KAJIMA CORPORATION CONTEST 2024(AtCoder Beginner Contest 340)ABCDEF 视频讲解

这场比较郁闷,C题短路,连续4次WA,导致罚时太多

A - Arithmetic Progression

Problem Statement

Print an arithmetic sequence with first term A A A, last term B B B, and common difference D D D.
You are only given inputs for which such an arithmetic sequence exists.

Constraints

1 ≤ A ≤ B ≤ 100 1 \leq A \leq B \leq 100 1AB100
1 ≤ D ≤ 100 1 \leq D \leq 100 1D100
There is an arithmetic sequence with first term A A A, last term B B B, and common difference D D D.
All input values are integers.

Input

The input is given from Standard Input in the following format:

A A A B B B D D D

Output

Print the terms of the arithmetic sequence with first term A A A, last term B B B, and common difference D D D, in order, separated by spaces.

Sample Input 1

3 9 2

Sample Output 1

3 5 7 9

The arithmetic sequence with first term 3 3 3, last term 9 9 9, and common difference 2 2 2 is ( 3 , 5 , 7 , 9 ) (3,5,7,9) (3,5,7,9).

Sample Input 2

10 10 1

Sample Output 2

10

The arithmetic sequence with first term 10 10 10, last term 10 10 10, and common difference 1 1 1 is ( 10 ) (10) (10).

Solution

具体见文末视频。


Code

#include <bits/stdc++.h>
#define int long longusing namespace std;typedef pair<int, int> PII;
typedef long long LL;signed main()
{cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);int A, B, C;cin >> A >> B >> C;for (int i = A; i <= B; i += C)cout << i << " ";return 0;
}

B - Append

Problem Statement

You have an empty sequence A A A. There are Q Q Q queries given, and you need to process them in the order they are given.

The queries are of the following two types:
1 x: Append x x x to the end of A A A.
2 k: Find the k k k-th value from the end of A A A. It is guaranteed that the length of A A A is at least k k k when this query is given.

Constraints

1 ≤ Q ≤ 100 1 \leq Q \leq 100 1Q100
In the first type of query, x x x is an integer satisfying 1 ≤ x ≤ 1 0 9 1 \leq x \leq 10^9 1x109.
In the second type of query, k k k is a positive integer not greater than the current length of sequence A A A.

Input

The input is given from Standard Input in the following format:

Q Q Q
q u e r y 1 \mathrm{query}_1 query1
q u e r y 2 \mathrm{query}_2 query2
⋮ \vdots
q u e r y Q \mathrm{query}_Q queryQ

Each query is in one of the following two formats:

1 1 1 x x x

2 2 2 k k k

Output

Print q q q lines, where q q q is the number of queries of the second type.

The i i i-th line should contain the answer to the i i i-th such query.

Sample Input 1

5
1 20
1 30
2 1
1 40
2 3

Sample Output 1

30
20

Initially, A A A is empty.
The first query appends 20 20 20 to the end of A A A, making A = ( 20 ) A=(20) A=(20).
The second query appends 30 30 30 to the end of A A A, making A = ( 20 , 30 ) A=(20,30) A=(20,30).
The answer to the third query is 30 30 30, which is the 1 1 1-st value from the end of A = ( 20 , 30 ) A=(20,30) A=(20,30).
The fourth query appends 40 40 40 to the end of A A A, making A = ( 20 , 30 , 40 ) A=(20,30,40) A=(20,30,40).
The answer to the fifth query is 20 20 20, which is the 3 3 3-rd value from the end of A = ( 20 , 30 , 40 ) A=(20,30,40) A=(20,30,40).

Solution

具体见文末视频。

Code

#include <bits/stdc++.h>
#define int long longusing namespace std;typedef pair<int, int> PII;
typedef long long LL;signed main()
{cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);int Q;cin >> Q;std::vector<int> S;while (Q --){int Op, X;cin >> Op >> X;if (Op == 1)S.push_back(X);elsecout << S[S.size() - X] << endl;}return 0;
}

C - Divide and Divide

Problem Statement

There is a single integer N N N written on a blackboard.

Takahashi will repeat the following series of operations until all integers not less than 2 2 2 are removed from the blackboard:
Choose one integer x x x not less than 2 2 2 written on the blackboard.
Erase one occurrence of x x x from the blackboard. Then, write two new integers ⌊ x 2 ⌋ \left \lfloor \dfrac{x}{2} \right\rfloor 2x and ⌈ x 2 ⌉ \left\lceil \dfrac{x}{2} \right\rceil 2x on the blackboard.
Takahashi must pay x x x yen to perform this series of operations.
Here, ⌊ a ⌋ \lfloor a \rfloor a denotes the largest integer not greater than a a a, and ⌈ a ⌉ \lceil a \rceil a denotes the smallest integer not less than a a a.
What is the total amount of money Takahashi will have paid when no more operations can be performed?

It can be proved that the total amount he will pay is constant regardless of the order in which the operations are performed.

Constraints

2 ≤ N ≤ 1 0 17 2 \leq N \leq 10^{17} 2N1017

Input

The input is given from Standard Input in the following format:

N N N

Output

Print the total amount of money Takahashi will have paid, in yen.

Sample Input 1

3

Sample Output 1

5

Here is an example of how Takahashi performs the operations:
Initially, there is one 3 3 3 written on the blackboard.
He chooses 3 3 3. He pays 3 3 3 yen, erases one 3 3 3 from the blackboard, and writes ⌊ 3 2 ⌋ = 1 \left \lfloor \dfrac{3}{2} \right\rfloor = 1 23=1 and ⌈ 3 2 ⌉ = 2 \left\lceil \dfrac{3}{2} \right\rceil = 2 23=2 on the blackboard.
There is one 2 2 2 and one 1 1 1 written on the blackboard.
He chooses 2 2 2. He pays 2 2 2 yen, erases one 2 2 2 from the blackboard, and writes ⌊ 2 2 ⌋ = 1 \left \lfloor \dfrac{2}{2} \right\rfloor = 1 22=1 and ⌈ 2 2 ⌉ = 1 \left\lceil \dfrac{2}{2} \right\rceil = 1 22=1 on the blackboard.
There are three 1 1 1s written on the blackboard.
Since all integers not less than 2 2 2 have been removed from the blackboard, the process is finished.
Takahashi has paid a total of 3 + 2 = 5 3 + 2 = 5 3+2=5 yen for the entire process, so print 5 5 5.

Sample Input 2

340

Sample Output 2

2888

Sample Input 3

100000000000000000

Sample Output 3

5655884811924144128

Solution

具体见文末视频。


Code

#include <bits/stdc++.h>using namespace std;inline __int128 read()
{char ch = getchar();__int128 x = 0, cf = 1;while(ch < '0' || ch > '9') {if(ch == '-') cf = -1;ch = getchar();}while(ch >= '0' && ch <= '9') {x = (x << 3) + (x << 1) + (ch ^ 48);ch = getchar();}return x * cf;
}void write(__int128 x)
{if(x<0)putchar('-'),x=-x;if(x>9)write(x/10);putchar(x%10+'0');return;
}__int128 Quick_Pow(__int128 a, __int128 b)
{__int128 Result = 1;while (b){if (b & 1) Result = Result * a;a = a * a;b >>= 1;}return Result;
}signed main()
{cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);__int128 N = read();__int128 T = 0, T2 = 0;while (Quick_Pow(2, T + 1) <= 4 * N) T ++;while (Quick_Pow(2, T2 + 1) <= 2 * N) T2 ++;__int128 Result = N * T - Quick_Pow(2, T2);write(Result);return 0;
}

D - Super Takahashi Bros.

Problem Statement

Takahashi is playing a game.
The game consists of N N N stages numbered 1 , 2 , … , N 1,2,\ldots,N 1,2,,N. Initially, only stage 1 1 1 can be played.
For each stage i i i ( 1 ≤ i ≤ N − 1 1\leq i \leq N-1 1iN1 ) that can be played, you can perform one of the following two actions at stage i i i:
Spend A i A_i Ai seconds to clear stage i i i. This allows you to play stage i + 1 i+1 i+1.
Spend B i B_i Bi seconds to clear stage i i i. This allows you to play stage X i X_i Xi.
Ignoring the times other than the time spent to clear the stages, how many seconds will it take at the minimum to be able to play stage N N N?

Constraints

2 ≤ N ≤ 2 × 1 0 5 2 \leq N \leq 2\times 10^5 2N2×105
1 ≤ A i , B i ≤ 1 0 9 1 \leq A_i, B_i \leq 10^9 1Ai,Bi109
1 ≤ X i ≤ N 1 \leq X_i \leq N 1XiN
All input values are integers.

Input

The input is given from Standard Input in the following format:

N N N
A 1 A_1 A1 B 1 B_1 B1 X 1 X_1 X1
A 2 A_2 A2 B 2 B_2 B2 X 2 X_2 X2
⋮ \vdots
A N − 1 A_{N-1} AN1 B N − 1 B_{N-1} BN1 X N − 1 X_{N-1} XN1

Output

Print the answer.

Sample Input 1

5
100 200 3
50 10 1
100 200 5
150 1 2

Sample Output 1

350

By acting as follows, you will be allowed to play stage 5 5 5 in 350 350 350 seconds.
Spend 100 100 100 seconds to clear stage 1 1 1, which allows you to play stage 2 2 2.
Spend 50 50 50 seconds to clear stage 2 2 2, which allows you to play stage 3 3 3.
Spend 200 200 200 seconds to clear stage 3 3 3, which allows you to play stage 5 5 5.

Sample Input 2

10
1000 10 9
1000 10 10
1000 10 2
1000 10 3
1000 10 4
1000 10 5
1000 10 6
1000 10 7
1000 10 8

Sample Output 2

90

Sample Input 3

6
1000000000 1000000000 1
1000000000 1000000000 1
1000000000 1000000000 1
1000000000 1000000000 1
1000000000 1000000000 1

Sample Output 3

5000000000

Solution

具体见文末视频。


Code

#include <bits/stdc++.h>
#define int long longusing namespace std;typedef pair<int, int> PII;
typedef long long LL;const int SIZE = 8e5 + 10;int N;
int A[SIZE], B[SIZE], X[SIZE];
int h[SIZE], w[SIZE], e[SIZE], ne[SIZE], idx;
bool st[SIZE];
int dist[SIZE];inline void add(int a, int b, int c)
{e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx ++;
}inline int Dijkstra(int start, int finish)
{memset(dist, 0x3f, sizeof dist);priority_queue<PII, vector<PII>, greater<PII>> heap;heap.push({0, start});st[start] = 1;while (heap.size()){auto t = heap.top();heap.pop();int u = t.second, dis = t.first;for (int i = h[u]; ~i; i = ne[i])if (dist[e[i]] > w[i] + dis)dist[e[i]] = w[i] + dis, heap.push({dist[e[i]], e[i]});}if (dist[finish] == 0x3f3f3f3f) return -1;return dist[finish];
}signed main()
{cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);memset(h, -1, sizeof h);cin >> N;for (int i = 1; i < N; i ++){cin >> A[i] >> B[i] >> X[i];add(i, i + 1, A[i]), add(i, X[i], B[i]);}cout << Dijkstra(1, N) << endl;return 0;
}

E - Mancala 2

Problem Statement

There are N N N boxes numbered 0 0 0 to N − 1 N-1 N1. Initially, box i i i contains A i A_i Ai balls.
Takahashi will perform the following operations for i = 1 , 2 , … , M i=1,2,\ldots,M i=1,2,,M in order:
Set a variable C C C to 0 0 0.
Take out all the balls from box B i B_i Bi and hold them in hand.
While holding at least one ball in hand, repeat the following process:
Increase the value of C C C by 1 1 1.
Put one ball from hand into box ( B i + C ) m o d N (B_i+C) \bmod N (Bi+C)modN.
Determine the number of balls in each box after completing all operations.

Constraints

1 ≤ N ≤ 2 × 1 0 5 1 \leq N \leq 2\times 10^5 1N2×105
1 ≤ M ≤ 2 × 1 0 5 1 \leq M \leq 2\times 10^5 1M2×105
0 ≤ A i ≤ 1 0 9 0 \leq A_i \leq 10^9 0Ai109
KaTeX parse error: Expected 'EOF', got '&' at position 12: 0 \leq B_i &̲lt; N
All input values are integers.

Input

The input is given from Standard Input in the following format:

N N N M M M
A 0 A_0 A0 A 1 A_1 A1 … \ldots A N − 1 A_{N-1} AN1
B 1 B_1 B1 B 2 B_2 B2 … \ldots B M B_M BM

Output

Let X i X_i Xi be the number of balls in box i i i after completing all operations. Print X 0 , X 1 , … , X N − 1 X_0,X_1,\ldots,X_{N-1} X0,X1,,XN1 in this order, separated by spaces.

Sample Input 1

5 3
1 2 3 4 5
2 4 0

Sample Output 1

0 4 2 7 2

The operations proceed as follows:
Figure

Sample Input 2

3 10
1000000000 1000000000 1000000000
0 1 0 1 0 1 0 1 0 1

Sample Output 2

104320141 45436840 2850243019

Sample Input 3

1 4
1
0 0 0 0

Sample Output 3

1

Solution

具体见文末视频。


Code

#include <bits/stdc++.h>
#define int long longusing namespace std;typedef pair<int, int> PII;
typedef long long LL;const int SIZE = 2e5 + 10;int N, M;
int A[SIZE], B[SIZE];
struct Segment
{struct Node{int l, r;LL Sum, Max, Min, Lazy;}Tree[SIZE << 2];void Pushup(int u){Tree[u].Sum = Tree[u << 1].Sum + Tree[u << 1 | 1].Sum;Tree[u].Max = max(Tree[u << 1].Max, Tree[u << 1 | 1].Max);Tree[u].Min = min(Tree[u << 1].Min, Tree[u << 1 | 1].Min);}void Pushdown(int u){if (Tree[u].Lazy){Tree[u << 1].Max += Tree[u].Lazy;Tree[u << 1].Min += Tree[u].Lazy;Tree[u << 1].Sum += (LL)(Tree[u << 1].r - Tree[u << 1].l + 1) * Tree[u].Lazy;Tree[u << 1].Lazy += Tree[u].Lazy;Tree[u << 1 | 1].Max += Tree[u].Lazy;Tree[u << 1 | 1].Min += Tree[u].Lazy;Tree[u << 1 | 1].Sum += (LL)(Tree[u << 1 | 1].r - Tree[u << 1 | 1].l + 1) * Tree[u].Lazy;Tree[u << 1 | 1].Lazy += Tree[u].Lazy;Tree[u].Lazy = 0;}}void Build(int u, int l, int r){Tree[u] = {l, r};if (l == r) return;int mid = l + r >> 1;Build(u << 1, l, mid), Build(u << 1 | 1, mid + 1, r);}void Modify(int u, int l, int r, int d){if (Tree[u].l >= l && Tree[u].r <= r){Tree[u].Sum += (LL)(Tree[u].r - Tree[u].l + 1) * d;Tree[u].Max += d, Tree[u].Min += d;Tree[u].Lazy += d;return;}Pushdown(u);int mid = Tree[u].l + Tree[u].r >> 1;if (mid >= l) Modify(u << 1, l, r, d);if (mid < r) Modify(u << 1 | 1, l, r, d);Pushup(u);}int Query(int u, int l, int r, int k){if (Tree[u].l >= l && Tree[u].r <= r){if (k == 1) return Tree[u].Sum;else if (k == 2) return Tree[u].Max;else return Tree[u].Min;}Pushdown(u);long long mid = Tree[u].l + Tree[u].r >> 1, Result;if (k == 1) Result = 0;else if (k == 2) Result = -1e18;else Result = 1e18;if (mid >= l) Result = Query(u << 1, l, r, k);if (mid < r){if (k == 1) Result += Query(u << 1 | 1, l, r, k);else if (k == 2) Result = max(Result, Query(u << 1 | 1, l, r, k));else Result = min(Result, Query(u << 1 | 1, l, r, k));}return Result;}int Sum(int l, int r) { return Query(1, l, r, 1); }int Max(int l, int r) { return Query(1, l, r, 2); }int Min(int l, int r) { return Query(1, l, r, 3); }
}Tool;signed main()
{cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);cin >> N >> M;Tool.Build(1, 1, N);for (int i = 1; i <= N; i ++)cin >> A[i], Tool.Modify(1, i, i, A[i]);for (int i = 1; i <= M; i ++)cin >> B[i], B[i] ++;for (int i = 1; i <= M; i ++){int X = Tool.Sum(B[i], B[i]), Turn = X / N, Rest = X % N;Tool.Modify(1, 1, N, Turn);if (Rest && B[i] + Rest > N){if (B[i] + 1 <= N) Tool.Modify(1, B[i] + 1, N, 1);Tool.Modify(1, 1, Rest - N + B[i], 1);}else if (Rest) Tool.Modify(1, B[i] + 1, B[i] + Rest, 1);Tool.Modify(1, B[i], B[i], -X);}for (int i = 1; i <= N; i ++)cout << Tool.Sum(i, i) << " ";return 0;
}

F - S = 1

Problem Statement

You are given integers X X X and Y Y Y, which satisfy at least one of X ≠ 0 X \neq 0 X=0 and Y ≠ 0 Y \neq 0 Y=0.

Find a pair of integers ( A , B ) (A, B) (A,B) that satisfies all of the following conditions. If no such pair exists, report so.
− 1 0 18 ≤ A , B ≤ 1 0 18 -10^{18} \leq A, B \leq 10^{18} 1018A,B1018
The area of the triangle with vertices at points ( 0 , 0 ) , ( X , Y ) , ( A , B ) (0, 0), (X, Y), (A, B) (0,0),(X,Y),(A,B) on the x y xy xy-plane is 1 1 1.

Constraints

− 1 0 17 ≤ X , Y ≤ 1 0 17 -10^{17} \leq X, Y \leq 10^{17} 1017X,Y1017
( X , Y ) ≠ ( 0 , 0 ) (X, Y) \neq (0, 0) (X,Y)=(0,0)
X X X and Y Y Y are integers.

Input

The input is given from Standard Input in the following format:

X X X Y Y Y

Output

If there is a pair of integers ( A , B ) (A, B) (A,B) that satisfies the conditions, print it in the following format:

A A A B B B

Otherwise, print -1.

Sample Input 1

3 5

Sample Output 1

1 1

The area of the triangle with vertices at points ( 0 , 0 ) , ( 3 , 5 ) , ( 1 , 1 ) (0, 0), (3, 5), (1, 1) (0,0),(3,5),(1,1) is 1 1 1. Thus, ( A , B ) = ( 1 , 1 ) (A, B) = (1, 1) (A,B)=(1,1) satisfies the conditions.

Sample Input 2

-2 0

Sample Output 2

0 1

Sample Input 3

8752654402832944 -6857065241301125

Sample Output 3

-1

Solution

具体见文末视频。


Code

#include <bits/stdc++.h>
#define int long longusing namespace std;typedef pair<int, int> PII;
typedef long long LL;int Exgcd(int a, int b, int &x, int &y)
{if (!b){x = 1, y = 0;return a;}int d = Exgcd(b, a % b, y, x);y -= a / b * x;return d;
}signed main()
{cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);int X, Y;cin >> X >> Y;if (((2 % __gcd(X, Y) + abs(__gcd(X, Y))) % __gcd(X, Y)) != 0)cout << -1 << endl;else{int A, B;int d = Exgcd(Y, X, A, B);cout << A * (2 / abs(d)) << " " << (-B) * (2 / abs(d)) << endl;}return 0;
}

G - Leaf Color

G题还没研究,等后面研究下。

视频题解

Atcoder Beginner Contest 340(A ~ F)


最后祝大家早日在这里插入图片描述

本文来自互联网用户投稿,该文观点仅代表作者本人,不代表本站立场。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如若转载,请注明出处:http://www.mzph.cn/news/678136.shtml

如若内容造成侵权/违法违规/事实不符,请联系多彩编程网进行投诉反馈email:809451989@qq.com,一经查实,立即删除!

相关文章

BootstrapBlazor 模板适配移动设备使用笔记

项目模板 Bootstrap Blazor App 模板 为了方便大家利用这套组件快速搭建项目&#xff0c;作者制作了 项目模板&#xff08;Project Templates&#xff09;&#xff0c;使用 dotnet new 命令行模式&#xff0c;使用步骤如下&#xff1a; 安装项目模板 dotnet new install Boo…

007集——数据存储的端序(大端序和小端序转换代码)——VB/VBA

VB/VBA存储的端序 1、要想制造高性能的VB/VBA代码&#xff0c;离了指针是很难办到的。 2、因为VB/VBA里&#xff0c;用Long来表示指针&#xff0c;而32位(包括64位兼容的)计算机里4字节整数的处理&#xff0c;是最快的方式&#xff01; 3、要想用指针来处理数据&#xff0c;…

fast.ai 深度学习笔记(六)

深度学习 2&#xff1a;第 2 部分第 12 课 原文&#xff1a;medium.com/hiromi_suenaga/deep-learning-2-part-2-lesson-12-215dfbf04a94 译者&#xff1a;飞龙 协议&#xff1a;CC BY-NC-SA 4.0 来自 fast.ai 课程的个人笔记。随着我继续复习课程以“真正”理解它&#xff0c;…

Vue-55、Vue技术vuex模块化

一、代码 1、store.js //改文件用于创建最为核心的store //引入vue import Vue from "vue"; //引入vuex import Vuex from vuex;//求和功能相关的配置 const countOptions{namespaced:true,actions:{jia:function (context,value) {console.log(action中的jia被调…

C语言什么是悬空指针?

一、问题 什么是悬空指针&#xff1f;为什么会出现&#xff1f;我们该如何避免悬空指针的出现&#xff1f; 二、解答 在C语言中&#xff0c;悬空指针指的是指向已删除&#xff08;或释放&#xff09;的内存位置的指针。如果一个指针指向的内存被释放&#xff0c;但指针本身并未…

如何实现视线(目光)的检测与实时跟踪

如何实现视线(目光)的检测与实时跟踪 核心步骤展示说明 找到人脸 检测人脸特征点 根据特征点找到人眼区域 高精度梯度算法检测瞳孔中心 根据眼睛周边特征点计算眼睛中心 瞳孔中心和眼睛中心基于视线模型计算视线方向 视线方向可视化 详细实现与说明&#xff1a; https://stud…

ubuntu20.04 安装mysql(8.x)

安装mysql命令 sudo apt-get install mysql-server安装完毕后&#xff0c;立即初始化密码 sudo mysql -u root # 初次进入终端无需密码ALTER USER rootlocalhost IDENTIFIED WITH caching_sha2_password BY yourpasswd; # 设置本地root密码设置mysql远程登录 设置远程登录账…

Linux目录的 /usr/bin 和 /usr/local/bin 的区别

Linux目录的 /usr/bin 和 /usr/local/bin 的区别 usr 是指 Unix System Resource&#xff0c;而不是User usr 是 Unix System Resource&#xff0c;而不是User /usr/bin下面的都是系统预装的可执行程序&#xff0c;系统升级有可能会被覆盖. /usr/local/bin 目录是给用户放置…

【深度学习】:实验6布置,图像自然语言描述生成(让计算机“看图说话”)

清华大学驭风计划 因为篇幅原因实验答案分开上传&#xff0c;深度学习专栏持续更新中&#xff0c;期待的小伙伴敬请关注 实验答案链接http://t.csdnimg.cn/bA48U 有任何疑问或者问题&#xff0c;也欢迎私信博主&#xff0c;大家可以相互讨论交流哟~~ 案例 6 &#xff1a;图像自…

Hadoop运行环境搭建

模板虚拟机环境准备 1&#xff09;准备一台模板虚拟机hadoop100&#xff0c;虚拟机配置要求如下&#xff1a; 模板虚拟机&#xff1a;内存4G&#xff0c;硬盘50G&#xff0c;安装必要环境&#xff0c;为安装hadoop做准备 [roothadoop100 ~]# yum install -y epel-release [r…

命令行随笔

1、xargs xargs命令是将 前一个命令的标准输出作为后一个命令的命令行参数&#xff0c;xargs的默认命令是echo&#xff0c;默认定界符是空格和回车。 而管道是将 前一个命令的标准输出作为后一个命令的标准输入 echo例子 # echo "apple banana orange" | xargs e…

MySQL篇----第十八篇

系列文章目录 文章目录 系列文章目录前言一、SQL 语言包括哪几部分?每部分都有哪些操作关键二、完整性约束包括哪些?三、什么是锁?四、什么叫视图?游标是什么?前言 前些天发现了一个巨牛的人工智能学习网站,通俗易懂,风趣幽默,忍不住分享一下给大家。点击跳转到网站,…

梯度提升树系列8——GBDT与其他集成学习方法的比较

目录 写在开头1. 主要集成学习算法对比1.1 GBDT1.2 随机森林1.3 AdaBoost1.4 整体对比2. 算法性能的比较分析2.1 准确率与性能2.2 训练时间和模型复杂度2.3 应用实例和案例研究3. 选择合适算法的标准3.1 数据集的特性3.1.1 数据规模与维度3.1.2 数据质量3.2 性能需求3.2.1 准确…

Unity报错Currently selected scripting backend (IL2CPP) is not installed

目录 什么是il2cpp il2cpp换mono Unity打包报错Currently selected scripting backend (IL2CPP) is not installed 什么是il2cpp Unity 编辑器模式下是采用.net 虚拟机解释执行.net 代码,发布的时候有两种模式,一种是mono虚拟机模式,一种是il2cpp模式。由于iOS AppStore…

pandas dataframe写入excel的多个sheet页面

pandas根据dataframe生成一个excel文件&#xff1a; Dataframe保存新文件 直接把dataframe格式的数据保存到多个sheet页程序如下&#xff1a; excel_file "导出excel文件.xlsx" if os.path.exists(excel_file):os.remove(excel_file)# 生成一个新文件 with pd.Ex…

怎么对接快团团团长?如何对接快团团团长?

1、首先来说&#xff0c;你要需要&#xff0c;树立良好的心态&#xff0c;拓展快团团大团长合作和开发传统渠道是一样的&#xff0c;能有10%的回复率就不错了&#xff0c;反复几次沟通也是非常有必要的。要有“大团长思维”&#xff0c;就是换位思考&#xff0c;他们是处于什么…

Unity类银河恶魔城学习记录3-6 Finalize BattleState源代码 P52

Alex教程每一P的教程原代码加上我自己的理解初步理解写的注释&#xff0c;可供学习Alex教程的人参考 此代码仅为较上一P有所改变的代码 【Unity教程】从0编程制作类银河恶魔城游戏_哔哩哔哩_bilibili Enemy.cs using System.Collections; using System.Collections.Generic; …

【Opencv学习】04-图像加法

文章目录 前言一、图像加法混合1.1 代码1.2 运行结果 二、图像的按位运算-组合相加2.1 代码2.2 运行结果示例&#xff1a;PPT平滑切换运行结果 总结 前言 简单说就是介绍了两张图如何组合在一起。 1、混合&#xff0c;透明度和颜色会发生改变 2、组合&#xff0c;叠加起来。可…

【讨论】C语言提高之指针表达式

在理解指针表达式之前先有一个概念就是“左值”和“右值”&#xff0c;对于左值就是可以出现在赋值符号左边的东西&#xff0c;右值就是那些可以出现在赋值符号右边的东西。进一步抽象可以这样理解&#xff1a;左值应该可以作为一个地址空间用来存放一个值&#xff0c;而右值可…

Unity学习笔记(零基础到就业)|Chapter03:C#核心

Unity学习笔记(零基础到就业)|Chapter03:C#核心 前言一、面向对象编程二、面向对象编程三大特性(一)封装1.类和对象(1)什么是类(2)类的声明(3)什么是(类)对象(4)实例化(类)对象的语法(5)类和结构体的区别2.成员变量和访问修饰符(1)成员变量基本规则(2)实…