Description
To improve the organization of his farm, Farmer John labels each of his N (1 <= N <= 5,000) cows with a distinct serial number in the range 1..20,000. Unfortunately, he is unaware that the cows interpret some serial numbers as better than others. In particular, a cow whose serial number has the highest prime factor enjoys the highest social standing among all the other cows.
(Recall that a prime number is just a number that has no divisors except for 1 and itself. The number 7 is prime while the number 6, being divisible by 2 and 3, is not).
Given a set of N (1 <= N <= 5,000) serial numbers in the range 1..20,000, determine the one that has the largest prime factor.
(Recall that a prime number is just a number that has no divisors except for 1 and itself. The number 7 is prime while the number 6, being divisible by 2 and 3, is not).
Given a set of N (1 <= N <= 5,000) serial numbers in the range 1..20,000, determine the one that has the largest prime factor.
Input
* Line 1: A single integer, N
* Lines 2..N+1: The serial numbers to be tested, one per line
* Lines 2..N+1: The serial numbers to be tested, one per line
Output
* Line 1: The integer with the largest prime factor. If there are more than one, output the one that appears earliest in the input file.
Sample Input
4 36 38 40 42
Sample Output
38
题意:给你n个数,让你求出其中质因数最大的数
思路:这是筛素数的思想,在筛素数的时候,记录每个数对应的最大质因数,这样在访问该数的时候可以以o(1)的复杂度找到它所对应的最大质因数。再加上贪心的想法,在每次访问的时候找到当前最大质因数,并且记录该数。
代码如下:
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <cmath> 6 using namespace std; 7 8 const int N=20010; 9 bool a[N]; 10 int ans[N]; 11 12 void isprime() //作用:找到每个数所对应的最大素数 13 { 14 int k=0; 15 memset(a,true,sizeof(a)); 16 a[0]=a[1]=false; 17 for(int i=2;i<N;i++) 18 { 19 if(a[i]) 20 { 21 ans[i]=i; 22 for(int j=i*2;j<N;j+=i) 23 ans[j]=i; 24 25 a[j]=false; 26 } 27 } 28 } 29 30 int main() 31 { 32 isprime(); 33 int m; 34 while(~scanf("%d",&m)) 35 { 36 int maxn=-N; 37 int x=-N; 38 while(m--) 39 { 40 int a; 41 scanf("%d",&a); 42 if(ans[a]>maxn) 43 { 44 maxn=ans[a]; 45 x=a; 46 } 47 } 48 printf("%d ",x); 49 } 50 return 0; 51 }