题意：有三个N*N的矩阵a,b,c,判断a*b是否等于c.

思路：暴力判断O(N*3),我没试能不能过。

正解是随机化算法，随机构造列向量p,然后分别计算a*(b*p)和c*p，比较之。

这个过程仅为O(N^2).

随机多组即可。

Code:

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <climits>
#include <cstdlib>
using namespace std;
  
#define N 1010
struct Matrix {
    int w, h, v[N][N];
    Matrix() {
        memset(v, 0, sizeof v);
    }
    bool operator == (const Matrix &B) const {
        if (w != B.w || h != B.h)
            return 0;
        register int i, j;
        for(i = 0; i < w; ++i)
            for(j = 0; j < h; ++j)
                if (v[i][j] != B.v[i][j])
                    return 0;
        return 1;
    }
    void operator = (const Matrix &B) {
        w = B.w, h = B.h;
        for(int i = 0; i < w; ++i)
            for(int j = 0; j < h; ++j)
                v[i][j] = B.v[i][j];
    }
    void operator *= (const Matrix &B);
}A, B, C, ran, tmp1, tmp2, get;
  
void Matrix::operator *= (const Matrix &B) {
    memset(get.v, 0, sizeof get.v);
    get.w = w, get.h = B.h;
    register int i, j, k;
    for(i = 0; i < get.w; ++i)
        for(k = 0; k < h; ++k)
            for(j = 0; j < get.h; ++j)
                get.v[i][j] += v[i][k] * B.v[k][j];
    *this = get;
}
int main() {
    int n;
    register int i, j;
    while(scanf("%d", &n) != EOF) {
        A.w = A.h = n;
        B.w = B.h = n;
        C.w = C.h = n;
        for(i = 0; i < n; ++i)
            for(j = 0; j < n; ++j)
                scanf("%d", &A.v[i][j]);
        for(i = 0; i < n; ++i)
            for(j = 0; j < n; ++j)
                scanf("%d", &B.v[i][j]);
        for(i = 0; i < n; ++i)
            for(j = 0; j < n; ++j)
                scanf("%d", &C.v[i][j]);
          
        int wrong = 0;
        for(int Case = 1; Case <= 1; ++Case) {
            ran.w = 1, ran.h = n;
            for(i = 0; i < n; ++i)
                ran.v[0][i] = rand();
            tmp1 = ran;
            tmp2 = ran;
            tmp1 *= A, tmp1 *= B;
            tmp2 *= C;
              
            if (!(tmp1 == tmp2)) {
                wrong = 1;
                break;
            }
        }
          
        if (wrong)
            puts("No");
        else
            puts("Yes");
    }
      
    return 0;
}

BZOJ2396 神奇的矩阵