def resolve():
N, M = LI()
LR = [LI_() for _ in range(M)]
# 余事象を考える
ans = M * (M - 1) // 2
# 1. 端点が一致する場合
cnt = collections.Counter()
for i, j in LR:
cnt[i] += 1
cnt[j] += 1
num1 = sum(i * (i - 1) // 2 for i in cnt.values())
# 2. 重複が無い場合
# 各Lに対して、R<LとなるRをカウントする
cnt = [0] * N
for _, i in LR:
cnt[i] += 1
acm = list(itertools.accumulate(cnt, initial=0))
# print(acm)
num2 = 0
for i, _ in LR:
num2 += acm[i]
# num2_ = 0
# for i in range(M - 1):
# for j in range(i + 1, M):
# aL, aR = LR[i]
# bL, bR = LR[j]
# if aR < bL or bR < aL:
# num2_ += 1
# 3. 包含関係にある場合
LR.sort(key=lambda x: x[1])
# Rの小さい順に線分を追加していって、追加した線分よりLが大きい線分を数える
# Rを小さい順に追加することで、Lだけを見ればよくなる
fwt = FenwickTree(N)
num3 = 0
for L, R in LR:
if L + 1 <= R - 1:
num3 += fwt.sum(L + 1, R - 1)
fwt.add(L, 1)
# print(num1, num2, num2_, num3)
ans = ans - num1 - num2 - num3
print(ans)
def test_add_when_negative_value_is_given(
self, fenwicktree: FenwickTree
) -> None:
'''
fenwicktree.add(p, x) is expected to add x to the p-th of the number
sequence.
GIVEN an initialized fenwicktree object
WHEN add negative value at arbitrary positions in the sequence
THEN fenwicktree.data has a partial sum at any position in the sequence
'''
for i in range(5):
fenwicktree.add(i, i + 1)
assert fenwicktree.data == [1, 3, 3, 10, 5]
fenwicktree.add(0, -5)
assert fenwicktree.data == [-4, -2, 3, 5, 5]
def test_add_multiple_times(self, fenwicktree: FenwickTree) -> None:
'''
fenwicktree.add(p, x) is expected to add x to the p-th of the number
sequence.
GIVEN an initialized fenwicktree object
WHEN add arbitrary value(s) at arbitrary positions in the sequence
THEN fenwicktree.data has a partial sum at any position in the sequence
'''
expected = [[1, 1, 0, 1, 0],
[1, 3, 0, 3, 0],
[1, 3, 3, 6, 0],
[1, 3, 3, 10, 0],
[1, 3, 3, 10, 5]
]
for i in range(5):
fenwicktree.add(i, i + 1)
assert fenwicktree.data == expected[i]
def test_initial_status(self, fenwicktree: FenwickTree) -> None:
'''
An initialized fenwicktree object is expected to have all the elements
0.
GIVEN an initialized fenwicktree object
WHEN before executing fenwicktree.add(p, x)
THEN all the elements are 0
'''
assert fenwicktree.data == [0, 0, 0, 0, 0]
pair_of_positions = self._generate_pair_of_positions()
for left, right in pair_of_positions:
assert fenwicktree.sum(left, right) == 0
def main() -> None:
n, q = map(int, sys.stdin.readline().split())
fenwick_tree = FenwickTree(n)
a = map(int, sys.stdin.readline().split())
for i, ai in enumerate(a):
fenwick_tree.add(i, ai)
for _ in range(q):
t, x, y = map(int, sys.stdin.readline().split())
if t == 0:
fenwick_tree.add(x, y)
if t == 1:
print(fenwick_tree.sum(x, y))
def main():
h, w, m = list(map(int, input().strip().split()))
x2y = [w] * h
y2x = [h] * w
obs_y2x = [[] for _ in range(h)]
for i in range(m):
xt, yt = list(map(int, input().strip().split()))
xt -= 1
yt -= 1
x2y[xt] = min(x2y[xt], yt)
y2x[yt] = min(y2x[yt], xt)
obs_y2x[xt].append(yt)
result = sum(x2y[: y2x[0]]) + sum(y2x[: x2y[0]])
tree = FenwickTree(w)
for j in range(x2y[0]):
tree.add(j, 1)
for i in range(y2x[0]):
for j in obs_y2x[i]:
tree.add(j, -tree.sum(j, j+1))
result -= tree.sum(0, x2y[i])
print(result)
def test_sum_when_additional_element_is_given(
self, fenwicktree: FenwickTree, left: int, right: int, expected: int
) -> None:
'''
fenwicktree.sum(left, right) is expected to calculate the sum of the
interval [left, right)
GIVEN With arbitrary value(s) at arbitrary positions in the sequence
added to an initialized fenwicktree object
WHEN specify the interval [left, right) after the additional element
is given
THEN returns the sum of interval [left, right)
'''
for i in range(5):
fenwicktree.add(i, i + 1)
assert fenwicktree.sum(0, 5) == 15
fenwicktree.add(2, 2)
assert fenwicktree.data == [1, 3, 5, 12, 5]
assert fenwicktree.sum(left, right) == expected
def fenwicktree() -> FenwickTree:
return FenwickTree(5)
#!/bin/python3
from atcoder.fenwicktree import FenwickTree
n = int(input().strip())
tree = FenwickTree(n)
for i in range(n):
tree.add(i, 1)
# AtCoder Library doesn't have get API.
# Insted you can use sum with span 1.
j = 3
print(tree.sum(j, j + 1))
# return the sum of a[l, r)
print(tree.sum(0, n))