Hdu2710_max Factor "Water Problem" "sieve method to find prime number"

Max Factor

Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others) total submission (s): 3966 Accepted S Ubmission (s): 1289
Problem Description
To improve the organization of he farm, Farmer John labels each of its n (1 <= n <= 5,000) cows with a distinct ser Ial number in the range 1..20,000. Unfortunately, he is unaware, the cows interpret some serial numbers as better than others. In particular, a cow whose serial number have the highest prime factor enjoys the highest social standing among all the oth ER cows.

(Recall that a prime number is just a number, the has no divisors except for 1 and itself. The number 7 is prime while the number 6, being divisible by 2 and 3, was not).

Given a set of n (1 <= n <= 5,000) serial numbers in the range 1..20,000, determine the one, and the largest PRI Me factor.

Input
* Line 1: A single integer, N

* Lines 2..n+1:the serial numbers to being tested, one per line

Output
* Line 1:the integers with the largest prime factor. If there is more than one, the output of the one that is appears earliest in the input file.

Sample Input
4
36
38
40
42

Sample Output
38

Source

Usaco 2005 October Bronze

The main topic: give you the number of N, to find out which number in the N number of the maximum factor,

Output this number, if there are multiple results, the output depends on the number of the front.

Idea: The Sieve method for the prime number change a bit. If I is a prime number, then I of 1, 2, 3 ... Times the

The maximum factor is I, when the sieve, the assignment is prime[j] = I, that is, the largest element of J

The child is I.

Note: Initialize all the numbers for the season to 0,prime[0] = prime[1] = 1.

That is, Prime[i] is a prime number of 0, Prime[i] is a prime number 1. After the change Prime[i]

maximum cause of I The child.

#include <stdio.h> #include <string.h> #include <math.h>int prime[20005];void isprime () {    prime[ 1] = 1;    for (int i = 2; I <= 20000; i++)    {        if (prime[i]==0)        {            prime[i]=i;            for (int j = i+i; J <= 20000; j+=i)                prime[j] = i;        }    } int main () {    int n,x;    IsPrime ();    while (~SCANF ("%d", &n))    {        int Max = 0;        while (n--)        {            scanf ("%d", &x);            if (Prime[x] > Prime[max])                Max = x;        }        printf ("%d\n", Max);    }    return 0;}

Hdu2710_max Factor "water Problem" "Sieve method for prime number"