矩阵的计算

C实现

#include <stdio.h>
#define N 100
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;
}
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
{
	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;
}
jz jz_mod(jz x, int m) //返回整体mod
{
	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] % m;
	return s;
}
void jz_cpy(jz *a, jz b)//复制到
{
	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实现

#include <bits/stdc++.h>
using namespace std;
constexpr int N = 100;
typedef struct matrix
{
	int h, w, aa[N][N];
	matrix() {}
	matrix(int x, int y) : h(x), w(y)
	{
		for (int i = 0; i < x; i++)
			for (int j = 0; j < y; j++)
				aa[i][j] = 0;
	}
	friend matrix operator*(matrix x, matrix 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)
	{
		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)
	{
		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) //bushi xor
	{//是乘方,不是XOR
		if (k == 1)
			return a;
		matrix aa = a ^ (k / 2);
		aa = aa * aa;
		if (k % 2 != 0)
			return aa * a;
		else
			return aa;
	}
	friend matrix operator%(matrix a, int k)
	{
		for (int i = 0; i < a.h; i++)
			for (int j = 0; j < a.w; j++)
				a.aa[i][j] %= k;
		return a;
	}
} jz;
jz a;
int main()
{

	return 0;
}

标签：aa,return,matrix,jz,int,代码,矩阵,++,CPP
来源： https://www.cnblogs.com/ssj233-aaa/p/16366234.html