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 :
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)
View More Similar Problems
Insert a Node at the head of a Linked List
Given a pointer to the head of a linked list, insert a new node before the head. The next value in the new node should point to head and the data value should be replaced with a given value. Return a reference to the new head of the list. The head pointer given may be null meaning that the initial list is empty. Function Description: Complete the function insertNodeAtHead in the editor below
View Solution →Insert a node at a specific position in a linked list
Given the pointer to the head node of a linked list and an integer to insert at a certain position, create a new node with the given integer as its data attribute, insert this node at the desired position and return the head node. A position of 0 indicates head, a position of 1 indicates one node away from the head and so on. The head pointer given may be null meaning that the initial list is e
View Solution →Delete a Node
Delete the node at a given position in a linked list and return a reference to the head node. The head is at position 0. The list may be empty after you delete the node. In that case, return a null value. Example: list=0->1->2->3 position=2 After removing the node at position 2, list'= 0->1->-3. Function Description: Complete the deleteNode function in the editor below. deleteNo
View Solution →Print in Reverse
Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing
View Solution →Reverse a linked list
Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty. Example: head references the list 1->2->3->Null. Manipulate the next pointers of each node in place and return head, now referencing the head of the list 3->2->1->Null. Function Descriptio
View Solution →Compare two linked lists
You’re given the pointer to the head nodes of two linked lists. Compare the data in the nodes of the linked lists to check if they are equal. If all data attributes are equal and the lists are the same length, return 1. Otherwise, return 0. Example: list1=1->2->3->Null list2=1->2->3->4->Null The two lists have equal data attributes for the first 3 nodes. list2 is longer, though, so the lis
View Solution →