当前位置:网站首页>慧通编程第4关 - 魔法学院第6课
慧通编程第4关 - 魔法学院第6课
2022-07-31 03:34:00 【中学生信竞】
梳子
int main(){
p.size(8).c(6);
for (int i = 1; i <= 6; i ++) {
p.fd(40).rt(90);
p.fd(40).rt(90);
p.fd(40).rt(180);
}
return 0;
}
锯齿
int main(){
for (int i = 1; i <= 7; i ++) {
p.rt(30).fd(30);
p.rt(120).fd(30);
p.lt(150);
}
return 0;
}
楼梯型
int main(){
p.c(11);
for (int i = 1; i <= 5; i ++) {
for (int j = 1; j <= 6; j ++) {
p.fd(30).rt(90);
}
p.lt(180);
}
return 0;
}
楼梯
int main(){
for (int i = 1; i <= 10; i ++) {
p.fd(20).rt(90);
p.fd(20).lt(90);
}
return 0;
}
实心矩形
int main(){
p.rr(100,80,14);
return 0;
}
2层实心矩形
int main(){
p.rr(100,80,14);
p.rr(80,50,0);
return 0;
}
6针形
int main(){
for (int i = 1; i <= 6; i ++) {
p.fd(100).bk(100).rt(60);
}
return 0;
}
彩色15针形
int main(){
pen.hide();
for (int i = 0; i < 15; i ++) {
pen.c(i);
pen.fd(100).bk(100) ;
pen.rt(360.0 / 15) ;
}
return 0;
}
4色30针形
int n = 30;
pen.size(5).hide();
for (int i = 0; i < n; i ++) {
pen.c(i % 4);
pen.fd(100).bk(100) ;
pen.rt(360.0 / n) ;
}
彩色环形
int main(){
int n = 10;
pen.size(8).hide();
for (int i = 0; i < n; i ++) {
pen.c(i);
pen.fd(50) ;
pen.rt(360.0 / n) ;
}
return 0;
}
边栏推荐
- 【异常】The field file exceeds its maximum permitted size of 1048576 bytes.
- Safety 20220722
- Difference between unallocated blocks and unused blocks in database files
- 分布式锁以及实现方式三种
- Point Cloud DBSCAN Clustering (MATLAB, not built-in function)
- Key Technologies of Interface Testing
- Good place to download jar packages
- CloudCompare & PCL calculate the degree of overlap between two point clouds
- LocalDate addition and subtraction operations and comparison size
- interprocess communication
猜你喜欢

"A daily practice, happy water problem" 1331. Array serial number conversion
![[C language] General method for finding the sum of the greatest common factor and the least common multiple of two integers m and n, the classical solution](/img/60/fa75e06af4d143ee3fb493221fa3d9.jpg)
[C language] General method for finding the sum of the greatest common factor and the least common multiple of two integers m and n, the classical solution

Web container and IIS --- Middleware penetration method 1

LocalDate addition and subtraction operations and comparison size

STM32 problem collection

IIR filter and FIR filter

分布式系统架构需要解决的问题

Detailed explanation of TCP (2)

Based on the local, linking the world | Schneider Electric "Industrial SI Alliance" joins hands with partners to go to the future industry

Key Technologies of Interface Testing
随机推荐
[C language] General method for finding the sum of the greatest common factor and the least common multiple of two integers m and n, the classical solution
Recursive query single table - single table tree structure - (self-use)
2022 Nioke Multi-School League Game 4 Solution
BP神经网络
安全20220722
大小端模式
Detailed explanation of TCP (3)
自己的一些思考
Redis实现分布式锁
《DeepJIT: An End-To-End Deep Learning Framework for Just-In-Time Defect Prediction》论文笔记
The Map Entry understanding and application
【Cocos Creator 3.5】缓动系统停止所有动画
Addition and Subtraction of Scores in LeetCode Medium Questions
Redis 使用 sorted set 做最新评论缓存
Redis 使用LIST做最新评论缓存
【动态规划】连续子数组的最大和
下载jar包的好地方
立足本土,链接全球 | 施耐德电气“工业SI同盟”携手伙伴共赴未来工业
【CocosCreator 3.5】CocosCreator 获取网络状态
Just debuted "Fight to Fame", safety and comfort are not lost