Description
A ring is composed of n (even number) circles as shown in diagram. Put natural numbers 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 ≤ 16)
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. You are to write a program that completes above process.
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
思路
题意：
输入正整数n，把整数1，2，3，……，n组成一个环，使得相邻两个整数之和均为素数。输出时序列头为1开始的序列，同一个环应恰好输出一次
题解:
#include<bits/stdc++.h>
using namespace std;
const int maxn = 40;
bool is_prime[maxn],vis[maxn];
int n,a[maxn];
void dfs(int cur)
{
if (cur == n && is_prime[a[0] + a[n-1]]) //递归边界，因为是环，所以还要测试第一个和最后一个
{
printf("%d",a[0]);
for (int i = 1;i < n;i++) printf(" %d",a[i]);
printf("\n");
}
else
{
for (int i = 2;i <= n;i++) //尝试放置每个数i
{
if (!vis[i] && is_prime[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;
vis[i] = true; //设置使用标志
dfs(cur+1);
vis[i] = false; //清楚标志
}
}
}
}
int main()
{
//提前预处理素数表
memset(is_prime,true,sizeof(is_prime));
is_prime[0] = is_prime[1] = false;
for (int i = 2;i < maxn;i++)
{
if (is_prime[i])
{
for (int j = 2 * i;j < maxn;j += i)
{
is_prime[j] = false;
}
}
}
int tcase = 0;
while (~scanf("%d",&n))
{
memset(vis,false,sizeof(vis));
memset(a,0,sizeof(a));
if (tcase) printf("\n");
printf("Case %d:\n",++tcase);
a[0] = 1;
dfs(1);
}
return 0;
}