The Strange Function


Problem Statement :


One of the most important skills a programmer needs to learn early on is the ability to pose a problem in an abstract way. This skill is important not just for researchers but also in applied fields like software engineering and web development.

You are able to solve most of a problem, except for one last subproblem, which you have posed in an abstract way as follows: Given an array consisting of n integers [ a1, a2, . . . , an ] .

Input Format

The first line contains a single integer n

The second  line contains n space-separated integers a1, a2, . . . , an .

Constraints

1  <=  n  <=  50000
- 10^6  <=   ai   <= 10^6

Output Format

Print a single integer denoting the answer



Solution :



title-img


                            Solution in C :

In C ++ :




#define _CRT_SECURE_NO_WARNINGS
#include <fstream>
#include <iostream>
#include <string>
#include <complex>
#include <math.h>
#include <set>
#include <vector>
#include <map>
#include <queue>
#include <stdio.h>
#include <stack>
#include <algorithm>
#include <list>
#include <ctime>

#include <memory.h>
#include <assert.h>

#define y0 sdkfaslhagaklsldk

#define y1 aasdfasdfasdf
#define yn askfhwqriuperikldjk
#define j1 assdgsdgasghsf
#define tm sdfjahlfasfh
#define lr asgasgash
#define norm asdfasdgasdgsd
#define have adsgagshdshfhds
#define ends asdgahhfdsfshdshfd

#define eps 1e-8
#define M_PI 3.141592653589793
#define bsize 512

#define ldouble long double
using namespace std;

#define bs 1000000007

const int N = 600031;

int n,ar[N];
long long S[N];

long long sparse_gcd[20][N],sparse_s[20][N],sparse_max[20][N];

int gcd(int a,int b){
	b=abs(b);
	while (a&&b)a>b?a%=b:b%=a;
	return a+b;
}

vector<pair<long long, long long> > order;

long long suf_max[N];

int main(){
//	freopen("apache.in","r",stdin);
//	freopen("apache.out","w",stdout);
	//freopen("input.txt", "r", stdin);
	//freopen("output.txt", "w", stdout);
	ios_base::sync_with_stdio(0);
//	cin.tie(0);

	cin>>n;
	long long ttl=0;

	for (int i=1;i<=n;i++){
		cin>>ar[i];
		S[i]=S[i-1]+ar[i];
	}

	suf_max[n]=S[n];
	for (int i=n-1;i>=0;--i)
		suf_max[i]=max(suf_max[i+1],S[i]);

	long long ans=-1e18;

	order.clear();
	for (int i=1;i<=n;i++){
		order.push_back(make_pair(S[i],i));
	}

	sort(order.begin(),order.end());

	long long MX=order.back().first;

	srand(time(NULL));

	if (rand()%2)
		random_shuffle(order.begin(),order.end());

	for (int ii=0;ii<order.size();ii++){
		int i=order[ii].second;
		ttl=MX-order[ii].first;
		long long cur_gcd=0;
		int cur_max=-1e9;
		long long cur_s=0;
		for (int j=i;j<=n;j++){
			cur_gcd=gcd(cur_gcd,ar[j]);
			cur_max=max(cur_max,ar[j]);
			cur_s+=ar[j];
			long long here=cur_gcd*(cur_s-cur_max);
			if (cur_gcd*1ll*(suf_max[j]-order[ii].first)<=ans)
				break;
			if (here>ans)
				ans=here;
		}
		if (clock()*1.0/CLOCKS_PER_SEC>1.98)
			break;
	}
	cout<<ans<<endl;

	cin.get(); cin.get();
	return 0;
}








In Java : 




//package hackerrank;

import java.util.ArrayDeque;
import java.util.Deque;
import java.util.Scanner;
import java.util.TreeMap;

public class Solution98 {
    static final Scanner scanner = new Scanner(System.in);
    static class Segment {
        int left, right, val;
        Segment(int left, int right, int val) {
            this.left = left;
            this.right = right;
            this.val = val;
        }
    }
    private static int gcd(int x, int y) {
        if (y == 0) {
            return x;
        }
        return gcd(y, x % y);
    }

    public static void main(String[] args) {
        int n = scanner.nextInt();
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = scanner.nextInt();
        }

        long ans = 0;
        long sum = 0;
        TreeMap<Integer, Integer> tmap = new TreeMap<>();
        SegmentTree tree = new SegmentTree(n);
        Deque<Segment> dq = new ArrayDeque<>();
        for (int i = 0; i < n; i++) {
            TreeMap<Integer, Integer> segmentTree = new TreeMap<>();
            for (int j : tmap.keySet()) {
                int k = gcd(j, Math.abs(a[i]));
                int idx = tmap.get(j);
                if (segmentTree.containsKey(k)) {
                    int cur = segmentTree.get(k);
                    segmentTree.put(k, Math.min(idx, cur));
                } else {
                    segmentTree.put(k, idx);
                }
            }
            if (segmentTree.containsKey(Math.abs(a[i]))) {
                int j = segmentTree.get(Math.abs(a[i]));
                segmentTree.put(Math.abs(a[i]), Math.min(i, j));
            } else {
                segmentTree.put(Math.abs(a[i]), i);
            }
            tmap = segmentTree;
            tree.add(i, i, sum);

            Segment s = new Segment(i, i, a[i]);
            while (!dq.isEmpty() && dq.getLast().val <= a[i]) {
                tree.add(dq.getLast().left, dq.getLast().right, -dq.getLast().val);
                s.left = dq.getLast().left;
                dq.removeLast();
            }
            tree.add(s.left, s.right, s.val);
            dq.addLast(s);
            sum += a[i];
            for (int j : tmap.keySet()) {
                int pos = tmap.get(j);
                ans = Math.max(ans, j * (sum - tree.get(pos, i)));
            }
        }

        System.out.println(ans);
    }
}

class SegmentTree {
    private int n;
    private long[] min, add;

    SegmentTree(int size) {
        n = 1;
        while (n < size) {
            n *= 2;
        }
        min = new long[2 * n];
        add = new long[2 * n];
    }

    private void clear(int i, int l, int r) {
        min[i] = 0;
        add[i] = 0;
        if (l == r - 1) {
            return;
        }
        int tm = (l + r) / 2;
        clear(2 * i + 1, l, tm);
        clear(2 * i + 2, tm, r);
    }

    void clear(int n) {
        this.n = n;
        clear(0, 0, n);
    }

    private void push(int i, int tl, int tr) {
        min[i] += add[i];
        if (tl != tr - 1) {
            add[2 * i + 1] += add[i];
            add[2 * i + 2] += add[i];
        }
        add[i] = 0;
    }

    private void add(int i, int lt, int rt, int l, int r, long diff) {
        push(i, lt, rt);
        if (l >= rt || r <= lt) {
            return;
        }

        if (l <= lt && rt <= r) {
            add[i] += diff;
            push(i, lt, rt);
            return;
        }

        int tm = (lt + rt) / 2;
        add(2 * i + 1, lt, tm, l, r, diff);
        add(2 * i + 2, tm, rt, l, r, diff);
        min[i] = Math.min(min[2 * i + 1], min[2 * i + 2]);
    }

    private long get(int i, int lt, int rt, int l, int r) {
        push(i, lt, rt);
        if (l >= rt || r <= lt) {
            return Long.MAX_VALUE;
        }

        if (l <= lt && rt <= r) {
            return min[i];
        }

        int tm = (lt + rt) / 2;
        return Math.min(get(2 * i + 1, lt, tm, l, r), get(2 * i + 2, tm, rt, l, r));
    }

    void add(int l, int r, long diff) {
        add(0, 0, n, l, r + 1, diff);
    }

    long get(int l, int r) {
        return get(0, 0, n, l, r + 1);
    }
}







In C :






#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define INF 200000
typedef struct _node{
  int gcd;
  int max;
  long long sum;
} node;
typedef struct _tree_node{
  enum {red,black} colour;
  int data;
  int real;
  struct _tree_node *left,*right,*parent;
}tree_node;
void solve(int start,int bs,int ns);
void copy_tree(tree_node **d,tree_node *r);
void build(int v,int tl,int tr);
long long query(int v,int tl,int tr,int l,int r,int *gcd,int *ma,long long *sum);
void getp(long long N,long long*prim);
long long CC(long long n, long long d);
int max(int x,int y);
int min(int x,int y);
int abss(int x);
int search(tree_node *root,int data);
void left_rotate(tree_node **root,tree_node *x);
void right_rotate(tree_node **root,tree_node *y);
void reconstruct(tree_node **root,tree_node *x);
int normal_insert(tree_node **root,tree_node *x);
void insert(tree_node **root,tree_node *x);
void sort_a(int*a,int size,int*new_size);
void merge(int*a,int*left,int*right,int left_size, int right_size,int*new_size);
void init( int n );
int range_sum( int i, int j );
void update( int i, long long val );
int a[50000],b[50000],nonzero[50000],rp[30],rc[30],treei[200000],rs,s,N;
long long ans,p[1000],tree[200000];
node t[200000];
tree_node *map[50000];

int main(){
  int n,x,ns,i,j,k,l;
  long long sum;
  tree_node *last_node,*p_node;
  getp(1000,p);
  scanf("%d",&n);
  for(i=0;i<n;i++)
    scanf("%d",a+i);
  build(1,0,n-1);
  init(n);
  for(i=sum=0;i<n;i++){
    sum+=a[i];
    update(i,sum);
  }
  for(i=n-1;i>=0;i--){
    if(i==n-1)
      last_node=NULL;
    else
      last_node=map[i+1];
    if(a[i]){
      nonzero[i]=i;
      for(j=rs=0,x=abss(a[i]);p[j] && p[j]*p[j]<=x;j++)
        if(x%p[j]==0){
          rp[rs]=p[j];
          rc[rs]=0;
          while(x%p[j]==0){
            rc[rs]++;
            x/=p[j];
          }
          rs++;
        }
      if(x!=1){
        rp[rs]=x;
        rc[rs]=1;
        rs++;
      }
      for(j=s=0;j<rs;j++)
        for(k=0,x=rp[j];k<rc[j];k++,x*=rp[j]){
          p_node=(tree_node*)malloc(sizeof(tree_node));
          p_node->data=x;
          p_node->left=p_node->right=p_node->parent=NULL;
          l=search(last_node,x);
          if(l!=INF){
            p_node->real=l;
            b[s++]=l+1;
          }
          else
            p_node->real=i;
          insert(&map[i],p_node);
        }
      b[s++]=i;
      sort_a(b,s,&ns);
      solve(i,ns,n);
    }
    else{
      nonzero[i]=INF;
      s=0;
      copy_tree(&map[i],last_node);
      if(i!=n-1 && nonzero[i+1]!=INF)
        b[s++]=nonzero[i+1];
      sort_a(b,s,&ns);
      solve(i,ns,n);
    }
    if(i!=n-1)
      nonzero[i]=min(nonzero[i],nonzero[i+1]);
  }
  printf("%lld",ans);
  return 0;
}
void solve(int start,int bs,int ns){
  int gcd,ma,i,j;
  long long t,sum;
  for(i=0;i<bs;i++){
    if(b[i]==ns)
      continue;
    if(i==bs-1)
      j=range_sum(b[i],ns-1);
    else
      j=range_sum(b[i],b[i+1]-1);
    t=query(1,0,ns-1,start,j,&gcd,&ma,&sum);
    if(t>ans)
      ans=t;
  }
  return;
}
void copy_tree(tree_node **d,tree_node *r){
  tree_node *p;
  if(!r)
    return;
  copy_tree(d,r->left);
  p=(tree_node*)malloc(sizeof(tree_node));
  p->data=r->data;
  p->real=r->real;
  p->left=p->right=p->parent=NULL;
  insert(d,p);
  b[s++]=r->real+1;
  copy_tree(d,r->right);
  return;
}
void build(int v,int tl,int tr){
  int tm;
  if(tl==tr){
    t[v].gcd=abss(a[tl]);
    t[v].max=t[v].sum=a[tl];
  }
  else{
    tm=(tl+tr)/2;
    build(2*v,tl,tm);
    build(2*v+1,tm+1,tr);
    t[v].gcd=CC(t[2*v].gcd,t[2*v+1].gcd);
    t[v].max=max(t[2*v].max,t[2*v+1].max);
    t[v].sum=t[2*v].sum+t[2*v+1].sum;
  }
  return;
}
long long query(int v,int tl,int tr,int l,int r,int *gcd,int *ma,long long *sum){
  int tm,g1,g2,m1,m2;
  long long s1,s2;
  if(tl>r || tr<l){
    *gcd=0;
    *ma=0;
    *sum=0;
    return 0;
  }
  if(tl>=l && tr<=r){
    *gcd=t[v].gcd;
    *ma=t[v].max;
    *sum=t[v].sum;
    return (*gcd)*((*sum)-(*ma));
  }
  tm=(tl+tr)/2;
  query(2*v,tl,tm,l,r,&g1,&m1,&s1);
  query(2*v+1,tm+1,tr,l,r,&g2,&m2,&s2);
  *gcd=CC(g1,g2);
  *ma=max(m1,m2);
  *sum=s1+s2;
  return (*gcd)*((*sum)-(*ma));
}
void getp(long long N,long long*prim)
{
  long long i,j,index=2,flag;
  if(N<=1){
    prim[0]=0;
    return;}
  if(N==2){
  prim[0]=2;
  prim[1]=0;
  return;}
  prim[0]=2;
  prim[1]=3;
  for(i=5;i<=N;i=i+2)
    {
      for(j=1,flag=1;prim[j]<=sqrt(i);j++)
        {
          if(i%prim[j]==0){
            flag=0;
            break;}
        }
      if(flag==1)
        {prim[index]=i;
          index++;}
    }
  prim[index]=0;
  return;
}
long long CC(long long n, long long d)
{
  if(n==0)
    return d;
  if(d==0)
    return n;
    while( 1 )
    {
        n = n % d;
	if( n == 0 )
		return d;
	d = d % n;
        if( d == 0 )
		return n;
    }
}
int max(int x,int y){
  return (x>y)?x:y;
}
int min(int x,int y){
  return (x<y)?x:y;
}
int abss(int x){
  return (x<0)?-x:x;
}
int search(tree_node *root,int data){
  if(!root)
    return INF;
  if(root->data==data)
    return root->real;
  if(data<root->data)
    return search(root->left,data);
  return search(root->right,data);
}
void left_rotate(tree_node **root,tree_node *x){
  tree_node *y;
  y=x->right;
  if(!y) return;
  x->right=y->left;
  if(y->left)
    y->left->parent=x;
  y->parent=x->parent;
  if(x->parent==NULL) *root=y;
  else
    if(x==x->parent->left)
      x->parent->left=y;
    else
      x->parent->right=y;
  y->left=x;
  x->parent=y;
  return;
}
void right_rotate(tree_node **root,tree_node *y){
  tree_node *x;
  x=y->left;
  if(!x) return;
  y->left=x->right;
  if(x->right)
    x->right->parent=y;
  x->parent=y->parent;
  if(y->parent==NULL) *root=x;
  else
    if(y==y->parent->right)
      y->parent->right=x;
    else
      y->parent->left=x;
  x->right=y;
  y->parent=x;
  return;
}
void reconstruct(tree_node **root,tree_node *x){
  tree_node *y,*z;
  y=x->parent;
  z=x->parent->parent;
  x->colour=black;
  z->colour=red;
  x->parent=z->parent;
  if(z->parent==NULL)
    *root=x;
  else if(z==z->parent->left)
    z->parent->left=x;
  else
    z->parent->right=x;
  if(z->left==y){
    x->left=y;
    x->right=z;
  }
  else{
    x->left=z;
    x->right=y;
  }
  y->parent=z->parent=x;
  y->left=y->right=z->left=z->right=NULL;
  return;
}
int normal_insert(tree_node **root,tree_node *x){
  if(*root==NULL)
    *root=x;
  else if((*root)->data==x->data)
    return 0;
  else{
    x->parent=*root;
    if((*root)->data>x->data)
      return normal_insert(&((*root)->left),x);
    else
      return normal_insert(&((*root)->right),x);
  }
  return 1;
}
void insert(tree_node **root,tree_node *x){
  if(!normal_insert(root,x))
    return;
  tree_node *y;
  x->colour=red;
  while(x!=*root && x->parent->colour==red){
    if(x->parent==x->parent->parent->left){
      y=x->parent->parent->right;
      if(!y)
        if(x==x->parent->left){
          x->parent->colour=black;
          x->parent->parent->colour=red;
          right_rotate(root,x->parent->parent);
        }
        else{
          y=x->parent;
          reconstruct(root,x);
          x=y;
        }
      else if(y->colour==red){
        x->parent->colour=black;
        y->colour=black;
        x->parent->parent->colour=red;
        x=x->parent->parent;
      }
      else{
        if(x==x->parent->right){
          x=x->parent;
          left_rotate(root,x);
        }
        x->parent->colour=black;
        x->parent->parent->colour=red;
        right_rotate(root,x->parent->parent);
      }
    }
    else{
      y=x->parent->parent->left;
      if(!y)
        if(x==x->parent->right){
          x->parent->colour=black;
          x->parent->parent->colour=red;
          left_rotate(root,x->parent->parent);
        }
        else{
          y=x->parent;
          reconstruct(root,x);
          x=y;
        }
      else if(y->colour==red){
        x->parent->colour=black;
        y->colour=black;
        x->parent->parent->colour=red;
        x=x->parent->parent;
      }
      else{
        if(x==x->parent->left){
          x=x->parent;
          right_rotate(root,x);
        }
        x->parent->colour=black;
        x->parent->parent->colour=red;
        left_rotate(root,x->parent->parent);
      }
    }
  }
  (*root)->colour=black;
  return;
}
void sort_a(int*a,int size,int*new_size){
  if (size < 2){
    (*new_size)=size;
    return;
  }
  int m = (size+1)/2,i;
  int *left,*right;
  left=(int*)malloc(m*sizeof(int));
  right=(int*)malloc((size-m)*sizeof(int));
  for(i=0;i<m;i++)
    left[i]=a[i];
  for(i=0;i<size-m;i++)
    right[i]=a[i+m];
  int new_l_size=0,new_r_size=0;
  sort_a(left,m,&new_l_size);
  sort_a(right,size-m,&new_r_size);
  merge(a,left,right,new_l_size,new_r_size,new_size);
  free(left);
  free(right);
  return;
}
void merge(int*a,int*left,int*right,int left_size, int right_size,int*new_size){
  int i = 0, j = 0,index=0;
  while (i < left_size|| j < right_size) {
    if (i == left_size) {
      a[index++] = right[j];
      j++;
    } else if (j == right_size) {
      a[index++] = left[i];
      i++;
    } else if (left[i] <= right[j]) {
      a[index++] = left[i];
      i++;
    } else {
      a[index++] = right[j];
      j++;
    }
    if(index>1&&a[index-2]==a[index-1])
      index--;
  }
  (*new_size)=index;
  return;
}
void init( int n )
{
  N = 1;
  while( N < n ) N *= 2;
  int i;
  for( i = 1; i < N + n; i++ ) tree[i] = -1000000000000000000LL;
}
int range_sum( int i, int j )
{
  int ansi;
  long long ans = -1000000000000000000LL;
  for( i += N, j += N; i <= j; i = ( i + 1 ) / 2, j = ( j - 1 ) / 2 )
  {
    if( i % 2 == 1 )
      if(tree[i]>ans){
        ans=tree[i];
        ansi=treei[i];
      }
    if( j % 2 == 0 )
      if(tree[j]>ans){
        ans=tree[j];
        ansi=treei[j];
      }
  }
  return ansi;
}
void update( int i, long long val )
{
  int j;
  for( j = i + N; j; j /= 2 )
    if(val>tree[j]){
      tree[j]=val;
      treei[j]=i;
    }
}









In Python3 :





from math import gcd

def parseInput(f):
    return [f(x) for x in input().split()]

n=int(input())
array=parseInput(int)
stack=[]
answer=float('-inf')
for number in array:
    for i in range(len(stack)):
        stack[i][0]=gcd(abs(stack[i][0]),abs(number))
        stack[i][1]+=number
        if number > stack[i][2]:
            stack[i][1]-=number-stack[i][2]
            stack[i][2]=number

    stack.append([number,0,number])
    newStack=[]
    for i in range(len(stack)):
        if newStack and newStack[-1][0] == stack[i][0]:
            if newStack[-1][1] <= stack[i][1]:
                if newStack[-1][1]+newStack[-1][2] > stack[i][1]+stack[i][2]:
                    newStack.append(stack[i])
                    continue
                newStack[-1][1]=stack[i][1]
                newStack[-1][2]=stack[i][2]
        else:
            newStack.append(stack[i])
    stack = newStack[:]
    answer=max(answer,max(abs(stack[i][0])*stack[i][1] for i in range(len(stack))))
print(answer)
                        








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Fibonacci Numbers Tree

Shashank loves trees and math. He has a rooted tree, T , consisting of N nodes uniquely labeled with integers in the inclusive range [1 , N ]. The node labeled as 1 is the root node of tree , and each node in is associated with some positive integer value (all values are initially ). Let's define Fk as the Kth Fibonacci number. Shashank wants to perform 22 types of operations over his tree, T

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Pair Sums

Given an array, we define its value to be the value obtained by following these instructions: Write down all pairs of numbers from this array. Compute the product of each pair. Find the sum of all the products. For example, for a given array, for a given array [7,2 ,-1 ,2 ] Note that ( 7 , 2 ) is listed twice, one for each occurrence of 2. Given an array of integers, find the largest v

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Lazy White Falcon

White Falcon just solved the data structure problem below using heavy-light decomposition. Can you help her find a new solution that doesn't require implementing any fancy techniques? There are 2 types of query operations that can be performed on a tree: 1 u x: Assign x as the value of node u. 2 u v: Print the sum of the node values in the unique path from node u to node v. Given a tree wi

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Ticket to Ride

Simon received the board game Ticket to Ride as a birthday present. After playing it with his friends, he decides to come up with a strategy for the game. There are n cities on the map and n - 1 road plans. Each road plan consists of the following: Two cities which can be directly connected by a road. The length of the proposed road. The entire road plan is designed in such a way that if o

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Heavy Light White Falcon

Our lazy white falcon finally decided to learn heavy-light decomposition. Her teacher gave an assignment for her to practice this new technique. Please help her by solving this problem. You are given a tree with N nodes and each node's value is initially 0. The problem asks you to operate the following two types of queries: "1 u x" assign x to the value of the node . "2 u v" print the maxim

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Number Game on a Tree

Andy and Lily love playing games with numbers and trees. Today they have a tree consisting of n nodes and n -1 edges. Each edge i has an integer weight, wi. Before the game starts, Andy chooses an unordered pair of distinct nodes, ( u , v ), and uses all the edge weights present on the unique path from node u to node v to construct a list of numbers. For example, in the diagram below, Andy

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