zoukankan      html  css  js  c++  java
  • Prime Ring Problem hdu-1016 DFS

    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

    给你一个数n 1-n组成一个环 相邻的两个数之和要是素数(所以同时要保证头和尾的和是素数)
    DFS 基础题
    核心代码就一丢丢 很容易看懂

     1 #include <iostream>
     2 using namespace std;
     3 #include<string.h>
     4 #include<set>
     5 #include<stdio.h>
     6 #include<math.h>
     7 #include<queue>
     8 #include<map>
     9 #include<algorithm>
    10 #include<cstdio>
    11 #include<cmath>
    12 #include<cstring>
    13 #include <cstdio>
    14 #include <cstdlib>
    15 #include<stack>
    16 int n;
    17 int a[25];
    18 int b[25];
    19 int sushu(int x)
    20 {
    21     int s=(int)sqrt(x);
    22     for(int i=2;i<=s;i++)
    23     {
    24         if(x%i==0)
    25             return 0;
    26     }
    27     return 1;
    28 }
    29 void dfs(int x)
    30 {
    31     if(x==n&&sushu(a[1]+a[n]))
    32     {
    33         int flag=0;
    34         for(int i=1;i<=n;i++)
    35         {
    36             if(!flag)
    37             {
    38                 flag=1;
    39                 printf("%d",a[i]);
    40             }
    41             else
    42             {
    43                 printf(" %d",a[i]);
    44             }
    45         }
    46         printf("
    ");
    47         return;
    48     }
    49     for(int i=2;i<=n;i++)
    50     {
    51         if(b[i]==0&&sushu(a[x]+i))
    52         {
    53             //cout<<i<<endl;
    54             b[i]=1;
    55             a[x+1]=i;
    56             dfs(x+1);
    57             b[i]=0;
    58         }
    59     }
    60 }
    61 int main()
    62 {
    63     int add=0;
    64     while(~scanf("%d",&n))
    65     {
    66         memset(a,0,sizeof(a));
    67         memset(b,0,sizeof(b));
    68         a[1]=1;
    69         printf("Case %d:
    ",++add);
    70         dfs(1);
    71         printf("
    ");
    72     }
    73     return 0;
    74 }
    View Code

    讲道理 输入是奇数的时候直接跳过dfs就好 所以我加了个判断,结果时间还变长了一点 ,编程就像女人,永远搞不懂她。

    贴上加判断的AC

    
    
  • 相关阅读:
    管理中的“变”与“不变”
    软件项目需求分析与管理的十大疑问
    小商家也要有O2O思维
    互联网时代CIO生存法则
    浅谈项目经理与部门经理之间的关系
    沃尔玛:“最后一公里”的致命伤
    大数据分析案例:永远别忘记天气这个变量
    IT项目中的6类知识转移
    C
    linu入门
  • 原文地址:https://www.cnblogs.com/dulute/p/7482517.html
Copyright © 2011-2022 走看看