Matrix Power Series - POJ 3233 - 矩阵快速幂

链接:

  http://poj.org/problem?id=3233

题目：

Description

Given a n × n matrix A and a positive integer k, find the sum S=A1+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

题意：

  给你一个矩阵

A

，让你求S=∑ni=1Ai。

思路：

  这是矩阵乘法中关于等比矩阵的求法，不难建立转移矩阵：

[EOAA]

  其中E为单位矩阵，0为0矩阵，然后我们求这个转移矩阵的n次幂，所得到的矩阵的右上角就是我们所要求的S。

实现：

#include <iostream>
#include <algorithm>
#include <set>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <stack>
#include <iomanip>
#include <functional>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>
#define il inline
#define ll long long
#define ull unsigned long long

using namespace std;
const int maxn = 107, Matmatsize = 2;
int Matrixsize = 2, mod = int(1e9)+7;
struct Matrix {
int m[maxn][maxn];
Matrix(int i = 0) {
memset(m, 0, sizeof m);
if (i == 1) {
for (int I = 0; I < Matrixsize; I++) m[I][I] = 1;
}
}
Matrix operator * (const Matrix tmp) const {
Matrix ret;
long long x;
for(int i=0 ; i<Matrixsize ; i++) {
for(int j=0 ; j<Matrixsize ; j++) {
x=0;
for(int k=0 ; k<Matrixsize ; k++) {
x+=((long long)m[i][k] * tmp.m[k][j]) % mod;
}
ret.m[i][j] = int(x % mod);
}
}
return ret;
}
Matrix operator+(const Matrix b) const {
Matrix a = *this;
for (int i = 0; i < Matrixsize; i++)
for (int j = 0; j < Matrixsize; j++)
a.m[i][j] = (a.m[i][j] + b.m[i][j]) % mod;
return a;
}
Matrix operator-(const Matrix b) const {
Matrix a = *this;
for (int i = 0; i < Matrixsize; i++)
for (int j = 0; j < Matrixsize; j++)
a.m[i][j] = (a.m[i][j] - b.m[i][j]) % mod;
return a;
}
Matrix operator-() const {
Matrix tmp = *this;
for (int i = 0; i < Matrixsize; i++)
for (int j = 0; j < Matrixsize; j++)
tmp.m[i][j] = -tmp.m[i][j];
return tmp;
}
Matrix qpow(long long n) {
Matrix ret = 1, tmp = *this;
while (n != 0) {
if (bool(n & 1)) ret = ret * tmp;
tmp = tmp * tmp;
n >>= 1;
}
return ret;
}
};

struct Matmat {
Matrix m[Matmatsize][Matmatsize];
Matmat (int i=0) {
for(int i=0 ; i<Matmatsize ; i++)
for(int j=0 ; j<Matmatsize ; j++)
m[i][j] = 0;
if(i == 1) for(int i=0 ; i<Matmatsize ; i++) m[i][i] = 1;
}
Matmat operator * (const Matmat tmp) const {
Matmat ret;
Matrix x;
for(int i=0 ; i<Matmatsize ; i++) {
for(int j=0 ; j<Matmatsize ; j++) {
x = 0;
for(int k=0 ; k<Matmatsize ; k++) {
x = x + m[i][k] * tmp.m[k][j];
}
ret.m[i][j] = x;
}
}
return ret;
}
Matmat qpow(long long n) {
Matmat ret = 1, tmp = *this;
while (n != 0) {
if (bool(n & 1))
ret = ret * tmp;
tmp = tmp * tmp;
n >>= 1;
}
return ret;
}
};

int main() {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
#endif
ios_base::sync_with_stdio(false);cin.tie(0);
int k;
while(cin >> Matrixsize >> k >> mod) {
Matrix E = 1, A = 0, ans;
for (int i = 0; i < Matrixsize; i++)
for (int j = 0; j < Matrixsize; j++) cin >> A.m[i][j];
Matmat tmp = 0;
tmp.m[0][0] = E, tmp.m[0][1] = tmp.m[1][1] = A;
ans = tmp.qpow(k).m[0][1];
for (int i = 0; i < Matrixsize; i++) {
for (int j = 0; j < Matrixsize - 1; j++) cout << ans.m[i][j] << ' ';
cout << ans.m[i][Matrixsize - 1] << '\n';
}
}
return 0;
}