题目
其中(A)是一个(n imes n)的矩阵，(S_k = A + A^2 + A^3 + … + A^k)，求(S_k)
按照数论出现和，那么构造一个和数论一起递推的式子(S_k = S_{k - 1} + A^k)
那么假设(A)不是矩阵
[left[egin{array}{l}
1 & 1\
0 & A
end{array}
ight] imes
left[egin{array}{l}
S_{k - 1}\
A^k
end{array}
ight] =
left[egin{array}{l}
S_k\
A^{k + 1}
end{array}
ight]]
用E代替1，用矩阵代替数字，转换一下就是
[left[egin{array}{l}
E & E\
0 & A
end{array}
ight] imes
left[egin{array}{l}
S_{k - 1}\
A^k
end{array}
ight] =
left[egin{array}{l}
S_k\
A^{k + 1}
end{array}
ight]
]
构建了个分块矩阵，大小是(2n imes 2n)
加速矩阵(left[egin{array}{l}
E & E\
0 & A
end{array}
ight]) 初始矩阵(left[egin{array}{l}
S_1\
A^2
end{array}
ight])
#include <iostream>
#include <cstdio>
#include <cstring>
#define ll long long
using namespace std;
const int N = 61;
int mod, k, n;
struct Matrix{//矩阵
int n,m;
int a[N][N];
Matrix(int x,int y):n(x),m(y){memset(a,0,sizeof(a));}
Matrix operator * (const Matrix &b){
Matrix ans(n,b.m);
for(int i = 0; i < n; i++){
for(int j = 0; j < b.m; j++){
for(int k = 0; k < m; k++){
ans.a[i][j] = (ans.a[i][j] + a[i][k] * b.a[k][j] % mod) % mod;
}
}
}
return ans;
}
};
Matrix ksm(Matrix a, ll b){
Matrix ans(a.n, a.m);
for(int i = 0; i <= max(a.n, a.m); i++)
ans.a[i][i] = 1;
while(b){
if(b & 1)ans = ans * a;
a = a * a;
b >>= 1;
}
return ans;
}
Matrix a(35, 35);
void solve(){
Matrix base(k * 2, k * 2);
for(int i = 0; i < k; i++)
base.a[i][i] = base.a[i][i + k] = 1;
for(int i = k; i < 2 * k; i++)
for(int j = k; j < 2 * k; j++)
base.a[i][j] = a.a[i - k][j - k];
base = ksm(base, n - 1);
Matrix ans(k * 2, k);
for(int i = 0; i < k; i++)
for(int j = 0; j < k; j++)
ans.a[i][j] = a.a[i][j];
a = a * a;
for(int i = k; i < 2 * k; i++)
for(int j = 0; j < k; j++)
ans.a[i][j] = a.a[i - k][j];
ans = base * ans;
for(int i = 0; i < k; i++){
for(int j = 0; j < k; j++)
printf("%d ", ans.a[i][j]);
putchar('
');
}
}
int main(){
scanf("%d%d%d", &k, &n, &mod);
for(int i = 0; i < k; i++)
for(int j = 0; j < k; j++)
scanf("%d", &a.a[i][j]);
solve();
return 0;
}