Max Factor

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2412    Accepted Submission(s): 734

Problem 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

 

Source

 

Recommend

teddy

 

#include
      
       #include
       
        #include
        
         using namespace std;const int N=20010;int prime[N];int isprime[N];void getPrime(){    //得到小于等于 N的素数，prime[0]存放的是个数    int i,j;    prime[0]=0;    for(i=0;i
        
       
      

 

 

 

 

 

#include
      
       #include
       
        #include
        
         using namespace std;const int N=20000+10;int prime[N];int isprime[N];void Init(){    int i,j;    int cnt=0;    for(i=0;i
         
          max){                max=tmp;                ans=num;            }        }        printf("%d\n",ans);    }    return 0;}
         
        
       
      

  

 

#include
      
       #include
       
        #include
        
         using namespace std;int prime[20010];int isPrime(int x){    for(int i=2;i*i<=x;i++)        if(x%i==0)            return 0;    return 1;}int main(){    //freopen("input.txt","r",stdin);    memset(prime,0,sizeof(prime));    prime[1]=1;    for(int i=2;i<=20000;i++)        if(isPrime(i))            prime[i]=1;    int n;    while(~scanf("%d",&n)){        int ans=1,tmp=1,max=1,res=1;        int num;        while(n--){            scanf("%d",&num);            int j=1;            while(1){                if(num%j==0 && prime[num/j]){                    tmp=num/j;                    res=num;                    break;                }                j++;            }            if(tmp>max){                max=tmp;                ans=num;            }        }        printf("%d\n",ans);    }    return 0;}
        
       
      

 

 

#include
      
       #include
       
        #include
        
         using namespace std;int prime[20010];void Init(){    memset(prime,0,sizeof(prime));    for(int i=2;i<=20000;i++)        if(!prime[i])            for(int j=i*i;j<=20000;j+=i)                prime[j]=1;}int main(){    //freopen("input.txt","r",stdin);    Init();    int n;    while(~scanf("%d",&n)){        int ans=1,tmp=1,max=1,res=1;        int num;        while(n--){            scanf("%d",&num);            int j=1;            while(1){                if(num%j==0 && !prime[num/j]){                    tmp=num/j;                    res=num;                    break;                }                j++;            }            if(tmp>max){                max=tmp;                ans=num;            }        }        printf("%d\n",ans);    }    return 0;}
        
       
      

 

 

下面这题，哎，，被蛋疼的 2  WA了两个小时：

 

#include
       
        #include
        
         #include
         
          using namespace std;const int N=1000000;int prime[N+10],isprime[N+10];int cnt,num[N+10];void getPrime(){    int i,j;    prime[0]=0;    for(i=0;i
          
           =2)    //2=1*1+1*1  你懂得            b++;                for(int i=1;i<=prime[0];i++)            if(prime[i]>=L && prime[i]<=U)                a++;        for(int i=0;i
           
            =L && num[i]<=U)                b++;        printf("%d %d %d %d\n",l,u,a,b);    }    return 0;}