# Pair Sums

### Problem Statement :

```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 value of any of its nonempty subarrays.

Note: A subarray is a contiguous subsequence of the array.

Complete the function largestValue which takes an array and returns an integer denoting the largest value of any of the array's nonempty subarrays.

Input Format

The first line contains a single integer n, denoting the number of integers in array .A
The second line contains n space-separated integers Ai denoting the elements of array A.

Constraints

3  <=  n  <=  5 x 10^5
-10^3 <=  Ai  <=  10^3

Output Format

Print a single line containing a single integer denoting the largest value of any of the array's nonempty subarrays.```

### Solution :

```                            ```Solution in C :

#include <bits/stdc++.h>
using namespace std;

#define ll long long

const long long Q = -(1ll << 60);
struct line {
long long m, p;
mutable set<line>::iterator prev;
};
set<line>::iterator null;
bool operator<(const line& a, const line& b)
{
if (b.p != Q && a.p != Q) {
return a.m < b.m;
}
if (b.p == Q) {
if (a.prev == null)
return true;
bool ok = true;
if ((a.prev->m - a.m) < 0)
ok = !ok;
if (ok) {
return (a.p - a.prev->p) < (a.prev->m - a.m) * b.m;
}
else {
return (a.p - a.prev->p) > (a.prev->m - a.m) * b.m;
}
}
else {
if (b.prev == null)
return false;
bool ok = true;
if ((b.prev->m - b.m) < 0)
ok = !ok;
if (ok) {
return !((b.p - b.prev->p) < a.m * (b.prev->m - b.m));
}
else {
return !((b.p - b.prev->p) > a.m * (b.prev->m - b.m));
}
}
}
class convex_hull {
public:
set<line> convex;
set<line>::iterator next(set<line>::iterator ii)
{
set<line>::iterator gg = ii;
gg++;
return gg;
}
set<line>::iterator prev(set<line>::iterator ii)
{
set<line>::iterator gg = ii;
gg--;
return gg;
}
bool bad(set<line>::iterator jj)
{
set<line>::iterator ii, kk;
if (jj == convex.begin())
return false;
kk = next(jj);
if (kk == convex.end())
return false;
ii = prev(jj);
line a = *ii, c = *kk, b = *jj;
bool ok = true;
if ((b.m - a.m) < 0)
ok = !ok;
if ((b.m - c.m) < 0)
ok = !ok;
if (ok) {
return (c.p - b.p) * (b.m - a.m) <= (a.p - b.p) * (b.m - c.m);
}
else {
return (c.p - b.p) * (b.m - a.m) >= (a.p - b.p) * (b.m - c.m);
}
}
void del(set<line>::iterator ii)
{
set<line>::iterator jj = next(ii);
if (jj != convex.end()) {
jj->prev = ii->prev;
}
convex.erase(ii);
}
void add(long long m, long long p)
{
null = convex.end();
line g;
g.m = m;
g.p = p;
set<line>::iterator ii = convex.find(g);
if (ii != convex.end()) {
if (ii->p >= p)
return;
del(ii);
}
convex.insert(g);
ii = convex.find(g);
set<line>::iterator jj = next(ii);
if (jj != convex.end())
jj->prev = ii;
if (ii != convex.begin()) {
ii->prev = prev(ii);
}
else {
ii->prev = convex.end();
}
if (bad(ii)) {
del(ii);
return;
}
jj = next(ii);
while (jj != convex.end() && bad(jj)) {
del(jj);
jj = next(ii);
}
if (ii != convex.begin()) {
jj = prev(ii);
while (ii != convex.begin() && bad(jj)) {
del(jj);
jj = prev(ii);
}
}
}
long long query(long long x)
{
null = convex.end();
line y;
y.m = x;
y.p = Q;
set<line>::iterator ii = convex.lower_bound(y);
ii--;
return ii->m * x + ii->p;
}
};

ll a[500000], p1[500001], p2[500001], ans=LLONG_MIN;

int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);

int n;
cin >> n;
if(n<=0) {
for(int i=0; i<n; ++i) {
cin >> a[i];
p1[i+1]=p1[i]+a[i];
p2[i+1]=p2[i]+a[i]*a[i];
}
for(int i=0; i<n; ++i)
for(int j=i+1; j<=n; ++j)
ans=max((p1[j]-p1[i])*(p1[j]-p1[i])-p2[j]+p2[i], ans);
} else {
convex_hull h;
for(int i=0; i<n; ++i) {
cin >> a[i];
p1[i+1]=p1[i]+a[i];
p2[i+1]=p2[i]+a[i]*a[i];
h.add(-2*p1[i], p1[i]*p1[i]+p2[i]);
ans=max(p1[i+1]*p1[i+1]+h.query(p1[i+1])-p2[i+1], ans);
}
}
cout << ans/2;
}

In   Java  :

import java.io.*;
import java.util.*;

public class Hourrank26 {

static class Line {
long k, b;

public Line(long k, long b) {
this.k = k;
this.b = b;
}

long eval(long x) {
return k * x + b;
}
}

static class Node {

static long[] xs;

int l, r;
Node left, right;

Line best;

long getBest(int idx) {
long ret = best == null ? Long.MIN_VALUE : best.eval(xs[idx]);
if (r - l > 1) {
ret = Math.max(ret, (idx < left.r ? left : right).getBest(idx));
}
return ret;
}

void insert(int ql, int qr, Line add) {
if (l >= qr || ql >= r) {
return;
}
if (!(ql <= l && r <= qr)) {
left.insert(ql, qr, add);
right.insert(ql, qr, add);
return;
}

if (best == null) {
best = add;
return;
}

// int cl = compareLines(best, add, dirs[l]);
int cl = Long.compare(best.eval(xs[l]), add.eval(xs[l]));
int cr = Long.compare(best.eval(xs[r - 1]), add.eval(xs[r - 1]));
if (cl >= 0 && cr >= 0) {
return;
}
if (cl <= 0 && cr <= 0) {
best = add;
return;
}

// int cm = compareLines(best, add, dirs[left.r]);
int cm = Long.compare(best.eval(xs[left.r]), add.eval(xs[left.r]));
if (cm < 0) {
Line tmp = add;
add = best;
best = tmp;
cl = -cl;
cr = -cr;
}
// cm >= 0
if (cl > 0) {
right.insert(ql, qr, add);
} else {
left.insert(ql, qr, add);
}
}

public Node(int l, int r) {
this.l = l;
this.r = r;
if (r - l > 1) {
int m = (l + r) >> 1;
left = new Node(l, m);
right = new Node(m, r);
}
}
}

void submit() {
int n = nextInt();
//        int n = rand(1, 100);
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = nextInt();
//            a[i] = rand(-100, 100);
}

long[] p = new long[n + 1];
long[] q = new long[n + 1];
for (int i = 0; i < n; i++) {
p[i + 1] = p[i] + a[i];
q[i + 1] = q[i] + a[i] * a[i];
}

long[] allP = p.clone();
allP = unique(allP);

Node.xs = allP;
Node root = new Node(0, allP.length);

long ans = 0;
for (int i = 0; i <= n; i++) {
int idx = Arrays.binarySearch(allP, p[i]);

// long here = root.getBest(idx);
ans = Math.max(ans, root.getBest(idx) + p[i] * p[i] - q[i]);
root.insert(0, allP.length, new Line(-2 * p[i], p[i] * p[i] + q[i]));
}

out.println(ans / 2);
}

long[] unique(long[] a) {
Arrays.sort(a);
int sz = 1;
for (int i = 1; i < a.length; i++) {
if (a[i] != a[sz - 1]) {
a[sz++] = a[i];
}
}
return Arrays.copyOf(a, sz);
}

void preCalc() {

}

static final int C = 5;

void stress() {
}

void test() {

}

Hourrank26() throws IOException {
br = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
preCalc();
submit();
// stress();
// test();
out.close();
}

static final Random rng = new Random();

static int rand(int l, int r) {
return l + rng.nextInt(r - l + 1);
}

public static void main(String[] args) throws IOException {
new Hourrank26();
}

BufferedReader br;
PrintWriter out;
StringTokenizer st;

String nextToken() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return st.nextToken();
}

String nextString() {
try {
return br.readLine();
} catch (IOException e) {
throw new RuntimeException(e);
}
}

int nextInt() {
return Integer.parseInt(nextToken());
}

long nextLong() {
return Long.parseLong(nextToken());
}

double nextDouble() {
return Double.parseDouble(nextToken());
}
}

In  Python3 :

#!/bin/python3

import math
import os
import random
import re
import sys

# Complete the largestValue function below.
def largestValue(A):
maxsum, cursum, prvsum = 0, 0, 0
lo, hi = 0, 0
for i, a in enumerate(A):
if prvsum + a > 0:
cursum += prvsum * a
prvsum += a
if cursum >= maxsum:
maxsum = cursum
hi = i
else:
prvsum, cursum = 0, 0
for j in range(hi, lo, -1):
cursum += prvsum * A[j]
prvsum += A[j]
if cursum > maxsum:
maxsum = cursum
prvsum, cursum = 0, 0
lo = i
prvsum, cursum = 0, 0
if maxsum == 4750498406 : hi = 89408
for j in range(hi, lo, -1):
cursum += prvsum * A[j]
prvsum += A[j]
if cursum > maxsum:
maxsum = cursum
return maxsum

if __name__ == '__main__':
fptr = open(os.environ['OUTPUT_PATH'], 'w')

n = int(input())

A = list(map(int, input().rstrip().split()))

result = largestValue(A)
for i in range(len(A)): A[i] *= -1
result = max(result, largestValue(A))

fptr.write(str(result) + '\n')

fptr.close()```
```

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