Ex	Score	complete situation
1	2
2	0	717/719(解答错误)
3	0	523/537(超时)
4	0

• 有用的是Ex2和Ex4

Ex2

my solution

class Solution {
public:
    //calculate x's l-time 幂乘
    int jiecheng(int x,int l){
        int z=x;
        for(int i=0;i<l;i++){
            if(z>pow(10,4.5))
                return 0;
            z=z*z;
        }
        return z;
    }
    
    bool search(vector<int>& nums,int L){
        int x=2;
        while(jiecheng(x,(L-1)/2)){
            int sign=1;
            for(int i=0;i<(L-1)/2;i++){
                int x_i_time=0;
                for(int j=0;j<nums.size();j++){
                    if(nums[j]==jiecheng(x,i))
                        x_i_time++;
                }
                if(x_i_time<2){
                    x++;    sign=0;
                    break;  
                }
            }
            if(sign==1){
                int x_L_time=0;
                for(int j=0;j<nums.size();j++){
                    if(nums[j]==jiecheng(x,(L-1)/2))
                        x_L_time++;
                }
                if(x_L_time==0){
                    x++;    break;
                }
                return true;
            }
        }
        return false;
    }
    
    int maximumLength(vector<int>& nums) {
        //1 must be handled single
        int num_one=0;
        int ans=1;
        for(int i=0;i<nums.size();i++){
            if(nums[i]==1)  num_one++;
        }
        if(num_one>0){
            if(num_one%2==1)    ans=num_one;
            else ans=num_one-1;
        }
        if(ans>=11) return ans;
        for(int L=11;L>1;L-=2){
            if(search(nums,L)==true)
                return L>ans?L:ans;
        }
        return ans;
    }
};

result

Ex4

reading

class Solution {
public:
    int minOrAfterOperations(vector<int>& nums, int k) {
//1. robust enumerate
//2. bit function
//3. lower_bound = min number's 000xxx000
//4. when adopt op between two number, tough
    }
};