1195: Prime Ring Problem
Result | TIME Limit | MEMORY Limit | Run Times | AC Times | JUDGE |
---|---|---|---|---|---|
10s | 8192K | 1973 | 499 | Standard |
A ring is composed of n 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 <cmath> using namespace std; const int maxn = 33; int isp[maxn], n, a[10000],vis[10000]; bool judge(int x) { int i; for (i=2; i<=sqrt((double)x); i++) { if (x % i == 0) return false; } return true; } void makeprime() { int i; for (i=2; i<=maxn; i++) if (judge(i)) isp[i] = 1; } void dfs(int cur) { if (cur == n && isp[a[0]+a[n-1]]==1) { int i; cout<<a[0]; for (i=1;i<n; i++) { cout<<" "<<a[i]; } cout<<endl; } else { int i; for (i=2; i<=n; i++) { if (!vis[i] && isp[i+a[cur-1]] == 1) { a[cur] = i; vis[i] = 1; dfs(cur+1); vis[i] = 0; } } } } int main() { makeprime(); int count=0; while (cin>>n) { count++; cout<<"Case "<<count<<":"<<endl; if(n%2==0) { memset (a,0,sizeof(a)); memset (vis,0,sizeof(vis)); vis[1] = 1; a[0] = 1; dfs(1); } cout<<endl; } return 0; }