Prime Ring problemTime limit:4000/2000 MS (java/others) Memory limit:65536/32768 K (java/others) Total submission (s): 40419 Accepted Submission (s): 17843
Problem Descriptiona 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 the adjacent circles should is a PR Ime.
Note:the number of first circle should always be 1.
INPUTN (0 < n < 20).
Outputthe 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 is to write a program, that completes above process.
Print a blank line after each case.
Sample Input68
Sample OutputCase 1:1 4 3 2 5 a 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
Sourceasia 1996, Shanghai (Mainland China) #include <iostream> #include <cstdio> #include <cstring>using namespace Std;int prime[] = { 3,5,7,11,13,17,19,23,29,31,37};//when n is less than 20 o'clock the sum of the maximum of two numbers is 37int primeflag[39];//is used to quickly determine whether a prime number, when the element value is 1 o'clock the current count is the prime of int visited[21 ];//is used to mark whether the current number has been used, a value of 1 is used, 0 means that an array with an int ans[21];//stored answer is not used, a void Dfs (int n)//Deep search and backtracking, in order to avoid timeouts using nonrecursive method {int flag = 0;// Represents the previous position of the current number of fills int i = 2;//from 2 to n searches for numbers that can be filled in while (flag >= 0) {///when flag<0 indicates that all solutions have been found while (i <= N) {//when i& Gt;n indicates the current position all numbers have been tried if (visited[i]! =1 && primeflag[ans[flag] + i] = = 1) {//When the current number has not been tried and satisfies the sum of the last number The number of primes is filled with ans[++flag] = i; Visited[i] =1; i = 2;//after each fill in a number, the loop variable is initialized to 2 to indicate the next position to restart the search} else i++; } if (flag = = = N1 && Primeflag[ans[flag] + 1] = = 1) {//when the last while loop has ended and flag has reached the final position and the first number meets the last number and is prime, indicates the current There is a solution for (int j = 0; J < N1; J + +) {cout << ans[j] << ""; } cout << ans[n1] << endl;//no space after last number} i = ans[flag];//The current position is saved when backtracking visited[i] = 0;//the Number remarked as not used i++;//number I have tried on the flag, the next cycle as long as the number from i++ to avoid repeating the dead loop flag;//back to a position to refill the number at the flag+1 locationbacktracking}}i NT main () {int n,ncase = 0; for (int i = 0; i < one; i++) {primeflag[prime[i]]= 1; } while (CIN >> N) {memset (visited,0,sizeof (visited)); memset (ans,0,sizeof (ans)); Ans[0] = 1; VISITED[1] =1; ncase++; cout << "Case" << ncase << ":" << Endl; if (n% 2 = = 0)//can satisfy test instructions produce output only if n is an even number, because the feasible solution must be odd and even alternate arrangement, when n is oddnumbered, there will be one more odd Dfs (n); cout <<endl; } return 0;}
