The suffix array has not been put in the title Blog. So dug up a write to enrich the next blog by the way to stay as a Board.

A string s in which the contents of different substrings have sigma{n-sa[i]+1-h[i]} (oh H[here) is known Height[])

So l=1,r= the above Sigma dichotomy is the number of substrings in the dictionary sequence.

Then ask S in the K-large substring w: because h[i] is related to i-1, so we have to subtract from N Downto 1,k-=n-sa[i]+1-h[i] to K.

Apparently sweep the substring dictionary order is decremented so you can get the K large substring W

And then greedy n downto 1 if you encounter a substring larger than w, it is not feasible to verify the current dichotomy of this k-large

Judging than w big or small use hash two cent to find LCP can

1#include <bits/stdc++.h>2 #defineN 2000053 #defineLL Long Long4 using namespacestd;5 ConstLL mo=1000000007;6 intf[ -],a[n],rank[n],sa[n],h[n],g1[n],g2[n],next[n],n,m,ans,k,t,l,r,mid,x,y,z;7 Chars[n]; LL f[n],w[n];8 intOhintKintTintPintQ) {9 intL=0, R=min (t-k,q-p) +1, j;Ten while(l<R) { onej=l+r+1>>1; a(f[k+j-1]-f[k-1]*w[j]%mo+mo)%mo== (f[p+j-1]-f[p-1]*w[j]%mo+mo)%mo? -l=j:r=j-1; - } the if(k+l>t)return 0; - if(p+l>q)return 1; - returna[k+l]>a[p+l]; - } + intJudintU) { - intp,q,k,t; + for(inti=n;i;--I) a if(n-sa[i]+1-h[i]<u) u-=n-sa[i]+1-h[i]; at Else{p=sa[i];q=n-u+1; break;} -t=n; k=1; - for(intI=n;i;) - if(oh (i,t,p,q)) { - if(i==t)return 0; -t=i; ++k; in}Else--i; - if(k>m)return 0;return 1; to } + intmain () { -scanf"%d", &m); scanf"%s", s+1); N=strlen (s+1); thew[0]=1; * for(intI=1; i<=n;++I) { $a[i]=s[i]-'a'+1;Panax Notoginsengf[a[i]]=1; -f[i]= (f[i-1]* in+a[i])%mo; thew[i]=w[i-1]* in%mo; + } a for(intI=2; i<= -; ++i) f[i]+=f[i-1]; the for(intI=1; I<=n;++i) rank[i]=f[a[i]]; t=f[ -]; + for(intm=1; m<n;m<<=1){ - for(intI=1; i<=n;++I) { $next[i]=g1[rank[i+m]]; $g1[rank[i+m]]=i; - } - for(inti=t;i>=0;--I) { the for(intj=g1[i];j;j=k) { -k=next[j]; next[j]=g2[rank[j]]; g2[rank[j]]=j;Wuyi } theg1[i]=0; - } wuz=0; - for(intI=1; i<=t;++I) { abouty=-1; $ for(intj=g2[i];j;j=k) { -k=next[j]; next[j]=0; - if(y!=rank[j+m]) y=rank[j+m],++z; -h[j]=z; a } +g2[i]=0; the } -t=z; $ for(intI=1; I<=n;++i) rank[i]=h[i]; the } the for(intI=1; I<=n;++i) sa[rank[i]]=i,h[i]=0; the for(intI=1; i<=n;++I) the if(rank[i]!=1){ -k=h[rank[i-1]];if(k)--k; in while(a[i+k]==a[sa[rank[i]-1]+k]) + +k; theh[rank[i]]=k; r+=n-i+1-k; the } aboutL=1; ++r; the while(l<R) { thek=l+r+1>>1; theJud (k)? l=k:r=k-1; + } - for(inti=n;i;--I) the if(n-sa[i]+1-h[i]<l) l-=n-sa[i]+1-h[i];Bayi Else{k=sa[i];t=n-l+1; break;} the for(intI=k;i<=t;++i) printf ("%c", a[i]+'a'-1); the return 0; -}

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Suffix array hash for LCP Bzoj 4310: Flea