IndiaHacks 2nd Elimination 2017A. Binary Blocks

https://codeforces.com/contest/838/problem/A
二维前缀和的应用，注意可能比较绕
然后注意边界可以拿min的替换就行

#define _CRT_SECURE_NO_WARNINGS 1
#include<iostream>
#include<vector>
#include<algorithm>
#include<math.h>
#include<sstream>
#include<string>
#include<string.h>
#include<iomanip>
#include<stdlib.h>
#include<map>
#include<queue>
#include<limits.h>
#include<climits>
#include<fstream>
#include<stack>

using namespace std;
typedef long long ll;

ll mp[2510][2510];
bool is_prime(ll x)
{
	for (ll a = 2; a * a <= x; a++)
		if (x % a == 0)
			return false;
	return true;
}
int main()
{
	ll n, m; cin >> n >> m; cin.get();
	string s;
	for (int i = 0; i < n; i++)
	{
		getline(cin, s);
		for (int j = 0; j < m; j++)
		{
			mp[i + 1][j + 1] = s[j] - '0';
		}
	}
	//前缀和
	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j <= m; j++)
		{
			mp[i][j] += mp[i - 1][j] + mp[i][j - 1] - mp[i - 1][j - 1];
		}
	}
	queue<ll>qu_pri;
	qu_pri.push(2);
	for (ll i = 3; i <= min(n, m); i += 2)
		if (is_prime(i))
			qu_pri.push(i);
	ll ans = LLONG_MAX;
	while (!qu_pri.empty())
	{
		ll now = qu_pri.front(); qu_pri.pop();
		ll ans_tmp = 0;
		for (ll i = 1; (i - 1) * now < n; i++)
		{
			for (ll j = 1; (j - 1) * now < m; j++)
			{
				ll tg = mp[min(i * now, n)][min(j * now, m)]
					+ mp[(i - 1) * now][(j - 1) * now]
					- mp[min(i * now, n)][(j - 1) * now]
					- mp[(i - 1) * now][min(j * now, m)];
				ans_tmp = ans_tmp +min(tg,now*now-tg);
			}
		}
		ans = min(ans, ans_tmp);
	}
	cout << ans;
	return 0;

}