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.
Inputn
(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
Title:
Put the natural numbers 1, 2, ..., n into a circle, and the sum of the numbers in two adjacent circles should be prime. The number of the first circle should always be 1.
Ideas:
First print out a list of prime numbers, without storing prime numbers, the prime numbers are marked as 1, not 0. Then the depth-first search is performed, and the two adjacent numbers in the exit are added to the prime number. Just type a deep search honestly and it's OK.
AC code
#include<stdio.h>
#include<string.h>
void dfs(int cur);
int prime(int x);
int n,a[20],b[41],c[20];
int main()
{
int i;
for(i=2;i<=40;i++)
b[i]=prime(i);//打表
int k=1;
while(scanf("%d",&n)!=EOF)
{
a[0]=1;
printf("Case %d:\n",k++);
if(n%2==0)
{
dfs(1);开始深搜
}
printf("\n");
}
return 0;
}
int prime(int x)
{
int i;
for(i=2;i*i<=x;i++)
if(x%i==0)
{
return 0;
break;
}
return 1;
}
void dfs(int cur)
{
int i;
if(cur==n&&b[a[0]+a[n-1]])
{
for(i=0;i<n-1;i++)
printf("%d ",a[i]);
printf("%d\n",a[n-1]);
}
else
for(i=2;i<=n;i++)
if(!c[i]&&b[i+a[cur-1]])
{
a[cur]=i;
c[i]=1;
dfs(cur+1);
c[i]=0;
}
}