How many integers can you find
Time Limit:5000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit Status Practice HDU 1796
Description
Now you get a number N, and a M-integers set, you should find out how many integers which are small than N, that they can divided exactly by any integers in the set. For example, N=12, and M-integer set is {2,3}, so there is another set {2,3,4,6,8,9,10}, all the integers of the set can be divided exactly by 2 or 3. As a result, you just output the number 7.
Input
There are a lot of cases. For each case, the first line contains two integers N and M. The follow line contains the M integers, and all of them are different from each other. 0<N<2^31,0<M<=10, and the M integer are non-negative and won’t exceed 20.
Output
For each case, output the number.
Sample Input
12 2 2 3
Sample Output
7
分析：这几天练容斥有感觉，知道是容斥，但是却有问题，容斥是 互质的数，然后对于2,4这样的数就不会做了，太肤浅了，直接求最小公倍数啊，对啊，互质的相乘就是因为最小公倍数就是乘积啊=_=，弱！
1 #include <iostream>
2 #include <cstring>
3 #include <cstdio>
4 #include <algorithm>
5 using namespace std;
6 typedef long long LL;
7 int num[30], n, m, a[30];
8 LL res;
9 LL gcd(LL a, LL b)
10 {
11 if (a == 0)
12 return b;
13 return gcd(b % a, a);
14 }
15 void dfs(int cur, int snum, int cnt)
16 {
17 if (snum == cnt)
18 {
19 int temp = n;
20 int mult = 1;
21 for (int i = 0; i < snum; i++)
22 mult = mult / gcd(mult, a[i]) * a[i]; // 防爆
23 if (mult == 0)
24 return;
25 if (temp % mult == 0)
26 res += temp / mult - 1;
27 else
28 res += temp / mult;
29 return;
30 }
31 for (int i = cur; i < m; i++)
32 {
33 a[snum] = num[i];
34 dfs(i + 1, snum + 1, cnt);
35 }
36 }
37 int main()
38 {
39 int tm;
40 while (scanf("%d%d", &n, &tm) != EOF)
41 {
42 m = 0;
43 for (int i = 0; i < tm; i++)
44 {
45 int temp;
46 scanf("%d", &temp); // 去0
47 if (temp)
48 num[m++] = temp;
49 }
50 LL sum = 0;
51 for (int i = 1; i <= m; i++)
52 {
53 res = 0;
54 dfs(0, 0, i);
55 if (i & 1)
56 sum += res;
57 else
58 sum -= res;
59 }
60 printf("%I64d\n", sum);
61 }
62 return 0;
63 }