题目:
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 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
题意:构造一个素数环,环要求相邻两数之和是素数。环的元素的个数小于等于20。输出素数环(字典序)。
方法:深搜
代码:
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int prime[12]={2,3,5,7,11,13,17,19,23,29,31,37};
int isprime[40];
int used[21],n,step;
int a[21];
int dfs(int step)
{
int i;
if(step==n+1&&isprime[a[1]+a[n]]==1)
{
for(i=1;i<n;i++)
{
printf("%d ",a[i]);
}
printf("%d\n",a[n]);
return 0;//跳出当前dfs函数
}
else
{
for(i=2;i<=n;i++)
{
if(!used[i]&&isprime[i+a[step-1]])
{
used[i]=1;
a[step]=i;
dfs(step+1);
used[i]=0;
}
}
}
return 0;//跳出当前dfs函数
}
int main()
{
memset(isprime,0,sizeof(isprime));
for(int i=0;i<12;i++)
{
isprime[prime[i]]=1;
}
int Case=1;
a[1]=1;
while(cin>>n)
{
memset(used,0,sizeof(used));
printf("Case %d:\n",Case++);
used[1]=1;
dfs(2);
cout<<endl;
}
return 0;
}