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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Kundu and Tree

Kundu is true tree lover. Tree is a connected graph having N vertices and N-1 edges. Today when he got a tree, he colored each edge with one of either red(r) or black(b) color. He is interested in knowing how many triplets(a,b,c) of vertices are there , such that, there is atleast one edge having red color on all the three paths i.e. from vertex a to b, vertex b to c and vertex c to a . Note that

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Super Maximum Cost Queries

Victoria has a tree, T , consisting of N nodes numbered from 1 to N. Each edge from node Ui to Vi in tree T has an integer weight, Wi. Let's define the cost, C, of a path from some node X to some other node Y as the maximum weight ( W ) for any edge in the unique path from node X to Y node . Victoria wants your help processing Q queries on tree T, where each query contains 2 integers, L and

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Contacts

We're going to make our own Contacts application! The application must perform two types of operations: 1 . add name, where name is a string denoting a contact name. This must store name as a new contact in the application. find partial, where partial is a string denoting a partial name to search the application for. It must count the number of contacts starting partial with and print the co

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No Prefix Set

There is a given list of strings where each string contains only lowercase letters from a - j, inclusive. The set of strings is said to be a GOOD SET if no string is a prefix of another string. In this case, print GOOD SET. Otherwise, print BAD SET on the first line followed by the string being checked. Note If two strings are identical, they are prefixes of each other. Function Descriptio

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Cube Summation

You are given a 3-D Matrix in which each block contains 0 initially. The first block is defined by the coordinate (1,1,1) and the last block is defined by the coordinate (N,N,N). There are two types of queries. UPDATE x y z W updates the value of block (x,y,z) to W. QUERY x1 y1 z1 x2 y2 z2 calculates the sum of the value of blocks whose x coordinate is between x1 and x2 (inclusive), y coor

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Direct Connections

Enter-View ( EV ) is a linear, street-like country. By linear, we mean all the cities of the country are placed on a single straight line - the x -axis. Thus every city's position can be defined by a single coordinate, xi, the distance from the left borderline of the country. You can treat all cities as single points. Unfortunately, the dictator of telecommunication of EV (Mr. S. Treat Jr.) do

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