极简题解|奔饭ing

递归会有性能损失，只能递推

#include<cstdio>
using namespace std;
unsigned long long n,f[10000010];
int main(){
    scanf("%llu",&n);
    f[1]=1;
    f[2]=2;
    f[3]=4;
    for(int i=4;i<=n;i++){
        f[i]=((f[i-1]+f[i-2]+f[i-3])%(unsigned long long)(1e9+7));
    }
    printf("%llu",f[n]);
    return 0;
}