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Every k-th, l-th, m-th, n-th dragon will be damaged,
How many dragons suffered damage , if the princess counted a total of D dragons?

1. Straightforward way

The number of dragons D can be quite small, so the problem can be solved in a straightforward way,
by iterating over dragons 1 through D and checking each dragon individually.

int f(vector<int> &nums ,int d){ //k平底锅脸、l门尾巴、m高跟鞋爪子、n叫妈妈
    int res = d;
    for(int i=1;i<=d;i++){//判断每一头龙
        bool flag = false;
        for(int j=0;j<nums.size();j++){
            if(i%nums[j]==0){
                flag = true;
                break;
            }
        }
        if(!flag) res--;
    }
    return res;
}

2. Inclusion-exclusion principle

Calculate LCM of the corresponding sets of numbers
Finally, the number of dragons that get damaged equals N1 - N2 + N3 - N4

int gcd(int a, int b) {
    return b?gcd(b, a % b):a;
}

int lcm(int a, int b) {
    return (a * b) / gcd(a, b);
}

int f(vector<int> &nums ,int d){ //k平底锅脸、l门尾巴、m高跟鞋爪子、n叫妈妈
    int res = 0;
    int n = nums.size();
    int flag;
    for (int mask = 1; mask < (1 << n); mask++) {//遍历所有子集 
        int cur_lcm = 1;
        for(int i=0;i<n;i++) //求当前集合最小公倍数
           if( (mask>> i) & 1 == 1)
                cur_lcm = lcm(cur_lcm,nums[i]);
        flag = __builtin_popcount(mask)%2==1?1:-1;
        res+= flag*(d/cur_lcm);
    }
    return res;
}