Description

Given an×nmatrixAand a positive integerk, find the sumS=A+A²+A³+ … +A^k.

Input

The input contains exactly one test case. The first line of input contains three positive integersn(n≤ 30),k(k≤ 10⁹) andm(m< 10⁴). Then follownlines each containingnnonnegative integers below 32,768, givingA

Output

Output the elements ofSmodulomin the same way asAis given.

Sample Input

2 2 4
0 1
1 1

Sample Output

1 2
2 3


題目大意：給出一個一個n*n的矩陣、k和m，求 S = A + A

²

 + A

³

 + ... + A

^k

.輸出矩陣S（每個元素對m取模）

AC程式碼：

#include<iostream>
#include<cstring>
using namespace std;
 
#define ll long long
const int maxn=30;
int mod=10000;
ll T,n,k;
struct matrix{
    ll a[maxn][maxn];
    matrix(){
        memset(a,0,sizeof(a));
    }
};
struct Matrix{
    matrix b[2][2];
    Matrix(){
        memset(b,0,sizeof(b));
    }
};
matrix mul(matrix a,matrix b){
    matrix res;
    for(int i=0;i<n;i++)
        
 
for(int j=0;j<n;j++)
            for(int k=0;k<n;k++)
                res.a[i][j] = (res.a[i][j] + a.a[i][k] * b.a[k][j])%mod;
    return res;
}
matrix add(matrix a,matrix b){
    matrix res;
    for(int i=0;i<n;i++)
        for(int j=0;j<n;j++)
                res.a[i][j] = (a.a[i][j] + b.a[i][j])%mod;
    return res;
}
Matrix mul(Matrix a,Matrix b){
    Matrix res;
    for(int i=0;i<2;i++)
        for(int j=0;j<2;j++)
            for(int k=0;k<2;k++)
                res.b[i][j] = add(res.b[i][j] , mul(a.b[i][k] , b.b[k][j]));
    return res;
}
matrix qpow(matrix A,ll m){//方陣A的m次冪
    matrix ans;
    for(int i=0;i<n;i++)
            ans.a[i][i]=1; //單位矩陣
    while(m){
        if(m&1)ans=mul(ans,A);
        A=mul(A,A);
        m>>=1;
    }
    return ans;
}
Matrix qpow(Matrix A,ll m){
    Matrix ans;
    for(int i=0;i<2;i++)
        for(int j=0;j<n;j++)
            ans.b[i][i].a[j][j]=1; //單位矩陣

    while(m){
        if(m&1)ans=mul(ans,A);
        A=mul(A,A);
        m>>=1;
    }
    return ans;
}
Matrix M;
int main()
{
    cin>>n>>k>>mod;
    matrix A;
    for(int i=0;i<n;i++)
        for(int j=0;j<n;j++)
            cin>>M.b[0][0].a[i][j];
    for(int i=0;i<n;i++){
        M.b[0][1].a[i][i]=1;
        M.b[1][1].a[i][i]=1;
    }
    Matrix res=qpow(M,k+1);
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            if(i==j){
                if(res.b[0][1].a[i][j]==0)
                    cout<<mod-1<<' ';
                else
                    cout<<res.b[0][1].a[i][j]-1<<' ';
            }else
                cout<<res.b[0][1].a[i][j]<<' ';
        }
        cout<<endl;
    }
}