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\n")
#define PN() printf("No\n")
#define PR(x) printf("%d\n", x)
#define PRR(a, b) printf("%d%d\n", 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;
}

原文地址：https://www.cnblogs.com/meditation5201314/p/12615002.html