HackerRank Algebra – Manasa and Pizza

Challenge: Manasa and Pizza

Subdomeniu: Algebra (algebra)

Scor cont: 80.0 / 80

Submission status: Accepted

Submission score: 1.0

Submission ID: 464735236

Limbaj: cpp14

Link challenge: https://www.hackerrank.com/challenges/manasa-and-pizza/problem

Cerinta

With the college fest approaching soon, Manasa is following a strict dieting regime . Today, she just cannot resist her temptation for having a pizza. An inner conflict ensues, and she decides that she will have a pizza, only if she comes up with a solution to the problem stated below. Help her get the pizza for herself.

Given a list _L_ of _N_ numbers, where  
_L = { a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>, a<sub>4</sub> ....  , a<sub>N</sub>}_  
Find the value of _M_ that is computed as described below.  

![im1](https://hr-challenge-images.s3.amazonaws.com/2434/2434.jpg)  


**Input Format**  
The first line contains an integer _N_ i.e. size of the list _L_.  
The next line contains _N_ space separated integers, each representing an element of the list _L_.  

**Output Format**  
Print the value of _M_ _modulo (10<sup>9</sup> + 7)_. 

**Constraints**  
1 ≤ _N_ ≤ 5100<br>
0 ≤ _a<sub>i</sub>_ ≤ 10<sup>15</sup> , where _i ∈ [1 .. N]_

**Sample Input 00**  

    3
    1 2 3

    
**Sample Output 00**  

    40392
    
**Explanation**

There are 8 subsets of given set,

1. S = {1,2,3} and L - S  ={0} value of F(6) = 19601
2. S = {1,2} and L - S  ={3} value of F(0) = 1
3. S = {1,3} and L - S  ={2} value of F(2) = 17
4. S = {2,3} and L - S  ={1} value of F(4) = 577
5. S = {1} and L - S  ={2,3} value of F(4) = 577
6. S = {2} and L - S  ={1,3} value of F(2) = 17
7. S = {3} and L - S  ={1,2} value of F(0) = 1
8. S = {} and L - S  ={1,2,3} value of F(6) = 19601

Adding all these values, we get M = 40392.

Cod sursa

//manasa-and-pizza.cpp
//Manasa and Pizza
//Ad Infinitum - Math Programming Contest June'14
//Author: derekhh

#include<iostream>
#include<cstring>
using namespace std;

const int MAXN = 3, MOD = 1000000007;

struct Matrix
{
	int v[MAXN][MAXN], row, col;
};

Matrix add(Matrix a, Matrix b)
{
	Matrix ans;
	ans.row = a.row, ans.col = a.col;
	for (int i = 1; i <= a.row; i++)
		for (int j = 1; j <= a.col; j++)
			ans.v[i][j] = (a.v[i][j] + b.v[i][j]) % MOD;
	return ans;
}

Matrix mul(Matrix a, Matrix b)
{
	Matrix ans;
	ans.row = a.row, ans.col = b.col;
	for (int i = 1; i <= a.row; i++)
	{
		for (int j = 1; j <= b.col; j++)
		{
			ans.v[i][j] = 0;
			for (int k = 1; k <= a.col; k++)
				ans.v[i][j] = (ans.v[i][j] + (long long)a.v[i][k] * b.v[k][j]) % MOD;
		}
	}
	return ans;
}

Matrix power(Matrix a, long long n)
{
	Matrix res;
	res.row = a.row, res.col = a.col;
	memset(res.v, 0, sizeof(res.v));
	for (int i = 1; i <= a.row; i++)
		res.v[i][i] = 1;
	while (n != 0)
	{
		if (n % 2 == 1)
			res = mul(res, a);
		a = mul(a, a);
		n /= 2;
	}
	return res;
}

long long a[5101];

Matrix f[5101];

int main()
{
	int n;
	cin >> n;

	long long sum = 0;
	for (int i = 1; i <= n; i++)
	{
		cin >> a[i];
		sum += a[i];
	}
	
	Matrix F;
	F.col = F.row = 2;
	F.v[1][1] = 6;
	F.v[1][2] = MOD - 1;
	F.v[2][1] = 1;
	F.v[2][2] = 0;

	Matrix tmp = power(F, sum - 1);
	Matrix F_base; 
	F_base.row = 2; F_base.col = 1;
	F_base.v[1][1] = 3; F_base.v[2][1] = 1;
	Matrix F_result = mul(tmp, F_base);
	int g0 = F_result.v[1][1];

	tmp = power(F, sum - 2);
	F_result = mul(tmp, F_base);
	int g1 = F_result.v[1][1];

	f[0].row = f[0].col = 2;
	f[0].v[1][1] = f[0].v[2][2] = 1;
	f[0].v[1][2] = f[0].v[2][1] = 0;
	
	for (int i = 1; i <= n; i++)
		f[i] = add(f[i - 1], mul(f[i - 1], power(F, 2 * a[i])));

	Matrix G_base;
	G_base.row = 2; G_base.col = 1;
	G_base.v[1][1] = g1; G_base.v[2][1] = g0;
	Matrix G_result = mul(f[n], G_base);
	cout << G_result.v[2][1] << endl;

	return 0;
}