HDU 1796 How many integers can you find

How many integers can you find

Time Limit: 5000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 1796
64-bit integer IO format: %I64d Java class name: Main

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

Source

2008 “Insigma International Cup” Zhejiang Collegiate Programming Contest - Warm Up（4）

解题：容斥原理，注意有0

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 const int maxn = 20;
 5 int a[maxn],n,m;
 6 LL LCM(LL a,LL b){
 7     return a/__gcd(a,b)*b;
 8 }
 9 int main(){
10     while(~scanf("%d%d",&n,&m)){
11         int tot = m;
12         for(int i = m = 0,tmp; i < tot; ++i){
13             scanf("%d",&tmp);
14             if(tmp) a[m++] = tmp;
15         }
16         LL ret = 0;
17         for(int i = 1; i < (1<<m); ++i){
18             LL lcm = 1;
19             int cnt = 0;
20             bool flag = false;
21             for(int j = 0; j < m; ++j){
22                 if((i>>j)&1){
23                     cnt++;
24                     lcm = LCM(lcm,a[j]);
25                     if(lcm < 0){
26                         flag = true;
27                         break;
28                     }
29                 }
30             }
31             if(flag) continue;
32             if(cnt&1) ret += (n - 1)/lcm;
33             else ret -= (n - 1)/lcm;
34         }
35         printf("%I64d
",ret);
36     }
37     return 0;
38 }

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