通过使用修改后的归并排序算法，可以使算法复杂度降为nlgn,但是对于此题，最好不要用，我用的后果就是tle,在如此小的规模下，函数调用所花的时间占程序总执行时间的比例太大，得不偿失。
int Merge_Inversion(char * A,int p, int q, int r){
	int result;
	int n1, n2;
	char L[50];
	char R[50];
	int i, j, k;
	result = 0;
	n1 = q - p + 1;
	n2 = r - q;
	for(i = 0; i < n1; i++){
		L[i] = A[p + i];
	}
	for(j = 0; j < n2; j++){
		R[j] = A[q + j + 1];
	}
	i = 0;
	j = 0;
	k = p;
	while(i < n1 && j < n2){
		if((int)R[j] < (int)L[i]){
			A[k++] = R[j++];
			result = result + n1 - i;
		}
		else{
			A[k++] = L[i++];
		}
	}
	for(; i < n1; i++)
		A[k++] = L[i];
	for(; j < n2; j++)
		A[k++] = R[j];
	return result;
}
int Total_Inversion(char * A,int low,int high){
	int mid, a, b, c;
	if(low == high)
		return 0;
	mid = (low + high) / 2;
	a = Total_Inversion(A, low, mid);
	b = Total_Inversion(A, mid + 1, high);
	c = Merge_Inversion(A, low, mid, high);
	return a + b + c;
}