Judge Info
- Memory Limit: 32768KB
- Case Time Limit: 10000MS
- Time Limit: 10000MS
- Judger: Number Only Judger
Description
Frog Frank likes 30 more than likes 40, yet he likes 12 and 39 equally. This is because he likes numbers that have a lot of different prime factors. For example, 30 have 3 prime factors (2,3 and 5) and 40 have 2(2 and 5) only. A prime number is a number that can be divided evenly only by itself and 1.
Task
You are given a list of numbers, find out which of the numbers Frank likes most. If there are more than one solutions, output the smallest.
Input
The first line of input contains 
, the number of test cases. First line of each test case contains a integers 
. The following line contains N positive integers, all of them are not greater than 100, 000.
Output
For each test case, print a line contains the solution.
Sample Input
2 10 3 5 7 9 11 13 15 17 19 21 11 2 4 6 8 10 13 39 105 200 201 143
Sample Output
15 105
解题思路：素数筛选
算法 from dd：
1 #include <stdio.h> 2 #include <string.h> 3 #include <math.h> 4 5 #define N 100000 6 #define yes '1' 7 #define no '0' 8 char flag[N+1]; 9 10 void is_prime(int n) 11 { 12 int i, j; 13 memset(flag, yes, sizeof(flag)); 14 flag[0]=flag[1]=no; 15 int len=sqrt(n)+1; 16 for(i=2; i<len; i++) 17 { 18 if(flag[i]==no) continue; 19 for(j=i+i; j<=n; j+=i) 20 flag[j]=no; 21 } 22 } 23 24 int main(void) 25 { 26 int n; 27 ///find the primes from 1 to n(n<=N) 28 scanf("%d", &n); 29 is_prime(n); 30 for(int i=1; i<=n; i++) 31 if(flag[i]==yes) 32 printf("%8d", i); 33 printf(" "); 34 return 0; 35 }
解题：
1 #include <stdio.h> 2 #include <string.h> 3 #include <math.h> 4 5 #define N 100050 6 #define yes '1' 7 #define no '0' 8 char flag[N]; 9 10 void is_prime(int n) 11 { 12 int i, j; 13 memset(flag, yes, sizeof(flag)); 14 flag[0]=flag[1]=no; 15 int len=sqrt(n)+1; 16 for(i=2; i<len; i++) 17 { 18 if(flag[i]==no) continue; 19 for(j=i+i; j<=n; j+=i) 20 flag[j]=no; 21 } 22 } 23 24 25 26 int main() 27 { 28 int n; 29 int count; 30 int t,i,k,num; 31 int max; 32 scanf("%d",&t); 33 while (t--) { 34 scanf("%d", &n); 35 max=0; 36 for (i=0;i<n;++i) { 37 count=0; 38 scanf("%d",&num); 39 40 is_prime(num); 41 for(i=1; i<=num; i++) 42 if(flag[i]==yes){ 43 if(num%i==0) 44 count++; 45 } 46 if(count>max) 47 { 48 max=count; 49 k=num; 50 } 51 if (count==max) 52 { 53 if (k>num) k=num; 54 } 55 } 56 printf("%d ",k); 57 } 58 return 0; 59 }
Winifred:
1 #include <stdio.h> 2 #include <string.h> 3 bool f[100025]; 4 int T,p[100025]; 5 6 void getprime() 7 { 8 int i,j; 9 memset(f,true,sizeof(f)); 10 f[1]=false; 11 for (i=2;i<=100000;i++) 12 if (f[i]) 13 { 14 for (j=i+i;j<=100000;j+=i) f[j]=false; 15 } 16 T=0; 17 for (i=2;i<=100000;i++) 18 if (f[i]) 19 { 20 T++; 21 p[T]=i; 22 } 23 } 24 25 int getdivnum(int x) 26 { 27 int ans=0,tmp=x; 28 for (int i=1;i<=T;i++) 29 { 30 if (tmp%p[i]==0) 31 { 32 ans++; 33 while (tmp%p[i]==0) tmp/=p[i]; 34 } 35 if (p[i]>tmp) break; 36 } 37 return ans; 38 } 39 int main() 40 { 41 getprime(); 42 int cas; 43 scanf("%d",&cas); 44 while (cas--) 45 { 46 int n; 47 scanf("%d",&n); 48 int Max=0,u; 49 for (int i=1;i<=n;i++) 50 { 51 int x; 52 scanf("%d",&x); 53 int t=getdivnum(x); 54 if (t>Max) 55 { 56 Max=t; 57 u=x; 58 } 59 else if (t==Max) 60 { 61 if (u>x) u=x; 62 } 63 } 64 printf("%d ",u); 65 } 66 getprime(); 67 }