[LightOJ1070]Algebraic Problem

题目:Algebraic Problem

链接：https://vjudge.net/problem/LightOJ-1070

分析：

1）$ a^n+b^n = ( a^{n-1}+b^{n-1} )*(a+b) - (a*b^{n-1}+a^{n-1}*b) $

构造矩阵: $ \left[ \begin{array}{cc} 0 & -1 \\ a*b & a+b \end{array} \right] $

$$ \left[ \begin{array}{cc} a*b^{n-1}+a^{n-1}*b  &   a^{n-1}+b^{n-1} \end{array} \right]  *  \left[ \begin{array}{cc} 0 & -1 \\ a*b & a+b \end{array} \right]  = \left[ \begin{array}{cc} a*b^n+a^n*b  &   a^n+b^n \end{array} \right] $$

2)注意特判0的情况，至于对$2^{64}$取模，开unsigned long long，自然溢出即可。

 #include <iostream>
 #include <cstring>
 using namespace std;
 typedef unsigned long long LLU;
 typedef unsigned int uint;
 struct Matrix{
     LLU a[][];
     Matrix(int f=){
         memset(a,,sizeof a);
         if(f==)for(int i=;i<;++i)a[i][i]=;
     }
 };
 Matrix operator*(Matrix& A,Matrix& B){
     Matrix C;
     for(int k=;k<;++k)
         for(int i=;i<;++i)
         for(int j=;j<;++j)
             C.a[i][j]+=A.a[i][k]*B.a[k][j];
     return C;
 }
 Matrix operator^(Matrix A,uint n){
     Matrix Rt();
     for(;n;n>>=){
         if(n&)Rt=Rt*A;
         A=A*A;
     }
     return Rt;
 }
 int main(){
     int T;scanf("%d",&T);
     Matrix A,ANS;LLU p,q;uint n;
     for(int i=;i<=T;++i){
         scanf("%llu%llu%u",&p,&q,&n);
         if(n==){
             printf("Case %d: 2\n",i);
             continue;
         }
         A.a[][]=;A.a[][]=-;
         A.a[][]=q;A.a[][]=p;
         ANS=A^(n-);
         LLU ans=*q*ANS.a[][]+ANS.a[][]*p;
         printf("Case %d: %llu\n",i,ans);
     }
     return ;
 }
         

3)$ a^n + b^n = (a^{n-1}+b^{n-1})*(a+b) - (a*b^{n-1}+a^{n-1}*b) = (a^{n-1}+b^{n-1})*(a+b)-a*b*(b^{n-2}+a^{n-2}) $

构造矩阵：$ \left[ \begin{array}{cc} a+b & -ab \\ 1 & 0 \end{array} \right] $

$$ \left[ \begin{array}{cc} a+b & -ab \\ 1 & 0 \end{array} \right] *  \left[ \begin{array}{c} a^{n-1}+b^{n-1}   \\   a^{n-2}+b^{n-2} \end{array} \right]  = \left[ \begin{array}{c} a^n+b^n \\   a^{n-1}+b^{n-1} \end{array} \right] $$