题解 | #斐波那契数列#
斐波那契数列
https://www.nowcoder.com/practice/ee5d403c1172487f8c7915b3c3d924c6
#include <stdio.h>
#define N 100000
int main(){
int n;
scanf("%d",&n);
long long int a[N];
a[1]=1;
a[2]=1;
for(int i=3;i<=n;i++)
a[i]=a[i-1]+a[i-2];
printf("%lld",a[n]);
return 0;
}
