Prime Ring Problem
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26913 Accepted Submission(s): 12008
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.
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.
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
/*
HDOJ 1016 DFS
*/
#include<iostream>
#include<stdio.h>
using namespace std;
int n,arr[50],visit[21];
int check(int x)
{
int i;
for(i=2;i<x;i++)
if(x%i==0)
return 0;
return 1;
}
void DFS(int x,int k)
{
int i;
arr[x]=k;visit[k]=1;//arr[1]=1代表第1个是1
if(x==n)
{
if(check(arr[x]+arr[1]))
{
printf("1");
for(i=2;i<=n;i++)
printf(" %d",arr[i]);
printf("\n");
}
return ;
}
for(i=1;i<=n;i++)
{
if(!visit[i]&&check(arr[x]+i))//已经访问的，满足条件就继续
{
DFS(x+1,i);
visit[i]=0;
}
}
return ;
}
int main(){
int k=1;
while(scanf("%d",&n)!=EOF)
{
printf("Case %d:\n",k++);
memset(visit,0,sizeof(visit));
DFS(1,1);
printf("\n");
}
return 0;
}