PROBLEM LINK
Panel Members
Problem Setter:Amit Pandey
Problem Tester:Pushkar Mishra
Editorialist:Prateek Gupta
Contest Admin:Praveen Dhinwa and Pushkar Mishra
Russian Translator:Sergey Kulik
Mandarin Translator:Hu Zecong
Vietnamese Translator: VNOI Team
Language Verifier:Rahul Arora
DIFFICULTY:
Easy-Medium
PREREQUISITES:
Basic Programming, Logic
PROBLEM:
Given a binary matrix having $1s$ and $0s$. Each $i$$th$row of the matrix contains $1s$ only in the given contiguous range [$l[i], r[i]$]. For each query, you need to find the sum of the matrix modulo $2$ if some given $x$$th$ row and $y$$th$ column are supposedly removed from the matrix. Also, each query does not modify the state of the matrix after its been processed.
EXPLANATION:
Let $initialSum$ be the sum of the original matrix.
Let $rowSum[i]$ and $colSum[i]$ denote the sum of the elements of $i$$th$ row and column respectively in the original matrix.
Let $A[i][j]$ denote the value at cell $(i,j)$ formed by intersection of $i$$th$ row and $i$$th$ column.
For each query, the sum of the final matrix for each query($x$,$y$) will be equal to $initialSum$ $-$ $rowSum[x]$ $-$ $colSum[y]$ $+$ $A[x][y]$
$initialSum$ of the matrix can be calculated as summation of $r[i]$ $-$ $l[i]$ $+$ $1$ for each $i$ $\mathcal\in$ $[1,N]$.
$rowSum[x]$ is already the number of ones in the $i$$th$ row i.e $r[x]$ $-$ $l[x]$ $+$ $1$.
$A[x][y]$ is $1$ if and only if $j$ $\mathcal\in$ [$l[x]$,$r[x]$].
Only task now left is to calculate $colSum[y]$.
Solution for Subtask 1:
To calculate the value of $colSum[y]$, we need to find out how many rows have $y$ $\mathcal\in$ [$l[i]$,$r[i]$] for each $i$ from $[1,N]$.
For this, we can iterate each row and find out. This approach will take $\mathcal{O}(N)$ per query which is infact slow for Subtask $2$ to pass.
Solution for Subtask 2:
We can calculate the values of $colSum[y]$ before hand for all $y$ $\mathcal\in$ [$1$,$N$]. For each $rowSum[i]$, we basically have to increase $colSum[j]$ $+=$ $1$ where $j$ $\mathcal\in$ $[l[i],r[i]]$.
In other words, we can do a range update in the following manner :-
for ( int i = 1; i <= N; i++ ) colSum[l[i]]++, colSum[r[i]+1]--;We can then do a cumulative sum to compute individual values for each column.
for ( int i = 1; i <= N; i++ ) colSum[i] += colSum[i-1];We now have all the necessary parameters that we need to answer each query in a constant number of operations.
COMPLEXITY
Overall Complexity for the solution to pass all the subtasks is $\mathcal{O}(N)$ precomputation and then $\mathcal{O}(1)$ per query. For details on the implementation, please have a look at the solutions below.