[
segment tree
]
CodeForces Segment Trees, Part 1, 3.3
Problem statement
https://codeforces.com/edu/course/2/lesson/4/3/practice/contest/274545/problem/C
Solution
The idea is the following: go from the left to the right and do the following:
- If we meet start of segment, save it.
- If we meet end of segment, set start of this segment as 1 and calculate number of ones inside.
So we will use problem 1.1.
Complexity
Standart compexities for segment tree.
Code
class SegmentTree:
def __init__(self, n, arr):
self.size = 1
while self.size < n:
self.size *= 2
self.T = [0] * (2 * self.size - 1)
self.arr = arr
def _build(self, x, lx, rx):
if rx - lx == 1:
if lx < len(self.arr):
self.T[x] = self.arr[lx]
else:
mx = (lx + rx)//2
self._build(2*x+1, lx, mx)
self._build(2*x+2, mx, rx)
self.T[x] = self.T[2*x+1] + self.T[2*x+2]
def build(self):
self._build(0, 0, self.size)
def _set(self, i, v, x, lx, rx):
if rx - lx == 1:
self.T[x] = v
return
mx = (lx + rx) // 2
if i < mx:
self._set(i, v, 2 * x + 1, lx, mx)
else:
self._set(i, v, 2 * x + 2, mx, rx)
self.T[x] = self.T[2 * x + 1] + self.T[2 * x + 2]
def set(self, i, v):
self._set(i, v, 0, 0, self.size)
def _sum(self, l, r, x, lx, rx):
if l >= rx or lx >= r:
return 0
if lx >= l and rx <= r:
return self.T[x]
mx = (lx + rx) // 2
s1 = self._sum(l, r, 2 * x + 1, lx, mx)
s2 = self._sum(l, r, 2 * x + 2, mx, rx)
return s1 + s2
def sum(self, l, r):
return self._sum(l, r, 0, 0, self.size)
# import io, os, sys
# input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
if __name__ == '__main__':
n = int(input())
arr = [int(i) for i in input().split()]
ans = [0]*n
d = {}
STree = SegmentTree(2*n, [0]*(2*n))
STree.build()
for i, num in enumerate(arr):
if num not in d:
d[num] = i
else:
start = d[num]
ans[num-1] = STree.sum(start+1, i)
STree.set(start, 1)
print(" ".join(str(x) for x in ans))