The 3n + 1 problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22148    Accepted Submission(s): 8276

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

Source

UVA

算法分析：

这道题拿出时间写过好几次，提交都WA了！ 以为自己写的算法真的有问题！后来在刘汝佳写的书上又看到了这道题，于是又敲，

最后终于AC了！

此题有两个坑： 1，读入的n和m， 大小不确定需要判断是否交换。   2，假设说 n>m, 这是要交换n和m， 但是在输出的时候仍要

先输出先读入的大数n, 再输出小的m， 也就是没交换钱的顺序！

就因为这两点，这道简单题的题，我不知道错了多少遍！   并且还要注意一点，就是n*3+1是否会超出int类型！

Accepted的代码：

#include <stdio.h>
#include <string.h>
#include <math.h>

long long cnt;
int max;

void jishu(int dd)
{
    long long ff=dd;
    cnt++;
    while(ff>1)
    {
        if(ff%2==1)
          ff=3*ff+1;
        else
          ff/=2;
        cnt++;
    }

    if(cnt>max)
    {
      max=cnt;
    }
    cnt=0;
}
int main()
{
    int n, m;
    int i, j;
    while(scanf("%d %d", &n, &m)!=EOF)
    {
        cnt=0;
        max=0;
        printf("%d %d ", n, m );
        if(n>m)
        {
            n=n^m; m=m^n; n=n^m;
        }
        for(i=n; i<=m; i++)
        {
            jishu(i);
        }
        printf("%d\n", max );
    }
    return 0;
}