Prime Ring Problem
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 61907 Accepted Submission(s): 26691
Total Submission(s): 61907 Accepted Submission(s): 26691
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.hdu.edu.cn/data/images/1016-1.gif)
Note: the number of first circle should always be 1.
![](http://acm.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
题解:dfs从1开始找,循环判断即可。
代码:
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <vector> #include <set> #include <map> using namespace std; int n,ans[21]; int vis[105]; int k=1; int is_Prime(int p){//素数判断 int a=sqrt(p); if(p==0||p==1) return 0; for(int i=2;i<=a;i++){ if(p%i==0) return 0; } return 1; } void dfs(int x){ if(k==n+1&&is_Prime(ans[n]+ans[1])){//如果已经找到n个数并且第n个与1的和是素数则足以构成环,输出。 for(int i=1;i<n;i++){ printf("%d ",ans[i]); } printf("%d ",ans[n]); } for(int i=2;i<=n;i++){ if(!vis[i]&&is_Prime(x+i)){//判断是否已经在素数序列中并且是否能够和前一个数构成素数 vis[i]=1; ans[k++]=i; dfs(i); vis[i]=0;//找其他数是否满足条件 k--; } } } int main() { int cnt=0; while(~scanf("%d",&n)&&n){ k=1; cnt++;//案例个数 printf("Case %d: ",cnt); memset(vis,0,sizeof vis); ans[k++]=1;//从1开始 vis[1]=1; dfs(1);//从1开始搜 printf(" "); } return 0; }