The Collatz Sequence
2013-12-07 17:34
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An algorithm given by Lothar Collatz produces sequences of integers, and is described as follows:
Step 1:Choose an arbitrary positive integer A as the first item in the sequence.
Step 2:If A = 1 then stop. Step 3:If A is even, then replace A by A / 2 and go to step 2.
Step 4:If A is odd, then replace A by 3 * A + 1 and go to step 2.
It has been shown that this algorithm will always stop (in step 2) for initial values of
A as large as 109, but some values of A encountered in the sequence may exceed the size of an integer on many computers. In this problem we want to determine the length of the sequence that includes all values produced until either
the algorithm stops (in step 2), or a value larger than some specified limit would be produced (in step 4).
A (for step 1) and the second giving L, the limiting value for terms in the sequence. Neither of these,
A or L, is larger than 2,147,483,647 (the largest value that can be stored in a 32-bit signed integer). The initial value of
A is always less than L. A line that contains two negative integers follows the last case.
A, the limiting value L, and the number of terms computed.
Step 1:Choose an arbitrary positive integer A as the first item in the sequence.
Step 2:If A = 1 then stop. Step 3:If A is even, then replace A by A / 2 and go to step 2.
Step 4:If A is odd, then replace A by 3 * A + 1 and go to step 2.
It has been shown that this algorithm will always stop (in step 2) for initial values of
A as large as 109, but some values of A encountered in the sequence may exceed the size of an integer on many computers. In this problem we want to determine the length of the sequence that includes all values produced until either
the algorithm stops (in step 2), or a value larger than some specified limit would be produced (in step 4).
Input
The input for this problem consists of multiple test cases. For each case, the input contains a single line with two positive integers, the first giving the initial value ofA (for step 1) and the second giving L, the limiting value for terms in the sequence. Neither of these,
A or L, is larger than 2,147,483,647 (the largest value that can be stored in a 32-bit signed integer). The initial value of
A is always less than L. A line that contains two negative integers follows the last case.
Output
For each input case display the case number (sequentially numbered starting with 1), a colon, the initial value forA, the limiting value L, and the number of terms computed.
Sample Input
3 100 34 100 75 250 27 2147483647 101 304 101 303 -1 -1
Sample Output
Case 1: A = 3, limit = 100, number of terms = 8 Case 2: A = 34, limit = 100, number of terms = 14 Case 3: A = 75, limit = 250, number of terms = 3 Case 4: A = 27, limit = 2147483647, number of terms = 112 Case 5: A = 101, limit = 304, number of terms = 26 Case 6: A = 101, limit = 303, number of terms = 1
#include <stdio.h> int main() { long i,j,k,a,l,terms,x; i = 0; while(scanf("%ld %ld",&a,&l),a >= 0 || l >= 0) { i++; if(a == l) break; terms = 0; x = a; while(x!=1 && x<=l) { if(x%2 == 1) x = 3*x+1; else x = x/2; if(x == 1) terms++; terms++; } printf("Case %ld: A = %ld, limit = %ld, number of terms = %ld\n",i,a,l,terms); } return 0; }
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