Prime Ring Problem
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18778 Accepted Submission(s): 8402
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.
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
Source
Recommend
JGShining
#include <stdio.h>
#include<string.h>
int a[41],visit[25],way[25];
int n,cw;
void print()
{
int i;
for(i=0;i<cw-1;i++)
printf("%d ",way[i]);
printf("%d\n",way[i]);
}
void dfs(int p,int pre)//p记录已经填到的位置
{
int i;
if(p==n-1)
{
for(i=1;i<=n;i++)
{
if(!visit[i]&&!a[i+1]&&!a[pre+i])//visit表示是否已填
{
way[cw++]=i;
print();
cw--;
return;
}
}
return;
}
for(i=1;i<=n;i++)
{
if(!visit[i]&&!a[i+pre])
{
visit[i]=1;
way[cw++]=i;
dfs(p+1,i);
visit[i]=0;
cw--;
}
}
}
int main()
{
int i,j,Case=1;
memset(a,0,sizeof a);
a[1]=1;
for(i=2;i<=40;i++)
if(!a[i])//用a来判断一个数是否为素数。题目数据较小所以很方便
for(j=2*i;j<=40;j+=i)
a[j]=1;
while(~scanf("%d",&n))
{
cw=1;
memset(visit,0,sizeof visit);
printf("Case %d:\n",Case++);
visit[1]=1;
way[0]=1;
dfs(1,1);
printf("\n");
}
return 0;
}