• hdu1016


    Prime Ring Problem

    Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 19946    Accepted Submission(s): 8935


    Problem Description
    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.

     
    Input
    n (0 < n < 20).
     
    Output
    The 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<cstring>
    #include<cstdio>
    #include<string>
    using namespace std;
    int sign[21],a[21],h=0;int n;
    int isprime(int n)
    {
        for(int i=2;i*i<=n;i++)
        {
            if(n%i==0)
                return 0;
        }
        return 1;
    }
    void dfs(int x,int y)
    {
        a[x] = y;
        sign[y] = 1;int i;
        if(x==n)
        {
            if(isprime(a[x]+a[1]))
            {
                printf("1");
                for(i=2;i<=n;i++)
                    printf(" %d",a[i]);
                printf("
    ");
            }
            return ;
        }
        for(i=1;i<=n;i++)
        {
            if(!sign[i] && isprime(a[x]+i))
            {
                dfs(x+1,i);
                sign[i] = 0;
            }
        }
        return ;
    }
    int main()
    {
        int flag = 1;
        while(scanf("%d",&n)!=EOF)
        {
            printf("Case %d:
    ",flag++);  
            memset(sign,0,sizeof(sign));
            dfs(1,1);
            printf("
    ");
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/Deng1185246160/p/3220782.html
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