The question : stay S Find out from the string X The number of different substrings that appear in the string ? among 1 <= |S| < $10^5$

Official explanation : Process the... In the suffix array sa[] Array and height[] Array . Without considering the inclusion of characters X Under the circumstances , The number of different substrings is

If characters are required X, Just record the distance sa[i] Recent characters X The location of ( use nxt[sa[i]] Express ) that will do , Number

understand : The suffix array height[i] Namely sa[i] And sa[i-1] Of LCP, Solve all the different substrings in the suffix array ( I've only written before SAM Dealing with all the different substrings ..) It's better to understand , There must be substrings in the string x when , It needs to be done first kmp Find all the matching positions , At the end of the process i When there is a suffix , take max It means that it must contain X, And different substrings ;

#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
#define rep0(i,l,r) for(int i = (l);i < (r);i++)
#define rep1(i,l,r) for(int i = (l);i <= (r);i++)
#define rep_0(i,r,l) for(int i = (r);i > (l);i--)
#define rep_1(i,r,l) for(int i = (r);i >= (l);i--)
#define MS0(a) memset(a,0,sizeof(a))
#define MS1(a) memset(a,-1,sizeof(a))
#define MSi(a) memset(a,0x3f,sizeof(a))
#define inf 0x3f3f3f3f
#define pb push_back
#define MK make_pair
#define A first
#define B second
#define eps 1e-8
#define zero(x) (((x)>0?(x):-(x))<eps)
#define bitnum(a) __builtin_popcount(a)
#define lowbit(x) (x&(-x))
#define clear0 (0xFFFFFFFE)
#define mod 1000000007
#define K(x) ((x)*(x))
typedef pair<int,int> PII;
typedef long long ll;
typedef unsigned long long ull;
template<typename T>
void read1(T &m)
T x = ,f = ;char ch = getchar();
while(ch <'' || ch >''){ if(ch == '-') f = -;ch=getchar(); }
while(ch >= '' && ch <= ''){ x = x* + ch - '';ch = getchar(); }
m = x*f;
template<typename T>
void read2(T &a,T &b){read1(a);read1(b);}
template<typename T>
void read3(T &a,T &b,T &c){read1(a);read1(b);read1(c);}
template<typename T>
void out(T a)
if(a>) out(a/);
inline ll gcd(ll a,ll b){ return b == ? a: gcd(b,a%b); }
template<class T1, class T2> inline void gmax(T1& a, T2 b){ if(a < b) a = b;}
template<class T1, class T2> inline void gmin(T1& a, T2 b){ if(a > b) a = b;}
const int maxn = ;
int sa[maxn], t[maxn],t2[maxn],c[maxn],w[maxn];
int cmp(int *r,int a,int b,int l)
return r[a] == r[b] && r[a+l] == r[b+l];
void build_sa(char *r,int n,int m)
int i, j, p, *x = t, *y = t2;
for(i = ; i < m;i++) c[i] = ;
for(i = ; i < n;i++) c[x[i] = r[i]]++;
for(i = ; i < m;i++) c[i] += c[i-];
for(i = n-; i >= ; i--) sa[--c[x[i]]] = i; for(j = , p = ; p < n; j <<= , m = p){
for(p = , i = n - j; i < n;i++) y[p++] = i;
for(i = ; i < n; i++) if(sa[i] >= j) y[p++] = sa[i]-j;
for(i = ; i < n; i++) w[i] = x[y[i]];
for(i = ; i < m; i++) c[i] = ;
for(i = ; i < n; i++) c[w[i]]++;
for(i = ; i < m; i++) c[i] += c[i-];
for(i = n-; i >= ; i--) sa[--c[w[i]]] = y[i];
for(p = , x[sa[]] = , i = ; i < n; i++)
x[sa[i]] = cmp(y, sa[i-], sa[i], j)? p-: p++;
if(p >= n) break;
m = p;
char S[maxn], X[maxn];
int height[maxn], rk[maxn];
void getHeight(int n)
for(int i = ; i<= n; i++) rk[sa[i]] = i;
for(int i = , j, k = ; i < n; height[rk[i++]] = k)
for(k? k--:, j = sa[rk[i] - ]; S[i+k] == S[j+k]; k++);
} int f[maxn];
void getfail(char *p)
f[] = f[] = ;
int n = strlen(p);
for(int i = ;i < n;i++){
int j = f[i];
if(j && p[i] != p[j]) j = f[j];
f[i+] = (p[i] == p[j] ?j+:);// i+1 It's going to recurs to the next n position
vector<int> vec;
void Find(char *T, char *p)
ll j = ,n = strlen(T),m = strlen(p);
for(int i = ;i < n;i++){
while(j && T[i] != p[j]) j = f[j];
if(T[i] == p[j]) j++;
if(j == m){
j = ;
i -= m-;
sort(vec.begin(), vec.end());
} int main()
int T, kase = ;
scanf("%s%s", X, S);
int len = strlen(S), m = strlen(X);
S[len] = '#';S[len+] = ;
getHeight(len); getfail(X);
ll ans = ;
if(sa[i]+m- > len) continue;
int nxt = lower_bound(vec.begin(), vec.end(), sa[i]+m-) - vec.begin();
ans += len - max(vec[nxt],sa[i] + height[i]);
printf("Case #%d: %lld\n", kase++, ans);
return ;

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