被51nod1653干翻
存个档省的自己忘掉。
对于一个线性递推数列$x_{n}=a*x_{n-1}+b*x_{n-2}$
我们设一个y，$x_{n}-y*x_{n-1}=(a-y)(x_{n-1}-b/(y-a)*x_{n-2})$
当$y=b/(y-a)$时$x_{n}-yx_{n-1}$会是一个等比数列，公比为$a-y$
设$y^{2}=ay+b$的解为r和s
当$r\neq s$时
$F(n)=x_{n}-r*x_{n-1}=F(1)*(a-r)^{n-1}=(x_{1}-r*x_{0})*(a-r)^{n-1}$
$G(n)=x_{n}-s*x_{n-1}=F(1)*(a-s)^{n-1}=(x_{1}-s*x_{0})*(a-s)^{n-1}$
$F(n)-G(n)=(s-r)*x_{n-1}$
根据韦达定理$r+s=a,rs=-b$
$x_{n-1}=(x_{1}-r*x_{0})*(a-r)^{n-1}/(s-r)-(x_{1}-s*x_{0})*(a-s)^{n-1}/(s-r)$
$x_{n-1}=(x_{1}-r*x_{0})*s^{n-1}/(s+b/s)+(x_{1}+s*x_{0})*r^{n-1}/(r+b/r)$
$x_{n}=c_{1}*s^n+c_{2}*r^n$
高斯消元解出$c_{1}$和$c_{2}$即可
以下是我自己的口胡
当$s=r$时$a=-2r$
$x_{1}-x_{0}=s*p$①
$x_{2}-x_{1}=s^2*p$②..
那么把①*s^n+②*s^(n-1)...加起来，就可以得到$x_{n}-s^n*x_{0}=p*n*s^{n}$就可以求解了
2022年8月31日 13:55
Bangladesh Education Board DPE has conducted the class 8th grade of Junior School Certificate Exam and Junior Dakhil Certificate Exam on 1st to 15th November 2022 at all centers in division wise under Ministry of Primary and Mass Education (MOPME), and the class 8th grade terminal examination tests are successfully conducted for all eligible JSC/JDC students for the academic year of 2022. JDC Result Dinajpur The Bangladesh government Minister of Secondary Education is going to announce the JSC Result 2022 in student wise for division students in education board wise, and the result documents will be submitted to the Prime Minister of the country after then the result with mark sheet will be announced to public to check the individual result.
2024年10月20日 00:29
韦达定理，终于看到见过的东西了，但为什么还是看不懂一点。（话说你真的不打算把我从小黑屋里放出来了嘛？还是好难过好难过(｡•́︿•̀｡)）
2024年10月20日 00:33
（你再不把我放出来的话，我最近打算要在心里默默开始讨厌你了，但总感觉应该不会成功……）
2024年10月20日 00:36
鬼知道这个Ophelia经历了什么