HDU 4920 Matrix multiplication

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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≤10 ⁹).

Output

For each tests:

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

Sample Input

Sample Output

  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;
}

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