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.

容斥原理裸题

 #include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
typedef long long ll; inline int gcd(int a,int b){
return b?gcd(b,a%b):a;
} int num[]; int main(){
int n,m;
while(scanf("%d%d",&n,&m)!=EOF){
n--;
for(int i=;i<=m;++i){
scanf("%d",&num[i]);
if(!num[i]){
i--;
m--;
}
}
ll ans=;
for(int i=;i<(<<m);++i){
int bit=;
int tmp=;
for(int j=;j<=m;++j){
if(i&(<<(j-))){
bit++;
tmp=tmp/gcd(tmp,num[j])*num[j];
}
}
if(bit%)ans+=n/tmp;
else ans-=n/tmp;
}
printf("%lld\n",ans);
}
return ;
}

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