先对每一行求一遍滑动窗口，列数变为(列数-k+1)

再对每一列求一遍滑动窗口，行数变为(行数-k+1)

剩下的就是每一个窗口里的最大值啦

#include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define endl '\n'
 
using namespace std;
 
typedef pair<int, int> PII;
typedef long long ll;
typedef long double ld;

const int N = 5010;

int n, m, k;
int a[N][N];
int q[N], hh, tt;

int main()
{
	IOS
	cin >> n >> m >> k;
	for(int i = 1; i <= n; i ++)
		for(int j = 1; j <= m; j ++)
			a[i][j] = i * j / __gcd(i, j);
			
	for(int i = 1; i <= n; i ++)
	{
		hh = 0, tt = -1;
		for(int j = 1; j <= m; j ++)
		{
			if(hh <= tt && q[hh] < j - k + 1)hh ++;
			while(hh <= tt && a[i][q[tt]] <= a[i][j])tt --;
			q[++ tt] = j;
			if(j >= k)a[i][j - k + 1] = a[i][q[hh]];
		}
	}
	m = m - k + 1;
	
	for(int j = 1; j <= m; j ++)
	{
		hh = 0, tt = -1;
		for(int i = 1; i <= n; i ++)
		{
			if(hh <= tt && q[hh] < i - k + 1)hh ++;
			while(hh <= tt && a[q[tt]][j] <= a[i][j])tt --;
			q[++ tt] = i;
			if(i >= k)a[i - k + 1][j] = a[q[hh]][j];
		}
	}
	n = n - k + 1;
	
	ll ans = 0;
	for(int i = 1; i <= n; i ++)
		for(int j = 1; j <= m; j ++)
			ans += a[i][j];
			
	cout << ans;
	
	return 0;
}