洛谷 P3390 【模板】矩阵快速幂 题解

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<cmath>
 4 #include<algorithm>
 5 
 6 const int MOD = 1e9+7;
 7 
 8 long long n,k;
 9 
10 struct Matrix
11 {
12     long long a[105][105];
13     
14     Matrix(){
15         memset(a,0,sizeof(a));
16     }
17     
18     Matrix operator * (const Matrix &x)const
19     {
20         Matrix r;
21         for(int i = 1;i <= n;++ i)
22             for(int j = 1;j <= n;++ j){
23                 r.a[i][j] = 0;
24                 for(int k = 1;k <= n;++ k)
25                     r.a[i][j] = (r.a[i][j]+a[i][k]*x.a[k][j]%MOD)%MOD;
26             }
27                 
28         return r;
29     }
30 }ans,base;
31 
32 void ksm()
33 {
34     ans = base;
35     for(long long b = k-1;b;b >>= 1LL)
36     {
37         if(b&1){
38             for(int i = 1;i <= n;++ i)
39             {
40                 for(int j = 1;j <= n;++ j)
41                     printf("%lld ",ans.a[i][j]);
42                 printf("\n");
43             }
44             ans = ans*base;
45         }
46             
47         base = base*base;
48     }
49 }
50 
51 int main()
52 {
53     scanf("%lld%lld",&n,&k);
54     for(int i = 1;i <= n;++ i)
55         for(int j = 1;j <= n;++ j)
56             scanf("%lld",&base.a[i][j]);
57     ksm();
58     for(int i = 1;i <= n;++ i)
59     {
60         for(int j = 1;j <= n;++ j)
61             printf("%lld ",ans.a[i][j]);
62         printf("\n");
63     }
64         
65     return 0;
66 }