Prime ring problem

最新推荐文章于 2019-05-21 16:55:45 发布

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最新推荐文章于 2019-05-21 16:55:45 发布

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原文链接：http://www.cnblogs.com/zsyacm666666/p/4680132.html

本文介绍了一种使用深搜求n个数的全排列，并结合素数表快速判断相邻数字之和是否为素数的方法来解决特定问题。通过实例展示了算法实现过程及输出格式要求。

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Description

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.

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

这个先用深搜求n个数的全排列，这样就枚举了每一种情况，在对素数打表，以进行快速判断。

这个UVA上需要注意输出的最后一组数据不能有多余的空行

#include"iostream"
#include"cstring"
#include"cstdio" using namespace std; int n,book[21],a[21]; int ca=1; int dic[52]={0,1,1,1,0,1, 0,1,0,0,0, 1,0,1,0,0, 0,1,0,1,0, 0,0,1,0,0, 0,0,0,1,0, 1,0,0,0,0, 0,1,0,0,0, 1,0,1,0,0, 0,1,0,0,0, 0 };

void dfs(int step)
{
    if ((step == n) && dic[a[1] + a[n]])
    {
        for(int i=1; i<n; i++)  cout<<a[i]<<' ';
        cout<<a[n]<<endl;
        return;
        return;
    }
    for(int i=2; i<=n; i++)
        if(!book[i] && dic[a[step] + i])
        {
            a[step+1]=i;
            book[i]=1;
            dfs(step+1);
            book[i]=0;
        }
}
int main()
{
    int x=1;
    while(scanf("%d",&n)!=EOF)
    {
        a[1]=1;
        if(x>1) cout<<endl;
        x++;
        cout<<"Case "<<ca++<<':'<<endl;
        dfs(1);
    }
    return 0;
}

View Code

也可以用暴力的方法求数全排，不过会超时，暂且贴上来吧

#include"iostream"
#include"cstring"
#include"cstdio"
using namespace std;
int n,book[21],a[21];
int ca=1;
int dic[52]={0,1,1,1,0,1,
0,1,0,0,0,
1,0,1,0,0,
0,1,0,1,0,
0,0,1,0,0,
0,0,0,1,0,
1,0,0,0,0,
0,1,0,0,0,
1,0,1,0,0,
0,1,0,0,0,
0
};

void print(int step)
{
if ((step == n+1))
{
int okk=1;
for(int k=1;k<n;k++) if(dic[a[k]+a[k+1]]==0) okk=0;
if(dic[a[n]+a[1]]==0) okk=0;
if(okk)
{
for(int i=1;i<n;i++) cout<<a[i]<<' ';
cout<<a[n]<<endl;
}
}

for(int i=1;i<=n;i++)
{
int ok=1;
for(int j=1;j<step;j++)
if(a[j]==i) ok=0;
if(ok)
{
a[step]=i;
print(step+1);
}
}
}

int main()
{
int x=1;
while(scanf("%d",&n)!=EOF)
{
memset(a,0,sizeof(a));
a[1]=1;
if(x>1) cout<<endl;x++;
cout<<"Case "<<ca++<<':'<<endl;
print(2);
}
return 0;
}