HackerEarth Fibonacci with GCD problem solution

In this HackerEarth Fibonacci with GCD problem solution, You are given N integers, A1, A2,…AN and Q queries. In each query, you are given two integers L and R(1 <= L <= R <= N). For each query, you have to find the value of:

GCD(F(Al),F(Al+1),F(Al+2)……F(AR))% 10^9 + 7

Can you do it?

HackerEarth Fibonacci with GCD problem solution.

#include <bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<stdlib.h>
#include<iomanip>
#include<math.h>
#include<limits.h>
#include<string.h>
#include <ctime>
#include <deque>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<bitset>
#include<set>
#include <sstream>
#include <list>
#define mod 1000000007
#define MAX 100000000

using namespace std;
#define scan(a) scanf("%d",&a);
#define print(a) printf("%dn",a);
#define mem(a,v) memset(a,v,sizeof(a))
#define clr(a) memset(a,0,sizeof(a))
#define pb(x) push_back(x)
#define sort(a) sort(a.begin(),a.end())
#define inf 1e9
#define mp(a,b) make_pair(a,b)
#define V vector
#define S string
#define Gu getchar_unlocked
#define Pu putchar_unlocked
#define Read(n) ch=Gu(); while (ch<'0') ch=Gu(); n=ch-'0'; ch=Gu(); while (!(ch<'0')) {n=10*n+ch-'0'; ch=Gu();}
#define Write(n) r=n; chptr=s; *chptr=r%10+'0'; r/=10; for (; r; r/=10) {++chptr; *chptr=r%10+'0'; } Pu(*chptr); for (; chptr!=s; ) Pu(*--chptr);
typedef long long LL;
typedef long double LD;
typedef long L;
typedef pair<int,int> pii;
const double pi=acos(-1.0);
int arr[100010];
struct T
{
    LL gc;
}tree[1000050];

LL gcd(LL a,LL b)
{
    if(b==0)    
        return a;
    else
        return gcd(b,a%b);
}
void build(int node,int a,int b)
{
    if(a==b)
    {
        tree[node].gc=arr[a];
        return;
    }
    int mid=(a+b)>>1;
    build(2*node,a,mid);
    build(2*node+1,mid+1,b);
    tree[node].gc = gcd(tree[2*node].gc,tree[2*node+1].gc);
    return;
}
LL query(int node,int i,int j,int a,int b)
{
    if(a<=i && b>=j)
    {
        return tree[node].gc;
    }
    if(b<i||a>j)
    {
        return -1;
    }
    int mid=(i+j)>>1;
    LL t1=query(2*node,i,mid,a,b);
    LL t2=query(2*node+1,mid+1,j,a,b);
    if(t1==-1 && t2==-1)
        return 1;
    else if(t1==-1)
        return t2;
    else if(t2==-1)
        return t1;
    else
        return gcd(t1,t2);
}
void multiply(LL f[2][2],LL m[2][2])
{
  LL x =  (f[0][0]*m[0][0] + f[0][1]*m[1][0])%mod;
  LL y =  (f[0][0]*m[0][1] + f[0][1]*m[1][1])%mod;
  LL z =  (f[1][0]*m[0][0] + f[1][1]*m[1][0])%mod;
  LL w =  (f[1][0]*m[0][1] + f[1][1]*m[1][1])%mod;
 
  f[0][0] = x;
  f[0][1] = y;
  f[1][0] = z;
  f[1][1] = w;
}
void power(LL f[2][2],LL n)
{
    if(n==0||n==1)
        return;               
    LL m[2][2]={{1,1},{1,0}};
    power(f,n/2);
    multiply(f,f);
    if(n%2!=0)
        multiply(f,m);  
}
LL fib(LL n)
{
    LL f[2][2]={{1,1},{1,0}};
    if(n==0)
        return 0;
    power(f,n-1);
    return (f[0][0]%mod);
}
int main()
{
    ios_base::sync_with_stdio(false);
    int n,q,a,b;
    scanf("%d %d",&n,&q);
    for(int i=0;i<n;i++)
        scanf("%d",&arr[i]);
    build(1,0,n-1);
    while(q--)
    {
        scanf("%d %d",&a,&b);
        a--;
        b--;
        int t1=query(1,0,n-1,a,b);
        cout<<fib(t1)%mod<<endl;
    }   
    return 0;
}

Second solution

#include<bits/stdc++.h>
#define MOD 1000000007
using namespace std;
int arr[100000];
int segtree[200000][3]={0};
int curmake;
int gcd(int a,int b)
{
    if(a==0)
        return b;
    return gcd(b%a,a);
}
void makesegment(int nodenow,int l, int r)
{
    if(l==r)
    {
        segtree[nodenow][0]=arr[l];
        segtree[nodenow][1]=-1;
        segtree[nodenow][2]=-1;
    }
    else
    {
        segtree[nodenow][1]=curmake;
        curmake++;
        makesegment(curmake-1,l,(l+r)/2);
        segtree[nodenow][2]=curmake;
        curmake++;
        makesegment(curmake-1,(l+r)/2+1,r);
        segtree[nodenow][0]=gcd(segtree[segtree[nodenow][1]][0],segtree[segtree[nodenow][2]][0]);
    }
}
int findgcd(int at,int atl,int atr,int l,int r)
{
    int a=-1,b=-1;
    if(atl==l  &&  atr==r)
        return segtree[at][0];
    if((atl+atr)/2>=l)
    a=findgcd(segtree[at][1],atl,(atl+atr)/2,l,min(r,(atl+atr)/2));
    if((atl+atr)/2+1<=r)
    b=findgcd(segtree[at][2],(atl+atr)/2+1,atr,max(l,(atl+atr)/2+1),r);
 
    if(a==-1)
        return b;
    if(b==-1)
        return a;
        
    return gcd(a,b);
}
void matmult(long long  a[][2],long long  b[][2],long long c[][2])
{
    int i,j,k;
    for(i=0;i<2;i++)
    {
        for(j=0;j<2;j++)
        {
            c[i][j]=0;
            for(k=0;k<2;k++)
            {
                c[i][j]+=(a[i][k]*b[k][j]);
                c[i][j]=c[i][j]%MOD;
            }
        }
    }

}
long long matpow(long long a[][2],int n)
{
    long long fib;
    long long ans[2][2]={{1,0},{0,1}},temp[2][2];
    int i,j;
    while(n>0)
    {
        if(n&1)
        {
            matmult(ans,a,temp);
            for(i=0;i<2;i++)
            {
                for(j=0;j<2;j++)
                {
                    ans[i][j]=temp[i][j];
                }
            }
        }
        matmult(a,a,temp);
        for(i=0;i<2;i++)
        {
                for(j=0;j<2;j++)
                {
                        a[i][j]=temp[i][j];

                }
        }
        n=n/2;
    }
    fib=(ans[0][0]*2+ans[0][1]);//modify here too
    fib=fib%MOD;
    return fib;
}

int main()
{
    
    ios::sync_with_stdio(false);
    int n,q;
    cin>>n>>q;
    assert(1<=n && n<=100000 && 1<=q && q<=100000);
    int i;
    for(i=0;i<n;i++)
    {
        cin>>arr[i];    
        assert(1<=arr[i] && arr[i]<=1000000000);
    }
    curmake=1;  
    makesegment(0,0,n-1);
    while(q--)
    {
        long long int a[2][2]={{1,1},{1,0}};
        int l,r;
        cin>>l>>r;
        assert(1<=l && l<=n && 1<=r && r<=n && l<=r);
        l--;
        r--;
        int g=findgcd(0,0,n-1,l,r);
        long long fib;
        g--;
         if(g>2)
            fib=matpow(a,g-2);
        else if(g>0)
            fib=g;
        else 
            fib=1;
        cout<<fib<<endl;
    }
    return 0;
}

coding problems solutions