Matrix Power Series

Matrix Power Series
时间限制(普通/Java):1000MS/3000MS 内存限制:65536KByte

描述

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

输入

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’s elements in row-major order.

输出

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

样例输入

2 2 4
0 1
1 1

样例输出

1 2
2 3

思路

快速幂 加 分治处理数列前一半和后一半

sk=sk/2+sk/2*a^k/2（偶数）

sk=sk/2+sk/2*a^k/2+ak（奇数）

AC代码

复制代码
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#include <bits/stdc++.h>
using namespace std;
int n,mod;
class Node
{
public:
    int a[40][40]={0};
    Node operator+(Node &temp)
    {
        Node res{};
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                res.a[i][j]= (this->a[i][j]+temp.a[i][j])%mod;
            }
        }
        return res;
    }
    Node operator*(Node &temp)
    {
        Node res{};
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
            {
                for(int k=1; k<=n; k++)
                    res.a[i][j]=(res.a[i][j]+ this->a[i][k]*temp.a[k][j])%mod;
            }
        return res;
    }
}st,yi;
Node quick_pow(int t)
{
    Node ans=yi;
    Node a=st;
    while(t)
    {
        if(t&1)ans=ans*a;
        a=a*a;
        t>>=1;
    }
    return ans;
}

Node dfs(int k)
{
    if(k==1)return st;
    Node res=dfs(k/2);
    Node temp= quick_pow(k/2);
    Node b=res;
    res=res * temp;
    res=res+b;
    if(k&1)
    {
        temp= quick_pow(k);
        res=res+temp;
    }
    return res;
}
void solve()
{
    int k;
    cin>>n>>k>>mod;
    for(int i=1; i<=n; i++)
    {
        for(int j=1; j<=n; j++)
        {
            cin>>st.a[i][j];
            st.a[i][j]%=mod;
        }
    }
    for(int i=1;i<=n;i++)yi.a[i][i]=1;
    Node ans;
    ans=dfs(k);
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            if(j!=1)cout<<" ";
            cout<<ans.a[i][j];
        }
        cout<<endl;
    }
}
signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    long long  _=1;
//    cin>>_;
    while(_--)
    {
        solve();
    }
}

本文作者：Minza

本文链接：https://www.cnblogs.com/minz-io/p/17632852.html