HackerEarth Manhattan distance problem solution

In this HackerEarth Manhattan distance problem solution, There are N towns in a coordinate plane. Town i is located at coordinate (xi,yi). The distance between town i and town j is |xi – xj| + |yi – yj|. Your task is to compute the sum of the distance between each pair of towns.

HackerEarth Manhattan distance problem solution.

#include<bits/stdc++.h>
using namespace std;

#define     F           first
#define     S           second
#define     pb          push_back
#define     lb          lower_bound
#define     ub          upper_bound
#define     vi          vector<int>
#define     all(x)      x.begin(),x.end()
#define     fix         fixed<<setprecision(10)
#define     rep(i,a,b)  for(int i=int(a);i<=int(b);i++)
#define     repb(i,b,a) for(int i=int(b);i>=int(a);i--)
#define     FastIO      ios_base::sync_with_stdio(0),cin.tie(0)

typedef double db;
typedef long long ll;

const int N=2e5+5;
const int mod=1e9+7;

int n,x[N],y[N];

void solve(){
    cin>>n;
    rep(i,1,n) cin>>x[i]>>y[i];
    sort(x+1,x+n+1);
    sort(y+1,y+n+1);
    ll ans=0,sumx=0,sumy=0;
    rep(i,1,n){
        ans+=1ll*x[i]*(i-1)-sumx;
        ans+=1ll*y[i]*(i-1)-sumy;
        sumx+=x[i];
        sumy+=y[i];
    }
    cout<<ans<<'n';
}

signed main(){
    FastIO;
    int t;
    cin>>t;
    while(t--) solve();
    return 0;
}

Second solution

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

const int N = 1e6 + 14;

int n, x[2][N];

int main() {
    ios::sync_with_stdio(0), cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
        cin >> n;
        ll ans = 0;
        for (int i = 0; i < n; ++i)
            cin >> x[0][i] >> x[1][i];
        for (auto &c : x) {
            sort(c, c + n);
            for (int i = 0; i < n; ++i)
                ans += c[i] * (ll) i - c[i] * ll(n - i - 1);
        }
        cout << ans << 'n';
    }
}

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