Ultraviolet (a) 100-The 3N + 1 problem

This question is very simple, but its background (the craz question, or angular conjecture) is not that simple. No one has yet provided a reliable proof.

Analysis:

You only need to enumerate and simulate each positive integer in the given range. However, this will definitely time out. We can use a memory-like idea to save the small number of results with an array, and then directly reference it when calculating a large number. However, not all data results can be saved. The experiment shows that the possible number in the calculation process is much larger than the upper limit of the given range. (See the attached table). Therefore, when calculating a large number (I took> = 10,000,000), we need to combine the direct enumeration with the memory search. I actually tried two methods. The first method is to enumerate the numbers greater than or equal to 10,000,000 in 0.068 seconds. The second method is to enumerate and memorize the numbers greater than or equal to 10,000,000, it takes 0.052 s, which is faster.

Note:

1. the number to be counted includes itself and the last generated one.

2. There is no limit on the relative size of the two numbers I and J in the question. That is to say, I> J may occur and need to be judged.

Code:

Method 1:

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<ctime>#include<memory.h>using namespace std;typedef long long LL;int a[10000005];int enumerate(LL x){    int counts=1;    while(x!=1)    {        if(x%2==0) x/=2;        else x=x*3+1;        counts++;    }    return counts;}int search(LL x){    if(x>=10000000) return enumerate(x);    if(a[x]!=-1) return a[x];    if(x%2==0)        return a[x]=search(x/2)+1;    else        return a[x]=search(3*x+1)+1;}int main(){    int m,n,i,ans;    memset(a,-1,sizeof(a));    a[1]=1;    while(cin>>m>>n)    {        cout<<m<<‘ ‘<<n;        ans=-1;        if(m>=n) swap(m,n);        for(i=m;i<=n;i++)            ans=max(ans,search(i));        cout<<‘ ‘<<ans<<endl;    }    return 0;}

 

Method 2:

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<ctime>#include<memory.h>using namespace std;typedef long long LL;int a[10000005];int enumerate(LL x){    int counts=1;    while(x>=10000000)    {        counts++;        x = x%2==0 ? x/2 : x*3+1;    }    if(a[x]!=-1) return a[x];    if(x%2==0)        return a[x]=enumerate(x/2)+counts;    else        return a[x]=enumerate(3*x+1)+counts;}int main(){    int m,n,i,ans;    memset(a,-1,sizeof(a));    a[1]=1;    while(cin>>m>>n)    {        cout<<m<<‘ ‘<<n;        ans=-1;        if(m>=n) swap(m,n);        for(i=m;i<=n;i++)            ans=max(ans,enumerate(i));        cout<<‘ ‘<<ans<<endl;    }    return 0;}

 

Appendix:

Range	Maximum value that may be encountered during Calculation	Answer
[1, 10]	52	20
[1,100]	9232	119
[1, 1000]	250504	179
[1, 10000]	27114424	262
[1, 100000]	1570824736	351
[1, 1000000]	56991483520	525

 

 

 

 

 

Ultraviolet (a) 100-The 3N + 1 problem