Now you have a string consists of uppercase letters, two integers
AA
A and
BB
B. We call a substring wonderful substring when the times it appears in that string is between
AA
A and
BB
B (
A≤times≤BA \le times \le B
A≤times≤B). Can you calculate the number of wonderful substrings in that string?
Input
Input has multiple test cases.
For each line, there is a string
SS
S, two integers
AA
A and
BB
B.
∑length(S)≤2×106\sum length(S) \le 2 \times 10^6
∑length(S)≤2×10
6
,
1≤A≤B≤length(S)1 \le A \le B \le length(S)
1≤A≤B≤length(S)
Output
For each test case, print the number of the wonderful substrings in a line.
樣例輸入
複製
AAA 2 3
ABAB 2 2
樣例輸出
複製
2
3

題意：

給你一個串，x,y，叫你求出現次數在x,y之間的不同子串個數

題解：

字尾陣列我並不會求至少出現k次的子串QAQ，看別人的題解發現原來是字尾自動機的模板題。。。

#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e5+1000;
char s[maxn];
int len,k,n,m;
char temp[5];
struct SAM{
	int last,cnt,nxt[maxn*2][26],fa[maxn*2],l[maxn*2],num[maxn*2];
	int ans;
	void init(){
		last = cnt=1;
		memset(nxt[1],0,sizeof nxt[1]);
		fa[1]=l[1]=num[1]=0;
		ans=0;
	}
	int inline newnode(){
		cnt++;
		memset(nxt[cnt],0,sizeof nxt[cnt]);
		fa[cnt]=l[cnt]=num[cnt]=0;
		return cnt;
	}
	void add(int c){
		int p = last;
        int np = newnode();
        last = np;
        l[np] =l[p]+1;
        while (p&&!nxt[p][c]){
            nxt[p][c] = np;
            p = fa[p];
        }
        if (!p){
            fa[np] =1;
        }else{
            int q = nxt[p][c];
            if (l[q]==l[p]+1){
                fa[np] =q;
            }else{
                int nq = newnode();
                memcpy(nxt[nq],nxt[q],sizeof nxt[q]);
                fa[nq] =fa[q];
                num[nq] = num[q];
                l[nq] = l[p]+1;
                fa[np] =fa[q] =nq;
                while (nxt[p][c]==q){
                    nxt[p][c]=nq;
                    p = fa[p];
                }
            }
        }
		int temp = last;
		while (temp){
			if (num[temp]>=k){
				break;
			}
			num[temp]++;
			if (num[temp]==k){
				ans+=l[temp]-l[fa[temp]];
			}
			temp = fa[temp];
		}
	}
}sam;
int main(){
    while(~scanf("%s",s))
    {

		len = strlen(s);
		sam.init();
        scanf("%d%d",&k,&m);
		for (int i=0;i<len;i++){
			sam.add(s[i]-'A');
		}
		int ans=sam.ans;
		k=m+1;
		sam.init();
		for (int i=0;i<len;i++){
			sam.add(s[i]-'A');
		}
		printf("%d\n",ans-sam.ans);
    }
	return 0;
}