1032 The 3n + 1 problem

The 3n + 1 problem

Problem Description

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.

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. 

 

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 opperation overflows a 32-bit integer.

 

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
  
  

 

最初用蛮力算法，效率是N的平方。输入 1 to 1,000,000就开始慢慢算了。

然后就想是不是要大数处理呢，还是又一个规律循环？或者就像之前的卡特兰数，另外有什么高效的算法。

另外没注意区间，sample中误导i小于j，没有多想。（sample input里面我看了几遍，虽然没说必定i<j，但也没看出哪里说i可以小于j。定向思维了还是英语不照啊？哪位童鞋可以留言说下啊？）

由于比较简单，最初没用函数，写一疙瘩里，自己测试没问题，但OJ老是wrong answer。想来可能是哪个变量没重置。但变量太多，晕掉了。直接提取到函数中，就AC了…

函数真是好东东，用起来多多益善啊(～ o ～)~zZ

/*
 *	hdu 1032 The 3n + 1 problem
 *	2011/07/25
 *	artwalk
 */

#include <iostream>

using namespace std;

int MAX(int i, int j);
int count(int k);

int main(int ac, char** av)
{
	int i, j;
	
	while ( cin >> i >> j ) {
		cout << i << " " << j << " ";

		if ( i > j ) {	// 注意 i j 之间
			int temp = i;
			i = j;
			j = temp;
		}
	
		cout << MAX(i, j) << endl;
			
	}

	return 0;
}

int MAX(int i, int j) {
	// i to j 中最大次数
	int maxnum = 0;
	
	for ( ; i <=j; ++i ) {
		int k = i;
		int temp = 0;
	
		temp = count(k);
		if ( maxnum < temp) {
			maxnum = temp;
		}
	}
	return maxnum;
}

int count(int k) {
	// 单个数的次数
	int num = 1;
	while ( k != 1 ) {
		if ( k % 2 != 0 ) {
			k = 3 * k + 1;
		} else {
			k = k >> 1;
		}
		++num;
	}

	return num;
}