思路

斐波那契数列，递推式为 $dp_0=dp_1=1,dp_i=dp_{i-1}+dp_{i-2}$。

code

#include<bits/stdc++.h>
using namespace std;
int main(){
   int n;
   cin>>n;
   long long dp[105];
   dp[0]=1;
   dp[1]=1;
   cout<<"1 1 ";
   for(int i=2;i<n;i++){
      dp[i]=dp[i-1]+dp[i-2];
      cout<<dp[i]<<" ";
   }
   return 0;
}

#include<bits/stdc++.h>
using namespace std;
int main()
{
	int n,i,a=1,b=0,c;
    cin>>n;
    cout<<"1"<<" ";
    for(i=2;i<=n;i++)
    {
    	c=a+b;
    	b=a;
    	a=c;
    	cout<<c<<" ";
	}
	return 0;
}

#include<bits/stdc++.h>
using namespace std;
int main(){
	int n,a=1,b=1,c;
	cin>>n;
	if(n==1){
		cout<<1;
	}
	if(n!=1){
		cout<<1<<" "<<1<<" ";
	}
	for(int i=3;i<=n;i++){
		c=a+b;
		a=b;
		b=c;
		cout<<c<<" ";
	}
	
}