【历年真题】斐波那契数列logn做法

描述

【题解】

用矩阵乘法加速递推
[0 1]
[1 1]

• [f[n-1]]
  [f[n-2]]
  =
  [f[n-1]]
  [f[n]]
  求A矩阵的n-2次幂然后再乘B矩阵。
  结果矩阵中的第二行第一列就是f[n]的结果了
  

【代码】

#include <cstdio>
#include <cstring>
#define ll long long
using namespace std;
const int N = 100;
const long long MOD = 1e9+7;
const int G = 2;

struct abc{
    ll v[G+10][G+10];
    void print(){
        for (int i = 1;i <= G;i++)
        {
            for (int j = 1;j <= G;j++)
                printf("%I64d ",v[i][j]);
            puts("");
        }
        puts("");
    }
    void E(){
        memset(v,0,sizeof v);
        for (int i = 1;i <= G;i++) v[i][i]=1;
    }
    void O(){
        memset(v,0,sizeof v);
    }
    abc operator * (const abc b) const{
        abc temp;temp.O();
        for (int i = 1;i <= G;i++)
            for (int j = 1;j <= G;j++){
                ll sum = 0;
                for (int k = 1;k<= G;k++){
                    sum=(sum+v[i][k]*b.v[k][j])%MOD;
                }
                temp.v[i][j] = sum;
            }
        return temp;
    }

    abc operator ^(int n)const{
        abc x;x.E();
        abc y;memcpy(y.v,v,sizeof(v));
        while (n>0){
            if (n&1) x=x*y;
            y = y*y;
            n/=2;
        }
        return x;
    }
};

abc a,b;
int n;

int main(){
    scanf("%d",&n);
    a.O();
    a.v[1][1] = 0;a.v[1][2] = 1;
    a.v[2][1] = 1;a.v[2][2] = 1;
    b.O();
    b.v[1][1] = 1;b.v[2][1] = 1;
    if (n<=2){
        printf("%d\n",1);
    }else{
        a = a^(n-2);
        a = a*b;
        printf("%I64d\n",a.v[2][1]);
    }
    return 0;
}

【历年真题】斐波那契数列logn做法

原文：https://www.cnblogs.com/AWCXV/p/11627373.html