Prime Ring Problem
Time Limit: 400
Total Submission(s): 28085 Accepted Submission(s): 12501
Prime Ring Problem
0/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 28085 Accepted Submission(s): 12501
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源码:#include <stdio.h> #include <string.h> #include <stdlib.h> int n; int prime[30]={1,1,0}; int way[30]; //存储最后结果序列 int visit[30]; //vis表示是否訪问 void isprime() //不是素数标记为1 { int i,j; for(i=2;i<30;i++) { for(j=2*i;j<30;j+=i){ prime[j]=1; } } /*for(i = 0;i < 30;i ++)printf("prime[%d]=%d\n",i,prime[i]);*/ } /* 素数序列的开头是这种: 2,3,5,7。11,13。17,19。23,29, 31,37,41。43。47。53,59,61,67。 71,73,79。83,89,97,101。103。107, 109,113,127,131。137,139,149,151 */ void dfs(int temp) //环当前位置 { int i; if(temp==n && prime[way[0]+way[n-1]]==0)//若到达边界且首尾两个数字相加满足素数 { for(i=0;i<n-1;i++)//打印方案 { printf("%d ",way[i]); } printf("%d\n",way[i]); } else { for(i=2;i<=n;i++) { if(visit[i]==0 && prime[i+way[temp-1]]==0) //i没有訪问过且与前一个数相加为素数 { way[temp]=i; visit[i]=1; //表示i訪问过dfs(temp+1); visit[i]=0; //i不满足条件, 之前做的标记要改动 } } } } int main() { int count=1; memset(visit,0,sizeof(visit)); //memset(way,0,sizeof(way)); way[0]=1; isprime(); while(~scanf("%d",&n)) {printf("Case %d:\n",count++); dfs(1); printf("\n"); } system("pause"); return 0; }