Description

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

Input

The input contains exactly one test case. The first line of input contains three positive integers n

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

Source

題目意思：

解題思路：

模板題，快速冪取模+二分優化。

補充一個關於等比序列分治的結論：

對於
Sn=(A^1+A^2+A^3+……+A^(n-1)+A^n) mod p
當n為偶數的時候
s[n]=(1+A^(n/2))* (A^1+A^2+A^3+……+A^(n/2)) =(1+A^(n/2))*S[n/2]
當n為奇數的時候
s[n]=(1+A^((n-1)/2+1))* (A^1+A^2+A^3+……+A^(n-1)/2)+A^((n-1)/2+1]
=(1+A^((n-1)/2+1))* S[(n-1)/2] + A^((n-1)/2+1) =(1+A^(n/2+1))*S[n/2] + A^(n/2+1)

#include <iostream>  
#include <cstdio>  
#include <cstring>  
#include <cmath>  
#include <vector>  
#include <queue>  
#include <map>  
#include <algorithm>  
#define INF 0xfffffff  
using namespace std;  
const long long MAXN=81;  
long long n,mod;  
struct Mat  
{  
    long long m[MAXN][MAXN];  
};  
Mat a,per;  
void init()  
{  
    long long i,j;  
    for(i=0; i<n; ++i)  
        for(j=0; j<n; ++j)  
        {  
            cin>>a.m[i][j];  
            a.m[i][j]%=mod;  
            per.m[i][j]=(i==j);  
        }  
}  
  
Mat mul(Mat A,Mat B)  
{  
    Mat ans;long long i,j,k;  
    for(i=0; i<n; i++)  
        for(j=0; j<n; j++)  
        {  
            ans.m[i][j]=0;  
            for(k=0; k<n; k++)  
                ans.m[i][j]+=(A.m[i][k]*B.m[k][j]);  
            ans.m[i][j]%=mod;  
        }  
    return ans;  
}  
Mat power(long long k)  
{  
    Mat p,ans=per;  
    p=a;  
    while(k)  
    {  
        if(k&1)  
        {  
            ans=mul(ans,p);  
            --k;  
        }  
        else  
        {  
            k/=2;  
            p=mul(p,p);  
        }  
    }  
    return ans;  
}  
  
Mat add(Mat a,Mat b)  
{  
    Mat c;long long i,j;  
    for(i=0; i<n; ++i)  
        for(j=0; j<n; ++j)  
            c.m[i][j]=(a.m[i][j]+b.m[i][j])%mod;  
    return c;  
}  
Mat sum(long long k)  
{  
    if(k==1) return a;  
    Mat temp,b;  
    temp=sum(k/2);  
    if(k&1)  
    {  
        b=power(k/2+1);  
        temp=add(temp,mul(temp,b));  
        temp=add(temp,b);  
    }  
    else  
    {  
        b=power(k/2);  
        temp=add(temp,mul(temp,b));  
    }  
    return temp;  
}  
  
int main()  
{  
    ios::sync_with_stdio(false);  
    cin.tie(0);  
    long long i,j,k;  
    while(cin>>n>>k>>mod)  
    {  
        init();  
        Mat ans=sum(k);  
        for(i=0; i<n; ++i)  
        {  
            for(j=0; j<n-1; ++j)  
                cout<<ans.m[i][j]<<" ";  
            cout<<ans.m[i][j]<<endl;  
        }  
    }  
    return 0;  
}

POJ 3233-Matrix Power Series( S = A + A^2 + A^3 + … + A^k 矩陣快速冪取模)