HDU 1016 Prime Ring Problem

最新推荐文章于 2021-06-01 18:39:59 发布

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最新推荐文章于 2021-06-01 18:39:59 发布

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本文介绍了一个使用深度搜索算法解决特定数列排列问题的方法，目标是在给定范围内找到所有相邻元素和为素数的环形排列。通过构建素数表并递归地尝试组合，实现问题的求解。

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

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

题意：给你一个数n，要输出一个环，要求为环的每一个相邻的2个数之和为素数。以1为环的起点，按逆时针方向排列，输出所有的可能！

思路：深度搜索！

AC代码

#include<stdio.h>
#include<string.h>
int flag[25],a[25],sumprim[25];
int n;
void makeprime(){			//打素数表
	int ok;
	memset(sumprim,0,sizeof(sumprim));
	for(int i=2; i<=(n<<1); i++){
		ok = 1;
		for(int j=2; j*j<=i; j++) if(i % j == 0) ok = 0;
		if(ok) sumprim[i] = 1;
	}
}
void dfs(int cur)
{
    if(cur==n&&sumprim[a[0]+a[n-1]])	//搜索树的树顶，还要判断第一个数与最后一个数的和
    {
        for(int i=0;i<n;i++){
            if(i>0)printf(" ");
            printf("%d",a[i]);
        }
        printf("\n");
    }
    else{
        for(int i=2;i<=n;i++){
            if(!flag[i]&&sumprim[i+a[cur-1]])	//i未使用并且与前一个数的和为素数
            {
                flag[i]=1;	//标记使用
                a[cur]=i;	//将符合条件的i赋值给a数组
                dfs(cur+1);
                flag[i]=0;
            }
        }
    }
}
int main()
{
    int cas=0;
    while(~scanf("%d",&n))
    {
		cas+=1;
		printf("Case %d:\n",cas);
        a[0]=1;
        makeprime();
        dfs(1);
		printf("\n");
    }
    return 0;
}