[矩阵乘法]裴波拉契数列III
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[ 矩 阵 乘 法 ] 裴 波 拉 契 数 列 I I I [矩阵乘法]裴波拉契数列III [矩阵乘法]裴波拉契数列III
Description
求数列f[n]=f[n-1]+f[n-2]+1的第N项.f[1]=1,f[2]=1.
Input
n(1<n<231-1)
Output
一个数为裴波拉契数列的第n项mod 9973;
Sample Input
12345
Sample Output
8932
题目解析
对于为什么用矩阵乘法来做,详见博客斐波那契数列II
我们考虑矩阵 ⊏ f [ n − 2 ] , f [ n − 1 ] , 1 ⊐ \sqsubset f[n - 2] , f[n - 1] , 1\sqsupset ⊏f[n−2],f[n−1],1⊐并利用斐波那契数列的递推关系来得到式子 ⊏ f [ n ] , f [ n − 1 ] ⊐ = ⊏ f [ n − 2 ] + f [ n − 1 ] , f [ n − 1 ] ⊐ \sqsubset f[n] , f[n - 1]\sqsupset = \sqsubset f[n - 2] + f[n - 1] , f[n - 1]\sqsupset ⊏f[n],f[n−1]⊐=⊏f[n−2]+f[n−1],f[n−1]⊐
然后可以构造出一个
3
∗
3
3 * 3
3∗3的矩阵
T
T
T
∣
0
1
0
1
1
0
0
1
1
∣
\begin{vmatrix} 0 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \\ \end{vmatrix}
∣∣∣∣∣∣010111001∣∣∣∣∣∣
然后可以通过
f
[
1
]
,
f
[
2
]
,
1
∗
T
=
f
[
2
]
,
f
[
3
]
,
1
f[1] , f[2] , 1* T = f[2] , f[3], 1
f[1],f[2],1∗T=f[2],f[3],1来实现代码了
Code
#include <cmath>
#include <cstdio>
#include <iostream>
using namespace std;
int nt;
const int MOD = 9973;
struct matrix
{
int n, m;
int t[10][10];
}t1, t2, t3;
matrix operator *(matrix t, matrix r)
{
matrix c;
c.n = t.n, c.m = r.m;
for (int i = 1; i <= c.n; ++ i)
for (int j = 1; j <= c.m; ++ j)
c.t[i][j]=0;
for (int k = 1; k <= t.m; ++ k)
for (int i = 1; i <= t.n; ++ i)
for (int j = 1; j <= r.m; ++ j)
c.t[i][j] = (c.t[i][j] + t.t[i][k] * r.t[k][j] % MOD) % MOD;
return c;
}
void rt (int k)
{
if (k == 1)
{
t2 = t1;
return;
}
rt (k / 2);
t2 = t2 * t2;
if (k & 1) t2 = t2 * t1;
}
int main()
{
scanf ("%d", &nt);
if (nt == 1)
{
printf ("1");
return 0;
}
t3.n = 1;
t1.n = t1.m = t3.m = 3;
t1.t[1][1] = 0, t1.t[1][2] = 1, t1.t[1][3] = 0;
t1.t[2][1] = 1, t1.t[2][2] = 1, t1.t[2][3] = 0;
t1.t[3][1] = 0, t1.t[3][2] = 1, t1.t[3][3] = 1;
t3.t[1][1] = t3.t[1][2] = t3.t[1][3] = 1;
rt (nt - 1);
t3 = t3 * t2;
printf ("%d", t3.t[1][1]);
return 0;
}
标签:波拉,数列,int,t2,t3,t1,III,matrix 来源: https://blog.csdn.net/qq_45519803/article/details/111290182