hdu 1016 Prime Ring 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.
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.
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<stdio.h> using namespace std; bool flag[100]={false}; int ans[25]; int n; int vis[25]={0}; int t; void dfs(int num) { if(num==n&&(flag[ans[0]+ans[num-1]]==0)) { for(int i=0;i<n-1;i++) printf("%d ",ans[i]); printf("%d\n",ans[n-1]); } else { if(num==n) return; for(int i=2;i<=n;i++) { // cout<<"i: "<<endl; if(!vis[i]&&(flag[ans[num-1]+i]==0)) { //cout<<"i: "<<i<<endl; vis[i]=1; ans[num]=i; dfs(num+1); vis[i]=0; } } } } int main() { ans[0]=1; for(int i=2;i<100;i++) { if(!flag[i]) { for(int j=i<<1;j<100;j+=i) flag[j]=true; } } t=1; // for(int i=0;i<23;i++) cout<<flag[i]<<endl; while(~scanf("%d",&n)) { for(int i=0;i<=n;i++) vis[i]=0; printf("Case %d:\n",t++); dfs(1); printf("\n"); } }