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  • Prime Ring Problem

    A ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n into each circle separately, and the sum of numbers in two adjacent circles should be a prime. 

    Note: the number of first circle should always be 1. 

     

    Inputn (0 < n < 20). 
    OutputThe output format is shown as sample below. Each row represents a series of circle numbers in the ring beginning from 1 clockwisely and anticlockwisely. The order of numbers must satisfy the above requirements. Print solutions in lexicographical order. 

    You are to write a program that completes above process. 

    Print a blank line after each case. 
    Sample Input

    6
    8

    Sample Output

    Case 1:
    1 4 3 2 5 6
    1 6 5 2 3 4
    
    Case 2:
    1 2 3 8 5 6 7 4
    1 2 5 8 3 4 7 6
    1 4 7 6 5 8 3 2
    1 6 7 4 3 8 5 2
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<string>
    #include<set>
    #include<vector>
    #include<stack>
    #include<queue>
    #include<algorithm>
    #include<cstdio>
    #include<algorithm>
    #include<functional>
    #include<sstream>
    using namespace std;
    int n,A[30];
    int vis[30];
    int isp[50];
    bool is_prime(int i)
    {
        int t=0;
        if (i == 2 || i == 1)
            return 1;
        for (int j = 2; j < i; j++)
        {
            if (i % j == 0)
                t++;
        }
        if (t != 0)
            return 0;
        else
            return 1;
    }
    void dfs(int cur)
    {
        if (cur == n && isp[A[0] + A[n - 1]])
        {
            for (int i = 0; i < n; i++)
            {
                if (i != n - 1)
                    printf("%d ", A[i]);
                else
                    printf("%d", A[i]);
            }
            printf("
    ");
        }
        else for(int i=2;i<=n;i++)
            if (!vis[i] && isp[i + A[cur - 1]])
            {
                A[cur] = i;
                vis[i] = 1;
                dfs(cur + 1);
                vis[i] = 0;
            }
    }
    int main()
    {
        int t=1;
        for (int i = 1; i <= 50; i++)
            isp[i] = is_prime(i);
        while (cin >> n)
        {
            printf("Case %d:
    ", t);
            memset(A, 0, sizeof(A));
            for (int i = 0; i <= n; i++)
                A[i] = i+1;
            memset(vis, 0, sizeof(vis));
            dfs(1);
            t++;
            printf("
    ");
        }
    }
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  • 原文地址:https://www.cnblogs.com/edych/p/7252234.html
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