```
COWA19B (Pongal Bunk)
concept related: prefix array sum
Approach:-
Maintain two arrays (array1,array2).
-> first array (array1) to know the number of queries having the specific index in their range of L,R.
for each query apply, 1)array1[L]=array1[L]+1
2) array1[R+1]=array1[R+1]-1
->second array (array2) to know the final value of element in the specific index of array
for each query to apply, 1) array2[R+1]=array2[L]-(R-L+1)
After Performing All the queries operations,
->Run an iteration for prefix array sum for array1 i.e , array1[i]=array[i]+array1[i-1] where 1<=i<=N
->Run other iteration for array2 similar to prefix sum doing following operations array2[i]=array2[i] + array2[i-1] + array1[i] where 1<=i<=N
Now, array2 contains the final values of elements present in array
Time-complexity: O(n+q+m)
reference-code for the problem is given below
__________________
#include <bits/stdc++.h>
#define ll long long int
#define mod 1000000007
using namespace std;
int main()
{
ll ans[1000002],ans1[1000002];
memset(ans1,0,sizeof(ans1));
memset(ans,0,sizeof(ans));
ll n,q,l,r;
cin>>n>>q;
while(q--)
{
cin>>l>>r;
ans[l]+=1;ans[r+1]-=1;
ans1[r+1]-=(r-l+1);
}
for(int i=1;i<=n;i++)
ans[i]+=ans[i-1];
for(int i=1;i<=n;i++)
ans1[i]+=ans1[i-1]+ans[i];
cin>>q;
while(q--)
{
cin>>l;
cout<<ans1[l]<<endl;
}
}
_________________
some-useful-links:-
https://www.geeksforgeeks.org/prefix-sum-array-implementation-applications-competitive-programming/
https://www.geeksforgeeks.org/constant-time-range-add-operation-array/
```