杭电ACM OJ 1032 The 3n + 1 problem 数据很弱，但是也需要思考一个方向

The 3n + 1 problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 41877    Accepted Submission(s): 15197


Problem Description

Problems in Computer Science are often classified as belonging to a certain class of problems (e.g., NP, Unsolvable, Recursive). In this problem you will be analyzing a property of an algorithm whose classification is not known for all possible inputs.

Consider the following algorithm: 

    1.      input n

    2.      print n

    3.      if n = 1 then STOP

    4.           if n is odd then n <- 3n + 1

    5.           else n <- n / 2

    6.      GOTO 2

Given the input 22, the following sequence of numbers will be printed 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1 

It is conjectured that the algorithm above will terminate (when a 1 is printed) for any integral input value. Despite the simplicity of the algorithm, it is unknown whether this conjecture is true. It has been verified, however, for all integers n such that
0 < n < 1,000,000 (and, in fact, for many more numbers than this.) 

Given an input n, it is possible to determine the number of numbers printed (including the 1). For a given n this is called the cycle-length of n. In the example above, the cycle length of 22 is 16. 

For any two numbers i and j you are to determine the maximum cycle length over all numbers between i and j. 

 

Input

The input will consist of a series of pairs of integers i and j, one pair of integers per line. All integers will be less than 1,000,000 and greater than 0. 

You should process all pairs of integers and for each pair determine the maximum cycle length over all integers between and including i and j. 

You can assume that no opperation overflows a 32-bit integer.

 

Output

For each pair of input integers i and j you should output i, j, and the maximum cycle length for integers between and including i and j. These three numbers should be separated by at least one space with all three numbers on one line and with one line of output
for each line of input. The integers i and j must appear in the output in the same order in which they appeared in the input and should be followed by the maximum cycle length (on the same line). 

 

Sample Input

1 10
100 200
201 210
900 1000

 

Sample Output

1 10 20
100 200 125
201 210 89
900 1000 174

 
翻译：给出一个数，如果是奇数，就把它乘3+1。如果是偶数，就除以2。直到除到1为止。（是肯定可以除到1的）
在除的这个过程中肯定有一个序列长度吧？
所以比如给你1 10.就是让你求1 2 3.。。10每个数都放进去，算算看哪个序列长度最长，输出来。
做法：数据很弱，可以硬做。但是也有个方向，从1开始把各种情况列出来，会发现一个循环，这一列的元素都会重复，所以才会能涵盖到所有数。

代码：

public class TheThreeNplusOneProblem1032 {
void calculate() {
// TODO: 2017/12/3 虽然数据弱，但是想了下，从1开始把各种情况列出来，会发现一个循环，这一列的元素都会重复，所以才会能涵盖到所有数

//answer = 174
int start = 900;
int end = 1000;

int max = Integer.MIN_VALUE;

for (int i = start; i <= end; i ++) {

int times = 0;

int t = i;
while (t != 1) {
times ++;

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

max = Math.max(max, times);
}

max ++;

System.out.print(max + " ");

}

public static void main(String[] args) throws Exception {
TheThreeNplusOneProblem1032 t = new TheThreeNplusOneProblem1032();
t.calculate();
}
}