zoukankan      html  css  js  c++  java
  • Project Euler Problem 14 Longest Collatz sequence

    Longest Collatz sequence

    Problem 14

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

    nn/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.


    C++:

    #include <iostream>
    #include <cstring>
    
    using namespace std;
    
    const int MAXN = 1000000;
    const int ONE_HUNDRED_MILLION = 100000000;
    
    int cs[ONE_HUNDRED_MILLION+1];
    
    // Collatz sequence count
    int cscount(long long x)
    {
        if(x <= ONE_HUNDRED_MILLION && cs[x])
            return cs[x];
    
        int count;
        if(x % 2 == 0)
            count = 1 + cscount(x / 2);
        else
            count =  1 + cscount(x * 3 + 1);
    
        if(x <= ONE_HUNDRED_MILLION)
            cs[x] = count;
    
        return count;
    }
    
    int main()
    {
        memset(cs, 0, sizeof(cs));
        cs[1] = 1;
    
        int n, ans;
        while(cin >> n && n <= MAXN) {
            ans = 1;
            for(int i=1; i<=n; i++) {
                cs[i] = cscount(i);
                if(cs[i] > cs[ans])
                    ans = i;
            }
            cout << ans << endl;
        }
    
        return 0;
    }

    Input data:

    999999



    C++(Too slow):

    #include <iostream>
    
    using namespace std;
    
    //#define DEBUG
    
    const int MAXN = 1000000;
    
    // Collatz sequence count
    int cscount(int start)
    {
    #ifdef DEBUG
            cout << start << ": ";
    #endif
    
        int count = 0;
        for(;;) {
    #ifdef DEBUG
            cout << start << " ";
    #endif
    
            count++;
            if(start == 1)
                break;
            if(start & 1)
                start = 3 * start + 1;
            else
                start >>= 1;
        }
    
    #ifdef DEBUG
        cout << endl;
    #endif
    
        return count;
    }
    
    int main()
    {
        int n, ans, maxcount=0, temp;
        while(cin >> n && n <= MAXN) {
            for(int i=1; i<=n; i++) {
                temp = cscount(i);
                if(temp > maxcount) {
                    maxcount = temp;
                    ans = i;
                }
            }
            cout << ans << endl;
        }
    
        return 0;
    }




  • 相关阅读:
    Java实现 LeetCode 416 分割等和子集
    Java实现 LeetCode 416 分割等和子集
    在Linux运行期间升级Linux系统(Uboot+kernel+Rootfs)
    【详解】嵌入式开发中固件的烧录方式
    嵌入式系统烧写uboot/bootloader/kernel的一般方法
    Linux下USB烧写uImage kernel
    Xmanager连接CentOS的远程桌面
    命令行利器Tmux
    u-boot中分区和内核MTD分区关系
    uboot环境变量与内核MTD分区关系
  • 原文地址:https://www.cnblogs.com/tigerisland/p/7564021.html
Copyright © 2011-2022 走看看