# HDU 6194 string string string （SAM）

## Description

Uncle Mao is a wonderful ACMER. One day he met an easy problem, but Uncle Mao was so lazy that he left the problem to you. I hope you can give him a solution.

Given a string s, we define a substring that happens exactly k times as an important string, and you need to find out how many substrings which are important strings.

## Input

The first line contains an integer T (T≤100) implying the number of test cases.

For each test case, there are two lines:

the first line contains an integer k (k≥1) which is described above;

the second line contain a string s (length(s)≤10^5).

## Output

For each test case, print the number of the important substrings in a line.

## Sample Input

2
2
abcabc
3
abcabcabcabc


## Sample Output

6
9


HDU 4641 一样的思路。

## AC 代码

#include <bits/stdc++.h>
using namespace std;
typedef __int64 LL;
const int maxn = 2e5 + 10;
LL ans;
int K;

struct SAM
{
int next[maxn][26];
int pre[maxn], step[maxn];
int last, id;
int num1[maxn], num2[maxn];
void init()
{
last = id = 0;
memset(next[0], -1, sizeof(next[0]));
pre[0] = -1;
step[0] = 0;
}
void insert(int c)
{
int p = last, np = ++id;
step[np] = step[p] + 1;
memset(next[np], -1, sizeof(next[np]));
num1[np] = num2[np] = 0;

while (p != -1 && next[p][c] == -1)
next[p][c] = np, p = pre[p];
if (p == -1)
pre[np] = 0;
else
{
int q = next[p][c];
if (step[q] != step[p] + 1)
{
int nq = ++id;
memcpy(next[nq], next[q], sizeof(next[q]));
num1[nq] = num1[q];
num2[nq] = num2[q];
step[nq] = step[p] + 1;
pre[nq] = pre[q];
pre[np] = pre[q] = nq;
while (p != -1 && next[p][c] == q)
next[p][c] = nq, p = pre[p];
}
else
pre[np] = q;
}
last = np;

/* diy */
bool f1 = false, f2 = false;
while (np != -1 && (!f1 || !f2))
{
if (!f1 && num1[np] < K)
{
num1[np]++;
if (num1[np] == K)
{
ans += step[np] - step[pre[np]];
f1 = true;
}
}
if (!f2 && num2[np] < K + 1)
{
num2[np]++;
if (num2[np] == K + 1)
{
ans -= step[np] - step[pre[np]];
f2 = true;
}
}
np = pre[np];
}
/* end diy */
}
} sam;

char str[maxn];
LL get_ans()
{
sam.init();
ans = 0;
int len = strlen(str);
for (int i = 0; i < len; i++)
sam.insert(str[i] - 'a');
return ans;
}

template <class T>
inline void scan_d(T &ret)
{
char c;
ret = 0;
while ((c = getchar()) < '0' || c > '9')
;
while (c >= '0' && c <= '9')
{
ret = ret * 10 + (c - '0'), c = getchar();
}
}
template <class T>
inline void print_d(T x)
{
if (x > 9)
{
print_d(x / 10);
}
putchar(x % 10 + '0');
}

int main()
{
int T;
scan_d(T);
while (T--)
{
scan_d(K);
gets(str);
print_d(get_ans());
putchar('\n');
}
return 0;
}