https://vjudge.net/contest/67836#problem/E

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

时间复杂度:$O(n!)$

题解:dfs 要先判断 N ,如果 N 是奇数的情况下不可能构成素数环,而且如果不先排除会报超时

代码:

#include <bits/stdc++.h>
using namespace std; int N;
int vis[30];
int a[30]; int prime(int x) {
for(int i = 2; i * i <= x; i ++)
if(x % i == 0)
return 0;
return 1;
} void dfs(int step) {
if(step == N + 1 && prime(a[1] + a[N])) {
for(int i = 1; i <= N; i ++) {
printf("%d", a[i]);
printf("%s", i != N ? " " : "\n");
}
return ;
}
for(int i = 2; i <= N; i ++) {
if(vis[i] == 0) {
if(prime(i + a[step - 1])) {
vis[i] = 1;
a[step] = i;
dfs(step + 1);
vis[i] = 0;
}
}
}
return ;
} int main() {
int cnt = 0;
while(~scanf("%d", &N)) {
memset(vis, 0, sizeof(vis));
memset(a, 0, sizeof(a));
printf("Case %d:\n", ++cnt);
if(N % 2) {
printf("\n");
continue;
}
a[1] = 1;
vis[1] = 1;
dfs(2);
printf("\n");
}
return 0;
}

  

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