poj3233

这道题其实算是把快速幂的思想用在多项式之中

A+A^2+A^3+…+A^n=(A+A^1…+A^[n/2])+A^[n/2](A+A+A^1…+A^[n/2])+n mod 2*A^n

然后就是打码的问题了

 var ans,a,b,c,f:array[..,..] of longint;
     d:array[..] of longint;
     i,j,n,m,p:longint;

 procedure mul;
   var i,j,k:longint;
   begin
     for i:= to n do
       for j:= to n do
       begin
         c[i,j]:=;
         for k:= to n do
           c[i,j]:=(c[i,j]+a[i,k]*b[k,j] mod p) mod p;
       end;
   end;

 procedure add;
   var i,j:longint;
   begin
     for i:= to n do
       for j:= to n do
         ans[i,j]:=(ans[i,j]+c[i,j]) mod p;
   end;

 procedure quick(x:longint);
   var i,j:longint;
   begin
     j:=;
     while x<> do
     begin
       inc(j);
       d[j]:=x mod ;
       x:=x shr ;
     end;
     fillchar(c,sizeof(c),);
     for i:= to n do
       c[i,i]:=;
     for i:=j downto  do
     begin
       a:=c;
       b:=c;
       mul;
       if d[i]= then
       begin
         a:=c;
         b:=f;
         mul;
       end;
     end;
   end;

 procedure work(x:longint);
   begin
     if x= then ans:=f
     else begin
       work(x shr );
       quick(x shr );
       a:=c;
       b:=ans;
       mul;
       add;
       if x mod = then
       begin
         quick(x);
         add;
       end;
     end;
   end;

 begin
   readln(n,m,p);
   for i:= to n do
     for j:= to n do
       read(f[i,j]);
   work(m);
   for i:= to n do
   begin
     for j:= to n do
     begin
       write(ans[i,j]);
       if j<>n then write(' ');
     end;
     writeln;
   end;
 end.