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
",m);
47 else
48 printf("1
");
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 :)