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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[1..100,1..10...

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 <>0 then for...

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 :pointer; procedu...

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?