用递归做是最简单的

#include <bits/stdc++.h>
using namespace std;
int dg(int n)
{
    if(n==1||n==2)
    {
        return 1;//如果是1/2，直接return1
    }
    else
    {
        return dg(n-1)+dg(n-2);//否则继续递归
    }
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    int n,m;
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>m;
        cout<<dg(m)<<'\n';
    }
    return 0;
}

#include<iostream>
using namespace std;
int a(int n){
	int a1=1,a2=1,a3;
    for(int i=0;i<n-2;i++){
    a3=a1+a2;
    a1=a2;
    a2=a3;}
    return a3;
}
int main(){
    int m;
    cin>>m;
    int n;
    for(int i=0;i<m;i++){
        cin>>n;
        if(n==1||n==2){
        	cout<<"1"<<endl;
		}
        else{
		n=a(n);
        cout<<n<<endl;}
    }
}

#include<stdio.h> int main() {int m,n,i,s1,s2; scanf("%d",&m); while(m--) {scanf("%d",&n); for(i=3,s1=s2=1; i<=n;i++) {s1=s1+s2; s2=s1-s2; } printf("%d\n",s1); } return 0; }

#include<stdio.h>
 int main() 
 {int m,n,i,s1,s2; 
scanf("%d",&m);//代表测试用例的数量
 while(m--)//当m小于等于零时停止循环
  {scanf("%d",&n); //输入用例
  for(i=3,s1=s2=1; i<=n;i++) //n小于三时直接输出s1=1
  {s1=s1+s2; s2=s1-s2; }//当n大于等于3时开始循环不断为s1，s2赋值
   printf("%d\n",s1); }//输出最后两项s1s2相加的结果s1
 return 0; }//得解

来个矩阵快速幂（虽然通常是需要取模的时候再用😋o(logn)）

#include <bits/stdc++.h>
using namespace std;
#define ll long long
typedef vector<vector<ll>> mat;

mat operator*(const mat &a, const mat &b) {
    mat c(2, vector<ll>(2));
    for (int i = 0; i < 2; ++i) {
        for (int j = 0; j < 2; ++j) {
            for (int k = 0; k < 2; ++k) {
                c[i][j] += a[i][k] * b[k][j];
            }
        }
    }
    return c;
}


mat operator^(mat a, ll b) {
    mat res = {{1, 0}, {0, 1}};
    while (b) {
        if (b & 1) res = res * a;
        a = a * a;
        b >>= 1;
    }
    return res;
}



void solve() {
    ll n;
    cin >> n;
    if (n == 1) {
        cout << 1 << '\n';
        return;
    }
    mat trans = {{1, 1}, {1, 0}};
    mat power = trans ^ (n - 2);
    mat init = {{1, 0}, {1, 0}};
    mat res = power * init;
    cout << res[0][0] << '\n';
}



int main() {
    int t;
    cin >> t;
    while (t--) {
        solve();
    }
    return 0;
}

#include<stdio.h>
int main()
{
    int m;
    scanf("%d",&m);
    int a[25]={0};
    a[1]=1;
    a[2]=1;
    int n;    
    for(int i=3;i<25;i++){
            a[i]=a[i-1]+a[i-2];
    }
    for(int j=1;j<=m;j++){
        scanf("%d",&n);
        printf("%d\n",a[n]);
    }
    return 0;
}