ABC286 D - Money in Hand

https://atcoder.jp/contests/abc286/tasks/abc286_d

提出

code: python

import itertools

n, x = list(map(int, input().split()))

ab = list(map(int, input().split())) for _ in range(n)

# print(pow(50, 50))

wa = []

for a, b in ab:

tmp = []

for i in range(b+1):

tmp.append(a*i)

wa.append(tmp)

# print(wa)

# 0, 2, 4, 6], [0, 5, 10, 15, 20, 25, 30

# 19 % 5 = 4

解答

code: python

n, x = map(int, input().split())

# 2 19

ab = list(map(int, input().split())) for _ in range(n)

# 2, 3], [5, 6

# dpij: i種類の硬貨でj円を払うことができるのか

dp = [False for _ in range(x+1) for __ in range(n+1)]

dp00 = True

for i in range(n):

for j in range(x+1):

a = abi0

b = abi1

for k in range(b+1):

if dpij and j + a*k <= x:

dpi+1j + a*k = True

if dpnx:

print("Yes")

else:

print("No")

テーマ

#dp

メモ

https://note.com/kai1023/n/nf116047654aa#52b97b9e-a796-4319-b055-3d8c2d8cc07b

提出

code: python

n, x = map(int, input().split())

ab = list(map(int, input().split())) for _ in range(n)

# print(ab)

# 2, 3], [5, 6

# dp = False * (pow(10, 4) + 5)

# dp0 = True

# for a, b in ab:

# for i in range(1, b+1):

# dpi*a = True

# i 個で

# dpij :=

c = []

for a, b in ab:

tmp = []

for i in range(1, b+1):

tmp.append(i*a)

c.append(tmp)

print(c)

# 2, 4, 6], [5, 10, 15, 20, 25, 30

# 1], 2, [100, 200, 300, 400, 500, 600, 700, 800, 900, 1000

# pow(2, bn) * n

# for v in c: