Prime Ring Problem
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34799 Accepted Submission(s): 15411
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.
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.
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
Source
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这是我第一道接触的搜索题,当时参考学长的代码写,却始终不能理解代码的意思,于是就一遍一遍第敲,直到整个代码都背下来了,后来接触的搜索逐渐增加,终于理解了这道题的代码,每次看到这道题,就能想起当年苦逼的敲代码。。。
题意:输入一个数n,代表从1到n个数,排序排成一个环,使得相邻两个数之和为素数(第一个和最后一个数也要满足),如果有多组情况,必须按从小到大的顺序输出,
附上代码:
1 #include <iostream> 2 using namespace std; 3 int a[30]= {0,1},b[30],sushu[50],n; //a表示素数环中的数字,b用来标记,sushu标记是否是素数 4 void DFS(int s,int k) //s代表当前循环到第几个数,k标记此时是否是第一次进人深搜函数 5 { 6 int i,j; 7 if(s>n) 8 { 9 if(sushu[a[1]+a[n]]) //最后一个数字和第一个数字的和 10 { 11 for(i=1; i<=n; i++) //全部满足,输出这组数据的全部数 12 { 13 if(i>1) cout<<" "; 14 cout<<a[i]; 15 } 16 cout<<endl; 17 } 18 else return; //若最后一组数据不满足,则返回上一次循环 19 } 20 for(i=2; i<=n; i++) 21 { 22 if(k) 23 for(j=1; j<30; j++) //第一次进来,所有的数字都可以使用 24 b[j]=1; 25 if(sushu[i+a[s-1]]&&b[i]) //i表示当前数字,a[s-1]表示上一个数字。两个数字之和为素数,且没有出现过i这个数字 26 { 27 a[s]=i; //s位置上的数字记录为i 28 b[i]=0; //0表示为已使用 29 DFS(s+1,0); //一个满足条件,进入下一个数字的选择 30 b[i]=1; //若返回上一层循环,标记已使用了的数字必须还原为未使用 31 } 32 } 33 } 34 int main() 35 { 36 int i,j,t=1; 37 for(i=0; i<50; i++) 38 sushu[i]=1; 39 sushu[0]=sushu[1]=0; 40 for(i=2; i<=25; i++) 41 if(sushu[i]) 42 for(j=i+i; j<50; j+=i) 43 sushu[j]=0; //素数打表,减少代码运行时间,素数为1,非素数为0 44 while(cin>>n) 45 { 46 cout<<"Case "<<t++<<":"<<endl; 47 DFS(2,1); 48 cout<<endl; 49 } 50 return 0; 51 }