Establishing suffix tree (i.e., a string in reverse order of insertion parent tree), can then be found in accordance with dfs arrangement which satisfies playing lexicographical ordering in ascending order, you can maintain the subtree every moment string length and sequence, and then directly dfs the binary sequence determination node, then interior node in messing to obtain the answer.
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define N 200005 4 #define ll long long 5 int V,last,ans,a[N<<1],sz[N<<1],mi[N<<1],id[N<<1],len[N<<1],fa[N<<1],ch[N<<1][31]; 6 ll k,mod,l[N<<1],sum[N<<1]; 7 char s[N]; 8 void add(int c){ 9 int p=last,np=last=++V; 10 len[np]=len[p]+1; 11 sz[np]=1; 12 for(;(p)&&(!ch[p][c]);p=fa[p])ch[p][c]=V; 13 if (!p)fa[np]=1; 14 else{ 15 int q=ch[p][c]; 16 if (len[q]==len[p]+1)fa[np]=q; 17 else{ 18 int nq=++V; 19 len[nq]=len[p]+1; 20 memcpy(ch[nq],ch[q],sizeof(ch[q])); 21 fa[nq]=fa[q]; 22 fa[np]=fa[q]=nq; 23 for(;(p)&&(ch[p][c]==q);p=fa[p])ch[p][c]=nq; 24 } 25 } 26 } 27 void dfs(int k){ 28 a[++a[0]]=k; 29 for(int i=0;i<26;i++) 30 if (ch[k][i])dfs(ch[k][i]); 31 } 32 int main(){ 33 int t; 34 scanf("%s%d",s,&t); 35 V=last=1; 36 int n=strlen(s); 37 for(int i=n-1;i>=0;i--)add(s[i]-'a'); 38 for(int i=1;i<=V;i++)a[len[i]]++; 39 for(int i=1;i<=n;i++)a[i]+=a[i-1]; 40 for(int i=1;i<=V;i++)id[a[len[i]]--]=i; 41 for(int i=1,j=n;i<=V;i++) 42 if (sz[i])mi[i]=--j; 43 memset(a,0,sizeof(a)); 44 for(int i=V;i;i--){ 45 sz[fa[id[i]]]+=sz[id[i]]; 46 mi[fa[id[i]]]=mi[id[i]]; 47 } 48 for(int i=V;i;i--)l[i]=(len[fa[i]]+len[i]+1LL)*(len[i]-len[fa[i]])/2*sz[i]; 49 memset(ch,0,sizeof(ch)); 50 for(int i=1;i<=V;i++)ch[fa[i]][s[mi[i]+len[fa[i]]]-'a']=i; 51 dfs(1); 52 for(int i=1;i<=V;i++)sum[i]=sum[i-1]+l[a[i]]; 53 while (t--){ 54 scanf("%lld%lld",&k,&mod); 55 k=k*ans%mod+1; 56 int p=lower_bound(sum+1,sum+V+1,k)-sum; 57 k-=sum[p-1]; 58 p=a[p]; 59 int x=len[fa[p]]+1,y=len[p]; 60 while (x<y){ 61 int mid=(x+y>>1); 62 if ((mid+len[fa[p]]+1LL)*(mid-len[fa[p]])/2*sz[p]>=k)y=mid; 63 else x=mid+1; 64 } 65 k-=(x+len[fa[p]])*(x-1LL-len[fa[p]])/2*sz[p]; 66 printf("%c\n",s[(k-1)%x+mi[p]]); 67 ans+=s[(k-1)%x+mi[p]]; 68 } 69 }