Now you have a string consists of uppercase letters, two integers AA and BB. We call a substring wonderful substring when the times it appears in that string is between AA and BB (A le times le BA≤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, two integers AA and BB.
sum length(S) le 2 imes 10^6∑length(S)≤2×106,
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.
样例输入
AAA 2 3 ABAB 2 2
样例输出
2 3
题目来源
题解：SAM模板题
参考代码：
1 //H 求子串出现次数在k1=<num<=k2; 2 #include <bits/stdc++.h> 3 using namespace std; 4 const int MAXN = 4e5+10; 5 char ss[200005]; 6 const int LetterSize = 26; 7 8 int tot, last,ch[MAXN][LetterSize],fa[MAXN],len[MAXN]; 9 int sum[MAXN],tp[MAXN],cnt[MAXN]; 10 11 void init() 12 { 13 last = tot = 1; 14 len[1] = 0; 15 memset(ch,0,sizeof ch); 16 memset(fa,0,sizeof fa); 17 memset(cnt,0,sizeof cnt); 18 } 19 20 void add( int x) 21 { 22 int p = last, np = last = ++tot; 23 len[np] = len[p] + 1, cnt[last] = 1; 24 while( p && !ch[p][x]) ch[p][x] = np, p = fa[p]; 25 if(p == 0) fa[np] = 1; 26 else 27 { 28 int q = ch[p][x]; 29 if( len[q] == len[p] + 1) 30 fa[np] = q; 31 else 32 { 33 int nq = ++tot; 34 memcpy( ch[nq], ch[q], sizeof ch[q]); 35 len[nq] = len[p] + 1, fa[nq] = fa[q], fa[q] = fa[np] = nq; 36 while( p && ch[p][x] == q) ch[p][x] = nq, p = fa[p]; 37 } 38 } 39 } 40 41 void toposort() 42 { 43 for(int i = 1; i <= len[last]; i++) sum[i] = 0; 44 for(int i = 1; i <= tot; i++) sum[len[i]]++; 45 for(int i = 1; i <= len[last]; i++) sum[i] += sum[i-1]; 46 for(int i = 1; i <= tot; i++) tp[sum[len[i]]--] = i; 47 } 48 49 50 int main() 51 { 52 53 int k1,k2; 54 while(scanf("%s",ss)!=EOF) 55 { 56 init(); 57 scanf("%d%d",&k1,&k2); 58 long long ans=0; 59 for(int i=0,len=strlen(ss);i<len;i++) add(ss[i]-'A'); 60 toposort(); 61 for(int i=tot;i;i--) 62 { 63 int p=tp[i],fp=fa[p]; 64 cnt[fp]+=cnt[p]; 65 if(cnt[p]>=k1 && cnt[p]<=k2) ans+=len[p]-len[fp]; 66 } 67 printf("%lld ",ans); 68 } 69 70 return 0; 71 }