Bella ciao long challenge What is wrong in my solution

help
, ,
1

#include
#include
#include
#include
#define ll long long
using namespace std;

int main(){
int t;
cin>>t;
while(t–){
int D,d,p,q,s,rate,count=0,total=0;
cin>>D>>d>>p>>q;
int arr[D];
for(int i=0;i<D;i++){
arr[i]=p;
}
//for(int i=d;i<D;i++){
// arr[i]+=q;
//}
for(int i=1;i<=D;i++){
s=i*d;
if(s<=D){
for(int i=s;i<D;i++){
arr[i]+=q;
}
}
else{
count++;
}
}
for(int i=0;i<D;i++){
total+=arr[i];
}
cout<<total<<endl;

}
return 0;
}

2

I did something just like this. Just try further and make up a formula

3

Getting TLE right?
Try to do in O(1) time complexity
By making formula (Sum of n terms of AP)

4

(post withdrawn by author, will be automatically deleted in 24 hours unless flagged)

5

Bro my all test cases were passed but while submitting it… It showed wrong submition

6

Try to dry run the code. You will catch exception cases easily. In problems like these, it is better to make a formula for different types of cases.

Hint : value of D/d plays an important role

7

Passing test cases doesn’t mean it will run with every possible value. Look for values that will generate a runtime error.

8

When will we get an updated rating?

9

Wait for a few hours

10

oh okay, thanks

11

Yes when I placed ll int instead of just int my submition passed partially… But later it showed TLE.

12

Actually brother there is no need for loops here.
I solved it like this

#include <bits/stdc++.h>
using namespace std;

int main() {
// your code goes here
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long t;
cin>>t;
while(t–){
long long int D,d,p,q;
cin>>D>>d>>p>>q;
long long di = D/d;
long long rem = D%d;
long long sum=0;
sum = (di*(2p+(di-1)q)(d))/2;
sum += rem(p+q*di);
cout<<sum<<endl;
}
return 0;
}

3 Likes
13

Your algorithm has a time complexity of O(n^2) I think, it’s hard to read the code. And since this problem can be solved with O(1) complexity, the time limit has been set very low.