容斥原理练习

class Solution {
public:
    typedef long long ll;
    ll gcd(ll a,ll b){
        return b == 0?a:gcd(b,a%b);
    }
    ll lcm(ll a,ll b){
        ll t = gcd(a,b);
        return a / t * b;
    }
    int bitcount(int v){
        int res = 0;
        while(v){
            if(v & 1)res++;
            v >>= 1;
        }
        return res;
    }
    ll findKthSmallest(vector<int>& coins, int k) {
        ll l = k - 1,r = *min_element(coins.begin(),coins.end()) * (ll)k,n = coins.size();
        // 计算小等于 mid的整数中，有几个整数能被集合里至少一种元素整除
        function<bool(ll mid)> check = [&](ll mid)->bool{
            ll res = 0;
            // 容斥原理，枚举所有子集
            for(int i = 1;i < 1 << n;i++){
                ll v = 1;
                for(int j = 0;j < n;j++){
                    if((i >> j) & 1){
                        v = lcm(v,coins[j]);
                        if(v > mid) break;
                    }
                }
                //奇正偶负
                res += (bitcount(i) % 2 == 0?-1:1) * (mid / v);
            }
            return res >= k;
        };

        
        while(l < r){
            ll mid = (l + r) >>1;
            if(check(mid)){
                r = mid;
            }else l = mid + 1;
        }
        return l;
    }
};