 Author: rocknot
Tester: yogesh_5326
Editorialist: rocknot

EASY-MEDIUM.

PREREQUISITES:

Query optimization

You are given length of array N you need to create array arr[] of N integers having all the elements {0} and perform Q queries on it. Each query consist of three integers l,r,A
you need to add value of A to the elements of the array arr[] from l(th) element to r(th) element.

for each querry if A is even add A to all elements of array arr[] from l to r
if A is odd add from l to r according to given pattern A,A+1,A,A-1,A,A+1…..

You need to print the array arr[] after performing all the queries.

EXPLANATION:

For each query add value of A to l(th) element and -A to r+1(th) element and iterate through all the elements of the array and add the value of [i]th element of array to [i+1]th element.

For odd values of A create an temp array with all values initialized to 0 and for each odd query add pattern 0,1,0,-1,0,1,-1… in all the elements of temp array and after performing all the queries add all the elements of temp array to the main array.

SOLUTIONS:

Setter's Solution
#include <bits/stdc++.h>
using namespace std;
int main(){
int T;
cin>>T;
while(T--){
int N,Q;
cin>>N;
cin>>Q;
int arr[N]={0},arr2[N]={0};
int l,r,A;
while(Q--){
cin>>l>>r>>A;
arr[l-1]+=A;
if(r<N){
arr[r]-=A;
}
if(A%2!=0){
bool flag=true;
for(int i=l;i<r;i+=2){
if(flag){
arr2[i]+=1;
flag=false;
}else{
arr2[i]-=1;
flag=true;
}
}
}
}
for(int i=1;i<N;i++){
arr[i]+=arr[i-1];
arr[i-1]+=arr2[i-1];
}
for(int i=0;i<N;i++){
cout<<arr[i]<<" ";
}
cout<<"\n";
}
return 0;

}