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): 25144    Accepted Submission(s): 11229

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.

 

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.

 

Sample Input

68

 

Sample Output

Case 1:1 4 3 2 5 61 6 5 2 3 4Case 2:1 2 3 8 5 6 7 41 2 5 8 3 4 7 61 4 7 6 5 8 3 21 6 7 4 3 8 5 2

 

<pre name="code" class="cpp">#include<stdio.h>#include<algorithm>#define max 100int v[max];int a[max];int vis[max];int i,n;void dfs(int cur){if(cur==n && v[a[0]+a[n-1]])  //递归边界，测试第一个数和最后一个数。{for(i=0;i<n;i++){if(i!=n-1) printf("%d ",a[i]);    //注意格式问题else printf("%d\n",a[i]);}}else{for(int i=2;i<=n;i++)                // 尝试放每一个数iif(!vis[i]&&v[i+a[cur-1]])       // 如果i没有用过，并且与前一个数之和为素数{a[cur]=i;    // i满足条件，则放入到数组a中vis[i]=1;    // 设置标记dfs(cur+1);vis[i]=0;    // 清除标记}}}int isprime(int n)   // 素数打表{int k=1;for(int i=2;i<n/2;i++){if(n%i==0)k=0;}return k;}int main(){   int  p=1;while(~scanf("%d",&n)){for(i=1;i<=2*n;i++)v[i]=isprime(i);  //判断 1 到 2*n 中哪些是素数(若是素数则赋值为1)。v[2]=0,v[7]=1.....for(i=0;i<n;i++)a[i]=i+1;          //将数 1 到 n 存到数组a里面。 a[0]=1,a[1]=2,a[2]=3,a[3]=4......printf("Case %d:\n",p);dfs(1);p++;printf("\n");}return 0;}