HackerEarth Dexter and Points problem solution YASH PAL, 31 July 202416 February 2026 In this HackerEarth Dexter and Points, problem solution Dexter has you on his kill table now. He gives you one last chance to survive. He gives you a problem to solve. If you solve the problem correctly, he will let you go, else he will kill you. You are given N integers a1,a2,,…,aN. Consider an N-dimensional hyperspace. Let (x1,x2,…,xN) be a point in this hyperspace and all xi for i related [1,N] are integers. Now, Dexter gives you a set which contains all the points such that 0 <= xi <= ai for i related [1,N]. Find the number of ways to select two points A and B from this set, such that the midpoint of A and B also lies in this set. As the required number can be really large, find the answer modulo 10^9 + 7. HackerEarth Dexter and Points problem solution.#include <bits/stdc++.h>using namespace std;#define TRACE#ifdef TRACE#define TR(...) __f(#__VA_ARGS__, __VA_ARGS__)template <typename Arg1>void __f(const char* name, Arg1&& arg1){ cerr << name << " : " << arg1 << std::endl;}template <typename Arg1, typename... Args>void __f(const char* names, Arg1&& arg1, Args&&... args){ const char* comma = strchr(names + 1, ',');cerr.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);}#else#define TR(...)#endiftypedef long long LL;typedef vector < int > VI;typedef pair < int,int > II;typedef vector < II > VII;#define MOD 1000000007#define EPS 1e-12#define N 200100#define PB push_back#define MP make_pair#define F first #define S second#define ALL(v) v.begin(),v.end()#define SZ(a) (int)a.size()#define FILL(a,b) memset(a,b,sizeof(a))#define SI(n) scanf("%d",&n)#define SLL(n) scanf("%lld",&n)#define PLLN(n) printf("%lldn",n)#define PIN(n) printf("%dn",n)#define REP(i,j,n) for(LL i=j;i<n;i++)#define PER(i,j,n) for(LL i=n-1;i>=j;i--)#define endl 'n'#define fast_io ios_base::sync_with_stdio(false);cin.tie(NULL)#define FILEIO(name) freopen(name".in", "r", stdin); freopen(name".out", "w", stdout); inline int mult(int a , int b) { LL x = a; x *= LL(b); if(x >= MOD) x %= MOD; return x; }inline int add(int a , int b) { return a + b >= MOD ? a + b - MOD : a + b; }inline int sub(int a , int b) { return a - b < 0 ? MOD - b + a : a - b; }LL powmod(LL a,LL b) { if(b==0)return 1; LL x=powmod(a,b/2); LL y=(x*x)%MOD; if(b%2) return (a*y)%MOD; return y%MOD; }int a[N];int inv;inline int nP2(int x) { return mult(x, x-1);}int main() { inv = powmod(2,MOD-2); int n; cin >> n; assert(n >= 1 && n <= 100000); for(int i = 0; i < n; i ++) { cin >> a[i]; assert(a[i] >= 0 && a[i] <= 1000000000); } int ans = 1; for(int i = 0; i < n; i ++) { int tmp = a[i]+1; if(a[i] & 1) { int x = (a[i]+1)/2; tmp = add(tmp, mult(2,nP2(x))); } else { int x = a[i]/2; tmp = add(tmp, nP2(x)); x ++; tmp = add(tmp, nP2(x)); } ans = mult(ans, tmp); } cout << ans << endl; return 0;} coding problems solutions HackerEarth HackerEarth