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 小范围数据，当然是递归+打表啦～这里掌握某种规律后也可以提高搜索效率，比如
素数环:给定n，1~n组成一个素数环，相邻两个数的和为素数。
首先偶数(2例外，但是本题不会出现两个数的和为2)不是素数，
所以素数环里奇偶间隔。如果n是奇数，必定有两个奇数相邻的情况。
所以当n为奇数时，输出“No Answer”。
当n == 1时只1个数，算作自环，输出1
所有n为偶数的情况都能变成奇偶间隔的环-----所以都有结果。
#include<stdio.h>
#include<string.h>
int n,c=,i;
int a[],b[];
int jo(int a)
{
return a%==?:;
}
int prime[]={,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,};
void dfs(int step)
{
int i;
if(step>n&&prime[a[]+a[n]]){
for(i=;i<=n;++i){
if(i==) printf("%d",a[i]);
else printf(" %d",a[i]);
}
printf("\n");
return;
}
if(jo(a[step-])){
for(i=;i<=n;i+=){
if(!b[i]&&prime[a[step-]+i]){
b[i]=;
a[step]=i;
dfs(step+);
b[i]=;
}
}
}
else{
for(i=;i<=n;i+=){
if(!b[i]&&prime[a[step-]+i]){
b[i]=;
a[step]=i;
dfs(step+);
b[i]=;
}
}
}
}
int main()
{
while(~scanf("%d",&n)){
memset(a,,sizeof(a));
memset(b,,sizeof(b));
a[]=;b[]=;
if(n==) printf("Case %d:\n1\n\n",++c);
else if(!jo(n)){
printf("Case %d:\n",++c);
dfs();
printf("\n");
}
}
return ;
}