Prime Ring Problem
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 74594 Accepted Submission(s): 31690
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
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
Asia 1996, Shanghai (Mainland China)
题意大致就是说要找到由1~n组成的环，且相邻的数两两相加都要是素数，输出所有满足条件的序列。
AC代码如下：
#include<stdio.h> #include<string.h> int f[100]={0}; int ans[21],v[21],n; void dfs(int i) { int m; if(i==n&&f[ans[i-1]+ans[0]]==0) //找到8个满足条件的数时输出 { for(m=0;m<n-1;m++) printf("%d ",ans[m]); printf("%d\n",ans[n-1]); } else { for(m=2;m<=n;m++) //找未用过的数 { if(v[m]==0) //找到还没有用过的数 { if(f[m+ans[i-1]]==0) //如果满足相邻数和为素数 { v[m]=-1; //标记已用过 ans[i++]=m; //将m放进数组 dfs(i); //找下一个数 v[m]=0; //接触标记 i--; //回溯 } } } } } int main() { int i,j,cases=0; //素数打表 for(i=2;i<100;i++) if(f[i]==0) for(j=i;i*j<100;j++) f[i*j]=1; while(scanf("%d",&n)!=EOF) { cases++; ans[0]=1; printf("Case %d:\n",cases); memset(v,0,sizeof(v)); dfs(1); printf("\n"); } return 0; }
