Résoudre AtCoder ABC168 avec python (A ~ D)

# A

N = int(input())
n = N % 10

if n in [2, 4, 5, 7, 9]:
    print('hon')
elif n in [0, 1, 6, 8]:
    print('pon')
else:
    print('bon')

# B

K = int(input())
S = input()

len_S = len(S)
if len_S <= K:
    print(S)
else:
    print(S[:K] + '...')

# C

A, B, H, M = list(map(int, input().split()))

import math
A_x = A * math.cos(2 * math.pi * (H + M / 60.) / 12.)
A_y = A * math.sin(2 * math.pi * (H + M / 60.) / 12.)
B_x = B * math.cos(2 * math.pi * M / 60.)
B_y = B * math.sin(2 * math.pi * M / 60.)

print('{:.15f}'.format(math.sqrt((A_x - B_x) ** 2 + (A_y - B_y) ** 2)))

# D
 
import numpy as np
from scipy.sparse import csr_matrix
from scipy.sparse.csgraph import breadth_first_order
 
N, M = list(map(int, input().split()))
ABs = np.array([list(map(int, input().split())) for i in range(M)])
 
row = ABs.T[0]-1
col = ABs.T[1]-1
data = [1] * (M)
 
csr = csr_matrix((data, (row, col)), shape=(N, N))
_, proc = breadth_first_order(csr, 0, directed=False)
 
if -9999 in proc[1:]:
  print('No')
else:
  print('Yes')
  print('\n'.join((proc[1:]+1).astype('str')))

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