「UVA524」 Prime Ring Problem 质数环

Description

---

输入正整数n,把整数1,2,…,n组成一个环，使得相邻两个整数之和均为素数。输出时，从整数1开始逆时针排列。同一个环恰好输出一次。n<=16.

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

Put natural numbers1,2……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<=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

题解

---

n<=16，数据很小，可以暴搜，途中用stack记录经历的点

这道题的难点是输出格式啊QAQ！！！

example:

Case 1:
1 4 3 2 5 6
1 6 5 2 3 4

Case 2://这里没有空格
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//这里也没有空格
//这里有回车

//luogu的原题中输出格式是崩坏的啊！！！QAQ

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int n;
bool sf[21];
int stack[21];
int flag;
bool prime(int x)
{
    for(int i=2;i<=sqrt(x);++i)
    if(x%i==0)return 0;
    return 1;
}
int tot=0;
void print()
{
    if(!prime(1+stack[n]))return;
    if(!flag){
        if(tot)cout<<endl;
        flag++;
        printf("Case %d:",++tot);
    }
    cout<<endl;
    cout<<1;
    for(int i=2;i<=n;++i)
    cout<<" "<<stack[i];
}
void search(int x)
{
    for(int i=2;i<=n;++i)
    if(!sf[i]&&prime(stack[x-1]+i))
    {
        stack[x]=i;
        sf[i]=1;
        if(x==n)print();
        else search(x+1);
        sf[i]=0;
    }
    return;
}
int main()
{
    while(cin>>n)
    {
    memset(sf,0,sizeof(sf));
    flag=0;
    stack[1]=1;
    sf[1]=1;
    search(2);
    cout<<endl;
    }
}