Title:
Given a string s[0..n-1] of length n, but the successor of i is no longer i+1, but (i*i+1)%n, find all "substrings of length n" ”, who is the largest lexicographically
n<=150000
analyze:
If it is a general string, then the suffix array can be directly obtained, but now the successor relationship has changed
In the process of multiplying the suffix array, we only care about the successor of the next 2^k of a certain position, so we can first multiply and preprocess the nx[i][j] of each position to represent the next 2^ of position i who is the successor of j
Time complexity O(nlogn)
1 #include<bits/stdc++.h> 2 using namespace std; 3 const int maxn=2e5; 4 char s[maxn+50]; 5 int sa[maxn+50],rk[maxn+50]; 6 int t[maxn+50],t2[maxn+50],c[maxn+50]; 7 int nx[maxn+5][19]; 8 int len,k; 9 queue<int> q[maxn+5]; 10 void getsa(int m)//m表示最大字符的编码 11 { 12 memset(t,-1,sizeof(t)); 13 memset(t2,-1,sizeof(t2)); 14 int *x=t,*y=t2; 15 for(int i=0;i<m;++i) c[i]=0; 16 for(int i=0;i<len;++i) c[x[i]=s[i]]++; 17 for(int i=1;i<m;++i) c[i]+=c[i-1]; 18 for(int i=len-1;i>=0;--i) sa[--c[x[i]]]=i; 19 for(int j=0,k=1;k<=len;k<<=1,++j) 20 { 21 /*int p=0; 22 for(int i=len-k;i<len;++i) y[p++]=i; 23 for(int i=0;i<len;++i) if(sa[i]>=k) y[p++]=sa[i]-k;*/ 24 25 int p=0; 26 for(int i=0;i<len;++i) q[nx[i][j]].push(i); 27 for(int i=0;i<len;++i) 28 while(!q[sa[i]].empty()) 29 { 30 y[p++]=q[sa[i]].front(); 31 q[sa[i]].pop(); 32 } 33 34 for(int i=0;i<m;++i) c[i]=0; 35 for(int i=0;i<len;++i) c[x[y[i]]]++; 36 for(int i=0;i<m;++i) c[i]+=c[i-1]; 37 for(int i=len-1;i>=0;--i) sa[--c[x[y[i]]]]=y[i]; 38 swap(x,y); 39 p=1,x[sa[0]]=0; 40 for(int i=1;i<len;++i) 41 if(y[sa[i-1]]==y[sa[i]]&&y[nx[sa[i-1]][j]]==y[nx[sa[i]][j]]) x[sa[i]]=p-1;else x[sa[i]]=p++; 42 if(p>=len) break; 43 m=p; 44 } 45 } 46 int main() 47 { 48 int T; 49 scanf("%d",&T); 50 for(int cas=1;cas<=T;++cas) 51 { 52 printf("Case #%d: ",cas); 53 scanf("%d",&len); 54 scanf("%s",s); 55 for(int i=0;i<len;++i) nx[i][0]=(1LL*i*i+1)%len; 56 for(int j=1;j<=18;++j) 57 for(int i=0;i<len;++i) nx[i][j]=nx[nx[i][j-1]][j-1]; 58 getsa('9'+1); 59 int pos=sa[len-1]; 60 for(int i=1;i<=len;++i,pos=nx[pos][0]) printf("%c",s[pos]); 61 printf("\n"); 62 } 63 64 // for(int i=0;i<n;++i) printf("%d ",sa[i]);printf("\n"); 65 // for(int i=0;i<n;++i) printf("%d ",rk[i]);printf("\n"); 66 // for(int i=0;i<n;++i) printf("%d ",height[i]);printf("\n"); 67 return 0; 68 69 }