题意：给定一个字符串，问你它有多少个子串恰好出现 k 次。
析：后缀数组，先把height 数组处理出来，然后每次取 k 个进行分析，假设取的是 i ~ i+k-1，那么就有重复的，一个是 i-1 ~ i+k-1，另一个是 i ~ i+k，但是这样就删多了，再加上 i - 1 ~ i+k，这样就OK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(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 = 1e5 + 50;
const int mod = 1000;
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 Array{
int s[maxn], sa[maxn], t[maxn], t2[maxn];
int h[maxn], r[maxn], c[maxn];
int n;
int dp[maxn][20]; void init(){ n = 0; memset(sa, 0, sizeof sa); }
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-1]] == y[sa[i]] && 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;
}
} void rmq_init(){
for(int i = 1; i <= n; ++i) dp[i][0] = h[i];
for(int j = 1; (1<<j) <= n; ++j)
for(int i = 1; i + (1<<j) <= n; ++i)
dp[i][j] = min(dp[i][j-1], dp[i+(1<<j-1)][j-1]);
} int query(int L, int R){
if(L == R) return L;
++L;
int k = int(log(R-L+1) / log(2.0));
return min(dp[L][k], dp[R-(1<<k)+1][k]);
}
};
char s[maxn];
Array arr; int main(){
int T; cin >> T;
while(T--){
int k;
arr.init();
scanf("%d", &k);
scanf("%s", s);
for(int i = 0; s[i]; ++i)
arr.s[arr.n++] = s[i] - 'a' + 1;
arr.s[arr.n++] = 0;
arr.build_sa(28);
arr.getHight();
arr.rmq_init();
int ans = 0;
for(int i = 1; i < arr.n; ++i){
int l = i+k-1 < arr.n ? arr.query(i, i+k-1) : 0;
int x = i-1 > 0 && i+k-1 < arr.n ? arr.query(i-1, i+k-1) :0;
int y = i+k < arr.n ? arr.query(i, i+k) : 0;
int z = i-1 > 0 && i+k < arr.n ? arr.query(i-1, i+k) : 0;
ans += l - x - y + z;
}
printf("%d\n", ans);
}
return 0;
}