lintcode Permutation Index

class Solution {  public:      /**      * @param A an integer array      * @return a long integer      */      long long permutationIndex(vector<int>& A) {          // Write your code here          // 找出A中每一个位置其后有多少个数比它小          // 然后再相加这些数与位对应的权          int len = A.size();          int c[len];          c[len - 1] = 0;// 最后一个数之后就没有比它小的数了          vector<int> a;          a.push_back(A[len - 1]);          for(int i = len - 2;i >= 0; --i){              auto iter = lower_bound(a.begin(), a.end(), A[i]);              c[i] = iter - a.begin();              a.insert(iter, A[i]);          }          long long ans = 1, fac = 1, cc = 1;          for(int i = len - 2;i >= 0; --i)              ans += (fac*=cc++) * c[i];          return ans;      }  };  

class Solution {public:    long getFacctor(int n){        long result = 1;        while(n>=1) {            result =result*n;            n--;        }        return result;    }public:    /**     * @param A an integer array     * @return a long integer     */    long long permutationIndex(vector<int>& A) {        // Write your code here        int len = A.size();        if(len==0)            return 0;        long number = 0;        for(int i = 0; i<len-1; i++) {            long lesscurrent =0;            for(int j = i+1;j<len; j++) {                if(A[i]>A[j])                 lesscurrent++;            }           number+=lesscurrent*getFactor(len-1-i);         }        return number+1;    }};