poj3233(Matrix Power Series)快速幂
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
题意：给定矩阵A，求 S = A + A2 + A3 + … + Ak
题解：设S = I + A + A2 + A3 + … + Ak-1
则
快速幂计算即可
#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <map>
#include <vector>
#include <list>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <functional>
#include <numeric>
#include <iomanip>
#include <limits>
#include <new>
#include <utility>
#include <iterator>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <ctime>
using namespace std;
typedef vector<int> vec;
typedef vector<vec> mat;
int n, k, m;
mat mul(mat& a, mat& b)
{
mat c(2*n, vec(2*n));
for (int i = 0; i < 2*n; ++i)
for (int k = 0; k < 2*n; ++k)
for (int j = 0; j < 2*n; ++j)
c[i][j] = (c[i][j] + a[i][k] * b[k][j]) % m;
return c;
}
mat pow(mat& a, int k)
{
mat b(2*n, vec(2*n));
for (int i = 0; i < 2*n; ++i)
b[i][i] = 1;
while (k > 0)
{
if (k&1)
b = mul(b, a);
a = mul(a, a);
k >>= 1;
}
return b;
}
int main()
{
cin >> n >> k >> m;
mat A(2*n, vec(2*n));
for (int i = 0; i < n; ++i)
{
for (int j = 0; j < n; ++j)
scanf("%d", &A[i][j]);
A[n+i][i] = A[n+i][n+i] = 1;
}
A = pow(A, k+1);
for (int i = 0; i < n; ++i)
for (int j = 0; j < n; ++j)
{
int a = A[n+i][j] % m;
if (i == j)
a = (a + m - 1) % m;
printf("%d%c", a, j+1==n?'\n':' ');
}
return 0;
}
算法码上来 文章被收录于专栏
公众号「算法码上来」。godweiyang带你学习算法，不管是编程算法，还是深度学习、自然语言处理算法都一网打尽，更有各种计算机新鲜知识和你分享。别急，算法码上来。
投递明阳智能等公司10个岗位 >