这是一道很坑爹的题，一定注意输入的两个数的大小，并且不能简单的交换，因为在最后的输出的时候还需要将原来的数按照原来的顺序和大小，这就是为什么还得开辟两个值得原因
Description
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 before the 1 is printed. 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
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.
Output
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
#include<iostream>
#include <cmath>
using namespace std;
int main ()
{
int m,n,w,j,i,max;
while(cin>>m>>n)
{
int x=m,y=n;
if(m>n)
{
int X=m;
m=n;
n=x;
}
max=-1;
for(i=m;i<=n;i++)
{
j=i;
w=0;
while(j>1)
{
if(j%2!=0)
j=3*j+1;
else
j=j/2;
w++;
//cout<<'*';
}
w++;
if(max==-1)
max=w;
else
if(max<w)
max=w;
}
cout<<x<<' '<<y<<' '<<max<<endl;
}
return 0;
}