HDU 4920 Matrix multiplication

HDU 4920 Matrix multiplication

Problem Description Given two matrices A and B of size n×n, find the product of them.

bobo hates big integers. So you are only asked to find the result modulo 3.
Input The input consists of several tests. For each tests:

The first line contains n (1≤n≤800). Each of the following n lines contain n integers -- the description of the matrix A. The j-th integer in the i-th line equals A_ij. The next n lines describe the matrix B in similar format (0≤A_ij,B_ij≤10⁹).
Output For each tests:

Print n lines. Each of them contain n integers -- the matrix A×B in similar format.
Sample Input

1
0
1
2
0 1
2 3
4 5
6 7

Sample Output

0
0 1
2 1

題意 ：矩陣相乘

思路：復雜度O(n^3)是接受不了的，但是出於題目的意思%3，所以我們可以不去處理0的情況，我們把原本最裡面的那層for（k ) 拿到最外面，然後當有一行的某列出現0 的時候，那麼對應的所有列對應的行也就不加入計算了，再加上出入外掛水過

#include 
#include 
#include 
#include 
typedef long long ll;
using namespace std;
const int maxn = 805;

int a[maxn][maxn], b[maxn][maxn], c[maxn][maxn];
int n;

int Scan() {
	int res = 0, ch, flag = 0;
	if ((ch = getchar()) == '-')
		flag = 1;
	else if (ch >= '0' && ch <= '9')
		res = ch - '0';
	while ((ch = getchar()) >= '0' && ch <= '9')
		res = res * 10 + ch - '0';
	return flag?-res:res;
}

int main() {
	while (scanf("%d", &n) != EOF) {
		for (int i = 1; i <= n; i++) 
			for (int j = 1; j <= n; j++) {
				a[i][j] = Scan() % 3;
				c[i][j] = 0;
			}
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= n; j++)
				b[i][j] = Scan() % 3;

		for (int k = 1; k <= n; k++)
			for (int i = 1; i <= n; i++)
				if (a[i][k])
					for (int j = 1; j <= n; j++)
						c[i][j] += a[i][k] * b[k][j];
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= n; j++)
				printf("%d%c", c[i][j]%3, (j==n)?'\n':' ');
	}	
	return 0;
}