Given a n × n matrix A and a positive integer k, find the sum S = A + A² + A³ + … + A^k.

Input

The input contains exactly one test case. The first line of input contains three positive integers n (n ≤ 30), k (k ≤ 10⁹) and m (m < 10⁴). Then follow n lines each containing n nonnegative integers below 32,768, giving A

Output

Output the elements of S modulo m in the same way as A is given.

Sample Input

2 2 4
0 1
1 1

Sample Output

1 2
2 3

矩陣乘法 快速冪 遞歸

可以把矩陣當成矩陣裏的元素，構建四個矩陣拼成的大矩陣進行快速冪遞推；

也可以普通地遞歸計算：

項數為偶數的時候類似於 (A^1+A^2+A^3+A^4)*(A^4 + I) = A^1+A^2+A^3+A^4+A^5+A^6+A^7+A^7

項數為奇數的時候要加上那個奇數項

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstring>
 4 #include<cstdio>
 5 #include<cmath>
 6 using namespace std;
 7 int read(){
 8     int x=0,f=1;char ch=getchar();
 9     while(ch<‘0‘ || ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
 
10     while(ch>=‘0‘ && ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
11     return x*f;
12 }
13 int n,m,K;
14 struct Mat{
15     int x[32][32];
16     void init(){
17         memset(x,0,sizeof x);
18         for(int i=1;i<=n;i++)x[i][i]=1;
19         return;
20     }
21     Mat operator + (const Mat &b){
22         Mat res;
23         for(int i=1;i<=n;i++)
24             for(int j=1;j<=n;j++)
25                 res.x[i][j]=(x[i][j]+b.x[i][j])%m;
26         return res;
27     }
28     Mat operator * (const Mat &b){
29         Mat res;
30         for(int i=1;i<=n;i++)
31             for(int j=1;j<=n;j++){
32                 res.x[i][j]=0;
33                 for(int k=1;k<=n;k++)
34                     (res.x[i][j]+=x[i][k]*b.x[k][j]%m)%=m;
35             }
36         return res;
37     }
38 }I,A;
39 inline Mat ksm(Mat a,int k){
40     Mat res;res.init();
41     while(k){
42         if(k&1)res=res*a;
43         a=a*a;
44         k>>=1;
45     }
46     return res;
47 }
48 Mat Bcalc(int k){
49     if(k==1)return A;
50     if(k&1)
51         return (ksm(A,k>>1)+I)*Bcalc(k>>1)+ksm(A,k);
52     return (ksm(A,k>>1)+I)*Bcalc(k>>1);
53 }
54 int main(){
55     int i,j;
56     n=read();K=read();m=read();
57     I.init();
58     for(i=1;i<=n;i++)
59         for(j=1;j<=n;j++)
60             A.x[i][j]=read();
61     Mat ans=Bcalc(K);
62     for(i=1;i<=n;i++){
63         for(j=1;j<=n;j++)
64             printf("%d ",ans.x[i][j]);
65         printf("\n");
66     }
67     return 0;
68 }

POJ3233 Matrix Power Series