Problem:
Score : 200 points

Problem Statement

There are 2000001 stones placed on a number line. The coordinates of these stones are -1000000, -999999, -999998, \ldots, 999999, 1000000.

Among them, some K consecutive stones are painted black, and the others are painted white.

Additionally, we know that the stone at coordinate X is painted black.

Print all coordinates that potentially contain a stone painted black, in ascending order.

Constraints

1 \leq K \leq 100

0 \leq X \leq 100

All values in input are integers.

Input

Input is given from Standard Input in the following format:

K X

Output

Print all coordinates that potentially contain a stone painted black, in ascending order, with spaces in between.

Sample Input 1

3 7

Sample Output 1

5 6 7 8 9

We know that there are three stones painted black, and the stone at coordinate 7 is painted black. There are three possible cases:

The three stones painted black are placed at coordinates 5, 6, and 7.

The three stones painted black are placed at coordinates 6, 7, and 8.

The three stones painted black are placed at coordinates 7, 8, and 9.

Thus, five coordinates potentially contain a stone painted black: 5, 6, 7, 8, and 9.

Sample Input 2

4 0

Sample Output 2

-3 -2 -1 0 1 2 3

Negative coordinates can also contain a stone painted black.

Sample Input 3

1 100

Sample Output 3

100

Solution in C (GCC 5.4.1) (judge verdict: Accepted)


#pragma warning(disable:4996)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <float.h>

#pragma region 定義
#define i_cin(X) scanf("%d", &X)
#define i_cin2(X,Y) scanf("%d %d",&X,&Y)
#define i_cin4(X,Y,Z,W) scanf("%d %d %d %d",&X,&Y,&Z,&W)
#define i_cin3(X,Y,Z) scanf("%d %d %d",&X,&Y,&Z)
#define l_cin(X) scanf("%ld",&X)
#define f_cin(X) scanf("%f",&X)
#define l_cin2(X,Y) scanf("%ld %ld",&X,&Y)
#define s_cin(X) scanf("%s",&X)
#define c_cin(X) scanf("%c",&X);
#define ll_cin(X) scanf("%lld", &X)
#define ull_cin(X) scanf("%llu", &X)
#define lf_cin(X) scanf("%lf", &X)

#define  i_cout(X) printf("%d\n",X)
#define  f_cout(X) printf("%g\n",X)
#define  i_cout2(X,Y) printf("%d %d\n",X,Y)
#define  l_cout(X) printf("%ld\n",X)
#define  s_cout(X) printf("%s\n",X)
#define  s_coutc(X) printf("%s",X)

#define  c_cout(X) printf("%c",X)


#define  ll_cout(X) printf("%lld\n",X)
#define  ull_cout(X) printf("%llu\n",X)



typedef long long  ll;
typedef unsigned long long  ull;

#define rept(x, s, n) for (int x = s; x < n; x++)

#define S_RTN(S)  s_cout(S);return 0


#define _itoa(A,N) sprintf(A, "%d", N);

int i_cins(int n, int* A);
int l_cins2(int n, long* A, long* B);
int s_dsort(const void* a, const void* b);
int s_asort(const void* a, const void* b);

int  _gcd(int a, int b);
int  _swp(int* a, int* b);
int _cknum(char* a, int n);
int _atoi(char* s, int len);


int s_asorts(const void* a, const void* b);

//昇順
int s_asorts(const void* a, const void* b) {
	return(strcmp((char*)a, (char*)b));
}
int s_dsort(const void* a, const void* b) {
	return(*(int*)b - *(int*)a);
}
int s_asort(const void* a, const void* b) {
	return(*(int*)a - *(int*)b);
}
int l_cins2(int n, long* a, long* b) {
	int i;
	rept(i, 0, n) {
		l_cin2(*(a + i), *(b + i));
	}
	return 0;
}
int i_cins(int n, int* a) {
	int i;
	for (i = 0; i < n; i++) {
		i_cin(*(a + i));
	}

	return 0;
}
int _gcd(int a, int b) {
	int r, tmp;
	if (a < b) { tmp = a; a = b; b = tmp; }
	r = a % b;
	while (r) {
		if (!r) break;
		a = b; b = r;
		r = a % b;
	}
	return(b);
}

//#define pai  3.1415926535      
#define deg_rad(X)  (X*(M_PI/180))

long _max(long a, long b) {
	return a > b ? a : b;
}
long _min(long a, long b) {
	return a < b ? a : b;
}
int _swp(int* a, int* b)
{
	int tmp;
	tmp = *b; *b = *a; *a = tmp;
	return 0;
}

int _cknum(char* a, int n) {
	int i;
	char t = '0';

	for (i = 0; i < n; i++) {
		if (a[i] < '0' || a[i]>'9') return 1;
	}
	return 0;
}


int _atoi(char* s, int len) {
	char tmp[20];
	memcpy(tmp, s, len);
	tmp[len] = 0x00;
	return (atoi(tmp));
}

//素因数分解　pf素数が入る　pc 素数の数
ll _pfact(ll a);
static ll pf[100000000];
static ll pc;

ll _pfact(ll a) {
	long i, r;
	if (a % 2 == 0) {
		pf[pc] = 2; pc++;
		return(_pfact(a / 2));
	}
	r = sqrt(a);
	for (i = 3; i <= r; i += 2) {
		if (a % i == 0) {
			pf[pc] = i; pc++;
			return(_pfact(a / i));
		}
	}
	if (a != 1) {
		pf[pc] = a; pc++;
	}
	return 0;

}


#pragma endregion
//150-C
//169-A

int main(void) {
	long k, x, i;
	
	scanf("%ld %ld",&k, &x);
	printf("\n");
	for (i = x - k + 1; i < x + k; i++) {
		printf("%ld ",i);
	}
	printf("\n");
	return 0;
}


//066-B

//066-B

//157-C
//int main(void) {
//	int n, m;
//	int s, c;
//	int a[3] = { -1,-1,-1 };
//	int i, ans=0;
//	
//	i_cin2(n, m);
//
//	if  (m == 0 && n!=1) { i_cout(-1); return 0; } //???2つ目の条件なし
//
//	for (i = 0; i < m; i++) {
//		i_cin2(s, c);
//		if ((a[s - 1] != -1) &&   //矛盾はないか
//			(a[s - 1] != c)  ) { i_cout(-1); return 0; }
//		switch (n) {
//		case 1:if (s != 3) { i_cout(-1); return 0; } 
//			   break;
//		case 2:if (s == 1) { i_cout(-1); return 0; }
//			  if (s == 2 && c == 0)
//			  {i_cout(-1); return 0;}
//			  break;
//		case 3:
//			if (s == 1 && c == 0)
//			{i_cout(-1); return 0;}
//			break;
//		}
//		a[s - 1] = c;
//	}
//
//	for (i = 0; i < 3; i++) {
//		if (a[i] == -1) a[i] = 0;
//	}
//	ans = a[0] * 100 + a[1] * 10 + a[2];
//
//	i_cout(ans);
//	return 0;
//}



 //ABC-07 D 幅優先探索
//#define Q_SIZE 100000
//static struct _queue {
//	char *d[Q_SIZE];
//	int h;
//	int t;
//} q;
//
//int  pushq(struct _queue* q, char *item);
//int  pushq(struct _queue* q, char *item) {
//	if (q->t == Q_SIZE) return -1;
//	q->d[q->t] = item;
//	q->t++;
//	return 0;
//}
//
//char* popq(struct _queue* q);
//char* popq(struct _queue* q) {
//	char* tmp;
//	if (q->h == q->t) return -1;
//	tmp = q->d[q->h];
//	q->h--;
//	return tmp;
//}
//
//static struct _p {
//	int x;
//	int y;
//	int cnt;
//} q;
//static char a[51][51];
//static struct _p p[50];
//static int    pcnt;
//
//int _enq(struct _p *p, int x, int y, int cnt);
//int _enq(struct _p *p,int x, int y, int cnt) {
//	p->x = x; p->y = y; p->cnt = cnt;
//	pushq(&q,p); pcnt++;
//}
//int _deq(struct _p* p);
//int _deq(struct _p* p) {
//	p=popq(&q);
//	return(p == -1 ? p : 0);
//}
//int sqr_chk(int x, int y, int cnt) {
//	int tmp;
//	tmp = pcnt;
//	if (a[x - 1][y - 1] == '.') 
//	{
//		_enq(&p[pcnt], x - 1, y - 1, cnt); a[x - 1][y - 1] = 'X';
//	}
//	if (a[x - 1][y] == '.')
//	{
//		_enq(&p[pcnt], x - 1, y, cnt); a[x - 1][y] = 'X';
//	}
//	if (a[x+1][y] == '.') 
//	{
//		_enq(&p[pcnt], x + 1, y, cnt); a[x + 1][y] = 'X';
//	}
//	if (a[x][y+1] == '.') 
//	{
//		_enq(&p[pcnt], x, y + 1, cnt); a[x][y + 1] = 'X';
//	}
//	if (pcnt == tmp) return -1;
//	else             return 0;
//}
//
//int main(void) {
//	int r, c;
//	int sy, sx, gy, gx;
//	int i = 0, j = 0, x = 0, y = 0, ans = 0;
//	char cnt = '0';
//
//	i_cin(r, c);
//	i_cin4(sx, sy, gx, gy);
//	if (sx == gx && sy == gy) {
//		i_cout(0); return 0;
//	}
//	for (i = 0; i < r; i++)
//		s_cin(a[i]);
//	sy--; sx--;
//	struct _p* pt;
//	pt = &p[cnt];
//	p->x = sx; p->y = sy; p->cnt = cnt;
//
//	int rc;
//	while (1) {
//		rc = sqr_chk(pt->x,pt->y,pt->cnt);
//		if (rc == -1) {
//			rc = _deq(pt);
//			if (rc == -1 || (pt->x == gx && pt->y == gy)) break;
//		}
//	}
//
//	if (rc == -1) i_cout(rc);
//	else i_cout(ans);
//	return 0;
//}