本节主要介绍快速排序、归并排序、二分查找。

一、快速排序

更详细解释可看此链接：AcWing

j划分

c++
#include <iostream>
using namespace std;

const int N = 1e6 + 10;

int n;
int q[N];

void quick_sort(int q[], int l, int r)
{
	if (l >= r) return; //递归的终止情况

	//第一步：分成子问题
	int x = q[(l + r) >> 1], i = l - 1, j = r + 1;
	while (i < j)
	{
		do i++; while (q[i] < x);
		do j--; while (q[j] > x);
		if (i < j) swap(q[i], q[j]);
	}

	//第二步：递归处理子问题
	quick_sort(q, l, j);
	quick_sort(q, j + 1, r);

	//第三步：子问题合并.快排这一步不需要操作，但归并排序的核心在这一步骤
}

int main()
{
	scanf("%d", &n);
	for (int i = 0; i < n; i++) scanf("%d", &q[i]);

	quick_sort(q, 0, n - 1);

	for (int i = 0; i < n; i++) printf("%d ", q[i]);

	return 0;
}

i划分

c++
#include <iostream>
using namespace std;

const int N = 1e6 + 10;

int n;
int q[N];

void quick_sort(int q[], int l, int r)
{
	if (l >= r) return; //递归的终止情况

	//第一步：分成子问题
	int x = q[(l + r + 1) >> 1], i = l - 1, j = r + 1;
	while (i < j)
	{
		do i++; while (q[i] < x);
		do j--; while (q[j] > x);
		if (i < j) swap(q[i], q[j]);
	}

	//第二步：递归处理子问题
	quick_sort(q, l, i - 1);
	quick_sort(q, i, r);

	//第三步：子问题合并.快排这一步不需要操作，但归并排序的核心在这一步骤
}

int main()
{
	scanf("%d", &n);
	for (int i = 0; i < n; i++) scanf("%d", &q[i]);

	quick_sort(q, 0, n - 1);

	for (int i = 0; i < n; i++) printf("%d ", q[i]);

	return 0;
}

二、归并排序

c++
#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
using namespace std;

const int N = 1000010;

int n;
int q[N], tmp[N];

void merge_sort(int q[], int l, int r)
{
	if (l >= r) return;

	int mid = l + r >> 1;

	merge_sort(q, l, mid), merge_sort(q, mid + 1, r);

	int k = 0, i = l, j = mid + 1;
	while (i <= mid && j <= r)
		if (q[i] <= q[j]) tmp[k++] = q[i++];
		else tmp[k++] = q[j++];
	while (i <= mid) tmp[k++] = q[i++];
	while (j <= r) tmp[k++] = q[j++];

	for (int i = l, j = 0; i <= r; i++, j++) q[i] = tmp[j];
}

int main()
{
	scanf("%d", &n);
	for (int i = 0; i < n; i++) scanf("%d", &q[i]);

	merge_sort(q, 0, n - 1);

	for (int i = 0; i < n; i++) printf("%d ", q[i]);

	return 0;
}

三、二分查找

整数二分

做法1

c++
#include <iostream>
using namespace std;

const int N = 100010;

int n, m;
int q[N];

int main()
{
	scanf("%d%d", &n, &m);
	for (int i = 0; i < n; i++) scanf("%d", &q[i]);//要求q[N]由小到大

	while (m--) {
		int x;
		scanf("%d", &x);

		int l = 0, r = n - 1;
		while (l < r) {
			int mid = l + r >> 1;
			if (q[mid] >= x) r = mid;//q[mid]>=x，则x在mid左边(包含mid)，所以r=mid
			else l = mid + 1;//否则，q[mid]<=x，则x在mid右边，所以l=mid+1
		}//多次二分循环，找出区间左端点

		if (q[l] != x) cout << "-1 -1" << endl;//如果q[l]!=x，说明不存在x
		else {
			cout << l << ' ';

			int l = 0, r = n - 1;
			while (l < r) {
				int mid = l + r + 1 >> 1;//+1避免陷入死循环
				if (q[mid] <= x) l = mid;//(包含mid)
				else r = mid - 1;
			}//多次二分循环，找出区间右端点

			cout << l << endl;
		}
	}

	return 0;
}

做法2

c++
#include <iostream>
using namespace std;

const int N = 1e5 + 10;
int q[N];

int SL(int l, int r, int x) {
	while (l < r) {
		int mid = l + r >> 1;
		if (q[mid] >= x) r = mid;
		else l = mid + 1;
	}
	return l;
}

int SR(int l, int r, int x) {
	while (l < r) {
		int mid = l + r + 1 >> 1;
		if (q[mid] <= x) l = mid;
		else r = mid - 1;
	}
	return r;
}

int main()
{
	int n, m;
	scanf("%d%d", &n, &m);
	for (int i = 0; i < n; ++i) scanf("%d", &q[i]);
	while (m--) {
		int x;
		scanf("%d", &x);
		int l = SL(0, n - 1, x);//查找左边界，并返回下标1
		if (q[l] != x) cout << "-1 -1" << endl;//如果找不到，返回-1 -1
		else {
			cout << l << ' ';//如果找到了，输出左下标
			cout << SR(0, n - 1, x) << endl;//输出右下标
		}
	}

	return 0;
}

浮点数二分

c++
#include <iostream>
using namespace std;

int main()
{
	double x;
	cin >> x;

	double l = 0, r = x;
	while (r - l > 1e-8)
	{
		double mid = (l + r) / 2;
		if (mid * mid >= x) r = mid;
		else l = mid;
	}

	printf("%lf\n", l);

	return 0;
}