整數二分模板

模板1：滿足條件的第一個數

int bianrysearch(int l, int r) {while (l < r) {int mid = l + r >> 1;if (check(mid)) r = mid;else  l = mid + 1;} return l;
}

模板2：滿足條件的最后一個數

int bianrysearch(int l, int r) {while (l < r) {int mid = l + r + 1 >> 1;if (check(mid)) l = mid;else  r = mid - 1;}return l;
}

浮點數二分模板

double bianrysearch(double l, double r) {double eps = 1e-6;   while (r - l > eps) {double mid = (l + r) / 2;if (check(mid)) r = mid;else l = mid;}return l;
}

一、Building an Aquarium

Building an Aquarium

思路分析

二分

具體代碼

#include <cstdio>
#include <iostream>using namespace std;
typedef long long ll;const int N = 2e5 + 10;
ll t, n, x;
ll a[N];bool check(ll mid) {ll sum = 0;for (int i = 1; i <= n; i++){if (a[i] < mid) sum += mid - a[i];}	return sum <= x;
}int main() {scanf("%lld", &t);while (t--) {scanf("%lld%lld", &n, &x);for (int i = 1; i <= n; i++) {scanf("%lld", &a[i]);}ll l = 1, r = 2e9 + 10;while (l < r) {ll mid = l + r + 1 >> 1;if (check(mid)) l = mid;else r = mid - 1;}printf("%lld\n", l);}return 0;
}

二、Tracking Segments

Tracking Segments

思路分析

二分+前綴和

具體代碼

#include <cstdio>
#include <iostream>
#include <cstring>using namespace std;typedef long long ll;const int N = 1e5 + 10;int t, n, m, q;
int s[N], pos[N];
ll preSum[N];struct node{int x, y;
}a[N];bool check(int mid) {for (int i = 1; i <= n; i++) s[i] = 0;for (int i = 1; i <= mid; i++)	s[pos[i]] = 1;for (int i = 1; i <= n; i++) {preSum[i] = preSum[i - 1] + s[i];}for (int i = 1; i <= m; i++) {if (2 * (preSum[a[i].y ] - preSum[a[i].x - 1]) > a[i].y - a[i].x + 1)	return true;}return false;
}int main() {cin >> t;while(t--) {cin >> n >> m;for (int i = 1; i <= m; i++) {cin >> a[i].x >> a[i].y;}cin >> q;for (int i = 1; i <= q; i++) {cin >> pos[i];}if (!check(q)) {puts("-1");continue;}	int l = 1, r = q;while (l < r) {int mid = l + r >> 1;if (check(mid)) r = mid;else l = mid + 1;}cout << l << endl;}return 0;
}

三、Wooden Toy Festival

Wooden Toy Festival

思路分析

二分

具體代碼

#include <cstdio>
#include <iostream>
#include <algorithm>using namespace std;const int N = 2e5 + 10;int t, n;
int a[N];bool check(int mid) {int idx = 0, t1 = a[0] + 2 * mid;while (a[idx] <= t1 && idx < n) idx++;t1 = a[idx] + 2 * mid;while (a[idx] <= t1 && idx < n) idx++;t1 = a[idx] + 2 * mid;while (a[idx] <= t1 && idx < n) idx++;if (idx == n) return true;return false;
}int main() {cin >> t;while (t--) {cin >> n;int l = 0, r; for (int i = 0; i < n; i++) {cin >> a[i];}sort(a, a + n);r = a[n - 1];while (l < r) {int mid = l + r >> 1;if (check(mid)) r = mid;else l = mid + 1;}		cout << l << endl;}return 0;
}

四、路標設置

路標設置

思路分析

二分，注意check函數寫法

具體代碼

#include <cstdio>
#include <iostream>
#include <algorithm>using namespace std;typedef long long ll;const int N = 100000 + 10;
ll len, n, k;
ll a[N];bool check(int mid) {ll dif = 0;for (int i = 1; i < n; i++) {dif += (a[i] - a[i - 1] - 1) / mid;}return dif <= k;
}int main() {cin >> len >> n >> k;ll l = 1, r = len;for(int i = 0; i < n; i++) {cin >> a[i];} sort(a, a + n);while (l < r) {ll mid = l + r >> 1;if (check(mid)) r = mid;else l = mid + 1;}cout << l << endl;return 0;
}

五、木材加工

木材加工

思路分析

二分

具體代碼

#include <cstdio>
#include <iostream>using namespace std;typedef long long ll;
const int N = 1e5 + 10;int n, k;
ll a[N];bool check(ll mid) {ll sum = 0;for (int i = 0; i < n; i++) sum += a[i] / mid;	return sum >= k;
}int main() {ll l = 0, r = 0;cin >> n >> k;for (int i = 0; i < n; i++) {scanf("%lld", &a[i]);r += a[i];} while(l < r) {ll mid = l + r + 1 >> 1;if (check(mid)) l = mid; else  r = mid - 1;}cout << l <<endl;return 0;
}