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。
View Code
1 program pku3318(input,output);
2 var
3 x,y,z : array[0..501,0..501] of int64;
4 left,right,answer1 : array[0..501] of int64;
5 n : longint;
6 rand : array[0..501] of longint;
7 procedure init;
8 var
9 i,j : longint;
10 begin
11 readln(n);
12 for i:=1 to n do
13 begin
14 for j:=1 to n do
15 read(x[i,j]);
16 readln;
17 end;
18 for i:=1 to n do
19 begin
20 for j:=1 to n do
21 read(y[i,j]);
22 readln;
23 end;
24 for i:=1 to n do
25 begin
26 for j:=1 to n do
27 read(z[i,j]);
28 readln;
29 end;
30 end; { init }
31 function check():boolean;
32 var
33 i : longint;
34 begin
35 for i:=1 to n do
36 if answer1[i]<>right[i] then
37 exit(false);
38 exit(true);
39 end; { check }
40 procedure main;
41 var
42 i,j : longint;
43 begin
44 randomize;
45 for i:=1 to n do
46 rand[i]:=trunc(random(20000));
47 fillchar(left,sizeof(left),0);
48 for i:=1 to n do
49 for j:=1 to n do
50 inc(left[i],y[i,j]*rand[j]);
51 fillchar(answer1,sizeof(answer1),0);
52 for i:=1 to n do
53 for j:=1 to n do
54 inc(answer1[i],left[j]*x[i,j]);
55 fillchar(right,sizeof(right),0);
56 for i:=1 to n do
57 for j:=1 to n do
58 inc(right[i],rand[j]*z[i,j]);
59 if not check() then
60 begin
61 writeln('NO');
62 exit;
63 end;
64 writeln('YES');
65 end; { main }
66 begin
67 while not eof do
68 begin
69 init;
70 main;
71 end;
72 end.