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A boy Petya loves chess very much. He even came up with a chess piece of his own, a semiknight. The semiknight can move in any of these four directions: 2 squares forward and 2 squares to the right, 2 squares forward and 2 squares to the left, 2 squares backward and 2 to the right and 2 squares backward and 2 to the left. Naturally, the semiknight cannot move beyond the limits of the chessboard.
Petya put two semiknights on a standard chessboard. Petya simultaneously moves with both semiknights. The squares are rather large, so after some move the semiknights can meet, that is, they can end up in the same square. After the meeting the semiknights can move on, so it is possible that they meet again. Petya wonders if there is such sequence of moves when the semiknights meet. Petya considers some squares bad. That is, they do not suit for the meeting. The semiknights can move through these squares but their meetings in these squares don't count.
Petya prepared multiple chess boards. Help Petya find out whether the semiknights can meet on some good square for each board.
Please see the test case analysis.
-----Input-----
The first line contains number t (1 ≤ t ≤ 50) — the number of boards. Each board is described by a matrix of characters, consisting of 8 rows and 8 columns. The matrix consists of characters ".", "#", "K", representing an empty good square, a bad square and the semiknight's position, correspondingly. It is guaranteed that matrix contains exactly 2 semiknights. The semiknight's squares are considered good for the meeting. The tests are separated by empty line.
-----Output-----
For each test, print on a single line the answer to the problem: "YES", if the semiknights can meet and "NO" otherwise.
-----Examples-----
Input
2
........
........
......#.
K..##..#
.......#
...##..#
......#.
K.......
........
........
..#.....
..#..#..
..####..
...##...
........
....K#K#
Output
YES
NO
-----Note-----
Consider the first board from the sample. We will assume the rows and columns of the matrix to be numbered 1 through 8 from top to bottom and from left to right, correspondingly. The knights can meet, for example, in square (2, 7). The semiknight from square (4, 1) goes to square (2, 3) and the semiknight goes from square (8, 1) to square (6, 3). Then both semiknights go to (4, 5) but this square is bad, so they move together to square (2, 7).
On the second board the semiknights will never meet.
|
N = range(8)
test = int(input())
for i_test in range(test):
if i_test:
input()
x1, y1, x2, y2 = 0, 0, 0, 0
map = [input() for i in N]
for i in N:
for j in N:
if map[i][j] == "K":
x1, y1, x2, y2 = x2, y2, i, j
if abs(x1 - x2) % 4 == 0 and abs(y1 - y2) % 4 == 0:
print("YES")
else:
print("NO")
|
ASSIGN VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR IF VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR VAR VAR NUMBER NUMBER NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR FOR VAR VAR FOR VAR VAR IF VAR VAR VAR STRING ASSIGN VAR VAR VAR VAR VAR VAR VAR VAR IF BIN_OP FUNC_CALL VAR BIN_OP VAR VAR NUMBER NUMBER BIN_OP FUNC_CALL VAR BIN_OP VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
A boy Petya loves chess very much. He even came up with a chess piece of his own, a semiknight. The semiknight can move in any of these four directions: 2 squares forward and 2 squares to the right, 2 squares forward and 2 squares to the left, 2 squares backward and 2 to the right and 2 squares backward and 2 to the left. Naturally, the semiknight cannot move beyond the limits of the chessboard.
Petya put two semiknights on a standard chessboard. Petya simultaneously moves with both semiknights. The squares are rather large, so after some move the semiknights can meet, that is, they can end up in the same square. After the meeting the semiknights can move on, so it is possible that they meet again. Petya wonders if there is such sequence of moves when the semiknights meet. Petya considers some squares bad. That is, they do not suit for the meeting. The semiknights can move through these squares but their meetings in these squares don't count.
Petya prepared multiple chess boards. Help Petya find out whether the semiknights can meet on some good square for each board.
Please see the test case analysis.
-----Input-----
The first line contains number t (1 ≤ t ≤ 50) — the number of boards. Each board is described by a matrix of characters, consisting of 8 rows and 8 columns. The matrix consists of characters ".", "#", "K", representing an empty good square, a bad square and the semiknight's position, correspondingly. It is guaranteed that matrix contains exactly 2 semiknights. The semiknight's squares are considered good for the meeting. The tests are separated by empty line.
-----Output-----
For each test, print on a single line the answer to the problem: "YES", if the semiknights can meet and "NO" otherwise.
-----Examples-----
Input
2
........
........
......#.
K..##..#
.......#
...##..#
......#.
K.......
........
........
..#.....
..#..#..
..####..
...##...
........
....K#K#
Output
YES
NO
-----Note-----
Consider the first board from the sample. We will assume the rows and columns of the matrix to be numbered 1 through 8 from top to bottom and from left to right, correspondingly. The knights can meet, for example, in square (2, 7). The semiknight from square (4, 1) goes to square (2, 3) and the semiknight goes from square (8, 1) to square (6, 3). Then both semiknights go to (4, 5) but this square is bad, so they move together to square (2, 7).
On the second board the semiknights will never meet.
|
3
n = int(input())
for _ in range(n):
k = []
b = []
for i in range(8):
s = input()
for j in range(8):
if s[j] == "K":
k.append((i, j))
print(
"YES"
if abs(k[0][0] - k[1][0]) % 4 == 0 and abs(k[0][1] - k[1][1]) % 4 == 0
else "NO"
)
if _ < n - 1:
input()
|
EXPR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER IF VAR VAR STRING EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER NUMBER NUMBER NUMBER BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER NUMBER NUMBER NUMBER STRING STRING IF VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR
|
A boy Petya loves chess very much. He even came up with a chess piece of his own, a semiknight. The semiknight can move in any of these four directions: 2 squares forward and 2 squares to the right, 2 squares forward and 2 squares to the left, 2 squares backward and 2 to the right and 2 squares backward and 2 to the left. Naturally, the semiknight cannot move beyond the limits of the chessboard.
Petya put two semiknights on a standard chessboard. Petya simultaneously moves with both semiknights. The squares are rather large, so after some move the semiknights can meet, that is, they can end up in the same square. After the meeting the semiknights can move on, so it is possible that they meet again. Petya wonders if there is such sequence of moves when the semiknights meet. Petya considers some squares bad. That is, they do not suit for the meeting. The semiknights can move through these squares but their meetings in these squares don't count.
Petya prepared multiple chess boards. Help Petya find out whether the semiknights can meet on some good square for each board.
Please see the test case analysis.
-----Input-----
The first line contains number t (1 ≤ t ≤ 50) — the number of boards. Each board is described by a matrix of characters, consisting of 8 rows and 8 columns. The matrix consists of characters ".", "#", "K", representing an empty good square, a bad square and the semiknight's position, correspondingly. It is guaranteed that matrix contains exactly 2 semiknights. The semiknight's squares are considered good for the meeting. The tests are separated by empty line.
-----Output-----
For each test, print on a single line the answer to the problem: "YES", if the semiknights can meet and "NO" otherwise.
-----Examples-----
Input
2
........
........
......#.
K..##..#
.......#
...##..#
......#.
K.......
........
........
..#.....
..#..#..
..####..
...##...
........
....K#K#
Output
YES
NO
-----Note-----
Consider the first board from the sample. We will assume the rows and columns of the matrix to be numbered 1 through 8 from top to bottom and from left to right, correspondingly. The knights can meet, for example, in square (2, 7). The semiknight from square (4, 1) goes to square (2, 3) and the semiknight goes from square (8, 1) to square (6, 3). Then both semiknights go to (4, 5) but this square is bad, so they move together to square (2, 7).
On the second board the semiknights will never meet.
|
t = int(input())
for k in range(t):
r8 = range(8)
a = [input() for i in r8]
ij = [(i, j) for i in r8 for j in r8]
x, y = ((i, j) for i, j in ij if a[i][j] == "K")
def s(p):
d = [
(p[0] - 2, p[1] - 2),
(p[0] + 2, p[1] - 2),
(p[0] - 2, p[1] + 2),
(p[0] + 2, p[1] + 2),
]
return set((i, j) for i, j in ij if (i, j) in d)
print("YES" if len(s(x) & s(y)) > 0 else "NO")
if k != t - 1:
input()
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR STRING FUNC_DEF ASSIGN VAR LIST BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER NUMBER RETURN FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER STRING STRING IF VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR
|
A boy Petya loves chess very much. He even came up with a chess piece of his own, a semiknight. The semiknight can move in any of these four directions: 2 squares forward and 2 squares to the right, 2 squares forward and 2 squares to the left, 2 squares backward and 2 to the right and 2 squares backward and 2 to the left. Naturally, the semiknight cannot move beyond the limits of the chessboard.
Petya put two semiknights on a standard chessboard. Petya simultaneously moves with both semiknights. The squares are rather large, so after some move the semiknights can meet, that is, they can end up in the same square. After the meeting the semiknights can move on, so it is possible that they meet again. Petya wonders if there is such sequence of moves when the semiknights meet. Petya considers some squares bad. That is, they do not suit for the meeting. The semiknights can move through these squares but their meetings in these squares don't count.
Petya prepared multiple chess boards. Help Petya find out whether the semiknights can meet on some good square for each board.
Please see the test case analysis.
-----Input-----
The first line contains number t (1 ≤ t ≤ 50) — the number of boards. Each board is described by a matrix of characters, consisting of 8 rows and 8 columns. The matrix consists of characters ".", "#", "K", representing an empty good square, a bad square and the semiknight's position, correspondingly. It is guaranteed that matrix contains exactly 2 semiknights. The semiknight's squares are considered good for the meeting. The tests are separated by empty line.
-----Output-----
For each test, print on a single line the answer to the problem: "YES", if the semiknights can meet and "NO" otherwise.
-----Examples-----
Input
2
........
........
......#.
K..##..#
.......#
...##..#
......#.
K.......
........
........
..#.....
..#..#..
..####..
...##...
........
....K#K#
Output
YES
NO
-----Note-----
Consider the first board from the sample. We will assume the rows and columns of the matrix to be numbered 1 through 8 from top to bottom and from left to right, correspondingly. The knights can meet, for example, in square (2, 7). The semiknight from square (4, 1) goes to square (2, 3) and the semiknight goes from square (8, 1) to square (6, 3). Then both semiknights go to (4, 5) but this square is bad, so they move together to square (2, 7).
On the second board the semiknights will never meet.
|
MOVS = [(2, -2), (-2, 2), (-2, -2), (2, 2)]
def check(a):
return 0 <= a < 8
set1 = set()
set2 = set()
dic1 = dict()
dic2 = dict()
def cango1(matrix, pos, lap):
for dx, dy in MOVS:
nx, ny = dx + pos[0], dy + pos[1]
if not check(nx) or not check(ny):
continue
if (nx, ny) in set1:
continue
dic1[nx, ny] = lap % 2
set1.add((nx, ny))
cango1(matrix, (nx, ny), lap + 1)
def cango2(matrix, pos, lap):
for dx, dy in MOVS:
nx, ny = dx + pos[0], dy + pos[1]
if not check(nx) or not check(ny):
continue
if (nx, ny) in set2:
continue
dic2[nx, ny] = lap % 2
set2.add((nx, ny))
cango2(matrix, (nx, ny), lap + 1)
q = int(input())
for ww in range(q):
matrix = [input().strip() for i in range(8)]
pos = []
bad = set()
for i in range(8):
for j in range(8):
if matrix[i][j] == "K":
pos.append((i, j))
if matrix[i][j] == "#":
bad.add((i, j))
set1, set2, dic1, dic2 = set(), set(), dict(), dict()
cango1(matrix, pos[0], 0)
cango2(matrix, pos[1], 0)
if ww != q - 1:
input()
sec = (set1 & set2) - bad
for x, y in sec:
if dic1[x, y] == dic2[x, y]:
print("YES")
break
else:
print("NO")
|
ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER FUNC_DEF RETURN NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_DEF FOR VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR NUMBER BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER FUNC_DEF FOR VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR NUMBER BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR NUMBER ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER IF VAR VAR VAR STRING EXPR FUNC_CALL VAR VAR VAR IF VAR VAR VAR STRING EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER NUMBER IF VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR FOR VAR VAR VAR IF VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Jzzhu has a big rectangular chocolate bar that consists of n × m unit squares. He wants to cut this bar exactly k times. Each cut must meet the following requirements:
each cut should be straight (horizontal or vertical); each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut); each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 × 6 chocolate for 5 times.
[Image]
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece is the number of unit squares in it.
-----Input-----
A single line contains three integers n, m, k (1 ≤ n, m ≤ 10^9; 1 ≤ k ≤ 2·10^9).
-----Output-----
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
-----Examples-----
Input
3 4 1
Output
6
Input
6 4 2
Output
8
Input
2 3 4
Output
-1
-----Note-----
In the first sample, Jzzhu can cut the chocolate following the picture below:
[Image]
In the second sample the optimal division looks like this:
[Image]
In the third sample, it's impossible to cut a 2 × 3 chocolate 4 times.
|
n, m, k = [int(s) for s in input().split()]
n, m = max(n, m), min(n, m)
if n + m - 2 < k:
print(-1)
elif n - 1 < k:
print(m // (k - n + 2))
elif m - 1 < k:
print(n // (k + 1) * m)
else:
print(max(n // (k + 1) * m, m // (k + 1) * n))
|
ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF BIN_OP BIN_OP VAR VAR NUMBER VAR EXPR FUNC_CALL VAR NUMBER IF BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER IF BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR
|
Jzzhu has a big rectangular chocolate bar that consists of n × m unit squares. He wants to cut this bar exactly k times. Each cut must meet the following requirements:
each cut should be straight (horizontal or vertical); each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut); each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 × 6 chocolate for 5 times.
[Image]
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece is the number of unit squares in it.
-----Input-----
A single line contains three integers n, m, k (1 ≤ n, m ≤ 10^9; 1 ≤ k ≤ 2·10^9).
-----Output-----
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
-----Examples-----
Input
3 4 1
Output
6
Input
6 4 2
Output
8
Input
2 3 4
Output
-1
-----Note-----
In the first sample, Jzzhu can cut the chocolate following the picture below:
[Image]
In the second sample the optimal division looks like this:
[Image]
In the third sample, it's impossible to cut a 2 × 3 chocolate 4 times.
|
n, m, k = map(int, input().split())
n, m = min(n, m), max(n, m)
def p(i, j):
return n // (i + 1) * (m // (j + 1))
if k < n:
print(max(p(k, 0), p(0, k)))
elif k < m:
print(p(0, k))
elif k <= n + m - 2:
print(p(k - (m - 1), m - 1))
else:
print(-1)
|
ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_DEF RETURN BIN_OP BIN_OP VAR BIN_OP VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR NUMBER VAR IF VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR NUMBER VAR IF VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER
|
Jzzhu has a big rectangular chocolate bar that consists of n × m unit squares. He wants to cut this bar exactly k times. Each cut must meet the following requirements:
each cut should be straight (horizontal or vertical); each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut); each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 × 6 chocolate for 5 times.
[Image]
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece is the number of unit squares in it.
-----Input-----
A single line contains three integers n, m, k (1 ≤ n, m ≤ 10^9; 1 ≤ k ≤ 2·10^9).
-----Output-----
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
-----Examples-----
Input
3 4 1
Output
6
Input
6 4 2
Output
8
Input
2 3 4
Output
-1
-----Note-----
In the first sample, Jzzhu can cut the chocolate following the picture below:
[Image]
In the second sample the optimal division looks like this:
[Image]
In the third sample, it's impossible to cut a 2 × 3 chocolate 4 times.
|
n, m, k = map(int, input().split())
if k > n + m - 2:
print(-1)
exit()
res = 0
if k < n:
res = max(res, n // (k + 1) * m)
else:
res = max(res, m // (k - (n - 1) + 1))
if k < m:
res = max(res, m // (k + 1) * n)
else:
res = max(res, n // (k - (m - 1) + 1))
print(res)
|
ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER NUMBER IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR
|
Jzzhu has a big rectangular chocolate bar that consists of n × m unit squares. He wants to cut this bar exactly k times. Each cut must meet the following requirements:
each cut should be straight (horizontal or vertical); each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut); each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 × 6 chocolate for 5 times.
[Image]
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece is the number of unit squares in it.
-----Input-----
A single line contains three integers n, m, k (1 ≤ n, m ≤ 10^9; 1 ≤ k ≤ 2·10^9).
-----Output-----
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
-----Examples-----
Input
3 4 1
Output
6
Input
6 4 2
Output
8
Input
2 3 4
Output
-1
-----Note-----
In the first sample, Jzzhu can cut the chocolate following the picture below:
[Image]
In the second sample the optimal division looks like this:
[Image]
In the third sample, it's impossible to cut a 2 × 3 chocolate 4 times.
|
n, m, k = map(int, input().split())
if k > n + m - 2:
print(-1)
else:
a, b = m * (n // (k + 1)), n * (m // (k + 1))
if a == 0:
a = m // (k - n + 2)
if b == 0:
b = n // (k - m + 2)
print(max(a, b))
|
ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR BIN_OP VAR NUMBER BIN_OP VAR BIN_OP VAR BIN_OP VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR
|
Jzzhu has a big rectangular chocolate bar that consists of n × m unit squares. He wants to cut this bar exactly k times. Each cut must meet the following requirements:
each cut should be straight (horizontal or vertical); each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut); each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 × 6 chocolate for 5 times.
[Image]
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece is the number of unit squares in it.
-----Input-----
A single line contains three integers n, m, k (1 ≤ n, m ≤ 10^9; 1 ≤ k ≤ 2·10^9).
-----Output-----
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
-----Examples-----
Input
3 4 1
Output
6
Input
6 4 2
Output
8
Input
2 3 4
Output
-1
-----Note-----
In the first sample, Jzzhu can cut the chocolate following the picture below:
[Image]
In the second sample the optimal division looks like this:
[Image]
In the third sample, it's impossible to cut a 2 × 3 chocolate 4 times.
|
def split1(m, n, k):
if m % (k + 1) == 0:
return n * (m // (k + 1))
if k < m:
return m // (k + 1) * n
else:
k2 = k - m + 1
return n // (k2 + 1)
def split(m, n, k):
r = max(split1(m, n, k), split1(n, m, k))
return r if r else -1
args = input().split()
print(split(*map(int, args)))
|
FUNC_DEF IF BIN_OP VAR BIN_OP VAR NUMBER NUMBER RETURN BIN_OP VAR BIN_OP VAR BIN_OP VAR NUMBER IF VAR VAR RETURN BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER RETURN BIN_OP VAR BIN_OP VAR NUMBER FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR RETURN VAR VAR NUMBER ASSIGN VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR
|
Jzzhu has a big rectangular chocolate bar that consists of n × m unit squares. He wants to cut this bar exactly k times. Each cut must meet the following requirements:
each cut should be straight (horizontal or vertical); each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut); each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 × 6 chocolate for 5 times.
[Image]
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece is the number of unit squares in it.
-----Input-----
A single line contains three integers n, m, k (1 ≤ n, m ≤ 10^9; 1 ≤ k ≤ 2·10^9).
-----Output-----
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
-----Examples-----
Input
3 4 1
Output
6
Input
6 4 2
Output
8
Input
2 3 4
Output
-1
-----Note-----
In the first sample, Jzzhu can cut the chocolate following the picture below:
[Image]
In the second sample the optimal division looks like this:
[Image]
In the third sample, it's impossible to cut a 2 × 3 chocolate 4 times.
|
a = input().split(" ")
n = int(a[0])
m = int(a[1])
k = int(a[2])
if k > m + n - 2:
print(-1)
else:
ans = 1
if k < n:
ans = max(ans, n // (k + 1) * m)
if k < m:
ans = max(ans, m // (k + 1) * n)
if k >= n:
ans = max(ans, m // (k - (n - 1) + 1))
if k >= m:
ans = max(ans, n // (k - (m - 1) + 1))
print(ans)
|
ASSIGN VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER IF VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER NUMBER IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR
|
Jzzhu has a big rectangular chocolate bar that consists of n × m unit squares. He wants to cut this bar exactly k times. Each cut must meet the following requirements:
each cut should be straight (horizontal or vertical); each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut); each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 × 6 chocolate for 5 times.
[Image]
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece is the number of unit squares in it.
-----Input-----
A single line contains three integers n, m, k (1 ≤ n, m ≤ 10^9; 1 ≤ k ≤ 2·10^9).
-----Output-----
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
-----Examples-----
Input
3 4 1
Output
6
Input
6 4 2
Output
8
Input
2 3 4
Output
-1
-----Note-----
In the first sample, Jzzhu can cut the chocolate following the picture below:
[Image]
In the second sample the optimal division looks like this:
[Image]
In the third sample, it's impossible to cut a 2 × 3 chocolate 4 times.
|
n, m, k = list(map(int, input().split()))
if n + m - 2 < k:
print("-1\n")
else:
x = min(n - 1, k)
y = max(0, k - x)
ans = n // (x + 1) * (m // (y + 1))
y = min(m - 1, k)
x = max(0, k - y)
ans = max(ans, n // (x + 1) * (m // (y + 1)))
print(ans, "\n")
|
ASSIGN VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF BIN_OP BIN_OP VAR VAR NUMBER VAR EXPR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR STRING
|
Jzzhu has a big rectangular chocolate bar that consists of n × m unit squares. He wants to cut this bar exactly k times. Each cut must meet the following requirements:
each cut should be straight (horizontal or vertical); each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut); each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 × 6 chocolate for 5 times.
[Image]
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece is the number of unit squares in it.
-----Input-----
A single line contains three integers n, m, k (1 ≤ n, m ≤ 10^9; 1 ≤ k ≤ 2·10^9).
-----Output-----
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
-----Examples-----
Input
3 4 1
Output
6
Input
6 4 2
Output
8
Input
2 3 4
Output
-1
-----Note-----
In the first sample, Jzzhu can cut the chocolate following the picture below:
[Image]
In the second sample the optimal division looks like this:
[Image]
In the third sample, it's impossible to cut a 2 × 3 chocolate 4 times.
|
def split1(m, n, k):
if k < m:
return m // (k + 1) * n
else:
return n // (k - m + 2)
def split(m, n, k):
r = max(split1(m, n, k), split1(n, m, k))
return r if r else -1
args = input().split()
print(split(*list(map(int, args))))
|
FUNC_DEF IF VAR VAR RETURN BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR RETURN BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR RETURN VAR VAR NUMBER ASSIGN VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR
|
Jzzhu has a big rectangular chocolate bar that consists of n × m unit squares. He wants to cut this bar exactly k times. Each cut must meet the following requirements:
each cut should be straight (horizontal or vertical); each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut); each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 × 6 chocolate for 5 times.
[Image]
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece is the number of unit squares in it.
-----Input-----
A single line contains three integers n, m, k (1 ≤ n, m ≤ 10^9; 1 ≤ k ≤ 2·10^9).
-----Output-----
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
-----Examples-----
Input
3 4 1
Output
6
Input
6 4 2
Output
8
Input
2 3 4
Output
-1
-----Note-----
In the first sample, Jzzhu can cut the chocolate following the picture below:
[Image]
In the second sample the optimal division looks like this:
[Image]
In the third sample, it's impossible to cut a 2 × 3 chocolate 4 times.
|
n, m, k = map(int, input().split())
if k <= n + m - 2:
x = min(k + 1, n)
y = k + 2 - x
a = n // x * (m // y)
y = min(k + 1, m)
x = k + 2 - y
b = n // x * (m // y)
print(max(a, b))
else:
print(-1)
|
ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER
|
Jzzhu has a big rectangular chocolate bar that consists of n × m unit squares. He wants to cut this bar exactly k times. Each cut must meet the following requirements:
each cut should be straight (horizontal or vertical); each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut); each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 × 6 chocolate for 5 times.
[Image]
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece is the number of unit squares in it.
-----Input-----
A single line contains three integers n, m, k (1 ≤ n, m ≤ 10^9; 1 ≤ k ≤ 2·10^9).
-----Output-----
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
-----Examples-----
Input
3 4 1
Output
6
Input
6 4 2
Output
8
Input
2 3 4
Output
-1
-----Note-----
In the first sample, Jzzhu can cut the chocolate following the picture below:
[Image]
In the second sample the optimal division looks like this:
[Image]
In the third sample, it's impossible to cut a 2 × 3 chocolate 4 times.
|
n, m, k = map(int, input().split())
if k > n + m - 2:
print(-1)
else:
if n < m:
n, m = m, n
if m % (k + 1) == 0:
print(m * n // (k + 1))
elif k <= n - 1:
if k > m - 1:
print(n // (k + 1) * m)
else:
print(max(n // (k + 1) * m, m // (k + 1) * n))
else:
print(m // (k - (n - 1) + 1))
|
ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR VAR ASSIGN VAR VAR VAR VAR IF BIN_OP VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER NUMBER
|
Jzzhu has a big rectangular chocolate bar that consists of n × m unit squares. He wants to cut this bar exactly k times. Each cut must meet the following requirements:
each cut should be straight (horizontal or vertical); each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut); each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 × 6 chocolate for 5 times.
[Image]
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece is the number of unit squares in it.
-----Input-----
A single line contains three integers n, m, k (1 ≤ n, m ≤ 10^9; 1 ≤ k ≤ 2·10^9).
-----Output-----
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
-----Examples-----
Input
3 4 1
Output
6
Input
6 4 2
Output
8
Input
2 3 4
Output
-1
-----Note-----
In the first sample, Jzzhu can cut the chocolate following the picture below:
[Image]
In the second sample the optimal division looks like this:
[Image]
In the third sample, it's impossible to cut a 2 × 3 chocolate 4 times.
|
n, m, k = list(map(int, input().split()))
ans, ans1 = 0, 0
if n + m - 2 < k:
print(-1)
return
else:
k1 = k
if k < n:
ans = n // (k + 1) * m
else:
k -= n - 1
ans = m // (k + 1)
if k1 < m:
ans1 = m // (k1 + 1) * n
else:
k1 -= m - 1
ans1 = n // (k1 + 1)
print(max(ans1, ans))
|
ASSIGN VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR NUMBER NUMBER IF BIN_OP BIN_OP VAR VAR NUMBER VAR EXPR FUNC_CALL VAR NUMBER RETURN ASSIGN VAR VAR IF VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR
|
Jzzhu has a big rectangular chocolate bar that consists of n × m unit squares. He wants to cut this bar exactly k times. Each cut must meet the following requirements:
each cut should be straight (horizontal or vertical); each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut); each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 × 6 chocolate for 5 times.
[Image]
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece is the number of unit squares in it.
-----Input-----
A single line contains three integers n, m, k (1 ≤ n, m ≤ 10^9; 1 ≤ k ≤ 2·10^9).
-----Output-----
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
-----Examples-----
Input
3 4 1
Output
6
Input
6 4 2
Output
8
Input
2 3 4
Output
-1
-----Note-----
In the first sample, Jzzhu can cut the chocolate following the picture below:
[Image]
In the second sample the optimal division looks like this:
[Image]
In the third sample, it's impossible to cut a 2 × 3 chocolate 4 times.
|
def cut(n, m, k):
if k > m + n - 2:
return -1
if k < max(m, n):
return max(n * (m // (k + 1)), m * (n // (k + 1)))
return max(n // (k - m + 2), m // (k - n + 2))
n, m, k = map(int, input().split())
print(cut(n, m, k))
|
FUNC_DEF IF VAR BIN_OP BIN_OP VAR VAR NUMBER RETURN NUMBER IF VAR FUNC_CALL VAR VAR VAR RETURN FUNC_CALL VAR BIN_OP VAR BIN_OP VAR BIN_OP VAR NUMBER BIN_OP VAR BIN_OP VAR BIN_OP VAR NUMBER RETURN FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR
|
Jzzhu has a big rectangular chocolate bar that consists of n × m unit squares. He wants to cut this bar exactly k times. Each cut must meet the following requirements:
each cut should be straight (horizontal or vertical); each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut); each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 × 6 chocolate for 5 times.
[Image]
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece is the number of unit squares in it.
-----Input-----
A single line contains three integers n, m, k (1 ≤ n, m ≤ 10^9; 1 ≤ k ≤ 2·10^9).
-----Output-----
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
-----Examples-----
Input
3 4 1
Output
6
Input
6 4 2
Output
8
Input
2 3 4
Output
-1
-----Note-----
In the first sample, Jzzhu can cut the chocolate following the picture below:
[Image]
In the second sample the optimal division looks like this:
[Image]
In the third sample, it's impossible to cut a 2 × 3 chocolate 4 times.
|
def read(mode=2):
inputs = input().strip()
if mode == 0:
return inputs
if mode == 1:
return inputs.split()
if mode == 2:
return [int(x) for x in inputs.split()]
def write(s="\n"):
if isinstance(s, list):
s = " ".join(map(str, s))
s = str(s)
print(s, end="")
n, m, k = read()
if n + m - 2 < k:
print(-1)
else:
mx = 0
if n > k:
mx = max(mx, n // (k + 1) * m)
else:
mx = max(mx, 1 * (m // (k - n + 2)))
if m > k:
mx = max(mx, m // (k + 1) * n)
else:
mx = max(mx, 1 * (n // (k - m + 2)))
print(mx)
|
FUNC_DEF NUMBER ASSIGN VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER RETURN VAR IF VAR NUMBER RETURN FUNC_CALL VAR IF VAR NUMBER RETURN FUNC_CALL VAR VAR VAR FUNC_CALL VAR FUNC_DEF STRING IF FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR STRING ASSIGN VAR VAR VAR FUNC_CALL VAR IF BIN_OP BIN_OP VAR VAR NUMBER VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR
|
Jzzhu has a big rectangular chocolate bar that consists of n × m unit squares. He wants to cut this bar exactly k times. Each cut must meet the following requirements:
each cut should be straight (horizontal or vertical); each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut); each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 × 6 chocolate for 5 times.
[Image]
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece is the number of unit squares in it.
-----Input-----
A single line contains three integers n, m, k (1 ≤ n, m ≤ 10^9; 1 ≤ k ≤ 2·10^9).
-----Output-----
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
-----Examples-----
Input
3 4 1
Output
6
Input
6 4 2
Output
8
Input
2 3 4
Output
-1
-----Note-----
In the first sample, Jzzhu can cut the chocolate following the picture below:
[Image]
In the second sample the optimal division looks like this:
[Image]
In the third sample, it's impossible to cut a 2 × 3 chocolate 4 times.
|
def main():
n, m, k = [int(i) for i in input().split()]
if k > n + m - 2:
print(-1)
return
if k > n - 1:
result1 = m // (k + 2 - n)
else:
result1 = n // (k + 1) * m
if k > m - 1:
result2 = n // (k + 2 - m)
else:
result2 = m // (k + 1) * n
print(max(result1, result2))
main()
|
FUNC_DEF ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER RETURN IF VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR IF VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR
|
Jzzhu has a big rectangular chocolate bar that consists of n × m unit squares. He wants to cut this bar exactly k times. Each cut must meet the following requirements:
each cut should be straight (horizontal or vertical); each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut); each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 × 6 chocolate for 5 times.
[Image]
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece is the number of unit squares in it.
-----Input-----
A single line contains three integers n, m, k (1 ≤ n, m ≤ 10^9; 1 ≤ k ≤ 2·10^9).
-----Output-----
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
-----Examples-----
Input
3 4 1
Output
6
Input
6 4 2
Output
8
Input
2 3 4
Output
-1
-----Note-----
In the first sample, Jzzhu can cut the chocolate following the picture below:
[Image]
In the second sample the optimal division looks like this:
[Image]
In the third sample, it's impossible to cut a 2 × 3 chocolate 4 times.
|
n, m, k = map(int, input().split())
if m > n:
n, m = m, n
ans = -1
if k < n:
ans = m * (n // (k + 1))
if k < m:
ans = max(ans, n * (m // (k + 1)))
elif k <= n - 1 + m - 1:
ans = m // (k + 1 - n + 1)
print(ans)
|
ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR BIN_OP VAR BIN_OP VAR NUMBER IF VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR
|
Jzzhu has a big rectangular chocolate bar that consists of n × m unit squares. He wants to cut this bar exactly k times. Each cut must meet the following requirements:
each cut should be straight (horizontal or vertical); each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut); each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 × 6 chocolate for 5 times.
[Image]
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece is the number of unit squares in it.
-----Input-----
A single line contains three integers n, m, k (1 ≤ n, m ≤ 10^9; 1 ≤ k ≤ 2·10^9).
-----Output-----
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
-----Examples-----
Input
3 4 1
Output
6
Input
6 4 2
Output
8
Input
2 3 4
Output
-1
-----Note-----
In the first sample, Jzzhu can cut the chocolate following the picture below:
[Image]
In the second sample the optimal division looks like this:
[Image]
In the third sample, it's impossible to cut a 2 × 3 chocolate 4 times.
|
l = input().split()
n = int(l[0])
m = int(l[1])
k = int(l[2])
if k <= n + m - 2:
if k < n:
outn = int(n / (k + 1)) * m
else:
outn = int(m / (k - n + 2))
if k < m:
outm = int(m / (k + 1)) * n
else:
outm = int(n / (k - m + 2))
print("", max(outn, outm), sep="")
else:
print("-1")
|
ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER IF VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR STRING FUNC_CALL VAR VAR VAR STRING EXPR FUNC_CALL VAR STRING
|
Jzzhu has a big rectangular chocolate bar that consists of n × m unit squares. He wants to cut this bar exactly k times. Each cut must meet the following requirements:
each cut should be straight (horizontal or vertical); each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut); each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 × 6 chocolate for 5 times.
[Image]
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece is the number of unit squares in it.
-----Input-----
A single line contains three integers n, m, k (1 ≤ n, m ≤ 10^9; 1 ≤ k ≤ 2·10^9).
-----Output-----
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
-----Examples-----
Input
3 4 1
Output
6
Input
6 4 2
Output
8
Input
2 3 4
Output
-1
-----Note-----
In the first sample, Jzzhu can cut the chocolate following the picture below:
[Image]
In the second sample the optimal division looks like this:
[Image]
In the third sample, it's impossible to cut a 2 × 3 chocolate 4 times.
|
def Calc(n, m, k):
if n + m - 2 < k:
return -1
if k <= n - 1:
return n // (k + 1) * m
return m // (k - n + 2)
n, m, k = [int(x) for x in input().split()]
print(max(Calc(n, m, k), Calc(m, n, k)))
|
FUNC_DEF IF BIN_OP BIN_OP VAR VAR NUMBER VAR RETURN NUMBER IF VAR BIN_OP VAR NUMBER RETURN BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR RETURN BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR
|
Jzzhu has a big rectangular chocolate bar that consists of n × m unit squares. He wants to cut this bar exactly k times. Each cut must meet the following requirements:
each cut should be straight (horizontal or vertical); each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut); each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 × 6 chocolate for 5 times.
[Image]
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece is the number of unit squares in it.
-----Input-----
A single line contains three integers n, m, k (1 ≤ n, m ≤ 10^9; 1 ≤ k ≤ 2·10^9).
-----Output-----
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
-----Examples-----
Input
3 4 1
Output
6
Input
6 4 2
Output
8
Input
2 3 4
Output
-1
-----Note-----
In the first sample, Jzzhu can cut the chocolate following the picture below:
[Image]
In the second sample the optimal division looks like this:
[Image]
In the third sample, it's impossible to cut a 2 × 3 chocolate 4 times.
|
n, m, k = map(int, input().split())
if n + m - 2 < k:
print(-1)
return
def divisors(n):
div = set()
for i in range(1, int(n ** (1 / 2)) + 1):
if n % i == 0:
div.add(i)
div.add(n // i)
return [*div]
def solve(n, m, k):
div = divisors(n)
ans = 0
for i in div:
if i - 1 > k:
continue
ans = max(ans, n // i * (m // (k - i + 2)))
return ans
print(max(solve(n, m, k), solve(m, n, k)))
|
ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF BIN_OP BIN_OP VAR VAR NUMBER VAR EXPR FUNC_CALL VAR NUMBER RETURN FUNC_DEF ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR BIN_OP VAR BIN_OP NUMBER NUMBER NUMBER IF BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR RETURN LIST VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR VAR IF BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER RETURN VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR
|
Jzzhu has a big rectangular chocolate bar that consists of n × m unit squares. He wants to cut this bar exactly k times. Each cut must meet the following requirements:
each cut should be straight (horizontal or vertical); each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut); each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 × 6 chocolate for 5 times.
[Image]
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece is the number of unit squares in it.
-----Input-----
A single line contains three integers n, m, k (1 ≤ n, m ≤ 10^9; 1 ≤ k ≤ 2·10^9).
-----Output-----
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
-----Examples-----
Input
3 4 1
Output
6
Input
6 4 2
Output
8
Input
2 3 4
Output
-1
-----Note-----
In the first sample, Jzzhu can cut the chocolate following the picture below:
[Image]
In the second sample the optimal division looks like this:
[Image]
In the third sample, it's impossible to cut a 2 × 3 chocolate 4 times.
|
def main():
a, b, c = map(int, input().split())
res, c = [-1], c + 1
if c < a + b:
for n, m, k in ((a, b, c), (b, a, c)):
if n < k:
k -= n - 1
n, m = m, 1
res.append(n // k * m)
print(max(res))
main()
|
FUNC_DEF ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR LIST NUMBER BIN_OP VAR NUMBER IF VAR BIN_OP VAR VAR FOR VAR VAR VAR VAR VAR VAR VAR VAR VAR IF VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR
|
Jzzhu has a big rectangular chocolate bar that consists of n × m unit squares. He wants to cut this bar exactly k times. Each cut must meet the following requirements:
each cut should be straight (horizontal or vertical); each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut); each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 × 6 chocolate for 5 times.
[Image]
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece is the number of unit squares in it.
-----Input-----
A single line contains three integers n, m, k (1 ≤ n, m ≤ 10^9; 1 ≤ k ≤ 2·10^9).
-----Output-----
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
-----Examples-----
Input
3 4 1
Output
6
Input
6 4 2
Output
8
Input
2 3 4
Output
-1
-----Note-----
In the first sample, Jzzhu can cut the chocolate following the picture below:
[Image]
In the second sample the optimal division looks like this:
[Image]
In the third sample, it's impossible to cut a 2 × 3 chocolate 4 times.
|
from itertools import *
def read_ints():
return map(int, input().strip().split())
n, m, k = read_ints()
def val(x, k, n, m):
Y = x + 1
Z = k - x + 1
if Y > 0 and Z > 0:
return n // (x + 1) * (m // (k - x + 1))
def sym(n, m, k):
x = min(n - 1, k)
return val(x, k, n, m)
def test(n, m, k):
if n + m + 2 < k:
return -1
answers = [sym(n, m, k), sym(m, n, k)]
return max(answers) or -1
print(test(n, m, k))
|
FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR FUNC_DEF ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR NUMBER VAR NUMBER RETURN BIN_OP BIN_OP VAR BIN_OP VAR NUMBER BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER FUNC_DEF ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR RETURN FUNC_CALL VAR VAR VAR VAR VAR FUNC_DEF IF BIN_OP BIN_OP VAR VAR NUMBER VAR RETURN NUMBER ASSIGN VAR LIST FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR RETURN FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR
|
Jzzhu has a big rectangular chocolate bar that consists of n × m unit squares. He wants to cut this bar exactly k times. Each cut must meet the following requirements:
each cut should be straight (horizontal or vertical); each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut); each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 × 6 chocolate for 5 times.
[Image]
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece is the number of unit squares in it.
-----Input-----
A single line contains three integers n, m, k (1 ≤ n, m ≤ 10^9; 1 ≤ k ≤ 2·10^9).
-----Output-----
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
-----Examples-----
Input
3 4 1
Output
6
Input
6 4 2
Output
8
Input
2 3 4
Output
-1
-----Note-----
In the first sample, Jzzhu can cut the chocolate following the picture below:
[Image]
In the second sample the optimal division looks like this:
[Image]
In the third sample, it's impossible to cut a 2 × 3 chocolate 4 times.
|
n, m, k = [int(x) for x in input().split()]
if k + 2 > n + m:
print(-1)
else:
if k >= n:
alpha = m // (k - n + 2)
else:
alpha = m * (n // (k + 1))
if k >= m:
beta = n // (k - m + 2)
else:
beta = n * (m // (k + 1))
print(max(alpha, beta))
|
ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR IF BIN_OP VAR NUMBER BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR
|
Jzzhu has a big rectangular chocolate bar that consists of n × m unit squares. He wants to cut this bar exactly k times. Each cut must meet the following requirements:
each cut should be straight (horizontal or vertical); each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut); each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 × 6 chocolate for 5 times.
[Image]
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece is the number of unit squares in it.
-----Input-----
A single line contains three integers n, m, k (1 ≤ n, m ≤ 10^9; 1 ≤ k ≤ 2·10^9).
-----Output-----
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
-----Examples-----
Input
3 4 1
Output
6
Input
6 4 2
Output
8
Input
2 3 4
Output
-1
-----Note-----
In the first sample, Jzzhu can cut the chocolate following the picture below:
[Image]
In the second sample the optimal division looks like this:
[Image]
In the third sample, it's impossible to cut a 2 × 3 chocolate 4 times.
|
inputx = input().split(" ")
n = int(inputx[0])
m = int(inputx[1])
k = int(inputx[2])
if n + m - 2 < k:
print(-1)
else:
if n > m:
n, m = m, n
if k < n:
print(max(n * int(m / (k + 1)), m * int(n / (k + 1))))
elif n <= k and k < m:
print(n * int(m / (k + 1)))
else:
print(int(n / (k - m + 2)))
|
ASSIGN VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER IF BIN_OP BIN_OP VAR VAR NUMBER VAR EXPR FUNC_CALL VAR NUMBER IF VAR VAR ASSIGN VAR VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER BIN_OP VAR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER IF VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER
|
Jzzhu has a big rectangular chocolate bar that consists of n × m unit squares. He wants to cut this bar exactly k times. Each cut must meet the following requirements:
each cut should be straight (horizontal or vertical); each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut); each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 × 6 chocolate for 5 times.
[Image]
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece is the number of unit squares in it.
-----Input-----
A single line contains three integers n, m, k (1 ≤ n, m ≤ 10^9; 1 ≤ k ≤ 2·10^9).
-----Output-----
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
-----Examples-----
Input
3 4 1
Output
6
Input
6 4 2
Output
8
Input
2 3 4
Output
-1
-----Note-----
In the first sample, Jzzhu can cut the chocolate following the picture below:
[Image]
In the second sample the optimal division looks like this:
[Image]
In the third sample, it's impossible to cut a 2 × 3 chocolate 4 times.
|
a, b, k = list(map(int, input().split()))
if a + b - 2 < k:
print(-1)
return
k += 2
if a > b:
a, b = b, a
i = 1
maxi = -1
while i * i <= a:
y = min(i, k - 1)
val1 = a // y * (b // (k - y))
if maxi < val1:
maxi = val1
j = min(a // i, k - 1)
val2 = a // j * (b // (k - j))
if maxi < val2:
maxi = val2
i += 1
print(maxi)
|
ASSIGN VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF BIN_OP BIN_OP VAR VAR NUMBER VAR EXPR FUNC_CALL VAR NUMBER RETURN VAR NUMBER IF VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR IF VAR VAR ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR
|
Hamming distance between strings a and b of equal length (denoted by h(a, b)) is equal to the number of distinct integers i (1 ≤ i ≤ |a|), such that ai ≠ bi, where ai is the i-th symbol of string a, bi is the i-th symbol of string b. For example, the Hamming distance between strings "aba" and "bba" equals 1, they have different first symbols. For strings "bbba" and "aaab" the Hamming distance equals 4.
John Doe had a paper on which four strings of equal length s1, s2, s3 and s4 were written. Each string si consisted only of lowercase letters "a" and "b". John found the Hamming distances between all pairs of strings he had. Then he lost the paper with the strings but he didn't lose the Hamming distances between all pairs.
Help John restore the strings; find some four strings s'1, s'2, s'3, s'4 of equal length that consist only of lowercase letters "a" and "b", such that the pairwise Hamming distances between them are the same as between John's strings. More formally, set s'i must satisfy the condition <image>.
To make the strings easier to put down on a piece of paper, you should choose among all suitable sets of strings the one that has strings of minimum length.
Input
The first line contains space-separated integers h(s1, s2), h(s1, s3), h(s1, s4). The second line contains space-separated integers h(s2, s3) and h(s2, s4). The third line contains the single integer h(s3, s4).
All given integers h(si, sj) are non-negative and do not exceed 105. It is guaranteed that at least one number h(si, sj) is positive.
Output
Print -1 if there's no suitable set of strings.
Otherwise print on the first line number len — the length of each string. On the i-th of the next four lines print string s'i. If there are multiple sets with the minimum length of the strings, print any of them.
Examples
Input
4 4 4
4 4
4
Output
6
aaaabb
aabbaa
bbaaaa
bbbbbb
|
def get_input():
a, b, d = map(int, input().split())
c, e = map(int, input().split())
f = int(input())
return [a, b, c, d, e, f]
def check_condition(a, b, c, d, e, f):
condition1 = (a + b + c) % 2 == 0
condition2 = (d + e + a) % 2 == 0
condition3 = (e + f + c) % 2 == 0
condition4 = (d + f + b) % 2 == 0
condition = condition1 and condition2 and condition3 and condition4
return condition
def find_t(a, b, c, d, e, f):
t_min1 = round((d + f - a - c) / 2)
t_min2 = round((e + f - a - b) / 2)
t_min3 = round((d + e - b - c) / 2)
t_min4 = 0
t_min = max(t_min1, t_min2, t_min3, t_min4)
t_max1 = round((d + e - a) / 2)
t_max2 = round((e + f - c) / 2)
t_max3 = round((d + f - b) / 2)
t_max = min(t_max1, t_max2, t_max3)
if t_min <= t_max:
return t_min
else:
return -1
def find_all(a, b, c, d, e, f, t):
x1 = round((a + c - d - f) / 2 + t)
x2 = round((d + f - b) / 2 - t)
y1 = round((a + b - e - f) / 2 + t)
y2 = round((e + f - c) / 2 - t)
z1 = round((b + c - d - e) / 2 + t)
z2 = round((d + e - a) / 2 - t)
return [x1, x2, y1, y2, z1, z2]
def generate_string(x1, x2, y1, y2, z1, z2, t):
n = x1 + x2 + y1 + y2 + z1 + z2 + t
s1 = "".join(["a"] * n)
s2 = "".join(["a"] * (z1 + z2 + t)) + "".join(["b"] * (x1 + x2 + y1 + y2))
s3 = (
"".join(["a"] * t)
+ "".join(["b"] * (y1 + y2 + z1 + z2))
+ "".join(["a"] * (x1 + x2))
)
s4 = (
"".join(["b"] * (t + z2))
+ "".join(["a"] * (z1 + y2))
+ "".join(["b"] * (y1 + x2))
+ "".join(["a"] * x1)
)
return [s1, s2, s3, s4]
def __main__():
fail_output = "-1"
a, b, c, d, e, f = map(int, get_input())
if not check_condition(a, b, c, d, e, f):
print(fail_output)
return False
t = find_t(a, b, c, d, e, f)
if t < 0:
print(fail_output)
return False
x1, x2, y1, y2, z1, z2 = map(int, find_all(a, b, c, d, e, f, t))
s1, s2, s3, s4 = map(str, generate_string(x1, x2, y1, y2, z1, z2, t))
print(str(x1 + x2 + y1 + y2 + z1 + z2 + t) + "\n")
print(s1 + "\n")
print(s2 + "\n")
print(s3 + "\n")
print(s4 + "\n")
__main__()
|
FUNC_DEF ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR RETURN LIST VAR VAR VAR VAR VAR VAR FUNC_DEF ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER NUMBER ASSIGN VAR VAR VAR VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR IF VAR VAR RETURN VAR RETURN NUMBER FUNC_DEF ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER VAR RETURN LIST VAR VAR VAR VAR VAR VAR FUNC_DEF ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL STRING BIN_OP LIST STRING VAR ASSIGN VAR BIN_OP FUNC_CALL STRING BIN_OP LIST STRING BIN_OP BIN_OP VAR VAR VAR FUNC_CALL STRING BIN_OP LIST STRING BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP FUNC_CALL STRING BIN_OP LIST STRING VAR FUNC_CALL STRING BIN_OP LIST STRING BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR FUNC_CALL STRING BIN_OP LIST STRING BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP FUNC_CALL STRING BIN_OP LIST STRING BIN_OP VAR VAR FUNC_CALL STRING BIN_OP LIST STRING BIN_OP VAR VAR FUNC_CALL STRING BIN_OP LIST STRING BIN_OP VAR VAR FUNC_CALL STRING BIN_OP LIST STRING VAR RETURN LIST VAR VAR VAR VAR FUNC_DEF ASSIGN VAR STRING ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR IF FUNC_CALL VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR RETURN NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR RETURN NUMBER ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR VAR VAR VAR STRING EXPR FUNC_CALL VAR BIN_OP VAR STRING EXPR FUNC_CALL VAR BIN_OP VAR STRING EXPR FUNC_CALL VAR BIN_OP VAR STRING EXPR FUNC_CALL VAR BIN_OP VAR STRING EXPR FUNC_CALL VAR
|
Hamming distance between strings a and b of equal length (denoted by h(a, b)) is equal to the number of distinct integers i (1 ≤ i ≤ |a|), such that ai ≠ bi, where ai is the i-th symbol of string a, bi is the i-th symbol of string b. For example, the Hamming distance between strings "aba" and "bba" equals 1, they have different first symbols. For strings "bbba" and "aaab" the Hamming distance equals 4.
John Doe had a paper on which four strings of equal length s1, s2, s3 and s4 were written. Each string si consisted only of lowercase letters "a" and "b". John found the Hamming distances between all pairs of strings he had. Then he lost the paper with the strings but he didn't lose the Hamming distances between all pairs.
Help John restore the strings; find some four strings s'1, s'2, s'3, s'4 of equal length that consist only of lowercase letters "a" and "b", such that the pairwise Hamming distances between them are the same as between John's strings. More formally, set s'i must satisfy the condition <image>.
To make the strings easier to put down on a piece of paper, you should choose among all suitable sets of strings the one that has strings of minimum length.
Input
The first line contains space-separated integers h(s1, s2), h(s1, s3), h(s1, s4). The second line contains space-separated integers h(s2, s3) and h(s2, s4). The third line contains the single integer h(s3, s4).
All given integers h(si, sj) are non-negative and do not exceed 105. It is guaranteed that at least one number h(si, sj) is positive.
Output
Print -1 if there's no suitable set of strings.
Otherwise print on the first line number len — the length of each string. On the i-th of the next four lines print string s'i. If there are multiple sets with the minimum length of the strings, print any of them.
Examples
Input
4 4 4
4 4
4
Output
6
aaaabb
aabbaa
bbaaaa
bbbbbb
|
h = [[0 in range(10)] for j in range(10)]
for i in range(1, 4):
h[i] = [(0) for j in range(i + 1)] + list(map(int, input().split()))
if h[1][2] + h[1][3] < h[2][3] or (h[1][2] + h[1][3] - h[2][3]) % 2 == 1:
print("-1")
exit(0)
BB = (h[1][2] + h[1][3] - h[2][3]) // 2
BA = h[1][2] - BB
AB = h[1][3] - BB
NowB = h[1][4]
NowLen = BB + AB + BA
Now24 = BA + NowB + BB
Now34 = AB + NowB + BB
BAB = 0
ABB = 0
BBB = 0
Dif = BA - AB - (h[2][4] - h[3][4])
if abs(Dif) % 2 == 1:
print("-1")
exit(0)
if Dif < 0:
ABB += abs(Dif) // 2
Now34 -= ABB * 2
if AB < ABB or NowB < ABB:
print("-1")
exit(0)
NowB -= ABB
else:
BAB += Dif // 2
Now24 -= BAB * 2
if BA < BAB or NowB < BAB:
print("-1")
exit(0)
NowB -= BAB
if Now24 < h[2][4] or (Now24 - h[2][4]) % 2 == 1:
print("-1")
exit(0)
for i in range(BB + 1):
if i > NowB:
break
Now = i * 2
if Now > Now24 - h[2][4]:
break
if min([(NowB - i) // 2, BA - BAB, AB - ABB]) * 2 >= Now24 - h[2][4] - Now:
BBB += i
BAB += (Now24 - h[2][4] - Now) // 2
ABB += (Now24 - h[2][4] - Now) // 2
NowB -= i + (Now24 - h[2][4] - Now)
print(NowLen + NowB)
print("".join(["a" for j in range(NowLen)] + ["a" for j in range(NowB)]))
print(
"".join(
["a" for j in range(AB)]
+ ["b" for j in range(BB + BA)]
+ ["a" for j in range(NowB)]
)
)
print(
"".join(
["b" for j in range(AB + BB)]
+ ["a" for j in range(BA)]
+ ["a" for j in range(NowB)]
)
)
print(
"".join(
["b" for j in range(ABB)]
+ ["a" for j in range(AB - ABB)]
+ ["b" for j in range(BBB)]
+ ["a" for j in range(BB - BBB)]
+ ["b" for j in range(BAB)]
+ ["a" for j in range(BA - BAB)]
+ ["b" for j in range(NowB)]
)
)
exit(0)
print("-1")
|
ASSIGN VAR LIST NUMBER FUNC_CALL VAR NUMBER VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR VAR BIN_OP NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF BIN_OP VAR NUMBER NUMBER VAR NUMBER NUMBER VAR NUMBER NUMBER BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER VAR NUMBER NUMBER VAR NUMBER NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER VAR NUMBER NUMBER VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER VAR ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER NUMBER IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER IF VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP VAR NUMBER IF VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER IF VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER VAR VAR IF VAR VAR NUMBER NUMBER BIN_OP BIN_OP VAR VAR NUMBER NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR VAR NUMBER NUMBER IF BIN_OP FUNC_CALL VAR LIST BIN_OP BIN_OP VAR VAR NUMBER BIN_OP VAR VAR BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER NUMBER VAR VAR VAR VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER NUMBER VAR NUMBER VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER NUMBER VAR NUMBER VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING BIN_OP STRING VAR FUNC_CALL VAR VAR STRING VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING BIN_OP BIN_OP STRING VAR FUNC_CALL VAR VAR STRING VAR FUNC_CALL VAR BIN_OP VAR VAR STRING VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING BIN_OP BIN_OP STRING VAR FUNC_CALL VAR BIN_OP VAR VAR STRING VAR FUNC_CALL VAR VAR STRING VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP STRING VAR FUNC_CALL VAR VAR STRING VAR FUNC_CALL VAR BIN_OP VAR VAR STRING VAR FUNC_CALL VAR VAR STRING VAR FUNC_CALL VAR BIN_OP VAR VAR STRING VAR FUNC_CALL VAR VAR STRING VAR FUNC_CALL VAR BIN_OP VAR VAR STRING VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR STRING
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = map(int, input().split())
ans = 0
c = b
if b < f:
ans = -1
else:
c -= f
for i in range(k):
if ans == -1:
break
if i % 2 == 0:
d = a - f
else:
d = f
if i < k - 1:
d *= 2
if b < d:
ans = -1
elif c < d:
ans += 1
c = b
c -= d
print(ans)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR VAR IF VAR VAR ASSIGN VAR NUMBER VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR IF VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
def solve(a, b, f, k):
origb = b
disa = 2 * f
disb = 2 * (a - f)
trava = False
counter = 0
b -= disa / 2
if b < 0:
return -1
while k > 0:
if trava:
trava = False
if k == 1:
disa = disa / 2
if disa > b:
counter += 1
b = origb - disa
if b < 0:
return -1
k -= 1
else:
b -= disa
if b < 0:
return -1
k -= 1
else:
trava = True
if k == 1:
disb = disb / 2
if disb > b:
counter += 1
b = origb - disb
if b < 0:
return -1
k -= 1
else:
b -= disb
if b < 0:
return -1
k -= 1
return counter
inputs = input("")
afind = inputs.find(" ")
a = int(inputs[0:afind])
inputs = inputs[afind + 1 :]
afind = inputs.find(" ")
b = int(inputs[0:afind])
inputs = inputs[afind + 1 :]
afind = inputs.find(" ")
f = int(inputs[0:afind])
inputs = inputs[afind + 1 :]
k = int(inputs[0:])
print(solve(a, b, f, k))
|
FUNC_DEF ASSIGN VAR VAR ASSIGN VAR BIN_OP NUMBER VAR ASSIGN VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER VAR BIN_OP VAR NUMBER IF VAR NUMBER RETURN NUMBER WHILE VAR NUMBER IF VAR ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER RETURN NUMBER VAR NUMBER VAR VAR IF VAR NUMBER RETURN NUMBER VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER RETURN NUMBER VAR NUMBER VAR VAR IF VAR NUMBER RETURN NUMBER VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = map(int, input().split())
c = b
cnt = 0
if cnt == 0:
for i in range(k):
if i & 1 == 0:
dis = a - f
else:
dis = f
c -= a - dis
if c < 0:
cnt = -1
break
if i < k - 1 and c < 2 * dis:
cnt += 1
c = b
if i == k - 1 and c < dis:
cnt += 1
c = b
c -= dis
if c < 0:
cnt = -1
break
print(cnt)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR BIN_OP NUMBER VAR VAR NUMBER ASSIGN VAR VAR IF VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = map(int, input().split())
ans = 0
cb = b
for i in range(k):
if i % 2 == 0:
if f > cb:
print(-1)
return
else:
cb -= f
if i != k - 1:
if cb < 2 * (a - f):
ans += 1
cb = b
cb -= a - f
else:
if cb < a - f:
ans += 1
cb = b
cb -= a - f
if cb < 0:
print(-1)
return
elif a - f > cb:
print(-1)
return
else:
cb -= a - f
if i != k - 1:
if cb < 2 * f:
ans += 1
cb = b
cb -= f
else:
if cb < f:
ans += 1
cb = b
cb -= f
if cb < 0:
print(-1)
return
print(ans)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR VAR FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER RETURN VAR VAR IF VAR BIN_OP VAR NUMBER IF VAR BIN_OP NUMBER BIN_OP VAR VAR VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR VAR IF VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER RETURN IF BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR NUMBER RETURN VAR BIN_OP VAR VAR IF VAR BIN_OP VAR NUMBER IF VAR BIN_OP NUMBER VAR VAR NUMBER ASSIGN VAR VAR VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER RETURN EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = map(int, input().split())
x, y = f, a - f
if k == 1:
if b < max(x, y):
print(-1)
quit()
elif k == 2:
if b < max(x, 2 * y):
print(-1)
quit()
elif b < max(2 * x, 2 * y):
print(-1)
quit()
gas = b
refill = 0
test = True
while k > 0:
if test:
if gas >= x + 2 * y:
gas -= a
elif gas >= x:
if k == 1:
if gas >= a:
break
gas = b - y
refill += 1
else:
print(-1)
quit()
test = not test
else:
if gas >= 2 * x + y:
gas -= a
elif gas >= y:
if k == 1:
if gas >= a:
break
gas = b - x
refill += 1
else:
print(-1)
quit()
test = not test
k -= 1
print(refill)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR BIN_OP VAR VAR IF VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR IF VAR NUMBER IF VAR FUNC_CALL VAR VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR IF VAR FUNC_CALL VAR BIN_OP NUMBER VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR NUMBER IF VAR IF VAR BIN_OP VAR BIN_OP NUMBER VAR VAR VAR IF VAR VAR IF VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR VAR IF VAR BIN_OP BIN_OP NUMBER VAR VAR VAR VAR IF VAR VAR IF VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = map(int, input().split())
g = a - f
t = b - f
s = 0
while t >= 0:
if t >= g:
k -= 1
t -= g
else:
t = b - g
s += 1
if t < 0:
break
k -= 1
if k == 0:
break
if t >= g:
t -= g
else:
t = b - 2 * g
s += 1
if t < 0:
break
if t >= f:
k -= 1
t -= f
else:
t = b - f
s += 1
if t < 0:
break
k -= 1
if k == 0:
break
if t >= f:
t -= f
else:
t = b - 2 * f
s += 1
if t < 0:
break
print(-1 if k else s)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER WHILE VAR NUMBER IF VAR VAR VAR NUMBER VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR NUMBER IF VAR NUMBER VAR NUMBER IF VAR NUMBER IF VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP NUMBER VAR VAR NUMBER IF VAR NUMBER IF VAR VAR VAR NUMBER VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR NUMBER IF VAR NUMBER VAR NUMBER IF VAR NUMBER IF VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP NUMBER VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = [int(z) for z in input().split()]
to_fuel = f
from_fuel = a - f
alrdy_fueled = 0
petrol = b
for i in range(k):
petrol -= to_fuel
if petrol < 0:
print(-1)
exit(0)
if petrol < 2 * from_fuel and i != k - 1:
petrol = b
alrdy_fueled += 1
elif i == k - 1:
if petrol < from_fuel:
petrol = b
alrdy_fueled += 1
petrol -= from_fuel
if petrol < 0:
print(-1)
exit(0)
to_fuel, from_fuel = from_fuel, to_fuel
print(alrdy_fueled)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR FOR VAR FUNC_CALL VAR VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR BIN_OP NUMBER VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER IF VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR VAR VAR NUMBER VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
import sys
def min_refills(end_point, capacity, fuel_point, journeys):
fuel = capacity
journey_count = 0
refills = 0
while True:
fuel -= fuel_point
if fuel < 0:
return -1
fuel_needed = (
2 * (end_point - fuel_point)
if journey_count < journeys - 1
else end_point - fuel_point
)
if fuel_needed > fuel:
refills += 1
fuel = capacity
if fuel_needed > capacity:
return -1
fuel -= end_point - fuel_point
journey_count += 1
if journey_count == journeys:
break
fuel -= end_point - fuel_point
fuel_needed = 2 * fuel_point if journey_count < journeys - 1 else fuel_point
if fuel_needed > fuel:
refills += 1
fuel = capacity
if fuel_needed > capacity:
return -1
fuel -= fuel_point
journey_count += 1
if journey_count == journeys:
break
return refills
line = sys.stdin.readline()
print(min_refills(*[int(x) for x in line.split()]))
|
IMPORT FUNC_DEF ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE NUMBER VAR VAR IF VAR NUMBER RETURN NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER BIN_OP NUMBER BIN_OP VAR VAR BIN_OP VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR IF VAR VAR RETURN NUMBER VAR BIN_OP VAR VAR VAR NUMBER IF VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER BIN_OP NUMBER VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR IF VAR VAR RETURN NUMBER VAR VAR VAR NUMBER IF VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
def main():
a, b, f, k = map(int, input().split())
if b < f or b < a - f:
print(-1)
elif k >= 2 and b < 2 * (a - f):
print(-1)
elif k > 2 and b < 2 * f:
print(-1)
else:
refuel, fuel = 0, b
for i in range(k - 1):
if i % 2:
if fuel < f + a:
refuel += 1
fuel = b - f
else:
fuel = fuel - a
elif fuel < 2 * a - f:
refuel += 1
fuel = b - (a - f)
else:
fuel = fuel - a
if fuel < a:
refuel += 1
print(refuel)
main()
|
FUNC_DEF ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR NUMBER VAR BIN_OP NUMBER BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR NUMBER VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR VAR NUMBER VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER IF VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR BIN_OP BIN_OP NUMBER VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = list(map(int, input().split()))
i = 0
bi = b
ans = 0
while i < k:
if f > bi:
print(-1)
return
bi -= f
if k - i == 1:
if a - f > bi:
bi = b
ans += 1
if a - f > b:
print(-1)
return
break
i += 1
if 2 * (a - f) > bi:
bi = b
ans += 1
if 2 * (a - f) > b:
print(-1)
return
bi -= 2 * (a - f)
if k - i == 1:
if f > bi:
bi = b
ans += 1
break
if 2 * f > bi:
bi = b
ans += 1
if 2 * f > b:
print(-1)
return
bi -= f
i += 1
print(ans)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR IF VAR VAR EXPR FUNC_CALL VAR NUMBER RETURN VAR VAR IF BIN_OP VAR VAR NUMBER IF BIN_OP VAR VAR VAR ASSIGN VAR VAR VAR NUMBER IF BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR NUMBER RETURN VAR NUMBER IF BIN_OP NUMBER BIN_OP VAR VAR VAR ASSIGN VAR VAR VAR NUMBER IF BIN_OP NUMBER BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR NUMBER RETURN VAR BIN_OP NUMBER BIN_OP VAR VAR IF BIN_OP VAR VAR NUMBER IF VAR VAR ASSIGN VAR VAR VAR NUMBER IF BIN_OP NUMBER VAR VAR ASSIGN VAR VAR VAR NUMBER IF BIN_OP NUMBER VAR VAR EXPR FUNC_CALL VAR NUMBER RETURN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
def solve(a, b, f, k):
cur_b = b
cur_k = 0
refueling_count = 0
can_0f = cur_b - f >= 0
if not can_0f:
return refueling_count, cur_k
else:
cur_b -= f
while cur_k < k:
faf_path = 2 * (a - f)
can_faf = cur_b - faf_path >= 0
can_faf_with_refueling = b - faf_path >= 0
can_fa = cur_b - (a - f) >= 0
can_fa_with_refueling = b - (a - f) >= 0
if can_faf and k - cur_k > 1:
cur_b = cur_b - faf_path
cur_k += 1
elif can_faf_with_refueling and k - cur_k > 1:
refueling_count += 1
cur_b = b - faf_path
cur_k += 1
elif can_fa:
cur_k += 1
return refueling_count, cur_k
elif can_fa_with_refueling:
cur_k += 1
refueling_count += 1
return refueling_count, cur_k
else:
return refueling_count, cur_k
if cur_k == k:
return refueling_count, cur_k
f0f_path = 2 * f
can_f0f = cur_b - f0f_path >= 0
can_f0f_with_refueling = b - f0f_path >= 0
can_f0 = cur_b - f >= 0
can_f0_with_refueling = b - f >= 0
if can_f0f and k - cur_k > 1:
cur_b = cur_b - f0f_path
cur_k += 1
elif can_f0f_with_refueling and k - cur_k > 1:
refueling_count += 1
cur_b = b - f0f_path
cur_k += 1
elif can_f0:
cur_k += 1
return refueling_count, cur_k
elif can_f0_with_refueling:
cur_k += 1
refueling_count += 1
return refueling_count, cur_k
else:
return refueling_count, cur_k
return refueling_count, cur_k
def main():
a, b, f, k = map(int, input().split())
refueling_count, cur_k = solve(a, b, f, k)
if cur_k == k:
print(refueling_count)
else:
print(-1)
main()
|
FUNC_DEF ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR NUMBER IF VAR RETURN VAR VAR VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR NUMBER IF VAR BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR NUMBER IF VAR BIN_OP VAR VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR NUMBER IF VAR VAR NUMBER RETURN VAR VAR IF VAR VAR NUMBER VAR NUMBER RETURN VAR VAR RETURN VAR VAR IF VAR VAR RETURN VAR VAR ASSIGN VAR BIN_OP NUMBER VAR ASSIGN VAR BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR NUMBER IF VAR BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR NUMBER IF VAR BIN_OP VAR VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR NUMBER IF VAR VAR NUMBER RETURN VAR VAR IF VAR VAR NUMBER VAR NUMBER RETURN VAR VAR RETURN VAR VAR RETURN VAR VAR FUNC_DEF ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = map(int, input().split())
ans = 0
t = b
l = a - f
x = 0
if b < max(f, l):
ans = -1
else:
while k > 0:
if x == 0:
t -= f
if t < 0:
ans = -1
break
t -= l
if k == 1:
l = 0
if b - l < 0:
ans = -1
break
if t - l < 0:
t = b - l
ans += 1
k -= 1
x = a
else:
t -= l
if t < 0:
ans = -1
break
t -= f
if k == 1:
f = 0
if b - f < 0:
ans = -1
break
if t - f < 0:
t = b - f
ans += 1
k -= 1
x = 0
print(ans)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER WHILE VAR NUMBER IF VAR NUMBER VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
import sys
a, b, f, k = map(int, input().split())
def game_over():
print(-1)
sys.exit()
fuel = b
add_fuel_count = 0
distance_to_stop = f
fuel_on_next_stop = fuel - distance_to_stop
if fuel_on_next_stop < 0:
game_over()
for i in range(1, k + 1):
fuel = fuel_on_next_stop
if i == k:
if i % 2:
distance_to_stop = a - f
else:
distance_to_stop = f
elif i % 2:
distance_to_stop = 2 * (a - f)
else:
distance_to_stop = 2 * f
if fuel < distance_to_stop:
fuel, add_fuel_count = b, add_fuel_count + 1
fuel_on_next_stop = fuel - distance_to_stop
if fuel_on_next_stop < 0:
game_over()
print(add_fuel_count)
|
IMPORT ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR IF VAR VAR IF BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR IF BIN_OP VAR NUMBER ASSIGN VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR BIN_OP NUMBER VAR IF VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
def predict(x, p):
if x == 1:
if p - a - f >= 0:
return "Go"
return "Fuel"
elif p - (a + a - f) >= 0:
return "Go"
else:
return "Fuel"
a, b, f, k = map(int, input().split())
t = 0
r = 0
www = b
for i in range(k):
if i == k - 1:
if b >= a:
continue
if i % 2 == 0:
d = predict(i % 2, b)
if d == "Go":
b -= a
elif b >= f and a - b <= a - f and b >= 0:
b = www
b -= a - f
if b < 0:
r = 1
break
t += 1
else:
r = 1
break
else:
d = predict(i % 2, b)
if d == "Go":
b -= a
elif b >= a - f:
b = www
b -= f
if b < 0:
r = 1
break
t += 1
else:
r = 1
break
if r == 1:
print("-1")
else:
print(t)
|
FUNC_DEF IF VAR NUMBER IF BIN_OP BIN_OP VAR VAR VAR NUMBER RETURN STRING RETURN STRING IF BIN_OP VAR BIN_OP BIN_OP VAR VAR VAR NUMBER RETURN STRING RETURN STRING ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR BIN_OP VAR NUMBER IF VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR IF VAR STRING VAR VAR IF VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR IF VAR STRING VAR VAR IF VAR BIN_OP VAR VAR ASSIGN VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = map(int, input().split())
x = 0
rf = 0
fuel = b
di = 0
j = 0
if fuel < f:
print(-1)
exit()
x = f
fuel -= f
while j != k:
if di == 0:
if j == k - 1:
if fuel >= a - f:
print(rf)
exit()
if fuel >= 2 * (a - f):
fuel = fuel - 2 * (a - f)
di = 1
else:
rf = rf + 1
if b >= a - f and j == k - 1:
print(rf)
exit()
fuel = b - 2 * (a - f)
if fuel < 0:
print(-1)
exit()
di = 1
j += 1
elif di == 1:
if j == k - 1:
if fuel >= f:
print(rf)
exit()
if fuel >= 2 * f:
x = f
di = 0
fuel -= 2 * f
else:
rf = rf + 1
if b >= f and j == k - 1:
print(rf)
exit()
fuel = b - 2 * f
if fuel < 0:
print(-1)
exit()
x = f
di = 0
j += 1
print(rf)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR VAR VAR VAR WHILE VAR VAR IF VAR NUMBER IF VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR IF VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR BIN_OP NUMBER BIN_OP VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR NUMBER VAR NUMBER IF VAR NUMBER IF VAR BIN_OP VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR IF VAR BIN_OP NUMBER VAR ASSIGN VAR VAR ASSIGN VAR NUMBER VAR BIN_OP NUMBER VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR BIN_OP NUMBER VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
def main():
a, b, f, k = list(map(int, input().split()))
fuels = 0
trips = 0
pos = 0
move = 1
gas = b
while trips < k:
if gas < 0:
print(-1)
return
if move == 1:
if pos == 0:
pos = f
gas -= f
elif pos == f:
needed_gas = a - f if trips == k - 1 else 2 * (a - f)
if gas < needed_gas:
gas = b
fuels += 1
gas -= a - f
pos = a
elif pos == a:
trips += 1
if trips == k:
break
move = -1
elif move == -1:
if pos == 0:
trips += 1
if trips == k:
break
move = 1
elif pos == f:
needed_gas = f if trips == k - 1 else 2 * f
if gas < needed_gas:
gas = b
fuels += 1
pos = 0
gas -= f
elif pos == a:
pos = f
gas -= a - f
print(fuels)
main()
|
FUNC_DEF ASSIGN VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR WHILE VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER RETURN IF VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER BIN_OP VAR VAR BIN_OP NUMBER BIN_OP VAR VAR IF VAR VAR ASSIGN VAR VAR VAR NUMBER VAR BIN_OP VAR VAR ASSIGN VAR VAR IF VAR VAR VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER IF VAR NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER VAR BIN_OP NUMBER VAR IF VAR VAR ASSIGN VAR VAR VAR NUMBER ASSIGN VAR NUMBER VAR VAR IF VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = map(int, input().split())
im = False
fuel = b
cnt = 0
for i in range(k):
if i % 2 == 0:
if fuel - f < 0:
im = True
break
if i < k - 1:
if fuel - (2 * a - f) < 0:
cnt += 1
fuel = b - (a - f)
else:
fuel -= a
elif fuel - a < 0:
cnt += 1
fuel = b - (a - f)
else:
fuel -= a
else:
if fuel - (a - f) < 0:
im = True
break
if i < k - 1:
if fuel - (a + f) < 0:
cnt += 1
fuel = b - f
else:
fuel -= a
elif fuel - a < 0:
cnt += 1
fuel = b - f
else:
fuel -= a
if fuel < 0:
im = True
break
print(-1 if im else cnt)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER IF BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR BIN_OP VAR NUMBER IF BIN_OP VAR BIN_OP BIN_OP NUMBER VAR VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR VAR VAR IF BIN_OP VAR VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR VAR VAR IF BIN_OP VAR BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR BIN_OP VAR NUMBER IF BIN_OP VAR BIN_OP VAR VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR IF BIN_OP VAR VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
from sys import stdin, stdout
a, b, f, k = map(int, stdin.readline().split())
ans, position = 0, 0
current = b
label = 1
for i in range(k):
if not position:
distance = a + a - f
curd = a - f
last = a
oil = f
else:
distance = a + f
curd = f
last = 0
oil = a - f
if distance <= current or a <= current and i == k - 1:
current -= a
elif oil <= current:
ans += 1
current = b - curd
else:
current = -1
position = last
if current < 0:
label = 0
if not label:
stdout.write("-1")
else:
stdout.write(str(ans))
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = map(int, input().split())
def distanceToFuel(pos):
return a - f if pos else f
if f > b:
print(-1)
else:
total = k * a
pos = 0
remaining = b
refuel = 0
while total > 0:
remaining -= distanceToFuel(pos)
total -= distanceToFuel(pos)
if k == 1:
if distanceToFuel(pos ^ 1) > b:
refuel = -1
break
elif remaining < distanceToFuel(pos ^ 1):
refuel = 1
break
else:
refuel = 0
break
if total == distanceToFuel(pos ^ 1):
if remaining >= distanceToFuel(pos ^ 1):
break
else:
refuel += 1
break
elif 2 * distanceToFuel(pos ^ 1) > remaining:
if 2 * distanceToFuel(pos ^ 1) > b:
refuel = -1
break
remaining = b
refuel += 1
pos ^= 1
total -= distanceToFuel(pos)
remaining -= distanceToFuel(pos)
print(refuel)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN VAR BIN_OP VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER WHILE VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER IF FUNC_CALL VAR BIN_OP VAR NUMBER VAR ASSIGN VAR NUMBER IF VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER IF BIN_OP NUMBER FUNC_CALL VAR BIN_OP VAR NUMBER VAR IF BIN_OP NUMBER FUNC_CALL VAR BIN_OP VAR NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR VAR VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
import sys
a, b, f, k = list(map(int, input().split()))
d1 = f
d2 = a - f
if k == 1:
if d1 > b or d2 > b:
print("-1")
sys.exit()
elif k == 2:
if d1 > b or 2 * d2 > b:
print("-1")
sys.exit()
elif 2 * d1 > b or 2 * d2 > b:
print("-1")
sys.exit()
rem = b
fill = 0
for i in range(k):
round = i + 1
if round % 2 != 0:
rem -= d1
if round == k and rem >= d2:
rem -= d2
elif round != k and rem >= 2 * d2:
rem -= d2
else:
fill += 1
rem = b
rem -= d2
else:
rem -= d2
if round == k and rem >= d1:
rem -= d1
elif round != k and rem >= 2 * d1:
rem -= d1
else:
fill += 1
rem = b
rem -= d1
print(fill)
|
IMPORT ASSIGN VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER IF VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF VAR NUMBER IF VAR VAR BIN_OP NUMBER VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF BIN_OP NUMBER VAR VAR BIN_OP NUMBER VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR VAR IF VAR VAR VAR VAR VAR VAR IF VAR VAR VAR BIN_OP NUMBER VAR VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR IF VAR VAR VAR VAR VAR VAR IF VAR VAR VAR BIN_OP NUMBER VAR VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
from sys import stdin, stdout
a, b, f, k = list(map(int, stdin.readline().rstrip().split()))
direction = 1
tank = b
refuels = 0
position = 0
if b < f or b < a - f:
possible = False
else:
possible = True
while k > 0 and possible == True:
if k == 1 and tank >= a:
k = 0
elif k == 1:
k = 0
refuels += 1
elif position == 0:
position = f
tank -= f
direction = 1
elif position == a:
position = f
tank -= a - f
direction = -1
elif direction == 1:
if tank < 2 * (a - f):
tank = b
refuels += 1
if b < 2 * (a - f):
possible = False
else:
position = a
direction = -1
tank -= a - f
k -= 1
elif direction == -1:
if tank < 2 * f:
refuels += 1
tank = b
if b < 2 * f:
possible = False
else:
position = 0
direction = 1
tank -= f
k -= 1
if not possible:
print(-1)
else:
print(refuels)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR NUMBER VAR NUMBER IF VAR NUMBER VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR VAR VAR ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER IF VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR VAR VAR NUMBER IF VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER VAR BIN_OP VAR VAR VAR NUMBER IF VAR NUMBER IF VAR BIN_OP NUMBER VAR VAR NUMBER ASSIGN VAR VAR IF VAR BIN_OP NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER VAR VAR VAR NUMBER IF VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
def bus():
[a, b, f, k] = map(int, input().split())
if b < f:
return -1
trips, fuel, pos, refill = 0, b - f, f, 0
while trips < k:
rounds = 2 * (fuel // (2 * a))
fuel -= rounds * a
if pos == f and fuel % (2 * a) >= 2 * (a - f):
rounds += 1
pos = a + f
fuel -= 2 * (a - f)
elif pos == a + f and fuel % (2 * a) >= 2 * f:
rounds += 1
pos = f
fuel -= 2 * f
trips += rounds
if trips == k - 1:
if pos == f and fuel >= a - f or pos == a + f and fuel >= f:
return refill
elif trips >= k:
return refill
if rounds == 0:
if fuel < b:
fuel, refill = b, refill + 1
else:
return -1
else:
fuel, refill = b, refill + 1
return refill
print(bus())
|
FUNC_DEF ASSIGN LIST VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR RETURN NUMBER ASSIGN VAR VAR VAR VAR NUMBER BIN_OP VAR VAR VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP NUMBER BIN_OP VAR BIN_OP NUMBER VAR VAR BIN_OP VAR VAR IF VAR VAR BIN_OP VAR BIN_OP NUMBER VAR BIN_OP NUMBER BIN_OP VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP NUMBER BIN_OP VAR VAR IF VAR BIN_OP VAR VAR BIN_OP VAR BIN_OP NUMBER VAR BIN_OP NUMBER VAR VAR NUMBER ASSIGN VAR VAR VAR BIN_OP NUMBER VAR VAR VAR IF VAR BIN_OP VAR NUMBER IF VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR VAR RETURN VAR IF VAR VAR RETURN VAR IF VAR NUMBER IF VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER RETURN NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER RETURN VAR EXPR FUNC_CALL VAR FUNC_CALL VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = map(int, input().split())
if (f > b or a - f > b) or k > 1 and 2 * (a - f) > b or k > 2 and 2 * f > b:
print(-1)
exit()
cur = b
ans = 0
for i in range(k):
if i & 1 == 0:
if cur < f:
cur = b - f
ans += 1
if cur >= a:
cur -= a
else:
ans += 1
cur = b - a + f
else:
if cur < a - f:
cur = b - a + f
ans += 1
if cur >= a:
cur -= a
else:
ans += 1
cur = b - f
print(ans)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR BIN_OP VAR VAR VAR VAR NUMBER BIN_OP NUMBER BIN_OP VAR VAR VAR VAR NUMBER BIN_OP NUMBER VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER IF VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR NUMBER IF VAR VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR IF VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR VAR NUMBER IF VAR VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
line = input()
thing = list(map(int, line.split()))
a = thing[0]
b = thing[1]
f = thing[2]
k = thing[3]
fuel = b - f
fa = True
refuels = 0
trips = 0
while True:
if fuel < 0:
refuels = -1
break
if fa:
if trips + 1 == k:
distance = a - f
if fuel < distance:
refuels += 1
fuel = b
if b < distance:
refuels = -1
break
break
distance = 2 * (a - f)
if fuel < distance:
refuels += 1
fuel = b
if b < distance:
refuels = -1
break
fuel = fuel - distance
trips += 1
fa = False
if not fa:
if trips + 1 == k:
distance = f
if fuel < distance:
refuels += 1
fuel = b
if b < distance:
refuels = -1
break
break
distance = 2 * f
if fuel < distance:
refuels += 1
fuel = b
if b < distance:
refuels = -1
break
fuel = fuel - distance
trips += 1
fa = True
print(refuels)
|
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE NUMBER IF VAR NUMBER ASSIGN VAR NUMBER IF VAR IF BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR IF VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP NUMBER BIN_OP VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR IF VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR IF BIN_OP VAR NUMBER VAR ASSIGN VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR IF VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP NUMBER VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR IF VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
def read_input():
input_str = str(input())
return map(int, input_str.split(" "))
def check(a: int, b: int, f: int, k: int) -> int:
if k == 1 and (a - f > b or f > b):
return -1
elif k == 2 and ((a - f) * 2 > b or f > b):
return -1
elif k > 2 and ((a - f) * 2 > b or f * 2 > b):
return -1
counter = 0
actual_b = b - f
for i in range(k - 1):
if i % 2 == 0:
if actual_b < (a - f) * 2:
counter += 1
actual_b = b
actual_b -= (a - f) * 2
else:
if actual_b < f * 2:
counter += 1
actual_b = b
actual_b -= f * 2
if k % 2 == 1:
if actual_b < a - f:
counter += 1
actual_b = b
elif actual_b < f:
counter += 1
actual_b = b
return counter
a, b, f, k = read_input()
result = check(a, b, f, k)
print(result)
|
FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR RETURN FUNC_CALL VAR VAR FUNC_CALL VAR STRING FUNC_DEF VAR VAR VAR VAR IF VAR NUMBER BIN_OP VAR VAR VAR VAR VAR RETURN NUMBER IF VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER VAR VAR VAR RETURN NUMBER IF VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER VAR BIN_OP VAR NUMBER VAR RETURN NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP BIN_OP VAR VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR RETURN VAR VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = map(int, input().split())
counter = 0
b2 = b - f
if b2 >= 0:
dist2 = (a - f) * 2
for i in range(k):
if i == k - 1:
dist2 /= 2
if b2 < dist2:
b2 = b
counter += 1
b2 -= dist2
if b2 < 0:
counter = -1
break
dist2 = 2 * a - dist2
print(counter)
else:
print(-1)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR VAR VAR NUMBER VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = [int(i) for i in input().split()]
if k >= 3:
if f > b or 2 * f > b or 2 * (a - f) > b:
print(-1)
else:
fuel = 0
fwd = 1
rem = b - f
while k != 1:
if fwd == 1:
if rem >= 2 * (a - f):
rem = rem - 2 * (a - f)
else:
rem = b - 2 * (a - f)
fuel += 1
k -= 1
fwd = 0
else:
if rem >= 2 * f:
rem = rem - 2 * f
else:
rem = b - 2 * f
fuel += 1
k -= 1
fwd = 1
if fwd == 1:
if rem >= a - f:
print(fuel)
else:
print(fuel + 1)
elif rem >= f:
print(fuel)
else:
print(fuel + 1)
elif k == 1:
if f > b or a - f > b:
print(-1)
elif b >= a:
print(0)
else:
print(1)
elif f > b or 2 * (a - f) > b:
print(-1)
else:
fuel = 0
rem = b - f
if rem < 2 * (a - f):
rem = b - 2 * (a - f)
fuel += 1
else:
rem = rem - 2 * (a - f)
if rem >= f:
print(fuel)
else:
print(fuel + 1)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER IF VAR VAR BIN_OP NUMBER VAR VAR BIN_OP NUMBER BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR WHILE VAR NUMBER IF VAR NUMBER IF VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP NUMBER BIN_OP VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER IF VAR BIN_OP NUMBER VAR ASSIGN VAR BIN_OP VAR BIN_OP NUMBER VAR ASSIGN VAR BIN_OP VAR BIN_OP NUMBER VAR VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER IF VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR NUMBER IF VAR VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR VAR BIN_OP NUMBER BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP NUMBER BIN_OP VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP NUMBER BIN_OP VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
DEBUG = False
def cover_distance():
a, b, f, k = map(int, input().split())
fuel = b
add_fuel_count = 0
fuel_on_next_stop = fuel - f
if fuel_on_next_stop < 0:
return -1
distance_to_destination = 0
distance_to_gas_station = 0
for i in range(1, k + 1):
if DEBUG:
print("i= {} : ".format(i))
print("\t fuel_on_station = {} ".format(fuel_on_next_stop))
if i == k:
if i % 2:
distance_to_destination = a - f
else:
distance_to_destination = f
fuel = fuel_on_next_stop
if fuel < distance_to_destination:
fuel = b
add_fuel_count += 1
fuel_on_next_stop = fuel - distance_to_destination
else:
if i % 2:
distance_to_gas_station = 2 * (a - f)
else:
distance_to_gas_station = 2 * f
fuel = fuel_on_next_stop
if fuel < distance_to_gas_station:
fuel = b
add_fuel_count += 1
fuel_on_next_stop = fuel - distance_to_gas_station
if DEBUG:
if distance_to_destination:
distance = distance_to_destination
else:
distance = distance_to_gas_station
print("\t distance = {} ".format(distance))
print("\t fuel_on_next_stop = {} ".format(fuel_on_next_stop))
if fuel == b:
print("\t YES")
if fuel_on_next_stop < 0:
return -1
return add_fuel_count
print(cover_distance())
|
ASSIGN VAR NUMBER FUNC_DEF ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER RETURN NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR IF VAR VAR IF BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR IF VAR VAR ASSIGN VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF BIN_OP VAR NUMBER ASSIGN VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR BIN_OP NUMBER VAR ASSIGN VAR VAR IF VAR VAR ASSIGN VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR IF VAR ASSIGN VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR IF VAR VAR EXPR FUNC_CALL VAR STRING IF VAR NUMBER RETURN NUMBER RETURN VAR EXPR FUNC_CALL VAR FUNC_CALL VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = map(int, input().split())
ff = 1
if k == 1:
if b >= a:
print(0)
elif b >= f and b >= a - f:
print(1)
else:
print(-1)
else:
p = b - f
ans = 0
for i in range(k - 1):
if i % 2 == 0:
if b < 2 * (a - f):
ff = 0
break
if p < 2 * (a - f):
p = b - 2 * (a - f)
ans += 1
else:
p -= 2 * (a - f)
else:
if b < 2 * f:
ff = 0
break
if p < 2 * f:
p = b - 2 * f
ans += 1
else:
p -= 2 * f
if k % 2 == 0:
if b < f:
ff = 0
if p < f:
ans += 1
elif k % 2 == 1:
if b < a - f:
ff = 0
if p < a - f:
ans += 1
if ff == 0:
print(-1)
else:
print(ans)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER IF VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR NUMBER IF VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP NUMBER BIN_OP VAR VAR VAR NUMBER VAR BIN_OP NUMBER BIN_OP VAR VAR IF VAR BIN_OP NUMBER VAR ASSIGN VAR NUMBER IF VAR BIN_OP NUMBER VAR ASSIGN VAR BIN_OP VAR BIN_OP NUMBER VAR VAR NUMBER VAR BIN_OP NUMBER VAR IF BIN_OP VAR NUMBER NUMBER IF VAR VAR ASSIGN VAR NUMBER IF VAR VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF VAR BIN_OP VAR VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = map(int, input().split())
s, t = 0, b - f
g = f = a - f
for i in range(k):
if t < 0:
break
if k - 1 - i:
g += f
if t < g:
s, t = s + 1, b
t -= g
g = f = a - f
print(-1 if t < 0 else s)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR NUMBER BIN_OP VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR NUMBER IF BIN_OP BIN_OP VAR NUMBER VAR VAR VAR IF VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR NUMBER NUMBER VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = map(int, input().split())
p = 0
r = 0
m = b
if k == 1:
if m < f or m < a - f:
print(-1)
quit()
d1, d2 = 2 * a - f, a + f
for _ in range(k - 1):
if p == 0:
if f > m:
r = -1
break
if m < d1:
r += 1
m = b - a + f
else:
m -= a
p = a
else:
if a - f > m:
r = -1
if m < d2:
r += 1
m = b - f
else:
m -= a
p = 0
if p == 0:
if m < f:
r = -1
elif m < a:
r += 1
elif m < a - f:
r = -1
elif m < a:
r += 1
print(r)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR IF VAR NUMBER IF VAR VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR VAR BIN_OP BIN_OP NUMBER VAR VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR VAR IF BIN_OP VAR VAR VAR ASSIGN VAR NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER IF VAR VAR VAR NUMBER IF VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, c, d = map(int, input().split())
x = b - c
f = 1
if b < c or b < a - c or d >= 2 and b < (a - c) * 2 or d > 2 and b < c * 2:
print(-1)
exit()
k = 0
for i in range(d):
if i + 1 == d:
if f == 1:
if x >= a - c:
print(k)
else:
print(k + 1)
elif x >= c:
print(k)
else:
print(k + 1)
exit()
if f == 1:
if x >= (a - c) * 2:
x = x - (a - c) * 2
else:
x = b - (a - c) * 2
k = k + 1
f = 2
else:
if x >= c * 2:
x = x - c * 2
else:
x = b - c * 2
k = k + 1
f = 1
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF VAR VAR VAR BIN_OP VAR VAR VAR NUMBER VAR BIN_OP BIN_OP VAR VAR NUMBER VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER VAR IF VAR NUMBER IF VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR IF VAR NUMBER IF VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
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a, b, f, k = map(int, input().split())
if max(2 * f, 2 * (a - f)) > b and k > 2:
print("-1")
elif k == 1 and max(f, a - f) > b:
print("-1")
elif k == 2 and (b < 2 * (a - f) or b < f):
print("-1")
else:
m = b
dir = 1
l = 0
pos = 1
x = 0
while l < k:
if pos == 1 and m >= f and dir == 1:
m -= f
dir = 1
pos = 2
elif pos == 2 and dir == 1:
if k - l > 1 and m >= 2 * (a - f):
dir = 2
pos = 2
l += 1
m -= 2 * (a - f)
elif k - l == 1 and m >= a - f:
l += 1
else:
m = b
x += 1
elif pos == 2 and dir == 2:
if k - l > 1 and m >= 2 * f:
dir = 1
pos = 2
l += 1
m -= 2 * f
elif k - l == 1 and m >= f:
l += 1
else:
m = b
x += 1
print(x)
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ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF FUNC_CALL VAR BIN_OP NUMBER VAR BIN_OP NUMBER BIN_OP VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR STRING IF VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR STRING IF VAR NUMBER VAR BIN_OP NUMBER BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR IF VAR NUMBER VAR VAR VAR NUMBER VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR NUMBER IF BIN_OP VAR VAR NUMBER VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER VAR BIN_OP NUMBER BIN_OP VAR VAR IF BIN_OP VAR VAR NUMBER VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER IF VAR NUMBER VAR NUMBER IF BIN_OP VAR VAR NUMBER VAR BIN_OP NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER VAR BIN_OP NUMBER VAR IF BIN_OP VAR VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
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a, b, f, k = tuple(map(int, input().split()))
d1 = f - 0
d2 = a - f
def sim():
x = b
c = 0
for i in range(1, k + 1, 2):
if x < d1:
return -1
x -= d1
if x < d2 or i < k and x < 2 * d2:
x = b
c += 1
if x < d2:
return -1
x -= d2
if i == k:
break
if x < d2:
return -1
x -= d2
if x < d1 or i + 1 < k and x < 2 * d1:
x = b
c += 1
if x < d1:
return -1
x -= d1
return c
print(sim())
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ASSIGN VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR FUNC_DEF ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER IF VAR VAR RETURN NUMBER VAR VAR IF VAR VAR VAR VAR VAR BIN_OP NUMBER VAR ASSIGN VAR VAR VAR NUMBER IF VAR VAR RETURN NUMBER VAR VAR IF VAR VAR IF VAR VAR RETURN NUMBER VAR VAR IF VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP NUMBER VAR ASSIGN VAR VAR VAR NUMBER IF VAR VAR RETURN NUMBER VAR VAR RETURN VAR EXPR FUNC_CALL VAR FUNC_CALL VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
import sys
l, tank, v, rep = (int(x) for x in input().split())
checkpoints = [(l * i + (v if i % 2 == 0 else l - v)) for i in range(rep)]
checkpoints.append(rep * l)
x = 0
gas = tank
cnt = 0
for y in checkpoints:
if y - x > gas:
if y - x > tank:
print(-1)
sys.exit()
cnt += 1
gas = tank
gas -= y - x
x = y
print(cnt)
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IMPORT ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR VAR IF BIN_OP VAR VAR VAR IF BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = map(int, input().split())
fuel = b
res = 0
for i in range(k):
fuel -= f
if fuel < 0:
print(-1)
exit(0)
if i == k - 1:
if a - f > fuel:
fuel = b
res += 1
fuel -= a - f
break
if (a - f) * 2 > fuel:
fuel = b
res += 1
fuel -= a - f
f = a - f
if fuel < 0:
print(-1)
exit(0)
print(res)
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ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR BIN_OP VAR NUMBER IF BIN_OP VAR VAR VAR ASSIGN VAR VAR VAR NUMBER VAR BIN_OP VAR VAR IF BIN_OP BIN_OP VAR VAR NUMBER VAR ASSIGN VAR VAR VAR NUMBER VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
import sys
a, b, f, k = map(int, input().split())
l1 = 2 * (a - f)
l2 = 2 * f
t = b
cnt = 0
r = True
if t >= f:
t -= f
for i in range(k - 1):
if r:
if t >= l1:
t -= l1
r = False
elif b >= l1:
cnt += 1
t = b - l1
r = False
else:
print(-1)
sys.exit()
elif t >= l2:
t -= l2
r = True
elif b >= l2:
cnt += 1
t = b - l2
r = True
else:
print(-1)
sys.exit()
if k % 2 == 1:
if t >= a - f:
print(cnt)
elif b >= a - f:
print(cnt + 1)
else:
print(-1)
elif t >= f:
print(cnt)
elif b >= f:
print(cnt + 1)
else:
print(-1)
else:
print(-1)
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IMPORT ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR BIN_OP NUMBER VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR IF VAR VAR VAR VAR ASSIGN VAR NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR IF VAR VAR VAR VAR ASSIGN VAR NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR IF BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR IF VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
def main():
a, b, f, k = input().split(" ")
endpoint = int(a)
capacity = int(b)
station = int(f)
journeys = int(k)
print(betterBus(endpoint, capacity, station, journeys))
def bus(endpoint, capacity, station, journeys):
refuels = 0
current = capacity
to = True
for j in range(journeys):
r = range(endpoint)
if not to:
r = range(endpoint, 0, -1)
to = True
else:
to = False
for e in r:
if e == station:
if j == journeys - 1:
if current - e < 0:
current = capacity
refuels += 1
elif current - (endpoint - station) * 2 < 0:
current = capacity
refuels += 1
elif current == 0:
return -1
current -= 1
return refuels
def betterBus(endpoint, capacity, station, journeys):
refuels = 0
current = capacity
to = True
for j in range(journeys):
if to:
current -= station
if current < 0:
return -1
if j == journeys - 1:
if current - (endpoint - station) < 0:
current = capacity
if current - (endpoint - station) < 0:
return -1
refuels += 1
elif current - (endpoint - station) * 2 < 0:
current = capacity
refuels += 1
current -= endpoint - station
to = False
else:
current -= endpoint - station
if current < 0:
return -1
if j == journeys - 1:
if current - station < 0:
current = capacity
if current - station < 0:
return -1
refuels += 1
elif current - station * 2 < 0:
current = capacity
refuels += 1
current -= station
to = True
return refuels
main()
|
FUNC_DEF ASSIGN VAR VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR IF VAR BIN_OP VAR NUMBER IF BIN_OP VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER IF BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR VAR VAR NUMBER IF VAR NUMBER RETURN NUMBER VAR NUMBER RETURN VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR IF VAR NUMBER RETURN NUMBER IF VAR BIN_OP VAR NUMBER IF BIN_OP VAR BIN_OP VAR VAR NUMBER ASSIGN VAR VAR IF BIN_OP VAR BIN_OP VAR VAR NUMBER RETURN NUMBER VAR NUMBER IF BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR VAR VAR NUMBER VAR BIN_OP VAR VAR ASSIGN VAR NUMBER VAR BIN_OP VAR VAR IF VAR NUMBER RETURN NUMBER IF VAR BIN_OP VAR NUMBER IF BIN_OP VAR VAR NUMBER ASSIGN VAR VAR IF BIN_OP VAR VAR NUMBER RETURN NUMBER VAR NUMBER IF BIN_OP VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR VAR NUMBER VAR VAR ASSIGN VAR NUMBER RETURN VAR EXPR FUNC_CALL VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
def calculate_minimal_refueling(obj_pos, fuel_capacity, gas_station_pos, trip_required):
direction = True
current_gas = fuel_capacity
refuel_count = 0
for trip in range(trip_required):
if direction:
current_gas -= gas_station_pos
if current_gas < 0:
return -1
elif (
trip + 1 != trip_required
and current_gas < 2 * (obj_pos - gas_station_pos)
or current_gas < obj_pos - gas_station_pos
):
current_gas = fuel_capacity
refuel_count += 1
current_gas -= obj_pos - gas_station_pos
if current_gas < 0:
return -1
direction = False
else:
current_gas -= obj_pos - gas_station_pos
if current_gas < 0:
return -1
elif (
trip + 1 != trip_required
and current_gas < 2 * gas_station_pos
or current_gas < gas_station_pos
):
current_gas = fuel_capacity
refuel_count += 1
current_gas -= gas_station_pos
if current_gas < 0:
return -1
direction = True
return refuel_count
def main():
a, b, f, k = map(int, input().split())
print(calculate_minimal_refueling(a, b, f, k))
main()
|
FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR IF VAR NUMBER RETURN NUMBER IF BIN_OP VAR NUMBER VAR VAR BIN_OP NUMBER BIN_OP VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR VAR NUMBER VAR BIN_OP VAR VAR IF VAR NUMBER RETURN NUMBER ASSIGN VAR NUMBER VAR BIN_OP VAR VAR IF VAR NUMBER RETURN NUMBER IF BIN_OP VAR NUMBER VAR VAR BIN_OP NUMBER VAR VAR VAR ASSIGN VAR VAR VAR NUMBER VAR VAR IF VAR NUMBER RETURN NUMBER ASSIGN VAR NUMBER RETURN VAR FUNC_DEF ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = [int(x) for x in input().split()]
oil = b
num = 0
oil -= f
if b < f or b < a - f or b < 2 * (a - f) and k >= 2 or b < 2 * f and k >= 3:
print(-1)
else:
for i in range(1, k):
if i % 2 == 1:
if oil >= 2 * (a - f):
oil -= 2 * (a - f)
else:
oil = b
num += 1
oil -= 2 * (a - f)
elif oil >= 2 * f:
oil -= 2 * f
else:
oil = b
num += 1
oil -= 2 * f
if k % 2 == 1:
if oil < a - f:
num += 1
elif oil < f:
num += 1
print(num)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR NUMBER VAR VAR IF VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP NUMBER BIN_OP VAR VAR VAR NUMBER VAR BIN_OP NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP NUMBER BIN_OP VAR VAR VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR VAR VAR NUMBER VAR BIN_OP NUMBER BIN_OP VAR VAR IF VAR BIN_OP NUMBER VAR VAR BIN_OP NUMBER VAR ASSIGN VAR VAR VAR NUMBER VAR BIN_OP NUMBER VAR IF BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP VAR VAR VAR NUMBER IF VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
info = [int(num) for num in input().split()]
a = info[0]
b = info[1]
f = info[2]
k = info[3]
gas_spots = [(a * i + (f if i % 2 == 0 else a - f)) for i in range(k)]
gas_spots.append(k * a)
curr_x = 0
curr_gas = b
result = 0
for spot in gas_spots:
to_travel = spot - curr_x
if to_travel > b:
print(-1)
break
elif to_travel > curr_gas:
result += 1
curr_gas = b
curr_gas -= to_travel
curr_x = spot
else:
print(result)
|
ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = list(map(int, input().split()))
refill = b
busPos = 0
counter = 0
shouldPrint = True
while True:
busPos += f
b -= f
if b < 0:
print(-1)
shouldPrint = False
break
if (2 if k > 1 else 1) * (a - f) > b:
b = refill
counter += 1
busPos += a - f
b -= a - f
if b < 0:
print(-1)
shouldPrint = False
break
k -= 1
if k == 0:
break
busPos -= a - f
b -= a - f
if b < 0:
print(-1)
shouldPrint = False
break
if (2 if k > 1 else 1) * f > b:
b = refill
counter += 1
busPos -= f
b -= f
if b < 0:
print(-1)
shouldPrint = False
break
k -= 1
if k == 0:
break
if shouldPrint:
print(counter)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE NUMBER VAR VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER NUMBER NUMBER BIN_OP VAR VAR VAR ASSIGN VAR VAR VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER NUMBER NUMBER VAR VAR ASSIGN VAR VAR VAR NUMBER VAR VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER IF VAR NUMBER IF VAR EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = map(int, input().split())
fuel = b - f
c, d = 0, [a - f << 1, f << 1]
flag = 1
for i in range(k):
if fuel < 0:
flag = 0
break
dis = d[i & 1] * (1 if i != k - 1 else 0.5)
if fuel < dis:
fuel = b
c += 1
fuel -= dis
if fuel < 0:
flag = 0
print(c if flag else "-1")
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR NUMBER LIST BIN_OP BIN_OP VAR VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR ASSIGN VAR VAR VAR NUMBER VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR VAR STRING
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
[a, b, f, k] = list(map(int, input().split()))
def fail():
print("-1")
exit(0)
def success(x):
print(x)
exit(0)
res = 0
r = b - f
while k > 0:
if r < 0:
fail()
if k == 1:
if r < a - f:
res += 1
r = b
if r < a - f:
fail()
success(res)
else:
if r < (a - f) * 2:
res += 1
r = b
r -= (a - f) * 2
if r < 0:
fail()
k -= 1
if k == 1:
if r < f:
res += 1
r = b
if r < f:
fail()
success(res)
else:
if r < f * 2:
res += 1
r = b
r -= f * 2
if r < 0:
fail()
k -= 1
|
ASSIGN LIST VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER FUNC_DEF EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR WHILE VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR IF VAR NUMBER IF VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR VAR IF VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR IF VAR BIN_OP BIN_OP VAR VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR VAR IF VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR IF VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
cin = input().split()
a = int(cin[0])
b = int(cin[1])
f = int(cin[2])
k = int(cin[3])
fuel = [0]
for h in range(k // 2):
fuel += [2 * a * h + f, 2 * a * (h + 1) - f]
if k % 2 == 1 and k != 1:
fuel += [2 * a * (k // 2) + f]
elif k == 1:
fuel += [f]
fuel += [a * k]
try:
d1 = fuel[2] - fuel[1]
d2 = fuel[3] - fuel[2]
except:
d1 = fuel[1] - fuel[0]
d2 = fuel[2] - fuel[1]
if d1 > b or d2 > b:
print(-1)
else:
fuelPointer = 0
gas = b
refuel = 0
while fuelPointer < len(fuel) - 1:
if gas < fuel[fuelPointer + 1] - fuel[fuelPointer]:
refuel += 1
gas = b
gas -= fuel[fuelPointer + 1] - fuel[fuelPointer]
fuelPointer += 1
print(refuel)
|
ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR LIST BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR BIN_OP VAR NUMBER VAR IF BIN_OP VAR NUMBER NUMBER VAR NUMBER VAR LIST BIN_OP BIN_OP BIN_OP NUMBER VAR BIN_OP VAR NUMBER VAR IF VAR NUMBER VAR LIST VAR VAR LIST BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR VAR VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
import sys
a, v, f, k = map(int, input().split())
st_v = v
s1 = f
s2 = a - f
res = 0
for i in range(2 * k):
if i % 2 == 0:
if i % 4 == 0:
v -= s1
if v < 0:
print(-1)
sys.exit()
if i != 2 * k - 2:
if v < 2 * s2:
v = st_v
res += 1
elif v < s2:
v = st_v
res += 1
else:
v -= s2
if v < 0:
print(-1)
sys.exit()
if i != 2 * k - 2:
if v < 2 * s1:
v = st_v
res += 1
elif v < s1:
v = st_v
res += 1
else:
if (i - 1) % 4 == 0:
v -= s2
elif (i - 1) % 4 == 2:
v -= s1
if v < 0:
print(-1)
sys.exit()
print(res)
|
IMPORT ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP NUMBER VAR IF BIN_OP VAR NUMBER NUMBER IF BIN_OP VAR NUMBER NUMBER VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR IF VAR BIN_OP BIN_OP NUMBER VAR NUMBER IF VAR BIN_OP NUMBER VAR ASSIGN VAR VAR VAR NUMBER IF VAR VAR ASSIGN VAR VAR VAR NUMBER VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR IF VAR BIN_OP BIN_OP NUMBER VAR NUMBER IF VAR BIN_OP NUMBER VAR ASSIGN VAR VAR VAR NUMBER IF VAR VAR ASSIGN VAR VAR VAR NUMBER IF BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER VAR VAR IF BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = map(int, input().split())
p = b
i = 0
direct = "p"
ans = 0
flag = 0
while k > 0:
if i == 0 and p < f or i == a and p < a - f:
flag = 1
break
elif i == 0:
p -= f
i = f
elif i == a:
p -= a - f
i = f
else:
if k == 1:
if direct == "p":
if p < a - f and b >= a - f:
p = b
ans += 1
elif b < a - f:
flag = 1
break
elif direct == "n":
if p < f and b >= f:
p = b
ans += 1
elif b < f:
flag = 1
break
elif direct == "p":
if p < 2 * (a - f) and b >= 2 * (a - f):
p = b
ans += 1
elif b < 2 * (a - f):
flag = 1
break
p -= 2 * (a - f)
direct = "n"
elif direct == "n":
if p < 2 * f and b >= 2 * f:
p = b
ans += 1
elif b < 2 * f:
flag = 1
break
p -= 2 * f
direct = "p"
k -= 1
if p < 0:
flag = 1
break
if flag == 1:
print(-1)
else:
print(ans)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR NUMBER IF VAR NUMBER VAR VAR VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER VAR VAR ASSIGN VAR VAR IF VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR IF VAR NUMBER IF VAR STRING IF VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR VAR NUMBER IF VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF VAR STRING IF VAR VAR VAR VAR ASSIGN VAR VAR VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER IF VAR STRING IF VAR BIN_OP NUMBER BIN_OP VAR VAR VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR VAR VAR NUMBER IF VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR NUMBER VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR STRING IF VAR STRING IF VAR BIN_OP NUMBER VAR VAR BIN_OP NUMBER VAR ASSIGN VAR VAR VAR NUMBER IF VAR BIN_OP NUMBER VAR ASSIGN VAR NUMBER VAR BIN_OP NUMBER VAR ASSIGN VAR STRING VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = map(int, input().split())
req = 0
live_fuel = b
while k > 0:
live_fuel -= f
k -= 1
if live_fuel < 0:
req = -1
break
else:
if live_fuel >= 2 * (a - f) or k == 0 and live_fuel >= a - f:
live_fuel -= a - f
else:
live_fuel = b
req += 1
live_fuel -= a - f
if live_fuel < 0:
req = -1
break
else:
if k <= 0:
break
k -= 1
live_fuel -= a - f
if live_fuel < 0:
req = -1
break
else:
if live_fuel >= 2 * f or k == 0 and live_fuel >= f:
live_fuel -= f
else:
live_fuel = b - f
req += 1
if live_fuel < 0:
req = -1
break
print(req)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR VAR WHILE VAR NUMBER VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER IF VAR BIN_OP NUMBER BIN_OP VAR VAR VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR VAR NUMBER VAR BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR NUMBER VAR BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR BIN_OP NUMBER VAR VAR NUMBER VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = map(int, input().split())
r, p, s, t = 0, b - f, 1, [2 * f, 2 * (a - f)]
if p < 0:
r -= 1
else:
for i in range(k - 1):
if p < t[s]:
p = b
r += 1
if p < t[s]:
r = -1
break
p -= t[s]
s = 1 - s
if t[s] // 2 > p:
r += 1
if t[s] // 2 > b:
r = -1
print(r)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR NUMBER BIN_OP VAR VAR NUMBER LIST BIN_OP NUMBER VAR BIN_OP NUMBER BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR VAR ASSIGN VAR VAR VAR NUMBER IF VAR VAR VAR ASSIGN VAR NUMBER VAR VAR VAR ASSIGN VAR BIN_OP NUMBER VAR IF BIN_OP VAR VAR NUMBER VAR VAR NUMBER IF BIN_OP VAR VAR NUMBER VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = map(int, input().split())
cur = b
good = 1
cnt = 0
for i in range(k):
if i % 2 == 0:
if i == k - 1:
if cur < f:
good = 0
elif cur >= a:
cur -= a
cnt += 0
elif b >= a - f:
cnt += 1
else:
good = 0
elif 2 * (a - f) > b:
good = 0
elif cur >= 2 * a - f:
cnt += 0
cur -= a
elif cur < f:
good = 0
else:
cnt += 1
cur = b - (a - f)
elif i == k - 1:
if cur < a - f:
good = 0
elif cur >= a:
cur -= a
cnt += 0
elif b >= f:
cnt += 1
else:
good = 0
elif 2 * f > b:
good = 0
elif cur >= a + f:
cur -= a
cnt += 0
elif cur < a - f:
good = 0
else:
cnt += 1
cur = b - f
if good == 1:
print(cnt)
else:
print(-1)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER IF VAR VAR VAR VAR VAR NUMBER IF VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP NUMBER BIN_OP VAR VAR VAR ASSIGN VAR NUMBER IF VAR BIN_OP BIN_OP NUMBER VAR VAR VAR NUMBER VAR VAR IF VAR VAR ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR IF VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF VAR VAR VAR VAR VAR NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP NUMBER VAR VAR ASSIGN VAR NUMBER IF VAR BIN_OP VAR VAR VAR VAR VAR NUMBER IF VAR BIN_OP VAR VAR ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = [int(i) for i in input().split(" ")]
x = f
y = a - f
oil = b
result = 0
alert = 0
if k == 1:
oil -= x
if oil < 0:
alert = 1
if oil < y:
result += 1
oil = b
oil -= y
if oil < 0:
alert = 1
else:
oil -= x
if oil < 0:
alert = 1
if oil < 2 * y:
result += 1
oil = b
if k == 2:
oil -= 2 * y
if oil < 0:
alert = 1
if oil < x:
result += 1
oil = b
else:
oil -= 2 * y
if oil < 0:
alert = 1
if oil < 2 * x:
result += 1
oil = b
if k > 2:
k -= 2
for ii in range(int((k - 1) / 2)):
oil -= 2 * x
if oil < 0:
alert = 1
if oil < 2 * y:
oil = b
result += 1
k -= 1
oil -= 2 * y
if oil < 0:
alert = 1
if oil < 2 * x:
oil = b
result += 1
k -= 1
if k == 1:
oil -= 2 * x
if oil < 0:
alert = 1
if oil < y:
oil = b
result += 1
if k == 2:
oil -= 2 * x
if oil < 0:
alert = 1
if oil < 2 * y:
oil = b
result += 1
oil -= 2 * y
if oil < x:
oil = b
result += 1
if alert == 1:
print(-1)
else:
print(result)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR BIN_OP NUMBER VAR VAR NUMBER ASSIGN VAR VAR IF VAR NUMBER VAR BIN_OP NUMBER VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR BIN_OP NUMBER VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR BIN_OP NUMBER VAR VAR NUMBER ASSIGN VAR VAR IF VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP NUMBER VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR BIN_OP NUMBER VAR ASSIGN VAR VAR VAR NUMBER VAR NUMBER VAR BIN_OP NUMBER VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR BIN_OP NUMBER VAR ASSIGN VAR VAR VAR NUMBER VAR NUMBER IF VAR NUMBER VAR BIN_OP NUMBER VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR VAR VAR NUMBER IF VAR NUMBER VAR BIN_OP NUMBER VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR BIN_OP NUMBER VAR ASSIGN VAR VAR VAR NUMBER VAR BIN_OP NUMBER VAR IF VAR VAR ASSIGN VAR VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = map(int, input().split())
x = 0
g = b
for kk in range(k - 1):
if kk % 2 == 0:
d = a + (a - f)
if g < d:
if g < f:
x = -1
break
x += 1
g = b - (a - f)
if g < 0:
x = -1
break
else:
g -= a
else:
d = a + f
if g < d:
if g < a - f:
x = -1
break
x += 1
g = b - f
if g < 0:
x = -1
break
else:
g -= a
if (k - 1) % 2 == 0:
if g < a:
if g < f:
x = -1
else:
x += 1
g = b - (a - f)
if g < 0:
x = -1
elif g < a:
if g < a - f:
x = -1
else:
x += 1
g = b - f
if g < 0:
x = -1
print(x)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR IF VAR VAR IF VAR VAR ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR VAR IF VAR BIN_OP VAR VAR ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER VAR VAR IF BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR IF VAR VAR ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR IF VAR BIN_OP VAR VAR ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = map(int, input().split())
if b < f or b < a - f:
print(-1)
exit()
com = b // a
ca = b % a
poi = com % 2
ans = 0
t = 0
while com < k:
t += 1
if poi == 0:
if ca < f:
ca = b - f
ans += 1
else:
ca = b - a + f
ans += 1
com += 1
elif ca < a - f:
ca = b - a + f
ans += 1
else:
ca = b - f
ans += 1
com += 1
com += ca // a
ca = ca % a
poi = com % 2
if t > 1000000:
print(-1)
exit()
print(ans)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR VAR NUMBER IF VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR VAR NUMBER VAR NUMBER IF VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR NUMBER VAR NUMBER VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = input().split()
a = int(a)
b = int(b)
f = int(f)
k = int(k)
i = 0
j = 1
fill = 0
store = b
direct = 1
while j <= k:
if i == 0:
n = f
direct = 1
if j > 1:
store -= abs(f)
elif i == a:
n = f
direct = 0
store -= abs(f - a)
elif direct == 0:
n = 0
store -= abs(f - a)
else:
n = a
store -= abs(f)
if store < 0:
fill = -1
break
if i == f:
if store < 2 * abs(n - i) and j < k:
fill += 1
store = b
elif store < abs(n - i) and j == k:
fill += 1
store = b
if b < abs(n - i):
fill = -1
break
if n == 0 or n == a:
j += 1
i = n
print(fill)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR IF VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER VAR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR IF VAR BIN_OP NUMBER FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR NUMBER ASSIGN VAR VAR IF VAR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR NUMBER ASSIGN VAR VAR IF VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
def solution(a, b, f, k):
d1 = 2 * (a - f)
d2 = 2 * f
if k == 1:
if f > b or a - f > b:
return -1
elif k == 2:
if f > b or d1 > b:
return -1
elif d1 > b or d2 > b:
return -1
fuels = 0
fue = b - f
for i in range(k - 1):
dist = d1 if i % 2 == 0 else d2
if fue < dist:
fuels += 1
fue = b
fue -= dist
dist = a - f if (k - 1) % 2 == 0 else f
if fue < dist:
fuels += 1
return fuels
a, b, f, k = map(int, input().split())
print(solution(a, b, f, k))
|
FUNC_DEF ASSIGN VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR BIN_OP NUMBER VAR IF VAR NUMBER IF VAR VAR BIN_OP VAR VAR VAR RETURN NUMBER IF VAR NUMBER IF VAR VAR VAR VAR RETURN NUMBER IF VAR VAR VAR VAR RETURN NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER BIN_OP VAR VAR VAR IF VAR VAR VAR NUMBER RETURN VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
l = list(map(int, input().split()))
a, fuel, f, k = l[0], l[1], l[2], l[3]
i = 1
b = fuel
c = 0
flag = 0
while i < k:
if i % 2 != 0:
if b >= 2 * a - f:
b -= a
elif b >= f:
b = fuel - (a - f)
c += 1
else:
print("-1")
flag = 1
break
elif b >= a + f:
b -= a
elif b >= a - f:
b = fuel - f
c += 1
else:
print("-1")
flag = 1
break
i += 1
if flag == 0:
if k % 2 == 0:
if b >= a:
b -= a
elif b >= a - f and fuel >= f:
c += 1
else:
flag = 1
print("-1")
elif b >= a:
b -= a
elif b >= f and fuel >= a - f:
c += 1
else:
flag = 1
print("-1")
if flag == 0:
print(c)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR IF BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP BIN_OP NUMBER VAR VAR VAR VAR IF VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR VAR NUMBER EXPR FUNC_CALL VAR STRING ASSIGN VAR NUMBER IF VAR BIN_OP VAR VAR VAR VAR IF VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR NUMBER EXPR FUNC_CALL VAR STRING ASSIGN VAR NUMBER VAR NUMBER IF VAR NUMBER IF BIN_OP VAR NUMBER NUMBER IF VAR VAR VAR VAR IF VAR BIN_OP VAR VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR STRING IF VAR VAR VAR VAR IF VAR VAR VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR STRING IF VAR NUMBER EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
def impossible():
print(-1)
exit()
dest, cap, station, journey = map(int, input().strip().split())
left, right = 2 * station, 2 * (dest - station)
if journey == 1 and (cap < station or cap < dest - station):
impossible()
if journey == 2 and (cap < station or cap < right):
impossible()
if journey >= 3 and (cap < left or cap < right):
impossible()
time = 0
consume = station
for i in range(journey - 1):
go = left if i & 1 else right
if consume + go > cap:
time += 1
consume = go
else:
consume += go
if (
journey & 1 == 0
and consume + station > cap
or journey & 1
and consume + dest - station > cap
):
print(time + 1)
else:
print(time)
|
FUNC_DEF EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR BIN_OP NUMBER VAR BIN_OP NUMBER BIN_OP VAR VAR IF VAR NUMBER VAR VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR IF VAR NUMBER VAR VAR VAR VAR EXPR FUNC_CALL VAR IF VAR NUMBER VAR VAR VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR VAR IF BIN_OP VAR VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER BIN_OP VAR VAR VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
init = list(map(int, input().split()))
finalPoint = init[0]
fuelCapacity = init[1]
refuelPoint = init[2]
voyageCount = init[3]
currentVoyageCount = 0
currentFuelCapacity = fuelCapacity
currentPosition = 0
refuelCount = 0
isEnd = False
error = False
stepSide = 1
def isHaveToRefuel():
global currentFuelCapacity
global refuelCount
if currentVoyageCount + 1 == voyageCount:
if stepSide == 1:
if currentFuelCapacity >= finalPoint - refuelPoint:
return
else:
currentFuelCapacity = fuelCapacity
refuelCount += 1
elif currentFuelCapacity >= refuelPoint - 0:
return
else:
currentFuelCapacity = fuelCapacity
refuelCount += 1
elif stepSide == 1:
if currentFuelCapacity >= (finalPoint - refuelPoint) * 2:
return
else:
currentFuelCapacity = fuelCapacity
refuelCount += 1
elif currentFuelCapacity >= (refuelPoint - 0) * 2:
return
else:
currentFuelCapacity = fuelCapacity
refuelCount += 1
def makeStep():
global currentPosition
global currentFuelCapacity
global error
if currentPosition == refuelPoint:
isHaveToRefuel()
if stepSide == 1:
currentFuelCapacity = currentFuelCapacity - (finalPoint - refuelPoint)
currentPosition = finalPoint
else:
currentFuelCapacity = currentFuelCapacity - (refuelPoint - 0)
currentPosition = 0
elif stepSide == 1:
currentFuelCapacity = currentFuelCapacity - (refuelPoint - 0)
currentPosition = refuelPoint
else:
currentFuelCapacity = currentFuelCapacity - (finalPoint - refuelPoint)
currentPosition = refuelPoint
if currentFuelCapacity < 0:
error = True
while currentVoyageCount < voyageCount:
makeStep()
if error == True:
isEnd = False
break
if currentPosition == 0 or currentPosition == finalPoint:
currentVoyageCount += 1
if currentPosition == 0:
stepSide = 1
elif currentPosition == finalPoint:
stepSide = -1
if currentVoyageCount == voyageCount:
isEnd = True
if isEnd == True:
print(refuelCount)
else:
print("-1")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FUNC_DEF IF BIN_OP VAR NUMBER VAR IF VAR NUMBER IF VAR BIN_OP VAR VAR RETURN ASSIGN VAR VAR VAR NUMBER IF VAR BIN_OP VAR NUMBER RETURN ASSIGN VAR VAR VAR NUMBER IF VAR NUMBER IF VAR BIN_OP BIN_OP VAR VAR NUMBER RETURN ASSIGN VAR VAR VAR NUMBER IF VAR BIN_OP BIN_OP VAR NUMBER NUMBER RETURN ASSIGN VAR VAR VAR NUMBER FUNC_DEF IF VAR VAR EXPR FUNC_CALL VAR IF VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR EXPR FUNC_CALL VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = tuple([int(i) for i in input().split(" ")])
pos = True
route = 0
petrol = 0
if abs(f) > b:
print(-1)
else:
tank = b
need = abs(f)
for i in range(k + 1):
tank -= need
route += need
if i % 2 == 1:
need = abs(f) * 2
else:
need = abs(a - f) * 2
if route + need > a * k:
need = a * k - route
if need > b:
pos = False
break
elif need > tank:
petrol += 1
tank = b
if pos:
print(petrol)
else:
print(-1)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR BIN_OP VAR VAR NUMBER IF BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR IF VAR VAR ASSIGN VAR NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR VAR IF VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = [int(i) for i in input().split()]
ff = [0]
for i in range(k):
d = f
if i % 2 == 1:
d = a - f
ff.append(i * a + d)
ff.append(k * a)
stops = 0
while len(ff) > 0:
next = 0
if ff[0] == k * a:
break
while ff[next + 1] - ff[0] <= b:
next += 1
if ff[next] == k * a:
break
if next == 0:
print("-1")
return
if ff[next] != k * a:
stops += 1
while next > 0:
ff.pop(0)
next -= 1
else:
break
print(stops)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR NUMBER WHILE FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER BIN_OP VAR VAR WHILE BIN_OP VAR BIN_OP VAR NUMBER VAR NUMBER VAR VAR NUMBER IF VAR VAR BIN_OP VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR STRING RETURN IF VAR VAR BIN_OP VAR VAR VAR NUMBER WHILE VAR NUMBER EXPR FUNC_CALL VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
import sys
DEBUG = False
a, b, f, k = map(int, input().split())
def game_over():
print(-1)
sys.exit()
fuel = b
add_fuel_count = 0
fuel_on_next_stop = fuel - f
if fuel_on_next_stop < 0:
game_over()
distance = 0
for i in range(1, k + 1):
fuel = fuel_on_next_stop
if DEBUG:
print("i= {} : ".format(i))
print("\t fuel_on_station = {} ".format(fuel))
if i == k:
if i % 2:
distance = a - f
else:
distance = f
elif i % 2:
distance = 2 * (a - f)
else:
distance = 2 * f
if fuel < distance:
fuel, add_fuel_count = b, add_fuel_count + 1
fuel_on_next_stop = fuel - distance
if fuel_on_next_stop < 0:
game_over()
if DEBUG:
print("\t distance = {} ".format(distance))
print("\t fuel_on_next_stop = {} ".format(fuel_on_next_stop))
if fuel == b:
print("\t YES")
print(add_fuel_count)
|
IMPORT ASSIGN VAR NUMBER ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR IF VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR IF VAR VAR IF BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR IF BIN_OP VAR NUMBER ASSIGN VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR BIN_OP NUMBER VAR IF VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR IF VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
l = [int(x) for x in input().split()]
a = []
b = l[1]
f = 0
c = 0
a.append(0)
for i in range(l[3]):
if i % 2 == 0:
a.append(i * l[0] + l[2])
else:
a.append((i + 1) * l[0] - l[2])
a.append(l[0] * l[3])
for j in range(1, l[3] + 2):
if b >= a[j] - a[j - 1]:
b = b - a[j]
b = b + a[j - 1]
else:
b = l[1]
if b >= a[j] - a[j - 1]:
f = f + 1
b = b - (a[j] - a[j - 1])
else:
c = -1
break
if c == -1:
print(c)
else:
print(f)
|
ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER IF VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = [int(x) for x in input().split()]
curpos = 0
possible = True
gas = b
refuelCount = 0
def dis(a, b):
return abs(a - b)
while k > 0:
if dis(curpos, f) > gas:
possible = False
break
dis_to_edge = dis(f, 0) if curpos == a else dis(f, a)
if k > 1 and dis(curpos, f) + dis_to_edge * 2 > gas:
gas -= dis(curpos, f)
if gas < 0:
possible = False
break
gas = b
refuelCount += 1
gas -= dis_to_edge
if gas < 0:
possible = False
break
curpos = a - curpos
elif k > 1:
curpos = a - curpos
gas -= a
if gas < 0:
possible = False
break
elif gas >= a:
curpos = a - curpos
gas -= a
elif dis(curpos, f) <= gas:
refuelCount += 1
curpos = a - curpos
gas = b
gas -= dis(curpos, f)
if gas < 0:
possible = False
break
else:
possible = False
break
if curpos == 0 or curpos == a:
k -= 1
if possible:
print(refuelCount)
else:
print(-1)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER FUNC_DEF RETURN FUNC_CALL VAR BIN_OP VAR VAR WHILE VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR VAR FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR VAR IF VAR NUMBER BIN_OP FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER VAR VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR VAR NUMBER VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR IF FUNC_CALL VAR VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR VAR VAR NUMBER IF VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = map(int, input().split())
fuel_count = 0
fuel = b
i = 0
fuel -= f
if fuel < 0:
print(-1)
exit(0)
while True:
if fuel >= a - f and i + 1 == k:
break
if b >= a - f and i + 1 == k:
fuel_count += 1
break
elif fuel < 2 * (a - f):
fuel = b
fuel_count += 1
if fuel < 2 * (a - f):
print(-1)
exit(0)
fuel -= 2 * (a - f)
i += 1
if i == k:
break
if fuel >= f and i + 1 == k:
break
if b >= f and i + 1 == k:
fuel_count += 1
break
elif fuel < 2 * f:
fuel = b
fuel_count += 1
if fuel < 2 * f:
print(-1)
exit(0)
fuel -= 2 * f
i += 1
if i == k:
break
print(fuel_count)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER WHILE NUMBER IF VAR BIN_OP VAR VAR BIN_OP VAR NUMBER VAR IF VAR BIN_OP VAR VAR BIN_OP VAR NUMBER VAR VAR NUMBER IF VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR VAR VAR NUMBER IF VAR BIN_OP NUMBER BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER VAR BIN_OP NUMBER BIN_OP VAR VAR VAR NUMBER IF VAR VAR IF VAR VAR BIN_OP VAR NUMBER VAR IF VAR VAR BIN_OP VAR NUMBER VAR VAR NUMBER IF VAR BIN_OP NUMBER VAR ASSIGN VAR VAR VAR NUMBER IF VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER VAR BIN_OP NUMBER VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = map(int, input().strip().split())
st = 0
mod = 0
fue = b
flg = 0
ans = 0
if fue < f:
flg = 1
fue -= f
mod = 1
for i in range(k):
if st == 0:
if mod == 0:
mod = 1
if fue < f:
fue = b
ans += 1
if fue < f:
flg = 1
break
fue -= f
else:
st = 1
mod = 1
if i == k - 1:
if fue < a - f:
fue = b
ans += 1
if fue < a - f:
flg = 1
break
fue -= a - f
continue
if fue < 2 * (a - f):
fue = b
ans += 1
if fue < 2 * (a - f):
flg = 1
break
fue -= 2 * (a - f)
else:
st = 0
mod = 1
if i == k - 1:
if fue < f:
fue = b
ans += 1
if fue < f:
flg = 1
break
fue -= f
continue
if fue < 2 * f:
fue = b
ans += 1
if fue < 2 * f:
flg = 1
break
fue -= 2 * f
if flg == 1:
print(-1)
else:
print(ans)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR VAR VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR VAR ASSIGN VAR VAR VAR NUMBER IF VAR BIN_OP VAR VAR ASSIGN VAR NUMBER VAR BIN_OP VAR VAR IF VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR VAR VAR NUMBER IF VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR NUMBER VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR VAR VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER VAR VAR IF VAR BIN_OP NUMBER VAR ASSIGN VAR VAR VAR NUMBER IF VAR BIN_OP NUMBER VAR ASSIGN VAR NUMBER VAR BIN_OP NUMBER VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = map(int, input().split())
curr, res = b, 0
for i in range(k):
prev = curr
curr -= a
if i % 2 == 0:
if curr < a - f and i < k - 1 or curr < 0:
curr = b - a + f
res += 1
if curr < a - f and i < k - 1 or curr < 0 or prev < f:
print(-1)
exit()
elif curr < f and i < k - 1 or curr < 0:
curr = b - f
res += 1
if curr < f and i < k - 1 or curr < 0 or prev < a - f:
print(-1)
exit()
print(res)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR VAR NUMBER IF VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR IF VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
R = lambda: list(map(int, input().split()))
a, b, f, k = R()
p = b
s = 0
if k == 1:
if f > b or a - f > b:
print(-1)
elif b >= a:
print(0)
else:
print(1)
elif k == 2:
if b < 2 * (a - f) or b < f:
print(-1)
elif b >= 2 * a:
print(0)
elif b >= a + a - f:
print(1)
elif b >= a + f:
print(1)
else:
print(2)
elif b < 2 * (a - f) or b < 2 * f:
print(-1)
else:
for i in range(k):
if i == k - 1:
if b < a:
s += 1
elif i % 2 == 0:
if b >= a + a - f:
b -= a
else:
s += 1
b = p - (a - f)
elif b >= a + f:
b -= a
else:
b = p - f
s += 1
print(s)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER IF VAR VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR NUMBER IF VAR BIN_OP NUMBER BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR NUMBER IF VAR BIN_OP BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR BIN_OP NUMBER BIN_OP VAR VAR VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR BIN_OP VAR NUMBER IF VAR VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP BIN_OP VAR VAR VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR IF VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
import sys
def solve():
end, full, real_mid, times = map(int, sys.stdin.readline().split())
oil = full
cnt = 0
for i in range(times):
if i % 2 == 0:
mid = real_mid
else:
mid = end - real_mid
if mid > oil:
return -1
oil -= mid
if i != times - 1:
if (end - mid) * 2 > full:
return -1
if (end - mid) * 2 > oil:
oil = full
cnt += 1
else:
if end - mid > full:
return -1
if end - mid > oil:
oil = full
cnt += 1
oil -= end - mid
return cnt
print(solve())
|
IMPORT FUNC_DEF ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR VAR RETURN NUMBER VAR VAR IF VAR BIN_OP VAR NUMBER IF BIN_OP BIN_OP VAR VAR NUMBER VAR RETURN NUMBER IF BIN_OP BIN_OP VAR VAR NUMBER VAR ASSIGN VAR VAR VAR NUMBER IF BIN_OP VAR VAR VAR RETURN NUMBER IF BIN_OP VAR VAR VAR ASSIGN VAR VAR VAR NUMBER VAR BIN_OP VAR VAR RETURN VAR EXPR FUNC_CALL VAR FUNC_CALL VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = map(int, input().split())
reFills = 0
journeys = 0
fuel = b - f
def nextStop():
if journeys % 2 == 0:
if journeys == k - 1:
return a - f
return (a - f) * 2
else:
if journeys == k - 1:
return f
return f * 2
if fuel < 0:
print(-1)
else:
while journeys < k:
next_stop = nextStop()
if fuel < next_stop:
reFills += 1
fuel = b
fuel -= next_stop
if fuel < 0:
reFills = -1
break
journeys += 1
print(reFills)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR FUNC_DEF IF BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP VAR NUMBER RETURN BIN_OP VAR VAR RETURN BIN_OP BIN_OP VAR VAR NUMBER IF VAR BIN_OP VAR NUMBER RETURN VAR RETURN BIN_OP VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER WHILE VAR VAR ASSIGN VAR FUNC_CALL VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = map(int, input().split())
t = 0
lt = b
ans = 0
while t < k:
if t & 1:
lt -= a - f
if lt < 0:
exit(print(-1))
if lt < f * 2:
ans += 1
if t == k - 1 and lt >= f:
ans -= 1
lt = b
lt -= f
if lt < 0:
exit(print(-1))
else:
lt -= f
if lt < 0:
exit(print(-1))
if lt < (a - f) * 2:
ans += 1
if t == k - 1 and lt >= a - f:
ans -= 1
lt = b
lt -= a - f
if lt < 0:
exit(print(-1))
t += 1
print(ans)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR IF BIN_OP VAR NUMBER VAR BIN_OP VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR NUMBER VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR NUMBER IF VAR BIN_OP BIN_OP VAR VAR NUMBER VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR
|
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
-----Input-----
The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
-----Output-----
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
-----Examples-----
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
-----Note-----
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
|
a, b, f, k = map(int, input().split())
d = 1
now = 0
res = 0
k += 1
g = b
bb = b
a, b = 0, a
while True:
if now == a or now == b:
k -= 1
g -= abs(now - f)
now = f
if not k or g < 0:
break
elif now == f:
c = b - f if d else f
dis = c if k == 1 else 2 * c
if g >= dis:
g -= c
now = b if d else a
d ^= 1
elif bb < dis:
break
else:
g = bb
res += 1
print(-1 if k else res)
|
ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR NUMBER VAR WHILE NUMBER IF VAR VAR VAR VAR VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR VAR IF VAR VAR NUMBER IF VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR VAR NUMBER VAR BIN_OP NUMBER VAR IF VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR VAR NUMBER IF VAR VAR ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER VAR
|
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