HDU 4920 Matrix multiplication

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

0
0 1
2 1

题意 ：矩阵相乘

思路：复杂度O(n^3)是接受不了的，但是出于题目的意思%3，所以我们可以不去处理0的情况，我们把原本最里面的那层for（k ) 拿到最外面，然后当有一行的某列出现0 的时候，那么对应的所有列对应的行也就不加入计算了，再加上出入外挂水过

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
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;
}

HDU 4920 Matrix multiplication,布布扣,bubuko.com

HDU 4920 Matrix multiplication

原文：http://blog.csdn.net/u011345136/article/details/38390839