题目描述
Consider the following algorithm to generate a sequence of numbers. Start with an integer n. If n is even, divide by 2. If n is odd, multiply by 3 and add 1. Repeat this process with the new value of n, terminating when n = 1. For example, the following sequence of numbers will be generated for n = 22: 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1 It is conjectured (but not yet proven) that this algorithm will terminate at n = 1 for every integer n. Still, the conjecture holds for all integers up to at least 1, 000, 000. For an input n, the cycle-length of n is the number of numbers generated up to and including the 1. In the example above, the cycle length of 22 is 16. Given any two numbers i and j, you are to determine the maximum cycle length over all numbers between i and j, including both endpoints.
输入
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.
输出
For each pair of input integers i and j, output i, j in the same order in which they appeared in the input and then the maximum cycle length for integers between and including i and j. These three numbers should be separated by one space, with all three numbers on one line and with one line of output for each line of input.
样例输入
1 10 100 200 201 210 900 1000
样例输出
1 10 20 100 200 125 201 210 89 900 1000 174
提示
来源
【题目翻译】：
描述
考虑下面的算法来生成一个数字序列。以整数n开始，如果n是偶数，除以2。如果n是奇数，乘以3，加1。用n的新值重复此过程，当n＝1时终止。例如，对于N＝22:22 11、34、17、52、26、13、40、20、10、5、16 、8、4、2、 1，将生成下列序列数。猜想（但尚未证明）该算法将终止于n＝α，对于每一个整数n。对于输入n，n的周期长度是生成并包括1的数字的数目。在上面的例子中，22的周期长度是16。给定两个数字i和j，你要确定在i和j之间的所有数字上的最大周期长度，包括两个端点。
输入
输入将由一系列整数i和j组成，每行一对整数。所有整数都小于1000000，大于0。
输出
对于每一对输入整数i和j，输出i，j以它们出现在输入中的相同顺序，然后是i和j之间整数的最大周期长度，这三个数字应该由一个空间分隔，所有三个数字在一行上，并且具有一行。用于输入的每一行。
注意：在输入时前面的数有可能小于后面的数，最后输出格式也要和题目中给出数据的先后顺序一致。
【代码】：
import java.util.Arrays;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner sc = new Scanner(System.in);
while (sc.hasNext()) {
int a = sc.nextInt();
int b = sc.nextInt();
int s[] = new int[Math.abs(b - a) + 1];
int x;
int m = a, n = b;
if (a > b) {
x = a;
a = b;
b = x;
}
for (int i = a; i <= b; i++) {
int k = i;
int sum = 1;
while (k != 1) {
if (k % 2 == 0)
k /= 2;
else
k = k * 3 + 1;
sum++;
}
s[i - a] = sum;
}
Arrays.sort(s);
if (m < n)
System.out.println(a + " " + b + " " + s[b - a]);
else
System.out.println(b + " " + a + " " + s[b - a]);
}
}
}