对第一个串s建立后缀自动机，用父亲指针求出每个节点代表的子串在s中出现的次数。对于串x，复制一个循环节加在后面。在后缀自动机上跑一遍，如果某个节点匹配长度大于x原来的长度，检查父节点是否也大于，是的话用父节点代替当前节点，把预处理出来的该节点的出现次数累加就是答案。
//Feb 21, 2013 7:32:16 AM bigoceanlhy 235C - Cyclical Quest GNU C++ Accepted 921 ms 240700 KB
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN = 1000000+5, MAXM = 2000000+5;
const int MAX_NODE = 2000000+5, MAX_CHD = 26;
int nv, chd[MAX_NODE][MAX_CHD], ml[MAX_NODE], fa[MAX_NODE], id[1<<8];
int fail[MAXN];
int N, cnt[MAX_NODE], r[MAX_NODE];
char s[MAXM];
void Add(int u, int _ml, int _fa, int v = -1)
{
ml[u] = _ml; fa[u] = _fa;
if (v == -1)
memset(chd[u], -1, sizeof(chd[u]));
else
memcpy(chd[u], chd[v], sizeof(chd[v]));
}
void Construct(char *str)
{
nv = 1; Add(0, 0, -1); cnt[0] = 1;
int cur = 0;
for (int i = 0; str[i]; i++)
{
int c = id[str[i]], p = cur;
cur = nv++; Add(cur, i+1, -1); cnt[cur] = 1;
cnt[cur] = 1;
for (; p != -1 && chd[p][c] == -1; p = fa[p])
chd[p][c] = cur;
if (p == -1)
fa[cur] = 0;
else
{
int q = chd[p][c];
if (ml[q] == ml[p]+1)
fa[cur] = q;
else
{
int r = nv++; Add(r, ml[q], fa[q], q); cnt[r] = 0;
ml[r] = ml[p]+1; fa[q] = fa[cur] = r;
for (; p != -1 && chd[p][c] == q; p = fa[p])
chd[p][c] = r;
}
}
}
}
bool cmp(const int &a, const int &b)
{
return ml[a] > ml[b];
}
void get_fail(char *pat)
{
fail[0] = -1;
for (int i = 1, j = -1; pat[i]; i++)
{
while (j != -1 && pat[j+1] != pat[i])
j = fail[j];
if (pat[j+1] == pat[i])
j++;
fail[i] = j;
}
}
int main()
{
for (int i = 0; i < MAX_CHD; i++)
id['a'+i] = i;
scanf("%s", s);
Construct(s);
for (int i = 0; i < nv; i++)
r[i] = i;
sort(r, r+nv, cmp);
for (int i = 0; i < nv; i++)
cnt[fa[r[i]]] += cnt[r[i]];
scanf("%d", &N);
while (N--)
{
scanf("%s", s);
get_fail(s);
int len = strlen(s), cir = len-fail[len-1]-1;
if (len%cir)
cir = len;
memcpy(s+len, s, sizeof(char)*cir);
s[len+cir-1] = 0;
int l = 0, u = 0, ans = 0;
for (int i = 0; s[i]; i++)
{
int c = id[s[i]];
if (chd[u][c] != -1)
l++, u = chd[u][c];
else
{
while (u != -1 && chd[u][c] == -1)
u = fa[u];
if (u != -1)
l = ml[u]+1, u = chd[u][c];
else
l = 0, u = 0;
}
if (l >= len)
{
if (ml[fa[u]] >= len)
u = fa[u];
ans += cnt[u];
}
}
printf("%d\n", ans);
}
return 0;
}