hdu 4920矩阵连乘-白红宇

hdu 4920矩阵连乘

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发布时间：2019-05-26

本文共 1448 字，大约阅读时间需要 4 分钟。

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

10120 12 34 56 7

Sample Output

00 1 2 1 ////  main.cpp//  160929////  Created by liuzhe on 17/3/30.//  Copyright © 2016年 my_code. All rights reserved.////#include 
    
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                  #define ll long long#define mod 3using namespace std;ll a[805][805],b[805][805],c[805][805];int main(){ int n; while(~scanf("%d",&n)) //while(cin>>n) { for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) { scanf("%lld",&a[i][j]); a[i][j]=a[i][j]%mod; } for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) { scanf("%lld",&b[i][j]); b[i][j]=b[i][j]%mod; } memset(c,0,sizeof(c)); for(int i=1;i<=n;i++) { for(int k=1;k<=n;k++) { for(int j=1;j<=n;j++) { c[i][j]+=(a[i][k]*b[k][j]); } //c[i][j]=c[i][j]%mod; } } for(int i=1;i<=n;i++) { for(int j=1;j