hdu1016 Prime Ring problem

Prime Ring problem

Time limit:4000/2000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 41154 Accepted Submission (s): 18221

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 6 5 2 3 4 case 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 Direct violence search, I guess. Need pruning, because there is no pruning, then test the n=20, half a day no results. So I thought about it. Pruning: 1 The starting number is not 1, indicating that all the combinations have been found, the rest is the rotation of the result of the Ring 2 if the current number and its previous addition is not a prime, you do not have to continue to look down.

#include <iostream>#include<stdio.h>using namespacestd;intN;intnum[ -];BOOLvis[ -];intprime[ -];voidDfsinta) {    if(a>n+1)return; if(num[1]!=1)return; if(a==n+1)    {         for(intI=1; i<n+1; i++)        {            if(!prime[num[i%n]+num[(i+1)%n])return; }        //if (num[1]==1)// {         for(intI=1; i<n; i++) printf ("%d", num[i%n]); printf ("%d", num[0]); printf ("\ n"); // }        return; }     for(intI=1; i<=n; i++)    {        if(!Vis[i]) {Num[a%n]=i; if(a>1)            {                if(!prime[num[a%n]+num[(A-1)%N]])Continue; }            //cout<<num[a%n]<< "" <<endl;vis[i]=true; DFS (a+1); Vis[i]=false; }    }}intMain () { for(intI=0; i< -; i++) prime[i]=true;  for(intI=2; i< -; i++)         for(intj=i<<1; j< -; j+=i) {if(Prime[j]) prime[j]=false; }    intt=1;  while((scanf ("%d", &n))! =EOF) {         for(intI=0; i< -; i++) {Num[i]=0; Vis[i]=false; } num[1]=1; printf ("Case %d:\n", t++); DFS (1); printf ("\ n"); }    return 0;}

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Sourceasia 1996, Shanghai (Mainland China)

Direct Violence Search 9

hdu1016 Prime Ring problem