uva 694



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 of A 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 of A (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 of A 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 for A, 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

题目大意：
实际上就是一个3n+1的问题，规则：
1.选择一个整数A；
2.如果A=1停止运算；
3.如果A是偶数，那么A=A/2转到第2步；
4.如果A是奇数，那么A = 3A+1转到第2步。
注意：如果m=1结束的时候，计数器要加一（count++），因为计数进行了第二步！

代码：
#include <stdio.h>
#include <stdlib.h>
int main()
{
    long long int m,n,p;
    int sum,term;
    sum=0;
    while(~scanf("%lld%lld",&m,&n)&&m>=0&&n>=0)
    {
        term=0;
        p=m;
        while(m>1&&m<=n)
        {
            term++;
            if(m&1)
                m=3*m+1;
            else
                m>>=1;
                if(m>n)
            term--;
        }
        printf("Case %d: A = %lld, limit = %lld, number of terms = %d\n",++sum,p,n,++term);
    }
    return 0;
}

要点1：选好适当的数据类型，此处应用long long int ,不要用unsigned long long int ,因为无符号，无法判断什么时候停止
要点2：此处灵活运用了位运算，位运算效率比较高