HDU 1796 容斥原理

表示今天累了。。就找了道容斥水题刷一下。。

  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.

Standard Input:

12 2

2 3

Output 

7

选择几个数，奇数个就加上，偶数个就减去。

#include <cstdio>
#include <cstdlib>
#include <iostream>
#define LL __int64
using namespace std;
const int N=15;
int a
,n,m;
LL ans;

LL gcd(LL a,LL b){
LL c;
while(b){
c=a%b;
a=b;
b=c;
}
return a;
}
void dfs(int beg,int step,LL lcm){
LL l;
l=a[beg]/gcd(lcm,a[beg])*lcm;
if(step&1)ans+=n/l;
else ans-=n/l;
for(int i=beg+1;i<=m;i++){
dfs(i,step+1,l);
}
}
void solve(){
int i,j;
for(i=1;i<=m;i++)
scanf("%d",&a[i]);
for(i=1,j=1;i<=m;i++)
if(a[i])
a[j++]=a[i];

ans=0;
m=j-1;
// //
// for(i=1;i<=m;i++)
// printf("%d\n",a[i]);
// //
for(int i=1;i<=m;i++){
dfs(i,1,a[i]);
}
cout<<ans<<endl;
}
int main(){
while(scanf("%d %d",&n,&m)!=EOF){
n--;
solve();
}
return 0;
}