[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[ egin{array}{cc} 0 & -1 \ a*b & a+b end{array} ight] $

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

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[2][2];
    Matrix(int f=0){
        memset(a,0,sizeof a);
        if(f==1)for(int i=0;i<2;++i)a[i][i]=1;
    }
};
Matrix operator*(Matrix& A,Matrix& B){
    Matrix C;
    for(int k=0;k<2;++k)
        for(int i=0;i<2;++i)
        for(int j=0;j<2;++j)
            C.a[i][j]+=A.a[i][k]*B.a[k][j];
    return C;
}
Matrix operator^(Matrix A,uint n){
    Matrix Rt(1);
    for(;n;n>>=1){
        if(n&1)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=1;i<=T;++i){
        scanf("%llu%llu%u",&p,&q,&n);
        if(n==0){
            printf("Case %d: 2
",i);
            continue;
        }
        A.a[0][0]=0;A.a[0][1]=-1;
        A.a[1][0]=q;A.a[1][1]=p;
        ANS=A^(n-1);
        LLU ans=2*q*ANS.a[0][1]+ANS.a[1][1]*p;
        printf("Case %d: %llu
",i,ans);
    }
    return 0;
}
        

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[ egin{array}{cc} a+b & -ab \ 1 & 0 end{array} ight] $

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