Max Factor(素数筛法)题解

Max Factor

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10245    Accepted Submission(s): 3304


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
436384042
 
Sample Output

38


思路:

忘记了memset()只能初始化0、-1,手贱赋了个1,然后疯狂WA

这里用到了素数筛选法:单击666查看大神详解

代码:

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<math.h>
#include<iostream>
#include<algorithm>
#define INF 0x3f3f3f3f
#define N 20010
using namespace std;
int vis[N];

void prime(){	//素数筛选 
	memset(vis,0,sizeof(vis));
	vis[0]=1,vis[1]=0;	//这里1都算素数 
	for(int i=2;i<=20000;i++){
		if(!vis[i]){
			for(int j=i*i;j<=20000;j+=i){
				vis[j]=1;
			}
		}
	}
}

int main(){
	int n,ans,maxp,i,a;
	prime();
	while(~scanf("%d",&n)){
			maxp=0;
			ans=0;
			while(n--){
				scanf("%d",&a);
				for(i=a;i>0;i--){
					if(!vis[i]){
						if((i>maxp) && a%i==0){
							maxp=i;
							ans=a;
							break;
						}		
					}
				}
			}
			printf("%d\n",ans);
	}
	return 0;
}


posted @ 2018-02-24 22:44  KirinSB  阅读(266)  评论(0编辑  收藏  举报