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.
You are to write a program that completes above process.
Print a blank line after each case.
Sample Input
6 8
思路：经典题目。
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 <stdio.h>
#include <math.h>
bool isprime[38],vis[20];
int n,A[20];
void trackback(int cur)
{
int i;
if(cur==n+1 && isprime[A[1]+A[n]])
{
for(i=1;i<=n;i++)
{
printf("%d",A[i]);
if(i<n) printf(" ");
else printf("\n");
}
}
else if(cur<=n)
{
for(i=2;i<=n;i++)
{
if(isprime[A[cur-1]+i] && !vis[i])
{
A[cur]=i;
vis[i]=1;
trackback(cur+1);
vis[i]=0;
}
}
}
}
int main()
{
int i,j,count=1;
for(i=2;i<=37;i++)//初始化素数表
{
isprime[i]=1;
for(j=2;j<=sqrt(i);j++)
{
if(i%j==0)
{
isprime[i]=0;
break;
}
}
}
for(i=1;i<=19;i++) vis[i]=0;//标志数组清零
A[1]=1;
while(~scanf("%d",&n))
{
printf("Case %d:\n",count++);
trackback(2);
printf("\n");
}
}