from turtle import *
speed(0)
for i in range(100):
    forward(200)
    left(89)
'''
by Ching-Shoei Chiang

#1 The sum from 0 to 10
total = 0
for i in range(11):
   total = total + i
   print('The sum from 0 to ', i, "is equal to", total)

#2 Print the final result only
total = 0
for i in range(11):
   total = total + i
print('The sum from 0 to ', 10, "is equal to", total)

#3 Function
def sum2n(n):
    total = 0
    for i in range(n+1):
        total = total + i
    return total

for i in [10,20,8,17]:
    print("The sum from 0 to", i, "is", sum2n(i))

#4 print the prime number from 2 to 100
def prime(n):
    for i in range(2,n):
        if n%i==0:
            return False
    return True

for i in range(2,101):
    if prime(i):
        print(i, end='; ')
        
#5 save the result in #4 into a list, tell me how many prime numbers
def prime(n):
    for i in range(2,n):
        if n%i==0:
            return False
    return True
L=[]
for i in range(2,101):
    if prime(i):
        L.append(i)
print(L)
print(len(L))


#6 Chinese remainder Theorem using for
for i in range(4, 100):
    if i%3==1 and i%4==2 and i%5==4:
        print(i)

#7 Chinese remainder Theorem using while
answer = False
i = 4
while not answer:
    if i%3==1 and i%4==2 and i%5==4:
        print(i)
        answer = True
    i = i+1 

#8 lcm(m,n)
def lcm(m,n):
    for i in range(m,m*n+1):
        if i%m==0 and i%n==0:
            return i
print(lcm(12,8))

#9 gcd(m,n)
def gcd(m,n):
    for i in range(m, 1, -1):
        if m%i==0 and n%i==0:
            return i
print(gcd(12,8))

#10 1000到2000之間，可被3整除，但不被5整除的數有幾個?
L=[]
for i in range(1000, 2001):
    if i%3==0 and i%5!=0:
        L.append(i)
print(L)
print(len(L))

#Recursive
#11 define a function to judge whether n is an even number
def evenp(n):
    if n%2==0:
        return True
    else:
        return False
print(evenp(6))
print(evenp(9))

#12 Solve the problem #11 recursively

def evenp(n):
    if n==0:
        return True
    if n==1:
        return False
    return evenp(n-2)

print(evenp(6))
print(evenp(9))
#13 Solve #10 by 輾轉相除法
# if a = b*d + m, then gcd(a,b)=gcd(b,m)

def gcd(m,n):
    if n == 0:
        return m
    return gcd(n, m%n)

print(gcd(12,8))
print(gcd(7,3))

#4 217 a prime number
m = 20
prime=True
for i in range(2, m):
    if m%i==0:
        prime=False
if prime:
    print(m, 'is a prime')
else:
    print(m, 'is not a prime')
'''