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.
Note: the number of first circle should always be 1.
Inputn (0 < n < 20).
OutputThe 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
#include<iostream> #include<cstdio> #include<cstring> #include<string> #include<set> #include<vector> #include<stack> #include<queue> #include<algorithm> #include<cstdio> #include<algorithm> #include<functional> #include<sstream> using namespace std; int n,A[30]; int vis[30]; int isp[50]; bool is_prime(int i) { int t=0; if (i == 2 || i == 1) return 1; for (int j = 2; j < i; j++) { if (i % j == 0) t++; } if (t != 0) return 0; else return 1; } void dfs(int cur) { if (cur == n && isp[A[0] + A[n - 1]]) { for (int i = 0; i < n; i++) { if (i != n - 1) printf("%d ", A[i]); else printf("%d", A[i]); } printf(" "); } 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() { int t=1; for (int i = 1; i <= 50; i++) isp[i] = is_prime(i); while (cin >> n) { printf("Case %d: ", t); memset(A, 0, sizeof(A)); for (int i = 0; i <= n; i++) A[i] = i+1; memset(vis, 0, sizeof(vis)); dfs(1); t++; printf(" "); } }