• 笔试-2020届春季校园网龙招聘在线笔试游戏开发程序员(C++/C#,感觉还行,不过有一些小问题,算是补充了C/C++知识了吧)


    感悟:一个小时,15个选择题,3个填空题,2道编程,还好编程题不难,就是写下伪代码,不然就裂开了。

    选择题:C/C++基本知识点吧,然后还有一些数据结构和算法知识。。。。有些考点都忘记了,列下来吧

    enum{
        ND, IS = 7, GREATER
    };
    cout<<ND<<GREATER<<endl;
    
    0,8:默认是0,然后赋值了的话就依次增加

    虚函数:父类指针指向子类对象,调用父类方法时,该父类方法就定义成虚函数

    填空题:

    1.list和vector和deque优缺点,使用环境

    list链表:非连续内存,适合插入删除,不适合随机访问
    vector动态数组:连续内存,适合随机访问,不适合插入删除,对存储数量没有要求
    deque双向队列:连续内存,适合首部的插入删除
    
    (写了下大致的区别,不过也差不多了,链表就是适合插入删除,不适合随机访问,数组反之)

    2.空指针和野指针还有什么巴拉巴拉。。。。

    空指针void * :可以指向任意对象,定义时声明
    野指针:delete或者free内存后没删除指针的值(就好比你亲戚搬家了,但你还保留原来地址的纸片)

    3.拷贝构造函数

    Class a = Class b的时候,就需要拷贝构造函数,将一一赋值

    编程题:

    1.一一个数,保证其二进制01010101这样10101001就是1,和0连续:不断/2 % 2判断就好了。

    2.LIS:

    2种解法:一种暴力O(n),就是二重循环判断

        2.二分法查找,但是不能保证1111这样答案是4(因为我是用lower_bound做的)

    /**
     
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                                                                            .                                                                             **/
                                                                            
    #include <bits/stdc++.h>
    using namespace std;
    
    #define FFI(a, b) for(int i = a; i < b; i++)
    #define FFJ(a, b) for(int j = a; j < b; j++)
    #define FFK(a, b) for(int k = a; k < b; k++)
    #define RR(a, b) for(int i = a; i > b; i++)
    #define ME(a, b) memset(a, b, sizeof(a))
    #define FILL(a, b, c) fill(a, a + b, c)
    #define SC(x) scanf("%d", &x)
    #define SCC(a, b) scanf("%d%d", &a, &b)
    #define SCCC(a, b, c) scanf("%d%d%d", &a, &b, &c)
    #define PY() printf("Yes
    ")
    #define PN() printf("No
    ")
    #define PR(x) printf("%d
    ", x)
    #define PRR(a, b) printf("%d%d
    ", a, b);
    #define INF 0x3f3f3f3f
    #define MAX 1001
    #define MOD 1000000007
    #define E 2.71828182845
    #define M 8
    #define N 6
    typedef long long LL;
    const double PI = acos(-1.0);
    typedef pair<int, int> Author;
    vector<pair<string, int> > VP; 
    
    
    /*******************************************************************************************************************************************************/
    int n = 6;
    int dp[MAX];
    
    void Solve1(){
        int res = 0;
        int a[] = {9, 8, 7, 6, 6, 7};
        FFI(0, n) dp[i] = 1;
            
        FFI(0, n){
            FFJ(0, i)
                if(a[j] <= a[i]) dp[i] = max(dp[i], dp[j] + 1);
            res = max(res, dp[i]);
        }
        cout<<"The max length is : "<<res<<endl;
    }
    void Solve2(){
        int a[] = {9, 8, 7, 6, 6, 7};
        
        fill(dp, dp + n, INF);
        FFI(0, n){
            *lower_bound(dp, dp + n, a[i]) = a[i];
            FFJ(0, n){
                cout<<dp[j]<<" ";
            } 
            cout<<endl;
        }
        cout<<"Ths max length is : "<<(lower_bound(dp, dp + n, INF) - dp)<<endl;
    }
    int main(void){    
        Solve1();
        Solve2();
        
        return EXIT_SUCCESS;
    }
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  • 原文地址:https://www.cnblogs.com/meditation5201314/p/12615002.html
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