首页 > ACM题库 > UVA > UVa OJ 100 – The 3n + 1 problem (3n + 1问题)
2013
12-30

UVa OJ 100 – The 3n + 1 problem (3n + 1问题)

Time limit: 3.000 seconds
时:3.000秒

 

Background
背景

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.
计算机科学中的问题常常用确定性来进行分类,(比如NP,无解,递归)。在这个问题中,你需要分析一个算法的性质,这个算法的分类对于全部可能的输入而言是未知的。

 

The Problem
问题

Consider the following algorithm:
考虑下面的算法:

input n
	print n
	if n = 1 then stop
		if n is odd then n <- 3n + 1
		else n <- n / 2
	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.
给定输入22,接下来会打印出的的数列是: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.)
猜想输入任何整输算法都会最终停止(到打印出1为止)。尽管算法很简单,但这个猜想是否正确仍不得而知。然而经过验证,算法对于0 < n < 1,000,000的所有整除n都成立(事实上还验证过更多的数)。

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.
对于给定的一个输入n,算法能够打印出数字(包括1)的数量是可以确定的。对于给定的n,这个数量称作n的“周期长度”。在上面的例子中,22的周期长度为16。

For any two numbers i and j you are to determine the maximum cycle length over all numbers between i and j.
给定任意的两个数i和j,你要确定在[i, j]的范围内的所有数中,最长的周期长度是多少。

 

The Input
输入

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.
输入一系列的整数对i和j,每一对整除独占一行。所有整除都小于1,000,000且大于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.
你要处理所有的整除对,并确定每对整除所限定范围内(含)的所有整数的最大周期长度。

You can assume that no operation overflows a 32-bit integer.
你可以假设不计算中不会出现32位整数的溢出错误。

 

The Output
输入

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).
对输入的每一对i和j应该输出i、j和[i, j]范围内的最大周期长度。这3个数字之间各由至少1个空格隔开,并在每一行输入之后,将这3个输全部输出在下一行内。输出整数i和j时必须按照他们在输入时的顺序输入,后面跟着输出最大周期长度(在同一行)。

 

Sample Input
输入示例

1 10
100 200
201 210
900 1000

 

Sample Output
输出示例

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

 

Analysis
分析

非常简单的起点题,算法题目中已经给出。注意测试数据中的i, j可能大小顺序相反,但输出时还要按原顺序输出,因此应另设变量存储原先的值,不可将原值丢失。

 

Solution
解答

#include <iostream>
#include <map>
using namespace std;

std::map<int, int> g_Tbl;

int CycleLength(int n) {
	int &cl = g_Tbl[n];
	if (cl == 0) {
		n = (n % 2 == 0) ? (n / 2) : (3 * n + 1);
		cl = 1 + CycleLength(n);
	}
	return cl;
}

int main(void) {
	g_Tbl[1] = 1;
	for (int nBeg, nEnd; cin >> nBeg >> nEnd;) {
		int nOrgBeg = nBeg, nOrgEnd = nEnd, nMaxLen = 1;
		if (nBeg > nEnd) {
			swap(nBeg, nEnd);
		}
		for (int i = nBeg, nLen; i <= nEnd; ++i, nLen = 1) {
			nLen = CycleLength(i);
			if (nLen > nMaxLen) {
				nMaxLen = nLen;
			}
		}
		cout << nOrgBeg << " " << nOrgEnd << " " << nMaxLen << endl;
	}
	return 0;
}

转自:http://www.cnblogs.com/devymex/archive/2010/08/04/1792114.html


  1. 很高兴你会喜欢这个网站。目前还没有一个开发团队,网站是我一个人在维护,都是用的开源系统,也没有太多需要开发的部分,主要是内容整理。非常感谢你的关注。

  2. 换句话说,A[k/2-1]不可能大于两数组合并之后的第k小值,所以我们可以将其抛弃。
    应该是,不可能小于合并后的第K小值吧

  3. 可以参考算法导论中的时间戳。就是结束访问时间,最后结束的顶点肯定是入度为0的顶点,因为DFS要回溯