AtCoder C problem summary that can be solved in high school mathematics

import math
a,b,x=[int(s) for s in input().split()]
if x>a*a*b/2:
  h=2*(a*a*b-x)/(a*a)
  print(math.degrees(math.atan(h/a)))
else:
  k=2*x/(a*b)
  print(math.degrees(math.atan(b/k)))

N,K=[int(s) for s in input().split()]
print((N+K-3)//(K-1))

The required number of intervals is $ \ left \ lceil \ dfrac {N-1} {K-1} \ right \ rceil $. Here, $ \ left \ lceil \ dfrac nm \ right \ rceil = \ left \ floorloor \ dfrac {n + m-1} n \ right \ rfloor $ is used and the ceiling function is not called. (I also learned this formula from the explanation of ABC.)

N=int(input())
ls=[]
while True:
  ls.append(N%2)
  N=-(N//2)
  if N==0:
    break
print(*reversed(ls),sep="")

The hardest part of this problem was to output a list of numbers like [1,1,0,1] in reverse order like 1011, and then connect them. I've previously told you that you can easily connect and output lists with print (* ls, sep = ""). Then, in detail, I lost when $ N = 0 $.

N,K=[int(s) for s in input().split()]
if K%2==1:
  print((N//K)**3)
else:
  print((N//K)**3+((N//(K//2)-N//K)**3))

import fractions
N,X=[int(s) for s in input().split()]
ls=[int(s)-X for s in input().split()]
a=0
for e in ls:
  a=fractions.gcd(a,e)
print(abs(a))

It can be seen that the sum of (absolute values) of the height difference between the adjacent flowers and the ground at both ends is twice the number of times of watering required.

More simply, just add the green part and you'll get the answer. This should be output.

N=int(input())
ls=[int(s) for s in input().split()]
ls.append(0)
a=0
for i in range(N):
  a+=max([0,ls[i]-ls[i+1]])
print(a)

N,M=[int(s) for s in input().split()]
if N>=M:
  print(0)
else:
  ls=[int(s) for s in input().split()]
  ls.sort()
  d=[ls[i+1]-ls[i] for i in range(M-1)]
  d.sort()
  print(sum(d[:M-N]))

The answer is automatically 0 for $ N \ ge M $. The d [: M-N] in the last line is the array d cut out in the middle (before M-N).

import fractions
N=int(input())
ls=[int(s) for s in input().split()]
a=0
for e in ls:
  a=fractions.gcd(a,e)
print(a)

If you ask for the time it takes to carry N people by the slowest means of transportation, that time + 4 minutes is the answer.

import math
N = int(input())
A = int(input())
B = int(input())
C = int(input())
D = int(input())
E = int(input())
X = min([A,B,C,D,E])
t=math.ceil(N/X)
print(t+4)

Or use $ \ left \ lceil \ dfrac NX \ right \ rceil = \ left \ floorloor \ dfrac {N + X-1} X \ right \ rfloor $

N = int(input())
ls = [int(input()) for i in range(5)]
X = min(ls)
print((N+X-1)//X+4)

import math
N,K=[int(s) for s in input().split()]
a=0
for i in range(N):
  a+=2**(-max([0,math.ceil(math.log2(K/(i+1)))]))
print(a/N)

Alternatively, you can combine a for statement and a while statement (double loop) and write as follows. In my case, there are fewer simple mistakes when using this "smart" method.

N,K=[int(s) for s in input().split()]
a=0
for i in range(N):
  x=i+1
  n=0
  while x<K:
    n+=1
    x*=2
  a+=2**(-n)
print(a/N)

b_{n}=  \begin{cases}
     b_{n-1}+b_{n-2} \quad (n\text{The steps are not broken})\\
     0\qquad (n\text{The steps are broken})
  \end{cases}

Holds for $ n \ ge2 $. The method of implementing this recursion is actually difficult for beginners, so I was wondering whether to deal with it in this article, but I dealt with it because it is too high school mathematics.

MOD=10**9+7
N,M=[int(s) for s in input().split()]
ls=[1 for _ in range(N+1)]
for i in range(M):
  ls[int(input())]=0
for n in range(2,N+1):
  if ls[n]!=0:
    ls[n]=(ls[n-1]+ls[n-2])%MOD
print(ls[N])

W,H,x,y = [int(s) for s in input().split()]
if 2*x==W and 2*y==H :
  print(W*H/2,1)
else:
  print(W*H/2,0)

The number of multiples of $ x $ that is greater than or equal to $ A $ and less than or equal to $ B $ is "the number of multiples of $ x $ that is less than $ B $"-"the multiple of $ x $ that is less than $ A-1 $". It is calculated by "number".

By the way, the fractions module does not have a function lcm to find the least common multiple. Therefore, I tried to find the least common multiple with $ C \ times D \ div gcd (C, D) $. This is the content of the "integer" field of Math A.

import fractions
A,B,C,D = [int(s) for s in input().split()]
L=C*D//fractions.gcd(C,D)
print(B-(A-1)
      -(B//C)+((A-1)//C)
      -(B//D)+((A-1)//D)
      +(B//L)-((A-1)//L))