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HDU4920妥妥的暴力

編輯：C++入門知識

HDU4920妥妥的暴力

原題http://acm.hdu.edu.cn/showproblem.php?pid=4920

Matrix multiplication

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 848 Accepted Submission(s): 339

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

Sample Output

0
0 1
2 1

Author Xiaoxu Guo (ftiasch)

Source 2014 Multi-University Training Contest 5

Recommend We have carefully selected several similar problems for you: 4919 4918 4917 4916 4915

//開始的時候還以為這題要用到什麼高深的算法。後來發現很多人過了，估計就是小技巧了
//首先。乘法的運算時間要比加法多。所以能避免乘法，盡量優化掉。
//經過這題，發現C++比G++跑的要更快，同樣的代碼差了差不多0.4秒。比賽的時候就是由於交了G++T了好多次
//最後，說一句，騷年，雖然暴力不是萬能的。但是沒有加上小技巧的暴力是萬萬不能的。
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define Max 800 + 5
int a[Max][Max];
int b[Max][Max];
int c[Max][Max];

int main(){
	int n,i,j;

	while(~scanf("%d",&n)){
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		memset(c,0,sizeof(c));
		for(i=0;i