Only one test case passed

#1

Hi,
I am trying to submit the code of divisible subarray but only one test case is passed , i am not getting the what’s wrong with my code , i am taking prefix sum of array and after that taking the modulo with n , and storing the frequency after taking modulo n and at last i am summing up all the values ans = pre[i]*pre[i-1]/2

#2

hey @deepu.kumar2222,what you are caluating as ans is correct . while storing frequency of sum initialize freq[0]=1 and then continue your logic it will work fine.

#3

i had done this still not getting the right answer

#4

Code:- #include<bits/stdc++.h> using namespace std; #define ll long ll a[1000005], pre[1000005]; int main() { int t; cin>>t; while(t–) { int n; cin>>n; int temp = 0; for(int i=0;i<n;i++) { cin>>a[i]; temp = temp + a[i]; a[i] = temp; } pre[0] = 1; for(int i=0;i<n;i++) { a[i] = a[i]%n; pre[a[i]]++; } int ans = 0; for(int i=0;i<n;i++) { int m = pre[i]; ans = ans + ((m)*(m-1))/2; } cout<<ans<<endl; } }

#5

@deepu.kumar2222,you have to update following things in your code:
1.since you have multiple cases initialize arrays inside while(t–) loop.
2.pre[] array should have all values zero initially except pre[0].
3.a[i] can be negative so after taking modulo, in next line write a[i]=(a[i]+n)%n;
4.take long long int everywhere.

#6

sir still getting wrong answer on test case 0,1,2,3

#7

@deepu.kumar2222,refer to this i have updated your code(https://ide.codingblocks.com/s/178790) .

#8

@deepu.kumar2222 i hope i have cleared your doubt if you have any else please let me know else mark this doubt as resolved.

#9

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