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E.cpp
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E.cpp
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#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("avx,avx2,fma")
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define dd double
#define ld long double
#define sl(n) scanf("%lld", &n)
#define si(n) scanf("%d", &n)
#define sd(n) scanf("%lf", &n)
#define pll pair <ll, ll>
#define pii pair <int, int>
#define mp make_pair
#define pb push_back
#define inf (1LL << 62)
#define loop(i, start, stop, inc) for(ll i = start; i <= stop; i += inc)
#define for1(i, stop) for(ll i = 1; i <= stop; i++)
#define for0(i, stop) for(ll i = 0; i < stop; i++)
#define rep1(i, start) for(ll i = start; i >= 1; i--)
#define rep0(i, start) for(ll i = (start-1); i >= 0; i--)
#define ms(n, i) memset(n, i, sizeof(n))
#define casep(n) printf("Case %lld:", ++n)
#define pn printf("\n")
#define pf printf
#define EL '\n'
#define fastio std::ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
const ll MAX_N = 1e6+10, mod = 2e9+63, base1 = 1e9+21, base2 = 1e9+181;
char s[MAX_N]; // 1-indexed
ll pw1[MAX_N], pw2[MAX_N], slen;
void pw_calc() {
pw1[0] = pw2[0] = 1;
for(int i = 1; i < MAX_N; i++) {
pw1[i] = (pw1[i-1] * base1) % mod;
pw2[i] = (pw2[i-1] * base2) % mod;
}
}
struct Hash {
ll h1[MAX_N], h2[MAX_N];
void init() {
h1[0] = h2[0] = 0;
for(int i = 1; i <= slen; i++) {
h1[i] = (h1[i-1] * base1 + s[i]) % mod;
h2[i] = (h2[i-1] * base2 + s[i]) % mod;
}
}
inline ll hashVal(int l, int r) {
ll hsh1 = (h1[r] - h1[l-1] * pw1[r-l+1]) % mod;
if(hsh1 < 0) hsh1 += mod;
ll hsh2 = (h2[r] - h2[l-1] * pw2[r-l+1]) % mod;
if(hsh2 < 0) hsh2 += mod;
return (hsh1 << 32) | hsh2;
}
inline ll hashOne(int l, int r) {
ll hsh1 = (h1[r] - h1[l-1] * pw1[r-l+1]) % mod;
if(hsh1 < 0) hsh1 += mod;
return hsh1;
}
inline ll hashTwo(int l, int r) {
ll hsh2 = (h2[r] - h2[l-1] * pw2[r-l+1]) % mod;
if(hsh2 < 0) hsh2 += mod;
return hsh2;
}
} fw;
/* call pw_calc() for calculating powers less than MAX_N
* slen = strlen(s+1); --> string length
* fw.init() will calculate the double hashes
* fw.hashVal(l,r) will return [l,,r] merged double hash value
* fw.hashOne(l, r) will return [l,,r] base1 hash
* fw.hashTwo(l, r) will return [l,,r] base2 hash
*/
ll zero, one;
char b[100009];
int main()
{
scanf("%s %s", b+1, s+1);
ll blen = strlen(b+1);
slen = strlen(s+1);
for1(i, blen)
b[i] == '0'? ++zero : ++one;
pw_calc();
fw.init();
ll ans = 0;
for1(i, slen) {
ll r0 = i, rest = slen - zero * r0;
if(rest <= 0 || rest % one != 0)
continue;
ll r1 = rest / one;
ll p0 = -1, p1 = -1, ptr = 1, rgt, chk = 1;
for1(j, blen) {
if(b[j] == '0') {
rgt = ptr + r0 - 1;
ll hsh = fw.hashVal(ptr, rgt);
if(p0 == -1) p0 = hsh;
else if(hsh != p0) {
chk = 0;
break;
}
if(hsh == p1) {
chk = 0;
break;
}
}
else {
rgt = ptr + r1 - 1;
ll hsh = fw.hashVal(ptr, rgt);
if(p1 == -1) p1 = hsh;
else if(hsh != p1) {
chk = 0;
break;
}
if(hsh == p0) {
chk = 0;
break;
}
}
ptr = rgt + 1;
}
if(chk) ans++;
}
cout << ans << endl;
return 0;
}