Max Factor
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2238 Accepted Submission(s): 684
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
一大清早起来，刷了道水题。。。找了下感觉，试了下素数分解的模板，求下午网络赛给力啊。
这个素数分解的模板用了很久了，真的是屡试不爽啊。。。哈哈
此题注意对1的处理。把1的最大素因子当成1就可以AC了。
#include#include #include #include using namespace std;const int MAXN=20000;int prime[MAXN+1];int getPrime()//得到小于等于MAXN的素数，prime[0]存放的是个数{ memset(prime,0,sizeof(prime)); for(int i=2;i<=MAXN;i++) { if(!prime[i]) prime[++prime[0]]=i; for(int j=1;j<=prime[0]&&prime[j]<=MAXN/i;j++) { prime[prime[j]*i]=1; if(i%prime[j]==0) break; } } return prime[0];}long long factor[100][2];int facCnt;int getFactors(long long x)//把x进行素数分解{ facCnt=0; long long tmp=x; for(int i=1;prime[i]<=tmp/prime[i];i++) { factor[facCnt][1]=0; if(tmp%prime[i]==0) { factor[facCnt][0]=prime[i]; while(tmp%prime[i]==0) { factor[facCnt][1]++; tmp/=prime[i]; } facCnt++; } } if(tmp!=1) { factor[facCnt][0]=tmp; factor[facCnt++][1]=1; } return facCnt;}int main(){ int n; getPrime(); int num; while(scanf("%d",&n)!=EOF) { int ans=0; int temp=0; for(int i=0;i