Prime Ring Problem
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 30395 Accepted Submission(s): 13525
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
Source
//第一次写dfs,既然发现和小白上的代码一样,开心 #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int prime[]= {0,0,1,1,0,1,0,1,0,0,0, 1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0, 0,0,1,0,1,0,0,0,0,0,1,0,0,0 };//方便判断一个数是否素数 int flag[25];//标志数组 int c[25];//结果数组 int n; void dfs(int cur) { int i,j; if(cur==n+1&&prime[1+c[cur-1]])//别忘记第一个数和最后一个数的和也应该是素数 { for(j=1; j<n; j++) printf("%d ",c[j]); printf("%d\n",c[n]); } else for(i=2; i<=n; i++)//每个数都应是从2——n选择 if(!flag[i]&&prime[i+c[cur-1]])//是否满足条件 { c[cur]=i;//尝试把i作为第cur个满足条件的数 flag[i]=1;//设置使用标志 dfs(cur+1); flag[i]=0;//谨记清除标记 } } int main() { int k=1; while(~scanf("%d",&n)) { printf("Case %d:\n",k++); memset(flag,0,sizeof(flag)); c[1]=1; dfs(2);//第一个数就是1,从第二个数开始 printf("\n"); } } 不知什么鬼,G++提交超时,c++ 256ms