UVA 694 The Collatz Sequence 题解

UVA 694 The Collatz Sequence 题解

  The Collatz Sequence
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).
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 of 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.
Output
For each input case display the case number (sequentially numbered starting with 1), a colon, the initial value for A, 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



本题其实简单，前面A,L用的都是int类型然后超时了，接着我用__int64，服务器不识别。然后又用long long 类型才秒过

题意：给你一个初始A递归：如果A==1则停止，如果A是偶数则A = A / 2 如果A是奇数 则 A = 3 * A + 1 

这里还有L的用处，L是限定A的大小，如果上述步骤中A超过了L也直接停止 输出 步骤数

UVA694.cpp
#include <iostream>
#include <cstdio>
using namespace std;

int main()
{
    long long A,L;
    int C = 0;
    long long a,b;
    while(scanf("%lld%lld",&A,&L),A>=0&&L>=0)
    {
        b = 0;
        a = A;
        while(b++,a != 1 && b < L)
        {
            if (a%2)
            {
                a = 3 * a + 1;
                if (a > L)
                {
                    break;
                }
            }
            else
            {
                a /= 2;
            }
        }
        printf("Case %d: A = %lld, limit = %lld, number of terms = %lld\n",++C,A,L,b);
    }
    return 0;
}