[2016-02-19] [UVA] [524] [Prime Ring problem]

[2016-02-19] [UVA] [524] [Prime Ring problem]

UVA-524Prime Ring problem Time Limit: 3000MS Memory Limit: Unknown 64bit IO Format: %lld &%llu Submit Status Description A Ring is composed of n (even number) circles as shown in diagram. Put natural numbers into each circle separately, and the sum of numbers in both adjacent circles should be a prime. Note: the number of first circle should always be 1. InputN (0 < n <=) 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. You is to write a program, that completes above process. Sample Input 68 Sample Output Case 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 Miguel A. Revilla 1999-01-11

• Time: 2016-02-19-21:16:47
  
• Topic Number: UVA 524
  
• Title: Given the number n, the number of output 1~n is arranged so that the sum of the adjacent numbers is the prime number (the first connection)
  
• Method: Dfs enumeration answer, backtracking optimization.
  
• Problems encountered during the problem solving process:
  
• Format problem: There is no blank line after the last set of data, otherwise WA
• There are no spaces after the last number in each line, otherwise the PE

#include <vector> #include <list> #include <map> #include <set> #include <deque> #include <queue> #include <stack> #include <bitset> #include <algorithm> #include <functional> #include <numeric> #include <utility> #include <sstream> #include <iostream> #include <iomanip> #include <cstdio> #include <cmath> #include <cstdlib> #include <cctype> #include <string> #include <cstring> #include <cstdio> #include <cmath> #include <cstdlib> #include <ctime> using namespace std; typedef Long long LL; #define The CLR (x , Y) memset ((×), (y),sizeof((x))) #define for (x, y , z) for (int.) = (y);(x) < (*);(x) + +) #define FORD ( z/y) for(int (x) = (y);(x) >= (*);(x)--) #define FOR2 ( z) for((x) = (y);(x) < (*);(x) + +) #define FORD2 ( x, Y, z) for((×) = (y);(x) >=;(x) const int maxn =; int ISPRI[MAXN*2],N,VIS[MAXN],RES[MAXN]; void Setpri () { ispri[1] = 1;CLR (ispri,-1);For (i,2,maxn*2) { if(Ispri[i]) {For (int j = i + i;j < maxn*2;j + = i)Ispri[j] = 0;                }        }}void dfs (int cur) { if(cur = = N) { if(!ispri[res[0] + res[n-1]]) return ;printf ("%d", Res[0]);For (i,1,n) {printf ("%d", Res[i]);                }puts (""); return ;        }For (i,2,n+1) { if(!vis[i] && ispri[res[cur-1] + i]) {Vis[i] = 1;Res[cur] = i;DFS (cur + 1);Vis[i] = 0;                }        }}int Main () { int cntcase = 0;Setpri (); while (~scanf ("%d", &n)) { if(cntcase) puts ("");printf ("Case%d:\n", ++cntcase);CLR (vis,0);vis[1] = res[0] = 1;DFS (1);        } return 0;}

From for notes (Wiz)

[2016-02-19] [UVA] [524] [Prime Ring problem]