Prime Ring Problem
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14061 Accepted Submission(s): 6390
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.
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
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
这道题可以通过深度搜索的办法来解决，下面是我的代码，已经通过了AC
//1016
#include <iostream>
#include <vector>
using namespace std;
int result[20];//存放计算的结果
vector<vector<int> > results;
int n;
//int primes[] = {2,3,5,7,11,13,17,19,23,29,31,37};//共有13个40以内的素数
bool prime[] = {0,0 ,1 ,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,1};
bool isPrime(int num)
{
return prime[num];
}
bool isIn(int list[], int len, int num)
{
for (int i = 0; i < len; ++ i)
if (list[i] == num)
return true;
return false;
}
void findCircle(int i)
{
int count =0;
for (int j = 2; j <= n; ++ j)
if (!isIn(result,i,j) && isPrime(result[i-1] + j))
{
result[i] = j;
count ++;
findCircle(i + 1);
}
if (count == 0)
return;
if (i == n - 1 && isPrime(1 + result[n - 1]))
{
vector<int > temp ;
for (int k = 0; k < n ; ++ k)
temp.push_back(result[k]);
results.push_back(temp);
}
}
int main()
{
int Case = 0;
while(cin >> n)
{
Case ++;
results.clear();
result[0] = 1;
findCircle(1);
cout << "Case "<<Case<<":"<<endl;
for (int i = 0; i < results.size(); ++ i )
{
cout << results[i][0]<< " ";
for (int j = 1; j < results[i].size() - 1; ++ j)
cout << results[i][j]<< " ";
cout << results[i][results[i].size() - 1] ;//这里特别注意，最后一个不输出空格
cout << endl;
}
cout << endl;
}
return 0;
}