题意：给定两个串A，B，问你A有多少不同的子串，并且不包含B。
析：首先A有多少个不同的子串，可以用后缀数组来解决，也就是 n - sa[i] - h[i] + 1。但是要是不包含B，可以先预处理A和B，把B在A中的位置都记录下来，然后在找不同子串的时候，走到匹配的位置就停止，如果再向后找就肯定包含B了。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
//#define all 1,n,1
#define FOR(i,x,n) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 5e4 + 20;
const int maxm = 100 + 10;
const ULL mod = 10007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
}
struct Suffix_array{
int s[maxn], sa[maxn], t[maxn];
int t2[maxn], h[maxn], r[maxn], c[maxn];
int n;
void init(){ n = 0; ms(sa, 0); }
void build_sa(int m){
int *x = t, *y = t2;
for(int i = 0; i < m; ++i) c[i] = 0;
for(int i = 0; i < n; ++i) ++c[x[i] = s[i]];
for(int i = 1; i < m; ++i) c[i] += c[i-1];
for(int i = n-1; i >= 0; --i) sa[--c[x[i]]] = i;
for(int k = 1; k <= n; k <<= 1){
int p = 0;
for(int i = n-k; i < n; ++i) y[p++] = i;
for(int i = 0; i < n; ++i) if(sa[i] >= k) y[p++] = sa[i] - k;
for(int i = 0; i < m; ++i) c[i] = 0;
for(int i = 0; i < n; ++i) ++c[x[y[i]]];
for(int i = 1; i < m; ++i) c[i] += c[i-1];
for(int i = n-1; i >= 0; --i) sa[--c[x[y[i]]]] = y[i];
swap(x, y);
p = 1; x[sa[0]] = 0;
for(int i = 1; i < n; ++i)
x[sa[i]] = y[sa[i]] == y[sa[i-1]] && y[sa[i-1]+k] == y[sa[i]+k] ? p-1 : p++;
if(p >= n) break;
m = p;
}
}
void getHight(){
int k = 0;
for(int i = 0; i < n; ++i) r[sa[i]] = i;
for(int i = 0; i < n; ++i){
if(k) --k;
int j = sa[r[i]-1];
while(s[i+k] == s[j+k]) ++k;
h[r[i]] = k;
}
}
};
Suffix_array arr;
char s[maxn], t[maxn];
int f[maxn];
void getFail(){
f[0] = f[1] = 0;
for(int i = 1; i < m; ++i){
int j = f[i];
while(j && s[i] != s[j]) j = f[j];
f[i+1] = s[i] == s[j] ? j + 1 : 0;
}
}
vector<int> v;
int main(){
int T; cin >> T;
for(int kase = 1; kase <= T; ++kase){
scanf("%s", t);
scanf("%s", s);
arr.init(); v.cl;
m = strlen(s);
getFail();
int j = 0;
for(int i = 0; t[i]; ++i){
arr.s[arr.n++] = t[i] - 'a' + 1;
while(j && s[j] != t[i]) j = f[j];
if(s[j] == t[i]) ++j;
if(j == m) v.pb(i), j = f[j];
}
arr.s[arr.n++] = 0;
v.pb(INF);
arr.build_sa(28);
arr.getHight();
int ans = 0; j = 0;
for(int i = 0; i < arr.n; ++i){
int x = *lower_bound(v.begin(), v.end(), arr.sa[i] + m - 1);
++x;
ans += max(0, min(x, arr.n) - arr.sa[i] - arr.h[i] - 1);
}
printf("Case %d: %d
", kase, ans);
}
return 0;
}