题解 | #称砝码#
称砝码
https://www.nowcoder.com/practice/f9a4c19050fc477e9e27eb75f3bfd49c?tpId=37&tqId=21264&rp=1&ru=/exam/oj/ta&qru=/exam/oj/ta&sourceUrl=%2Fexam%2Foj%2Fta%3Fdifficulty%3D4%26page%3D1%26pageSize%3D50%26search%3D%26tpId%3D37%26type%3D37&difficulty=undefined&judgeStatus=undefined&tags=&title=
一眼过去就是递归，看完题解发现还是大佬们的方法快，给跪了
from functools import cache
@cache
def dfs(i, total):
if i == n:
ans.add(total)
return
for j in range(X[i]+1):
v = j*M[i]
total += v
dfs(i+1, total)
total -= v
while True:
try:
n = int(input().strip())
M = list(map(int, input().strip().split()))
X = list(map(int, input().strip().split()))
ans = set()
dfs(0, total=0)
print(len(ans))
except:
break
