HDU

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思路：经典dfs问题，注意1和最后一个数的和也必须是质数。对于n是奇数的情况，根本不会有结果，因为必定会有两个奇数相邻，他们的和是大于2的偶数就不可能是质数。

AC代码

#include <cstdio>#include <cmath>#include <algorithm>#include <cstring>#include <utility>#include <string>#include <iostream>#include <map>#include <set>#include <vector>#include <queue>#include <stack>using namespace std;#pragma comment(linker, "/STACK:1024000000,1024000000") #define eps 1e-10#define inf 0x3f3f3f3f#define PI pair<int, int> typedef long long LL;const int maxn = 20 + 5;int vis[maxn], ans[maxn], d[maxn], n;void init(int n) {int m = sqrt(n+0.5);memset(vis, 0, sizeof(vis));for(int i = 2; i <= m; ++i) if(!vis[i])for(int j = i*i; j <= n; j+=i) vis[j] = 1;}void dfs(int cnt) {if(cnt == n) {if(!vis[ans[n-1]+1]) for(int i = 0; i < n; ++i) {if(i==n-1) printf("%d\n", ans[i]);else printf("%d ", ans[i]);}return;}for(int i = 2; i <= n; ++i) {if(!d[i] && !vis[ans[cnt-1]+i]) {d[i] = 1;ans[cnt] = i;dfs(cnt+1);d[i] = 0;}}}int main() {init(22);int kase = 0;while(scanf("%d", &n) == 1) {memset(d, 0, sizeof(d));printf("Case %d:\n", ++kase);ans[0] = 1;d[1] = 1;dfs(1);printf("\n");}return 0;}

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