Prime Ring Problem
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19946 Accepted Submission(s): 8935
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<cstring> #include<cstdio> #include<string> using namespace std; int sign[21],a[21],h=0;int n; int isprime(int n) { for(int i=2;i*i<=n;i++) { if(n%i==0) return 0; } return 1; } void dfs(int x,int y) { a[x] = y; sign[y] = 1;int i; if(x==n) { if(isprime(a[x]+a[1])) { printf("1"); for(i=2;i<=n;i++) printf(" %d",a[i]); printf(" "); } return ; } for(i=1;i<=n;i++) { if(!sign[i] && isprime(a[x]+i)) { dfs(x+1,i); sign[i] = 0; } } return ; } int main() { int flag = 1; while(scanf("%d",&n)!=EOF) { printf("Case %d: ",flag++); memset(sign,0,sizeof(sign)); dfs(1,1); printf(" "); } return 0; }