UVA Problem 100 The 3n + 1 problem Solution

Problem Statement 
Problem Type : Adhoc/Dynamic Programming
Problem Difficulty : Easy
Solution Available : Dynamic Programming

Problems in Computer Science are often classified as belonging to a certain
class of problems (e.g., NP, Unsolvable, Recursive). In this problem you will be
analyzing a property of an algorithm whose classification is not known for all
possible inputs.

The Problem

Consider the following algorithm:

1. input n

2. print n

3. if n = 1 then STOP

4. if n is odd then n<-3n+1

5. else n<-n/2

6. GOTO 2

Given the input 22, the following sequence of numbers will be printed 22 11 34
17 52 26 13 40 20 10 5 16 8 4 2 1

It is conjectured that the algorithm above will terminate (when a 1 is printed)
for any integral input value. Despite the simplicity of the algorithm, it is
unknown whether this conjecture is true. It has been verified, however, for all
integers n such that 0 < n < 1,000,000 (and, in fact, for many more numbers than
this.)

Given an input n, it is possible to determine the number of numbers printed
(including the 1). For a given n this is called the cycle-length of n. In the
example above, the cycle length of 22 is 16.

For any two numbers i and j you are to determine the maximum cycle length over
all numbers between i and j.

The Input

The input will consist of a series of pairs of integers i and j, one pair of
integers per line. All integers will be less than 1,000,000 and greater than 0.

You should process all pairs of integers and for each pair determine the maximum
cycle length over all integers between and including i and j.

You can assume that no operation overflows a 32-bit integer.

The Output

For each pair of input integers i and j you should output i, j, and the maximum
cycle length for integers between and including i and j. These three numbers
should be separated by at least one space with all three numbers on one line and
with one line of output for each line of input. The integers i and j must appear
in the output in the same order in which they appeared in the input and should
be followed by the maximum cycle length (on the same line).

Sample Input

1 10
100 200
201 210
900 1000

Sample Output

1 10 20
100 200 125
201 210 89
900 1000 174



Dynamic Programming Approach

#include <stdio.h>
//Time : 0.032 s 
#define sz 1000000

/*#define LL __int64*/
#define UL unsigned long
UL tb[sz+1];
UL calc(UL n){
 if(n < sz){
  if( tb[n] )
   return tb[n];
  if(n & 1) 
   return tb[n] = 1 + calc( n + (n << 1) + 1);
  return tb[n] = 1 + calc( n >> 1);
  
 }
 if(n & 1)
  return 1 + calc( n + (n << 1) + 1); 
 return 1 + calc( n >> 1);
}
int main(){
 UL max,a,b,i;

 max = 0;

 tb[1] = 1;
 while(scanf("%lu %lu",&a,&b) == 2){
  printf("%lu %lu ",a,b);
  if( a > b){
   a ^= b ^= a ^= b;
  }
  max = 0;
  for( i = a; i <= b; i++){
   if( max < calc(i)){
    max = tb[i];
   }
  }
  printf("%lu\n",max);

 }

 return 0;
}