Leetcode-923 3Sum With Multiplicity(三数之和的多种可能)

class Solution
{
    private:
        int hashList[102];
    public:
        long long int get(long long int n,long long int m)
        {
            if(m==1)
                return n;
            else if(m==2)
                return (n*(n-1))/2;
            else if(m==3)
                return (n*(n-1)*(n-2))/6;
        }
        int threeSumMulti(vector<int>& A, int target)
        {
            memset(hashList,0,sizeof(hashList));

            for(auto d:A)
                hashList[d] ++;

            long long int result = 0;
            for(int i = 0; i <= target&&i<=100; i ++)
            {
                if(hashList[i]>=3 && 3*i==target)
                {
                    result += get(hashList[i],3);
                    result %= 1000000007;
                }
                if(2*i<target && hashList[i]>=2)
                {
                    if(target-2*i > i && target-2*i <= 100 && hashList[target-2*i])
                    {
                        result += hashList[target-2*i]*get(hashList[i],2);
                        result %= 1000000007;
                    //    cout << i << " " << i << " " << target-2*i << endl;
                    }
                }
                if(hashList[i])
                {
                    for(int j = i+1; j <= target && j<=100; j ++)
                    {
                        if(hashList[j]>=2 && 2*j+i==target)
                        {
                            result += hashList[i]*get(hashList[j],2);
                            result %= 1000000007;
                        //    cout << i << " " << j << " " << j << endl;
                        }
                        if(hashList[j])
                        {
                            if(target-i-j > j && target-i-j <= 100 && hashList[target-i-j])
                            {
                                result += hashList[i]*hashList[j]*hashList[target-i-j];
                                result %= 1000000007;
                            //    cout << i << " " << j << " " << target-i-j << endl;
                            }
                        }
                    }
                }
            }
            result %= 1000000007;
            return (int)result;
        }
};