PROBLEM LINK:
Author: Sunita Sen
Tester: Arnab Chanda
Editorialist: Sunita Sen
DIFFICULTY:
SIMPLE
PREREQUISITES:
None
PROBLEM:
Given an array of N integers and an integer value K, find the sum of remainders when each element of A is divided by K.
EXPLANATION:
Iterate through the given array, and for each element find the remainder when the element is divided by K.
The remainder can be calculated using the MOD(%) operator.
Lastly, add all the remainders and print it.
sum = \sum_{i=1}^{N} (A_i \% K)
SOLUTIONS:
Cpp Solution
#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(ll i=0;i<(n);i++)
#define FOR(i,a,b) for(ll i=(a);i<(b);i++)
#define dfor(i,a,b) for(ll i=(a);i>=(b);i--)
#define mem(u, vis) memset(u, vis, sizeof(u))
#define ll long long
#define ull unsigned long long
#define MOD 1000000007
#define MAX 1e9
#define pb push_back
#define F first
#define S second
int main() {
int t=1;
cin>>t;
while(t--){
ll n,k;
cin>>n>>k;
ll a[n],sum=0;
for(int i=0;i<n;++i){
cin>>a[i];
sum += a[i]%k;
}
cout<<sum<<endl;
}
}