Brownie Slicing

通过分析可以发现这是个二分。对于答案ans，如果满足，那么小于的也一定满足，我们要检查大于ans的是否满足

然后就是检查，通过模拟几次，可以发现，确定行，然后检查列就行了。

// Problem: Brownie Slicing
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/22353/P
// Memory Limit: 65536 MB
// Time Limit: 2000 ms
// 2022-05-28 18:02:12
//
// Powered by CP Editor (https://cpeditor.org)#include<bits/stdc++.h>
using namespace std;#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pairtypedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> Vll;
typedef vector<pair<int, int> > vpii;
typedef vector<pair<ll, ll> > vpll;                        const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const double pi  = acos(-1.0);
inline ll ksc(ll x,ll y,ll mod)
{ll ans = 0;while (y) {if (y & 1)ans = (ans + x) %mod;y >>= 1;x = (x + x) %mod;}return ans;
}
inline ll qmi (ll a, ll b) {ll ans = 1;while (b) {if (b & 1) ans = ans * a;a = a * a;b >>= 1;}return ans;
}
inline int read () {int x = 0, f = 0;char ch = getchar();while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();return f?-x:x;
}
template<typename T> void print(T x) {if (x < 0) putchar('-'), x = -x;if (x >= 10) print(x/10);putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {return (a + b) %mod;
}
// inline ll inv (ll a) {// return qmi(a, mod - 2);
// }
#define int long long
int n, m;
int a, b;
int s[550][550];bool get (int u, int d, int tar) {int c = 1;int cnt = 0;for (int j = c; j <= m; j ++) {int x1, y1, x2, y2;x2 = d, y2 = j;x1 = u, y1 = c;int sum = s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1];if (sum >= tar) {c = j + 1;cnt ++;}}return cnt >= b;
}
bool check (int tar) {int cnt = 0;int last = 1;for (int i = 1; i <= n; i ++) {if (get(last, i, tar)) {cnt ++;last = i + 1;}}return cnt >= a;
}
void solve() {cin >> n >> m >> a >> b;for (int i = 1; i <= n; i ++) {for (int j = 1; j <= m; j ++){cin >> s[i][j];s[i][j] += s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1];}}int l = 0, r = 1e9;while (l < r) {int mid = l + r + 1>> 1;if (check(mid))l = mid;else r = mid - 1;}cout << l << endl;
}
signed main () {// ios::sync_with_stdio(0),cin.tie(0), cout.tie(0);int t;t =1;//cin >> t;while (t --) solve();return 0;
}

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