51Nod 1284 2 3 5 7的倍数（容斥）

题目传送门
代码：

#include<bits/stdc++.h>
using namespace std;

#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define ls (rt<<1)
#define rs (rt<<1|1)
#define pb push_back
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)
#define MEM(a,b,start,end) for(int ii=start;ii<=end;ii++) a[ii]=b
#define low(x) (x&(-x))
#define v(x) vector<x>
#define Map1(a,b) map<a,b>
#define Map2(a,b) unordered_map<a,b>
#define pqg(x) priority_queue<x,vector<x>,greater<x> >
#define pql(x) priority_queue<x,vector<x>,less<x> >
typedef unsigned long long ULL;
typedef long long LL;
typedef double D;
typedef complex<double> COMD;
const int maxn=10000+100;

int res[4]={2,3,5,7};

int main(){

    LL n;
    scanf("%lld",&n);
    LL num=n;
    for(int i=1;i<16;i++){

        int mul=1;
        int cnt=0;
        for(int j=0;j<4;j++){

            if((i>>j)&1) mul*=res[j],cnt++;
        }
        if(cnt%2) num-=n/mul;
        else num+=n/mul;
    }
    printf("%lld\n",num);
}