Fibonacci sequence 纪中 1240 矩阵乘法

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这里写图片描述

分析

这就是裸的矩阵乘法啊！！！
数列f[n]=f[n-1]+f[n-2],f[1]=f[2]=1的前n项和s[n]的快速求法
0 1 0
1 1 1
0 0 1

代码

const  a1:array[1..3,1..3] of int64=((0,1,0),                                (1,1,1),                                (0,0,1));  f:array[1..3] of int64=(1,1,1);type  arr=array[1..3,1..3] of int64;var  a:arr;  b:array[1..10] of int64;  i,j,k:longint;  n,n1:int64;  m:longint;  ans:longint;function cheng(x,x1:arr):arr;var  i,j,k:longint;begin  fillchar(cheng,sizeof(cheng),0);  for i:=1 to 3 do    for j:=1 to 3 do      for k:=1 to 3 do        cheng[i,j]:=(cheng[i,j]+x[i,k]*x1[k,j]) mod 10000;end;function seach(n:int64):arr;var  i,j,k:longint;  an:arr;begin  fillchar(an,sizeof(an),0);  if n<>1    then      if n mod 2=0        then begin               an:=seach(n div 2);               seach:=cheng(an,an);             end        else begin               an:=seach(n div 2);               an:=cheng(an,an);               seach:=cheng(an,a1);             end    else      seach:=a1;end;begin  readln(m);  for i:=1 to m do begin    read(n,n1);    n:=n-2; n1:=n1-1;    ans:=0;    fillchar(b,sizeof(b),0);    if n>0 then begin      a:=seach(n);      for j:=1 to 3 do        for k:=1 to 3 do          b[j]:=(b[j]+f[k]*a[k,j]) mod 10000;      ans:=b[3] mod 10000;    end    else if n<>-1 then ans:=1;    n:=n1;    fillchar(b,sizeof(b),0);    if n>0 then begin      a:=seach(n);      for j:=1 to 3 do        for k:=1 to 3 do          b[j]:=(b[j]+f[k]*a[k,j]) mod 10000;      ans:=b[3] mod 10000-ans;    end    else if n<>-1 then ans:=1-ans;    if ans<0 then ans:=ans+10000;    writeln(ans);  end;end.

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