zoukankan      html  css  js  c++  java
  • (Problem 14)Longest Collatz sequence

    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.

    方法1:
     1 #include<stdio.h>
     2 #include<math.h>
     3 #include<string.h>
     4 #include<ctype.h>
     5 #include<stdlib.h>
     6 #include<stdbool.h>
     7 
     8 int powcount(long long n)  //计算2的幂数
     9 {
    10     int count=0;
    11     while(n>>=1) count++;
    12     return count;
    13 }
    14 
    15 bool ispower(long long v)  //判断n是否为2的幂
    16 {
    17     if(((v & (v - 1)) == 0))  return true;
    18     else return false;
    19 }
    20 
    21 int length(long long n)
    22 {
    23     int sum=1;
    24     while(1)
    25     {
    26         if(n==1) break;
    27         if((n & 1)==0)
    28         {
    29             if(ispower(n)) return sum+powcount(n);
    30             else n=n/2;
    31         }
    32         else n=3*n+1;
    33         sum++;
    34     }
    35     return sum;
    36 }
    37 
    38 int main()
    39 {
    40     int i,t,k,max=0;
    41     for(i=2; i<1000000; i++)
    42     {
    43         t=length(i);
    44         if(t>max)
    45         {
    46             max=t;
    47             k=i;
    48         }
    49     }
    50     printf("%lld\n",k);
    51     return 0;
    52 }

    方法2:

     1 #include<stdio.h>
     2 #include<math.h>
     3 #include<string.h>
     4 #include<ctype.h>
     5 #include<stdlib.h>
     6 #include<stdbool.h>
     7 
     8 int a[1000001];
     9 
    10 void find()
    11 {
    12     long long i,j,k,f,sum,max=0;
    13     a[1]=1,a[2]=2;
    14     for(j=3; j<1000000; j++)
    15     {
    16         sum=1,k=i=j;
    17         while(1)
    18         {
    19             if((i & 1)==0)
    20             {
    21                 i=i/2;
    22                 if(i<k)
    23                 {
    24                     a[k]=sum+a[i];
    25                     break;
    26                 }
    27             }
    28             else
    29             {
    30                 i=3*i+1;
    31             }
    32             sum++;
    33         }
    34         if(a[k]>max)
    35         {
    36             max=a[k];
    37             f=k;
    38         }
    39     }
    40     printf("%d\n",f);
    41 }
    42 
    43 int main()
    44 {
    45     find();
    46     return 0;
    47 }
    Answer:
    837799
  • 相关阅读:
    TF-IDF与余弦类似性的应用(一):自己主动提取关键词
    三层中的大学问
    浅析JavaBean
    查看和改动MySQL数据库表存储引擎
    菜鸟之路--线性表__链表实现
    STL_算法_元素计数(count、count_if)
    ZOJ 3691 Flower(最大流+二分)
    字符的编码与解码
    主动訪问用户数据的背后是品牌战略
    输入n,求1~n累加
  • 原文地址:https://www.cnblogs.com/cpoint/p/3367342.html
Copyright © 2011-2022 走看看