#include<bits/stdc++.h>
using namespace std;
long long n,a[100000],k;
int main()
{
	cin>>n;
	if (n==1)
	{
		cout<<"YES";
		return 0;
	}
	a[1]=1;
	a[2]=1;
	k=2;
	while (1)
	{
		if (a[k]>n) break;
		if (a[k]==n) 
		{
			cout<<"YES";
			return 0;
		}
		k++;
		a[k]=a[k-1]+a[k-2];
	}
	cout<<"NO";
	return 0;
}

#include<bits/stdc++.h> using namespace std; int main(){ long long n,a=1,b=1,c; scanf("%lld",&n); if(na) { printf("YES"); exit(0); } while(1) { c=a+b; a=b; b=c; if(nc) { printf("YES"); exit(0); } if(c>n) { printf("NO"); exit(0); } } return 0; }

这道题目比较简单，用一个死循环枚举斐波那契数就能解决，如果枚举到就退出，并输出“YES”，如果大于也退出， 并输出“NO”。 代码如下：

#include <bits/stdc++.h>

using namespace std;

int main ()
{
	long long n, a = 1, b = 1, c;
	scanf ("%lld", &n);
	if (n == a)
	{
		printf ("YES");
		exit (0);
	}
	while (1)
	{
		c = a + b;
		a = b;
		b = c;
		if (n == c)
		{
			printf ("YES");
			exit (0);
		}
		if (c > n)
		{
			printf ("NO");
			exit (0);
		}
	}
	return 0;
}

#include<bits/stdc++.h>
using namespace std;
long long a=1,b=1,c=2,n;
int main(){
	cin>>n;
	while(n>c){
		a=b;
		b=c;
		c=a+b;
	}if(n==c)cout<<"YES";
	else cout<<"NO";
}

#include<bits/stdc++.h> using namespace std; int s,h,a[100010]={0,1}; int main() { ios::sync_with_stdio(0); cin.tie(0); cout.tie(0); int n; cin>>n; for(int i=2;a[i-1]<n;i++) { a[i]=a[i-1]+a[i-2]; if(a[i]==n) { cout<<"YES"; return 0; } } cout<<"NO"; return 0; }

#include<bits/stdc++.h>
using namespace std;
int s,h,a[100010]={0,1};
int main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int n;
	cin>>n;
	for(int i=2;a[i-1]<n;i++)
	{
		a[i]=a[i-1]+a[i-2];
		if(a[i]==n)
		{
			cout<<"YES";
			return 0;
		}
	}
	cout<<"NO";
	return 0;
}

#include<bits/stdc++.h> using namespace std; int n,a,b,c; int main(){ cin>>n; a=1; b=1; while(1){ c=a+b; a=b; b=c; if(c==n){ cout<<"YES"; return 0; } if(c>n){ cout<<"NO"; return 0; } } return 0; }