TLE in Pongal Bunk problem code wars 1.0

Link to my submission : https://www.codechef.com/viewsolution/29029523

Link to the question: https://www.codechef.com/COWR2020/problems/COWA19B/

I don’t think this can be optimized further, and there must be a different way of approach… Can someone please share how to approach this…

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
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#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;
    }
} 
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some-useful-links:-

https://www.geeksforgeeks.org/prefix-sum-array-implementation-applications-competitive-programming/
https://www.geeksforgeeks.org/constant-time-range-add-operation-array/

Try changing cin/cout to scanf/printf. Worked for me.