51nod 1140 矩阵相乘结果的判断

Posted 2020-09-12 ws_ccd

弱弱的买了随机算法的视频水了一下2333

真的是好神

大概就是判AB=C,这样的话再等式两边同乘一个1*n的矩阵H（貌似有个专业的名字），这样矩阵乘法的复杂度就是n^2的。

因为矩阵乘法是有结合律的，所以就是先算出HA（蛤？？）,再算(HA)*B，然后和HC看是不是相等就好

 

get高端暴力姿势

 1 #include<bits/stdc++.h>
 2 #define LL long long
 3 using namespace std;
 4 
 5 const int QWQ=505;
 6 
 7 LL a[QWQ][QWQ],b[QWQ][QWQ],c[QWQ][QWQ];
 8 int n;
 9 LL rnd[QWQ],ans1[QWQ],ans2[QWQ],ans3[QWQ];
10 
11 int main()
12 {
13     srand(time(0));
14     cin>>n;
15     for (int i=1; i<=n; i++) for (int j=1; j<=n; j++) scanf("%lld",&a[i][j]);
16     for (int i=1; i<=n; i++) for (int j=1; j<=n; j++) scanf("%lld",&b[i][j]);
17     for (int i=1; i<=n; i++) for (int j=1; j<=n; j++) scanf("%lld",&c[i][j]);
18     int cnt=50;
19     while (cnt--)
20     {
21         memset(ans1,0,sizeof(ans1));
22         memset(ans2,0,sizeof(ans2));
23         memset(ans3,0,sizeof(ans3));
24         for (int i=1; i<=n; i++) rnd[i]=(LL)rand()%16;
25     //    for (int i=1; i<=n; i++) printf("%d ",rnd[i]); system("pause");
26         
27         for (int i=1; i<=n; i++)
28             for (int j=1; j<=n; j++)
29                 ans1[i]+=rnd[j]*a[j][i];
30     //    for (int i=1; i<=n; i++) printf("%d ",ans1[i]); cout<<endl;
31         for (int i=1; i<=n; i++)
32             for (int j=1; j<=n; j++)
33                 ans2[i]+=ans1[j]*b[j][i];
34     //    for (int i=1; i<=n; i++) printf("%d ",ans2[i]); cout<<endl;
35         for (int i=1; i<=n; i++)
36             for (int j=1; j<=n; j++)
37                     ans3[i]+=rnd[j]*c[j][i];
38     //    for (int i=1; i<=n; i++) printf("%d ",ans3[i]); cout<<endl;
39         
40         for (int i=1; i<=n; i++)
41             if (ans2[i]!=ans3[i]) 
42             {
43                 puts("No");
44                 return 0;
45             }
46     }
47     puts("Yes");
48     return 0;
49 }