开贴做题，囧~~~

ygcUID: 2396629

Problem 14.  Find the longest sequence using a starting number under one million.
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.


这个题，照着问题写代码就可以了，不过算得比较慢，我没能想出什么好的方法来加速它。

=============================================================================================


n <- 1000000
i <- 1
count <- 0
long_chain_num <- 0      
while(i <n){
    old.i <- i
    cnt <- 2      ## i and 1 are not count in the while loop...
    while( i != 1 ){
        if(i %% 2) {
            i <- 3*i +1
        } else{
            i <- i /2

        }
        cnt = cnt + 1
    }
    if (count < cnt){
        count <- cnt
        long_chain_num <- old.i
    }
    i <- old.i + 1
}

cat (long_chain_num,"\n")

Answer: 837799

SpectoriaUID: 2488330

大叔在干什么？:funk:

ygcUID: 2396629

做数学题。。。。

第一贴上有链接。。。

SpectoriaUID: 2488330

:eek: :funk: 赞，勤奋的大叔

ygcUID: 2396629

偶不来更新了。。

以后贴blog里。:o

ygcUID: 2396629

:(

SpectoriaUID: 2488330

大叔终于做完鸟？:confused: :rolleyes: ;d:

ygcUID: 2396629

早不来贴了。。

没做完的说。。。
只做了40多道而已。

SpectoriaUID: 2488330

40多道:eek:
勤奋的大叔Orz

ygcUID: 2396629

前面的题好简单，越来越难。。后面的超出我的能力了。。。囧。。。

SpectoriaUID: 2488330

;d: patpat :handshake