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矩阵快速幂
首先定义两个矩阵想成的运算函数mul
typedef vector<vector<int> > Mat;
const int M = 1e4+10; // 容易溢出,给一个小于int开根号的数来模
Mat mul(Mat &A, Mat &B)
{
int m = A.size(), n = B[0].size(), l = B.size();
Mat C(m, vector<int>(n));
for(int i = 0; i < m; i++)
for(int k = 0; k < l; k++)
for(int j = 0; j < n; j++)
C[i][j] = (C[i][j]+A[i][k]*B[k][j])%M;
return C;
}
接下来用快速幂就可以啦
Mat pow(Mat A, int n)
{
int m = A.size();
Mat B(m, vector<int>(m));
for(int i = 0; i < m; i++) B[i][i] = 1;
while(n>0)
{
if(n%2==1) B = mul(B, A);
A = mul(A, A);
n /= 2;
}
return B;
}
二、斐波那契数列快速计算
因此,计算出来
,那么
就是斐波那契数列第n项的值。求解程序如下:
#include <iostream>
#include <vector>
using namespace std;
typedef vector<vector<int> > Mat;
const int M = 1e4+10; // 容易溢出,给一个小于int开根号的数来模
Mat mul(Mat &A, Mat &B)
{
int m = A.size(), n = B[0].size(), l = B.size();
Mat C(m, vector<int>(n));
for(int i = 0; i < m; i++)
for(int k = 0; k < l; k++)
for(int j = 0; j < n; j++)
C[i][j] = (C[i][j]+A[i][k]*B[k][j])%M;
return C;
}
Mat pow(Mat A, int n)
{
int m = A.size();
Mat B(m, vector<int>(m));
for(int i = 0; i < m; i++) B[i][i] = 1;
while(n>0)
{
if(n%2==1) B = mul(B, A);
A = mul(A, A);
n /= 2;
}
return B;
}
int main(int argc, char const *argv[])
{
Mat F = Mat({{1, 1}, {1, 0}});
int n; cin >> n;
Mat res = pow(F, n);
cout << res[0][0] << endl;
return 0;
}