(Analysis by Benjamin Qi)

The first step is to use complementary counting. The number of rectangular sub-grids with minimum equal to $100$ is equal to the number of rectangular sub-grids with minimum at least $100$ minus the number of rectangular sub-grids with minimum at least $101$.

To count the number of rectangular sub-grids with minimum at least $m$, create an $N\times N$ boolean array $ok$ such that $ok[i][j]=1$ if $G[i][j]\ge m$. We want to count the number of rectangular sub-grids in $ok$ that consist solely of ones.

If $ok$ was an $N\times 1$ rectangle rather than an $N\times N$ rectangle, the following loop would suffice to compute the answer:

int run = 0;
for (int i = 0; i < N; ++i) {
if (ok[i][0]) ans += ++run;
else run = 0;
}


Each run of $l$ consecutive ones contributes $\frac{l(l+1)}{2}$ to the answer.

Define $\texttt{all_ones}_{i,j}[k]$ to be true if all of the cells from $(i,k)$ to $(j,k)$ contain ones, and false otherwise. It suffices to iterate over $(i,j)$, compute $\texttt{all_ones}_{i,j}[k]$ for all $0\le k<N$, and then apply the 1D solution to $\texttt{all_ones}$. This takes $\mathcal{O}(N^4)$ time since there are $\mathcal{O}(N^3)$ triples $(i,j,k)$ and for each one, we do $\mathcal{O}(N)$ work to compute $\texttt{all_ones}_{i,j}[k]$.

To speed this up to $\mathcal{O}(N^3)$ time, we can use 1D prefix sums to compute $\texttt{all_ones}_{i,j}[k]$ in $\mathcal{O}(1)$ time.

Danny Mittal's code:

import java.io.BufferedReader;
import java.io.IOException;
import java.util.StringTokenizer;

public class JustGreenEnough2 {

public static void main(String[] args) throws IOException {
int[][] pasture = new int[n][n];
for (int y = 0; y < n; y++) {
for (int x = 0; x < n; x++) {
pasture[y][x] = Integer.parseInt(tokenizer.nextToken());
}
}
int[][] sumsBelow = new int[n][n + 1];
int[][] sumsAtMost = new int[n][n + 1];
for (int y = 0; y < n; y++) {
for (int x = 0; x < n; x++) {
sumsBelow[y][x + 1] = sumsBelow[y][x] + (pasture[y][x] < 100 ? 1 : 0);
sumsAtMost[y][x + 1] = sumsAtMost[y][x] + (pasture[y][x] <= 100 ? 1 : 0);
}
}
for (int x1 = 0; x1 < n; x1++) {
for (int x2 = x1 + 1; x2 <= n; x2++) {
int y1 = -1;
int y2 = -1;
for (int y0 = 0; y0 < n; y0++) {
while (y1 < n && (y1 < y0 || sumsAtMost[y1][x2] - sumsAtMost[y1][x1] == 0)) {
y1++;
}
while (y2 < n && (y2 < y0 || sumsBelow[y2][x2] - sumsBelow[y2][x1] == 0)) {
y2++;
}
}
}
}
}
}


Alternatively, note that $\texttt{all_ones}_{i,j}[k]=\texttt{all_ones}_{i,j-1}[k]\& ok[j][k]$. So let's fix $i$ and compute

$$\texttt{all_ones}_{i,i},\texttt{all_ones}_{i,i+1},\ldots,\texttt{all_ones}_{i,N-1}$$
in order.

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

using ll = long long;

int N;
bool ok[1000][1000];

ll solve() {
ll ans = 0;
for (int i = 0; i < N; ++i) {
vector<bool> all_ones(N,true);
for (int j = i; j < N; ++j) {
// add rectangles with upper row i and lower row j
int run = 0;
for (int k = 0; k < N; ++k) {
// all_ones_{i,j-1}[k] -> all_ones_{i,j}[k]
all_ones[k] = all_ones[k]&ok[j][k];
if (all_ones[k]) ans += ++run; // update answer
else run = 0;
}
}
}
return ans;
}

int main() {
cin >> N;
vector<vector<int>> pasture(N,vector<int>(N));
for (vector<int>& a: pasture)
for (int& b: a) cin >> b;

for (int i = 0; i < N; ++i)
for (int j = 0; j < N; ++j)
ok[i][j] = pasture[i][j] >= 100;
ll ans = solve();

for (int i = 0; i < N; ++i)
for (int j = 0; j < N; ++j)
ok[i][j] = pasture[i][j] > 100;
ans -= solve();

cout << ans << "\n";
}


It was possible (but not necessary) to solve this problem in $\mathcal{O}(N^2)$ time. In the code below, for a fixed $i$, I iterate over all $j$ in decreasing (rather than increasing order as the solution above does) and maintain the sum of the contributions of all runs in $\texttt{all_ones}_{i,j}$ in $\texttt{sum_comb}$. When $j$ decreases by one, I update $\texttt{sum_comb}$ accordingly for each $k$ such that $\texttt{all_ones}_{i,j+1}[k]=0$ and $\texttt{all_ones}_{i,j}[k]=1$.

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

using ll = long long;

int N;
bool ok[1000][1000];

ll solve() {
ll ans = 0;
vector<int> lst(N,N-1);
for (int i = N-1; i >= 0; --i) {
for (int j = i; j < N; ++j) to_add[j].clear();
for (int k = 0; k < N; ++k) {
if (ok[i][k] == 0) lst[k] = i-1;
}
int sum_comb = 0;
vector<int> lef(N,-1), rig(N,-1);
for (int j = N-1; j >= i; --j) {
// all_ones_{i,j+1}[k] = 0, all_ones_{i,j}[k] = 1
int l = k, r = k;
auto c2 = [](int x) { return (x+1)*(x+2)/2; };
if (k && lef[k-1] != -1) {
l = lef[k-1];
sum_comb -= c2(k-1-l);
}
if (k+1 < N && rig[k+1] != -1) {
r = rig[k+1];
sum_comb -= c2(r-k-1);
}
lef[r] = l, rig[l] = r;
sum_comb += c2(r-l);
}
ans += sum_comb;
}
}
return ans;
}

int main() {
cin >> N;
vector<vector<int>> pasture(N,vector<int>(N));
for (vector<int>& a: pasture)
for (int& b: a) cin >> b;

for (int i = 0; i < N; ++i)
for (int j = 0; j < N; ++j)
ok[i][j] = pasture[i][j] >= 100;
ll ans = solve();

for (int i = 0; i < N; ++i)
for (int j = 0; j < N; ++j)
ok[i][j] = pasture[i][j] > 100;
ans -= solve();

cout << ans << "\n";
}


For an additional challenge, try Maximum Building II.