Ex	Score	complete extent	mean
1	2		char op
2	0	717/719(解答错误,迭代条件写错)	按长度枚举
3	0	523/537(超时,弱智题已AC)	枚举
4	0

• 有用的是Ex2和Ex4

Experience++

1. each problem can be submitted infinitely with negligible punishment time
2. some cases of the problem is hidden.
3. each kernel function can be utilized without citing header file.
4. notice to discriminate errors ,be them wrong solution(usually on wrong iteration condition &grammar error) or runtime exceeding.

Ex2：按长度暴力枚举

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

更正

• 36行后，删去"break;"：若不行，x++后进入下一个循环判断，break就忽略了可能的取值（break是跳出当前循环）
  

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
    }
};