The 3n + 1 problem
</pre>The 3n + 1 problem</div><div class="plm" style="text-align: center;font-size:14px;"><table align="center"><tbody><tr><td><strong>Time Limit:</strong> 1000MS</td><td width="10px"> </td><td><strong>Memory Limit:</strong> 10000K</td></tr><tr><td><strong>Total Submissions:</strong> 53927</td><td width="10px"> </td><td><strong>Accepted:</strong> 17142</td></tr></tbody></table></div><p class="pst" style="font-size: 18pt; font-weight: bold; color: blue;">Description</p><div class="ptx" lang="en-US" style="font-family: 'Times New Roman', Times, serif;font-size:14px;">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: <pre> 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
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 10,000 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 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
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
#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int a[100010];
int main()
{
ios::sync_with_stdio(false);
long long ans;
int m,sum,n;
for(int i=1; i<100010; i++)
{
ans=i;
sum=1;
while(ans!=1)
{
if(ans%2==1)
{
ans=ans*3+1;
}
else
{
ans>>=1;
}
sum++;
}
a[i]=sum;
}
while(cin>>n>>m)
{
sum=0;
int flag=1;
if(n>m)
{
flag=0;
swap(n,m); //n,m的关系；
}
for(int i=n;i<=m;i++)
{
if(sum<a[i])
sum=a[i];
}
if(flag)
cout<<n<<" "<<m<<" ";
else
cout<<m<<" "<<n<<" ";
cout<<sum<<endl;
}
return 0;
}
思考：n到1000000?暴力还行么？虽然能过，但时间长；
#include <iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> using namespace std; int a[100010]; int main() { ios::sync_with_stdio(false); long long ans; int m,sum,n; for(int i=1; i<100010; i++) { ans=i; sum=1; while(ans!=1) { if(ans%2==1) { ans=ans*3+1; } else { ans>>=1; } sum++; if(ans<i) { sum+=a[ans]-1; //减少时间； break; } } a[i]=sum; } while(cin>>n>>m) { sum=0; int flag=1; if(n>m) { flag=0; swap(n,m); } for(int i=n;i<=m;i++) { if(sum<a[i]) sum=a[i]; } if(flag) cout<<n<<" "<<m<<" "; else cout<<m<<" "<<n<<" "; cout<<sum<<endl; } return 0; }