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;
} 

5

5=5

6

8=2*2*2

#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;
}