POJ 3233-Matrix Power Series( S = A + A^2 + A^3 + … + A^k 矩阵快速幂取模)

最新推荐文章于 2020-12-12 12:21:52 发布

原创最新推荐文章于 2020-12-12 12:21:52 发布 · 1.6k 阅读

1 ·

CC 4.0 BY-SA版权

文章标签：

#POJ 3233 #Matrix Power Series #快速幂取模 #矩阵快速幂 #C++

POJ 初级计划同时被 2 个专栏收录

179 篇文章

订阅专栏

ACM_数论

25 篇文章

订阅专栏

本文介绍了一道经典的算法竞赛题目，要求计算给定矩阵A的幂级数求和S = A + A^2 + ... + A^k，并提供了一个高效的解决方案，利用快速幂和二分法优化算法。

Matrix Power Series

Time Limit: 3000MS		Memory Limit: 131072K
Total Submissions: 20309		Accepted: 8524

Description

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’s elements in row-major order.

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

POJ Monthly--2007.06.03, Huang, Jinsong

题目意思：

已知n，k，m，S = A + A ² + A ³ + … + A^k . 求：S%m

解题思路：

模板题，快速幂取模+二分优化。

#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;
}
/**
2 2 4
0 1
1 1
**/