CodeForces 1056E - Check Transcription - [字符串hash]

题目链接：https://codeforces.com/problemset/problem/1056/E

One of Arkady's friends works at a huge radio telescope. A few decades ago the telescope has sent a signal $s$ towards a faraway galaxy. Recently they've received a response $t$ which they believe to be a response from aliens! The scientists now want to check if the signal $t$ is similar to $s$.

The original signal $s$ was a sequence of zeros and ones (everyone knows that binary code is the universe-wide language). The returned signal $t$, however, does not look as easy as $s$, but the scientists don't give up! They represented $t$ as a sequence of English letters and say that $t$ is similar to $s$ if you can replace all zeros in $s$ with some string $r_0$ and all ones in $s$ with some other string $r_1$ and obtain $t$. The strings $r_0$ and $r_1$ must be different and non-empty.

Please help Arkady's friend and find the number of possible replacements for zeros and ones (the number of pairs of strings $r_0$ and $r_1$) that transform $s$ to $t$.

Input
The first line contains a string $s (2 \le |s| \le 10^5)$ consisting of zeros and ones — the original signal.

The second line contains a string $t (1 \le |t| \le 10^6)$ consisting of lowercase English letters only — the received signal.

It is guaranteed, that the string $s$ contains at least one '0' and at least one '1'.

Output
Print a single integer — the number of pairs of strings $r_0$ and $r_1$ that transform $s$ to $t$.

In case there are no such pairs, print 0.

Examples
Input
01
aaaaaa
Output
4

Input
001
kokokokotlin
Output
2

Note
In the first example, the possible pairs $(r_0,r_1)$ are as follows:

"a", "aaaaa"
"aa", "aaaa"
"aaaa", "aa"
"aaaaa", "a"
The pair "aaa", "aaa" is not allowed, since $r_0$ and $r_1$ must be different.

In the second example, the following pairs are possible:

"ko", "kokotlin"
"koko", "tlin"

题意：

给定一个 $0,1$ 字符串 $s$，和一个小写字母组成的字符串 $t$。

现在要找出两个互不相同且非空的字符串 $r_0, r_1$，使得将 $s$ 中的 $0$ 全部替换成 $r_0$，将 $1$ 全部替换成 $r_1$ 后，即可得到 $t$。

要求输出不同二元组 $(r_0,r_1)$ 的数量。

题解：

不妨枚举 $r_0$ 的长度，这样一来，根据 $0,1$ 的数目，可以直接算出 $r_1$ 的长度。

然后遍历 $s$，看此时假设的 $r_0$ 和 $r_1$ 是否可以成立，这个可以对 $t$ 字符串hash之后 $O(1)$ 地求出每个 $0,1$ 转换出来的 $r_0,r_1$ 符不符合要求。

关于时间复杂度，表面上看起来时间复杂度为 $O(|s||t|)$ 的，但实际上由于要满足条件 $|t| = cnt_0 \cdot |r_0| + cnt_1 \cdot |r_1|$，可供选择的 $(|r_0|,|r_1|)$ 比 |t| 少很多。

而且，对于很多的 $(|r_0|,|r_1|)$，若其对应的 $(r_0,r_1)$ 并非可行解，其实很快就能判定出来，不需要 $O(|s|)$ 次hash。

AC代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int P=;
const int M=1e7+;
const int maxs=1e5+;
const int maxt=1e6+;

int lens,lent;
char s[maxs],t[maxt];
int tot0,tot1;

ll pre[maxt],Ppow[maxt];
void prework(int len)
{
    pre[]=;
    Ppow[]=;
    for(int i=;i<=len;i++)
    {
        pre[i]=pre[i-]*P+(t[i]-'a'+), pre[i]%=M;
        Ppow[i]=Ppow[i-]*P%M;
    }
}
inline ll Hash(int l,int r) {
    return (pre[r]-pre[l-]*Ppow[r-(l-)]%M+M)%M;
}
int main()
{
    scanf("%s%s",s+,t+); lens=strlen(s+), lent=strlen(t+);
    tot0=tot1=;
    for(int i=;i<=lens;i++) tot0+=(s[i]==''), tot1+=(s[i]=='');

    int ans=;
    int len0,len1;
    prework(lent);
    for(len0=;len0<lent;len0++)
    {
        if(tot0*len0>=lent) break;
        if((lent-tot0*len0)%tot1>) continue;
        len1=(lent-tot0*len0)/tot1;

        bool ok=;
        int cnt0=, cnt1=;
        ll hash_r0=, hash_r1=;
        for(int i=;i<=lens;i++)
        {
            if(s[i]=='')
            {
                int l=cnt0*len0+cnt1*len1+, r=l+len0-;
                ll has=Hash(l,r);
                if(hash_r0==) {
                    hash_r0=has;
                } else {
                    if(hash_r0!=has) {
                        ok=; break;
                    }
                }
                cnt0++;
            }
            if(s[i]=='')
            {
                int l=cnt0*len0+cnt1*len1+, r=l+len1-;
                ll has=Hash(l,r);
                if(hash_r1==) {
                    hash_r1=has;
                } else {
                    if(hash_r1!=has) {
                        ok=; break;
                    }
                }
                cnt1++;
            }
            if(hash_r0!= && hash_r1!= && hash_r0==hash_r1) {
                ok=;
                break;
            }
        }
        if(ok) ans++;
    }
    printf("%d\n",ans);
}

注意：不要使用unsigned long long类型的溢出自动取模，会被卡掉；改为对大质数取模就能通过了。