HDU 1016 Prime Ring Problem（深搜）

#include<iostream>#include<cstdio>#include<cmath>#include<cstdlib>using namespace std;int fun1(int a,int b[]){    int flag=1,k;    for(k=0;k<a;k++)    {        if(b[k]==b[a])//出现过了        {            flag=0;            break;        }    }    return flag;//返回是否出现过}int fun2(int a,int c,int b[]){    int k,flag=1;    for(k=2;k<=sqrt(b[a]+b[a-1]);k++)    {        if((b[a]+b[a-1])%k==0)//判断素数        {            flag=0;            break;        }    }    if(flag&&(a==c-1))    {        for(k=2;k<=sqrt(b[a]+b[0]);k++)        {            if((b[a]+b[0])%k==0)            {                flag=0;                break;            }        }    }    return flag;//返回是否是素数的标记}void fun3(int a,int b_[]){    int k;    printf("1 ");    for(k=1;k<a-1;k++)        printf("%d ",b_[k]);    printf("%d\n",b_[k]);}int main(){    int i=1,n,a[20],c_=0,x,m=0;    memset(a,1,sizeof(a));    while(scanf("%d",&n)!=EOF)    {        i=1,c_=0;        while(1)        {            while(1)            {                a[i]++;                x=1;                if(a[i]==n+1)                {                    a[i]=1;                    i--;                    x=0;                }                if(x&&fun1(i,a))//                {                    if(fun2(i,n,a))                    {                        if(i==n-1)                        {                            if(c_==0)                                printf("Case %d:\n",++m);                            fun3(n,a);                            c_=1;                            i--;                        }                        i++;                        if(a[0]>=2)break;                    }                    if(a[0]>=2)break;                }                if(a[0]>=2)break;            }             if(a[0]>=2)            {                a[0]=1;                printf("\n");                break;            }        }    }    return 0;}

题目概述:

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.

Simple Input:

6

8

Simple 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

题目分析：给定数n(n<20)，在n的所有排列中，记a[1],a]2],...,a[n]，求满足a[i]+a[i+1](1<=i<n)是素数并且a[1]+a[n]也是素数的所有排列。排列按字典序输出。