Description
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.
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.
Output
For each pair of input integers i and j,
output i, j in the same order in which they appeared in the input
and then the maximum cycle length for integers between and
including i and j. These three numbers should be separated by one
space, with all three numbers on one line and with one line of
output for each line of input.
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>
int
f(
int
n);
int
l;
int
f(
int
n)
- {
if
(n==1)
return
l;
if
(n%2==0)
{
n/=2;
l++;
f(n);
}
else
{
n=3*n+1;
l++;
f(n);
}
- }
int
main()
- {
int
i,a,b,n,s,t,p=0;
while
(
scanf
(
"%d %d"
,&a,&b)!=EOF)
{
if
(a>b)
{
t=a;
a=b;
b=t;
p=1;
}
s=0;
n=0;
for
(i=a;i<=b;i++)
{
l=1;
if
(s
{
s=l;
}
}
if
(p==0)
printf
(
"%d %d %d\n"
,a,b,s);
else
printf
(
"%d %d %d\n"
,b,a,s);
}
return
0;
- }