POJ-2034 Anti-prime Sequences
题意:给出三个数n,m,d,求n到m间满足相邻2~d个数的和是非素数的序列。
思路:dfs。
题目链接:http://poj.org/problem?id=2034
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <cstdio> 2 #include <cstring> 3 #include <cstdlib> 4 #include <cmath> 5 #include <string> 6 #include <algorithm> 7 #include <iostream> 8 using namespace std; 9 const int N=10010; 10 11 bool prime[N],flag,vis[N]; 12 int b[N]; 13 int n,m,d; 14 15 void dfs(int cnt,int t){ 16 if(flag) return ; 17 if(cnt>=2){ 18 int sum=0; 19 int i,k; 20 for(i=cnt,k=0;k<d&&i>=1;k++,i--){ 21 sum+=b[i]; 22 if(k!=0&&!prime[sum]) return ; 23 if(k==d-1) t++; 24 } 25 } 26 if(t==m-n-d+2){ 27 flag=true; 28 for(int i=1;i<=m-n+1;i++) printf(i==m-n+1?"%d\n":"%d,",b[i]); 29 return ; 30 } 31 for(int i=n;i<=m;i++){ 32 if(!vis[i]){ 33 vis[i]=true; 34 b[cnt+1]=i; 35 dfs(cnt+1,t); 36 vis[i]=false; 37 } 38 } 39 } 40 41 int main(){ 42 43 // freopen("data.in","r",stdin); 44 // freopen("data.out","w",stdout); 45 46 memset(prime,false,sizeof(prime)); 47 for(int i=2;i<=N;i++) 48 if(!prime[i]) 49 for(int j=2;i*j<=N;j++) 50 prime[i*j]=true; 51 while(scanf("%d%d%d",&n,&m,&d),n||m||d){ 52 memset(b,0,sizeof(b)); 53 memset(vis,false,sizeof(vis)); 54 flag=false; 55 dfs(0,0); 56 if(!flag) puts("No anti-prime sequence exists."); 57 } 58 return 0; 59 }