• Project Euler Problem 92 Square digit chains


    Square digit chains

    Problem 92

    A number chain is created by continuously adding the square of the digits in a number to form a new number until it has been seen before.

    For example,

    44 → 32 → 13 → 10 → 11
    85 → 89 → 145 → 42 → 20 → 4 → 16 → 37 → 58 → 89

    Therefore any chain that arrives at 1 or 89 will become stuck in an endless loop. What is most amazing is that EVERY starting number will eventually arrive at 1 or 89.

    How many starting numbers below ten million will arrive at 89?


    C++(Faster):

    #include <iostream>
    #include <cstring>
    #include <stack>
    
    using namespace std;
    
    const int MAXN = 10000000;
    
    int arrive[MAXN], chainno[MAXN];
    int no;
    
    int nextval(int n)
    {
        int sum, v;
    
        sum = 0;
        while(n) {
            v = n % 10;
            sum += v * v;
            n /= 10;
        }
    
        return sum;
    }
    
    void makechain(int n)
    {
        no++;
    
        while(n != 1 && n != 89) {
            if(chainno[n] > 0) {
                n = arrive[chainno[n]];
                break;
            }
            chainno[n] = no;
            n = nextval(n);
        }
    
        arrive[no] = n;
    }
    
    int main()
    {
        int n;
    
        while(cin >> n && n <= MAXN) {
            memset(arrive, 0, sizeof(arrive));
            memset(chainno, 0, sizeof(chainno));
            chainno[1] = 1;
            arrive[1] = 1;
            chainno[89] = 2;
            arrive[2] = 89;
            no = 2;
    
            for(int i=1; i<n; i++)
                if(i != 1 && i != 89)
                    makechain(i);
    
            int count = 0;
            for(int i=1; i<n; i++)
                if(arrive[chainno[i]] == 89)
                    count++;
    
            cout << count << endl;
        }
    
        return 0;
    }




    C++:

    #include <iostream>
    #include <cstring>
    #include <stack>
    
    using namespace std;
    
    const int MAXN = 10000000;
    
    int arrive[MAXN];
    
    int nextval(int n)
    {
        int sum, v;
    
        sum = 0;
        while(n) {
            v = n % 10;
            sum += v * v;
            n /= 10;
        }
    
        return sum;
    }
    
    void makechain(int n)
    {
        stack<int> s;
    
        while(n != 1 && n != 89) {
            if(arrive[n] > 0) {
                n = arrive[n];
                break;
            }
            s.push(n);
            n = nextval(n);
        }
    
        while(!s.empty()) {
            int val = s.top();
            s.pop();
            arrive[val] = n;
        }
    }
    
    int main()
    {
        int n;
    
        while(cin >> n && n <= MAXN) {
            memset(arrive, 0, sizeof(arrive));
            arrive[1] = 1;
            arrive[89] = 89;
    
            for(int i=1; i<n; i++)
                makechain(i);
    
            int count = 0;
            for(int i=1; i<n; i++)
                if(arrive[i] == 89)
                    count++;
    
            cout << count << endl;
        }
    
        return 0;
    }






  • 相关阅读:
    Android培训准备资料之Android开发环境的搭建
    第二十天
    第十九天
    第十八天
    第十七天
    第十六天
    第十四天
    第十三天
    十二天
    十一天
  • 原文地址:https://www.cnblogs.com/tigerisland/p/7564005.html
Copyright © 2020-2023  润新知