The Collatz Sequence

  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

是不是又觉得自己的英语水平瘸腿了....大概的意思是说，输入一个数将会经历有四个步骤，这个数要不变成一结束，要不超出最后一个数结束，最后输出一共经历了几步结束的。understand？开始去做吧，唯一一点要注意的是，要用long long，（主要怕数据溢出）。其他的没什么好错的了，别忘了超出lim不算一步，等于一算一步。

#include<stdio.h>int main(){long long n,l,i,count=0,p=0;p=0;while(scanf("%lld%lld",&n,&l)!=EOF){p++;i=n;if(n==-1&&l==-1)return 0;count=0;while(1){if(n==1){count++;break;}else if(n>l)break;else if(!(n%2)){n=n/2;count++;}else if(n%2){n=3*n+1;count++;}}printf("Case %lld: A = %lld, limit = %lld, number of terms = %lld\n",p,i,l,count);}return 0;}