dfs算法1.2(3n+1)

Consider the following algorithm to generate a sequence of numbers. Start with an integer n. If n is even, divide by 2. If n is odd, multiply by 3 and add 1. Repeat this process with the new value of n, terminating when n = 1. For example, the following sequence of numbers will be generated for n = 22: 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1 It is conjectured (but not yet proven) that this algorithm will terminate at n = 1 for every integer n. Still, the conjecture holds for all integers up to at least 1, 000, 000. For an input n, the cycle-length of n is the number of numbers generated up to and including the 1. In the example above, the cycle length of 22 is 16. Given any two numbers i and j, you are to determine the maximum cycle length over all numbers between i and j, including both endpoints.

就是给一个数把它通过变化变成1需要几步

#include<bits/stdc++.h>
using namespace std;
int n,m;
int k;
int dfs(int x)
{
    ++k;
    if(x==1)return k;
    if(x%2==0)
    {
        dfs(x/2);
    }
    else if(x%2==1)
    {
        dfs(3*x+1);
    }

}
int main()
{

    while(cin>>n>>m)
    {//没说n,m哪个大！
        int oo=0;
        if(n>m){
            swap(n,m);
            oo=1;
        }
        int mm=-1e9;
        for(int i=n; i<=m; i++)
        {
            k=0;
            dfs(i);
            mm=max(mm,k);
        }
        if(oo==0)cout<<n<<" "<<m<<" "<<mm<<endl;
        else cout<<m<<" "<<n<<" "<<mm<<endl;
    }
}