#include <bits/stdc++.h>
const int n = 512, N = n;
void multiply (int c[N][N], int a[N][N], int b[N][N]) {
	for(int i = 0; i < n; i ++)
		for(int l = 0; l < n; l ++)
			for(int j = 0; j < n; j ++)
			c[i][j] += (a[i][l] * b[l][j]);
}
int c[N][N], a[N][N], b[N][N];
int main() {
	int seedA, seedB;
	scanf("%d%d", &seedA, &seedB);
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            a[i][j] = ((i | j) + j) ^ seedA;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            b[i][j] = ((i & j) + i) ^ seedB;
    multiply(c, a, b);
    for (int i = 0; i < n; i++) {
        int sum = 0;
        for (int j = 0; j < n; j++) sum ^= c[i][j];
        printf("%d\n", sum);
    }
}

这份代码远比下面的代码快，可它们只是交换了循环顺序，都是 $O(n^3)$ 的，这是为什么？

#include <bits/stdc++.h>
const int n = 512, N = n;
void multiply (int c[N][N], int a[N][N], int b[N][N]) {
	for(int i = 0; i < n; i ++)
      for(int j = 0; j < n; j ++)
			for(int l = 0; l < n; l ++)
			c[i][j] += (a[i][l] * b[l][j]);
}
int c[N][N], a[N][N], b[N][N];
int main() {
	int seedA, seedB;
	scanf("%d%d", &seedA, &seedB);
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            a[i][j] = ((i | j) + j) ^ seedA;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            b[i][j] = ((i & j) + i) ^ seedB;
    multiply(c, a, b);
    for (int i = 0; i < n; i++) {
        int sum = 0;
        for (int j = 0; j < n; j++) sum ^= c[i][j];
        printf("%d\n", sum);
    }
}