[Project Euler]加入欧拉 Problem 14

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

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

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

13 40 20 10 5 16 8 4 2 1

It 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.

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

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

 

Collatz问题。 这个我还是用暴力的方法解决了，还没发现好的规律。

写程序过程中小总结：

※ 对于变量定义的时候最好要初始化，不然会有意想不到的结果。 好比改程序里面的n 和 max。 C语言里面不初始化，系统是不会给变量自动赋0的。这点要额外小心注意。

欧拉项目第十四题#include <stdio.h>

#include <stdlib.h>



int lenCollatz(unsigned long n); // 输入一个数，进行Collatz迭代， 最终返回链长 



int main(int argc, char *argv[])

{

    unsigned long n = 13, max = 1;

    int lenmax = 1;

    while(n < 1000000)

    {

        if(lenmax < lenCollatz(n))

        {

            max = n;

            lenmax = lenCollatz(n);

        }        

        n++;        

    }       

    printf("Max numer is %lu, max length is %d", max, lenmax);

    

  system("PAUSE");	

  return 0;

}



int lenCollatz(unsigned long n)

{

    int len = 1;

    while(n != 1)

    {

        if(n % 2 == 0)

        n /= 2;

        else

        n = 3*n + 1;

        len ++;

    }

    return len;

}