UVA524 素数环 Prime Ring Problem

题目OJ地址：

https://www.luogu.org/problemnew/show/UVA524

hdu oj 1016：  https://vjudge.net/problem/HDU-1016

zoj 1457  ：https://vjudge.net/problem/ZOJ-1457

题意翻译

输入正整数n,把整数1,2,...,n组成一个环，使得相邻两个整数之和均为素数。输出时，从整数1开始逆时针排列。同一个环恰好输出一次。.

多组数据，读入到EOF结束。

第i组数据输出前加上一行Case i:

相邻两组数据中间加上一个空行。

Prime Ring Problem

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.

Inputn (0 < n < 20). 
OutputThe 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

 #include<stdio.h>
 #include<stdlib.h>
 #include<math.h>
 int N;
 int b[]={};
 int total=,a[]={};
 int search(int);        //回溯过程
 int print();            //输出方案
 int pd(int x,int y);    //判断素数x+y是否质数

 int main()
 {
     int t=;
     while(scanf("%d",&N)!=EOF)
     {
         a[]=;
         for(int i=;i<;i++) a[i]=;
         for(int i=;i<;i++) b[i]=;
         t++;
         printf("Case %d:\n",t);
         search();
         printf("\n");
     }
     //printf("%d\n",total);                    //输出总方案数
 }
 int search(int t)
 {
     int i;
     for(i=;i<=N;i++)           //有20个数可选
      if((!b[i])&&pd(a[t-],i))  //判断与前一个数是否构成素数及该数是否可用
      {
          a[t]=i;
          b[i]=;
          if (t==N) { if(pd(a[N],a[])==) print();}
          else search(t+);
          b[i]=;
      }
 }
 int print()
 {
    int j;
    total++;
    //printf("<%d>",total);
    printf("%d",a[]);
    for(j=;j<=N;j++)
        printf(" %d",a[j]);
    printf("\n");
 }
 int pd(int x,int y)
 {
     int k=,i=x+y;
     while(k<=sqrt(i)&&i%k!=) k++;
     if(k>sqrt(i)) return ;
     else return ;
 }

本题目分析见  https://www.cnblogs.com/huashanqingzhu/p/4747009.html

这里要注意：洛谷OJ的测试数据比较弱，n最大是16.  hdu oj和uva oj原题的n是到20的。