香槟塔

n, m = map(int, input().split())
a = [int(i) for i in input().split()]
b = [0 for i in range(n)] for i in range(m):
    order = [int(j) for j in input().split()]  if order[0] == 1:  print(b[order[1]-1])  else:
        k = 0   while b[order[1]+k-1]+order[2] > a[order[1]+k-1] and order[1]+k < n:
            order[2] = order[2] - (a[order[1]+k-1]-b[order[1]+k-1])
            b[order[1]+k-1] = a[order[1]+k-1]
            k += 1   b[order[1] + k-1] += min(order[2], (a[order[1]+k-1]-b[order[1] + k-1]))

Python Solution
假设第 层初始状态值为 ，则装满的情况为该层状态值达到 。

假如第1层容积为2，则初始状态值为-2，倒入体积3的酒，则状态值归0，剩余体积1的酒向下流动。

n, m = [int(i) for i in  input().split()]
a = [int(i) for i in input().split()]
st = [-i for i in a]
all_ml = [[int(i) for i in input().split()] for i in range(m)]


def pour(st, ml):
    '''香槟塔的第x层到体积v的酒'''
    x = ml[1]-1
    v = ml[2]
    while True:
        v = v + st[x]
        if v < 0:
            st[x] = v
            break
        else:
            st[x] = 0
            x += 1
            if x == len(st):
                break
    return st

def ask(a, st, ml):
    '''k层装酒量=k层容积+k层状态值'''
    k = ml[1]
    print(a[k-1]+st[k-1])


for ml in all_ml:
    if ml[0] == 2:
        st = pour(st, ml)
    else:
        ask(a, st, ml)