hdu 3117 Fibonacci Numbers 数学 矩阵快速幂

题目

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3117

题目来源：随便兜到的。

简要题意：求斐波那契第nn项，大于八位输出前后四位。

数据范围：0⩽n⩽1080\leqslant n \leqslant 10^8

题解

首先斐波那契足够大了之后fibx≈ϕfibx−1fib_x\approx \phi fib_{x-1}其中ϕ=1+5√2\phi=\frac{1+\sqrt{5}}{2}，也就是黄金数。

取3939为足够大则fibx=fib39⋅ϕn−39fib_x={fib_{39}\cdot \phi^{n-39}}

令bit=log10fib39+(n−39)⋅log10ϕbit = \log_{10}{fib_{39}}+(n-39)\cdot \log_{10}\phi

则fibx=10bitfib_x=10^{bit}其位数为⌊bit⌋\lfloor bit \rfloor

前四位就是10bit/10⌊bit⌋−3=10bit−⌊bit⌋+310^{bit}/10^{\lfloor bit \rfloor -3}=10^{bit-\lfloor bit \rfloor +3}

后四位的话矩阵快速幂递推就行了，好像也有做法是找循环节的。

实现

这类题实现的话要用log10这个函数比较好，不然会容易被卡掉精度。

代码

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <stack>
#include <queue>
#include <string>
#include <vector>
#include <set>
#include <map>

#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define fi first
#define se second
using namespace std;
typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
LL powmod(LL a,LL b, LL MOD) {LL res=1;a%=MOD;for(;b;b>>=1){if(b&1)res=res*a%MOD;a=a*a%MOD;}return res;}
// head
int fib[55];
double phi = (1.0+sqrt(5.0))/2;

const int N = 2;
const LL MOD = 10000;
struct Matrix {
LL grid

;
int row, col;

Matrix() {
row = col = N;
memset(grid, 0, sizeof(grid));
}
Matrix operator *(const Matrix &b);
Matrix operator ^(int exp);
void setSize(int row, int col) {
this->row = row, this->col = col;
}
void setValue(int row, int col, LL val) {
grid[row][col] = val;
}
void clear() {
memset(grid, 0, sizeof(grid));
}
LL getValue(int row, int col) {
return grid[row][col];
}
};
Matrix Matrix::operator *(const Matrix &b) {
Matrix res;
res.setSize(row, b.col);
for (int i = 0; i < res.row; i++) {
for (int j = 0; j < res.col; j++) {
for (int k = 0; k < col; k++)
res.grid[i][j] = (res.grid[i][j] + (grid[i][k] * b.grid[k][j]) % MOD + MOD) % MOD;
}
}
return res;
}

Matrix Matrix::operator ^(int exp) {
Matrix res, temp;
for (int i = 0; i < N; i++) res.grid[i][i] = 1;
temp = *this;
while (exp > 0) {
if (exp & 1) res = temp * res;
exp >>= 1;
temp = temp * temp;
}
return res;
}

int getFib(int n) {
Matrix init, trans;
init.setSize(1, 2);
init.setValue(0, 1, 1);
trans.setValue(0, 1, 1);
trans.setValue(1, 0, 1);
trans.setValue(1, 1, 1);
init = init*(trans^n);
return init.getValue(0, 0);
}
int main()
{
fib[1] = 1;
for (int i = 2; i <= 39; i++) {
fib[i] = fib[i-1]+fib[i-2];
}
int n;
while (scanf("%d", &n) == 1) {
if (n < 40) {
printf("%d\n", fib
);
continue;
}
double x = log10(fib[39])+(n-39)*log10(phi);
printf("%d...", (int)pow(10, x-(floor(x)-3)));
printf("%04d\n", getFib(n));
}
return 0;
}