## Editorial for SPBINARY2

Remember to use this editorial only when stuck, and not to copy-paste code from it. Please be respectful to the problem author and editorialist.
Submitting an official solution before solving the problem yourself is a bannable offence.

Lưu ý: Các code mẫu dưới đây chỉ mang tính tham khảo và có thể không AC được bài tập này

#### Code mẫu của RR

#include <sstream>
#include <iomanip>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <string>
#include <deque>
#include <complex>

#define FOR(i,a,b) for(int i=(a),_b=(b); i<=_b; i++)
#define FORD(i,a,b) for(int i=(a),_b=(b); i>=_b; i--)
#define REP(i,a) for(int i=0,_a=(a); i<_a; i++)
#define FORN(i,a,b) for(int i=(a),_b=(b);i<_b;i++)
#define DOWN(i,a,b) for(int i=a,_b=(b);i>=_b;i--)
#define SET(a,v) memset(a,v,sizeof(a))
#define sqr(x) ((x)*(x))
#define ll long long
#define F first
#define S second
#define PB push_back
#define MP make_pair

#define DEBUG(x) cout << #x << " = "; cout << x << endl;
#define PR(a,n) cout << #a << " = "; FOR(_,1,n) cout << a[_] << ' '; cout << endl;
#define PR0(a,n) cout << #a << " = "; REP(_,n) cout << a[_] << ' '; cout << endl;
using namespace std;

int INP,AM,REACHEOF;
#define BUFSIZE (1<<12)
char BUF[BUFSIZE+1], *inp=BUF;
#define GETCHAR(INP) { \
if(!*inp) { \
if (REACHEOF) return 0;\
memset(BUF,0,sizeof BUF);\
if (inpzzz != BUFSIZE) REACHEOF = true;\
inp=BUF; \
} \
INP=*inp++; \
}
#define DIG(a) (((a)>='0')&&((a)<='9'))
#define GN(j) { \
AM=0;\
GETCHAR(INP); while(!DIG(INP) && INP!='-') GETCHAR(INP);\
if (INP=='-') {AM=1;GETCHAR(INP);} \
j=INP-'0'; GETCHAR(INP); \
while(DIG(INP)){j=10*j+(INP-'0');GETCHAR(INP);} \
if (AM) j=-j;\
}

const long double PI = acos((long double) -1.0);
const int MOD = 1000000000;

int f[1000111], s[1000111];

int main() {
int n, k; scanf("%d%d", &n, &k);
f[0] = s[0] = 1;
FOR(i,1,k) {
f[i] = (f[i-1] * 2) % MOD;
s[i] = (s[i-1] + f[i]) % MOD;
}

FOR(i,k+1,n) {
f[i] = (s[i-1] - s[i-k-1] + MOD) % MOD;
s[i] = (s[i-1] + f[i]) % MOD;
}

cout << f[n] << endl;
return 0;
}


#### Code mẫu của hieult

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cmath>
//#include <conio.h>
#define ep 0.00001
#define oo 1000000010
#define mod 1000000000
double const PI=4*atan(1);

int f[1111111],n,k;

using namespace std;

int main(){
// freopen("A.in","r",stdin);
// freopen("A.out","w",stdout);
scanf("%d %d",&n,&k);
f[0] = 2; f[1] = 2;
for(int i = 2;i<=n;i++){
if(i<=k) f[i] = (f[i-1]*2)%mod;
else f[i] = (f[i-1]*2-f[i-k-1])%mod;
if(f[i]<0) f[i]+=mod;
}
printf("%d\n",f[n]);
//  getch();
}