[
segment tree
]
CodeForces Segment Trees, Part 1, 3.2
Problem statement
https://codeforces.com/edu/course/2/lesson/4/3/practice/contest/274545/problem/B
Solution
This is problem, is opposite to previous, given answer, return permutation. The idea is to use problem 2.2 where we need to find k-th one.
Imagine, that we have [4 1 3 5 2]
then in the beginning we have [1 1 1 1 1] -> [1 0 1 1 1] -> [1 0 1 1 0] -> ...
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 find_k_(self, k, 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 = int(input())
arr = [int(i) for i in input().split()][::-1]
ans = []
STree = SegmentTree(n, [1]*n)
STree.build()
for num in arr:
k = STree.find_k(num)
STree.set(k, 0)
ans += [n-k]
print(" ".join(str(x) for x in ans[::-1]))