();
    }
    if(c == EOF)
        exit(0);
    while(isdigit(c)){
        a = (a << 3) + (a << 1) + (c ^ ‘0‘);
        c = getchar();
    }
    return f ? -a : a;
}

const int seed = 13331 , MOD1 = 1e9 + 7 , MOD2 = 1e9 + 9;
char s1[100010] , s2[1000010];
int L1 , L2 , cnt0 , cnt1 , cnt;
ll Hash[1000010][2] , poww[1000010][2];

void input(){
    scanf("%s %s" , s1 + 1 , s2 + 1);
    L1 = strlen(s1 + 1);
    L2 = strlen(s2 + 1);
}

void init(){
    for(int i = 1 ; i <= L1 ; ++i)
        s1[i] == ‘0‘ ? ++cnt0 : ++cnt1;
    poww[0][0] = poww[0][1] = 1;
    for(int i = 1 ; i <= L2 ; ++i){
        Hash[i][0] = (Hash[i - 1][0] * seed + s2[i]) % MOD1;
        Hash[i][1] = (Hash[i - 1][1] * seed + s2[i]) % MOD2;
        poww[i][0] = poww[i - 1][0] * seed % MOD1;
        poww[i][1] = poww[i - 1][1] * seed % MOD2;
    }
}

inline PLL calcHash(int i , int j){
    return PLL((Hash[i + j - 1][0] - Hash[i - 1][0] * poww[j][0] % MOD1 + MOD1) % MOD1 , (Hash[i + j - 1][1] - Hash[i - 1][1] * poww[j][1] % MOD2 + MOD2) % MOD2);
}

void work(){
    for(int i = 1 ; i * cnt0 < L2 ; ++i)
        if(!((L2 - i * cnt0) % cnt1)){
            int p = 1 , j = (L2 - i * cnt0) / cnt1;
            bool f = 1;
            PLL Hash0 = PLL(-1,0) , Hash1 = PLL(-1,0);
            for(int k = 1 ; f && k <= L1 ; ++k)
                if(s1[k] == ‘0‘){
                    if(Hash0.first == -1)
                        Hash0 = calcHash(p , i);
                    else
                        if(Hash0 != calcHash(p , i))
                            f = 0;
                    p += i;
                }
                else{
                    if(Hash1.first == -1)
                        Hash1 = calcHash(p , j);
                    else
                        if(Hash1 != calcHash(p , j))
                            f = 0;
                    p += j;
                }
            cnt += f && (Hash0 != Hash1);
        }
}

void output(){
    cout << cnt;
}

int main(){
#ifndef ONLINE_JUDGE
    freopen("in" , "r" , stdin);
    //freopen("out" , "w" , stdout);
#endif
    input();
    init();
    work();
    output();
    return 0;
}