51 nod 1140

第1行，1个数N。(0 <= N <= 500)
第2 - N + 1行：每行N个数，对应矩阵A的元素。（0 <= M[i] <= 16)
第N + 2 - 2N + 1行：每行N个数，对应矩阵B的元素。（0 <= M[i] <= 16)
第2N + 2 - 3N + 1行：每行N个数，对应矩阵C的元素。

如果相等输出Yes，否则输出No。

2
1 0
0 1
0 1
1 0
0 1
1 0

Yes

H*A*B=H*C

随机几个H看能不能满足这个式子就好啦。

#include <bits/stdc++.h>//#include <ext/pb_ds/tree_policy.hpp>//#include <ext/pb_ds/assoc_container.hpp>//using namespace __gnu_pbds;using namespace std;#define pi acos(-1)#define endl '\n'#define me(x) memset(x,0,sizeof(x));#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)#define close() ios::sync_with_stdio(0);#define srand() srand(time(0));typedef long long LL;const int INF=0x3f3f3f3f;const LL LINF=0x3f3f3f3f3f3f3f3fLL;//const int dx[]={-1,0,1,0,-1,-1,1,1};//const int dy[]={0,1,0,-1,1,-1,1,-1};const int maxn=5e2+5;const int maxx=1e5+100;const double EPS=1e-9;const int MOD=1000000007;#define mod(x) ((x)%MOD);template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}//typedef tree<pt,null_type,less< pt >,rb_tree_tag,tree_order_statistics_node_update> rbtree;/*lch[root] = build(L1,p-1,L2+1,L2+cnt);rch[root] = build(p+1,R1,L2+cnt+1,R2);中前*//*lch[root] = build(L1,p-1,L2,L2+cnt-1);rch[root] = build(p+1,R1,L2+cnt,R2-1);中后*/long long gcd(long long a , long long b){if(b==0) return a;a%=b;return gcd(b,a);}int a[maxn][maxn],b[maxn][maxn],c[maxn][maxn];int ans1[maxn],ans2[maxn],ans3[maxn],rnd[maxn];int n;inline int Scan(){int res=0,ch,flag=0;if((ch=getchar())=='-')flag=1;else if(ch>='0' && ch<='9')res=ch-'0';while((ch=getchar())>='0'&&ch<='9')res=res*10+ch-'0';return flag ? -res : res;}int main(){srand();cin>>n;for(int i=1;i<=n;i++)for (int j=1; j<=n; j++)a[i][j]=Scan();for(int i=1;i<=n;i++)for (int j=1; j<=n; j++)b[i][j]=Scan();for(int i=1;i<=n;i++)for (int j=1; j<=n; j++)c[i][j]=Scan();int cnt=20;while(cnt--){me(ans1);me(ans2);me(ans3);for(int i=1;i<=n;i++)rnd[i]=(LL)rand()%16;for(int i=1;i<=n;i++)for (int j=1; j<=n; j++)ans1[i]+=rnd[j]*a[j][i];for(int i=1;i<=n;i++)for (int j=1; j<=n; j++)ans2[i]+=ans1[j]*b[j][i];for(int i=1;i<=n;i++)for (int j=1; j<=n; j++)ans3[i]+=rnd[j]*c[j][i];for(int i=1;i<=n;i++)if(ans2[i]!=ans3[i])4000{puts("No");return 0;}}puts("Yes");}