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=131; const int M=1e7+7; const int maxs=1e5+5; const int maxt=1e6+5; int lens,lent; char s[maxs],t[maxt]; int tot0,tot1; ll pre[maxt],Ppow[maxt]; void prework(int len) { pre[0]=0; Ppow[0]=1; for(int i=1;i<=len;i++) { pre[i]=pre[i-1]*P+(t[i]-'a'+1), pre[i]%=M; Ppow[i]=Ppow[i-1]*P%M; } } inline ll Hash(int l,int r) { return (pre[r]-pre[l-1]*Ppow[r-(l-1)]%M+M)%M; } int main() { scanf("%s%s",s+1,t+1); lens=strlen(s+1), lent=strlen(t+1); tot0=tot1=0; for(int i=1;i<=lens;i++) tot0+=(s[i]=='0'), tot1+=(s[i]=='1'); int ans=0; int len0,len1; prework(lent); for(len0=1;len0<lent;len0++) { if(tot0*len0>=lent) break; if((lent-tot0*len0)%tot1>0) continue; len1=(lent-tot0*len0)/tot1; bool ok=1; int cnt0=0, cnt1=0; ll hash_r0=0, hash_r1=0; for(int i=1;i<=lens;i++) { if(s[i]=='0') { int l=cnt0*len0+cnt1*len1+1, r=l+len0-1; ll has=Hash(l,r); if(hash_r0==0) { hash_r0=has; } else { if(hash_r0!=has) { ok=0; break; } } cnt0++; } if(s[i]=='1') { int l=cnt0*len0+cnt1*len1+1, r=l+len1-1; ll has=Hash(l,r); if(hash_r1==0) { hash_r1=has; } else { if(hash_r1!=has) { ok=0; break; } } cnt1++; } if(hash_r0!=0 && hash_r1!=0 && hash_r0==hash_r1) { ok=0; break; } } if(ok) ans++; } printf("%d ",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。