HDU 1005-优快云博客

本文链接：https://blog.youkuaiyun.com/RedPolya/article/details/53428396

本文介绍了一种使用矩阵快速幂的方法来高效计算特定数列f(n)的值，该数列定义为f(1)=1,f(2)=1,f(n)=(A*f(n-1)+B*f(n-2))mod7。通过构建特定矩阵并利用快速幂运算，可以显著减少计算复杂度。

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A number sequence is defined as follows:

f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.

Given A, B, and n, you are to calculate the value of f(n).

N很大，有种方法找寻环节，毕竟循环节小于等于49个。

我这里为了用下矩阵乘法的模板，就直接用矩阵加速了。

就是要求

[a b 10] n - 1

${\begin{bmatrix} a & 1\\ b & 0 \end{bmatrix}}^{n-1}$
快速幂就好啦，最后答案就是base[0][1]+base[1][1]

#include<cstdio> 
#include<cstring>
#include<iostream>
using namespace std;
typedef long long ll;
const int P = 7;
const int N=13;
ll n,m;

struct matrix{
    ll a[N][N];
    int row,col;
    matrix():row(N),col(N){memset(a,0,sizeof(a));}
    matrix(int x,int y):row(x),col(y){memset(a,0,sizeof(a));}
    ll* operator [] (int x){return a[x];}
    matrix operator * (matrix x){
        matrix tmp ;
        for (int i=0;i<=n+1;i++)
            for (int j=0;j<=n+1;j++){
                tmp[i][j]=0;
                for (int k=0;k<=n+1;k++)
                    tmp[i][j]=(tmp[i][j]+a[i][k]*x[k][j])%P;
            }
        return tmp;
    }   
    void operator *= (matrix x){*this = *this * x;}
    matrix operator ^ (ll x){
        matrix ret;
        for (int i=0;i<=n+1;i++)ret[i][i]=1;
        matrix tmp = *this;
        for (;x;x>>=1,tmp*=tmp){if(x&1)ret *=tmp;}
        return ret;
    }
};

int main(){
    int a,b,num;
    n=2;
    while(cin>>a>>b>>num,(a||b||num)){
        matrix base;
        base[0][0]=a;base[0][1]=1;base[1][0]=b;base[1][1]=0;
        base = base ^ (num-1);
        printf("%d\n",(base[0][1]+base[1][1])%7);
    }
    return 0;
}