UVA - 524 Prime Ring Problem

Description





A ring is composed of n (even number) circles as shown in diagram. Put natural numbers

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 <= 16)

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.

You are to write a program that completes above process.

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

分析：本题可采用回溯发求解，注意把素数的判断用数组提前计算出来，另外注意输出格式。

#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>

using namespace std;
const int maxn = 17;
int isp[maxn*2];
int A[maxn];
int vis[maxn];
int n;

int is_prime(int n)
{
if(n<2) return 0;
for(int i=2; i<=sqrt(n); i++)
if(n%i==0) return 0;
return 1;
}

void dfs(int cur)
{
if(cur==n && isp[A[0]+A[n-1]]) {
for(int i=0; i<n; i++) { if(i>0) cout << ' '; cout << A[i]; }
cout << endl;
}else{
for(int i=2; i<=n; i++) {
if(!vis[i] && isp[i+A[cur-1]]) {
vis[i] = 1;
A[cur] = i;
dfs(cur+1);
vis[i] = 0;
}
}
}
}

int main()
{
for(int i=1; i<maxn*2; i++) isp[i] = is_prime(i);
int kase = 0;
while(cin >> n) {
if(kase) cout << endl;
cout << "Case " << ++kase << ":\n";
memset(vis, 0, sizeof(vis));
A[0] = 1;
dfs(1);
}
return 0;
}