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.
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 4Case 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
题意:
将n个数来放到n个位置上,相邻的两个数和一定要是素数,输出符合的序列,序列一定要是从1为开头。
放第n个数时,要记得和第一个数进行判断和是否为素数
代码:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n;
int vis[27]; //vis数组用来标记数是否用了
int num[27]; //num来存放序列
int cnt=0;
int isp(int x) //判断素数
{
for(int i=2;i*i<=x;i++)
{
if(x%i==0)
return 0;
}
return 1;
}
void dfs(int step)
{
//如果step=n说明n个数都找到,然后要判断最后一个数和第一个数的和是否符合素数
if(step==n&&isp(num[0]+num[step-1])==1)
{
printf("%d",num[0]);
for(int i=1;i<n;i++)
printf(" %d",num[i]);
printf("\n");
return ;
}
for(int i=2;i<=n;i++)
{
if(vis[i]==0&&isp(i+num[step-1])==1)
{
num[step]=i;
vis[i]=1;
dfs(step+1);
vis[i]=0;
}
}
}
int main()
{
num[0]=1; //第一个数永远是1
while(scanf("%d",&n)!=EOF)
{
memset(vis,0,sizeof(vis));
cnt++;
printf("Case %d:\n",cnt);
dfs(1);
printf("\n");
}
return 0;
}