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
#include <iostream>
#include <cstdio>
using namespace std;
const int maxn = 50;
int n,A[maxn] = {1},isp[maxn], vis[maxn];
void dfs(int cur)
{
if(cur == n&& isp[A[0] + A[n - 1]]) {
for(int i = 0; i < n; i++) {
i ? printf(" %d", A[i]) : printf("%d", A[i]);
}
printf("\n");
} else for(int i = 2; i <= n; i++) {
if(!vis[i]&& isp[i + A[cur - 1]]) {
A[cur] = i;
vis[i] = 1;
dfs(cur + 1);
vis[i] = 0;
}
}
}
int main()
{
for(int i =2;i<=50;i++)
{
isp[i] = 1;
}
for(int i = 2;i<=50;i++)
{
for(int j = i + i; j + i <= 50; j +=i)
{
isp[j] = 0;
}
}
int kase = 0;
while(cin>>n)
{
if(kase++)
{
cout<<"\n";
}
cout<<"Case "<<kase<<":\n";
dfs(1);
}
return 0;
}