题目:
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.
Inputn (0 < n < 20).
Note: the number of first circle should always be 1.
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 8Sample 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
思路:
DFS。输出要空一行,每行最后没有空格。
代码:
#include<iostream> #include<cstdio> #include<cmath> using namespace std; int n; int vis[30]; int tot[30]; int ans; int prime(int a) { int flag=0; for(int i=2;i<=sqrt(a)+1;i++) { if(a%i==0)flag=1; } if(flag==1)return 0; else return 1; } void dfs(int num,int ans) { tot[ans]=num; vis[num]=1; int i; if(ans==n-1&&prime(tot[ans]+tot[0])==1) { cout<<tot[0]; for(i=1;i<n;i++) { cout<<" "<<tot[i]; } cout<<endl; } for(i=2;i<=n;i++) { if(prime(i+num)==1&&vis[i]==0) { vis[i]=1; dfs(i,ans+1); vis[i]=0; } } } int main() { int t=0; while(scanf("%d",&n)!=EOF) { t++; printf("Case %d:\n",t); ans=0; dfs(1,0); cout<<endl; } return 0; }