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
Solution
題目大意:輸入一個數字n,輸出滿足相鄰的數組相加爲素數的序列 序列長度爲n。
解題思路:DFS,n不超過20,先找出50以內的素數,用DFS遍歷。
Code
#include <stdio.h>
#include <string.h>
#define maxn 50
int is_prime[maxn], a[maxn], n, vis[maxn];
void get_prime(int n) {
memset(is_prime, 0, sizeof(is_prime));
is_prime[0] = is_prime[1] = 1;
for (int i = 2;i <= n;i++)
if (!is_prime[i])
{
for (int j = 2;j*i <= n;j++) is_prime[j*i] = 1;
}
}
void dfs(int step)
{
int i;
if (step == n + 1 && !is_prime[a[n] + a[1]])//結束
{
for (i = 1; i < n; i++)
printf("%d ", a[i]);
printf("%d\n", a[n]);
return;
}
for (i = 2; i <= n; i++)
{
if (!vis[i] && !is_prime[i + a[step - 1]])
{
a[step] = i;
vis[i] = 1;
dfs(step + 1);
vis[i] = 0;//回溯
}
}
}
int main(int argc, char const *argv[])
{
int cas = 1;
get_prime(maxn);
while (~scanf_s("%d", &n))
{
memset(vis, 0, sizeof(vis));
printf("Case %d:\n", cas++);
a[1] = 1;
dfs(2);
printf("\n");
}
return 0;
}