Project Euler 问题 14(Collatz 问题)

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问题描述

为正整数集合定义了以下迭代序列:

The following iterative sequence is defined for the set of positive integers:

n ->n/2 (n 是偶数)n ->3n + 1(n 是奇数)

n ->n/2 (n is even)n ->3n + 1 (n is odd)

使用上面的规则并从 13 开始，我们生成以下序列:

Using the rule above and starting with 13, we generate the following sequence:

13 40 20 10 5 16 8 4 2 1可以看出，这个序列(从13开始到1结束)包含10项.虽然还没有被证明(Collatz 问题)，但认为所有的起始数都以 1 结束.

13 40 20 10 5 16 8 4 2 1It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.

100 万以下的哪个起始数字产生最长的链?

Which starting number, under one million, produces the longest chain?

注意:一旦链开始，条款就可以超过一百万.

NOTE: Once the chain starts the terms are allowed to go above one million.

我尝试使用蛮力方法在 C 中编写解决方案.但是，我的程序在尝试计算 113383 时似乎卡住了.请指教 :)

I tried coding a solution to this in C using the bruteforce method. However, it seems that my program stalls when trying to calculate 113383. Please advise :)

#include <stdio.h>
#define LIMIT 1000000

int iteration(int value)
{
 if(value%2==0)
  return (value/2);
 else
  return (3*value+1);
}

int count_iterations(int value)
{
 int count=1;
 //printf("%d
", value);
 while(value!=1)
 {
  value=iteration(value);
  //printf("%d
", value);
  count++;
 }
 return count;
}

int main()
{
 int iteration_count=0, max=0;
 int i,count;


 for (i=1; i<LIMIT; i++)
 {
  printf("Current iteration : %d
", i);
  iteration_count=count_iterations(i);
  if (iteration_count>max)
   {
   max=iteration_count;
   count=i;
   }

 }

 //iteration_count=count_iterations(113383);
 printf("Count = %d
i = %d
",max,count);

}

euler