Find the Seed


Problem Statement :


A company needs random numbers for its operation. N random numbers have been generated using  N numbers as seeds and the following recurrence formula:

The numbers used as seeds are F(N-1),F(N-2),...,F(1),F(0). F(K) is the Kth term of the recurrence.

Due to a failure on the servers, the company lost its seed numbers. Now they just have the recurrence formula and the previously generated N random numbers.

The company wants to recover the numbers used as seeds, so they have hired you for doing this task.

Input Format

The first line contains two space-separated integers, N and K, respectively.
The second line contains the space-separated integers describing   F(K),F(K-1),...,F(K-N+2),F(K-N+1). F(K)(all these numbers are non-negative integers < 10^9).
The third line contains the space-separated coefficients of the recurrence formula, C(1),C(2),...,C(N-1),C(N). All of these coefficients are positive integers <10^9.

Constraints

1 <= N <= 50
1 <= K <=10^9
0 <= K-N+1



Solution :



title-img


                            Solution in C :

In C++ :





#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long ll;

ll MOD = 1000000007;
ll powe(ll a, ll b) {
    if (b == 0) return 1;
    if (b % 2 == 0) return powe( (a*a)%MOD, b/2);
    return (a*powe(a, b-1)) % MOD;
}

ll inv(ll a) {
    return powe(a, MOD-2);
}

struct matrix {
    vector<vector<ll>> M;
    matrix I() const {
        matrix ans;
        ans.M = M;
        for (int i = 0; i < M.size(); ++i)
            for (int j = 0; j < M.size(); ++j)
            ans.M[i][j] = !!(i==j);
        return ans;
    }
    matrix operator*(const matrix& rhs) const {
        matrix ans;
        ans.M = M;
        int N = M.size();
        for (int i = 0; i < N; ++i) for (int j = 0; j < N; ++j) {
            ans.M[i][j] = 0;
            for (int k = 0; k < N; ++k)
                ans.M[i][j] = (ans.M[i][j] + M[i][k]*rhs.M[k][j]) % MOD;
        }
        return ans;
    }
};

matrix powm(const matrix& M, int b) {
    if (b == 0) return M.I();
    if (b % 2 == 0) return powm( M*M, b/2);
    return M*powm(M, b-1);
}

int main() {
    ll N, K;
    cin >> N >> K;
    vector<ll> F(N);
    for (ll &x : F) cin >> x;
    vector<ll> C(N);
    for (ll &x : C) cin >> x;
    matrix M;
    M.M = vector<vector<ll>>(N, vector<ll>(N));
    for (int i = 0; i < N-1; ++i)
        for (int j = 0; j < N; ++j)
            M.M[i][j] = !!((i+1) == j);
    ll cinv = inv(C.back());
    M.M[N-1][0] = cinv;
    for (int j = 1; j < N; ++j)
        M.M[N-1][j] = ((MOD - C[j-1])*cinv)%MOD;
    auto M2 = powm(M, K-N+1);
    vector<ll> ans(N);
    for (int i = 0; i < N; ++i) {
        ans[i] = 0;
        for (int j = 0; j < N; ++j)
            ans[i] = (ans[i] + F[j]*M2.M[i][j]) % MOD;
    }
    for (int i = 0; i < N; ++i) {
        if (i) printf(" ");
        printf("%lld", ans[i]);
    }
    printf("\n");
    return 0;
}








In Java :





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

public class Solution {

    static final int Q = 1000000007;

    /** Assume 1 <= a < M and (a, M) = 1.  Return b such that 1 <= b < M and a*b = 1(mod M). */
    static int inv(int a, int M) {
        //assert a >= 1 && a < M;
        int t = 0;
        int r = M;
        int newt = 1;
        int newr = (int) a;
        while (newr != 0) {
            int q = r / newr;
            int temp = t - q * newt;
            t = newt;
            newt = temp;
            temp = r - q * newr;
            r = newr;
            newr = temp;
        }
        int result = (t < 0) ? t + M : t;
        //assert (a * result) % M == 1 && result >= 1 && result < M;
        return result;
    }

    static long[][] multiply(long[][] m1, long[][] m2) {
        int N = m1.length;
        long[][] m3 = new long[N][N];
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
                for (int k = 0; k < N; k++) {
                    m3[i][j] = (m3[i][j] + ((m1[i][k] * m2[k][j]) % Q)) % Q;
                }
            }
        }
        return m3;
    }

    /** Assume b >= 1 and each a[i][j] * (Q - 1) is in long range. */
    static long[][] pow(long[][] a, long b) {
        if (b == 1L) {
            return a;
        }
        long[][] half = pow(a, b / 2);
        long[][] almost = multiply(half, half);
        return (b % 2 == 0) ? almost : multiply(almost, a);
    }
    
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        long k = in.nextLong();
        long[] f = new long[n];
        int i;
        for (i = 0; i < n; i++) {
            f[i] = in.nextLong();
        }
        int[] c = new int[n];
        for (i = 0; i < n; i++) {
            c[i] = in.nextInt();
        }

        StringBuilder builder = new StringBuilder();
        if (k - n + 1 == 0L) {
            for (i = 0; i < n; i++) {
                if (i != 0) {
                    builder.append(' ');
                }
                builder.append(f[i]);
            }
        }
        else {
            long[][] mat = new long[n][n];
            for (i = 0; i < n - 1; i++) {
                mat[i + 1][i] = 1L;
            }
            mat[0][n - 1] = (long) inv(c[n - 1], Q);
            for (i = 1; i < n; i++) {
                mat[i][n - 1] = (Q - ((c[i - 1] * mat[0][n - 1]) % Q)) % Q;
            }
            long[][] power = pow(mat, k - n + 1);
            int j;
            long a;
            for (i = 0; i < n; i++) {
                a = 0L;
                for (j = 0; j < n; j++) {
                    a = (a + f[j] * power[j][i]) % Q;
                }
                if (i != 0) {
                    builder.append(' ');
                }
                builder.append(a);
            }
        }
        System.out.println(builder);
    }
}








In C :





#include <stdio.h>
#include <stdlib.h>
#define MOD 1000000007
long long modInverse(long long a,long long mod);
void one(long long*a,int SIZE);
void mul(long long*a,long long*b,int SIZE);
void powm(long long*a,int n,long long*res,int SIZE);
int F[50],C[50];
long long a[50][50],ans[50][50],A[50];

int main(){
  int N,K,i,j;
  scanf("%d%d",&N,&K);
  for(i=0;i<N;i++)
    scanf("%d",F+i);
  for(i=0;i<N;i++)
    scanf("%d",C+i);
  for(i=0;i<N-1;i++)
    for(j=0;j<N;j++)
      if(i==j-1)
        a[i][j]=1;
      else
        a[i][j]=0;
  a[N-1][0]=modInverse(C[N-1],MOD);
  for(i=1;i<N;i++)
    a[N-1][i]=(MOD-C[i-1])*a[N-1][0]%MOD;
  powm(&a[0][0],K-N+1,&ans[0][0],50);
  for(i=0;i<N;i++)
    for(j=0,A[i]=0;j<N;j++)
      A[i]=(A[i]+F[j]*ans[i][j])%MOD;
  for(i=0;i<N;i++)
    printf("%lld ",A[i]);
  return 0;
}
long long modInverse(long long a,long long mod){
	long long b0 = mod, t, q;
	long long x0 = 0, x1 = 1;
	while (a > 1) {
		q = a / mod;
		t = mod; mod = a % mod; a = t;
		t = x0; x0 = x1 - q * x0; x1 = t;
	}
	if (x1 < 0) x1 += b0;
	return x1;
}
void one(long long*a,int SIZE){
    int i,j;
    for (i = 0; i < SIZE; i++)
        for (j = 0; j < SIZE; j++)
            a[i*SIZE+j] = (i == j);
    return;
}
void mul(long long*a,long long*b,int SIZE){
    int i,j,k;
    long long res[SIZE][SIZE];
    for(i=0;i<SIZE;i++)
      for(j=0;j<SIZE;j++)
        res[i][j]=0;
    for (i = 0; i < SIZE; i++)
        for (j = 0; j < SIZE; j++)
            for (k = 0; k < SIZE; k++)
                res[i][j] =(res[i][j] + a[i*SIZE+k] * b[k*SIZE+j])%MOD;
    for (i = 0; i < SIZE; i++)
        for (j = 0; j < SIZE; j++)
            a[i*SIZE+j] = res[i][j];
    return;
}
void powm(long long*a,int n,long long*res,int SIZE){
    one(res,SIZE);
    while (n > 0) {
        if (n % 2 == 0)
        {
            mul(a, a,SIZE);
            n /= 2;
        }
        else {
            mul(res, a,SIZE);
            n--;
        }
    }
}








In Python3 :





#http://stackoverflow.com/questions/24701490/modular-matrix-inversion-with-large-number?lq=1
MOD = 1000000007

def generalizedEuclidianAlgorithm(a, b):
    if b > a:
        return generalizedEuclidianAlgorithm(b,a);
    elif b == 0:
        return (1, 0);
    else:
        (x, y) = generalizedEuclidianAlgorithm(b, a % b);
        return (y, x - (a // b) * y)

def inversemodp(a, p):
    a = a % p
    if (a == 0):
        # print "a is 0 mod p"
        return 0
    (x,y) = generalizedEuclidianAlgorithm(p, a % p);
    return y % p

def identitymatrix(n):
    return [[int(x == y) for x in range(0, n)] for y in range(0, n)]

def multiply_vector_scalar(vector, scalar, q):
    kq = []
    for i in range (0, len(vector)):
        kq.append (vector[i] * scalar %q)
    return kq

def minus_vector_scalar1(vector1, scalar, vector2, q):
    kq = []
    for i in range (0, len(vector1)):
        kq.append ((vector1[i] - scalar * vector2[i]) %q)
    return kq

def inversematrix1(matrix, q):
    n = len(matrix)

    A =[]
    for j in range (0, n):
        temp = []
        for i in range (0, n):
            temp.append (matrix[j][i])
        A.append(temp)

    Ainv = identitymatrix(n)

    for i in range(0, n):
        factor = inversemodp(A[i][i], q)
        A[i] = multiply_vector_scalar(A[i],factor,q)
        Ainv[i] = multiply_vector_scalar(Ainv[i],factor,q)
        for j in range(0, n):
            if (i != j):
                factor = A[j][i]
                A[j] = minus_vector_scalar1(A[j],factor,A[i],q)
                Ainv[j] = minus_vector_scalar1(Ainv[j],factor,Ainv[i],q)
    return Ainv

def mult(x, y):
	c = [[0 for _ in range(len(y[0]))] for _ in range(len(x))]

	for i in range(len(x)):
		for j in range(len(y[0])):
			for k in range(len(x)):
				c[i][j] += x[i][k] * y[k][j]
				c[i][j] = c[i][j] % MOD
	return c

def matpow(b, p):
	if p == 0: return identitymatrix(n)
	if p == 1: return b
	if p % 2 == 1:
		return mult(b, matpow(b, p - 1))
	ret = matpow(b, p // 2)
	return mult(ret, ret)

n, k = map(int, input().split())
arrk = list(map(int, input().split()))
arrc = list(map(int, input().split()))

left = [[x] for x in arrk];
middle = [[0 for _ in range(n)] for _ in range(n)]
middle[0] = list(arrc)
for i in range(1,n):
	middle[i][i-1] = 1

inv = inversematrix1(middle, MOD)
inv = [[int(x) for x in y] for y in inv]

ans = matpow(inv, k - n + 1)
ans = mult(ans, left)

print(' '.join(map(lambda x : str(x[0]), ans)))
                        








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