Problem:
Score : 200 points

Problem Statement

Takahashi's house has only one socket.

Takahashi wants to extend it with some number of power strips, each with A sockets, into B or more empty sockets.

One power strip with A sockets can extend one empty socket into A empty sockets.

Find the minimum number of power strips required.

Constraints

All values in input are integers.

2 \leq A \leq 20

1 \leq B \leq 20

Input

Input is given from Standard Input in the following format:

A B

Output

Print the minimum number of power strips required.

Sample Input 1

4 10

Sample Output 1

3

3 power strips, each with 4 sockets, extend the socket into 10 empty sockets.

Sample Input 2

8 9

Sample Output 2

2

2 power strips, each with 8 sockets, extend the socket into 15 empty sockets.

Sample Input 3

8 8

Sample Output 3

1

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)


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


#define i_cin(X) scanf("%d", &X)
#define i_cin2(X,Y) scanf("%d %d",&X,&Y)
#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  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  ll_cout(X) printf("%lld\n",X)

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

#define Yes_RTN s_cout("Yes");return 0
#define No_RTN  s_cout("No");return 0
#define YES_RTN s_cout("YES");return 0
#define NO_RTN  s_cout("NO");return 0


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


int main(void) {
	int  A, B, i,cnt = 0;
	int tmp, r;
	i_cin2(A,B);

	if (B<=A) { i_cout(1); return 0; }
	tmp = B / (A-1);
	if (A == 2) {
		if (B % 2) {
			i_cout(tmp); return 0;
		}
		else {
			i_cout(tmp-1);     return 0;
		}
	}
	r = B % (A - 1);
	if (r <= 1) i_cout(tmp);
	else i_cout(tmp + 1);
	return 0;

}
	
//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);
}