hdu 1016 Prime Ring Problem
Prime Ring Problem
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 53931 Accepted Submission(s):
23862
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.
![](http://acm.split.hdu.edu.cn/data/images/1016-1.gif)
Note: the number of first circle should always be 1.
![](http://acm.split.hdu.edu.cn/data/images/1016-1.gif)
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
直接dfs就可以了
#include <stdio.h> #include <stdlib.h> #include <iostream> #include <cstring> #include <cmath> using namespace std; int vis[25],n; bool prime(int k) { for(int i=2;i<=sqrt(k);i++) { if(k%i==0) return false; } return true; } void dfs(int k) { if(vis[k]==n&&prime(k+1)) { printf("1"); for(int i=2;i<=n;i++) { for(int j=1;j<=n;j++) { if(vis[j]==i) printf(" %d",j); } } printf("\n"); return ; } for(int i=1;i<=n;i++) { if(prime(i+k)&&!vis[i]) { vis[i]=vis[k]+1; dfs(i); vis[i]=0; } } } int main() { int Case=0; while(~scanf("%d",&n)) { printf("Case %d:\n",++Case); memset(vis,0,sizeof(vis)); vis[1]=1; dfs(1); printf("\n"); } return 0; }
作者:红雨
出处:https://www.cnblogs.com/52why
微信公众号: 红雨python