Find Terrorists

Time limit: 5 seconds

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The Prime Minister and his Accumulated Council of Ministers(ACM) are trying hard to find all possible terrorist locations. In his dream, the Prime Minister gets a message from God suggesting that the answer to all terrorist problems are numbers(say one such number is X) such that the number of factors of X(including 1 and X) is prime. These numbers supposedly contain the encrypted locations of terrorists. Since the ACM has no programmer, the Prime Minister needs your help in finding out such numbers.

Note:- 1 is not considered a prime number .

Input

The first line of input will contain an integer T <= 20 denoting the number of test cases.

T lines follow, one per test case.

Each test case will be a line formatted as "L H" where L and H are integers and 0

Output

Output one line per case a space separated list of all integers(sorted ascending) lying between L and H(both inclusive) such that the number of factors of each integer is prime.In case no such integer exist output -1.

Sample Input

31 11 22 5

Sample Output

-122 3 4 5

一个素数的因子个数必定是2，2是素数，所以凡是素数都符合条件。对于合数，由于合数能分解为素因子的幂的乘积，比如15 = 3^1 * 5^1；18 = 2^1 * 3^2。分解式中次数加1的乘积就是合数的因子个数，比如15的因子个数是（1 + 1）*（1 + 1） = 4，他的因子是1， 3， 5， 15；18的因子个数是（1 + 1）*（2 + 1） = 6，它的因子是1， 2， 3， 6， 9， 18。利用素因子分解的方法就能统计合数的因子个数。

 

1 #include 
     
       2 #include 
      
        3 #include 
       
         4 #include 
        
          5 #include 
         
           6 #include 
          
            7 #include 
           
             8 #define LL long long 9 #define MAXI 214748364710 #define MAXL 922337203685477580711 #define dg(i) cout << "*" << i << endl;12 using namespace std;13 14 bool prime[2000000] = { 1, 1};15 16 //素数筛选17 void init_Prime()18 {19 int i, j;20 for(i = 2; i < 1000000; i++)21 {22 if(!prime[i])23 {24 for(j = 2; j * i < 2000000; j++)25 {26 prime[i * j] = 1;27 }28 }29 }30 }31 32 //素因子分解33 bool Count(int x)34 {35 if(x == 1) return false;36 if(!prime[x]) return true;37 int cnt = 1, e, i;38 vector
            
              v;39 vector
             
              ::iterator it;40 for(i = 2; i <= x / 2; i++)41 {42 if(!prime[i] && x % i == 0)43 {44 e = 0;45 int tx = x;46 do47 {48 e++;49 tx /= i;50 }while(tx % i == 0);51 v.push_back(e);52 }53 }54 if(!v.empty())55 {56 for(it = v.begin(); it != v.end(); it++)57 {58 cnt *= (*it + 1);59 }60 }61 if(prime[cnt]) return false;62 else return true;63 }64 65 int main()66 {67 vector
              
                num;68 vector
               
                ::iterator it;69 int T, L, H, i;70 init_Prime();71 scanf("%d", &T);72 while(T--)73 {74 num.clear();75 scanf("%d %d", &L, &H);76 for(i = L; i <= H; i++)77 {78 if(Count(i))79 {80 num.push_back(i);81 }82 }83 if(num.empty()) printf("-1\n");84 else85 {86 it = num.begin();87 printf("%d", *it);88 for(it++; it != num.end(); it++)89 printf(" %d", *it);90 printf("\n");91 }92 }93 return 0;94 }