Hdoj 1016 HDU 1016 prime ring problem ACM 1016 in HDU

Question address: Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 1016 Description:
Prime ring Problem

Time Limit: 4000/2000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)
Total submission (s): 6150 accepted submission (s): 2745

Problem descriptiona ring is compose of N circles as shown in digoal. put Natural Number 1, 2 ,..., N into each circle separately, and the sum of numbers in two adjacent circles shoshould be a prime.

Note: The number of First Circle shoshould 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 are to write a program that completes abve 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 4 case 2:1 2 3 8 5 6 7 41 2 5 3 4 7 61 4 6 5 8 3 21 6 7 4 3 8 5 2 Question Analysis: Typical DFS questions, no pruning required, direct The question output is Bt. Generally, the carriage return is returned, and the first vehicle return is if (n = 1; after several PE operationsProgramLastOutput two carriage returns to a, ym ...........  Code As follows:/* miyu original, Please Note: Reproduced from __________ White House http://www.cppblog.com/MiYuAuthor by: miyutest: Program: */# include <iostream> using namespace STD; bool prim [25]; int res [25]; bool hash [25]; int N; void setprim () {memset (Prim, 0, sizeof (prim )); prim [2] = prim [3] = prim [5] = prim [7] = prim [11] = prim [13] = prim [17] = prim [19] = prim [23] = true ;} bool DFS (INT num, int N) {res [N] = num; If (n> N) {Return false;} If (n = n-1) {for (INT I = 2; I <= N; ++ I) {If (prim [num + I] & prim [I + 1] &! Hash [I]) {res [n + 1] = I; for (INT I = 1; I <= N; ++ I) printf (I = 1? "% D": "% d", Res [I]); putchar ('\ n') ;}}for (INT I = 2; I <= N; ++ I) {If (prim [num + I] &! Hash [I]) {hash [I] = true; DFS (I, n + 1); hash [I] = false ;}} return false ;} int main () {setprim (); int CA = 1; while (CIN> N) {sizeof (hash, 0, sizeof (hash); printf ("case % d: \ n ", CA ++); hash [1] = true; DFS (1, 1); putchar ('\ n');} return 0 ;}