class Solution:
def handleQuery(self, nums1: List[int], nums2: List[int], queries: List[List[int]])
-> List[int]:
n = len(nums1)
cnt1 = [0] * (4 * n)
flip = [False] * (4 * n)
# 维护区间 1 的个数
def maintain(o: int) -> None:
cnt1[o] = cnt1[o * 2] + cnt1[o * 2 + 1]
# 执行区间反转
def do(o: int, l: int, r: int) -> None:
cnt1[o] = r - l + 1 - cnt1[o]
flip[o] = not flip[o]
# 初始化线段树 o,l,r=1,1,n
def build(o: int, l: int, r: int) -> None:
if l == r:
cnt1[o] = nums1[l - 1]
return
m = (l + r) // 2
build(o * 2, l, m)
build(o * 2 + 1, m + 1, r)
maintain(o)
# 反转区间 [L,R] o,l,r=1,1,n
def update(o: int, l: int, r: int, L: int, R: int) -> None:
if L <= l and r <= R:
do(o, l, r)
return
m = (l + r) // 2
if flip[o]:
do(o * 2, l, m)
do(o * 2 + 1, m + 1, r)
flip[o] = False
if m >= L: update(o * 2, l, m, L, R)
if m < R: update(o * 2 + 1, m + 1, r, L, R)
maintain(o)
build(1, 1, n)
ans, s = [], sum(nums2)
for op, l, r in queries:
if op == 1: update(1, 1, n, l + 1, r + 1)
elif op == 2: s += l * cnt1[1]
else: ans.append(s)
return ans