HDU 2710 Max Factor

题目：

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. 

Input

* Line 1: A single integer, N  * 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 #include <iostream> #include <stdio.h> #include <algorithm> #include <string.h> #include <math.h> #include <queue> #include <vector> #include <map> using namespace std; typedef long long LL; typedef unsigned long long UL; #define MS(x,y) memset(x,y,sizeof(x)) #define rpt(i,l,r) for(int i=l;i<=r;i++) #define rpd(i,r,l) for(int i=r;i>=l;i--) LL gcd(LL a,LL b){ return b==0?a:gcd(b,a%b);} #define N 1000005 int prime[N]; void init(){ int sum=0; prime[1]=sum; for(int i=2;i<N;i++){ if(!prime[i]){ sum++; prime[i]=sum; for(int j=i+i;j<N;j+=i){ prime[j]=sum; } } } } int main() { init(); int n; while(scanf("%d",&n)!=EOF) { int temp; scanf("%d",&temp); int max=prime[temp]; int mark=temp; n--; while(n--){ scanf("%d",&temp); if(prime[temp]>max){ max=prime[temp]; mark=temp; } } printf("%d\n",mark); } return 0; }

题意：农场主给牛编号了，哪头牛的最大素数因子在牛群中最大，那他的地位就最高，求出地位最高的牛。有n头牛 1<=n<=5000; 编号小于20000；

思路：用素数筛选法 ：首先将20000以内的数都处理了，求出每个数的最大素数因子的处于素数的第几位。  每次输入的时候保留最大素数因子的数，最后输出那个数。