LintCode "Permutation Index"

class Solution {
    long long w(int n)
    {
        long long ret = 1;
        while(n > 1)
        {
            ret *= n;
            n --;
        }
        return ret;
    }
    long long calc(vector<int>& A, int i, long long pwr, int last)
    {
        if (last == 0) return 0;

        int d = A[i] - 1;
        long long curr = d * pwr;
        for(int j = i + 1; j < A.size(); j ++)
        {
            if(A[j] > A[i])
                A[j] --;
        }
        long long next = calc(A, i + 1, pwr / last, last - 1);
        return curr + next;
    }
public:
    /**
     * @param A an integer array
     * @return a long integer
     */
    long long permutationIndex(vector<int>& A) {
        int n = A.size();

        vector<int> b = A;
        sort(b.begin(), b.end());
        unordered_map<int, int> hm;
        int i = 1;
        for(auto v : b)
            hm[v] = i ++;

        for(int i = 0; i < A.size(); i ++)
        {
            A[i] = hm[A[i]];
        }

        long long base = w(n - 1);
        return calc(A, 0, base, n - 1) + 1;
    }
};