Hangdian 2710 (water question)

Max Factor

<span size="+0"><strong><span style="font-family:Arial;font-size:12px;color:green;FONT-WEIGHT: bold">Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3693    Accepted Submission(s): 1181</span></strong></span>

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.

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Input

* Line 1: A single integer, N* Lines 2..N+1: The serial numbers to be tested, one per line

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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.

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Sample Input

436384042

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Sample Output

38

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Source

USACO 2005 October Bronze 

<Span style = "font-size: 14px;"> A water question is filtered out by prime numbers or directly, but it is a waste of afternoon due to a small detail. </Span>

<Span style = "font-size: 14px;"> the code is as follows: </span>

<Span style = "font-size: 14px;" >#include <stdio. h> int prime (int n) {int I; for (I = 2; I <n-1; I ++) if (N % I = 0) return 0; return 1;} int main () {int I, j, n, m, Max, max1; while (~ Scanf ("% d", & N) {max = 0; for (I = 0; I <n; I ++) {scanf ("% d ", & M); For (j = m; j> 0; j --) {If (M % J = 0 & prime (j )) // The Code will not change the order of the two items, and it will time out. Do not pay attention when doing so {If (j> = max) max = J, max1 = m; break ;}} printf ("% d \ n", max1) ;}return 0 ;}</span>