题解报告：hdu 1032 The 3n + 1 problem（克拉兹问题）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1032

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.

译文：计算机科学中的问题通常被归类为属于某类问题（例如，NP，Unsolvable，Recursive）。在这个问题中，您将分析一个算法的属性，该算法的分类对于所有可能的输入都是未知的。

考虑以下算法：

1.输入n

2.打印n

3.如果n = 1则停止

4.如果n为奇数，则n <-3n + 1

5.否则n < - n / 2

6.转到2 

给定输入22，将打印下列数字序列22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1

据推测，上述算法将终止（当打印1时）任何积分输入值。尽管算法很简单，但是这个猜想是否属实却是未知数。然而，已经证实，对于所有整数n使得0 <n <1,000,000（事实上，对于比这更多的数字）

给定输入n，可以确定打印的数字的数量（包括1）。对于给定的n，这称为n的周期长度。在上面的例子中，22的周期长度是16. 

对于任何两个数字i和j，您应确定i和j之间所有数字的最大周期长度。

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.

译文：输入将包含一系列的整数i和j对，每行一对整数。所有整数将小于1,000,000且大于0. 
您应该处理所有整数对，并为每对确定i和j之间（包括i和j之间的所有整数）的最大周期长度。
您可以假定没有操作溢出32位整数。

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). 

译文：对于每对输入整数i和j，您应该输出i，j以及i和j之间（包括i和j）之间的整数的最大周期长度。这三个数字应该由至少一个空格分隔，所有三个数字在一行上并且每行输入使用一行输出。整数i和j必须以它们出现在输入中的相同顺序出现在输出中，并且后面应跟着最大循环长度（在同一行上）。

Sample Input

1 10

100 200

201 210

900 1000

Sample Output

1 10 20

100 200 125

201 210 89

900 1000 174

解题思路：这道题的意思就是输入一个区间i,j找出这里面的最长周期；周期是这样子：当这个数（不为1）是奇数时就变为3*n+1,为偶数时就变为n/2,并且用sum来计数周期,直到n为1就跳出。（水题！！！注意杭电oj出题的一些坑）

AC代码：

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int a,b,sum,t,maxn;
    bool f=false;//标记是否交换了
    while(cin>>a>>b){
        if(a>b){
            swap(a,b);//题目的陷井
            f=true;
        }
        maxn=0;
        for(int i=a;i<=b;i++){
            sum=1,t=i;
            while(t!=1){
                if(t%2)t=3*t+1;
                else t/=2;
                sum++;
            }
            maxn=max(sum,maxn);
        }
        if(f){
            swap(a,b);//题目有说保持原来的数据输出,所以还得交换过来
            f=false;//同时置f为false
        }
        cout<<a<<' '<<b<<' '<<maxn<<endl;
    }
    return 0;
}