【POJ 3233】Matrix Power Series

【题目链接】

点击打开链接

【算法】

要求 A^1 + A^2 + A^3 + ... + A^k

考虑通过二分来计算这个式子 ：

令f(k) = A^1 + A^2 + A ^ 3 + ... + A^k

那么，当k为奇数时,f(k) = f(k-1) + A ^ k

当k为偶数时,f(k) = f(n/2) + A ^ (n/2) * f(n/2）

因此，可以通过二分 + 矩阵乘法快速幂的方式，在O(n^3log(n)^2)的时间内解决此题

【代码】

#include <algorithm>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <clocale>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <limits>
#include <list>
#include <map>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <utility>
#include <vector>
#include <cwchar>
#include <cwctype>
#include <stack>
#include <limits.h>
using namespace std;
#define MAXN 35

int i,j,n,k,m;
struct Matrix
{
        int mat[MAXN][MAXN];
} a,ans;

inline Matrix add(Matrix a,Matrix b)
{
        int i,j;
        Matrix ans;
        memset(ans.mat,,sizeof(ans.mat));
        for (i = ; i <= n; i++)
        {
                for (j = ; j <= n; j++)
                {
                        ans.mat[i][j] = (a.mat[i][j] + b.mat[i][j]) % m;
                }
        }
        return ans;
}
inline Matrix mul(Matrix a,Matrix b)
{
        int i,j,k;
        Matrix ans;
        memset(ans.mat,,sizeof(ans.mat));
        for (i = ; i <= n; i++)
        {
                for (j = ; j <= n; j++)
                {
                        for (k = ; k <= n; k++)
                        {
                                ans.mat[i][j] = (ans.mat[i][j] + a.mat[i][k] * b.mat[k][j]) % m;
                        }
                }
        }
        return ans;
}
inline Matrix power(Matrix a,int m)
{
        Matrix ans,p = a;
        for (i = ; i <= n; i++)
        {
                for (j = ; j <= n; j++)
                {
                        ans.mat[i][j] = (i == j);
                }
        }
        while (m > )
        {
                if (m & ) ans = mul(ans,p);
                p = mul(p,p);
                m >>= ;
        }
        return ans;
}
Matrix solve(int n)
{
        Matrix tmp;
        if (n == ) return a;
        if (n %  == )
        {
                tmp = solve(n/);
                return add(tmp,mul(power(a,n/),tmp));
        } else return add(solve(n-),power(a,n));
 }

int main()
{

        scanf("%d%d%d",&n,&k,&m);
        for (i = ; i <= n; i++)
        {
                for (j = ; j <= n; j++)
                {
                        scanf("%d",&a.mat[i][j]);
                }
        }
        ans = solve(k);
        for (i = ; i <= n; i++)
        {
                for (j = ; j < n; j++)
                {
                        printf("%d ",ans.mat[i][j]);
                }
                printf("%d\n",ans.mat[i][n]);
        }

        return ;

}