By
yusijia
Updated:
题目来源:《算法竞赛入门经典》 作者:刘汝佳
题目:
把1~n,连续的放到一个环里,使相邻的数字和为素数,输出所有结果。
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#include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> #include <cctype> using namespace std;
int is_prime(int n) { int i; int k = sqrt(n); for(i = 2; i <= k; i++){ if(n % i == 0) break; } if(i <= k || n == 1) return 0; else return 1; }
int main() { int a[30] = {0}, isp[30] = {0}; int n; scanf("%d", &n); for(int i = 2; i <= 2 * n; i++){ isp[i] = is_prime(i); } for(int i = 0; i < n; i++) a[i] = i + 1; do{ int ok = 1; for(int i = 0; i < n; i++) if(!isp[a[i] + a[(i + 1) % n]]){ ok = 0; break; } if(ok){ for(int i = 0; i < n; i++) printf("%d ", a[i]); printf("\n"); }
}while(next_permutation(a + 1, a + n)); return 0; }
#include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> #include <cctype> using namespace std;
int vis[30] = {0}, n; int a[30] = {0}, isp[30] = {0};
void dfs(int cur) { if(cur == n && isp[a[0] + a[n - 1]]){ for(int i = 0; i < n; i++) printf("%d ", a[i]); printf("\n"); } else{ for(int i = 2; i <= n; i++){ if(!vis[i] && isp[i + a[cur - 1]]){ a[cur] = i; vis[i] = 1; dfs(cur + 1); vis[i] = 0; } } } }
int is_prime(int n) { int i; int k = sqrt(n); for(i = 2; i <= k; i++){ if(n % i == 0) break; } if(i <= k || n == 1) return 0; else return 1; }
int main() { scanf("%d", &n); for(int i = 1; i <= 2 * n; i++){ isp[i] = is_prime(i); } a[0] = 1; vis[1] = 1; dfs(1); return 0; }
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