In this HackerEarth Special Matrix problem solution, you have an N x M matrix, where 1 to N are rows and 1 to M are columns.

Element at the intersection of ith row and jth column is equal to F(i + j), where F(x) is equal to the number of prime divisors of x. Determine the sum of all the elements of the matrix.


hackerEarth Special matrix problem solution


HackerEarth Special matrix problem solution.

#include "bits/stdc++.h"
#define debug(...) fprintf(stderr, __VA_ARGS__), fflush(stderr)
#define time__(d) \
    for ( \
        auto blockTime = make_pair(chrono::high_resolution_clock::now(), true); \
        blockTime.second; \
        debug("%s: %d ms\n", d, (int)chrono::duration_cast<chrono::milliseconds>(chrono::high_resolution_clock::now() - blockTime.first).count()), blockTime.second = false \
    )
#ifdef LOCAL
#include "pprint.hpp"
#endif
#define endl '\n';
#define pb push_back
#define mod 1000000007
#define ll long long int
#define prn(x) cerr<<x<<endl;
#define all(x) x.begin(),x.end()
using namespace std;
#define test_case 1

const int Mx = 2e6+5;
vector<int> v, p;

void solution(){
    int n,m;
    cin>>n>>m;
    assert(n>=1 && n<=1000000);
    assert(m>=1 && m<=1000000);
    if(n<m)
        swap(n,m);
    ll ans=0;
    for(int i=1;i<=n;++i){
        int tot=min(i,m);
        ans += (ll)v[i+1]*(ll)tot;
    }
    int tot=m-1;
    for(int i=2;i<=m;++i){
        ans += (ll)v[n+i]*(ll)tot;
        tot--;
    }
    cout<<ans<<endl;
}

int main(int argc, char *argv[])
{
    #ifdef LOCAL
        freopen("in.txt" , "r" , stdin);
    #else
        ios_base::sync_with_stdio(false);
        cin.tie(NULL); cout.tie(NULL);
    #endif
    p.assign(Mx+5,1);
    v.assign(Mx+5,0);
    p[0]=p[1]=0;
    for(int i=2;i<Mx;++i)
        if(p[i])
            for(int j=i;j<Mx;j+=i){
                v[j]++;
                p[j]=0;
            }
    int t=1;
    if(test_case)
        cin>>t;
    assert(t>=1 && t<=10);
    time__("Solution Time"){
        while(t--){
            solution();
        }
    }

    return 0;
}