本文将不会更新视频题解.
A. 输入输出实验——scanf和printf的
def main():
a, b = map(int, input().split())
if a == 1:
year, month, day = map(int, input().split("-"))
elif a == 2:
month, day, year = map(int, input().split("/"))
else:
day, month, year = map(int, input().split("/"))
if b == 1:
print("%04d" % year, "%02d" % month, "%02d" % day, sep="-")
elif b == 2:
print("%02d" % month, "%02d" % day, "%04d" % year, sep="/")
else:
print("%02d" % day, "%02d" % month, "%04d" % year, sep="/")
return
main()
B. 简单思维训练——这是一道$A+B$
def main():
a, b = map(int, input().split())
c = sum(map(int, input().split()))
d, e = min(a, c), min(b, 10 - c)
f, g = max(a, 10 - c), max(b, c)
print(d + e, f + g - 10)
return
main()
C. 初遇高精度——这还是一道$A+B$
def main():
input()
a, b = int(input().split()[-1][-1]), int(input().split()[-1][-1])
print(str(a + b)[-1])
return
main()
D. 时间复杂度入门——从$l$加到$r$
def main():
n = int(input())
for _ in range(n):
l, r = map(int, input().split())
print((l + r) * (r - l + 1) // 2)
return
main()
E. 空间复杂度入门——计算空间开销
def main():
n, m = map(int, input().split())
s = [0, 0]
for _ in range(n):
a, b = map(int, input().split())
s[1] += a * b
s = [s[1] // 1024, s[1] % 1024]
if s[0] > m or (s[0] == m and s[1] != 0):
print("MLE")
else:
print("OK\n{}KB {}B".format(*s))
return
main()
F. 考拉兹猜想——怎么又是你
def main():
k = int(input())
match k % 3:
case 1:
print(2)
case 2:
print(4)
case 0:
print(1)
return
main()
G. 子串计数I
def main():
n = int(input())
s = input()
l = [0] * 4
for i in range(n - 1):
l[{"00": 0, "01": 1, "10": 2, "11": 3}[s[i:i + 2]]] += 1
print(*l)
return
main()
H. 子串计数II
def main():
n, q = map(int, input().split())
s = input()
p, a = 0, [0]
for i in range(n - 1):
p += {"00": 0, "01": 1, "10": -1, "11": 0}[s[i:i + 2]]
a.append(p)
while q:
l, r = map(int, input().split())
print("No" if a[r - 1] - a[l - 1] else "Yes")
q -= 1
return
main()
I. 寻找消失的集合I
def main():
input()
a = set(map(int, input().split()))
c = set(map(int, input().split()))
for b in map(int, input().split()):
if b in a and b in c:
print("Yes")
elif b not in a and b not in c:
print("Yes")
else:
print("No")
return
main()
J. 寻找消失的集合II
def main():
n, m, k = map(int, input().split())
a = set(map(int, input().split()))
c = set(map(int, input().split()))
b = c.difference(a)
p = list(a.difference(b))
for i in range(k - m + n):
b.add(p[i])
print(*b)
return
main()
K. 后缀表示法
class Rpn:
stack = []
@staticmethod
def construct(s):
n, *r = s.split(".")
n = int(n) * 10 ** 4
if r:
n += int(r[0]) * 10 ** (4 - len(r[0]))
return n
def push(self, x):
self.stack.append(x)
return
def pop(self):
return self.stack.pop()
def addition(self):
r, l = self.pop(), self.pop()
self.push(l + r)
return
def subtraction(self):
r, l = self.pop(), self.pop()
self.push(l - r)
return
def multiplication(self):
r, l = self.pop(), self.pop()
self.push(l * r // 10 ** 4)
return
def division(self):
r, l = self.pop(), self.pop()
self.push(l * 10 ** 4 // r)
return
def main():
n = int(input())
rpn = Rpn()
q = input().split()
for i in range(n):
a = q[i]
if a == "+":
rpn.addition()
elif a == "-":
rpn.subtraction()
elif a == "*":
rpn.multiplication()
elif a == "/":
rpn.division()
else:
rpn.push(rpn.construct(a))
r = str(rpn.pop())
print("{}.{}".format(r[:-4], r[-4:]))
return
main()
L. 能做到这题
def main():
n = int(input())
p, q, r = [], [], []
for _ in range(n):
a, b, c = input().split()
c = int(c)
if c == 2:
p.append(a)
p.append(b)
elif c == -2:
q.append(a)
q.append(b)
else:
r.append((a, b))
for i in r:
if i[0] in p:
q.append(i[1])
r.remove(i)
break
elif i[0] in q:
p.append(i[1])
r.remove(i)
break
elif i[1] in p:
q.append(i[0])
r.remove(i)
break
elif i[1] in q:
p.append(i[0])
r.remove(i)
break
print(1, p[0], sep="\n")
return
main()
M. 注意力惊人A
def main():
n = int(input())
print(*[5 * i for i in range(1, n + 1)])
return
main()
N. FBI #2
def main():
fib = [0, 1]
n = int(input())
l = [int(input()) for _ in range(n)]
for i in range(2, max(l) + 1):
fib.append(fib[i - 1] + fib[i - 2])
print(*[fib[i] for i in l], sep="\n")
return
main()