POJ 3318 Matrix Multiplication

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

2
1 0
2 3
5 1
0 8
5 1
10 26

Sample Output

YES

Hint

Multiple inputs will be tested. So O(n³) algorithm will get TLE.

设v是一个随机生成的n行1列的矩阵，取值范围为0～1

由A×B=C可得：

A×B×v=C×v

假如A×B!=C,那么有：(A×B-C)×v!=0

因为和只有1列的矩阵相乘都是O(n^2)，且结果为n×1的矩阵，所以时间上是可以的

但是本身A×B×v=C×v不能推出A×B=C，但A×B×v!=C×v一定能推出A×B!=C

如：1 2 3  1       3 1 2  1    6

　　3 2 1×1   =  1 2 3×1 = 6

　　2 1 3  1       2 3 1  1    6

所以该算法判断不成立是正确的，但判断成立是不确定的

但是只要不是“不成立”，就是“成立”，所以问题不大，所以可以多次判断来决定是否“成立”

而且假设算出的矩阵第i行不为0，那么可能v[i][1]=0，使得判断失误

所以至少有1/2的几率判断出“不成立”

要在原有算法上多次判定，大概60次就行，错误率为2^-60

其实v的取值范围扩大和判断次数增加都能增加准确度，这里v用值0/1只是方便分析

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<ctime>
using namespace std;
int n,flag,a[501][501],b[501][501],c[501][501],A[501],B[501],C[501],v[501];
int main()
{int i,j,k;
  while (cin>>n)
    {
      flag=0;
      for (i=1;i<=n;i++)
    {
      for (j=1;j<=n;j++)
        scanf("%d",&a[i][j]);
    }
      for (i=1;i<=n;i++)
    {
      for (j=1;j<=n;j++)
        scanf("%d",&b[i][j]);
    }
      for (i=1;i<=n;i++)
    {
      for (j=1;j<=n;j++)
        scanf("%d",&c[i][j]);
    }
      srand(time(0));
      for (k=1;k<=60;k++)
    {
      for (i=1;i<=n;i++)
        v[i]=rand()%2;
      for (i=1;i<=n;i++)
        {
          B[i]=0;
          for (j=1;j<=n;j++)
        B[i]+=b[i][j]*v[j];
        }
      for (i=1;i<=n;i++)
        {
          C[i]=0;
          for (j=1;j<=n;j++)
        C[i]+=c[i][j]*v[j];
        }
      for (i=1;i<=n;i++)
        {
          A[i]=0;
          for (j=1;j<=n;j++)
        A[i]+=a[i][j]*B[j];
        }
      for (i=1;i<=n;i++)
        if (A[i]!=C[i])
          {
        flag=1;
        break;
          }
      if (flag) break;
    }
      if (flag) printf("NO
");
      else printf("YES
");
    }
}