Fibonacci Numbers HDU3117

最新推荐文章于 2020-07-23 15:35:33 发布

KLFTESPACE

最新推荐文章于 2020-07-23 15:35:33 发布

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分类专栏：快速幂 HDU

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本文介绍了一种使用矩阵快速幂算法高效计算大数Fibonacci数列的方法，通过提取大数前四位来解决特定问题。文章详细展示了算法实现过程，包括矩阵乘法、快速幂运算等关键步骤。

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用了提取大数前四位和矩阵快速幂。

/*矩阵快速幂*/

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
//const int mod = 10000;
typedef long long LL;
const int maxn = 20;
const int M[maxn+5] = {1, 10, 100, 1000, 10000};

LL mod;
int N = 40;

int fib[45];

struct Matrix {
    LL mat[maxn][maxn];
    int x, y;
    Matrix() {
        memset(mat, 0, sizeof(mat));
        for (int i = 1; i <= maxn - 5; i++)
            mat[i][i] = 1;
    }
};

void maketable()
{
    fib[0] = 0;
    fib[1] = 1;
    for(int i = 2; i < N; i++)
        fib[i] = fib[i - 2] + fib[i - 1];
}

inline void mat_mul(Matrix a, Matrix b, Matrix &c) {
    memset(c.mat, 0, sizeof(c.mat));
    c.x = a.x; c.y = b.y;
/*    for (int i = 1; i <= c.x; i++) {
        for (int j = 1; j <= c.y; j++) {
            for (int k = 1; k <= a.y; k++) {
                c.mat[i][j] += (a.mat[i][k] * b.mat[k][j]) % mod;
                c.mat[i][j] %= mod;
            }
        }
    }
    */

    //仅适用于2*2
        for (int i = 1; i <= 2; i++)
            for (int j = 1; j <= 2; j++)
                c.mat[i][j] = (a.mat[i][1] * b.mat[1][j] + a.mat[i][2] * b.mat[2][j]) % mod;
    return ;
}

//快速幂
inline void mat_pow(Matrix &a, int z) {
    Matrix ans, base = a;
    ans.x = a.x; ans.y = a.y;
    while (z) {
        if (z & 1 == 1) mat_mul(ans, base, ans);
        mat_mul(base, base, base);
        z >>= 1;
    }
    a = ans;
}
int main() {
    maketable();
  /*  int t;
    cin >> t;*/
    LL a, b, n, m;

    while(~scanf("%lld", &n))
    {

        a = 0, b = 1;


        Matrix A, B;
        A.x = 2; A.y = 2;
        A.mat[1][1] = 1; A.mat[1][2] = 1;
        A.mat[2][1] = 1; A.mat[2][2] = 0;

        B.x = 2; B.y = 1;

        B.mat[1][1] = b; B.mat[2][1] = a;

        if(n < N){
/*
            mod = 1;


            mat_pow(A, n - 1);
            mat_mul(A, B, B);

            printf("%lld\n", B.mat[1][1]);*/

            cout << fib[n] << endl;

            continue;
        }

        else {
            double x = log10(1 / sqrt(5)) + n * log10((1 + sqrt(5)) / 2.0);
            double y = x - (int)(x) + 3;
            int ans = (int)pow(10.0, y);
            cout << ans << "...";
        mod = 10000;

        mat_pow(A, n - 1);
        mat_mul(A, B, B);


        printf("%04d\n", B.mat[1][1]);
        }
    }
    return 0;
}