模板矩阵快速幂

Posted 2020-10-31 Driver_Lao

洛谷 3390

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cstring>
 4 #define rg register
 5 #define N 110
 6 #define LL long long 
 7 using namespace std;
 8 const LL Mod=1e9+7;
 9 LL n,m,tmp[N][N];
10 struct Mat{LL m[N][N];}a,t,ans;
11 inline LL read(){
12     LL k=0,f=1; char c=getchar();
13     while(c<‘0‘||c>‘9‘)c==‘-‘&&(f=-1),c=getchar();
14     while(‘0‘<=c&&c<=‘9‘)k=k*10+c-‘0‘,c=getchar();
15     return k*f;
16 }
17 inline void mul(Mat &a,Mat b){
18     memset(tmp,0,sizeof(tmp));
19     for(rg int i=1;i<=n;i++)
20     for(rg int j=1;j<=n;j++)
21     for(rg int k=1;k<=n;k++)
22     tmp[i][j]+=a.m[i][k]*b.m[k][j],tmp[i][j]%=Mod;
23     for(rg int i=1;i<=n;i++)
24     for(rg int j=1;j<=n;j++)
25     a.m[i][j]=tmp[i][j];
26 }
27 int main(){
28     n=read(); m=read();
29     for(rg int i=1;i<=n;i++) 
30     for(rg int j=1;j<=n;j++) t.m[i][j]=read();
31     for(rg int i=1;i<=n;i++) ans.m[i][i]=1;
32     while(m){
33         if(m&1) mul(ans,t);
34         mul(t,t);
35         m>>=1;
36     }
37     for(rg int i=1;i<=n;i++){
38         for(rg int j=1;j<=n;j++) printf("%lld ",ans.m[i][j]);
39         puts("");
40     }
41     return 0;
42 } 

View Code