矩阵快速幂

Posted 2021-11-25 darlingroot

传送门

 

emm

没错这就是矩阵快速幂的裸体板子题

所以

没什么可说的哇qwq

 

矩阵乘 + 快速幂 （+ 预处理）

 

 1 #include<cstdio>
 2 #include<cstring>
 3 #define ll long long
 4 using namespace std;
 5 const ll mod = 1e9+7;
 6 ll n,k;
 7 struct mat
 8 {
 9     ll m[105][105];
10 }l,r;
11 mat mul(mat x,mat y)
12 {
13     mat sum;
14     for(int i = 1;i <= n;i++)
15         for(int j = 1;j <= n;j++)
16             sum.m[i][j] = 0;
17     for(int i = 1;i <= n;i++)
18         for(int j = 1;j <= n;j++)
19             for(int k = 1;k <= n;k++)
20                 sum.m[i][j] = (sum.m[i][j]%mod +(x.m[i][k] * y.m[k][j])% mod)% mod;
21     return sum;
22 }
23 mat qpow(mat x,ll y)
24 {
25     mat ans = r;
26     while(y)
27     {
28         if(y & 1)
29             ans = mul(ans,x);
30         x = mul(x,x);
31         y >>= 1;
32     }
33     return ans;
34 }
35 int main()
36 {
37     scanf("%lld%lld",&n,&k);
38     for(int i = 1;i <= n;i++)
39         for(int j = 1;j <= n;j++)
40             scanf("%lld",&l.m[i][j]);
41     for(int i = 1;i <= n;i++)
42         r.m[i][i] = 1;
43     mat ans = qpow(l,k);
44     for(int i = 1;i <= n;i++)
45     {
46         for(int j = 1;j <= n;j++)
47             printf("%lld ",ans.m[i][j]);
48         printf("\n");
49     }
50     return 0;
51 }