The 3n + 1 problem
|
Background
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.
The Problem
Consider
the following algorithm:
1. input n
2.
print n
3.
if n = 1 then STOP
4.
if n is odd then
5.
else
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.
The 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 operation overflows a 32-bit integer.
The 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<stdio.h> void funtion(int n1,int n2); int main() { int n1,n2; while(scanf("%d %d",&n1,&n2)==2) { funtion(n1,n2); } return 0; } void funtion(int n1,int n2) { int i,m1,m2,j,k,l,m=0,n=1,c=0; m1=n1; m2=n2; if(m1>m2) { i=m1; m1=m2; m2=i; } for(i=m1;i<=m2;i++) { k=i;n=1; while(k!=1) { if(k%2==0) { k=k/2; } else { k=k*3+1; } n++; } if(n>m) { m=n; } } printf("%d %d %d\n",n1,n2,m); }
No comments:
Post a Comment