HackerRank City


Problem Statement :


HackerRank-city is an acyclic connected graph (or tree). Its not an ordinary place, the construction of the whole tree takes place in N steps. The process is described below:

It initially has 1 node.
At each step, you must create 3 duplicates of the current tree, and create 2 new nodes to connect all 4 copies in the following H shape:

At each ith step, the tree becomes 4 times bigger plus 2 new nodes, as well as 5 new edges connecting everything together. The length of the new edges being added at step i is denoted by input Ai.

Calculate the sum of distances between each pair of nodes; as these answers may run large, print your answer modulo 1000000007.

Input Format

The first line contains an integer, N (the number of steps). The second line contains N space-separated integers describing A0,A1,...,An-2,An-1 .

Constraints
1 <= N <= 10^6
1 <= Ai <= 9

Subtask
For 50% score 1 <= N <=10

Output Format

Print the sum of distances between each pair of nodes modulo 1000000007.



Solution :



title-img


                            Solution in C :

In C++ :





#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <cmath>
using namespace std;
#define For(i,a,b) for(int i=a;i<=b;i++)
#define Ford(i,a,b) for(int i=a;i>=b;i--)
#define fi first
#define se second
#define sr(x) (int)x.size()
#define BUG(x) {cout << #x << " = " << x << endl;}
#define PR(x,a,b) {cout << #x << " = "; For(_,a,b) cout << x[_] << ' '; cout << endl;}
#define Bit(s,i) (((s)&(1<<(i)))>0)
#define Two(x) (1<<(x))
const int MOD = 1000000007;
const int nmax = 1000100;
const double E = 1e-8;
const double PI = acos(-1);
typedef long long LL;
typedef pair<int,int> pii;
int N,m,stest;

LL s[nmax], n[nmax];
LL a, f[nmax];
LL l[nmax];

int main() {
  //freopen("input.txt","r",stdin);
  s[0] = 0;
  n[0] = 1;
  l[0] = 0;
  f[0] = 0;
  cin >> N;
  For(i,1,N) {
    cin >> a;

    n[i] = (4 * n[i-1] + 2) % MOD;

    l[i] = ( 2 * l[i-1] + 3 * a ) % MOD;

    f[i] = (
      f[i-1] +
      l[i-1] + a +
      (f[i-1] + ( 2 * a + l[i-1] ) * n[i-1]) % MOD +
      l[i-1] + 2 * a +
      ( f[i-1] + ( 3 * a + l[i-1] ) * n[i-1] ) * 2 % MOD
    ) % MOD;

    s[i] = (
      4 * s[i-1] +
      ( f[i-1] * n[i-1] * 2 % MOD + n[i-1] * n[i-1] * 2 % MOD * a % MOD) * 2 +
      ( f[i-1] * n[i-1] * 2 % MOD + n[i-1] * n[i-1] * 3 % MOD * a % MOD) * 4 +
      ( f[i-1] + a * n[i-1] ) * 4 +
      ( f[i-1] + a * 2 * n[i-1] ) * 4 +
      a
    ) % MOD;
  }
  cout << s[N] << endl;
  return 0;
}








In Java :





import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static final long mod = (int) 1e9+7;
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int[] a = new int[n + 1];
        for (int i = 1; i <= n; i++) {
            a[i] = in.nextInt();
        }
        long b = 1, c = 0, d = 0, e = 0;
        for (int i = 1; i <= n; i++) {
            c = (a[i] +
                    8 * d +
                    12 * a[i] * b +
                    12 * ((b * d) % mod) +
                    16 * a[i] * ((b * b) % mod) +
                    (c * 4) % mod) % mod;
            d = (4 * d +
                    8 * a[i] * b +
                    3 * a[i] +
                    ((3 * b + 2)  * e) % mod) % mod;
            b = (4 * b + 2) % mod;
            e = (3 * a[i] + 2 * e) % mod;
        }
        System.out.println(c);
    }
}








In C :





#include <assert.h>
#include <limits.h>
#include <math.h>
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

const long long int P=1000000007;

int main() {
    int n;
    scanf("%d",&n);
    long long x=0,y=1,z=0,ans=0;
    int a;
    for (int i=0;i<n;i++) {
        scanf("%d",&a);
        ans=ans*4+(y*12+8)%P*x%P+(y*12+8)%P*y%P*a+(y*2+1)%P*(y*2+1)%P*a;
        x=  x*4+(z+a*2)%P*y%P+(z+a*3)%P*y%P*2+z*2+a*3;
        y=  y*4+2; 
        z=  z*2+a*3;
        ans%=P;
        x%=P;
        y%=P;
        z%=P;
    }
    printf("%lld",ans);
    return 0;
}








In Python3 :





N=int(input())
A=[int(x) for x in input().split()]
S=29*A[0]
L=11*A[0]
LL=3*A[0]
M=1000000007
n=6
for i in range(1,N):
    na=n*A[i]
    na2=na+na
    L2=(L+L)%M
    na2L=(na2+L2)%M
    S=(4*(S+na+na2L+n*(na2L+na))+2*n*na2L+A[i])%M
    L=(L2+2*(3*na+n*LL)+na2L+n*LL+3*A[i]+2*LL)%M
    LL= (LL+LL+3*A[i])%M
    n = (4 * n + 2)%M
print(S)
                        








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