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,无论在之间加多少个括号,x1总是作为被除数,x2总是作为除数,因此结果最大的做法是将x3到xn的所有除法转换为乘法,即x1/(x2/.../xn)=x1/x2*x3*...*xn.

string optimalDivision(vector<int>& nums) {
    string res;
    int n = nums.size();
    //特殊情况处理
    if (n == 1) return to_string(nums[0]);
    if (n == 2) return to_string(nums[0]) + "/" + to_string(nums[1]);
    //一般情况
    res = to_string(nums[0]) + "/(";
    for (int i = 1; i < n - 1; i++) {
        res += to_string(nums[i]) + "/"; //注意运算符优先级,+比+=要优先,这样的用法是可以的
    }
    res += to_string(nums[n - 1]) + ")";
    return res;
}