POJ 3318 Matrix Multiplication(随机化算法)

给你三个矩阵A，B，C。让你判断A*B是否等于C。

随机一组数据，然后判断乘以A，B之后是否与乘C之后相等。

很扯淡的啊，感觉这种算法不严谨啊、、、

Matrix Multiplication

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 16255 Accepted: 3515

Description

You are given three n × n matrices A, B and C. Does the equation A × B = C hold true?

Input

The first line of input contains a positive integer n (n ≤ 500) followed by the the three matrices A, B and C respectively. Each matrix's description is a block of n × n integers.

It guarantees that the elements of A and B are less than 100 in absolute value and elements of C are 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;}