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I've considered your opinion and changed my program. But still I got WA.
[pascal]

program k331;

type
Tpoint=^Tr;
Tr=record
a :integer;
next :Tpoint;
end;

var
v,e,i,j :integer;
a,b,c :integer;
link,dist,tt :array[1..100,1..100] of integer;
so :array[1..10000] of integer;
path,hash :array ...

I have no idea what formular to use to solve this math problem. Please give me a hint. Thank you!

Oh, I see... Thank you!

But the correct answer for this test case is (0,2,3) and (0,1,3) is not correct. (I took a look at the problem index and this problem is single-answered.) How to explain this?

Thank you for your help! I made so foolish mistake I got AC now!

I have a question with the problem: In the sample #1 (0,2,3) and (0,1,3) are both solutions .Why say (0,2,3) is the correct solution and (0,1,3) is the 'image' ? Why cannot (0,2,3) be the 'image' of (0,1,3)?

Here's my code. I can't find the bugs. :( Help me! Thank you!
[pascal]
program k326;

type
Tshi=array[0..50] of extended;

var
shi :array[1..50] of Tshi;
n,i :integer;
a :extended;

procedure Pmul(a,b:Tshi; var c:Tshi);

var
i,j :integer;

begin
fillchar(c,sizeof(c),0);
for i:=0 to n do if a ...

I guess that Dynamic Programming should be used to solve this problem, but I don't know the details. Please help me and thank you very much!

This problem seems so hard. I don't know what algorithm to use. Please help me! Thanks!

Why did my code get WA? Help me, please!

[pascal]
program k331;

type
Tpoint=^Tr;
Tr=record
a :integer;
next :Tpoint;
end;

var
v,e,i :integer;
a,b,c :integer;
hash,link,dist,tt :array[1..100,1..100] of integer;
so :array[1..10000] of integer;
path :array[1..100,1..100] of Tpoint;
ma ...

Thank you!

Will you please show me your program? Thanks a lot!

I have a correct output here, you can compare it with the output which your program makes:
6 8
35 49
204 288
1189 1681
6930 9800
40391 57121
235416 332928
1372105 1940449
7997214 11309768
46611179 65918160

(The output format is not as the problem demands.)

I don't think it is enough by trying y either. I saw someone sloved it in math may -- just create a list of numbers: A(1)=0 , A(2)=1, and then A(n)=6*A(n-1)-A(n-2), and A(3)~A(12) are the first ten y's. This is extramely fast! But working out the formula was too hard!

I cannot find one.Can u help me?