Triangles on a Rectangle.cpp

/* 
    Solution by Rahul Surana
    
    ***********************************************************


You are given an array a1,a2,…,an consisting of integers from 0 to 9. A subarray al,al+1,al+2,…,ar−1,ar is good if the sum of elements of this subarray is equal to the length of this subarray (∑i=lrai=r−l+1).

For example, if a=[1,2,0], then there are 3 good subarrays: a1…1=[1],a2…3=[2,0] and a1…3=[1,2,0].

Calculate the number of good subarrays of the array a.

Input
The first line contains one integer t (1≤t≤1000) — the number of test cases.

The first line of each test case contains one integer n (1≤n≤105) — the length of the array a.

The second line of each test case contains a string consisting of n decimal digits, where the i-th digit is equal to the value of ai.

It is guaranteed that the sum of n over all test cases does not exceed 105.

Output
For each test case print one integer — the number of good subarrays of the array a.



    ***********************************************************
*/


#include <bits/stdc++.h>
#define ll          long long
#define vl          vector<ll>
#define vi          vector<int>
#define pi          pair<int,int>
#define pl          pair<ll,ll>
#define all(a)      a.begin(),a.end()
#define mem(a,x)    memset(a,x,sizeof(a))
#define pb          push_back
#define mp          make_pair
#define F           first
#define S           second
#define FOR(i,a)     for(int i = 0; i < a; i++)
#define trace(x)    cerr<<#x<<" : "<<x<<endl;
#define trace2(x,y) cerr<<#x<<" : "<<x<<" | "<<#y<<" : "<<y<<endl;
#define trace3(x,y,z) cerr<<#x<<" : "<<x<<" | "<<#y<<" : "<<y<<" | "<<#z<<" : "<<z<<endl;
#define fast_io 	std::ios::sync_with_stdio(false),cin.tie(NULL),cout.tie(NULL)
 
using namespace std;


int main()
{
	fast_io;
    int t;
    cin >> t;
    while(t--) {
        int w,h;
        cin >> w >> h;
        vector<ll> cord[4];
        int x;
        cin >> x;
        FOR(i,x) { ll a; cin >> a; cord[0].pb(a); }
        cin >> x;
        FOR(i,x) { ll a; cin >> a; cord[1].pb(a); }
        cin >> x;
        FOR(i,x) { ll a; cin >> a; cord[2].pb(a); }
        cin >> x;
        FOR(i,x) { ll a; cin >> a; cord[3].pb(a); }

        ll ans = 0;
        ans = max(ans,h*(cord[0][cord[0].size()-1] - cord[0][0]));
        ans = max(ans,h*(cord[1][cord[1].size()-1] - cord[1][0]));
        ans = max(ans,w*(cord[2][cord[2].size()-1] - cord[2][0]));
        ans = max(ans,w*(cord[3][cord[3].size()-1] - cord[3][0]));

        // if(c != 1) cout << "YES\n";
        // else cout << "NO\n";
        cout << ans <<"\n";
    }
}