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斐波那契数列的通项公式x+洛谷P2626x

#include<cstdio>
#include<iostream>
#include<cmath>

using namespace std;

int main()
{
    int n;
    scanf("%d",&n);
    n--;
    double q=sqrt(5.0);
    int ans;
    ans=((pow((1+q)/2.0,n)/q-(pow((1-q)/2.0,n)/n)));
    cout<<ans<<endl;
    return 0;
} 

洛谷P2626

题目背景

大家都知道,斐波那契数列是满足如下性质的一个数列: ? f(1) = 1 ? f(2) = 1 ? f(n) = f(n-1) + f(n-2) (n ≥ 2 且 n 为整数)。

题目描述

请你求出第n个斐波那契数列的数mod(或%)2^31之后的值。并把它分解质因数。

输入输出格式

输入格式:

 

n

 

输出格式:

 

把第n个斐波那契数列的数分解质因数。

 

输入输出样例

输入样例#1:
5
输出样例#1:
5=5
输入样例#2:
6
输出样例#2:
8=2*2*2

说明

n<=48


#include <iostream> #include <cstdio> #include <cmath> #define Max 300 const long long Mod = pow (2, 31); void read (long long &now) { now = 0; char word = getchar (); while (word < 0 || word > 9) word = getchar (); while (word >= 0 && word <= 9) { now = now * 10 + word - 0; word = getchar (); } } long long fibonacii[Max]; int main (int argc, char *argv[]) { register long long N; fibonacii[1] = 1; fibonacii[2] = 1; read (N); for (long long i = 3; i <= N; i++) fibonacii[i] = (fibonacii[i - 1] + fibonacii[i - 2]) % Mod; long long Count = 0; long long number = 2; register long long now = fibonacii[N]; printf ("%lld=", now); while (now != 1) { if (now % number) number++; else { Count++; if (Count == 1) printf ("%lld", number); else printf ("*%lld", number); now /= number; } } return 0; }

 

斐波那契数列的通项公式x+洛谷P2626x