Hangzhou power 1016 prime ring Problem

Prime ring Problem Time Limit: 4000/2000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)
Total submission (s): 25187 accepted submission (s): 11246


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 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) Pay attention to the boundary and end-to-end inference to determine whether the conditions are met.

# Include <iostream> # include <cmath> # include <cstring> # include <algorithm> # define m 25 using namespace STD; int vis [m], n, ans [m], T = 0; int sum (int A, int B) // function for determining prime numbers {int K = 0, sum, I; sum = A + B; for (I = 2; I <sum; I ++) {If (sum % I = 0) {k = 1; return 0 ;}} if (! K) return 1;} void DFS (int x) {int I; If (x = N & sum (ANS [0], ANS [n-1]) // determine whether the first and end of a recursive boundary meet the condition {for (I = 0; I <n; I ++) {if (I = 0) cout <ans [0]; else cout <"<ans [I];} cout <Endl;} else {for (I = 2; I <= N; I ++) {If (! Vis [I] & sum (I, ANS [x-1]) {ans [x] = I; vis [I] = 1; // mark used count DFS (x + 1); vis [I] = 0; // restore used count }}} int main () {int I, j; while (CIN> N) {T ++; memset (VIS, 0, sizeof (VIS )); cout <"case" <t <":" <Endl; vis [1] = 1; ans [0] = 1; DFS (1 ); cout <Endl; // There is an empty line between each example} return 0 ;}

Hangzhou power 1016 prime ring Problem