[原创]POJ 3233 Matrix Power Series [矩阵快速幂]【数论】[水]

2016-09-13 17:00:13 Tabris_ 阅读数：278

博客爬取于2020-06-14 22:43:20
以下为正文

版权声明：本文为Tabris原创文章，未经博主允许不得私自转载。
https://blog.csdn.net/qq_33184171/article/details/52527503

题目链接 : http://poj.org/problem?id=3233

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Matrix Power Series
Time Limit: 3000MS Memory Limit: 131072K
Total Submissions: 20930 Accepted: 8760
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 (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
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题目大意:
不用解释了吧 就是求Sn
Sn = A + A^2 + A^3 + … + A^k.

解题思路：
这种题目一定想到矩阵快速幂
然后就是怎么构造矩阵了

[A O] 乘 [A E] 等 [A^2 S1]
[O O] 号 [O E]号 [O S0]

矩阵大致就是这么构造出来的

然后注意的事E只有主对角线是1 剩下的都是0 (Sb的我全写成E然后WA的都怀疑人生了)

附本题代码
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# include <stdio.h>
# include <string.h>
# include <iostream>
# include <algorithm>
# include <vector>
# include <queue>
# include <set>
# include <map>
# include <string>
# include <math.h>
# include <stdlib.h>
# include <time.h>
using namespace std;
using namespace std;
typedef long long int LL ;
# define INF 0x3f3f3f3f
# define pb push_back

# define lalal puts("*******");
/*************************************/

int MOD;
const int M = 32*2;

struct Matrix
{
    LL m[M][M];
    void display(int N)
    {
        for(int i=0; i<N; i++)
        {
            for(int j=0; j<N; j++)
            {
                if(j) printf(" ");
                printf("%I64d",m[i][j]);
            }
            puts("");
        }
    }
    void clearI()
    {
        for(int i=0; i<M; i++)
            for(int j=0; j<M; j++)
                m[i][j]=(i==j);
    }
    void clearO()
    {
        for(int i=0; i<M; i++)
            for(int j=0; j<M; j++)
                m[i][j]=0;
    }
};
Matrix operator * (Matrix &a,Matrix &b)
{
    Matrix c;
    c.clearO();

    for(int k=0; k<M; k++)
        for(int i=0; i<M; i++)
        {
            if(a.m[i][k]==0) continue;
            for(int j=0; j<M; j++)
            {
                if(b.m[k][j]==0) continue;
                c.m[i][j]=(c.m[i][j]+a.m[i][k]*b.m[k][j])%MOD;
            }
        }
    return c;
}

Matrix operator ^ (Matrix &a,LL b)
{
    Matrix c;
    c.clearI();
    while(b)
    {
        if(b&1) c=c*a;
        b>>=1;
        a=a*a;
    }
    return c;
}

int main()
{
    LL n,k,m;
    while(~scanf("%I64d%I64d%I64d",&n,&k,&m))
    {
        MOD = m;
        Matrix a,b;
        a.clearO(),b.clearO();
        for(int i=0; i<n; i++)
        {
            b.m[i][n+i]=b.m[n+i][n+i]=1;
            for(int j=0; j<n; j++)
            {
                scanf("%I64d",&a.m[i][j]);
                b.m[i][j]=a.m[i][j];

            }
        }

        b=b^(k);
        a=a*b;

        for(int i=0; i<n; i++)
        {
            for(int j=0; j<n; j++)
            {
                if(j) printf(" ");
                printf("%I64d",a.m[i][j+n]);
            }
            puts("");
        }

    }
    return 0;
}