AOJ - 0009 Prime Number (素数筛法） && AOJ - 0005 (求最大公约数和最小公倍数)

http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=34870

求n内的素数个数。

 1 /* ***********************************************
 2 Author        : zch
 3 Created Time  :2015/5/19 8:46:16
 4 File Name     :a.cpp
 5  ************************************************ */
 6
 7 #include <cstdio>
 8 #include <cstring>
 9 #include <iostream>
10 #include <algorithm>
11 #include <vector>
12 #include <queue>
13 #include <set>
14 #include <map>
15 #include <string>
16 #include <cmath>
17 #include <cstdlib>
18 #include <ctime>
19 using namespace std;
20 typedef long long ll;
21 const int maxn = 1000005;
22 int prime[maxn];
23 bool is_prime[maxn];
24
25 int solve(int n) {
26     int p=0;
27     for(int i=0;i<=n;i++) is_prime[i]=true;
28     is_prime[0]=is_prime[1]=false;
29     for(int i=2;i<=n;i++) {
30         if(is_prime[i])
31         {
32             prime[p++]=i;
33             for(int j=2*i;j<=n;j+=i)
34                 is_prime[j]=false;
35         }
36     }
37     return p;
38 }
39
40 int main()
41 {
42     //freopen("a.txt","r",stdin);
43     //freopen("b.txt","w",stdout);
44     int n;
45     while(~scanf("%d",&n)) {
46         printf("%d\n",solve(n));
47     }
48     return 0;
49 }

http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=30390

 1 /* ***********************************************
 2 Author        : zch
 3 Created Time  :2015/5/19 9:53:16
 4 File Name     :a.cpp
 5  ************************************************ */
 6
 7 #include <cstdio>
 8 #include <cstring>
 9 #include <iostream>
10 #include <algorithm>
11 #include <vector>
12 #include <queue>
13 #include <set>
14 #include <map>
15 #include <string>
16 #include <cmath>
17 #include <cstdlib>
18 #include <ctime>
19 using namespace std;
20 typedef long long ll;
21 ll gcd(ll a,ll b) {
22     return b==0?a:gcd(b,a%b);
23 }
24
25
26 int main()
27 {
28     //freopen("a.txt","r",stdin);
29     //freopen("b.txt","w",stdout);
30     ll x,y;
31     while(~scanf("%lld%lld",&x,&y)) {
32         printf("%lld %lld\n",gcd(x,y),x/gcd(x,y)*y);
33     }
34     return 0;
35 }