HDU 1016 Prime Ring Problem(简单回溯)
2014-05-27 16:35
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Problem B
Time Limit : 4000/2000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 11 Accepted Submission(s) : 8
Problem Description
A 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 two adjacent circles should be a prime. Note: the number of first circle should always be 1.
Input
n (0 < n < 20).
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. Print solutions in lexicographical order. You
are to write a program that completes above process. Print a blank line after each case.
Sample Input
6 8
Sample Output
Case 1: 1 4 3 2 5 6 1 6 5 2 3 4 Case 2: 1 2 3 8 5 6 7 4 1 2 5 8 3 4 7 6 1 4 7 6 5 8 3 2 1 6 7 4 3 8 5 2
#include<iostream> #include<cstdio> #include<cmath> #include<cstring> using namespace std; const int MAXN=25; int f[MAXN]; bool done[MAXN]; int csc=0; int is_prim(int x) { int i; int k=sqrt(x); for(i=2;i<=k;i++) if(x%i==0)break; if(i>k)return 1; return 0; } void select(int n,int cur) { int i; if(cur==n+1&&is_prim(f[1]+f )) { for(i=1;i<=n;i++) if(i!=n)printf("%d ",f[i]); else printf("%d\n",f[i]); } if(cur==n+1) return; for(i=1;i<=n;i++) { if(!done[i]&&is_prim(f[cur-1]+i)) { f[cur]=i; done[i]=1; select(n,cur+1); done[i]=0; } } } int main() { // freopen("123.txt","r",stdin); int n,i; while(~scanf("%d",&n)) { printf("Case %d:\n",++csc); memset(done,0,sizeof(done)); f[1]=1; done[1]=1; select(n,2); done[1]=0; printf("\n"); } return 0; }
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