HDU2710 Max Factor
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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
436384042
Sample Output
38
题目大意:就是求所给数的素因子最大的数,如果存在多个这样的数,那么输出他们中最小的就可以了:
#include<iostream>#include<cstring>#include<cmath>using namespace std;const int M=200005;int prime[M],pe[M];void Init(){memset(prime,0,sizeof(prime));for(int i=2;i<M;i++){if(!prime[i]){for(int j=i+i;j<M;j+=i){prime[j]=1;}}}int j=0;for(int i=2;i<M;i++){if(!prime[i])//cout<<i<<" ";pe[j++]=i;}}int solve(int a){int t=0;for(int i=0;pe[i]<=a;i++){if(a%pe[i]==0)t=pe[i];}return t;}int main(){Init();int N,a;while(cin>>N){int pmax=-1;int res;for(int i=0;i<N;i++){cin>>a;int temp=solve(a);//cout<<"temp="<<temp<<endl;if(pmax<temp){res=a;pmax=temp;}}cout<<res<<endl;}return 0;}
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