hdu1016

Prime Ring Problem

Time Limit: 2000 MS Memory Limit: 32768 KB

64-bit integer IO format: %I64d , %I64u Java class name: Main

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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

Source

Asia 1996, Shanghai (Mainland China)

#include<iostream>#include<cstdio>#include<algorithm>#include<queue>#include<string.h>#include<math.h>using namespace std;int vis[25];int a[25];int n;int sushu(int su){    for(int i=2;i<=sqrt(su);i++)    {        if(su%i==0) return 0;    }    return 1;}void dfs(int step){    if(step==n+1)    {        if(sushu(a[1]+a[n])){        for(int i=1;i<=n;i++){        if(i==1)        printf("%d",a[i]);        else printf(" %d",a[i]);        }        printf("\n");        }        return ;    }    for(int i=1;i<=n;i++)    {        if(!vis[i]&&sushu(i+a[step-1]))        {            a[step] = i;            vis[i] = 1;            dfs(step+1);            if(i!=1)            vis[i] = 0;        }    }}int main(){    int d=1;    while(~scanf("%d",&n))    {        memset(vis,0,sizeof(vis));        memset(a,0,sizeof(a));        printf("Case %d:\n",d++);        dfs(1);        printf("\n");    }    return 0;}

不要在找到一个排列是去判断相邻的两个数的和是不是素数，这样会超时，虽然只有20个树，应该在dfs 的时候就判断i+a【step-1】是不是素数，如果是素数就继续往下搜。