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.
nums <- 1:1000000
Nums <- 1:1000000
i <- 1
while(!(is.na(i)))
{
i <- which.max(Nums[Nums>0])
nums[i] <- 1
j <- i
repeat
{
if(j==1) break
if((j %% 2)==0)
{
Nums[j/2] <-0
j <- j/2
}
else if((3*j+1)<=1000000)
{
Nums[3*j+1] <- 0
j <- 3*j+1
}
nums[i] <- nums[i] + 1
}
Nums[i] <- -Nums[i]
}
print(which.max(nums))
#999847
len <- 1e6-1
from <- 1:len
seq.length <- c(-1,rep(0,len-1))
remain <- 2:len # 还未抵1
while(length(remain)!=0){
from[remain] <- ifelse(from[remain]%%2==0,from[remain]/2,from[remain]*3+1)
done <- from[remain]<=len & seq.length[from[remain]]<0 #下一步过后就重复计算了
done.index <- remain[done]
seq.length[done.index] <- seq.length[from[done.index]]-1-seq.length[done.index]
remain <- remain[!done]
seq.length[remain] <- seq.length[remain]+1
}
print(which.min(seq.length))
which.max(unlist(llply(1:len,function(x){
l <- 0
while(x!=1){
x <- ifelse(x%%2==0,x/2,x*3+1)
l <- l+1
}
l
})))
#include <iostream>
#include <vector>
using namespace std;
int main(int argc, const char *argv[])
{
typedef long INT;
const INT len=999999;
INT v[len];
for (INT i = 0; i < len; i++) {
v[i]=i;
}
INT largest_remain=len-1;
INT longest=largest_remain;
INT llength=-0;
vector<INT> index;
while(largest_remain!=0){
index.push_back(largest_remain);
do{
index.push_back(index.back()%2==1?(index.back()-1)/2:(index.back()*3+3));//C++从0开始算，为方便把题目做个转化
}while(index.back()!=0 && (index.back()>=len || v[index.back()]>0));
for (INT i = index.size()-2; i >= 0; i--) {
if(index[i]<len){
v[index[i]] = v[index.back()]-index.size()+i+1;
}
}
index.clear();
if(llength>v[largest_remain]){//存储的长度是负数
llength=v[largest_remain];
longest=largest_remain;
}
do{
largest_remain--;
}while(v[largest_remain]<0);
}
cout<<longest+1<<endl;
return 0;
}
