The 3n+1 Problem(求各位高手修改)

最新推荐文章于 2020-12-22 19:44:23 发布

原创最新推荐文章于 2020-12-22 19:44:23 发布 · 372 阅读

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#numbers #input #algorithm #integer #output #pair

本文探讨了一个算法，用于生成从任意整数开始的一系列数字，通过奇偶判断进行运算，直到序列终止于1。对于给定的两个整数范围，计算并输出该范围内所有数字的最大循环长度。

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 ofn, terminating when n = 1. For example, the following sequence ofnumbers 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 willterminate at n = 1 for every integer n. Still, the conjecture holdsfor all integers up to at least 1, 000, 000.

For an input n, the cycle-length of n is the number of numbersgenerated up to and including the 1. In the example above, thecycle length of 22 is 16. Given any two numbers i and j, you are todetermine the maximum cycle length over all numbers between i andj, including both endpoints.

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 than1,000,000 and greater than 0.

Output
For each pair of input integers i and j, output i, j in the sameorder in which they appeared in the input and then the maximumcycle length for integers between and including i and j. Thesethree numbers should be separated by one space, with all threenumbers on one line and with one line of output for each line ofinput.

Sample Input

1 10
100 200
201 210
900 1000

Sample Output

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

#include<stdio.h>
int func(int n)
{

intcount=1;
   while(n!=1)
   {
      if(n%2==0)
         n=n/2;
      else
         n=n*3+1;
      count++;
   }
   returncount;
}
int main()
{
   inti,j,max,t;
   scanf("%d%d",&i,&j);
   max=func(i);
   for(t=i+1;t<=j;t++)
   {
      if(max<func(t))
         max=func(t);
   }
   printf("%d%d %d\n",i,j,max);

}