Description
A ring is composed of n (even number) circles as shown in diagram. Put natural numbers 1,2,3,...,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.
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 【代码】:
/*素数环*/ #include <bits/stdc++.h> using namespace std; int n,prime[1005]={1},v[1005],a[1005]; void get_prime() { int m=sqrt(2*n+0.5);//环-倍增 memset(prime,1,sizeof(prime));//0为素数 prime[0]=prime[1]=0;//非素数 for(int i=2;i<=m;i++) { if(prime[i])//素数 { for(int j=i+i;j<=2*n;j+=i) { prime[j]=0;//非素数 } } } } void dfs(int cur) { if(cur == n && prime[ a[0]+a[n-1] ]) { for(int i=0; i<n; i++) { printf("%d%c", a[i], i == n - 1 ? ‘\n‘ : ‘ ‘); } // printf("\n"); } else { for(int i=2; i<=n; i++) { if( !v[i] && prime[ i + a[cur-1] ] ) { a[cur] = i; v[i]=1; dfs(cur+1); v[i]=0; } } } } int main() { scanf("%d",&n); //初始化 memset(v,0,sizeof(v)); a[0]=1; get_prime(); dfs(1); }
时间: 2024-10-11 16:19:43