斐波那契数列 f[0]=1,f[1]=1,f[n]=f[n-1]+f[n-2](x>=2)
1 #define mm(a) memset(a,0,sizeof(a)); 2 #define max(x,y) (x)>(y)?(x):(y) 3 #define min(x,y) (x)<(y)?(x):(y) 4 #define Fopen freopen("1.in","r",stdin); freopen("m.out","w",stdout); 5 #define rep(i,a,b) for(ll i=(a);i<=(b);i++) 6 #define per(i,b,a) for(ll i=(b);i>=(a);i--) 7 #include<bits/stdc++.h> 8 typedef long long ll; 9 #define PII pair<ll,ll> 10 using namespace std; 11 const int INF=0x3f3f3f3f; 12 const int MAXN=(int)2e6+5; 13 14 ll dp[MAXN]; 15 int main() 16 { 17 int n; 18 dp[1]=1; 19 dp[0]=1; 20 scanf("%d",&n); 21 for(int i=2;i<=n;i++)dp[i]=dp[i-1]+dp[i-2]; 22 printf("%lld ",dp[n]); 23 24 return 0; 25 }