斐波那契数列编程c++

代码如下：

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int i;
    int n;
    int *p;
    scanf("%d",&n);

    if(n == 1)
    {
        printf("1\n");
        return 0;
    }else if(n == 2)
    {
        printf("1 1\n");
        return 0;
    }

    p = (int*)malloc(sizeof(int)*n);
    
    p[0]= 1;
    p[1]=1;
    printf("1 1 ");
    for (i=2;i<=n;i++)
    {
        p[i] = p[i-1] + p[i-2];
        if(i<n)
            printf("%d ",p[i]);
        else
            printf("%d\n",p[i]);
    }
    free(p);
    return 0;
}

#include <iostream>
using namespace std;
 
int main()
{
    int n=12; 
    int a=0,b=1;
    int sum;
    for (int i=1;i<=n;i++)
    {
        sum=a+b;
        a=b;
        cout<<a<<" ";
        b=sum;
    }
    cout<<endl;
    return 0;
}

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

int Fib(int n)
{
if(n <= 1)
return 1;

    int f=0, g=1;
    int tmp=0;    //上图蓝色曲线所使用的临时变量
    while(--n)
    {
        tmp = g;    //临时存储本次的g
        g = f+g;    //计算f(n)，并传递给下一行的g
        f = tmp;    //将保存的g传递给f，用以下次计算
    }
    return g;
}

int main(){
int c=Fib( 2);
int n=0;
scanf("%d",&n);
for(int i=1;i<=n;i++){
printf("%d ",Fib(i));

}
return 1; 

}

#include<iostream>
#include<cmath>
using namespace std;
int main(){
    int i;
    int n=12; 
        for(i=1;i<=n;i++){
            cout<<1/sqrt(5)*(pow(((1+sqrt(5))/2),i)-pow(((1-sqrt(5))/2),i))<<" ";
        }
    return 0;
}

#include <stdio.h>
int main()
{
    int a1=1,a2=1,s;
   
    printf("%d %d",a1,a2);
    for (int i=3;i<=12;i++)
    {
        s=a1+a2;
        printf("%d ",s);
        a1=a2;
        a2=s;
    }

    return 0;
}

#include<bits/stdc++.h>
using namespace std;
int  n,p[50];
int f(int x)
{
    if(p[x])return p[x];
    if(x<=2)return x-1;
    return p[x]=f(x-1)+f(x-2);
}
int main()
{
    cin>>n;
    cout<<f(n);
    return 0;
}