Divisible Subarrays: Please help me with this

#1

#include
using namespace std;
#define ll long long
int main(){

int t;
ll n;
ll arr[10000005];
cin>>t;
for(int k=0;k<t;k++){
	
	ll prefix_sum[10000005] = {0};
	int sum=0;
	cin>>n;
    prefix_sum[0] = 1;
	for(ll i=0;i<n;i++){
		cin>>arr[i];
		sum+=arr[i];
		
        if(sum<0){
            sum = (sum+n)%n;
        }
        else{
            sum%=n;
        }
        
		prefix_sum[sum]++;
	}

	ll ans=0;
	for(ll i=0;i<n;i++){
		long long int number = prefix_sum[i];
		ans+= (number*(number-1))/2;
	}
	cout<<ans<<endl;
}


return 0;

}

#2

hello @Aditikumari

use modulo property here to avoid overflow.
sum=(sum+arr[i])%n;

#3

#include
using namespace std;
#define ll long long
int main(){

int t;
ll n;
ll arr[10000005];
cin>>t;
for(int k=0;k<t;k++){
	
	ll prefix_sum[10000005] = {0};
	int sum=0;
	cin>>n;
    prefix_sum[0] = 1;
	for(ll i=0;i<n;i++){
		cin>>arr[i];

sum=(sum+arr[i])%n;

        if(sum<0){
            sum = (sum+n)%n;
        }
        else{
            sum%=n;
        }
        
		prefix_sum[sum]++;
	}

	ll ans=0;
	for(ll i=0;i<n;i++){
		long long int number = prefix_sum[i];
		ans+= (number*(number-1))/2;
	}
	cout<<ans<<endl;
}


return 0;

} Sir still showing error

#4

Sir can u provide me the correct code… so that i may know my mistake

#5

ok pls save ur code here-> https://ide.codingblocks.com/
and share the link with me

#6

https://ide.codingblocks.com/s/240972 saved sir, please check

#7

@Aditikumari

checker is showing runerror even for correct solutions. may be there is some issue is with server.
pls try again after sometime,

#8

sure sir, sir my solution is correct??

#9

yeah . . . . . . . .

#10

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