本节主要介绍高精度加/减/乘/除以及位运算。


# 高精度
## 高精度加法
```c++
#include <iostream>
#include <vector>
using namespace std;

//C = A + B
vector<int> add(vector<int>& A, vector<int>& B)//使用&，避免再次拷贝
{
	vector<int> C;

	int t = 0; //进位
	for (int i = 0; i < A.size() || i < B.size(); i++)
	{
		if (i < A.size()) t += A[i];
		if (i < B.size()) t += B[i];
		C.push_back(t % 10);
		t /= 10; //如果t>=10，进位
	}

	if (t) C.push_back(1); //进位
	return C;
}

int main()
{
	string a, b;
	vector<int> A, B;

	//表示数字时，倒着表示
	//'0'的ASCII码为48，如'9'-'0'=57-48=9
	cin >> a >> b; //a="123456"
	for (int i = a.size() - 1; i >= 0; i--) A.push_back(a[i] - '0'); // A = [6, 5, 4, 3, 2, 1]
	for (int i = b.size() - 1; i >= 0; i--) B.push_back(b[i] - '0'); // -'0'将字符转化成整数


	auto C = add(A, B);//auto == vector<int>

	for (int i = C.size() - 1; i >= 0; i--) printf("%d", C[i]);
	return 0;
}
```
## 高精度减法
```c++
#include <iostream>
#include <vector>
using namespace std;

//判断是否有 A >= B
bool cmp(vector<int>& A, vector<int>& B)
{
	if (A.size() != B.size()) return A.size() > B.size();
	for (int i = A.size() - 1; i >= 0; i--)
		if (A[i] != B[i])
			return A[i] > B[i];
	return true;
}

//C = A - B
vector<int> sub(vector<int>& A, vector<int>& B)//使用&，避免再次拷贝
{
	vector<int> C;
	for (int i = 0,t=0; i < A.size(); i++)
	{
		t = A[i] - t;//减去借位
		if (i < B.size()) t -= B[i];//判断B[i]是否存在
		C.push_back((t + 10) % 10);//如果t>=0，则为t；如果t<0，则为t+10。用(t+10)%10考虑

		//判断是否需要借位
		if (t < 0) t = 1;
		else t = 0;
	}

	while (C.size() > 1 && C.back() == 0) C.pop_back(); //去掉前导0

	return C;
}

int main()
{
	string a, b;
	vector<int> A, B;

	//表示数字时，倒着表示
	cin >> a >> b; //a="123456"
	for (int i = a.size() - 1; i >= 0; i--) A.push_back(a[i] - '0'); // A = [6, 5, 4, 3, 2, 1]
	for (int i = b.size() - 1; i >= 0; i--) B.push_back(b[i] - '0');

	if (cmp(A, B))
	{
		auto C = sub(A, B);

		for (int i = C.size() - 1; i >= 0; i--) printf("%d", C[i]);
	}
	else
	{
		auto C = sub(B, A);

		printf("-");
		for (int i = C.size() - 1; i >= 0; i--) printf("%d", C[i]);
	}

	return 0;
}
```
## 高精度乘法
```c++
#include <iostream>
#include <vector>
using namespace std;

// C = A * b 高精度*低精度
vector<int> mul(vector<int>& A, int b)
{
	vector<int> C;

	int t = 0; //进位
	for (int i = 0; i < A.size() || t; i++)
	{
		if (i < A.size()) t += A[i] * b;
		C.push_back(t % 10);
		t /= 10;
	}

	return C;
}

int main()
{
	string a;
	vector<int> A;
	int b;

	cin >> a >> b;
	for (int i = a.size() - 1; i >= 0; i--) A.push_back(a[i] - '0');

	auto C = mul(A, b);

	for (int i = C.size() - 1; i >= 0; i--) printf("%d", C[i]);

	return 0;
}
```
## 高精度除法 
```c++
#include <iostream>
#include <vector>
using namespace std;

// C = A / b 商是C，余数是r
vector<int> div(vector<int>& A, int b, int& r) //r是引用
{
	vector<int> C;
	r = 0;
	for (int i = A.size()-1; i >=0; i--)
	{
		r = r * 10 + A[i];
		C.push_back(r / b);
		r %= b; //余数是模除数
	}

	reverse(C.begin(), C.end());

	while (C.size() > 1 && C.back() == 0) C.pop_back(); //

	return C;
}

int main()
{
	string a;
	vector<int> A;
	int b;

	cin >> a >> b;
	for (int i = a.size() - 1; i >= 0; i--) A.push_back(a[i] - '0');

	int r;
	auto C = div(A, b, r);

	for (int i = C.size() - 1; i >= 0; i--) printf("%d", C[i]);
	cout << endl << r << endl;

	return 0;
}
```
# 位运算
```c++
#include <iostream>
using namespace std;

int lowbit(int x)
{
    return x & -x;
}

int main()
{
    int n;
    cin >> n;
    int x, res;
    while (n--) {
        cin >> x;
        res = 0;
        while (x) x -= lowbit(x), res++;
        cout << res << ' ';
    }
    return 0;
}
```

算法基础-1 第二部分

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](https://www.acwing.com/solution/content/16777/)

## 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;
}
```