《只言片语》

并查集模板

#include <bits/stdc++.h>
using namespace std;
typedef int ll;
class bcj
{
private:
	vector<ll> fa;

public:
	bcj() {}
	bcj(ll n)
	{
		for (auto i = 0; i <= n + 1; i++)
			fa.push_back(i);
	}
	ll& operator[](ll x) { return fa[x]; }
	ll& at(ll x) { return fa[x]; }
	ll size() const { return fa.size(); }
	ll ff(ll x);
	void mg(ll x, ll y) { fa[ff(y)] = ff(x); }
};
ll bcj::ff(ll x)
{
	return (fa[x] == x) ? x : (fa[x] = ff(fa[x]));
}
int main()
{
	bcj a;
	return 0;
}

矩阵

C Version

#include <stdio.h>
#define N 100
//N 矩阵最大长和宽
typedef struct matrix
{
	int h, w, aa[N][N];
	//h:长 ， w:宽,  aa 矩阵
} jz;
void initjz(jz *a, int x, int y) //初始化,行x列y,全部0
{
	a->h = x, a->w = y;
	for (int i = 0; i < x; i++)
		for (int j = 0; j < y; j++)
			a->aa[i][j] = 0;
}
void initjz(jz *a, int x, int y, int z) //初始化,行x列y,全部赋值z
{
	a->h = x, a->w = y;
	for (int i = 0; i < x; i++)
		for (int j = 0; j < y; j++)
			a->aa[i][j] = z;
}
jz jz_cheng(jz x, jz y) //返回x*y
{
	jz s;
	initjz(&s, x.h, y.w);
	for (int i = 0; i < x.h; i++)
		for (int j = 0; j < y.w; j++)
			for (int k = 0; k < x.w; k++)
				s.aa[i][j] += x.aa[i][k] * y.aa[k][j];
	return s;
}
jz jz_jia(jz x, jz y) //返回x+y
{
	jz s;
	initjz(&s, x.h, x.w);
	for (int i = 0; i < s.h; i++)
		for (int j = 0; j < s.w; j++)
			s.aa[i][j] = x.aa[i][j] + y.aa[i][j];
	return s;
}
jz jz_jian(jz x, jz y) //返回x-y
{
	jz s;
	initjz(&s, x.h, x.w);
	for (int i = 0; i < s.h; i++)
		for (int j = 0; j < s.w; j++)
			s.aa[i][j] = x.aa[i][j] - y.aa[i][j];
	return s;
}
jz jz_chengf(jz a, int k) //返回a的k次，以快速幂实现
{
	if (k == 1)
		return a;
	jz aa = jz_chengf(a, k / 2);
	aa = jz_cheng(aa, aa);
	if (k % 2 != 0)
		return jz_cheng(aa, a);
	else
		return aa;
}
void jz_cpy(jz *a, jz b) //复制b到a
{
	a->h = b.h;
	a->w = b.w;
	for (int i = 0; i < a->h; i++)
		for (int j = 0; j < a->w; j++)
			a->aa[i][j] = b.aa[i][j];
	return;
}
int main()
{
	jz a, b;
	initjz(&a, 23, 31);
	b = a; //或jz_cpy(&b,a);
	jz c = jz_jia(a, b);
	return 0;
}

CPP Version

#include <bits/stdc++.h>
using namespace std;
constexpr int N = 100;
typedef struct matrix
{
	int h, w, aa[N][N];
	//h:长 ， w:宽,  aa 矩阵
	matrix() {}
	matrix(int x, int y) : h(x), w(y) //建立长x宽y的矩阵全部赋值0
	{
		for (int i = 0; i < x; i++)
			for (int j = 0; j < y; j++)
				aa[i][j] = 0;
	}
	matrix(int x, int y, int z) : h(x), w(y) //建立长x宽y的矩阵全部赋值z
	{
		for (int i = 0; i < x; i++)
			for (int j = 0; j < y; j++)
				aa[i][j] = z;
	}
	friend matrix operator*(matrix x, matrix y) //矩阵x*矩阵y
	{
		matrix s(x.h, y.w);
		for (int i = 0; i < x.h; i++)
			for (int j = 0; j < y.w; j++)
				for (int k = 0; k < x.w; k++)
					s.aa[i][j] += x.aa[i][k] * y.aa[k][j];
		return s;
	}
	friend matrix operator+(matrix x, matrix y) //矩阵x+矩阵y
	{
		matrix s(x.h, x.w);
		for (int i = 0; i < s.h; i++)
			for (int j = 0; j < s.w; j++)
				s.aa[i][j] = x.aa[i][j] + y.aa[i][j];
		return s;
	}
	friend matrix operator-(matrix x, matrix y) //矩阵x-矩阵y
	{
		matrix s(x.h, x.w);
		for (int i = 0; i < s.h; i++)
			for (int j = 0; j < s.w; j++)
				s.aa[i][j] = x.aa[i][j] - y.aa[i][j];
		return s;
	}
	friend matrix operator^(matrix a, int k) //是乘方,不是XOR,用快速幂+递归实现
	{
		if (k == 1)
			return a;
		matrix aa = a ^ (k / 2);
		aa = aa * aa;
		if (k % 2 != 0)
			return aa * a;
		else
			return aa;
	}
} jz;
jz a;
int main()
{
	return 0;
}