UVa694 The Collatz Sequence

 The Collatz Sequence 

An algorithm given by Lothar Collatz produces sequences of integers, and isdescribed as follows:

Step 1:

Choose an arbitrary positive integer A as the first item inthe 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 initialvalues ofA as large as 10⁹, but some values of A encountered inthe sequence may exceed the size of an integer on many computers. In thisproblem we want to determine the length of the sequence that includes allvalues produced until either the algorithm stops (in step 2), or a valuelarger 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 givingthe initial value ofA (for step 1) and the second giving L, the limitingvalue for terms in the sequence. Neither of these,A or L, is largerthan 2,147,483,647 (the largest value that can be stored in a 32-bit signedinteger). The initial value ofA is always less than L. A line thatcontains two negative integers follows the last case.

Output 

For each input case display the case number (sequentially numbered startingwith 1), a colon, the initial value forA, the limiting value L, and thenumber 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

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Miguel Revilla
2000-08-14

这题就是按照题目中的算法，求出算法的次数，要注意的就是数大于1时进入循环，超出极限的时候跳出。这题有个陷阱就是题中每行输出前有个空格，实际上这空格是不需要的，让我WA了挺多次的。

#include <iostream>#include <cstdio>using namespace std;int main() {long long number;long long limit;int Case = 1;while (cin >> number >> limit) {if (number < 0 && limit < 0)break;int current = 1;printf("Case %d: A = %lld, limit = %lld, number of terms = ",Case,number,limit);Case++;while (number > 1) {if (number % 2)number = number * 3 + 1;elsenumber = number / 2;if (number > limit)break;current++;}cout << current << endl;}return 0;}