2013
12-30

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

Time limit: 3.000 seconds

## 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.

## The Problem问题

Consider the following algorithm:

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.

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.

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).

1 10
100 200
201 210
900 1000

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

## 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;
}

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