标签:杭电
<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.
<span style="font-size:14px;"> </span>
Input
* Line 1: A single integer, N
* Lines 2..N+1: The serial numbers to be tested, one per line
<span style="font-size:14px;"> </span>
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.
<span style="font-size:14px;"> </span>
Sample Input
4 36 38 40 42
<span style="font-size:14px;"> </span>
Sample Output
38
<span style="font-size:14px;"> </span>
Source
USACO 2005 October Bronze
<span style="font-size:14px;">一道水题,用筛选素数或者直接就出来了,但就是因为一个小细节,浪费了一下午。</span>
<span style="font-size:14px;">代码如下:</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))//代码都不变将这两项次序发生变化,就会超时 诶,做的时候没注意啊
{
if(j>=max)
max=j,max1=m;
break;
}
}
}
printf("%d\n",max1);
}
return 0;
}</span>标签:杭电
原文地址:http://blog.csdn.net/ice_alone/article/details/38495215