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  • UVA

    原题

    Description

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    A ring is composed of n (even number) circles as shown in diagram. Put natural numbers $1, 2, dots, n$ 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 <= 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;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/shawn-ji/p/4696557.html
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