100-The 3N + 1 problem

A simple question can be directly simulated according to the question requirements. Note that the intermediate results can be saved to increase the running time by an order of magnitude.

There should be a faster method, but for the scale given in this question, thisAlgorithmThe results can be obtained quickly.

 

Returns the volume I index.

Returns the total index.

 

// 100_3n_plus_1.cpp: defines the entry point for the console application. <br/> // </P> <p> # include <math. h> </P> <p> # ifndef online_judge <br/> # include <fstream> <br/> STD: ifstream CIN ("in.txt "); <br/> STD: ofstream cout ("out.txt "); <br/> # else <br/> # include <iostream> <br/> # endif </P> <p> using namespace STD; </P> <p> # define Max 1000000 <br/> int * seqlen; </P> <p> int getseqlen (int x) <br/> {<br/> If (seqlen [x]> 0) <br/>{< br/> return seqlen [X]; <br/>}</P> <p> int nseq = 1; </P> <p> int y = x; <br/> while (X! = 1) <br/>{< br/> If (x <= max & seqlen [x]> 0) <br/>{< br/> nseq + = (seqlen [x]-1); <br/> break; <br/>}</P> <p> If (X % 2 = 0) <br/>{< br/> X/= 2; <br/>}< br/> else <br/> {<br/> X = 3 * x + 1; <br/>}</P> <p> ++ nseq; <br/>}</P> <p> seqlen [y] = nseq; </P> <p> return nseq; <br/>}</P> <p> int getmax (int I, Int J) <br/> {<br/> int base = I <= J? I: J; <br/> int bound = I >= J? I: J; </P> <p> int Nmax = 0; <br/> int nlen = 0; </P> <p> for (INT x = base; x <= bound; ++ X) <br/>{< br/> nlen = getseqlen (x); <br/> If (nlen> Nmax) <br/>{< br/> Nmax = nlen; <br/>}</P> <p> return Nmax; <br/>}</P> <p> int main () <br/> {<br/> seqlen = new int [Max + 1]; </P> <p> memset (seqlen, 0, sizeof (INT) * (MAX + 1); </P> <p> int I, J; </P> <p> while (CIN> I> J) <br/> {<br/> cout <I <''<j <'' <getmax (I, j) <'/N '; <br/>}</P> <p>

 

100-The 3N + 1 problem