Prime Ring (C-Brute force solution)

1#include <cstdio>2#include <iostream>3 using namespacestd;4 5 intN, a[ -], isp[ -], vis[ -];6 7 int  is(intx)8 {   9      for(inti =2; I*i <= x; i++)  Ten         if(x% i = =0)  One             return 0;  A         return 1;  - } -  the  -  - voidDfsintcur) - { +     if(cur==n&&isp[a[0]+a[n-1]]) -     { +          for(intI=0; i<n;i++) Aprintf"%d", A[i]); atprintf"\ n"); -     } -     Else  for(intI=2; i<=n;i++) -         if(!vis[i]&&isp[i+a[cur-1]]) -         { -A[cur] =i; inVis[i] =1;  -DFS (cur+1);  toVis[i] =0; +         } - } the  * intMain () $ {Panax Notoginseng     //int n; -     intm=0;  the      while(SCANF ("%d", &n)! =EOF) +     { A      for(intI=2; i<=n*2; i++) theIsp[i] = is(i);  +memset (Vis,0,sizeof(VIS));  -a[0] =1; $m++; $printf"Case %d:\n", m); -Dfs1);  -printf"\ n"); the     } -     return 0;Wuyi}

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Prime Circle (Violence) (Purple book 194 pages)

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.

Input N (0 < n <=)

N (0 < n <= 16)

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.

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

Test instructions
Enter a positive integer n to make the 1-n a ring, making the adjacent two integers a prime number. The output starts with an integer of 1 and is arranged counterclockwise. Just can make a ring output once,N (0 < n <=)


Analysis:
Each ring corresponds to an arrangement of 1--n, but the total number of permutations is up to 16! =2*10^13, the normal notation will time out and apply backtracking.
Dfs. Depth-first traversal.

Code:

1#include <cstdio>2#include <iostream>3 using namespacestd;4 5 intN, a[ -], isp[ -], vis[ -];6 7 int  is(intx)8 {   9      for(inti =2; I*i <= x; i++)  Ten         if(x% i = =0)  One             return 0;  A         return 1;  - } -  the  -  - voidDfsintcur) - { +     if(cur==n&&isp[a[0]+a[n-1]]) -     { +          for(intI=0; i<n;i++) Aprintf"%d", A[i]); atprintf"\ n"); -     } -     Else  for(intI=2; i<=n;i++) -         if(!vis[i]&&isp[i+a[cur-1]]) -         { -A[cur] =i; inVis[i] =1;  -DFS (cur+1);  toVis[i] =0; +         } - } the  * intMain () $ {Panax Notoginseng     //int n; -     intm=0;  the      while(SCANF ("%d", &n)! =EOF) +     { A      for(intI=2; i<=n*2; i++) theIsp[i] = is(i);  +memset (Vis,0,sizeof(VIS));  -a[0] =1; $m++; $printf"Case %d:\n", m); -Dfs1);  -printf"\ n"); the     } -     return 0;Wuyi}

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Prime Ring (C-Brute force solution)