pku3318 Matrix Multiplication

有三个矩阵A，B，C，问A*B是否C，(n^3)的算法会超时。

构造一个n*1的矩阵，由

A*B=C

A*B*X=C*X

A*(B*X)=C*X

那么在(n^2)的时间内就能判定一次。

program pku3318(input,output);
var
   x,y,z          : array[0..501,0..501] of int64;
   left,right,answer1 : array[0..501] of int64;
   n              : longint;
   rand              : array[0..501] of longint;
procedure init;
var
   i,j : longint;
begin
   readln(n);
   for i:=1 to n do
   begin
      for j:=1 to n do
     read(x[i,j]);
      readln;
   end;
   for i:=1 to n do
   begin
      for j:=1 to n do
     read(y[i,j]);
      readln;
   end;
   for i:=1 to n do
   begin
      for j:=1 to n do
     read(z[i,j]);
      readln;
   end;
end; { init }
function check():boolean;
var
   i : longint;
begin
   for i:=1 to n do
      if answer1[i]<>right[i] then
     exit(false);
   exit(true);
end; { check }
procedure main;
var
   i,j     : longint;
begin
   randomize;
   for i:=1 to n do
      rand[i]:=trunc(random(20000));
   fillchar(left,sizeof(left),0);
   for i:=1 to n do
      for j:=1 to n do
     inc(left[i],y[i,j]*rand[j]);
   fillchar(answer1,sizeof(answer1),0);
   for i:=1 to n do
      for j:=1 to n do
     inc(answer1[i],left[j]*x[i,j]);
   fillchar(right,sizeof(right),0);
   for i:=1 to n do
      for j:=1 to n do
     inc(right[i],rand[j]*z[i,j]);
   if not check() then
   begin
      writeln('NO');
      exit;
   end;
   writeln('YES');
end; { main }
begin
   while not eof do
   begin
      init;
      main;
   end;
end.