Ex	Score	complete situation
1	2
2	0	717/719(解答错误)
3	0	523/537(超时,弱智题已AC)
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 singleint 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}
};