UVA

原题

Description

 

A ring is composed of n (even number) circles as shown in diagram. Put natural numbers  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

 分析：

　　　　题目可以使用dfs求解

这一编写一个高效的判断素数的函数   如下

int is_prime(int x) {
    for (int i = 2; i*i  <= x; i++)
    if (x % i == 0) return 0;
    return 1;
}

我的代码

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int is_prime(int x) {
    for (int i = 2; i*2 <= x; i++)
    if (x % i == 0) return 0;
    return 1;
}

int n, A[50], isp[50], vis[50];
void dfs(int cur) {
    if (cur == n && isp[A[0] + A[n - 1]]) {
        for (int i = 0; i < n; i++) {
            if (i != 0) printf(" ");
            printf("%d", A[i]);
        }
        printf("
");
    }
    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 main() {
    int kase = 0;
    while (scanf("%d", &n) == 1 && n > 0) {
        if (kase > 0) printf("
");
        printf("Case %d:
", ++kase);
        for (int i = 2; i <= n * 2; i++) isp[i] = is_prime(i);
        memset(vis, 0, sizeof(vis));
        A[0] = 1;
        dfs(1);
    }
    return 0;
}