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


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("%d\n",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;
}