Given a list of positive integers, the adjacent integers will perform the float division. For example, [2,3,4] -> 2 / 3 / 4.

However, you can add any number of parenthesis at any position to change the priority of operations. You should find out how to add parenthesis to get the maximum result, and return the corresponding expression in string format. Your expression should NOT contain redundant parenthesis.

Example:

Input: [1000,100,10,2]
Output: "1000/(100/10/2)"
Explanation:
1000/(100/10/2) = 1000/((100/10)/2) = 200
However, the bold parenthesis in "1000/((100/10)/2)" are redundant, 
since they don't influence the operation priority. So you should return "1000/(100/10/2)". 

Other cases:
1000/(100/10)/2 = 50
1000/(100/(10/2)) = 50
1000/100/10/2 = 0.5
1000/100/(10/2) = 2

Note:

The length of the input array is [1, 10].
Elements in the given array will be in range [2, 1000].
There is only one optimal division for each test case.

思路:x1/x2/.../xn,無論在之間加多少個括號,x1總是作為被除數,x2總是作為除數,因此結果最大的做法是將x3到xn的所有除法轉換為乘法,即x1/(x2/.../xn)=x1/x2*x3*...*xn.

string optimalDivision(vector<int>& nums) {
    string res;
    int n = nums.size();
    //特殊情況處理
    if (n == 1) return to_string(nums[0]);
    if (n == 2) return to_string(nums[0]) + "/" + to_string(nums[1]);
    //一般情況
    res = to_string(nums[0]) + "/(";
    for (int i = 1; i < n - 1; i++) {
        res += to_string(nums[i]) + "/"; //注意運算符優(yōu)先級,+比+=要優(yōu)先,這樣的用法是可以的
    }
    res += to_string(nums[n - 1]) + ")";
    return res;
}