# 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 :

```                            ```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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