ACM: 二分又二分 数论题 poj 3233

Matrix Power Series

Description

Given a n × nmatrix A and a positive integer k, find the sumS = A + A² + A³ +… + A^k.

Input

The input contains exactlyone test case. The first line of input contains three positiveintegers n (n ≤ 30), k (k ≤10⁹) and m (m <10⁴). Then follow n lines each containingn nonnegative integers below 32,768, giving A’selements in row-major order.

Output

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

Sample Input

2 2 4
0 1
1 1

Sample Output

1 2
2 3

题意: 计算矩阵A的和, sum = (A + A^2 + A^3 + ... + A^k)%m

解题思路:
       1. 式子: sum = A+A^2+A^3+...+A^k
       当k是偶数时: 1表示单位矩阵(对角线全为1,其余为0)
               sum = (A+A^2+...+A^k)*(1+A^k/2); 
       当k是奇数时:
               sum = (A+A^2+...+A^k/2)*(1+A^k/2+1)+A^k/2+1
       2. A^k也是用二分方法求解.
       当k是偶数时:
                A^k = A^(k/2)*A^(k/2);
       当k是奇数时:
                A^k = A^(k/2)*A^(k/2)*A;

代码:
#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
#define MAX 35

struct matrix
{
    int row, column;
    int g[MAX][MAX];
};

int n, K, m;
matrix a, temp;

inline matrix matrix_add(matrix A, matrix B, int mod)
{
    matrix c;
    c.row = c.column = A.row;
    for(int i = 0; i < c.row; ++i)
    {
        for(int j = 0; j < c.column; ++j)
        {
            c.g[i][j] = (A.g[i][j]+B.g[i][j])%mod;
        }
    }
    return c;
}

inline matrix multiply(matrix A, matrix B, int mod)
{
    matrix c;
    c.row = A.row;
    c.column = B.column;
    memset(c.g,0,sizeof(c.g));
    for(int i = 0; i < A.row; ++i)
    {
        for(int k = 0; k < A.column; ++k)
        {
            if(A.g[i][k] != 0)
            {
                for(int j = 0; j < B.column; ++j)
                {
                    c.g[i][j] = (c.g[i][j]+A.g[i][k]*B.g[k][j])%mod;
                }
            }
        }
    }
    return c;
}

matrix matrix_pow(matrix A, int n, int mod)
{
    if(n == 1) return a;
    matrix t = matrix_pow(A,n/2,mod);
    matrix ans = multiply(t,t,mod);
    if(n % 2 == 1) ans = multiply(ans,A,mod);
    return ans;
}

matrix sum(matrix A, int n, int mod)
{
    if(n == 1) return a;
    if(n % 2 == 0)
        return multiply(sum(A,n/2,mod),matrix_add(temp,matrix_pow(A,n/2,mod),mod),mod);
    else
        return matrix_add(
            multiply(sum(A,n/2,mod),matrix_add(temp,matrix_pow(A,n/2+1,mod),mod),mod),
            matrix_pow(A,n/2+1,mod),
            mod
        );
}

int main()
{
//    freopen("input.txt","r",stdin);
    while(scanf("%d %d %d",&n, &K, &m) != EOF)
    {
        temp.row = temp.column = n;
        memset(temp.g,0,sizeof(temp.g));
        for(int i = 0; i < n; ++i)
            temp.g[i][i] = 1;
        
        a.row = a.column = n;
        memset(a.g,0,sizeof(a.g));
        for(int i = 0; i < n; ++i)
        {
            for(int j = 0; j < n; ++j)
                scanf("%d",&a.g[i][j]);
        }
        matrix result = sum(a,K,m);
        for(int i = 0; i < result.row; ++i)
        {
            for(int j = 0; j < result.column; ++j)
            {
                if(j == 0) printf("%d",result.g[i][j]);
                else printf(" %d",result.g[i][j]);
            }
            printf("\n");
        }
    }
    return 0;
}