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
题意:给定一个数作为长度,排成一个圆环状的素数环,其中数不能重复,求出所有可能性
题解:一道很简单的DFS题目
#include<bits/stdc++.h> using namespace std; int x,mark[25],a[25]; int check(int n){ if(n==1) return 0; if(n==2) return 1; for(int i=2;i<n;i++) if(n%i==0) return 0; return 1; } void dfs(int step){ if(step>x&&check(mark[x]+mark[1])){ int i; for(i=1;i<=x-1;i++) cout<<mark[i]<<" "; cout<<mark[i]<<endl; } for(int i=2;i<=x;i++){ mark[step]=i; if(check(mark[step]+mark[step-1])&&!a[i]){ a[i]=1; dfs(step+1); a[i]=0; } } } int main() { int k=1; while(cin>>x){ cout<<"Case "<<k++<<":"<<endl; memset(a,0,sizeof(a)); mark[1]=1; dfs(2); cout<<endl; } return 0; }