UVA 694 - The Collatz Sequence

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最新推荐文章于 2024-12-28 17:21:35 发布

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本文介绍了Collatz算法，一种生成整数序列的过程，并详细解释了如何计算包含所有产生的值直到算法停止或超过指定限制的序列长度。通过输入初始值和限制值，输出序列的长度。案例展示了算法的应用。

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

#define RUN
#ifdef RUN

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <vector>
#include <list>
#include <cctype> 
#include <algorithm>
#include <utility>
#include <math.h>

using namespace std;

#define MAXN 105

long long A, limit;
int cnt;
int casenum = 1;

void play(int casenum){
	cnt = 1;

	long long aA = A;

	while(true){
		if(aA == 1){
			printf("Case %d: A = %lld, limit = %lld, number of terms = %d\n", casenum, A, limit, cnt);
			return;
		}

		if(aA%2 == 0){
			aA = aA/2;
		}
		else{
			aA = 3*aA + 1;
		}

		if(aA > limit){
			printf("Case %d: A = %lld, limit = %lld, number of terms = %d\n", casenum, A, limit, cnt);
			return;
		}

		++cnt;
	}

	
}


int main(){

#ifndef ONLINE_JUDGE
	freopen("694.in", "r", stdin);
	freopen("694.out", "w", stdout); 
#endif


	while(scanf("%lld%lld", &A, &limit)==2 && A!=-1 && limit!=-1){
		play(casenum);
		++casenum;
	}


}


#endif