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  • 杭电 oj 1016 Prime Ring Problem

    **

    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
    深搜 递归
    全排列加判断条件

    #include<iostream>
    #include<algorithm>
    #include<cstdio>
    #include<cstring>
    using namespace std;
    int a[22];
    int b[22];
    int n,m;
    int sushu(int t)
    {
        for(int i=2;i*i<=t;i++)
        {
            if(t%i==0)
                return 0;
        }
        return 1;
    }
    void dfs(int y)
    {
        if(y==n+1&&sushu(a[1]+a[n])==1)//最后一个与第一个和是不是素数
        {
            for(int i=1;i<n;i++)
            {
                printf("%d ",a[i]);
            }
            printf("%d
    ",a[n]);
            return ;
        }
        for(int i=2;i<=n;i++)//全排列
        {
            if(b[i]==0&&sushu(i+a[y-1])==1)//判断 如果相邻两个的和是不是素数
            {
                a[y]=i;
                b[i]=1;
                dfs(y+1);
                b[i]=0;//归零 方便下一个递归 使用
            }
        }
    }
    int main()
    {
        int k=0;
        while(scanf("%d",&n)!=EOF)
        {
            memset(b,0,sizeof(b));
            memset(a,0,sizeof(a));
            printf("Case %d:
    ",++k);
            if(n==1)
            {
                printf("1
    ");
                continue;
            }
            if(n%2!=0)
            {
                printf("No Answer
    ");
                continue;
            }
            a[1]=1;
            dfs(2);
            printf("
    ");
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/nanfenggu/p/7900174.html
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