HDU - 1032 The 3n + 1 problem

题目：

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

题目简述：

对于3n+1问题，22的结果是：22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1 ，一共16个数

所以22的cycle length是16

现在，输入x,y，要求输出x和y之间所有数的cycle length的最大值。

我的第一个程序是，先记忆化搜索求出1到1000000的cycle length

然后用max型线段树，来减少查询时间。

代码：

#include<iostream>
using namespace std;

int num[1000000];
int max[4000000];

int init(long long k)
{
	if (k<1000000 && num[k])return num[k];
	int r;
	if (k % 2)r = init(k * 3 + 1) + 1;
	else r = init(k / 2) + 1;
	if (k < 1000000)num[k] = r;
	return r;
}

void build(int key, int low, int high)
{
	if (low == high)
	{
		max[key] = num[low];
		return;
	}
	int mid = (low + high) / 2;
	build(key * 2, low, mid);
	build(key * 2 + 1, mid + 1, high);
	max[key] = (max[key * 2] > max[key * 2 + 1]) ? max[key * 2] : max[key * 2 + 1];
}

int query(int key, int low, int high, int x, int y)
{
	if (x > y)return query(key, low, high, y, x);
	if (low == x && high == y)return max[key];
	int mid = (low + high) / 2;
	if (mid < x)return query(key * 2 + 1, mid + 1, high, x, y);
	if (mid >= y)return query(key * 2, low, mid, x, y);
	int a = query(key * 2, low, mid, x, mid);
	int b = query(key * 2 + 1, mid + 1, high, mid + 1, y);
	return  (a>b) ? a : b;
}

int main()
{
	for (int i = 1; i < 1000000; i++)num[i] = 1 - (i>1);
	for (int i = 1; i < 1000000; i++)init(i);
	build(1, 1, 1000000);
	int x, y;
	while (scanf("%d%d", &x, &y) != -1)printf("%d %d %d\n", x, y, query(1, 1, 1000000, x, y));
	return 0;
}

最开始超时了，因为没有判断x和y谁大谁小，这个代码在query里面加了一行，就AC了。

然而我发现有个人的代码特别快，而且特别短，我就看了一下。

居然是！！不用记忆华搜索也不用线段树，根本连数组都没有出现！

于是我也试了一下这个方法。

代码：

#include<iostream>
using namespace std;

int l(int k)
{
	if (k == 1)return 1;
	if (k % 2)return l(k * 3 + 1) + 1;
	return l(k / 2) + 1;
}

int maxs(int x, int y)
{
	if (x > y)return maxs(y, x);
	int maxx = 0;
	for (int i = x; i <= y; i++)if (maxx < l(i))maxx = l(i);
	return maxx;
}

int main()
{	
	int x, y;
	while (scanf("%d%d", &x, &y) != -1)printf("%d %d %d\n", x, y, maxs(x,y));
	return 0;
}

果然很短，而且还是0ms。。。

关键还是要看给的数据。

在上面的init函数里面，我专门做了判断，判断参数k有没有超过1000000

实际上，我的k是long long的。

这是因为，运行init函数的时候就发现了，在计算1到1000000的cycle length的时候，会出现很大的数。

到底有多大我没有去追究，反正已经超过int了，所以我计算1到1000000的cycle length才会很慢。

题目已经说明了不会超过int，也就是说输入是有限制的。