题目链接http://120.78.128.11/Problem.jsp?pid=2332

容斥原理：AUBUC=A+B+C-AB-AC-BC+ABC

输入T表示T组测试数据（1<=T<=100）

接下来T组测试数据

每组第一行一个n表示灯泡个数（1<=n<=10^9）

第二行三个数a,b,c表示V_Dragon每次选择的数（1<=a,b,c<=10^6）（a,b,c全为质数且a,b,c两两互不相等）

不懂格式的同学可以参考以下格式：

Output

数组最后亮着的灯的个数

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<math.h>
#include<set>
#include<vector>
#include<sstream>
#include<queue>
#define ll long long
#define PI 3.1415926535897932384626
#define inf 0x3f3f3f3f
using namespace std;
const ll maxn=200010;
int main()
{
    ll t;
    scanf("%d",&t);
    while(t--){
        ll n;
        scanf("%lld",&n);
        ll num1,num2,num3;
        scanf("%lld%lld%lld",&num1,&num2,&num3);
        ll a=n/num1;
        ll b=n/num2;
        ll c=n/num3;
        ll abc=n/(num1*num2*num3);
        ll ab=n/(num1*num2);
        ll bc=n/(num2*num3);
        ll ac=n/(num1*num3);
        //prllf("a:%d b:%d c:%d abc:%d ab:%d ac:%d bc:%d",a,b,c,abc,ab,ac,bc);
        printf("%lld\n",a+b+c-2*ab-2*ac-2*bc+4*abc);
    }
}