POJ 3318 Matrix Multiplication(隨機化演算法)

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給你三個矩陣A，B，C。讓你判斷A*B是否等於C。

隨機一組資料，然後判斷乘以A，B之後是否與乘C之後相等。

很扯淡的啊，感覺這種演算法不嚴謹啊、、、

Matrix Multiplication

Time Limit: 2000MS		Memory Limit: 65536K
Total 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;}