初学算法之筛选素数法

筛选求素数

理解：把从1开始的、某一范围内的正整数从小到大顺序排列， 1不是素数，首先把它筛掉。剩下的数中选择最小的数是素数，然后去掉它的倍数。依次类推，直到筛子为空时结束。

做题时的一些技巧：

1.设求1….n中的素数

只需使1到sqrt（n）的倍数去掉即可。

2．设数组时，因为判断是否为素数只需要用到0和1两个值，则定义数组为bool型

即 bool （只占一位）nu[111]；

用bool型可扩大数组范围。（int 32位）

例题：

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 lineOutput* 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

这题有两个坑。
1.数值1在题中默认为素数。
2.多样例。
我的代码：

 1 #include<stdio.h>
 2 #include<math.h>
 3 #include<string.h>
 4 #define y 22222
 5 int nu[y];
 6 int u=0;
 7 int main()
 8 {
 9     int t,n,f,m=0;
10     int a;
11     while(~(scanf("%d",&n)))
12     {memset(nu, 0, sizeof(nu));
13         f=0,m=0,a=0,u=0;
14         while(n--)
15         {
16             
17             scanf("%d",&t);
18             for(int i=2;i<=sqrt(t);i++)
19             {
20                 a=i;
21                 if(nu[a]==0)
22                     while(a<=t)
23                     {
24                         a+=i;
25                         nu[a]=1;
26                         
27                     }
28             }
29             for(int i=t;i>1;i--)
30             {
31                 if(nu[i]==0&&t%i==0)
32                 {
33                     f=i;
34                     
35                     m=(f>u)?t:m;
36                     u=(f>u)?f:u;
37                     break;
38                     
39                 }
40                 
41                 
42             }
43             
44         }
45         if(m!=0)
46             printf("%d\n",m);
47         else
48             printf("1\n");
49     }
50     return 0;
51 }

后来算算这个代码的时间复杂度，羞愧的无地自容。

但还有好多没有掌握，明天又要学新知识，且存，日后修改。

欢迎各位在评论提出自己的思路 :)