Best Cow Fences G

#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
const double eps = 1e-6;
int n, m;
double a[N];
double b[N], sum[N];

//  看能否凑出长度大于等于m 的字段平均数是 x
bool check(double x) {
	for (int i = 1; i <= n; i++) {
		b[i] = a[i] - x;
		sum[i] = sum[i - 1] + b[i];
	}
	double minn = 1e9;
	for (int i = m; i <= n; i++) {
		minn = min(minn, sum[i - m]);
		if (sum[i] - minn >= 0)
			return 1;
	}
	return 0;
}


int main() {
	cin >> n >> m;
	for (int i = 1; i <= n; i++)
		cin >> a[i];
	double l = 0, r = 2000;
	while (l + eps < r) {
		double mid = (l + r) / 2;
		if (check(mid))
			l = mid;
		else
			r  = mid;
	}
	cout << int(r * 1000);
	return 0;
}

防线

#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
#define int long long
struct node {
	int s, e, d;
};
node a[N];
int T, n;

// 到x 位置结束的防具和
int check(int x) {
	int tot = 0;
	for (int i = 1; i <= n; i++) {
		if (a[i].s > x)
			continue;
		// tot += (min(x, a[i].e) - a[i].s) / a[i].d + 1;
		tot += (min(x, a[i].e) + a[i].d - a[i].s ) / a[i].d;
	}
	return tot;
}

signed main() {
	cin >> T;
	while (T--) {
		cin >> n;
		for (int i = 1; i <= n; i++)
			cin >> a[i].s >> a[i].e >> a[i].d;
		int l = 0, r = 2147483647, mid, ans = -1;
		while (l <= r) {
			//mid = (l + r) / 2; 
			mid = l + (r-l)/2;
			if (check(mid) % 2 == 1)
				ans = mid, r = mid - 1;
			else
				l = mid + 1;
		}
		if (ans == - 1)
			cout << "There's no weakness.\n";
		else {
			int tot = check(ans) - check(ans - 1);
			printf("%lld %lld\n", ans, tot);
		}
	}
	return 0;
}

一元三次方程

#include <bits/stdc++.h>
using namespace std;
double a, b, c, d;
const double eps = 1e-3;

double f(double x) {
	return a * x * x * x + b * x * x + c * x + d;
}

int main() {
	cin >> a >> b >> c >> d;
	for (double i = -100 ; i <= 100; i++) {
		double l = i, r = i + 1;
		if (f(l) == 0) {
			printf("%.2lf ", l);
		} else if (f(l)*f(r) < 0) { // 意味着 (l,r) 之间有跟
			while (l + eps < r) {
				double mid = (l + r) / 2;
				if (f(l)*f(mid) <= 0)
					r = mid;
				else
					l  = mid;
			}
			printf("%.2lf ", l);
		}
	}
	return 0;
}

数列分段

#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
#define int long long
int n, m, a[N];
int l, r, mid, ans = -1;

// 检查在当前的段落和为 x 的情况下是否能凑出不超过m段
bool check(int x) {
	int cnt = 1;
	int sum = a[1];
	for (int i = 2; i <= n; i++) {
		if (a[i] + sum <= x)
			sum += a[i];
		else
			sum = a[i], cnt++;
	}
	return cnt <= m;
}

signed main() {
	cin >> n >> m;
	for (int i = 1; i <= n; i++)
		cin >> a[i], l = max(l, a[i]);
	r = 1e9;
	// 最大值最小
	while (l <= r) {
		mid  = (l + r) / 2;
		if (check(mid))
			ans = mid, r = mid - 1;
		else
			l = mid + 1;
	}
	cout << ans;
	return 0;
}

进击的奶牛

#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int n, m, a[N];

// 检查在距离 为 x 的时候是否能满足放置 m 头牛
bool check(int x) {
	int cnt = 1;
	int last = a[1];  // 第一个位置肯定放牛
	for (int i = 2; i <= n; i++)
		if (a[i] - last >= x)
			last = a[i], cnt++;
	return cnt >= m;
}

int main() {
	cin >> n >> m;
	for (int i = 1; i <= n; i++)
		cin >> a[i];
	sort(a + 1, a + 1 + n);
	int l = 1, r  = 1e9, mid, ans = -1;
	// 最小值最大模板
	while (l <= r) {
		mid = (l + r) / 2;
		if (check(mid))
			ans = mid, l  = mid + 1;
		else
			r = mid - 1;
	}
	cout << ans;
	return 0;
}

汽车拉力赛

// 最大值最小
int cnt;

void f(int x, int y, int mid) {
	vis[x][y] = 1;
	if (a[x][y] == 1)
		cnt++;
	for (四个方向) {
		if ()
		}
}

bool  check(int mid) {
	cnt = 0; // 标记走了几个路标
	memset(vis, 0, sizeof vis);
	dfs(x1, y1, mid);
	if (cnt == tot)
		return 1;
	else
		return 0;
}