简单递推

from math import sqrt,ceil,gcd,log;re=lambda:map(int,input().strip().split())
a = [1, 1]
for i in range(2, 21):
    a.append(a[i - 1] + a[i - 2])
print(a[int(input())])

定义数组时，因该是a[n+1],因为这样当n=0的时候数组才存在，a[0]=1;

#include <stdio.h>

int main()

{

int n;

scanf("%d",&n);

int a[n+1];

a[0]=1;

a[1]=1;

a[2]=2;

for(int i=2;i<=n;i++){

a[i]=a[i-1]+a[i-2];

}

printf("%d\n",a[n]);

return 0;

}

递归解法 ：

#include<stdio.h>
int f[100];
int fib(int n)
{
    if(f[n] != 0)
    {
        return f[n];
    }
    else if(n == 0)
    {
        return 1;
    }
    else if(n == 1)
    {
        return 1;
    }
    else 
    {
        return (f[n - 1] = fib(n - 1)) + (f[n - 2] = fib(n - 2));
    }
}
int main()
{
    int n;
    scanf("%d",&n);
    int ans = fib(n);
    printf("%d",ans);
    return 0;
}

#include<stdio.h>
int main(){
    int a1,a2,b,s,n;
    a1=0,a2=1,s=1;
    scanf("%d",&n);
    for(b=1;b<=n;b++)
    {
        s=a1+a2;
        a1=a2;
        a2=s;
    }
    printf("%d",s);
    return 0;
}