Theory:
The idea for solving the problem is: for any fibonacci word F(n) = F(n-1)+F(n-2) = F(n-2)+F(n-3)+F(n-2), so word F(n-2) is both suffix and prefix of word F(n). After observing a little more we can find that F(n-2) will always be the prefix for all the rest of the fibonacci word.Now find the minimum F(n-3) that is longer than the pattern.
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#include <iostream> #include <cstdio> #include <cstdio> #include <cstring> #include <vector> #include <cmath> #include <cstdlib> #define INF 1000000 #define ll long long using namespace std; ll fib[105]; string Fib[105]; vector<int> vb; int *failureFunc; void kmpsearch(string p,string s,int min_index) { int i=1,len=0; int length = p.size(); //int failureFunc[length]; failureFunc=(int *)malloc(sizeof(int)*length); failureFunc[0]=0; while(i<length) { if(p[i]==p[len]) { len+=1; failureFunc[i]=len; i+=1; } else { if(len!=0) { len=failureFunc[len-1]; } else { failureFunc[i]=0; i+=1; } } } int I = s.size(); i=0;len=0; vb.clear(); while(i+len<I) { if(i>fib[min_index-1]) break; if(p[len]==s[i+len]) { if(len==length-1) { vb.push_back(i); } else { len+=1; continue; } } if(len) { i=i+len-failureFunc[len-1]; len=failureFunc[len-1]; } else { i+=1; } } } void findAt(int min_indx,string p) { kmpsearch(p,Fib[min_indx],min_indx); } int main() { int n; int kase=1; string pat; // string prev,prev1,tmp; fib[0]=1;fib[1]=1; Fib[0]="0";Fib[1]="1"; for(int i=2;i<104;i++) fib[i]=min(fib[i-1]+fib[i-2],9223372036854775805LL); int ii; for(ii=1;ii<3 || Fib[ii-3].size()<100000;ii++) { Fib[ii+1]=Fib[ii]; Fib[ii+1].append(Fib[ii-1]); } while(scanf("%d",&n)!=EOF) { cin>>pat; if(pat=="0") { if(n==1) cout<<"Case "<<kase++<<": "<<0<<"\n"; else if(n==0) cout<<"Case "<<kase++<<": "<<1<<"\n"; else cout<<"Case "<<kase++<<": "<<fib[n-2]<<"\n"; continue; } else if(pat=="1") { if(n==1) cout<<"Case "<<kase++<<": "<<1<<"\n"; else if(n==0) cout<<"Case "<<kase++<<": "<<0<<"\n"; else cout<<"Case "<<kase++<<": "<<fib[n-1]<<"\n"; continue; } else { int min_idx=3; while(fib[min_idx-3]<=pat.size()) min_idx+=1; if(min_idx>n) min_idx=n; findAt(min_idx,pat); ll prev=0,suf=0; ll overlap[2]={0,0}; int j,k; int sz=pat.size(); for(int i=0;i<vb.size();i++) { j=vb[i]; k=j+sz-1; if(j<fib[min_idx-2]) { if(k<fib[min_idx-2]) { prev+=1; } else overlap[0]+=1; } else if(j<fib[min_idx-1]) { if(k<fib[min_idx-1]) suf+=1; else overlap[1]+=1; } } int f=1; ll tmp; for(int i=0;i<=n-min_idx+1;i++)//for(int i=min_idx-1;i<=n;i++) { f=1-f; tmp=prev;//initially f(n-2) prev=prev+suf+overlap[f];//initially this f(n-1) suf=tmp; } cout<<"Case "<<kase++<<": "<<prev<<"\n"; } } return 0; } |