[
segment tree
]
CodeForces Segment Trees, Part 2, 3.2
Problem statement
https://codeforces.com/edu/course/2/lesson/5/3/practice/contest/280799/problem/B
Solution
See comments in code.
Complexity
Standart compexities for segment tree.
Code
class SegmentTree:
def __init__(self, n):
self.size = 1
while self.size < n:
self.size *= 2
self.T = [0] * (2 * self.size - 1) # keep sums
self.L = [0] * (2 * self.size - 1) # for each node 1 if we need to flip
self.ZERO = 0 # neutral element, for min it is +inf
self.NO_OPERATION = 0 # this is for our propagate to understand that we do not need it
def op_modify(self, a, len_):
return len_ - a
def op_sum(self, a, b):
return a + b
def propagate(self, x, lx, rx):
if self.L[x] % 2 == self.NO_OPERATION or rx - lx == 1:
return
mx = (lx + rx)//2
self.L[2 * x + 1] += 1
self.T[2 * x + 1] = self.op_modify(self.T[2 * x + 1], mx - lx)
self.L[2 * x + 2] += 1
self.T[2 * x + 2] = self.op_modify(self.T[2 * x + 2], rx - mx)
self.L[x] = self.NO_OPERATION
def _update(self, l, r, x, lx, rx):
self.propagate(x, lx, rx)
if l >= rx or lx >= r:
return
if lx >= l and rx <= r:
self.L[x] = 1
self.T[x] = self.op_modify(self.T[x], rx - lx)
return
mx = (lx + rx)//2
self._update(l, r, 2*x+1, lx, mx)
self._update(l, r, 2*x+2, mx, rx)
self.T[x] = self.op_sum(self.T[2*x+1], self.T[2*x+2])
def update(self, l, r):
return self._update(l, r, 0, 0, self.size)
def find_k_(self, k, x, lx, rx):
self.propagate(x, lx, rx)
if rx - lx == 1:
return lx
mx = (lx + rx) // 2
if k < self.T[2 * x + 1]:
return self.find_k_(k, 2 * x + 1, lx, mx)
else:
return self.find_k_(k - self.T[2 * x + 1], 2 * x + 2, mx, rx)
def find_k(self, k):
return self.find_k_(k, 0, 0, self.size)
if __name__ == '__main__':
n, m = [int(i) for i in input().split()]
MOD = 10**9 + 7
STree = SegmentTree(n)
for i in range(m):
t = [int(i) for i in input().split()]
if t[0] == 1:
STree.update(t[1], t[2])
else:
print(STree.find_k(t[1]))