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using namespace std; int main(){ int n; int k; cin >> n >> k; int a[n]; int m[k]; for(int i=0; i<k; i++) m[i]=0; for(int i = 0; i < n; i++){ cin >> a[i]; m[a[i]%k]++; } int sum=0; sum+=(m[0]*(m[0]-1))/2; for(int i=1; i<=k/2 && i!=k-i; i++){ sum+=m[i]*m[k-i]; } if(k%2==0) sum+=(m[k/2]*(m[k/2]-1))/2; cout<<sum; return 0; }
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Divisible Sum Pairs
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I did this in O(n)
If any suggestions please let me know