Prime Ring Problem
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. 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: 素数环 每一个相邻相加都得为素数,输出序列。
用的 dfs
#include <bits/stdc++.h>
using namespace std;
int n,f[25]={0},num[25]={0},p[45];
void dfs(int t)
{
if(t>n&&p[num[n]+num[1]])
{
cout<<num[1];
for(int i=2; i<=n; i++)
printf(" %d",num[i]);
printf("\n");
}
else
{
for(int i=2;i<=n;i++)
{
if( !f[i] && p[i+num[t-1]] )
{
f[i]=1;
num[t]=i;
dfs(t+1);
f[i]=0;
}
}
}
}
int main()
{
int a=1;
for(int k=2;k<44;k++) ///素数的预处理 存相应位置 是否为素数
{
p[k]=1;
for(int i=2; i<=sqrt(k); i++)
if(k%i==0)
p[k]=0;
}
while(cin>>n)
{
printf("Case %d:\n",a++);
num[1]=1;
dfs(2);
printf("\n");
}
return 0;
}