Poj 3318 matrix multiplication (randomization algorithm)

Here are three matrices A, B, and C. Let you determine whether a * B is equal to C.

A random group of data, and then determine whether it is equal after multiplying A and B with multiplication C.

I think this algorithm is not rigorous ,,,

Matrix Multiplication

Time limit:2000 ms		Memory limit:65536 K
Total submissions:16255		Accepted:3515

Description

You are given threeN×NMatricesA,BAndC. Does the equationA×B=CHold true?

Input

The first line of input contains a positive integerN(NLess than or equal to 500) followed by the three matricesA,BAndCRespectively. Each matrix's description is a block of n × n integers.

It guarantees that the elementsAAndBAre less than 100 in absolute value and elementsCAre less than 10,000,000 in absolute value.

Output

Output "yes" if the equation holds true, otherwise "no ".

Sample Input

21 02 35 10 85 110 26

Sample output

YES

Hint

Multiple inputs will be tested. So O (N3) algorithm will get TLE.

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <time.h>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#define max( x, y )  ( ((x) > (y)) ? (x) : (y) )#define min( x, y )  ( ((x) < (y)) ? (x) : (y) )#define Mod 1000000007#define LL long longusing namespace std;const int maxn = 510;int a[maxn][maxn];int b[maxn][maxn];int c[maxn][maxn];int f[maxn];int aa[maxn];int bb[maxn];int cc[maxn];int main(){    int n;    cin >>n;    memset(aa, 0, sizeof(aa));    memset(bb, 0, sizeof(bb));    memset(cc, 0, sizeof(cc));    for(int i = 0; i < n; i++)        for(int j = 0; j < n; j++) scanf("%d",&a[i][j]);    for(int i = 0; i < n; i++)        for(int j = 0; j < n; j++) scanf("%d",&b[i][j]);    for(int i = 0; i < n; i++)        for(int j = 0; j < n; j++) scanf("%d",&c[i][j]);    srand((unsigned int)time(0));    for(int i = 0; i < n; i++)        f[i] = rand()%100;    for(int i = 0; i < n; i++)    {        for(int j = 0; j < n; j++)        {            bb[i] += b[i][j]*f[j];            cc[i] += c[i][j]*f[j];        }    }    for(int i = 0; i < n; i++)        for(int j = 0; j < n; j++) aa[i] += a[i][j]*bb[j];    int flag = 0;    for(int i = 0; i < n; i++)    {        if(aa[i] != cc[i])        {            flag = 1;            break;        }    }    if(flag)        cout<<"NO"<<endl;    else        cout<<"YES"<<endl;}