hdu 1016 Prime Ring Problem

题目链接

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 33354    Accepted Submission(s): 14760

Problem Description

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

68

 

Sample Output

Case 1:1 4 3 2 5 61 6 5 2 3 4Case 2:1 2 3 8 5 6 7 41 2 5 8 3 4 7 61 4 7 6 5 8 3 21 6 7 4 3 8 5 2

<span style="font-size:18px;color:#006600;background-color: rgb(255, 255, 255);">#include<iostream>#include<cstdio>#include<string>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#include<algorithm>using namespace std;#define PI acos-1.0#define N 30int n, ans, flag, prime[N];bool vis[N];int isprime (int x, int y);void DFS (int cur);int main (){    ans = 0;    while (cin >> n)    {        ans++;        flag = 0;        memset (vis, false, sizeof (vis));        prime[1] = 1;        DFS (2);        cout << endl;    }}void DFS (int cur){    vis[1] = true;    if (cur > n && isprime (prime[1], prime[n]))    {        if (!flag)        {            cout << "Case " << ans << ':' << endl;            flag = 1;        }        for (int i=1; i<n; i++)            cout << prime[i] << ' ';        cout << prime[n] << endl;    }    else    {        for (int i=2; i<=n; i++)        {            prime[cur] = i;            if (!vis[i] && isprime (prime[cur], prime[cur-1]))            {                vis[i] = true;                DFS (cur + 1);                vis[i] = false;            }        }    }}int isprime (int x, int y){    int sum = x + y;    for (int i=2; i<sum; i++)        if (sum % i == 0)            return 0;    return 1;}</span>