矩阵既可以看成是一张数位表，也可以看成是若干个行向量或者若干个列向量的向量表

神奇的矩阵

description

solution

暴力做$A\ast B$会达到$n^3$的复杂度，难以接受

考虑，如果对于矩阵$A,B,C$满足$A\ast B=C$，显然有$A\ast B\ast R=C\ast R$

于是有随机一个$1\times n$的向量$R$，然后check等式是否成立，$A\ast R\ast B$就会降成$n^2$的复杂度

随机多次都无法满足这个式子，$A\ast B=C$的概率就微乎其微（除非你是非酋）

code

#include <bits/stdc++.h>
using namespace std;
#define int long long
int n;struct matrix {int n, m;int c[1000][1000];matrix() {memset( c, 0, sizeof( c ) );}matrix operator * ( matrix &t ) {matrix ans;ans.n = n, ans.m = t.m;for( int i = 0;i <= n;i ++ )for( int j = 0;j <= t.m;j ++ )for( int k = 0;k <= m;k ++ )ans.c[i][j] += c[i][k] * t.c[k][j];return ans;}
}A, B, C, R, ans1, ans2;signed main() {srand( time( 0 ) );next :while( ~ scanf( "%lld", &n ) ) {n --;A.n = A.m = B.n = B.m = C.n = C.m = n;for( int i = 0;i <= n;i ++ )for( int j = 0;j <= n;j ++ )scanf( "%lld", &A.c[i][j] );for( int i = 0;i <= n;i ++ )for( int j = 0;j <= n;j ++ )scanf( "%lld", &B.c[i][j] );for( int i = 0;i <= n;i ++ )for( int j = 0;j <= n;j ++ )scanf( "%lld", &C.c[i][j] );int t = 30;again :while( t -- ) {R.n = 0, R.m = n;for( int i = 0;i <= n;i ++ )R.c[0][i] = rand();ans1 = R * A * B;ans2 = R * C;for( int i = 0;i <= n;i ++ )if( ans1.c[0][i] != ans2.c[0][i] )goto again;printf( "Yes\n" );goto next;}printf( "No\n" );}return 0;
} 

[NOI2013]向量内积

description

solution

• k=2

  • 求出矩阵两两内积$(\mathrm{m}\mathrm{o}\mathrm{d}2)$ ，即$Y=A\ast A^T$
  • 接下来就是判断$Y=E,E$为全$1$矩阵（$Y_{i,j}$代表着$A$的$i$行向量与$A^T$的$j$列向量也就是原来的$A_j$行向量的内积）因为如果全$1$代表着每两个向量的内积都为$1(\mathrm{m}\mathrm{o}\mathrm{d}2)$
    • 判断方法就是上一题的随机化
    • 只要不等，就会有一个$0$向量，找到其位置$pos$，最后暴力求每个向量与其的内积是否整除$k$即可

• k=3，此时$A_{i,j}=0/1/2$，不能在使用上述$E$来判断了

  • 转换一下即可，$Z_{i,j}=Y_{i,j}^2(\mathrm{m}\mathrm{o}\mathrm{d}3)$，有$1^2\equiv 2^3\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}3)$，只有$0^2\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}3)$

  • 再次使用$E$来进行判断

  • 问题在于，$Z$是$Y$每个单项的平方，不是整体的平方，不能使用矩阵快速得到

    • 设$\alpha$是随机的一个$1\times n$向量

    • $(Z\ast \alpha {)}_i={\sum }_{j=1}^n{Z}_{i,j}\ast {\alpha }_j={\sum }_{j=1}^n{Y}_{i,j}^2\ast {\alpha }_j={\sum }_{j=1}^n{\alpha }_j({\sum }_{k=1}^n{A}_{i,k}{A}_{k,j}^T{)}^2$

      $={\sum }_{j=1}^n{\alpha }_j{\sum }_{k_1=1}^n{A}_{i,k_1}{A}_{k_1,j}^T\ast {\sum }_{j=1}^n{\alpha }_j{\sum }_{k_2=1}^n{A}_{i,k_2}{A}_{k_2,j}^T$

    • 发现可以变为${\sum }_{k_1,k_2}{A}_{i,k_1}{A}_{i,k_2}\ast {\sum }_{j=1}^n{\alpha }_j{A}_{k_1,j}{A}_{k_2,j}^T$

      预处理出和$i$无关部分，设$g_{k_1,k_2}={\sum }_{j=1}^n{\alpha }_j{A}_{k_1,j}{A}_{k_2,j}^T$

code

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define maxn 100005
#define maxd 105
int n, d, k, sum, pos, flag;
int x[maxn][maxd], g[maxn][maxd];
int ret[maxd], r[maxn];void calc2() {for( int i = 1;i <= d;i ++ ) ret[i] = 0;for( int i = 1;i <= d;i ++ )for( int j = 1;j <= n;j ++ )ret[i] = ( ret[i] + r[j] * x[j][i] ) % k;for( int i = 1;i <= n;i ++ ) {int ans = 0;for( int j = 1;j <= d;j ++ )ans = ( ans + ret[j] * x[i][j] ) % k;if( ans != sum ) {pos = i, flag = 1;break;}}
}void calc3() {for( int k1 = 1;k1 <= d;k1 ++ )for( int k2 = 1;k2 <= d;k2 ++ ) {g[k1][k2] = 0;for( int j = 1;j <= n;j ++ )g[k1][k2] = ( g[k1][k2] + r[j] * x[j][k1] % k * x[j][k2] ) % k;}for( int i = 1;i <= n;i ++ ) {int ans = 0;for( int k1 = 1;k1 <= d;k1 ++ )for( int k2 = 1;k2 <= d;k2 ++ )ans = ( ans + x[i][k1] * x[i][k2] % k * g[k1][k2] ) % k;if( ans != sum ) {pos = i, flag = 1;break;}}
}signed main() {srand( time( 0 ) );scanf( "%lld %lld %lld", &n, &d, &k );for( int i = 1;i <= n;i ++ )for( int j = 1;j <= d;j ++ )scanf( "%lld", &x[i][j] );	for( int T = 1;T <= 6;T ++ ) {sum = 0;for( int i = 1;i <= n;i ++ )r[i] = rand() % k, sum = ( sum + r[i] ) % k;if( k == 2 ) calc2();else calc3();if( flag ) break;}if( ! flag ) return ! printf( "-1 -1\n" );else {for( int i = 1;i <= n;i ++ )if( i ^ pos ) {int ans = 0;for( int j = 1;j <= d;j ++ )ans = ( ans + x[i][j] * x[pos][j] ) % k;if( ! ans ) return ! printf( "%lld %lld\n", min( i, pos ), max( i, pos ) );}}return 0;
}