Matrix multiplication Fast Power Codevs 1732 Fibonacci sequence 2

1732 Fibonacci Series 2

time limit: 1 sspace limit: 128000 KBtitle level: Diamonds Diamond SolvingView Run ResultsTitle Description Description

In the "1250 Fibonacci series", we find the value of the nth Fibonacci series. But in 1250, n<=109. Now, your task is still to find the value of the nth Fibonacci sequence, but note that n is an integer and 1 <= n <= 100000000000000

Enter a description Input Description

Enter multiple sets of data, one row for each group of data, and an integer n (1 <= n <= 100000000000000)

Output description Output Description

Outputs several rows. Output per line (corresponding input) n Fibonacci number (considering the number will be very large, mod 1000000007)

Sample input Sample Input

3
4
5

Sample output Sample Output

2
3
5

Data range and Tips Data Size & Hint

1 <= N <= 100000000000000

Although the data is large, but the matrix fast power can still be solved, and fast, as long as the attention to prevent longlong multiplication overflow can

1 #defineN 32 #definell Long Long3#include <iostream>4 using namespacestd;5#include <cstdio>6#include <cstring>7 structjz{8     intCal,line;9     Long LongJz[n][n];Ten }; One Long Longq=1000000007; A intRead () - { -     Chars; the     intans=0, ff=1; -s=GetChar (); -      while(s<'0'|| S>'9') -     { +         if(s=='-') ff=-1; -s=GetChar (); +     } A      while('0'<=s&&s<='9') at     { -ans=ans*Ten+s-'0'; -s=GetChar (); -     } -     returnans*ff; - } in Long LongQuick_mod (ll A,ll b)/*slow multiplication prevents overflow*/ - { toa%=q;b%=Q; +ll ans=0; -      while(b) the     { *         if(b&1) $         {Panax Notoginsengans+=A; -ans%=q;// the         } +b>>=1; Aa<<=1; thea%=Q; +     } -     returnans; $ } $ Jz martax (Jz x,jz y) - { - Jz ans; theAns.line=X.line; -Ans.cal=y.cal;Wuyi      for(intI=1; i<=ans.line;++i) the        for(intj=1; j<=ans.cal;++j) -       { Wuans.jz[i][j]=0; -            for(intk=1; k<=x.cal;++k) AboutAns.jz[i][j]= (Ans.jz[i][j]+quick_mod (x.jz[i][k],y.jz[k][j]))%Q; $       } -     returnans; - } - Long LongFast_martax (Long LongN) A { +     if(n==1|| n==2)return 1; then-=2; - Jz ans,a; $A.cal=a.line=2; thea.jz[1][1]=0; a.jz[1][2]=1; thea.jz[2][1]=1; a.jz[2][2]=1; theAns.line=2; ans.cal=1; theans.jz[1][1]=1; ans.jz[2][1]=1; -      while(n) in     { the         if(n&1) the         { Aboutans=Martax (A,ans); the         } then>>=1; theA=Martax (a,a); +     } -     returnans.jz[2][1]%Q; the }Bayi intMain () the { the     Long LongN; -      while(SCANF ("%lld", &n) = =1) -     { theprintf"%lld\n", Fast_martax (n)); the     } the     return 0; the}

Matrix multiplication Fast Power Codevs 1732 Fibonacci sequence 2