description stringlengths 171 4k | code stringlengths 94 3.98k | normalized_code stringlengths 57 4.99k |
|---|---|---|
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | gans = []
for _ in range(int(input())):
y, x = list(map(int, input().split()))
c2, c3, c4, c5, c6, c1 = list(map(int, input().split()))
for i in range(6):
c1 = min(c1, c6 + c2)
c2 = min(c2, c1 + c3)
c3 = min(c3, c2 + c4)
c4 = min(c4, c3 + c5)
c5 = min(c5, c4 + c6)
c6 = min(c6, c5 + c1)
if x == 0 and y == 0:
gans.append(0)
continue
elif x == 0:
if y > 0:
ans = y * c1
else:
ans = -y * c4
gans.append(ans)
continue
elif y == 0:
if x > 0:
ans = x * c3
else:
ans = -x * c6
gans.append(ans)
continue
if x * y > 0:
if x > 0:
a, b, c = c1, c2, c3
else:
a, b, c = c4, c5, c6
x = -x
y = -y
ans = min(x, y) * b
if x > y:
ans += (x - y) * c
else:
ans += (y - x) * a
gans.append(ans)
else:
if x < 0:
ans = -c6 * x + c1 * y
else:
ans = -c4 * y + c3 * x
gans.append(ans)
print("\n".join(map(str, gans))) | ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR IF VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR IF BIN_OP VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR VAR IF VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | import sys
reader = (s.rstrip() for s in sys.stdin)
input = reader.__next__
def gift():
for _ in range(t):
tx, ty = list(map(int, input().split()))
c1, c2, c3, c4, c5, c6 = list(map(int, input().split()))
dic = [[1, 1], [0, 1], [-1, 0], [-1, -1], [0, -1], [1, 0]]
if tx > 0:
if ty > 0:
if tx > ty:
cost1 = c2 * ty + c6 * tx
cost2 = (tx - ty) * c5 + c1 * tx
cost3 = ty * c1 + (tx - ty) * c6
elif tx < ty:
cost1 = (ty - tx) * c2 + tx * c1
cost2 = ty * c2 + tx * c6
cost3 = ty * c1 + (ty - tx) * c3
else:
cost1 = tx * c1
cost2 = tx * c2 + tx * c6
cost3 = tx * c2 + tx * c6
elif ty == 0:
cost1 = tx * c6
cost2 = tx * c1 + tx * c5
cost3 = tx * c1 + tx * c5
else:
cost1 = abs(ty) * c5 + tx * c6
cost2 = c5 * (tx - ty) + c1 * tx
cost3 = c6 * (tx - ty) + c4 * abs(ty)
elif tx < 0:
if ty > 0:
cost1 = c2 * ty + abs(tx) * c3
cost2 = (ty - tx) * c2 + abs(tx) * c4
cost3 = c1 * ty + (ty - tx) * c3
elif ty == 0:
cost1 = abs(tx) * c3
cost2 = abs(tx) * (c2 + c4)
cost3 = abs(tx) * (c2 + c4)
elif tx > ty:
cost1 = c6 * (tx - ty) + -ty * c4
cost2 = (tx - ty) * c5 + -tx * c4
cost3 = -ty * c5 + c3 * -tx
elif tx < ty:
cost1 = (ty - tx) * c3 + -ty * c4
cost2 = -tx * c3 + -ty * c5
cost3 = -tx * c4 + (ty - tx) * c2
else:
cost1 = -tx * c4
cost2 = -tx * (c3 + c5)
cost3 = -tx * (c3 + c5)
elif ty > 0:
cost1 = c2 * ty
cost2 = (c1 + c3) * ty
cost3 = (c1 + c3) * ty
elif ty == 0:
cost1 = 0
cost2 = 0
cost3 = 0
else:
cost1 = c5 * -ty
cost2 = (c4 + c6) * -ty
cost3 = (c4 + c6) * -ty
yield min(cost1, cost2, cost3)
t = int(input())
ans = gift()
print(*ans, sep="\n") | IMPORT ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR FUNC_DEF FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER IF VAR NUMBER IF VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR IF VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP VAR VAR BIN_OP VAR FUNC_CALL VAR VAR IF VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR IF VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR IF VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR STRING |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | def sol(a1, a2, b1, b2, c1, c2):
d = -b1 * a2 + b2 * a1
d1 = c1 * b2 - c2 * b1
d2 = -c1 * a2 + c2 * a1
x1 = d1 // d
x2 = d2 // d
return x1, x2
m = [[1, 1], [0, 1], [-1, 0], [-1, -1], [0, -1], [1, 0]]
for qq in range(int(input())):
f = [int(s) for s in input().split()]
cost = [int(s) for s in input().split()]
ans = 10**30
for i in range(6):
for j in range(6):
if (j - i) % 3 != 0:
x = sol(m[i][0], m[i][1], m[j][0], m[j][1], f[0], f[1])
if x[0] >= 0 and x[1] >= 0:
t = x[0] * cost[i] + x[1] * cost[j]
ans = min(ans, t)
print(ans) | FUNC_DEF ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR RETURN VAR VAR ASSIGN VAR LIST LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER IF BIN_OP BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER VAR VAR NUMBER VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER VAR NUMBER IF VAR NUMBER NUMBER VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | t = int(input())
while t:
t += -1
x, y = map(int, input().split())
c1, c2, c3, c4, c5, c6 = map(int, input().split())
ans, t2 = 999999999999999999999, 0
if y > 0:
t1 = min(c2 * y, (c1 + c3) * y)
if x > 0:
t2 = min(c6 * x, (c5 + c1) * x)
else:
t2 = min(c3, c4 + c2) * abs(x)
ans = min(ans, t1 + t2)
t1 = min(c1, c6 + c2) * y
if x > y:
t2 = min(c6, c5 + c1) * (x - y)
else:
t2 = min(c3, c4 + c2) * (y - x)
ans = min(ans, t1 + t2)
else:
t1 = min(c5, c4 + c6) * abs(y)
if x > 0:
t2 = min(c6 * x, (c5 + c1) * x)
else:
t2 = min(c3, c4 + c2) * abs(x)
ans = min(ans, t1 + t2)
t1 = min(c4, c5 + c3) * abs(y)
if x > y:
t2 = min(c6, c5 + c1) * (x - y)
else:
t2 = min(c3, c4 + c2) * (y - x)
ans = min(ans, t1 + t2)
if x > 0:
t1 = min(c6, c5 + c1) * x
if y > 0:
t2 = min(c2, c3 + c1) * y
else:
t2 = min(c5, c4 + c6) * abs(y)
ans = min(ans, t1 + t2)
t1 = min(c1, c6 + c2) * x
if y > x:
t2 = min(c2, c3 + c1) * (y - x)
else:
t2 = min(c5, c4 + c6) * (x - y)
ans = min(ans, t1 + t2)
else:
t1 = min(c3, c4 + c2) * abs(x)
if y > 0:
t2 = min(c2, c3 + c1) * y
else:
t2 = min(c5, c4 + c6) * abs(y)
ans = min(ans, t1 + t2)
t1 = min(c4, c5 + c3) * abs(x)
if y > x:
t2 = min(c2, c3 + c1) * (y - x)
else:
t2 = min(c5, c4 + c6) * (x - y)
ans = min(ans, t1 + t2)
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR NUMBER NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR VAR IF VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR VAR IF VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | for nt in range(int(input())):
y, x = list(map(int, input().split()))
c1, c2, c3, c4, c5, c6 = list(map(int, input().split()))
for i in range(10):
c2 = min(c2, c1 + c3)
c3 = min(c3, c2 + c4)
c4 = min(c4, c3 + c5)
c5 = min(c5, c4 + c6)
c6 = min(c6, c5 + c1)
c1 = min(c1, c6 + c2)
if x >= 0:
if y >= 0:
m = min(x, y)
x -= m
y -= m
print(m * c1 + c2 * x + c6 * y)
else:
y = abs(y)
print(x * c2 + y * c3)
else:
x = abs(x)
if y >= 0:
print(x * c5 + y * c6)
else:
y = abs(y)
m = min(x, y)
x -= m
y -= m
print(m * c4 + x * c5 + y * c3) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR IF VAR NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | import sys
input = sys.stdin.readline
t = int(input())
for i in range(t):
x, y = map(int, input().split())
lis = list(map(int, input().split()))
costs = [0] * 6
costs[0] = min(lis[0], lis[1] + lis[5])
costs[1] = min(lis[1], lis[0] + lis[2])
costs[2] = min(lis[2], lis[1] + lis[3])
costs[3] = min(lis[3], lis[2] + lis[4])
costs[4] = min(lis[4], lis[3] + lis[5])
costs[5] = min(lis[5], lis[4] + lis[0])
ans = 0
if x >= 0 and y >= 0:
ans += min(x, y) * costs[0]
if x >= y:
ans += (x - y) * costs[5]
else:
ans += (y - x) * costs[1]
elif x <= 0 and y <= 0:
ans += min(abs(x), abs(y)) * costs[3]
if x <= y:
ans += (y - x) * costs[2]
else:
ans += (x - y) * costs[4]
elif x >= 0 and y <= 0:
ans += x * costs[5]
ans += abs(y) * costs[4]
elif x <= 0 and y >= 0:
ans += abs(x) * costs[2]
ans += abs(y) * costs[1]
print(ans) | IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR VAR VAR NUMBER IF VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR NUMBER VAR BIN_OP BIN_OP VAR VAR VAR NUMBER IF VAR NUMBER VAR NUMBER VAR BIN_OP FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR NUMBER IF VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR NUMBER VAR BIN_OP BIN_OP VAR VAR VAR NUMBER IF VAR NUMBER VAR NUMBER VAR BIN_OP VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | def func(x, c2, c5):
if x < 0:
return -x * c5
else:
return x * c2
def func_2(x, c4, c1):
if x < 0:
return -x * c4
else:
return x * c1
def func_3(x, c6, c3):
if x < 0:
return -x * c3
else:
return x * c6
def main():
y, x = map(int, input().split())
c1, c2, c3, c4, c5, c6 = map(int, input().split())
lu = 1, 0
ru = 1, 1
rd = -1, 0
r = 0, 1
l = 1, 0
ld = -1, -1
a = func_3(y, c6, c3) + func(x, c2, c5)
b = func_2(y, c4, c1) + func(x - y, c2, c5)
c = func_2(x, c4, c1) + func_3(y - x, c6, c3)
print(min(a, b, c))
def __starting_point():
t = int(input())
for i in range(t):
main()
__starting_point() | FUNC_DEF IF VAR NUMBER RETURN BIN_OP VAR VAR RETURN BIN_OP VAR VAR FUNC_DEF IF VAR NUMBER RETURN BIN_OP VAR VAR RETURN BIN_OP VAR VAR FUNC_DEF IF VAR NUMBER RETURN BIN_OP VAR VAR RETURN BIN_OP VAR VAR FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER NUMBER ASSIGN VAR NUMBER NUMBER ASSIGN VAR NUMBER NUMBER ASSIGN VAR NUMBER NUMBER ASSIGN VAR NUMBER NUMBER ASSIGN VAR NUMBER NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | t = int(input())
for you in range(t):
l = input().split()
x = int(l[0])
y = int(l[1])
l = input().split()
c1 = int(l[0])
c2 = int(l[1])
c3 = int(l[2])
c4 = int(l[3])
c5 = int(l[4])
c6 = int(l[5])
if x >= 0 and y >= 0:
mina = c6 * x + c2 * y
if y <= x:
mina = min(mina, c1 * y + c6 * (x - y))
mina = min(mina, c1 * x + c5 * (x - y))
print(mina)
else:
mina = min(mina, c1 * y + c3 * (y - x))
mina = min(mina, c1 * x + c2 * (y - x))
print(mina)
continue
if x <= 0 and y <= 0:
c1, c2, c3, c4, c5, c6 = c4, c5, c6, c1, c2, c3
x = -x
y = -y
mina = c6 * x + c2 * y
if y <= x:
mina = min(mina, c1 * y + c6 * (x - y))
mina = min(mina, c1 * x + c5 * (x - y))
print(mina)
else:
mina = min(mina, c1 * y + c3 * (y - x))
mina = min(mina, c1 * x + c2 * (y - x))
print(mina)
continue
if x <= 0 and y >= 0:
x = -x
mina = c3 * x + c2 * y
mina = min(mina, c1 * y + c3 * (y + x))
mina = min(mina, c4 * x + c2 * (y + x))
print(mina)
continue
y = -y
mina = c6 * x + c5 * y
mina = min(mina, c1 * x + c5 * (y + x))
mina = min(mina, c4 * y + c6 * (y + x))
print(mina) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR 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 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 FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | t = int(input())
for _ in range(t):
x, y = map(int, input().split())
c = list(map(int, input().split()))
for i in range(6):
c[i] = min(c[i], c[(i - 1) % 6] + c[(i + 1) % 6])
if x > 0 and y > 0:
cnt = min(x, y)
print(cnt * c[0] + (x - cnt) * c[5] + (y - cnt) * c[1])
elif x < 0 and y < 0:
x, y = -x, -y
cnt = min(x, y)
print(cnt * c[3] + (x - cnt) * c[2] + (y - cnt) * c[4])
else:
ans = 0
if x < 0:
ans += -x * c[2]
else:
ans += x * c[5]
if y < 0:
ans += -y * c[4]
else:
ans += y * c[1]
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR VAR VAR NUMBER BIN_OP BIN_OP VAR VAR VAR NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR VAR VAR NUMBER BIN_OP BIN_OP VAR VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR VAR NUMBER VAR BIN_OP VAR VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR VAR NUMBER VAR BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | import sys
input = sys.stdin.readline
T = int(input())
for _ in range(T):
n, m = map(int, input().split())
c1, c2, c3, c4, c5, c6 = map(int, input().split())
c1 = min(c1, c2 + c6)
c2 = min(c2, c1 + c3)
c3 = min(c3, c2 + c4)
c4 = min(c4, c3 + c5)
c5 = min(c5, c4 + c6)
c6 = min(c6, c5 + c1)
ans = 0
if m > 0:
if n <= 0:
ans = c3 * -n + c2 * m
elif n <= m:
ans = c1 * n + c2 * (m - n)
else:
ans = c1 * m + c6 * (n - m)
elif m < 0:
if n <= 0:
if n <= m:
ans = c4 * -m + c3 * (-n + m)
else:
ans = c4 * -n + c5 * (n - m)
else:
ans = c6 * n + c5 * -m
elif n <= 0:
ans = c3 * -n
else:
ans = c6 * n
print(ans) | IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR IF VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR IF VAR NUMBER IF VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | T = int(input())
for t in range(T):
x, y = tuple(map(int, input().split()))
c1, c2, c3, c4, c5, c6 = tuple(map(int, input().split()))
h_p = min(c2, c1 + c3)
h_n = min(c5, c6 + c4)
v_p = min(c6, c1 + c5)
v_n = min(c3, c2 + c4)
d_p = min(c1, c2 + c6)
d_n = min(c4, c3 + c5)
h = h_p if y > 0 else h_n
v = v_p if x > 0 else v_n
cost = abs(x * v) + abs(y * h)
d = d_p if x > 0 else d_n
h = h_p if x < y else h_n
cost = min(cost, abs(x * d) + abs(abs(x - y) * h))
d = d_p if y > 0 else d_n
v = v_p if y < x else v_n
cost = min(cost, abs(y * d) + abs(abs(x - y) * v))
print(cost) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR NUMBER VAR VAR ASSIGN VAR VAR NUMBER VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR BIN_OP VAR VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR VAR NUMBER VAR VAR ASSIGN VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR BIN_OP VAR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR BIN_OP VAR VAR VAR ASSIGN VAR VAR NUMBER VAR VAR ASSIGN VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR BIN_OP VAR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | t = int(input())
def process():
x, y = map(int, input().split())
c = list(map(int, input().split()))
ans = 0
if x > 0 and y > 0:
d = min(x, y)
if c[0] < c[5] + c[1]:
ans += d * c[0]
x -= d
y -= d
if x < 0 and y < 0:
d = abs(max(x, y))
if c[3] < c[4] + c[2]:
ans += d * c[3]
x += d
y += d
if y > 0:
if c[1] < c[0] + c[2]:
ans += abs(y) * c[1]
else:
ans += abs(y) * (c[0] + c[2])
if y < 0:
if c[4] < c[3] + c[5]:
ans += abs(y) * c[4]
else:
ans += abs(y) * (c[3] + c[5])
if x < 0:
if c[2] < c[3] + c[1]:
ans += abs(x) * c[2]
else:
ans += abs(x) * (c[3] + c[1])
if x > 0:
if c[5] < c[4] + c[0]:
ans += abs(x) * c[5]
else:
ans += abs(x) * (c[4] + c[0])
print(ans)
for it in range(t):
process() | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER VAR BIN_OP VAR VAR NUMBER VAR VAR VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER VAR BIN_OP VAR VAR NUMBER VAR VAR VAR VAR IF VAR NUMBER IF VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR NUMBER IF VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR NUMBER IF VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR NUMBER IF VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | t = int(input())
for _ in range(t):
x, y = map(int, input().split())
cost = list(map(int, input().split()))
truecost = [0] * 6
for i in range(6):
truecost[i] = min(cost[i], cost[(i - 1) % 6] + cost[(i + 1) % 6])
ans = 0
if x > 0 and y > 0:
w = min(x, y)
ans += w * truecost[0]
x -= w
y -= w
if x < 0 and y < 0:
w = min(-x, -y)
ans += w * truecost[3]
x += w
y += w
if x < 0:
ans += -x * truecost[2]
x = 0
if y < 0:
ans += -y * truecost[4]
y = 0
if x > 0:
ans += x * truecost[5]
x = 0
if y > 0:
ans += y * truecost[1]
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR VAR NUMBER VAR VAR VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR VAR NUMBER VAR VAR VAR VAR IF VAR NUMBER VAR BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | t = int(input())
while t > 0:
y, x = map(int, input().split())
c1, c2, c3, c4, c5, c6 = map(int, input().split())
rh = min(c2, c1 + c3)
lh = min(c5, c4 + c6)
uv = min(c6, c1 + lh)
lv = min(c3, rh + c4)
ud = min(c1, uv + rh)
ld = min(c4, lv + lh)
a = 0
if y > 0:
rs = y * ud
ls = y * uv
dl = x - y
if dl > 0:
rs += dl * rh
else:
rs += abs(dl) * lh
if x > 0:
ls += x * rh
else:
ls += abs(x) * lh
if x <= y and x >= 0:
print(min(rs, ls, x * ud + (y - x) * uv))
else:
print(min(rs, ls))
else:
rs = abs(y) * ld
ls = abs(y) * lv
dl = x - y
if dl > 0:
rs += dl * rh
else:
rs += abs(dl) * lh
if x > 0:
ls += x * rh
else:
ls += abs(x) * lh
if x <= 0 and x >= y:
print(min(rs, ls, abs(x) * ld + (abs(y) - abs(x)) * lv))
else:
print(min(rs, ls))
t -= 1 | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR IF VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR IF VAR VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR IF VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR IF VAR NUMBER VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR BIN_OP BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR NUMBER |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | import sys
input = sys.stdin.readline
def main():
t = int(input())
for _ in range(t):
X, Y = [int(x) for x in input().split()]
C1, C2, C3, C4, C5, C6 = [int(x) for x in input().split()]
ans = float("inf")
if X == 0 and Y == 0:
print(0)
continue
elif X == 0:
if Y > 0:
ans = min(ans, C2 * Y)
ans = min(ans, C3 * Y + C1 * Y)
else:
Y = abs(Y)
ans = min(ans, C5 * Y)
ans = min(ans, C4 * Y + C6 * Y)
elif Y == 0:
if X > 0:
ans = min(ans, C6 * X)
ans = min(ans, C1 * X + C5 * X)
else:
X = abs(X)
ans = min(ans, C3 * X)
ans = min(ans, C2 * X + C4 * X)
elif X > 0 and Y > 0:
ans = min(ans, C6 * X + C2 * Y)
if X == Y:
ans = min(ans, C1 * X)
elif X > Y:
ans = min(ans, C1 * X + C5 * (X - Y))
ans = min(ans, C1 * Y + C6 * (X - Y))
else:
ans = min(ans, C1 * X + C2 * (Y - X))
ans = min(ans, C1 * Y + C3 * (Y - X))
elif X > 0 and Y < 0:
X = abs(X)
Y = abs(Y)
ans = min(ans, C6 * X + C5 * Y)
ans = min(ans, C1 * X + C5 * (Y + X))
ans = min(ans, C4 * Y + C6 * (Y + X))
elif X < 0 and Y > 0:
X = abs(X)
Y = abs(Y)
ans = min(ans, C3 * X + C2 * Y)
ans = min(ans, C4 * X + C2 * (Y + X))
ans = min(ans, C1 * Y + C3 * (Y + X))
elif X < 0 and Y < 0:
X = abs(X)
Y = abs(Y)
ans = min(ans, C3 * X + C5 * Y)
if X == Y:
ans = min(ans, C4 * X)
elif X > Y:
ans = min(ans, C4 * X + C2 * (X - Y))
ans = min(ans, C4 * Y + C3 * (X - Y))
else:
ans = min(ans, C4 * X + C5 * (Y - X))
ans = min(ans, C4 * Y + C6 * (Y - X))
print(ans)
main() | IMPORT ASSIGN VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR STRING IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR IF VAR NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | def main():
n = int(input())
for i in range(n):
x, y = map(int, input().split())
c1, c2, c3, c4, c5, c6 = map(int, input().split())
answer = 0
if x > 0:
answer += x * c6
else:
answer += -x * c3
if y > 0:
answer += y * c2
else:
answer += -y * c5
m = min(x, y)
ma = max(x, y)
if x > 0 or y > 0:
if m > 0:
answer = min(answer, m * c1 + (x - m) * c6 + (y - m) * c2)
answer = min(answer, ma * c1 + (ma - x) * c3 + (ma - y) * c5)
if x < 0 or y < 0:
if ma < 0:
answer = min(answer, -ma * c4 + (ma - x) * c3 + (ma - y) * c5)
answer = min(answer, -m * c4 + (x - m) * c6 + (y - m) * c2)
print(answer)
main() | FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR IF VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER VAR NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR IF VAR NUMBER VAR NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | import sys
def minp():
return sys.stdin.readline().strip()
def mint():
return int(minp())
def mints():
return map(int, minp().split())
masks = []
dirs = [(1, 1), (0, 1), (-1, 0), (-1, -1), (0, -1), (1, 0)]
for i in range(1 << 6):
bad = False
for j in range(3):
if (i >> j & 1) + (i >> j + 3 & 1) != 1:
bad = True
if not bad:
if i & 1 != 0:
i1 = 0
else:
i1 = 3
if i & 2 != 0:
i2 = 1
else:
i2 = 1 + 3
if i & 4 != 0:
i3 = 2
else:
i3 = 2 + 3
masks.append((i, dirs[i1], dirs[i2], dirs[i3], i1, i2, i3))
def solve():
x, y = mints()
c = list(mints())
best = int(1e30)
for i, v1, v2, v3, i1, i2, i3 in masks:
d1 = v2[0] * v3[1] - v2[1] * v3[0]
tx, ty = x, y
k1 = (tx * v3[1] - ty * v3[0]) // d1
k2 = (ty * v2[0] - tx * v2[1]) // d1
tx, ty = x - v1[0], y - v1[1]
k1v = (tx * v3[1] - ty * v3[0]) // d1
k2v = (ty * v2[0] - tx * v2[1]) // d1
dk1 = k1v - k1
dk2 = k2v - k2
ck1 = -k1 // dk1
ck2 = -k2 // dk2
if dk1 > 0:
mk1 = ck1
Mk1 = int(2e20)
else:
mk1 = int(0)
Mk1 = ck1
if dk2 > 0:
mk2 = ck2
Mk2 = int(2e20)
else:
mk2 = int(0)
Mk2 = ck2
m = max(max(mk1, mk2), 0)
M = min(Mk1, Mk2)
if m <= M:
tx, ty = x - v1[0] * m, y - v1[1] * m
k1 = (tx * v3[1] - ty * v3[0]) // d1
k2 = (ty * v2[0] - tx * v2[1]) // d1
best = min(best, m * c[i1] + k1 * c[i2] + k2 * c[i3])
tx, ty = x - v1[0] * M, y - v1[1] * M
k1 = (tx * v3[1] - ty * v3[0]) // d1
k2 = (ty * v2[0] - tx * v2[1]) // d1
best = min(best, M * c[i1] + k1 * c[i2] + k2 * c[i3])
print(best)
for i in range(mint()):
solve() | IMPORT FUNC_DEF RETURN FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER IF BIN_OP BIN_OP BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER IF VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR NUMBER FOR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER BIN_OP VAR VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER BIN_OP VAR VAR NUMBER VAR ASSIGN VAR VAR BIN_OP VAR VAR NUMBER BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER BIN_OP VAR VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER BIN_OP VAR VAR NUMBER VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER BIN_OP VAR VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER BIN_OP VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER BIN_OP VAR VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER BIN_OP VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | import sys
sys.setrecursionlimit(10**8)
def solve(curx, cury, cost):
if curx == tx and cury == ty:
return cost
elif curx == tx:
if ty > cury:
return cost + c2 * (ty - cury)
else:
return cost + c5 * (cury - ty)
elif cury == ty:
if tx > curx:
return cost + c6 * (tx - curx)
else:
return cost + c3 * (curx - tx)
elif tx > curx and ty > cury:
diff = min(tx - curx, ty - cury)
return solve(curx + diff, cury + diff, cost + c1 * diff)
elif tx > curx and ty < cury:
return cost + c6 * (tx - curx) + c5 * (cury - ty)
elif tx < curx and ty < cury:
diff = min(curx - tx, cury - ty)
return solve(curx - diff, cury - diff, cost + c4 * diff)
elif tx < curx and ty > cury:
return cost + c3 * (curx - tx) + c2 * (ty - cury)
for _ in range(int(input())):
tx, ty = map(int, input().split())
c1, c2, c3, c4, c5, c6 = map(int, input().split())
c1 = min(c1, c2 + c6)
c2 = min(c2, c1 + c3)
c3 = min(c3, c2 + c4)
c4 = min(c4, c3 + c5)
c5 = min(c5, c6 + c4)
c6 = min(c6, c1 + c5)
print(solve(0, 0, 0)) | IMPORT EXPR FUNC_CALL VAR BIN_OP NUMBER NUMBER FUNC_DEF IF VAR VAR VAR VAR RETURN VAR IF VAR VAR IF VAR VAR RETURN BIN_OP VAR BIN_OP VAR BIN_OP VAR VAR RETURN BIN_OP VAR BIN_OP VAR BIN_OP VAR VAR IF VAR VAR IF VAR VAR RETURN BIN_OP VAR BIN_OP VAR BIN_OP VAR VAR RETURN BIN_OP VAR BIN_OP VAR BIN_OP VAR VAR IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR BIN_OP VAR VAR RETURN FUNC_CALL VAR BIN_OP VAR VAR BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR IF VAR VAR VAR VAR RETURN BIN_OP BIN_OP VAR BIN_OP VAR BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR BIN_OP VAR VAR RETURN FUNC_CALL VAR BIN_OP VAR VAR BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR IF VAR VAR VAR VAR RETURN BIN_OP BIN_OP VAR BIN_OP VAR BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR NUMBER NUMBER NUMBER |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | for _ in range(int(input())):
y, x = map(int, input().split())
d = abs(x - y)
ans = 10000000000000000000000000000
c = [0] + list(map(int, input().split()))
if x >= 0 and y >= 0:
path1 = c[2] * x + c[6] * y
path2 = c[1] * x + c[3] * d
path3 = c[1] * y + c[2] * d
if x < y:
path2 = c[1] * y + c[5] * d
path3 = c[1] * x + c[6] * d
ans = min(ans, path1)
ans = min(ans, path2)
ans = min(ans, path3)
elif x < 0 and y >= 0:
path1 = c[5] * -x + c[6] * y
path2 = c[1] * y + c[5] * d
path3 = c[4] * -x + c[6] * d
ans = min(ans, path1)
ans = min(ans, path2)
ans = min(ans, path3)
elif x < 0 and y < 0:
path1 = c[5] * -x + c[3] * -y
path2 = c[4] * -x + c[6] * d
path3 = c[4] * -y + c[5] * d
if x > y:
path2 = c[4] * -y + c[2] * d
path3 = c[4] * -x + c[3] * d
ans = min(ans, path1)
ans = min(ans, path2)
ans = min(ans, path3)
elif x >= 0 and y < 0:
path1 = c[2] * x + c[3] * -y
path2 = c[1] * x + c[3] * d
path3 = c[4] * -y + c[2] * d
ans = min(ans, path1)
ans = min(ans, path2)
ans = min(ans, path3)
print(ans) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR IF VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR IF VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | def fun(x, y):
global c1, c2, c3, c4, c5, c6
val = 0
if x >= 0:
val += c6 * x
else:
val += c3 * abs(x)
if y >= 0:
val += c2 * y
else:
val += c5 * abs(y)
return val
def fun1(x):
global c1, c2, c3, c4, c5, c6
val = 0
if x >= 0:
val += c1 * x
else:
val += c4 * abs(x)
return val
for _ in range(int(input())):
x, y = map(int, input().split())
c1, c2, c3, c4, c5, c6 = map(int, input().split())
ans = []
tem = 0
tem += fun(x, y)
ans.append(tem)
tem = 0
tem += fun1(x)
dis = y - x
tem += fun(0, dis)
ans.append(tem)
tem = 0
tem += fun1(y)
dis = x - y
tem += fun(dis, 0)
ans.append(tem)
print(min(ans)) | FUNC_DEF ASSIGN VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR FUNC_CALL VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR FUNC_CALL VAR VAR RETURN VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR FUNC_CALL VAR NUMBER VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | def calc():
global c1, c2, c3, c4, c5, c6, x1, x2, x3, x4, x5, x6
return x1 * c1 + x2 * c2 + x3 * c3 + x4 * c4 + x5 * c5 + x6 * c6
for _ in range(int(input())):
x, y = map(int, input().split())
c1, c2, c3, c4, c5, c6 = map(int, input().split())
ans = []
if x > 0 and y > 0:
x1, x2, x3, x4, x5, x6 = 0, y, 0, 0, 0, x
elif x > 0:
x1, x2, x3, x4, x5, x6 = 0, 0, 0, 0, -y, x
elif y > 0:
x1, x2, x3, x4, x5, x6 = 0, y, -x, 0, 0, 0
else:
x1, x2, x3, x4, x5, x6 = 0, 0, -x, 0, -y, 0
ans.append(calc())
if x > 0 and y - x > 0:
x1, x2, x3, x4, x5, x6 = x, y - x, 0, 0, 0, 0
elif y - x > 0:
x1, x2, x3, x4, x5, x6 = 0, y - x, 0, -x, 0, 0
elif x > 0:
x1, x2, x3, x4, x5, x6 = x, 0, 0, 0, x - y, 0
else:
x1, x2, x3, x4, x5, x6 = 0, 0, 0, -x, x - y, 0
ans.append(calc())
if y > 0 and x - y > 0:
x1, x2, x3, x4, x5, x6 = y, 0, 0, 0, 0, x - y
elif x - y > 0:
x1, x2, x3, x4, x5, x6 = 0, 0, 0, -y, 0, x - y
elif y > 0:
x1, x2, x3, x4, x5, x6 = y, 0, y - x, 0, 0, 0
else:
x1, x2, x3, x4, x5, x6 = 0, 0, y - x, -y, 0, 0
ans.append(calc())
print(min(ans)) | FUNC_DEF RETURN BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST IF VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER NUMBER NUMBER VAR IF VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR NUMBER NUMBER NUMBER NUMBER VAR VAR IF VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR NUMBER VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR VAR VAR VAR VAR NUMBER NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR IF VAR NUMBER BIN_OP VAR VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR NUMBER NUMBER NUMBER NUMBER IF BIN_OP VAR VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR NUMBER BIN_OP VAR VAR NUMBER VAR NUMBER NUMBER IF VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR VAR NUMBER NUMBER NUMBER BIN_OP VAR VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR NUMBER NUMBER NUMBER VAR BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR IF VAR NUMBER BIN_OP VAR VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR VAR NUMBER NUMBER NUMBER NUMBER BIN_OP VAR VAR IF BIN_OP VAR VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR NUMBER NUMBER NUMBER VAR NUMBER BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR VAR NUMBER BIN_OP VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR VAR VAR VAR VAR NUMBER NUMBER BIN_OP VAR VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | n = int(input())
for _ in range(n):
y, x = map(int, input().split())
costs = list(map(int, input().split()))
for i in range(36):
costs[i % 6] = min(costs[i % 6], costs[(i + 5) % 6] + costs[(i + 1) % 6])
ans = 0
if x >= 0:
if y >= 0:
if x >= y:
ans = (x - y) * costs[1] + y * costs[0]
else:
ans = x * costs[0] + (y - x) * costs[5]
else:
ans = x * costs[1] - y * costs[2]
elif y >= 0:
ans = -x * costs[4] + y * costs[5]
elif x <= y:
ans = (y - x) * costs[4] - y * costs[3]
else:
ans = -x * costs[3] + (x - y) * costs[2]
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER IF VAR NUMBER IF VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP VAR VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | def cost(d, c1, c2):
return d * (c1 if d > 0 else -c2)
for _ in range(int(input())):
x, y = map(int, input().split())
c = [0] + list(map(int, input().split()))
opt = []
opt.append(cost(x - y, c[6], c[3]) + cost(y, c[1], c[4]))
opt.append(cost(x, c[6], c[3]) + cost(y, c[2], c[5]))
opt.append(cost(x, c[1], c[4]) + cost(y - x, c[2], c[5]))
print(min(opt)) | FUNC_DEF RETURN BIN_OP VAR VAR NUMBER VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR BIN_OP VAR VAR VAR NUMBER VAR NUMBER FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER FUNC_CALL VAR BIN_OP VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | import sys
input = sys.stdin.readline
for f in range(int(input())):
x, y = map(int, input().split())
c1, c2, c3, c4, c5, c6 = map(int, input().split())
c2 = min(c2, c1 + c3)
c5 = min(c5, c4 + c6)
c6 = min(c6, c1 + c5)
c3 = min(c3, c2 + c4)
c1 = min(c1, c2 + c6)
c4 = min(c4, c3 + c5)
if x * y > 0:
if x > 0:
d = min(x, y)
x -= d
y -= d
print(x * c6 + y * c2 + d * c1)
else:
x = -x
y = -y
d = min(x, y)
x -= d
y -= d
print(x * c3 + y * c5 + d * c4)
else:
cost = 0
if x > 0:
cost += x * c6
else:
cost -= x * c3
if y > 0:
cost += y * c2
else:
cost -= y * c5
print(cost) | IMPORT ASSIGN VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR IF BIN_OP VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR IF VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | t = int(input())
for _ in range(t):
x, y = map(int, input().split(" "))
c = [int(val) for val in input().split(" ")]
if x == 0 and y == 0:
print(0)
elif x == 0:
if y > 0:
print(min(c[1] * y, (c[0] + c[2]) * y))
else:
print(min(c[4] * -y, (c[3] + c[5]) * -y))
elif y == 0:
if x > 0:
print(min(c[5] * x, (c[0] + c[4]) * x))
else:
print(min(c[2] * -x, (c[1] + c[3]) * -x))
elif x > 0:
if x == y:
print(min(c[0] * x, (c[1] + c[5]) * x))
elif y > x:
print(
min(
c[0] * x + c[1] * (y - x),
c[1] * y + c[5] * x,
c[0] * y + c[2] * (y - x),
)
)
elif y < 0:
print(
min(
c[5] * x + c[4] * -y,
c[5] * (x - y) + c[3] * -y,
c[4] * (x - y) + c[0] * x,
)
)
else:
print(
min(
c[5] * (x - y) + c[0] * y,
c[1] * y + c[5] * x,
c[4] * (x - y) + c[0] * x,
)
)
elif x < 0:
if x == y:
print(min(c[3] * -x, (c[2] + c[4]) * -x))
elif x > y:
print(
min(
c[3] * -x + c[4] * (x - y),
c[4] * -y + c[2] * -x,
c[3] * -y + c[5] * (x - y),
)
)
elif y > 0:
print(
min(
c[2] * -x + c[1] * y,
c[2] * (y - x) + c[0] * y,
c[1] * (y - x) + c[3] * -x,
)
)
else:
print(
min(
c[2] * (y - x) + c[3] * -y,
c[4] * -y + c[2] * -x,
c[1] * (y - x) + c[3] * -x,
)
) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR NUMBER VAR EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR NUMBER VAR IF VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR NUMBER VAR EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR NUMBER VAR IF VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR NUMBER VAR IF VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER BIN_OP VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR VAR BIN_OP VAR NUMBER VAR IF VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR NUMBER VAR IF VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER BIN_OP VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR VAR BIN_OP VAR NUMBER VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | t = int(input())
for _ in range(0, t):
x, y = map(int, input().split())
c1, c2, c3, c4, c5, c6 = map(int, input().split())
c11 = min(c1, c2 + c6)
c22 = min(c2, c3 + c1)
c33 = min(c3, c4 + c2)
c44 = min(c4, c5 + c3)
c55 = min(c5, c6 + c4)
c66 = min(c6, c1 + c5)
ss = 0
if x >= 0 and y >= 0:
mm = min(x, y)
ss += mm * c11
x -= mm
y -= mm
if x > 0:
ss += x * c66
else:
ss += y * c22
elif x <= 0 and y >= 0:
ss += -1 * x * c33
ss += y * c22
elif x <= 0 and y <= 0:
x, y = -1 * x, -1 * y
mm = min(x, y)
ss += mm * c44
x -= mm
y -= mm
if x > 0:
ss += x * c33
else:
ss += y * c55
elif x >= 0 and y <= 0:
ss += x * c66
ss += -1 * y * c55
print(ss) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR VAR IF VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR VAR VAR BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR VAR BIN_OP NUMBER VAR BIN_OP NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR VAR IF VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP BIN_OP NUMBER VAR VAR EXPR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | for _ in range(int(input())):
x, y = map(int, input().split())
l = list(map(int, input().split()))
c1, c2, c3, c4, c5, c6 = l[0], l[5], l[4], l[3], l[2], l[1]
if x == 0 and y == 0:
print(0)
elif y == 0:
if x > 0:
if c1 + c3 < c2:
print(x * (c1 + c3))
else:
print(x * c2)
elif c4 + c6 < c5:
print(abs(x) * (c4 + c6))
else:
print(abs(x) * c5)
elif x == 0:
if y > 0:
if c1 + c5 < c6:
print(y * (c1 + c5))
else:
print(c6 * y)
elif c4 + c2 < c3:
print(abs(y) * (c4 + c2))
else:
print(abs(y) * c3)
elif x > 0 and y > 0:
if x > y:
li = []
li.append(c2 * x + c6 * y)
li.append(c1 * x + c3 * (x - y))
li.append(c1 * y + c2 * (x - y))
print(min(li))
else:
li = []
li.append(c2 * x + c6 * y)
li.append(c1 * x + c6 * (y - x))
li.append(c1 * y + c5 * (y - x))
print(min(li))
elif x < 0 and y < 0:
x = abs(x)
y = abs(y)
c1, c4 = c4, c1
c2, c5 = c5, c2
c3, c6 = c6, c3
if x > y:
li = []
li.append(c2 * x + c6 * y)
li.append(c1 * x + c3 * (x - y))
li.append(c1 * y + c2 * (x - y))
print(min(li))
else:
li = []
li.append(c2 * x + c6 * y)
li.append(c1 * x + c6 * (y - x))
li.append(c1 * y + c5 * (y - x))
print(min(li))
elif y > 0 and x < 0:
x = abs(x)
li = []
li.append(c6 * y + x * c5)
li.append(y * c1 + (x + y) * c5)
li.append(x * c4 + (y + x) * c6)
print(min(li))
else:
y = abs(y)
li = []
li.append(c3 * y + c2 * x)
li.append(x * c1 + (x + y) * c3)
li.append(y * c4 + (x + y) * c2)
print(min(li)) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR NUMBER IF VAR NUMBER IF BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR IF BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR IF VAR NUMBER IF VAR NUMBER IF BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR IF BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR IF VAR NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR LIST EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR LIST EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR IF VAR VAR ASSIGN VAR LIST EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR LIST EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | def xyc(x, y):
an1 = 0
if x > 0:
an1 += cos[5] * x
else:
an1 += cos[2] * -x
if y > 0:
an1 += cos[1] * y
else:
an1 += cos[4] * -y
return an1
t = int(input())
for _ in range(t):
p = list(map(int, input().split()))
x, y = p[0], p[1]
cos = list(map(int, input().split()))
for a in range(len(cos)):
if cos[(a + 1) % 6] + cos[(a - 1) % 6] < cos[a]:
cos[a] = cos[(a + 1) % 6] + cos[(a - 1) % 6]
an1 = xyc(x, y)
an2 = 0
if x * y < 0:
an2 = an1
elif x > 0 and x > y:
an2 += cos[0] * y
an2 += xyc(x - y, 0)
elif x > 0:
an2 += cos[0] * x
an2 += xyc(0, y - x)
elif x < 0 and x < y:
an2 += cos[3] * -y
an2 += xyc(x - y, 0)
else:
an2 += cos[3] * -x
an2 += xyc(0, y - x)
print(min(an1, an2)) | FUNC_DEF ASSIGN VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR IF VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR VAR ASSIGN VAR VAR BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER IF BIN_OP VAR VAR NUMBER ASSIGN VAR VAR IF VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR VAR FUNC_CALL VAR BIN_OP VAR VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR IF VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR VAR FUNC_CALL VAR BIN_OP VAR VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | import sys
def inp():
return int(input())
def inlt():
return list(map(int, input().split()))
def insr():
s = input()
return list(s[: len(s) - 1])
def invr():
return map(int, input().split())
def soln(targetxy, costs):
assert len(costs) == 6
finalcost = 999
for _ in range(6):
costs[0] = min(costs[0], costs[5] + costs[1])
costs[1] = min(costs[1], costs[0] + costs[2])
costs[2] = min(costs[2], costs[3] + costs[1])
costs[3] = min(costs[3], costs[2] + costs[4])
costs[4] = min(costs[4], costs[3] + costs[5])
costs[5] = min(costs[5], costs[4] + costs[0])
if targetxy[1] < 0:
targetxy[0] *= -1
targetxy[1] *= -1
costs = costs[3:6] + costs[0:3]
if targetxy[0] < 0:
diagstep = targetxy[0] * -1 * costs[4] + targetxy[1] * costs[5]
otherstep1 = targetxy[0] * -1 * costs[4] + targetxy[1] * (costs[0] + costs[4])
otherstep2 = targetxy[1] * costs[5] + targetxy[0] * -1 * (costs[5] + costs[3])
finalcost = min(min(diagstep, otherstep1), otherstep2)
else:
diagstep = targetxy[0] * costs[1] + targetxy[1] * costs[5]
common = min(targetxy[0], targetxy[1])
straightstep = (
common * costs[0]
+ (targetxy[0] - common) * costs[1]
+ (targetxy[1] - common) * costs[5]
)
finalcost = min(diagstep, straightstep)
return finalcost
tests = inp()
for test in range(tests):
targetxy = inlt()
costs = inlt()
targetxy = [targetxy[1], targetxy[0]]
print(soln(targetxy, costs)) | IMPORT FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR RETURN FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER IF VAR NUMBER NUMBER VAR NUMBER NUMBER VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER NUMBER IF VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER VAR NUMBER BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR NUMBER BIN_OP BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR NUMBER VAR VAR NUMBER BIN_OP BIN_OP VAR NUMBER VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | def find_shortest_route(c1, c2, c3, c4, c5, c6, x, y):
costs = []
n1 = y
n2 = x - n1
if n1 >= 0 and n2 >= 0:
costs.append(c1 * n1 + c2 * n2)
n1 = x
n3 = n1 - y
if n1 >= 0 and n3 >= 0:
costs.append(c1 * n1 + c3 * n3)
n1 = y
n5 = n1 - x
if n1 >= 0 and n5 >= 0:
costs.append(c1 * n1 + c5 * n5)
n1 = x
n6 = y - n1
if n1 >= 0 and n6 >= 0:
costs.append(c1 * n1 + c6 * n6)
n2 = x
n3 = -y
if n2 >= 0 and n3 >= 0:
costs.append(c2 * n2 + c3 * n3)
n4 = -y
n2 = x + n4
if n2 >= 0 and n4 >= 0:
costs.append(c2 * n2 + c4 * n4)
n2 = x
n6 = y
if n2 >= 0 and n6 >= 0:
costs.append(c2 * n2 + c6 * n6)
n4 = -x
n3 = -y - n4
if n3 >= 0 and n4 >= 0:
costs.append(c3 * n3 + c4 * n4)
n5 = -x
n3 = -y
if n3 >= 0 and n5 >= 0:
costs.append(c3 * n3 + c5 * n5)
n4 = -y
n5 = -x - n4
if n4 >= 0 and n5 >= 0:
costs.append(c4 * n4 + c5 * n5)
n4 = -x
n6 = y + n4
if n4 >= 0 and n6 >= 0:
costs.append(c4 * n4 + c6 * n6)
n5 = -x
n6 = y
if n5 >= 0 and n6 >= 0:
costs.append(c5 * n5 + c6 * n6)
print(min(*costs))
n_t = int(input())
for t in range(n_t):
y, x = map(int, input().split(" "))
c1, c2, c3, c4, c5, c6 = map(int, input().split(" "))
find_shortest_route(c1, c2, c3, c4, c5, c6, x, y) | FUNC_DEF ASSIGN VAR LIST ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | import sys
input = sys.stdin.readline
def main():
x, y = map(int, input().split())
clst = list(map(int, input().split()))
clst[0] = min(clst[0], clst[5] + clst[1])
clst[1] = min(clst[1], clst[0] + clst[2])
clst[2] = min(clst[2], clst[1] + clst[3])
clst[3] = min(clst[3], clst[2] + clst[4])
clst[4] = min(clst[4], clst[3] + clst[5])
clst[5] = min(clst[5], clst[4] + clst[0])
directions = [(1, 1), (0, 1), (-1, 0), (-1, -1), (0, -1), (1, 0)]
ans = 10**20
for i in range(6):
dx1, dy1 = directions[i]
dx2, dy2 = directions[(i + 1) % 6]
a = (x * dy2 - y * dx2) // (dx1 * dy2 - dy1 * dx2)
b = (x * dy1 - y * dx1) // (dx2 * dy1 - dy2 * dx1)
if a < 0 or b < 0:
continue
ans = min(ans, clst[i] * a + clst[(i + 1) % 6] * b)
print(ans)
for _ in range(int(input())):
main() | IMPORT ASSIGN VAR VAR FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | for _ in range(int(input())):
x, y = map(int, input().split())
c = list(map(int, input().split()))
for i in range(6):
c[i] = min(c[i], c[(i + 1) % 6] + c[(i - 1) % 6])
if x >= 0:
if y >= x:
print(x * c[0] + (y - x) * c[1])
elif y <= 0:
print(x * c[5] + abs(y) * c[4])
else:
print((x - y) * c[5] + y * c[0])
elif y <= x:
print((x - y) * c[4] + abs(x) * c[3])
elif y >= 0:
print(abs(x) * c[2] + y * c[1])
else:
print(abs(y) * c[3] + (y - x) * c[2]) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER BIN_OP VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR NUMBER BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR NUMBER BIN_OP BIN_OP VAR VAR VAR NUMBER |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | import sys
from sys import stdin
tt = int(stdin.readline())
for loop in range(tt):
X, Y = map(int, stdin.readline().split())
c = [None] + list(map(int, stdin.readline().split()))
ans = 0
x, y = X, Y
if x >= 0:
ans += x * c[6]
else:
ans += abs(x) * c[3]
if y >= 0:
ans += y * c[2]
else:
ans += abs(y) * c[5]
x, y = X, Y
tmp = 0
if x >= 0:
tmp += c[1] * x
y -= x
else:
tmp += c[4] * abs(x)
y -= x
if y >= 0:
tmp += y * c[2]
else:
tmp += abs(y) * c[5]
ans = min(ans, tmp)
x, y = X, Y
tmp = 0
if y >= 0:
tmp += c[1] * y
x -= y
else:
tmp += c[4] * abs(y)
x -= y
if x >= 0:
tmp += x * c[6]
else:
tmp += abs(x) * c[3]
ans = min(ans, tmp)
print(ans) | IMPORT ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NONE FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR VAR VAR VAR IF VAR NUMBER VAR BIN_OP VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR VAR VAR IF VAR NUMBER VAR BIN_OP VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR VAR VAR IF VAR NUMBER VAR BIN_OP VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | import sys
input = sys.stdin.buffer.readline
def oneLineArrayPrint(arr):
print(" ".join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
print("\n".join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
print("\n".join([" ".join([str(x) for x in y]) for y in arr]))
allAns = []
t = int(input())
for _ in range(t):
x, y = [int(z) for z in input().split()]
c1, c2, c3, c4, c5, c6 = [int(z) for z in input().split()]
ans = float("inf")
t = y
s = x - t
ans2 = 0
if t >= 0:
ans2 += c1 * abs(t)
else:
ans2 += c4 * abs(t)
if s >= 0:
ans2 += c6 * abs(s)
else:
ans2 += c3 * abs(s)
ans = min(ans, ans2)
s = x
t = y
ans2 = 0
if s >= 0:
ans2 += c6 * abs(s)
else:
ans2 += c3 * abs(s)
if t >= 0:
ans2 += c2 * abs(t)
else:
ans2 += c5 * abs(t)
ans = min(ans, ans2)
t = x
s = y - t
ans2 = 0
if s >= 0:
ans2 += c2 * abs(s)
else:
ans2 += c5 * abs(s)
if t >= 0:
ans2 += c1 * abs(t)
else:
ans2 += c4 * abs(t)
ans = min(ans, ans2)
allAns.append(ans)
multiLineArrayPrint(allAns) | IMPORT ASSIGN VAR VAR FUNC_DEF EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR FUNC_DEF EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR FUNC_DEF EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR VAR VAR ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR FUNC_CALL VAR VAR VAR BIN_OP VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR BIN_OP VAR FUNC_CALL VAR VAR VAR BIN_OP VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR FUNC_CALL VAR VAR VAR BIN_OP VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR BIN_OP VAR FUNC_CALL VAR VAR VAR BIN_OP VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR FUNC_CALL VAR VAR VAR BIN_OP VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR BIN_OP VAR FUNC_CALL VAR VAR VAR BIN_OP VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | import sys
input = sys.stdin.readline
I = lambda: list(map(int, input().split()))
(t,) = I()
for _ in range(t):
x, y = I()
l = I()
a = min(l[2], l[3] + l[1])
b = min(l[5], l[4] + l[0])
c = min(l[4], l[3] + l[5])
d = min(l[1], l[0] + l[2])
e = min(l[1] + l[5], l[0])
f = min(l[2] + l[4], l[3])
pk = 0
if x > 0 and y > 0:
pk = min(x, y)
x -= pk
y -= pk
pk *= e
elif x < 0 and y < 0:
pk = max(x, y)
x -= pk
y -= pk
pk *= -f
an = pk
if x < 0:
an += -x * a
else:
an += x * b
if y < 0:
an += -y * c
else:
an += y * d
print(an) | IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR IF VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR IF VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | for _ in range(int(input())):
repl = True
x, y = map(int, input().split())
c1, c2, c3, c4, c5, c6 = map(int, input().split())
while repl:
repl = False
if c1 + c3 < c2:
c2 = c1 + c3
repl = True
if c2 + c4 < c3:
c3 = c2 + c4
repl = True
if c3 + c5 < c4:
c4 = c3 + c5
repl = True
if c4 + c6 < c5:
c5 = c4 + c6
repl = True
if c5 + c1 < c6:
c6 = c5 + c1
repl = True
if c6 + c2 < c1:
c1 = c6 + c2
repl = True
if x >= 0 and y >= 0:
pivot = min(x, y)
print(c1 * pivot + c2 * (y - pivot) + c6 * (x - pivot))
elif x <= 0 and y <= 0:
pivot = min(-x, -y)
print(c4 * pivot + c5 * (-y - pivot) + c3 * (-x - pivot))
elif x >= 0 and y <= 0:
print(c6 * x + c5 * -y)
elif x <= 0 and y >= 0:
print(c3 * -x + c2 * y) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR WHILE VAR ASSIGN VAR NUMBER IF BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | def optimize_costs(c1, c2, c3, c4, c5, c6):
c1 = min(c1, c2 + c6)
c2 = min(c2, c1 + c3)
c3 = min(c3, c2 + c4)
c4 = min(c4, c3 + c5)
c5 = min(c5, c4 + c6)
c6 = min(c6, c5 + c1)
return c1, c2, c3, c4, c5, c6
def main(input_f):
t = int(input_f())
for _ in range(t):
x, y = map(int, input_f().split(" "))
c1, c2, c3, c4, c5, c6 = map(int, input_f().split(" "))
for _ in range(10):
c1, c2, c3, c4, c5, c6 = optimize_costs(c1, c2, c3, c4, c5, c6)
if x < 0 and y < 0:
if x < y:
print(-y * c4 + (y - x) * c3)
else:
print(-x * c4 + (x - y) * c5)
elif x < 0 and y >= 0:
print(-x * c3 + y * c2)
elif x >= 0 and y < 0:
print(x * c6 - y * c5)
elif x < y:
print(x * c1 + (y - x) * c2)
else:
print(y * c1 + (x - y) * c6)
main(input) | FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR RETURN VAR VAR VAR VAR VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR IF VAR NUMBER VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR IF VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | t = int(input())
for i in range(t):
x, y = list(map(int, input().split()))
c = list(map(int, input().split()))
turns = [(1, 1), (0, 1), (-1, 0), (-1, -1), (0, -1), (1, 0)]
true_paths = [None, None, None, None, None, None]
min_in_c = min(c)
for j in range(6):
if c[j] == min_in_c:
true_paths[j] = c[j]
for g in range(6):
sum_x = turns[j][0] + turns[g][0]
sum_y = turns[j][1] + turns[g][1]
for k in range(6):
if turns[k][0] == sum_x and turns[k][1] == sum_y:
sum1 = c[k]
sum2 = c[j] + c[g]
true_paths[k] = min(sum1, sum2)
for j in range(6):
if true_paths[j] is None:
true_paths[j] = c[j]
changes = True
while changes:
changes = False
for j in range(6):
for g in range(6):
sum_x = turns[j][0] + turns[g][0]
sum_y = turns[j][1] + turns[g][1]
for k in range(6):
if turns[k][0] == sum_x and turns[k][1] == sum_y:
if true_paths[j] + true_paths[g] < true_paths[k]:
true_paths[k] = true_paths[j] + true_paths[g]
changes = True
ans = 0
if x > 0 and y > 0:
if true_paths[0] < true_paths[1] + true_paths[5]:
ans += min(x, y) * true_paths[0]
min_a = min(x, y)
x -= min_a
y -= min_a
elif x < 0 and y < 0:
if true_paths[3] < true_paths[2] + true_paths[4]:
ans += min(-x, -y) * true_paths[3]
min_a = min(-x, -y)
x += min_a
y += min_a
if x < 0:
ans += true_paths[2] * -x
if x > 0:
ans += true_paths[5] * x
if y < 0:
ans += true_paths[4] * -y
if y > 0:
ans += true_paths[1] * y
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER ASSIGN VAR LIST NONE NONE NONE NONE NONE NONE ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER IF VAR VAR VAR ASSIGN VAR VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP VAR VAR NUMBER VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR NUMBER VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER IF VAR VAR NUMBER VAR VAR VAR NUMBER VAR ASSIGN VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER IF VAR VAR NONE ASSIGN VAR VAR VAR VAR ASSIGN VAR NUMBER WHILE VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP VAR VAR NUMBER VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR NUMBER VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER IF VAR VAR NUMBER VAR VAR VAR NUMBER VAR IF BIN_OP VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR NUMBER IF VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR IF VAR NUMBER VAR NUMBER IF VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR IF VAR NUMBER VAR BIN_OP VAR NUMBER VAR IF VAR NUMBER VAR BIN_OP VAR NUMBER VAR IF VAR NUMBER VAR BIN_OP VAR NUMBER VAR IF VAR NUMBER VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | for _ in range(int(input())):
x0, y0 = input().split()
x0, y0 = int(x0), int(y0)
c = [0] + list(map(int, input().split()))
c[0] = c[-1]
c.append(c[1])
for i in range(1, 7):
c[i] = min(c[i], c[i - 1] + c[i + 1])
c[7] = c[1]
s = None, [1, 1], [0, 1], [-1, 0], [-1, -1], [0, -1], [1, 0], [1, 1]
ans = 1e30
for i in range(1, 7):
x1, y1 = s[i]
x2, y2 = s[i + 1]
a = (x0 * y2 - y0 * x2) // (x1 * y2 - y1 * x2)
b = (x0 * y1 - y0 * x1) // (x2 * y1 - y2 * x1)
if a >= 0 and b >= 0:
ans = min(ans, a * c[i] + b * c[i + 1])
print(ans) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR NONE LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | for _ in range(int(input())):
x, y = map(int, input().split())
c1, c2, c3, c4, c5, c6 = map(int, input().split())
arr = [0] * 7
cost = 0
if x >= 0 and y >= 0:
arr[1] = min(x, y)
x -= arr[1]
y -= arr[1]
cost += arr[1] * min(c2 + c6, c1)
if x:
arr[6] = x
cost += arr[6] * min(c1 + c5, c6)
if y:
arr[2] = y
cost += arr[2] * min(c1 + c3, c2)
elif x >= 0 and y <= 0:
arr[6] = x
cost += arr[6] * min(c6, c5 + c1)
arr[5] = -y
cost += arr[5] * min(c4 + c6, c5)
elif x <= 0 and y >= 0:
arr[2] = y
cost += arr[2] * min(c1 + c3, c2)
arr[3] = -x
cost += arr[3] * min(c2 + c4, c3)
elif x < 0 and y < 0:
arr[4] = min(-x, -y)
x += arr[4]
y += arr[4]
cost += arr[4] * min(c3 + c5, c4)
if x:
arr[3] = -x
cost += arr[3] * min(c2 + c4, c3)
if y:
arr[5] = -y
cost += arr[5] * min(c4 + c6, c5)
print(cost) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR VAR NUMBER VAR BIN_OP VAR NUMBER FUNC_CALL VAR BIN_OP VAR VAR VAR IF VAR ASSIGN VAR NUMBER VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR BIN_OP VAR VAR VAR IF VAR ASSIGN VAR NUMBER VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR BIN_OP VAR VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR NUMBER VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR BIN_OP VAR VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR BIN_OP VAR VAR VAR ASSIGN VAR NUMBER VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR BIN_OP VAR VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR VAR NUMBER VAR BIN_OP VAR NUMBER FUNC_CALL VAR BIN_OP VAR VAR VAR IF VAR ASSIGN VAR NUMBER VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR BIN_OP VAR VAR VAR IF VAR ASSIGN VAR NUMBER VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | from sys import stdin
t = int(stdin.readline())
for _ in range(t):
x, y = tuple(int(x) for x in stdin.readline().split())
lst = list(int(x) for x in stdin.readline().split())
flag = True
while flag:
flag = False
for i in range(6):
if lst[i - 1] + lst[(i + 1) % 6] < lst[i]:
lst[i] = lst[i - 1] + lst[(i + 1) % 6]
flag = True
if x >= 0 and y >= 0:
if x >= y:
print(y * lst[0] + (x - y) * lst[5])
else:
print(x * lst[0] + (y - x) * lst[1])
elif x <= 0 and y <= 0:
if x >= y:
print(-x * lst[3] + (x - y) * lst[4])
else:
print(-y * lst[3] + (y - x) * lst[2])
elif x >= 0 and y <= 0:
print(x * lst[5] + -y * lst[4])
elif x <= 0 and y >= 0:
print(-x * lst[2] + y * lst[1]) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER WHILE VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER IF BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR VAR ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR VAR VAR NUMBER IF VAR NUMBER VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR VAR VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP VAR VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP VAR VAR NUMBER |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | t = int(input())
for _ in range(t):
x, y = map(int, input().split())
a, b, c, d, e, f = map(int, input().split())
c = [f, a, b, c, d, e]
m = 0
for i in range(6):
if c[i] < c[m]:
m = i
j = (m + 1) % 6
while j != m:
c[j] = min(c[(j + 1) % 6] + c[(j - 1 + 6) % 6], c[j])
j = (j + 1) % 6
j = (m - 1 + 6) % 6
while j != m:
c[j] = min(c[(j + 1) % 6] + c[(j - 1 + 6) % 6], c[j])
j = (j - 1 + 6) % 6
if x == 0 and y >= 0:
print(y * c[2])
elif x == 0 and y <= 0:
print(abs(y) * c[5])
elif y == 0 and x >= 0:
print(abs(x) * c[0])
elif y == 0 and x <= 0:
print(abs(x) * c[3])
elif x > 0 and y > 0:
if y > x:
print(x * c[1] + (y - x) * c[2])
else:
print(y * c[1] + (x - y) * c[0])
elif x < 0 and y < 0:
if abs(y) > abs(x):
print(abs(x) * c[4] + (abs(y) - abs(x)) * c[5])
else:
print(abs(y) * c[4] + (abs(x) - abs(y)) * c[3])
else:
ans = 0
if x > 0:
ans = ans + x * c[0]
if y > 0:
ans = ans + y * c[2]
if x < 0:
ans = ans + abs(x) * c[3]
if y < 0:
ans = ans + abs(y) * c[5]
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST VAR VAR VAR VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER IF VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER WHILE VAR VAR ASSIGN VAR VAR FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER WHILE VAR VAR ASSIGN VAR VAR FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR VAR VAR NUMBER IF VAR NUMBER VAR NUMBER IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR NUMBER BIN_OP BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR NUMBER BIN_OP BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | t = int(input())
for _ in range(t):
x, y = map(int, input().split())
c1, c2, c3, c4, c5, c6 = map(int, input().split())
if x == y == 0:
print(0)
continue
if x == 0:
one_line = True
l = abs(y)
if y > 0:
cs1, cm, cs2 = c1, c2, c3
else:
cs1, cm, cs2 = c4, c5, c6
elif y == 0:
one_line = True
l = abs(x)
if x > 0:
cs1, cm, cs2 = c5, c6, c1
else:
cs1, cm, cs2 = c2, c3, c4
elif x == y:
one_line = True
l = abs(x)
if x > 0:
cs1, cm, cs2 = c6, c1, c2
else:
cs1, cm, cs2 = c3, c4, c5
else:
one_line = False
if x > 0:
if x < y:
cs1, cm1, cm2, cs2 = c6, c1, c2, c3
a, b = x, y - x
elif y > 0:
cs1, cm1, cm2, cs2 = c5, c6, c1, c2
a, b = x - y, y
else:
cs1, cm1, cm2, cs2 = c4, c5, c6, c1
a, b = -y, x
elif x > y:
cs1, cm1, cm2, cs2 = c3, c4, c5, c6
a, b = -x, x - y
elif y < 0:
cs1, cm1, cm2, cs2 = c2, c3, c4, c5
a, b = y - x, -y
else:
cs1, cm1, cm2, cs2 = c1, c2, c3, c4
a, b = y, -x
if one_line:
print(min(cm * l, (cs1 + cs2) * l))
else:
print(
min(cm1 * a + cm2 * b, cm1 * a + (cm1 + cs2) * b, (cs1 + cm2) * a + cm2 * b)
) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR IF VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER IF VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR IF VAR EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | t = int(input())
def neg(i):
return (i + 3) % 6
for case in range(t):
x, y = map(int, input().split())
x *= 1
y *= 1
C = [int(x) for x in input().split()]
steps = [0, 0, 0, 0, 0, 0]
if x < 0:
steps[2] = -x
else:
steps[5] = x
if y < 0:
steps[4] = -y
else:
steps[1] = y
def normalize():
for i in range(3):
mn = min(steps[i], steps[i + 3])
steps[i] -= mn
steps[i + 3] -= mn
moves = [
(0, (5, 1)),
(1, (0, 2)),
(2, (1, 3)),
(3, (2, 4)),
(4, (3, 5)),
(5, (4, 0)),
]
res = 10**100
while True:
for i, (j, k) in moves:
if (
C[i]
> C[j]
+ C[k]
- (steps[neg(j)] > 0) * C[neg(j)]
- (steps[neg(k)] > 0) * C[neg(k)]
):
mn = min(steps[i], steps[neg(j)], steps[neg(k)])
steps[i] -= mn
steps[j] += mn
steps[k] += mn
normalize()
if C[i] > C[j] + C[k] - (steps[neg(j)] > 0) * C[neg(j)]:
mn = min(steps[i], steps[neg(j)])
steps[i] -= mn
steps[j] += mn
steps[k] += mn
normalize()
if C[i] > C[j] + C[k] - (steps[neg(k)] > 0) * C[neg(k)]:
mn = min(steps[i], steps[neg(k)])
steps[i] -= mn
steps[j] += mn
steps[k] += mn
normalize()
if C[i] > C[j] + C[k]:
mn = steps[i]
steps[i] -= mn
steps[j] += mn
steps[k] += mn
normalize()
if C[i] - (steps[neg(i)] > 0) * C[neg(i)] < C[j] + C[k]:
mn = min(steps[j], steps[k], steps[neg(i)])
steps[i] += mn
steps[j] -= mn
steps[k] -= mn
normalize()
if C[i] < C[j] + C[k]:
mn = min(steps[j], steps[k])
steps[i] += mn
steps[j] -= mn
steps[k] -= mn
normalize()
s = sum(s * c for s, c in zip(steps, C))
if s < res:
res = s
else:
break
print(res) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN BIN_OP BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER IF VAR NUMBER ASSIGN VAR NUMBER VAR ASSIGN VAR NUMBER VAR IF VAR NUMBER ASSIGN VAR NUMBER VAR ASSIGN VAR NUMBER VAR FUNC_DEF FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER WHILE NUMBER FOR VAR VAR VAR VAR IF VAR VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR BIN_OP VAR FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP VAR FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR IF VAR VAR BIN_OP BIN_OP VAR VAR VAR VAR BIN_OP VAR FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR IF VAR VAR BIN_OP BIN_OP VAR VAR VAR VAR BIN_OP VAR FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR IF VAR VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR IF BIN_OP VAR VAR BIN_OP VAR FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR IF VAR VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | for _ in range(int(input())):
x, y = map(int, input().split())
c1, c2, c3, c4, c5, c6 = map(int, input().split())
c2 = min(c2, c1 + c3)
c3 = min(c3, c2 + c4)
c4 = min(c4, c3 + c5)
c5 = min(c5, c4 + c6)
c6 = min(c6, c5 + c1)
c1 = min(c1, c6 + c2)
if x * y < 0:
if x < 0:
print(c3 * -x + c2 * y)
else:
print(c6 * x + c5 * -y)
elif x + y > 0:
if x > y:
print(y * c1 + (x - y) * c6)
else:
print(x * c1 + (y - x) * c2)
elif x > y:
print(-x * c4 + (x - y) * c5)
else:
print(-y * c4 + (y - x) * c3) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR IF BIN_OP VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR IF BIN_OP VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | import sys
input = sys.stdin.readline
t = int(input())
for _ in range(t):
x, y = map(int, input().split())
c1, c2, c3, c4, c5, c6 = map(int, input().split())
d1 = min(c1, c2 + c6)
d2 = min(c2, c1 + c3)
d3 = min(c3, c2 + c4)
d4 = min(c4, c3 + c5)
d5 = min(c5, c4 + c6)
d6 = min(c6, c5 + c1)
if x >= 0 and y >= 0:
a1 = x * d6 + y * d2
a2 = x * d1 + (y - x) * d2 if y - x >= 0 else x * d1 + (x - y) * d5
a3 = (x - y) * d6 + y * d1 if x - y >= 0 else (y - x) * d3 + y * d1
print(min(a1, a2, a3))
elif x >= 0 and y < 0:
y = -y
a1 = x * d6 + y * d5
print(a1)
elif x < 0 and y >= 0:
x = -x
a1 = x * d3 + y * d2
print(a1)
else:
x = -x
y = -y
a1 = x * d3 + y * d5
a2 = x * d4 + (y - x) * d5 if y - x >= 0 else x * d4 + (x - y) * d2
a3 = (x - y) * d3 + y * d4 if x - y >= 0 else (y - x) * d6 + y * d4
print(min(a1, a2, a3)) | IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR NUMBER BIN_OP BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR NUMBER BIN_OP BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | import sys
def input():
return sys.stdin.readline().strip()
def list2d(a, b, c):
return [[c for j in range(b)] for i in range(a)]
def list3d(a, b, c, d):
return [[[d for k in range(c)] for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e):
return [
[[[e for l in range(d)] for k in range(c)] for j in range(b)] for i in range(a)
]
def ceil(x, y=1):
return int(-(-x // y))
def INT():
return int(input())
def MAP():
return map(int, input().split())
def LIST(N=None):
return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes():
print("Yes")
def No():
print("No")
def YES():
print("YES")
def NO():
print("NO")
INF = 10**19
MOD = 10**9 + 7
EPS = 10**-10
for _ in range(INT()):
gy, gx = MAP()
C = LIST()
C[5] = min(C[5], C[0] + C[4])
C[1] = min(C[1], C[0] + C[2])
C[2] = min(C[2], C[3] + C[1])
C[4] = min(C[4], C[3] + C[5])
C[5] = min(C[5], C[0] + C[4])
C[1] = min(C[1], C[0] + C[2])
C[2] = min(C[2], C[3] + C[1])
C[4] = min(C[4], C[3] + C[5])
ans = 0
if gx > 0 and gy > 0:
mn = min(gx, gy)
ans += mn * min(C[1] + C[5], C[0])
if gx > gy:
ans += (gx - mn) * C[1]
elif gx < gy:
ans += (gy - mn) * C[5]
elif gx < 0 and gy < 0:
gx, gy = abs(gx), abs(gy)
mn = min(gx, gy)
ans += mn * min(C[2] + C[4], C[3])
if gx > gy:
ans += (gx - mn) * C[4]
elif gx < gy:
ans += (gy - mn) * C[2]
else:
if gx < 0:
ans += abs(gx) * C[4]
elif gx > 0:
ans += gx * C[1]
if gy < 0:
ans += abs(gy) * C[2]
elif gy > 0:
ans += gy * C[5]
print(ans) | IMPORT FUNC_DEF RETURN FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR FUNC_DEF RETURN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR FUNC_DEF RETURN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR FUNC_DEF NUMBER RETURN FUNC_CALL VAR BIN_OP VAR VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF NONE RETURN VAR NONE FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_DEF EXPR FUNC_CALL VAR STRING FUNC_DEF EXPR FUNC_CALL VAR STRING FUNC_DEF EXPR FUNC_CALL VAR STRING FUNC_DEF EXPR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER IF VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR NUMBER IF VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER IF VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR NUMBER IF VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR NUMBER IF VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR VAR NUMBER IF VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | n = int(input())
def calc(x, y, z):
if x >= 0:
return x * y
else:
return abs(x) * z
for j in range(n):
x, y = map(int, input().split())
c1, c2, c3, c4, c5, c6 = map(int, input().split())
best = 9223372036854775807
best = min(best, calc(x, c6, c3) + calc(y, c2, c5))
best = min(best, calc(y, c1, c4) + calc(x - y, c6, c3))
best = min(best, calc(x, c1, c4) + calc(y - x, c2, c5))
print(best) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF IF VAR NUMBER RETURN BIN_OP VAR VAR RETURN BIN_OP FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | from sys import gettrace, stdin
if gettrace():
inputi = input
else:
def input():
return next(stdin)[:-1]
def inputi():
return stdin.buffer.readline()
def solve():
r, c = map(int, inputi().split())
costs = [int(a) for a in inputi().split()]
if r >= 0:
if c <= 0:
o1 = costs[4] * -c + costs[5] * r
o2 = costs[4] * (r - c) + costs[0] * r
o3 = costs[3] * -c + costs[5] * (r - c)
elif c <= r:
o1 = costs[5] * (r - c) + costs[0] * c
o2 = costs[5] * r + costs[1] * c
o3 = costs[4] * (r - c) + costs[0] * r
else:
o1 = costs[0] * r + costs[1] * (c - r)
o2 = costs[0] * c + costs[2] * (c - r)
o3 = costs[5] * r + costs[1] * c
elif c >= 0:
o1 = costs[1] * c + costs[2] * -r
o2 = costs[1] * (c - r) + costs[3] * -r
o3 = costs[0] * c + costs[2] * (c - r)
elif c >= r:
o1 = costs[2] * (c - r) + costs[3] * -c
o2 = costs[2] * -r + costs[4] * -c
o3 = costs[1] * (c - r) + costs[3] * -r
else:
o1 = costs[3] * -r + costs[4] * (r - c)
o2 = costs[3] * -c + costs[5] * (r - c)
o3 = costs[2] * -r + costs[4] * -c
print(min(o1, o2, o3))
def main():
t = int(inputi())
for _ in range(t):
solve()
main() | IF FUNC_CALL VAR ASSIGN VAR VAR FUNC_DEF RETURN FUNC_CALL VAR VAR NUMBER FUNC_DEF RETURN FUNC_CALL VAR FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER BIN_OP VAR VAR IF VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR IF VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER BIN_OP VAR VAR IF VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | def read_int():
return int(input())
def read_ints():
return map(int, input().split(" "))
d = [(1, 1), (0, 1), (-1, 0), (-1, -1), (0, -1), (1, 0)]
t = read_int()
for case_num in range(t):
x, y = read_ints()
c = list(read_ints())
ans = -1
for i in range(6):
dx = d[i][0]
dy = d[i][1]
if dx != 0 and x // dx >= 0:
t = x // dx
cost = t * c[i]
y1 = d[i][1] * t
for j in range(6):
if j == i or d[j][0] != 0:
continue
t1 = (y - y1) // d[j][1]
if t1 >= 0:
nc = cost + t1 * c[j]
if ans == -1 or ans > nc:
ans = nc
if dy != 0 and y // dy >= 0:
t = y // dy
cost = t * c[i]
x1 = d[i][0] * t
for j in range(6):
if j == i or d[j][1] != 0:
continue
t1 = (x - x1) // d[j][0]
if t1 >= 0:
nc = cost + t1 * c[j]
if ans == -1 or ans > nc:
ans = nc
print(ans) | FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER IF VAR NUMBER BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR NUMBER VAR FOR VAR FUNC_CALL VAR NUMBER IF VAR VAR VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR VAR IF VAR NUMBER VAR VAR ASSIGN VAR VAR IF VAR NUMBER BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR NUMBER VAR FOR VAR FUNC_CALL VAR NUMBER IF VAR VAR VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR VAR IF VAR NUMBER VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | for t in range(int(input())):
x, y = map(int, input().split())
c1, c2, c3, c4, c5, c6 = map(int, input().split())
def cal(x, y):
if x == 0 and y == 0:
return 0
elif x > 0 and y == 0:
return min(c1 + c5, c6) * x
elif x < 0 and y == 0:
return min(c2 + c4, c3) * x * -1
elif x == 0 and y > 0:
return min(c1 + c3, c2) * y
elif x == 0 and y < 0:
return min(c4 + c6, c5) * y * -1
elif x == y and x > 0:
return min(c2 + c6, c1) * x
elif x == y and x < 0:
return min(c3 + c5, c4) * x * -1
else:
temp = cal(x, 0) + cal(0, y)
if abs(x) > abs(y):
diff = abs(x) - abs(y)
if x < 0:
diff *= -1
temp = min(temp, cal(diff, 0) + cal(x - diff, y))
elif abs(y) > abs(x):
diff = abs(y) - abs(x)
if y < 0:
diff *= -1
temp = min(temp, cal(0, diff) + cal(x, y - diff))
return temp
print(cal(x, y)) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF IF VAR NUMBER VAR NUMBER RETURN NUMBER IF VAR NUMBER VAR NUMBER RETURN BIN_OP FUNC_CALL VAR BIN_OP VAR VAR VAR VAR IF VAR NUMBER VAR NUMBER RETURN BIN_OP BIN_OP FUNC_CALL VAR BIN_OP VAR VAR VAR VAR NUMBER IF VAR NUMBER VAR NUMBER RETURN BIN_OP FUNC_CALL VAR BIN_OP VAR VAR VAR VAR IF VAR NUMBER VAR NUMBER RETURN BIN_OP BIN_OP FUNC_CALL VAR BIN_OP VAR VAR VAR VAR NUMBER IF VAR VAR VAR NUMBER RETURN BIN_OP FUNC_CALL VAR BIN_OP VAR VAR VAR VAR IF VAR VAR VAR NUMBER RETURN BIN_OP BIN_OP FUNC_CALL VAR BIN_OP VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR NUMBER VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR BIN_OP VAR VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP VAR VAR RETURN VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | def inp():
return int(input())
def inlt():
return list(map(int, input().split()))
def insr():
s = input()
return list(s[: len(s)])
def invr():
return map(int, input().split())
n = inp()
for _ in range(n):
target = inlt()
l = inlt()
if target[0] >= 0 and target[1] >= 0:
path1 = target[0] * l[5] + target[1] * l[1]
if target[0] > target[1]:
path2 = target[0] * l[0] + abs(target[0] - target[1]) * l[4]
path3 = target[1] * l[0] + abs(target[0] - target[1]) * l[5]
else:
path2 = target[1] * l[0] + abs(target[1] - target[0]) * l[2]
path3 = target[0] * l[0] + abs(target[0] - target[1]) * l[1]
print(min(path1, path2, path3))
elif target[0] <= 0 and target[1] >= 0:
path1 = abs(target[0] * l[2]) + abs(target[1] * l[1])
path2 = abs(target[1] * l[0]) + (abs(target[0]) + abs(target[1])) * l[2]
path3 = abs(target[0] * l[3]) + (abs(target[0]) + target[1]) * l[1]
print(min(path1, path2, path3))
elif target[0] >= 0 and target[1] <= 0:
path1 = target[0] * l[5] + abs(target[1]) * l[4]
path2 = target[0] * l[0] + (target[0] + abs(target[1])) * l[4]
path3 = abs(target[1]) * l[3] + (target[0] + abs(target[1])) * l[5]
print(min(path1, path2, path3))
else:
path1 = abs(target[0]) * l[2] + abs(target[1]) * l[4]
if target[0] > target[1]:
path2 = abs(target[0]) * l[3] + abs(target[0] - target[1]) * l[4]
path3 = abs(target[1]) * l[3] + abs(target[0] - target[1]) * l[5]
else:
path2 = abs(target[1]) * l[3] + abs(target[1] - target[0]) * l[2]
path3 = abs(target[0]) * l[3] + abs(target[0] - target[1]) * l[1]
print(min(path1, path2, path3)) | FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR RETURN FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR IF VAR NUMBER NUMBER VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR NUMBER BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR NUMBER BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR NUMBER BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR NUMBER BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR IF VAR NUMBER NUMBER VAR NUMBER NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR IF VAR NUMBER NUMBER VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR NUMBER BIN_OP BIN_OP VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER VAR NUMBER BIN_OP BIN_OP VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER VAR NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER VAR NUMBER BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER VAR NUMBER BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER VAR NUMBER BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER VAR NUMBER BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | T = int(input())
for _ in range(T):
r, c = map(int, input().split())
UR, R, D, LD, L, U = map(int, input().split())
UR, R, D, LD, L, U = (
min(UR, U + R),
min(R, UR + D),
min(D, R + LD),
min(LD, L + D),
min(L, U + LD),
min(U, L + UR),
)
cc = 0
if r > 0 and c > 0:
mn = min(r, c)
cc += mn * UR
r, c = r - mn, c - mn
if r < 0 and c < 0:
mn = min(-r, -c)
cc += mn * LD
r, c = r + mn, c + mn
if r >= 0:
cc += U * r
else:
cc += D * -r
if c >= 0:
cc += R * c
else:
cc += L * -c
print(cc) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR IF VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR IF VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | import sys
input = sys.stdin.readline
(T,) = map(int, input().split())
for _ in range(T):
y, x = map(int, input().split())
X = [0] + list(map(int, input().split()))
if x >= 0 and y >= 0:
a, b, c = X[2], X[6], X[1]
a = min(a, c + X[3])
b = min(b, c + X[5])
elif x > 0 and y < 0:
a, b = min(X[1] + X[3], X[2]), min(X[4] + X[2], X[3])
c = a + b
y = -y
elif x <= 0 and y <= 0:
a, b, c = X[5], X[3], X[4]
a = min(a, c + X[6])
b = min(b, c + X[2])
x, y = -x, -y
elif x < 0 and y > 0:
a, b = min(X[4] + X[6], X[5]), min(X[1] + X[5], X[6])
c = a + b
x = -x
c = min(a + b, c)
R = min(x, y) * c
if x > y:
R += (x - y) * a
else:
R += (y - x) * b
print(R) | IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR NUMBER ASSIGN VAR VAR VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR VAR IF VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | import itertools
import sys
(*data,) = map(int, sys.stdin.read().split()[::-1])
def inp():
return data.pop()
def app(c, speeds, combi, d):
x = 0
y = 0
cost = 0
for dire, dist in zip(combi, d):
y += speeds[dire][0] * dist
x += speeds[dire][1] * dist
cost += c[dire] * dist
return y, x, cost
res = []
for _ in range(inp()):
y, x = inp(), inp()
(*c,) = [inp() for _ in range(6)]
speeds = [(1, 1), (0, 1), (-1, 0), (-1, -1), (0, -1), (1, 0)]
distances = [abs(x), abs(y), abs(y - x), abs(y + x)]
ans = float("inf")
for r in range(4):
for combi in itertools.combinations(range(6), r):
is_ok = True
for i in range(3):
if i in combi and i + 3 in combi:
is_ok = False
break
if not is_ok:
continue
for d in itertools.permutations(distances, r):
yy, xx, cost = app(c, speeds, combi, d)
if xx == x and yy == y:
ans = min(ans, cost)
res.append(ans)
print("\n".join(map(str, res))) | IMPORT IMPORT ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR NUMBER FUNC_DEF RETURN FUNC_CALL VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR VAR NUMBER VAR VAR BIN_OP VAR VAR NUMBER VAR VAR BIN_OP VAR VAR VAR RETURN VAR VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER ASSIGN VAR LIST FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR BIN_OP VAR VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER IF VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR NUMBER IF VAR FOR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | t = int(input())
for _ in range(t):
x, y = map(int, input().split())
c = list(map(int, input().split()))
UR = 10**9
UL = 10**9
R = 10**9
L = 10**9
DL = 10**9
DR = 10**9
UR = min(c[0], c[1] + c[5])
UL = min(c[5], c[0] + c[4])
R = min(c[1], c[2] + c[0])
L = min(c[4], c[5] + c[3])
DL = min(c[3], c[4] + c[2])
DR = min(c[2], c[1] + c[3])
ans = 0
if x == 0:
if y >= 0:
ans += y * R
else:
ans += -y * L
elif x > 0:
if y == 0:
ans += x * UL
elif y > 0:
if x >= y:
ans += y * UR
ans += (x - y) * UL
else:
ans += x * UR
ans += (y - x) * R
else:
ans = x * UL + -y * L
elif y == 0:
ans = DR * -x
elif y > 0:
ans = R * y + -x * DR
elif x >= y:
ans += -x * DL
ans += -(y - x) * L
else:
ans += -y * DL
ans += -(x - y) * DR
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR IF VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR VAR IF VAR NUMBER IF VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR IF VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | def det(x, y, z, w):
return x * w - y * z
t = int(input())
for test in range(t):
[x, y] = list(map(int, input().rstrip().split()))
c = list(map(int, input().rstrip().split()))
for j in range(6):
for i in range(6):
c[i] = min(c[i], c[(i - 1) % 6] + c[(i + 1) % 6])
direction = [(1, 1), (0, 1), (-1, 0), (-1, -1), (0, -1), (1, 0)]
cost = float("inf")
for i in range(6):
for j in range(i + 1, 6):
d = direction[i]
e = direction[j]
dete = det(d[0], e[0], d[1], e[1])
if dete != 0:
a = det(x, e[0], y, e[1]) // dete
b = det(d[0], x, d[1], y) // dete
if (
a * d[0] + b * e[0] == x
and a * d[1] + b * e[1] == y
and a > 0
and b > 0
):
cost = min(cost, a * c[i] + b * c[j])
if x == 0:
if y > 0:
cost = min(cost, y * c[1])
elif y == 0:
cost = 0
else:
cost = min(cost, -y * c[4])
elif y == x:
if y > 0:
cost = min(cost, y * c[0])
else:
cost = min(cost, -y * c[3])
elif y == 0:
if x > 0:
cost = min(cost, x * c[5])
else:
cost = min(cost, -x * c[2])
print(cost) | FUNC_DEF RETURN BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN LIST VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR VAR NUMBER VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR NUMBER VAR VAR IF BIN_OP BIN_OP VAR VAR NUMBER BIN_OP VAR VAR NUMBER VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR IF VAR NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR NUMBER IF VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR NUMBER IF VAR NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | t = int(input())
for _ in range(t):
x, y = map(int, input().split())
c = list(map(int, input().split()))
for i in range(6):
c[i] = min(c[i], c[int((i - 1) % 6)] + c[int(i + 1) % 6])
answer = 0
path = []
if x >= 0 and y >= 0:
distance = min(x, y)
answer += distance * c[0]
path.append("C1")
x -= distance
y -= distance
if x <= 0 and y <= 0:
distance = max(x, y)
distance *= -1
x += distance
y += distance
answer += distance * c[3]
path.append("C4")
if x <= 0:
x *= -1
answer += c[2] * x
path.append("C3")
else:
answer += c[5] * x
path.append("C4")
if y <= 0:
y *= -1
answer += c[4] * y
path.append("C3")
else:
answer += c[1] * y
path.append("C2")
print(answer) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR STRING VAR VAR VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR NUMBER VAR VAR VAR VAR VAR BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR STRING IF VAR NUMBER VAR NUMBER VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR STRING VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR STRING IF VAR NUMBER VAR NUMBER VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR STRING VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | for _ in range(int(input())):
x, y = list(map(int, input().split()))
c = list(map(int, input().split()))
for i in range(50):
c[0] = min(c[0], c[5] + c[1])
c[1] = min(c[1], c[0] + c[2])
c[2] = min(c[2], c[1] + c[3])
c[3] = min(c[3], c[2] + c[4])
c[4] = min(c[4], c[3] + c[5])
c[5] = min(c[5], c[4] + c[0])
ans = 10**30
if x >= 0 and y >= 0:
ans = min(ans, c[1] * y + c[5] * x)
mn = min(x, y)
cost = c[0] * mn
x -= mn
y -= mn
if x >= 0:
cost += c[5] * x
if y >= 0:
cost += c[1] * y
ans = min(ans, cost)
elif x <= 0 and y >= 0:
x = abs(x)
ans = min(ans, c[2] * x + c[1] * y)
elif x <= 0 and y <= 0:
x = abs(x)
y = abs(y)
ans = min(ans, y * c[4] + c[2] * x)
mn = min(x, y)
cost = mn * c[3]
x -= mn
y -= mn
if x >= 0:
cost += x * c[2]
if y >= 0:
cost += y * c[4]
ans = min(ans, cost)
else:
y = abs(y)
ans = min(ans, x * c[5] + c[4] * y)
print(ans) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR IF VAR NUMBER VAR BIN_OP VAR NUMBER VAR IF VAR NUMBER VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR NUMBER VAR VAR VAR VAR IF VAR NUMBER VAR BIN_OP VAR VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR |
Lindsey Buckingham told Stevie Nicks "Go your own way". Nicks is now sad and wants to go away as quickly as possible, but she lives in a 2D hexagonal world.
Consider a hexagonal tiling of the plane as on the picture below. [Image]
Nicks wishes to go from the cell marked $(0, 0)$ to a certain cell given by the coordinates. She may go from a hexagon to any of its six neighbors you want, but there is a cost associated with each of them. The costs depend only on the direction in which you travel. Going from $(0, 0)$ to $(1, 1)$ will take the exact same cost as going from $(-2, -1)$ to $(-1, 0)$. The costs are given in the input in the order $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ as in the picture below. [Image]
Print the smallest cost of a path from the origin which has coordinates $(0, 0)$ to the given cell.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^{4}$). Description of the test cases follows.
The first line of each test case contains two integers $x$ and $y$ ($-10^{9} \le x, y \le 10^{9}$) representing the coordinates of the target hexagon.
The second line of each test case contains six integers $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $c_6$ ($1 \le c_1, c_2, c_3, c_4, c_5, c_6 \le 10^{9}$) representing the six costs of the making one step in a particular direction (refer to the picture above to see which edge is for each value).
-----Output-----
For each testcase output the smallest cost of a path from the origin to the given cell.
-----Example-----
Input
2
-3 1
1 3 5 7 9 11
1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
18
1000000000000000000
-----Note-----
The picture below shows the solution for the first sample. The cost $18$ is reached by taking $c_3$ 3 times and $c_2$ once, amounting to $5+5+5+3=18$. [Image] | t = int(input())
while t > 0:
x, y = map(int, input().split())
lst = list(map(int, input().split()))
cost = lst.copy()
for i in range(1, 7):
cost[i % 6] = min(cost[i % 6], cost[(i - 1) % 6] + cost[(i + 1) % 6])
cnt = [0] * 6
if x == 0 and y == 0:
pass
elif x >= 0 and y >= 0:
if x == y:
cnt[0] = x
elif x > y:
cnt[0] = y
cnt[5] = x - y
elif x < y:
cnt[0] = x
cnt[1] = y - x
elif x <= 0 and y <= 0:
if x == y:
cnt[3] = -x
elif x > y:
cnt[3] = -x
cnt[4] = x - y
elif x < y:
cnt[3] = -y
cnt[2] = y - x
elif x >= 0 and y <= 0:
cnt[4] = -y
cnt[5] = x
else:
cnt[2] = -x
cnt[1] = y
ans = 0
for i in range(6):
ans += cnt[i] * cost[i]
print(ans)
t -= 1 | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP LIST NUMBER NUMBER IF VAR NUMBER VAR NUMBER IF VAR NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER VAR IF VAR VAR ASSIGN VAR NUMBER VAR ASSIGN VAR NUMBER BIN_OP VAR VAR IF VAR VAR ASSIGN VAR NUMBER VAR ASSIGN VAR NUMBER BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER VAR IF VAR VAR ASSIGN VAR NUMBER VAR ASSIGN VAR NUMBER BIN_OP VAR VAR IF VAR VAR ASSIGN VAR NUMBER VAR ASSIGN VAR NUMBER BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER VAR ASSIGN VAR NUMBER VAR ASSIGN VAR NUMBER VAR ASSIGN VAR NUMBER VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER |
T is playing a game with his friend, HL.
There are n piles of stones, the i-th pile initially has a_i stones.
T and HL will take alternating turns, with T going first. In each turn, a player chooses a non-empty pile and then removes a single stone from it. However, one cannot choose a pile that has been chosen in the previous turn (the pile that was chosen by the other player, or if the current turn is the first turn then the player can choose any non-empty pile). The player who cannot choose a pile in his turn loses, and the game ends.
Assuming both players play optimally, given the starting configuration of t games, determine the winner of each game.
Input
The first line of the input contains a single integer t (1 ≤ t ≤ 100) — the number of games. The description of the games follows. Each description contains two lines:
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of piles.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100).
Output
For each game, print on a single line the name of the winner, "T" or "HL" (without quotes)
Example
Input
2
1
2
2
1 1
Output
T
HL
Note
In the first game, T removes a single stone from the only pile in his first turn. After that, although the pile still contains 1 stone, HL cannot choose from this pile because it has been chosen by T in the previous turn. Therefore, T is the winner. | t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
if n == 1:
print("T")
elif n == 2:
if a[0] == a[1]:
print("HL")
else:
print("T")
elif sum(a) % 2 == 1 or max(a) > sum(a) - max(a):
print("T")
else:
print("HL") | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER EXPR FUNC_CALL VAR STRING IF VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
T is playing a game with his friend, HL.
There are n piles of stones, the i-th pile initially has a_i stones.
T and HL will take alternating turns, with T going first. In each turn, a player chooses a non-empty pile and then removes a single stone from it. However, one cannot choose a pile that has been chosen in the previous turn (the pile that was chosen by the other player, or if the current turn is the first turn then the player can choose any non-empty pile). The player who cannot choose a pile in his turn loses, and the game ends.
Assuming both players play optimally, given the starting configuration of t games, determine the winner of each game.
Input
The first line of the input contains a single integer t (1 ≤ t ≤ 100) — the number of games. The description of the games follows. Each description contains two lines:
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of piles.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100).
Output
For each game, print on a single line the name of the winner, "T" or "HL" (without quotes)
Example
Input
2
1
2
2
1 1
Output
T
HL
Note
In the first game, T removes a single stone from the only pile in his first turn. After that, although the pile still contains 1 stone, HL cannot choose from this pile because it has been chosen by T in the previous turn. Therefore, T is the winner. | for _ in range(int(input())):
n = int(input())
l1 = list(map(int, input().split()))
l1.sort()
sum1 = sum(l1) - l1[-1]
if l1[-1] > sum1:
print("T")
else:
sum1 += l1[-1]
if sum1 % 2 == 0:
print("HL")
else:
print("T") | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER VAR EXPR FUNC_CALL VAR STRING VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
T is playing a game with his friend, HL.
There are n piles of stones, the i-th pile initially has a_i stones.
T and HL will take alternating turns, with T going first. In each turn, a player chooses a non-empty pile and then removes a single stone from it. However, one cannot choose a pile that has been chosen in the previous turn (the pile that was chosen by the other player, or if the current turn is the first turn then the player can choose any non-empty pile). The player who cannot choose a pile in his turn loses, and the game ends.
Assuming both players play optimally, given the starting configuration of t games, determine the winner of each game.
Input
The first line of the input contains a single integer t (1 ≤ t ≤ 100) — the number of games. The description of the games follows. Each description contains two lines:
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of piles.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100).
Output
For each game, print on a single line the name of the winner, "T" or "HL" (without quotes)
Example
Input
2
1
2
2
1 1
Output
T
HL
Note
In the first game, T removes a single stone from the only pile in his first turn. After that, although the pile still contains 1 stone, HL cannot choose from this pile because it has been chosen by T in the previous turn. Therefore, T is the winner. | import sys
input = sys.stdin.readline
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
if n == 1:
print("T")
continue
a = sorted(a, key=lambda x: -x)
partialsum = 0
for i in range(1, n):
partialsum += a[i]
totalsum = a[0] + partialsum
if a[0] > partialsum:
print("T")
elif totalsum % 2 == 0:
print("HL")
else:
print("T") | IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER EXPR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER VAR IF VAR NUMBER VAR EXPR FUNC_CALL VAR STRING IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
T is playing a game with his friend, HL.
There are n piles of stones, the i-th pile initially has a_i stones.
T and HL will take alternating turns, with T going first. In each turn, a player chooses a non-empty pile and then removes a single stone from it. However, one cannot choose a pile that has been chosen in the previous turn (the pile that was chosen by the other player, or if the current turn is the first turn then the player can choose any non-empty pile). The player who cannot choose a pile in his turn loses, and the game ends.
Assuming both players play optimally, given the starting configuration of t games, determine the winner of each game.
Input
The first line of the input contains a single integer t (1 ≤ t ≤ 100) — the number of games. The description of the games follows. Each description contains two lines:
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of piles.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100).
Output
For each game, print on a single line the name of the winner, "T" or "HL" (without quotes)
Example
Input
2
1
2
2
1 1
Output
T
HL
Note
In the first game, T removes a single stone from the only pile in his first turn. After that, although the pile still contains 1 stone, HL cannot choose from this pile because it has been chosen by T in the previous turn. Therefore, T is the winner. | for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
m = 0
s = 0
for i in a:
m = max(m, i)
s += i
if m > s // 2:
print("T")
elif s % 2 == 0:
print("HL")
else:
print("T") | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR IF VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
T is playing a game with his friend, HL.
There are n piles of stones, the i-th pile initially has a_i stones.
T and HL will take alternating turns, with T going first. In each turn, a player chooses a non-empty pile and then removes a single stone from it. However, one cannot choose a pile that has been chosen in the previous turn (the pile that was chosen by the other player, or if the current turn is the first turn then the player can choose any non-empty pile). The player who cannot choose a pile in his turn loses, and the game ends.
Assuming both players play optimally, given the starting configuration of t games, determine the winner of each game.
Input
The first line of the input contains a single integer t (1 ≤ t ≤ 100) — the number of games. The description of the games follows. Each description contains two lines:
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of piles.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100).
Output
For each game, print on a single line the name of the winner, "T" or "HL" (without quotes)
Example
Input
2
1
2
2
1 1
Output
T
HL
Note
In the first game, T removes a single stone from the only pile in his first turn. After that, although the pile still contains 1 stone, HL cannot choose from this pile because it has been chosen by T in the previous turn. Therefore, T is the winner. | for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
s = sum(a)
if s % 2 != 0:
print("T")
else:
k = s // 2
t = 0
for i in range(n):
if a[i] > k:
t = 1
break
if t == 1:
print("T")
else:
print("HL") | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
T is playing a game with his friend, HL.
There are n piles of stones, the i-th pile initially has a_i stones.
T and HL will take alternating turns, with T going first. In each turn, a player chooses a non-empty pile and then removes a single stone from it. However, one cannot choose a pile that has been chosen in the previous turn (the pile that was chosen by the other player, or if the current turn is the first turn then the player can choose any non-empty pile). The player who cannot choose a pile in his turn loses, and the game ends.
Assuming both players play optimally, given the starting configuration of t games, determine the winner of each game.
Input
The first line of the input contains a single integer t (1 ≤ t ≤ 100) — the number of games. The description of the games follows. Each description contains two lines:
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of piles.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100).
Output
For each game, print on a single line the name of the winner, "T" or "HL" (without quotes)
Example
Input
2
1
2
2
1 1
Output
T
HL
Note
In the first game, T removes a single stone from the only pile in his first turn. After that, although the pile still contains 1 stone, HL cannot choose from this pile because it has been chosen by T in the previous turn. Therefore, T is the winner. | from sys import stdin, stdout
mp = lambda: map(int, stdin.readline().split())
li = lambda: list(map(int, stdin.readline().split()))
for _ in range(int(input())):
n = int(input())
a = li()
mx, sm = max(a), sum(a)
if mx * 2 > sm or sm & 1:
print("T")
else:
print("HL") | ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
T is playing a game with his friend, HL.
There are n piles of stones, the i-th pile initially has a_i stones.
T and HL will take alternating turns, with T going first. In each turn, a player chooses a non-empty pile and then removes a single stone from it. However, one cannot choose a pile that has been chosen in the previous turn (the pile that was chosen by the other player, or if the current turn is the first turn then the player can choose any non-empty pile). The player who cannot choose a pile in his turn loses, and the game ends.
Assuming both players play optimally, given the starting configuration of t games, determine the winner of each game.
Input
The first line of the input contains a single integer t (1 ≤ t ≤ 100) — the number of games. The description of the games follows. Each description contains two lines:
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of piles.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100).
Output
For each game, print on a single line the name of the winner, "T" or "HL" (without quotes)
Example
Input
2
1
2
2
1 1
Output
T
HL
Note
In the first game, T removes a single stone from the only pile in his first turn. After that, although the pile still contains 1 stone, HL cannot choose from this pile because it has been chosen by T in the previous turn. Therefore, T is the winner. | t = int(input())
for q in range(t):
n = int(input())
mas = list(map(int, input().split()))
if max(mas) > sum(mas) - max(mas):
print("T")
elif sum(mas) % 2 == 0:
print("HL")
else:
print("T") | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
T is playing a game with his friend, HL.
There are n piles of stones, the i-th pile initially has a_i stones.
T and HL will take alternating turns, with T going first. In each turn, a player chooses a non-empty pile and then removes a single stone from it. However, one cannot choose a pile that has been chosen in the previous turn (the pile that was chosen by the other player, or if the current turn is the first turn then the player can choose any non-empty pile). The player who cannot choose a pile in his turn loses, and the game ends.
Assuming both players play optimally, given the starting configuration of t games, determine the winner of each game.
Input
The first line of the input contains a single integer t (1 ≤ t ≤ 100) — the number of games. The description of the games follows. Each description contains two lines:
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of piles.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100).
Output
For each game, print on a single line the name of the winner, "T" or "HL" (without quotes)
Example
Input
2
1
2
2
1 1
Output
T
HL
Note
In the first game, T removes a single stone from the only pile in his first turn. After that, although the pile still contains 1 stone, HL cannot choose from this pile because it has been chosen by T in the previous turn. Therefore, T is the winner. | for t in range(int(input())):
n = int(input())
a = sorted(list(map(int, input().split())))
if a[-1] > sum(a[:-1]):
print("T")
elif sum(a) % 2 == 0:
print("HL")
else:
print("T") | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR STRING IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
T is playing a game with his friend, HL.
There are n piles of stones, the i-th pile initially has a_i stones.
T and HL will take alternating turns, with T going first. In each turn, a player chooses a non-empty pile and then removes a single stone from it. However, one cannot choose a pile that has been chosen in the previous turn (the pile that was chosen by the other player, or if the current turn is the first turn then the player can choose any non-empty pile). The player who cannot choose a pile in his turn loses, and the game ends.
Assuming both players play optimally, given the starting configuration of t games, determine the winner of each game.
Input
The first line of the input contains a single integer t (1 ≤ t ≤ 100) — the number of games. The description of the games follows. Each description contains two lines:
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of piles.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100).
Output
For each game, print on a single line the name of the winner, "T" or "HL" (without quotes)
Example
Input
2
1
2
2
1 1
Output
T
HL
Note
In the first game, T removes a single stone from the only pile in his first turn. After that, although the pile still contains 1 stone, HL cannot choose from this pile because it has been chosen by T in the previous turn. Therefore, T is the winner. | def solve(n, piles):
mx = max(piles)
sm = sum(piles)
if mx > sm - mx:
return "T"
if sm % 2 == 1:
return "T"
return "HL"
t = int(input())
for t in range(t):
n = int(input())
piles = [int(s) for s in input().split()]
result = solve(n, piles)
print(result) | FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR BIN_OP VAR VAR RETURN STRING IF BIN_OP VAR NUMBER NUMBER RETURN STRING RETURN STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR |
T is playing a game with his friend, HL.
There are n piles of stones, the i-th pile initially has a_i stones.
T and HL will take alternating turns, with T going first. In each turn, a player chooses a non-empty pile and then removes a single stone from it. However, one cannot choose a pile that has been chosen in the previous turn (the pile that was chosen by the other player, or if the current turn is the first turn then the player can choose any non-empty pile). The player who cannot choose a pile in his turn loses, and the game ends.
Assuming both players play optimally, given the starting configuration of t games, determine the winner of each game.
Input
The first line of the input contains a single integer t (1 ≤ t ≤ 100) — the number of games. The description of the games follows. Each description contains two lines:
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of piles.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100).
Output
For each game, print on a single line the name of the winner, "T" or "HL" (without quotes)
Example
Input
2
1
2
2
1 1
Output
T
HL
Note
In the first game, T removes a single stone from the only pile in his first turn. After that, although the pile still contains 1 stone, HL cannot choose from this pile because it has been chosen by T in the previous turn. Therefore, T is the winner. | for _ in range(int(input())):
n = int(input())
l = list(map(int, input().split()))
l.sort()
c = sum(l)
if l[-1] > c - l[-1]:
print("T")
elif sum(l) % 2 == 1:
print("T")
elif len(l) == 1:
print("T")
else:
print("HL") | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR STRING IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
You are given a binary string $s$ of length $n$.
Let's define $d_i$ as the number whose decimal representation is $s_i s_{i+1}$ (possibly, with a leading zero). We define $f(s)$ to be the sum of all the valid $d_i$. In other words, $f(s) = \sum\limits_{i=1}^{n-1} d_i$.
For example, for the string $s = 1011$:
$d_1 = 10$ (ten);
$d_2 = 01$ (one)
$d_3 = 11$ (eleven);
$f(s) = 10 + 01 + 11 = 22$.
In one operation you can swap any two adjacent elements of the string. Find the minimum value of $f(s)$ that can be achieved if at most $k$ operations are allowed.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). Description of the test cases follows.
First line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^5$, $0 \le k \le 10^9$) — the length of the string and the maximum number of operations allowed.
The second line of each test case contains the binary string $s$ of length $n$, consisting of only zeros and ones.
It is also given that sum of $n$ over all the test cases doesn't exceed $10^5$.
-----Output-----
For each test case, print the minimum value of $f(s)$ you can obtain with at most $k$ operations.
-----Examples-----
Input
3
4 0
1010
7 1
0010100
5 2
00110
Output
21
22
12
-----Note-----
For the first example, you can't do any operation so the optimal string is $s$ itself. $f(s) = f(1010) = 10 + 01 + 10 = 21$.
For the second example, one of the optimal strings you can obtain is "0011000". The string has an $f$ value of $22$.
For the third example, one of the optimal strings you can obtain is "00011". The string has an $f$ value of $12$. | t = int(input())
ans = list()
for i in range(0, t):
n, k = [int(x) for x in input().split()]
arr = [int(x) for x in input()]
arr_sum = 0
j = n - 1
if k >= n:
d = 0
else:
d = n - k - 1
while j >= d:
if arr[j] == 1:
arr[j] = 0
arr_sum += 1
k -= n - 1 - j
break
j -= 1
else:
arr_sum += arr[-1]
for j in range(0, min([k + 1, n])):
if arr[j] == 1:
arr[j] = 0
arr_sum += 10
break
else:
arr_sum += 10 * arr[0]
for i in range(1, n - 1):
arr_sum += 11 * arr[i]
ans.append(arr_sum)
for s in ans:
print(s) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER WHILE VAR VAR IF VAR VAR NUMBER ASSIGN VAR VAR NUMBER VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR VAR NUMBER VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR LIST BIN_OP VAR NUMBER VAR IF VAR VAR NUMBER ASSIGN VAR VAR NUMBER VAR NUMBER VAR BIN_OP NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER VAR BIN_OP NUMBER VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR |
You are given a binary string $s$ of length $n$.
Let's define $d_i$ as the number whose decimal representation is $s_i s_{i+1}$ (possibly, with a leading zero). We define $f(s)$ to be the sum of all the valid $d_i$. In other words, $f(s) = \sum\limits_{i=1}^{n-1} d_i$.
For example, for the string $s = 1011$:
$d_1 = 10$ (ten);
$d_2 = 01$ (one)
$d_3 = 11$ (eleven);
$f(s) = 10 + 01 + 11 = 22$.
In one operation you can swap any two adjacent elements of the string. Find the minimum value of $f(s)$ that can be achieved if at most $k$ operations are allowed.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). Description of the test cases follows.
First line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^5$, $0 \le k \le 10^9$) — the length of the string and the maximum number of operations allowed.
The second line of each test case contains the binary string $s$ of length $n$, consisting of only zeros and ones.
It is also given that sum of $n$ over all the test cases doesn't exceed $10^5$.
-----Output-----
For each test case, print the minimum value of $f(s)$ you can obtain with at most $k$ operations.
-----Examples-----
Input
3
4 0
1010
7 1
0010100
5 2
00110
Output
21
22
12
-----Note-----
For the first example, you can't do any operation so the optimal string is $s$ itself. $f(s) = f(1010) = 10 + 01 + 10 = 21$.
For the second example, one of the optimal strings you can obtain is "0011000". The string has an $f$ value of $22$.
For the third example, one of the optimal strings you can obtain is "00011". The string has an $f$ value of $12$. | for i in range(int(input())):
n, k = map(int, input().split())
s = input()
num = s.count("1")
tot = 0
if num == 1:
i = s.index("1")
j = n - s.rfind("1") - 1
if k >= j:
k -= j
tot += 1
num -= 1
elif k >= i:
k -= i
tot += 10
num -= 1
tot += num * 11
print(tot)
elif num > 1:
i = s.index("1")
j = n - s.rfind("1") - 1
if k >= j:
k -= j
tot += 1
num -= 1
if k >= i:
k -= i
tot += 10
num -= 1
tot += num * 11
print(tot)
else:
print(0) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR FUNC_CALL VAR STRING NUMBER IF VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR FUNC_CALL VAR STRING NUMBER IF VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER |
You are given a binary string $s$ of length $n$.
Let's define $d_i$ as the number whose decimal representation is $s_i s_{i+1}$ (possibly, with a leading zero). We define $f(s)$ to be the sum of all the valid $d_i$. In other words, $f(s) = \sum\limits_{i=1}^{n-1} d_i$.
For example, for the string $s = 1011$:
$d_1 = 10$ (ten);
$d_2 = 01$ (one)
$d_3 = 11$ (eleven);
$f(s) = 10 + 01 + 11 = 22$.
In one operation you can swap any two adjacent elements of the string. Find the minimum value of $f(s)$ that can be achieved if at most $k$ operations are allowed.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). Description of the test cases follows.
First line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^5$, $0 \le k \le 10^9$) — the length of the string and the maximum number of operations allowed.
The second line of each test case contains the binary string $s$ of length $n$, consisting of only zeros and ones.
It is also given that sum of $n$ over all the test cases doesn't exceed $10^5$.
-----Output-----
For each test case, print the minimum value of $f(s)$ you can obtain with at most $k$ operations.
-----Examples-----
Input
3
4 0
1010
7 1
0010100
5 2
00110
Output
21
22
12
-----Note-----
For the first example, you can't do any operation so the optimal string is $s$ itself. $f(s) = f(1010) = 10 + 01 + 10 = 21$.
For the second example, one of the optimal strings you can obtain is "0011000". The string has an $f$ value of $22$.
For the third example, one of the optimal strings you can obtain is "00011". The string has an $f$ value of $12$. | t = int(input())
for _ in range(t):
n, k = map(int, input().split())
s = input()
back, front = -1, -1
cnt = -1
s = list(s)
for i in range(n - 1, -1, -1):
cnt += 1
if s[i] == "1":
ind1 = i
front = cnt
break
cnt = -1
for i in range(n):
cnt += 1
if s[i] == "1":
ind2 = i
back = cnt
break
if front == -1:
print(0)
continue
elif ind1 == ind2:
if k >= front:
print(1)
continue
elif k >= back:
print(10)
continue
else:
print(11)
else:
if k >= front:
s[ind1], s[-1] = s[-1], s[ind1]
k -= front
if k >= back:
s[ind2], s[0] = s[0], s[ind2]
res = 0
if s[0] == "1":
res += 10
if s[-1] == "1":
res += 1
for i in range(1, n - 1):
if s[i] == "1":
res += 11
print(res) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER VAR NUMBER IF VAR VAR STRING ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR NUMBER IF VAR VAR STRING ASSIGN VAR VAR ASSIGN VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR VAR IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR VAR ASSIGN VAR VAR VAR NUMBER VAR NUMBER VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR VAR NUMBER VAR NUMBER VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER STRING VAR NUMBER IF VAR NUMBER STRING VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR STRING VAR NUMBER EXPR FUNC_CALL VAR VAR |
You are given a binary string $s$ of length $n$.
Let's define $d_i$ as the number whose decimal representation is $s_i s_{i+1}$ (possibly, with a leading zero). We define $f(s)$ to be the sum of all the valid $d_i$. In other words, $f(s) = \sum\limits_{i=1}^{n-1} d_i$.
For example, for the string $s = 1011$:
$d_1 = 10$ (ten);
$d_2 = 01$ (one)
$d_3 = 11$ (eleven);
$f(s) = 10 + 01 + 11 = 22$.
In one operation you can swap any two adjacent elements of the string. Find the minimum value of $f(s)$ that can be achieved if at most $k$ operations are allowed.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). Description of the test cases follows.
First line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^5$, $0 \le k \le 10^9$) — the length of the string and the maximum number of operations allowed.
The second line of each test case contains the binary string $s$ of length $n$, consisting of only zeros and ones.
It is also given that sum of $n$ over all the test cases doesn't exceed $10^5$.
-----Output-----
For each test case, print the minimum value of $f(s)$ you can obtain with at most $k$ operations.
-----Examples-----
Input
3
4 0
1010
7 1
0010100
5 2
00110
Output
21
22
12
-----Note-----
For the first example, you can't do any operation so the optimal string is $s$ itself. $f(s) = f(1010) = 10 + 01 + 10 = 21$.
For the second example, one of the optimal strings you can obtain is "0011000". The string has an $f$ value of $22$.
For the third example, one of the optimal strings you can obtain is "00011". The string has an $f$ value of $12$. | t = int(input())
for _ in range(t):
n, k = map(int, input().split())
s = input()
f = 999999999
l = -999999999
ans = 0
for i in range(n):
if s[i] == "1":
f = min(f, i)
l = max(l, i)
ans += 11
l = n - 1 - l
used = 0
if k >= l:
ans -= 10
k -= l
used = 1
if k >= f and (used == 0 or ans > 11):
ans -= 1
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR NUMBER IF VAR VAR VAR NUMBER VAR VAR ASSIGN VAR NUMBER IF VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
You are given a binary string $s$ of length $n$.
Let's define $d_i$ as the number whose decimal representation is $s_i s_{i+1}$ (possibly, with a leading zero). We define $f(s)$ to be the sum of all the valid $d_i$. In other words, $f(s) = \sum\limits_{i=1}^{n-1} d_i$.
For example, for the string $s = 1011$:
$d_1 = 10$ (ten);
$d_2 = 01$ (one)
$d_3 = 11$ (eleven);
$f(s) = 10 + 01 + 11 = 22$.
In one operation you can swap any two adjacent elements of the string. Find the minimum value of $f(s)$ that can be achieved if at most $k$ operations are allowed.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). Description of the test cases follows.
First line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^5$, $0 \le k \le 10^9$) — the length of the string and the maximum number of operations allowed.
The second line of each test case contains the binary string $s$ of length $n$, consisting of only zeros and ones.
It is also given that sum of $n$ over all the test cases doesn't exceed $10^5$.
-----Output-----
For each test case, print the minimum value of $f(s)$ you can obtain with at most $k$ operations.
-----Examples-----
Input
3
4 0
1010
7 1
0010100
5 2
00110
Output
21
22
12
-----Note-----
For the first example, you can't do any operation so the optimal string is $s$ itself. $f(s) = f(1010) = 10 + 01 + 10 = 21$.
For the second example, one of the optimal strings you can obtain is "0011000". The string has an $f$ value of $22$.
For the third example, one of the optimal strings you can obtain is "00011". The string has an $f$ value of $12$. | t = int(input())
for _ in range(t):
n, k = map(int, input().split())
s = input()
count = 0
for i in range(n - 1):
count += int(s[i] + s[i + 1])
if count == 0:
print(0)
continue
if s[n - 1] != "1":
right = n - 1
while s[right] != "1":
right -= 1
k -= n - 1 - right
if k >= 0:
count -= 10
if right == 0:
count += 1
else:
k += n - 1 - right
if s[0] != "1" and count != 1:
left = 0
while s[left] != "1":
left += 1
k -= left
if left != n - 1 and k >= 0:
count -= 1
print(count) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR BIN_OP VAR NUMBER STRING ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR STRING VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR IF VAR NUMBER VAR NUMBER IF VAR NUMBER VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR IF VAR NUMBER STRING VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR STRING VAR NUMBER VAR VAR IF VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
You are given a binary string $s$ of length $n$.
Let's define $d_i$ as the number whose decimal representation is $s_i s_{i+1}$ (possibly, with a leading zero). We define $f(s)$ to be the sum of all the valid $d_i$. In other words, $f(s) = \sum\limits_{i=1}^{n-1} d_i$.
For example, for the string $s = 1011$:
$d_1 = 10$ (ten);
$d_2 = 01$ (one)
$d_3 = 11$ (eleven);
$f(s) = 10 + 01 + 11 = 22$.
In one operation you can swap any two adjacent elements of the string. Find the minimum value of $f(s)$ that can be achieved if at most $k$ operations are allowed.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). Description of the test cases follows.
First line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^5$, $0 \le k \le 10^9$) — the length of the string and the maximum number of operations allowed.
The second line of each test case contains the binary string $s$ of length $n$, consisting of only zeros and ones.
It is also given that sum of $n$ over all the test cases doesn't exceed $10^5$.
-----Output-----
For each test case, print the minimum value of $f(s)$ you can obtain with at most $k$ operations.
-----Examples-----
Input
3
4 0
1010
7 1
0010100
5 2
00110
Output
21
22
12
-----Note-----
For the first example, you can't do any operation so the optimal string is $s$ itself. $f(s) = f(1010) = 10 + 01 + 10 = 21$.
For the second example, one of the optimal strings you can obtain is "0011000". The string has an $f$ value of $22$.
For the third example, one of the optimal strings you can obtain is "00011". The string has an $f$ value of $12$. | for h in range(int(input())):
n, k = map(int, input().split())
u = input()
x = []
for i in u:
x.append(i)
c = x.count("1") * 11
if c == 0:
print(c)
continue
if x[-1] == "0":
m = 0
for i in range(n):
if x[i] == "1":
m = i
if k >= n - m - 1:
x[m], x[-1] = x[-1], x[m]
k = k - (n - m - 1)
c -= 10
else:
c -= 10
if x[0] == "0":
m = n
for i in range(n):
if x[i] == "1":
break
if k >= i and i != n - 1:
k -= i
c -= 1
else:
c -= 1
print(c) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR STRING NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR VAR IF VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER VAR NUMBER VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER VAR NUMBER VAR NUMBER IF VAR NUMBER STRING ASSIGN VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
You are given a binary string $s$ of length $n$.
Let's define $d_i$ as the number whose decimal representation is $s_i s_{i+1}$ (possibly, with a leading zero). We define $f(s)$ to be the sum of all the valid $d_i$. In other words, $f(s) = \sum\limits_{i=1}^{n-1} d_i$.
For example, for the string $s = 1011$:
$d_1 = 10$ (ten);
$d_2 = 01$ (one)
$d_3 = 11$ (eleven);
$f(s) = 10 + 01 + 11 = 22$.
In one operation you can swap any two adjacent elements of the string. Find the minimum value of $f(s)$ that can be achieved if at most $k$ operations are allowed.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). Description of the test cases follows.
First line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^5$, $0 \le k \le 10^9$) — the length of the string and the maximum number of operations allowed.
The second line of each test case contains the binary string $s$ of length $n$, consisting of only zeros and ones.
It is also given that sum of $n$ over all the test cases doesn't exceed $10^5$.
-----Output-----
For each test case, print the minimum value of $f(s)$ you can obtain with at most $k$ operations.
-----Examples-----
Input
3
4 0
1010
7 1
0010100
5 2
00110
Output
21
22
12
-----Note-----
For the first example, you can't do any operation so the optimal string is $s$ itself. $f(s) = f(1010) = 10 + 01 + 10 = 21$.
For the second example, one of the optimal strings you can obtain is "0011000". The string has an $f$ value of $22$.
For the third example, one of the optimal strings you can obtain is "00011". The string has an $f$ value of $12$. | for _ in range(int(input())):
n, k = map(int, input().split())
s = input()
ans, f, l, o = 0, n, -1, 0
for i in range(n):
if s[i] == "0":
continue
o += 1
if f == n:
f = i
l = i
if o and n - l - 1 <= k:
ans += 1
o -= 1
k -= n - l - 1
if o and f <= k:
ans += 10
o -= 1
ans += 11 * o
print(ans) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR VAR VAR NUMBER VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR IF VAR BIN_OP BIN_OP VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR VAR NUMBER VAR NUMBER VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR VAR |
You are given a binary string $s$ of length $n$.
Let's define $d_i$ as the number whose decimal representation is $s_i s_{i+1}$ (possibly, with a leading zero). We define $f(s)$ to be the sum of all the valid $d_i$. In other words, $f(s) = \sum\limits_{i=1}^{n-1} d_i$.
For example, for the string $s = 1011$:
$d_1 = 10$ (ten);
$d_2 = 01$ (one)
$d_3 = 11$ (eleven);
$f(s) = 10 + 01 + 11 = 22$.
In one operation you can swap any two adjacent elements of the string. Find the minimum value of $f(s)$ that can be achieved if at most $k$ operations are allowed.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). Description of the test cases follows.
First line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^5$, $0 \le k \le 10^9$) — the length of the string and the maximum number of operations allowed.
The second line of each test case contains the binary string $s$ of length $n$, consisting of only zeros and ones.
It is also given that sum of $n$ over all the test cases doesn't exceed $10^5$.
-----Output-----
For each test case, print the minimum value of $f(s)$ you can obtain with at most $k$ operations.
-----Examples-----
Input
3
4 0
1010
7 1
0010100
5 2
00110
Output
21
22
12
-----Note-----
For the first example, you can't do any operation so the optimal string is $s$ itself. $f(s) = f(1010) = 10 + 01 + 10 = 21$.
For the second example, one of the optimal strings you can obtain is "0011000". The string has an $f$ value of $22$.
For the third example, one of the optimal strings you can obtain is "00011". The string has an $f$ value of $12$. | t = int(input())
for _ in range(t):
n, k = map(int, input().split())
s = list(input())
tot = 0
f = k
last = first = 0
if s[-1] != "1":
for i in range(n - 2, -1, -1):
if k == 0:
break
if s[i] == "1":
s[i], s[-1] = s[-1], s[i]
last = 1
k -= 1
break
k -= 1
if not last:
k = f
if s[0] != "1":
for i in range(1, n - 1):
if k == 0:
break
if s[i] == "1":
s[0], s[i] = s[i], s[0]
break
k -= 1
for i in range(n):
num = s[i]
if num == "0":
continue
if i == 0:
tot += 10
elif i == n - 1:
tot += 1
else:
tot += 11
print(tot) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR NUMBER IF VAR NUMBER STRING FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR NUMBER IF VAR VAR STRING ASSIGN VAR VAR VAR NUMBER VAR NUMBER VAR VAR ASSIGN VAR NUMBER VAR NUMBER VAR NUMBER IF VAR ASSIGN VAR VAR IF VAR NUMBER STRING FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR NUMBER IF VAR VAR STRING ASSIGN VAR NUMBER VAR VAR VAR VAR VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF VAR STRING IF VAR NUMBER VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
You are given a binary string $s$ of length $n$.
Let's define $d_i$ as the number whose decimal representation is $s_i s_{i+1}$ (possibly, with a leading zero). We define $f(s)$ to be the sum of all the valid $d_i$. In other words, $f(s) = \sum\limits_{i=1}^{n-1} d_i$.
For example, for the string $s = 1011$:
$d_1 = 10$ (ten);
$d_2 = 01$ (one)
$d_3 = 11$ (eleven);
$f(s) = 10 + 01 + 11 = 22$.
In one operation you can swap any two adjacent elements of the string. Find the minimum value of $f(s)$ that can be achieved if at most $k$ operations are allowed.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). Description of the test cases follows.
First line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^5$, $0 \le k \le 10^9$) — the length of the string and the maximum number of operations allowed.
The second line of each test case contains the binary string $s$ of length $n$, consisting of only zeros and ones.
It is also given that sum of $n$ over all the test cases doesn't exceed $10^5$.
-----Output-----
For each test case, print the minimum value of $f(s)$ you can obtain with at most $k$ operations.
-----Examples-----
Input
3
4 0
1010
7 1
0010100
5 2
00110
Output
21
22
12
-----Note-----
For the first example, you can't do any operation so the optimal string is $s$ itself. $f(s) = f(1010) = 10 + 01 + 10 = 21$.
For the second example, one of the optimal strings you can obtain is "0011000". The string has an $f$ value of $22$.
For the third example, one of the optimal strings you can obtain is "00011". The string has an $f$ value of $12$. | t = int(input())
def solve(t):
n, k = map(int, input().split())
s = input()
first, last = 10**10, -(10**10)
num = s.count("1")
sum_ = num * 11
for i in range(n):
if s[i] == "1":
first = i + 1
break
for i in range(n):
if s[n - i - 1] == "1":
last = n - i
break
if first == 1:
sum_ -= 1
if last == n:
sum_ -= 10
flag = True
if k >= n - last and last != n:
k -= n - last
sum_ -= 10
if last == 1:
sum_ += 1
if first == last:
flag = False
if k >= first - 1 and first != 1 and flag and first != n:
sum_ -= 1
print(sum_)
for i in range(t):
solve(i + 1) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR BIN_OP NUMBER NUMBER BIN_OP NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR BIN_OP BIN_OP VAR VAR NUMBER STRING ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR BIN_OP VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR NUMBER IF VAR NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR NUMBER VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER |
You are given a binary string $s$ of length $n$.
Let's define $d_i$ as the number whose decimal representation is $s_i s_{i+1}$ (possibly, with a leading zero). We define $f(s)$ to be the sum of all the valid $d_i$. In other words, $f(s) = \sum\limits_{i=1}^{n-1} d_i$.
For example, for the string $s = 1011$:
$d_1 = 10$ (ten);
$d_2 = 01$ (one)
$d_3 = 11$ (eleven);
$f(s) = 10 + 01 + 11 = 22$.
In one operation you can swap any two adjacent elements of the string. Find the minimum value of $f(s)$ that can be achieved if at most $k$ operations are allowed.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). Description of the test cases follows.
First line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^5$, $0 \le k \le 10^9$) — the length of the string and the maximum number of operations allowed.
The second line of each test case contains the binary string $s$ of length $n$, consisting of only zeros and ones.
It is also given that sum of $n$ over all the test cases doesn't exceed $10^5$.
-----Output-----
For each test case, print the minimum value of $f(s)$ you can obtain with at most $k$ operations.
-----Examples-----
Input
3
4 0
1010
7 1
0010100
5 2
00110
Output
21
22
12
-----Note-----
For the first example, you can't do any operation so the optimal string is $s$ itself. $f(s) = f(1010) = 10 + 01 + 10 = 21$.
For the second example, one of the optimal strings you can obtain is "0011000". The string has an $f$ value of $22$.
For the third example, one of the optimal strings you can obtain is "00011". The string has an $f$ value of $12$. | def count(l):
ans = 0
n = len(l)
for i in range(n - 1):
if l[i] == "0" and l[i + 1] == "0":
continue
elif l[i] == "0" and l[i + 1] == "1":
ans += 1
continue
elif l[i] == "1" and l[i + 1] == "0":
ans += 10
continue
else:
ans += 11
return ans
def ss(n, l, k):
f = 0
if l[-1] == "0":
for i in range(n - 2, -1, -1):
if l[i] == "1":
s = n - i - 1
if s <= k:
k -= s
l[i] = "0"
l[-1] = "1"
f = 1
break
if l[-1] == "1":
f = 1
if f == 1 and l.count("1") == 1:
return count(l)
if l[0] == "0":
for i in range(1, n):
if l[i] == "1":
s = i
if s <= k:
k -= s
l[i] = "0"
l[0] = "1"
f = 1
break
return count(l)
for _ in range(int(input())):
n, k = map(int, input().split())
l = list(str(input()))
print(ss(n, l, k)) | FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR STRING VAR BIN_OP VAR NUMBER STRING IF VAR VAR STRING VAR BIN_OP VAR NUMBER STRING VAR NUMBER IF VAR VAR STRING VAR BIN_OP VAR NUMBER STRING VAR NUMBER VAR NUMBER RETURN VAR FUNC_DEF ASSIGN VAR NUMBER IF VAR NUMBER STRING FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR VAR ASSIGN VAR VAR STRING ASSIGN VAR NUMBER STRING ASSIGN VAR NUMBER IF VAR NUMBER STRING ASSIGN VAR NUMBER IF VAR NUMBER FUNC_CALL VAR STRING NUMBER RETURN FUNC_CALL VAR VAR IF VAR NUMBER STRING FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR STRING ASSIGN VAR VAR IF VAR VAR VAR VAR ASSIGN VAR VAR STRING ASSIGN VAR NUMBER STRING ASSIGN VAR NUMBER RETURN FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR |
You are given a binary string $s$ of length $n$.
Let's define $d_i$ as the number whose decimal representation is $s_i s_{i+1}$ (possibly, with a leading zero). We define $f(s)$ to be the sum of all the valid $d_i$. In other words, $f(s) = \sum\limits_{i=1}^{n-1} d_i$.
For example, for the string $s = 1011$:
$d_1 = 10$ (ten);
$d_2 = 01$ (one)
$d_3 = 11$ (eleven);
$f(s) = 10 + 01 + 11 = 22$.
In one operation you can swap any two adjacent elements of the string. Find the minimum value of $f(s)$ that can be achieved if at most $k$ operations are allowed.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). Description of the test cases follows.
First line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^5$, $0 \le k \le 10^9$) — the length of the string and the maximum number of operations allowed.
The second line of each test case contains the binary string $s$ of length $n$, consisting of only zeros and ones.
It is also given that sum of $n$ over all the test cases doesn't exceed $10^5$.
-----Output-----
For each test case, print the minimum value of $f(s)$ you can obtain with at most $k$ operations.
-----Examples-----
Input
3
4 0
1010
7 1
0010100
5 2
00110
Output
21
22
12
-----Note-----
For the first example, you can't do any operation so the optimal string is $s$ itself. $f(s) = f(1010) = 10 + 01 + 10 = 21$.
For the second example, one of the optimal strings you can obtain is "0011000". The string has an $f$ value of $22$.
For the third example, one of the optimal strings you can obtain is "00011". The string has an $f$ value of $12$. | def fun(l):
n = len(l)
c = 0
for i in range(n - 1):
c += int(l[i] + l[i + 1])
return c
t = int(input())
for _ in range(t):
n, k = map(int, input().split())
s = input().rstrip()
l = [x for x in s]
for i in range(n - 1, -1, -1):
if l[i] == "1":
if k >= n - 1 - i:
l[i], l[n - 1] = l[n - 1], l[i]
k -= n - 1 - i
break
for i in range(n - 1):
if l[i] == "1":
if k >= i:
l[0], l[i] = l[i], l[0]
k -= i
break
ans2 = fun(l)
print(ans2) | FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR STRING IF VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP BIN_OP VAR NUMBER VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR STRING IF VAR VAR ASSIGN VAR NUMBER VAR VAR VAR VAR VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR |
You are given a binary string $s$ of length $n$.
Let's define $d_i$ as the number whose decimal representation is $s_i s_{i+1}$ (possibly, with a leading zero). We define $f(s)$ to be the sum of all the valid $d_i$. In other words, $f(s) = \sum\limits_{i=1}^{n-1} d_i$.
For example, for the string $s = 1011$:
$d_1 = 10$ (ten);
$d_2 = 01$ (one)
$d_3 = 11$ (eleven);
$f(s) = 10 + 01 + 11 = 22$.
In one operation you can swap any two adjacent elements of the string. Find the minimum value of $f(s)$ that can be achieved if at most $k$ operations are allowed.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). Description of the test cases follows.
First line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^5$, $0 \le k \le 10^9$) — the length of the string and the maximum number of operations allowed.
The second line of each test case contains the binary string $s$ of length $n$, consisting of only zeros and ones.
It is also given that sum of $n$ over all the test cases doesn't exceed $10^5$.
-----Output-----
For each test case, print the minimum value of $f(s)$ you can obtain with at most $k$ operations.
-----Examples-----
Input
3
4 0
1010
7 1
0010100
5 2
00110
Output
21
22
12
-----Note-----
For the first example, you can't do any operation so the optimal string is $s$ itself. $f(s) = f(1010) = 10 + 01 + 10 = 21$.
For the second example, one of the optimal strings you can obtain is "0011000". The string has an $f$ value of $22$.
For the third example, one of the optimal strings you can obtain is "00011". The string has an $f$ value of $12$. | for _ in range(int(input())):
n, k = map(int, input().split())
s = input()
beg = s.find("1")
if beg == -1:
print(0)
continue
endix = s.rfind("1")
end = n - endix - 1
s = list(s)
if k >= end:
if s.count("1") == 1:
print(1)
continue
s[endix] = "0"
s[n - 1] = "1"
k -= end
if k >= beg:
s[beg] = "0"
s[0] = "1"
ans = 0
s = "".join(s)
for i in range(n - 1):
ans += int(s[i : i + 2])
print(ans) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR STRING IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR IF FUNC_CALL VAR STRING NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER STRING VAR VAR IF VAR VAR ASSIGN VAR VAR STRING ASSIGN VAR NUMBER STRING ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL STRING VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR |
You are given a binary string $s$ of length $n$.
Let's define $d_i$ as the number whose decimal representation is $s_i s_{i+1}$ (possibly, with a leading zero). We define $f(s)$ to be the sum of all the valid $d_i$. In other words, $f(s) = \sum\limits_{i=1}^{n-1} d_i$.
For example, for the string $s = 1011$:
$d_1 = 10$ (ten);
$d_2 = 01$ (one)
$d_3 = 11$ (eleven);
$f(s) = 10 + 01 + 11 = 22$.
In one operation you can swap any two adjacent elements of the string. Find the minimum value of $f(s)$ that can be achieved if at most $k$ operations are allowed.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). Description of the test cases follows.
First line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^5$, $0 \le k \le 10^9$) — the length of the string and the maximum number of operations allowed.
The second line of each test case contains the binary string $s$ of length $n$, consisting of only zeros and ones.
It is also given that sum of $n$ over all the test cases doesn't exceed $10^5$.
-----Output-----
For each test case, print the minimum value of $f(s)$ you can obtain with at most $k$ operations.
-----Examples-----
Input
3
4 0
1010
7 1
0010100
5 2
00110
Output
21
22
12
-----Note-----
For the first example, you can't do any operation so the optimal string is $s$ itself. $f(s) = f(1010) = 10 + 01 + 10 = 21$.
For the second example, one of the optimal strings you can obtain is "0011000". The string has an $f$ value of $22$.
For the third example, one of the optimal strings you can obtain is "00011". The string has an $f$ value of $12$. | from sys import stdin, stdout
I = stdin.readline
O = stdout.write
for _ in range(int(I())):
n, k = list(map(int, I().split()))
st = I()
la = -k
st = list(st)
ans = 0
if st.count("1") == 0:
print(0)
continue
for j in range(n - 1, -1, -1):
if st[j] == "1":
la = j
break
if k >= n - 1 - la:
st[la] = "0"
st[n - 1] = "1"
k -= n - 1 - la
s = 0
for i in range(n - 1):
if st[i] == "1":
now = i - s
if k >= now:
k -= now
st[i] = "0"
st[s] = "1"
s += 1
for i in range(n - 1):
ans += int(st[i] + st[i + 1])
print(ans) | ASSIGN VAR VAR ASSIGN VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER IF FUNC_CALL VAR STRING NUMBER EXPR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR STRING ASSIGN VAR VAR IF VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER STRING VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR STRING ASSIGN VAR BIN_OP VAR VAR IF VAR VAR VAR VAR ASSIGN VAR VAR STRING ASSIGN VAR VAR STRING VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR |
You are given a binary string $s$ of length $n$.
Let's define $d_i$ as the number whose decimal representation is $s_i s_{i+1}$ (possibly, with a leading zero). We define $f(s)$ to be the sum of all the valid $d_i$. In other words, $f(s) = \sum\limits_{i=1}^{n-1} d_i$.
For example, for the string $s = 1011$:
$d_1 = 10$ (ten);
$d_2 = 01$ (one)
$d_3 = 11$ (eleven);
$f(s) = 10 + 01 + 11 = 22$.
In one operation you can swap any two adjacent elements of the string. Find the minimum value of $f(s)$ that can be achieved if at most $k$ operations are allowed.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). Description of the test cases follows.
First line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^5$, $0 \le k \le 10^9$) — the length of the string and the maximum number of operations allowed.
The second line of each test case contains the binary string $s$ of length $n$, consisting of only zeros and ones.
It is also given that sum of $n$ over all the test cases doesn't exceed $10^5$.
-----Output-----
For each test case, print the minimum value of $f(s)$ you can obtain with at most $k$ operations.
-----Examples-----
Input
3
4 0
1010
7 1
0010100
5 2
00110
Output
21
22
12
-----Note-----
For the first example, you can't do any operation so the optimal string is $s$ itself. $f(s) = f(1010) = 10 + 01 + 10 = 21$.
For the second example, one of the optimal strings you can obtain is "0011000". The string has an $f$ value of $22$.
For the third example, one of the optimal strings you can obtain is "00011". The string has an $f$ value of $12$. | x = int(input())
ret = []
for z in range(x):
n, k = input().split()
k = int(k)
a = input()
spl = False
lst = a.split("1")
l, r, x = 0, 0, 0
l = len(lst[0])
r = len(lst[-1])
ones = len(lst) - 1
min = ones * 11
if l == len(a):
ret.append(min)
continue
if k >= l + r:
x = 11
spl = True
elif k >= r:
x = 10
elif k >= l:
x = 1
if ones == 1 and spl:
x = 10
min = min - x
ret.append(min)
for i in ret:
print(i) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF VAR BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR |
You are given a binary string $s$ of length $n$.
Let's define $d_i$ as the number whose decimal representation is $s_i s_{i+1}$ (possibly, with a leading zero). We define $f(s)$ to be the sum of all the valid $d_i$. In other words, $f(s) = \sum\limits_{i=1}^{n-1} d_i$.
For example, for the string $s = 1011$:
$d_1 = 10$ (ten);
$d_2 = 01$ (one)
$d_3 = 11$ (eleven);
$f(s) = 10 + 01 + 11 = 22$.
In one operation you can swap any two adjacent elements of the string. Find the minimum value of $f(s)$ that can be achieved if at most $k$ operations are allowed.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). Description of the test cases follows.
First line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^5$, $0 \le k \le 10^9$) — the length of the string and the maximum number of operations allowed.
The second line of each test case contains the binary string $s$ of length $n$, consisting of only zeros and ones.
It is also given that sum of $n$ over all the test cases doesn't exceed $10^5$.
-----Output-----
For each test case, print the minimum value of $f(s)$ you can obtain with at most $k$ operations.
-----Examples-----
Input
3
4 0
1010
7 1
0010100
5 2
00110
Output
21
22
12
-----Note-----
For the first example, you can't do any operation so the optimal string is $s$ itself. $f(s) = f(1010) = 10 + 01 + 10 = 21$.
For the second example, one of the optimal strings you can obtain is "0011000". The string has an $f$ value of $22$.
For the third example, one of the optimal strings you can obtain is "00011". The string has an $f$ value of $12$. | for _ in range(int(input())):
n, k = [int(x) for x in input().split()]
s = list(input())
l = []
ans = 0
t = s.count("1")
if t == 0:
print(0)
continue
for i in range(n):
if s[i] == "1":
l.append(i)
ans += 11
used = 0
if k >= n - l[-1] - 1:
k -= n - l[-1] - 1
used = 1
ans -= 10
if k >= l[0] and (used == 0 or t > 1):
ans -= 1
print(ans) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING EXPR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR BIN_OP BIN_OP VAR VAR NUMBER NUMBER VAR BIN_OP BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER VAR NUMBER IF VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
You are given a binary string $s$ of length $n$.
Let's define $d_i$ as the number whose decimal representation is $s_i s_{i+1}$ (possibly, with a leading zero). We define $f(s)$ to be the sum of all the valid $d_i$. In other words, $f(s) = \sum\limits_{i=1}^{n-1} d_i$.
For example, for the string $s = 1011$:
$d_1 = 10$ (ten);
$d_2 = 01$ (one)
$d_3 = 11$ (eleven);
$f(s) = 10 + 01 + 11 = 22$.
In one operation you can swap any two adjacent elements of the string. Find the minimum value of $f(s)$ that can be achieved if at most $k$ operations are allowed.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). Description of the test cases follows.
First line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^5$, $0 \le k \le 10^9$) — the length of the string and the maximum number of operations allowed.
The second line of each test case contains the binary string $s$ of length $n$, consisting of only zeros and ones.
It is also given that sum of $n$ over all the test cases doesn't exceed $10^5$.
-----Output-----
For each test case, print the minimum value of $f(s)$ you can obtain with at most $k$ operations.
-----Examples-----
Input
3
4 0
1010
7 1
0010100
5 2
00110
Output
21
22
12
-----Note-----
For the first example, you can't do any operation so the optimal string is $s$ itself. $f(s) = f(1010) = 10 + 01 + 10 = 21$.
For the second example, one of the optimal strings you can obtain is "0011000". The string has an $f$ value of $22$.
For the third example, one of the optimal strings you can obtain is "00011". The string has an $f$ value of $12$. | for _ in range(int(input())):
a, b = map(int, input().split())
c = input()
c = list(c)
if c.count("1"):
y = c.index("1")
s = c.count("1") * 11
c = c[::-1]
x = c.index("1")
if x <= b:
b = b - x
s -= 10
if y <= b and c.count("1") != 1:
s -= 1
elif y <= b:
s -= 1
print(s)
else:
print(0) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR STRING NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING IF VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR NUMBER IF VAR VAR FUNC_CALL VAR STRING NUMBER VAR NUMBER IF VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER |
You are given a binary string $s$ of length $n$.
Let's define $d_i$ as the number whose decimal representation is $s_i s_{i+1}$ (possibly, with a leading zero). We define $f(s)$ to be the sum of all the valid $d_i$. In other words, $f(s) = \sum\limits_{i=1}^{n-1} d_i$.
For example, for the string $s = 1011$:
$d_1 = 10$ (ten);
$d_2 = 01$ (one)
$d_3 = 11$ (eleven);
$f(s) = 10 + 01 + 11 = 22$.
In one operation you can swap any two adjacent elements of the string. Find the minimum value of $f(s)$ that can be achieved if at most $k$ operations are allowed.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). Description of the test cases follows.
First line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^5$, $0 \le k \le 10^9$) — the length of the string and the maximum number of operations allowed.
The second line of each test case contains the binary string $s$ of length $n$, consisting of only zeros and ones.
It is also given that sum of $n$ over all the test cases doesn't exceed $10^5$.
-----Output-----
For each test case, print the minimum value of $f(s)$ you can obtain with at most $k$ operations.
-----Examples-----
Input
3
4 0
1010
7 1
0010100
5 2
00110
Output
21
22
12
-----Note-----
For the first example, you can't do any operation so the optimal string is $s$ itself. $f(s) = f(1010) = 10 + 01 + 10 = 21$.
For the second example, one of the optimal strings you can obtain is "0011000". The string has an $f$ value of $22$.
For the third example, one of the optimal strings you can obtain is "00011". The string has an $f$ value of $12$. | def substringSum(s):
sum = 0
for i in range(1, len(s)):
sum += int(s[i - 1] + s[i])
return sum
def solution(n, s, k):
min_idx = s.index("1")
max_idx = -1
for idx, x in enumerate(s):
if s[idx] == "1":
max_idx = idx
sum = substringSum(s)
if k == 0:
print(sum)
else:
rightShift = False
if k >= n - 1 - max_idx and max_idx != n - 1:
sum -= 10 - (max_idx == 0)
k -= n - 1 - max_idx
rightShift = True
if (
min_idx != max_idx or not rightShift and min_idx != n - 1
) and 1 <= min_idx <= k:
sum -= 1
k -= min_idx
print(sum)
T = int(input())
for t in range(T):
n, k = [int(x) for x in input().split()]
s = input()
if "1" not in s:
print("0")
else:
solution(n, s, k) | FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER IF VAR BIN_OP BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR BIN_OP NUMBER VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR NUMBER IF VAR VAR VAR VAR BIN_OP VAR NUMBER NUMBER VAR VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR IF STRING VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR VAR VAR |
You are given a binary string $s$ of length $n$.
Let's define $d_i$ as the number whose decimal representation is $s_i s_{i+1}$ (possibly, with a leading zero). We define $f(s)$ to be the sum of all the valid $d_i$. In other words, $f(s) = \sum\limits_{i=1}^{n-1} d_i$.
For example, for the string $s = 1011$:
$d_1 = 10$ (ten);
$d_2 = 01$ (one)
$d_3 = 11$ (eleven);
$f(s) = 10 + 01 + 11 = 22$.
In one operation you can swap any two adjacent elements of the string. Find the minimum value of $f(s)$ that can be achieved if at most $k$ operations are allowed.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). Description of the test cases follows.
First line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^5$, $0 \le k \le 10^9$) — the length of the string and the maximum number of operations allowed.
The second line of each test case contains the binary string $s$ of length $n$, consisting of only zeros and ones.
It is also given that sum of $n$ over all the test cases doesn't exceed $10^5$.
-----Output-----
For each test case, print the minimum value of $f(s)$ you can obtain with at most $k$ operations.
-----Examples-----
Input
3
4 0
1010
7 1
0010100
5 2
00110
Output
21
22
12
-----Note-----
For the first example, you can't do any operation so the optimal string is $s$ itself. $f(s) = f(1010) = 10 + 01 + 10 = 21$.
For the second example, one of the optimal strings you can obtain is "0011000". The string has an $f$ value of $22$.
For the third example, one of the optimal strings you can obtain is "00011". The string has an $f$ value of $12$. | tc = int(input())
for t in range(tc):
n, k = map(int, input().split())
st = input()
one = 0
s = n + 100
e = 0
for i in range(n):
if st[i] == "1":
one = one + 1
if s == n + 100:
s = i
e = i
sm = 0
if one > 0 and n - 1 - e <= k:
sm = 1
one = one - 1
k = k - (n - 1 - e)
if one > 0 and s <= k:
sm = sm + 10
one = one - 1
print(sm + 11 * one) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER BIN_OP BIN_OP VAR NUMBER VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR NUMBER VAR IF VAR NUMBER VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP NUMBER VAR |
You are given a binary string $s$ of length $n$.
Let's define $d_i$ as the number whose decimal representation is $s_i s_{i+1}$ (possibly, with a leading zero). We define $f(s)$ to be the sum of all the valid $d_i$. In other words, $f(s) = \sum\limits_{i=1}^{n-1} d_i$.
For example, for the string $s = 1011$:
$d_1 = 10$ (ten);
$d_2 = 01$ (one)
$d_3 = 11$ (eleven);
$f(s) = 10 + 01 + 11 = 22$.
In one operation you can swap any two adjacent elements of the string. Find the minimum value of $f(s)$ that can be achieved if at most $k$ operations are allowed.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). Description of the test cases follows.
First line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^5$, $0 \le k \le 10^9$) — the length of the string and the maximum number of operations allowed.
The second line of each test case contains the binary string $s$ of length $n$, consisting of only zeros and ones.
It is also given that sum of $n$ over all the test cases doesn't exceed $10^5$.
-----Output-----
For each test case, print the minimum value of $f(s)$ you can obtain with at most $k$ operations.
-----Examples-----
Input
3
4 0
1010
7 1
0010100
5 2
00110
Output
21
22
12
-----Note-----
For the first example, you can't do any operation so the optimal string is $s$ itself. $f(s) = f(1010) = 10 + 01 + 10 = 21$.
For the second example, one of the optimal strings you can obtain is "0011000". The string has an $f$ value of $22$.
For the third example, one of the optimal strings you can obtain is "00011". The string has an $f$ value of $12$. | for _ in range(int(input())):
n, k = [int(x) for x in input().split(" ")]
s = input()
cnt = 0
for c in s:
if c == "1":
cnt += 1
if cnt == 0:
print(0)
continue
ans = cnt * 11
left, right = 0, n - 1
while s[left] == "0":
left += 1
while s[right] == "0":
right -= 1
if k >= left + n - 1 - right and left != right:
ans -= 11
elif k >= n - 1 - right:
ans -= 10
elif k >= left:
ans -= 1
print(ans) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR VAR IF VAR STRING VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER BIN_OP VAR NUMBER WHILE VAR VAR STRING VAR NUMBER WHILE VAR VAR STRING VAR NUMBER IF VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER VAR VAR VAR VAR NUMBER IF VAR BIN_OP BIN_OP VAR NUMBER VAR VAR NUMBER IF VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
You are given a binary string $s$ of length $n$.
Let's define $d_i$ as the number whose decimal representation is $s_i s_{i+1}$ (possibly, with a leading zero). We define $f(s)$ to be the sum of all the valid $d_i$. In other words, $f(s) = \sum\limits_{i=1}^{n-1} d_i$.
For example, for the string $s = 1011$:
$d_1 = 10$ (ten);
$d_2 = 01$ (one)
$d_3 = 11$ (eleven);
$f(s) = 10 + 01 + 11 = 22$.
In one operation you can swap any two adjacent elements of the string. Find the minimum value of $f(s)$ that can be achieved if at most $k$ operations are allowed.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). Description of the test cases follows.
First line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^5$, $0 \le k \le 10^9$) — the length of the string and the maximum number of operations allowed.
The second line of each test case contains the binary string $s$ of length $n$, consisting of only zeros and ones.
It is also given that sum of $n$ over all the test cases doesn't exceed $10^5$.
-----Output-----
For each test case, print the minimum value of $f(s)$ you can obtain with at most $k$ operations.
-----Examples-----
Input
3
4 0
1010
7 1
0010100
5 2
00110
Output
21
22
12
-----Note-----
For the first example, you can't do any operation so the optimal string is $s$ itself. $f(s) = f(1010) = 10 + 01 + 10 = 21$.
For the second example, one of the optimal strings you can obtain is "0011000". The string has an $f$ value of $22$.
For the third example, one of the optimal strings you can obtain is "00011". The string has an $f$ value of $12$. | def solve():
n, k = map(int, input().split())
s = list(input())
indl = -1
indr = -1
for i in range(n):
if s[i] == "1":
indl = i
break
for i in range(n - 1, -1, -1):
if s[i] == "1":
indr = i
break
ans = 0
if indl != -1:
if n - 1 - indr <= k:
k -= n - 1 - indr
ans += 1
del s[indr]
if s.count("1") == 0:
print(1)
return
k -= indl
if k >= 0:
ans += 10
del s[indl]
ans += s.count("1") * 11
print(ans)
for _ in range(int(input())):
solve() | FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR STRING ASSIGN VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER IF BIN_OP BIN_OP VAR NUMBER VAR VAR VAR BIN_OP BIN_OP VAR NUMBER VAR VAR NUMBER VAR VAR IF FUNC_CALL VAR STRING NUMBER EXPR FUNC_CALL VAR NUMBER RETURN VAR VAR IF VAR NUMBER VAR NUMBER VAR VAR VAR BIN_OP FUNC_CALL VAR STRING NUMBER EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR |
You are given a binary string $s$ of length $n$.
Let's define $d_i$ as the number whose decimal representation is $s_i s_{i+1}$ (possibly, with a leading zero). We define $f(s)$ to be the sum of all the valid $d_i$. In other words, $f(s) = \sum\limits_{i=1}^{n-1} d_i$.
For example, for the string $s = 1011$:
$d_1 = 10$ (ten);
$d_2 = 01$ (one)
$d_3 = 11$ (eleven);
$f(s) = 10 + 01 + 11 = 22$.
In one operation you can swap any two adjacent elements of the string. Find the minimum value of $f(s)$ that can be achieved if at most $k$ operations are allowed.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). Description of the test cases follows.
First line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^5$, $0 \le k \le 10^9$) — the length of the string and the maximum number of operations allowed.
The second line of each test case contains the binary string $s$ of length $n$, consisting of only zeros and ones.
It is also given that sum of $n$ over all the test cases doesn't exceed $10^5$.
-----Output-----
For each test case, print the minimum value of $f(s)$ you can obtain with at most $k$ operations.
-----Examples-----
Input
3
4 0
1010
7 1
0010100
5 2
00110
Output
21
22
12
-----Note-----
For the first example, you can't do any operation so the optimal string is $s$ itself. $f(s) = f(1010) = 10 + 01 + 10 = 21$.
For the second example, one of the optimal strings you can obtain is "0011000". The string has an $f$ value of $22$.
For the third example, one of the optimal strings you can obtain is "00011". The string has an $f$ value of $12$. | for y in [0] * int(input()):
n, k = map(int, input().split())
s = input()
i, j = (x.find("1") % 8**10 for x in (s, s[::-1]))
print(11 * s.count("1") - (k >= i + j < n - 1 or i <= k < j) - 10 * (j <= k)) | FOR VAR BIN_OP LIST NUMBER FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR BIN_OP FUNC_CALL VAR STRING BIN_OP NUMBER NUMBER VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP NUMBER FUNC_CALL VAR STRING VAR BIN_OP VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP NUMBER VAR VAR |
You are given a binary string $s$ of length $n$.
Let's define $d_i$ as the number whose decimal representation is $s_i s_{i+1}$ (possibly, with a leading zero). We define $f(s)$ to be the sum of all the valid $d_i$. In other words, $f(s) = \sum\limits_{i=1}^{n-1} d_i$.
For example, for the string $s = 1011$:
$d_1 = 10$ (ten);
$d_2 = 01$ (one)
$d_3 = 11$ (eleven);
$f(s) = 10 + 01 + 11 = 22$.
In one operation you can swap any two adjacent elements of the string. Find the minimum value of $f(s)$ that can be achieved if at most $k$ operations are allowed.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). Description of the test cases follows.
First line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^5$, $0 \le k \le 10^9$) — the length of the string and the maximum number of operations allowed.
The second line of each test case contains the binary string $s$ of length $n$, consisting of only zeros and ones.
It is also given that sum of $n$ over all the test cases doesn't exceed $10^5$.
-----Output-----
For each test case, print the minimum value of $f(s)$ you can obtain with at most $k$ operations.
-----Examples-----
Input
3
4 0
1010
7 1
0010100
5 2
00110
Output
21
22
12
-----Note-----
For the first example, you can't do any operation so the optimal string is $s$ itself. $f(s) = f(1010) = 10 + 01 + 10 = 21$.
For the second example, one of the optimal strings you can obtain is "0011000". The string has an $f$ value of $22$.
For the third example, one of the optimal strings you can obtain is "00011". The string has an $f$ value of $12$. | loopcount = int(input())
for loop in range(loopcount):
x = input()
x = x.split(" ")
x = list(map(int, x))
a = input()
output = 0
first = -1
last = -1
for i in range(x[0]):
if a[i] == "1":
output += 11
last = i
if first == -1:
first = i
if first == -1:
print(output)
continue
if first == last:
if x[0] - 1 - last <= x[1]:
output -= 10
x[1] -= x[0] - 1 - last
elif last <= x[1]:
output -= 1
else:
if x[0] - 1 - last <= x[1]:
output -= 10
x[1] -= x[0] - 1 - last
if first <= x[1]:
output -= 1
print(output) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR NUMBER IF VAR VAR STRING VAR NUMBER ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR VAR IF BIN_OP BIN_OP VAR NUMBER NUMBER VAR VAR NUMBER VAR NUMBER VAR NUMBER BIN_OP BIN_OP VAR NUMBER NUMBER VAR IF VAR VAR NUMBER VAR NUMBER IF BIN_OP BIN_OP VAR NUMBER NUMBER VAR VAR NUMBER VAR NUMBER VAR NUMBER BIN_OP BIN_OP VAR NUMBER NUMBER VAR IF VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
You are given a binary string $s$ of length $n$.
Let's define $d_i$ as the number whose decimal representation is $s_i s_{i+1}$ (possibly, with a leading zero). We define $f(s)$ to be the sum of all the valid $d_i$. In other words, $f(s) = \sum\limits_{i=1}^{n-1} d_i$.
For example, for the string $s = 1011$:
$d_1 = 10$ (ten);
$d_2 = 01$ (one)
$d_3 = 11$ (eleven);
$f(s) = 10 + 01 + 11 = 22$.
In one operation you can swap any two adjacent elements of the string. Find the minimum value of $f(s)$ that can be achieved if at most $k$ operations are allowed.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). Description of the test cases follows.
First line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^5$, $0 \le k \le 10^9$) — the length of the string and the maximum number of operations allowed.
The second line of each test case contains the binary string $s$ of length $n$, consisting of only zeros and ones.
It is also given that sum of $n$ over all the test cases doesn't exceed $10^5$.
-----Output-----
For each test case, print the minimum value of $f(s)$ you can obtain with at most $k$ operations.
-----Examples-----
Input
3
4 0
1010
7 1
0010100
5 2
00110
Output
21
22
12
-----Note-----
For the first example, you can't do any operation so the optimal string is $s$ itself. $f(s) = f(1010) = 10 + 01 + 10 = 21$.
For the second example, one of the optimal strings you can obtain is "0011000". The string has an $f$ value of $22$.
For the third example, one of the optimal strings you can obtain is "00011". The string has an $f$ value of $12$. | t = int(input())
for _ in range(t):
n, k = (int(number) for number in input().split(" "))
item = 0
zero_count_right = 0
zero_count_left = 0
ones = 0
zeros = 0
for number in input():
if item == 0:
if number == "1":
left = 1
count_left = False
else:
left = 0
count_left = True
if number == "0":
zeros += 1
zero_count_right += 1
if count_left is True:
zero_count_left += 1
else:
ones += 1
zero_count_right = 0
count_left = False
item += 1
sum_row = ones * 11
if ones >= 1 and zero_count_right <= k:
sum_row -= 10
ones -= 1
k -= zero_count_right
if ones >= 1 and zero_count_left <= k:
sum_row -= 1
print(sum_row) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR IF VAR NUMBER IF VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR NUMBER VAR VAR VAR NUMBER VAR NUMBER VAR VAR IF VAR NUMBER VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
You are given a binary string $s$ of length $n$.
Let's define $d_i$ as the number whose decimal representation is $s_i s_{i+1}$ (possibly, with a leading zero). We define $f(s)$ to be the sum of all the valid $d_i$. In other words, $f(s) = \sum\limits_{i=1}^{n-1} d_i$.
For example, for the string $s = 1011$:
$d_1 = 10$ (ten);
$d_2 = 01$ (one)
$d_3 = 11$ (eleven);
$f(s) = 10 + 01 + 11 = 22$.
In one operation you can swap any two adjacent elements of the string. Find the minimum value of $f(s)$ that can be achieved if at most $k$ operations are allowed.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). Description of the test cases follows.
First line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^5$, $0 \le k \le 10^9$) — the length of the string and the maximum number of operations allowed.
The second line of each test case contains the binary string $s$ of length $n$, consisting of only zeros and ones.
It is also given that sum of $n$ over all the test cases doesn't exceed $10^5$.
-----Output-----
For each test case, print the minimum value of $f(s)$ you can obtain with at most $k$ operations.
-----Examples-----
Input
3
4 0
1010
7 1
0010100
5 2
00110
Output
21
22
12
-----Note-----
For the first example, you can't do any operation so the optimal string is $s$ itself. $f(s) = f(1010) = 10 + 01 + 10 = 21$.
For the second example, one of the optimal strings you can obtain is "0011000". The string has an $f$ value of $22$.
For the third example, one of the optimal strings you can obtain is "00011". The string has an $f$ value of $12$. | for t in range(int(input())):
n, k = [int(i) for i in input().split()]
s = input()
first = -1
last = n
ctr = 0
for i in range(n):
if s[i] == "1":
if first == -1:
first = i
last = i
ctr += 1
if ctr == 0:
print(0)
elif ctr == 1:
if n - 1 - last <= k:
print(1)
elif first <= k:
print(10)
else:
print(11)
else:
total = 0
if n - 1 - last <= k:
total += 1
ctr -= 1
k -= n - 1 - last
if first <= k:
total += 10
ctr -= 1
total += ctr * 11
print(total) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR NUMBER IF BIN_OP BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP BIN_OP VAR NUMBER VAR VAR VAR NUMBER VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR IF VAR VAR VAR NUMBER VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR |
You are given a binary string $s$ of length $n$.
Let's define $d_i$ as the number whose decimal representation is $s_i s_{i+1}$ (possibly, with a leading zero). We define $f(s)$ to be the sum of all the valid $d_i$. In other words, $f(s) = \sum\limits_{i=1}^{n-1} d_i$.
For example, for the string $s = 1011$:
$d_1 = 10$ (ten);
$d_2 = 01$ (one)
$d_3 = 11$ (eleven);
$f(s) = 10 + 01 + 11 = 22$.
In one operation you can swap any two adjacent elements of the string. Find the minimum value of $f(s)$ that can be achieved if at most $k$ operations are allowed.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). Description of the test cases follows.
First line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^5$, $0 \le k \le 10^9$) — the length of the string and the maximum number of operations allowed.
The second line of each test case contains the binary string $s$ of length $n$, consisting of only zeros and ones.
It is also given that sum of $n$ over all the test cases doesn't exceed $10^5$.
-----Output-----
For each test case, print the minimum value of $f(s)$ you can obtain with at most $k$ operations.
-----Examples-----
Input
3
4 0
1010
7 1
0010100
5 2
00110
Output
21
22
12
-----Note-----
For the first example, you can't do any operation so the optimal string is $s$ itself. $f(s) = f(1010) = 10 + 01 + 10 = 21$.
For the second example, one of the optimal strings you can obtain is "0011000". The string has an $f$ value of $22$.
For the third example, one of the optimal strings you can obtain is "00011". The string has an $f$ value of $12$. | def task(n, k, s):
total = 0
for i in range(n - 1):
value = int(s[i : i + 2])
total += value
ones = 0
for c in s:
if c == "1":
ones += 1
startzeros = 0
for c in s:
if c == "0":
startzeros += 1
else:
break
endzeros = 0
for c in s[::-1]:
if c == "0":
endzeros += 1
else:
break
if ones > 0:
if endzeros > 0:
if endzeros <= k:
k -= endzeros
if ones == 1 and startzeros == 0:
total -= 9
else:
total -= 10
ones -= 1
else:
ones -= 1
if ones > 0 and startzeros > 0 and startzeros <= k:
total -= 1
print(total)
t = int(input())
for i in range(0, t):
n, k = map(int, input().split())
s = input()
task(n, k, s) | FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR NUMBER FOR VAR VAR IF VAR STRING VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR STRING VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR NUMBER IF VAR STRING VAR NUMBER IF VAR NUMBER IF VAR NUMBER IF VAR VAR VAR VAR IF VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER IF VAR NUMBER VAR NUMBER VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR VAR VAR |
You are given a binary string $s$ of length $n$.
Let's define $d_i$ as the number whose decimal representation is $s_i s_{i+1}$ (possibly, with a leading zero). We define $f(s)$ to be the sum of all the valid $d_i$. In other words, $f(s) = \sum\limits_{i=1}^{n-1} d_i$.
For example, for the string $s = 1011$:
$d_1 = 10$ (ten);
$d_2 = 01$ (one)
$d_3 = 11$ (eleven);
$f(s) = 10 + 01 + 11 = 22$.
In one operation you can swap any two adjacent elements of the string. Find the minimum value of $f(s)$ that can be achieved if at most $k$ operations are allowed.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). Description of the test cases follows.
First line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^5$, $0 \le k \le 10^9$) — the length of the string and the maximum number of operations allowed.
The second line of each test case contains the binary string $s$ of length $n$, consisting of only zeros and ones.
It is also given that sum of $n$ over all the test cases doesn't exceed $10^5$.
-----Output-----
For each test case, print the minimum value of $f(s)$ you can obtain with at most $k$ operations.
-----Examples-----
Input
3
4 0
1010
7 1
0010100
5 2
00110
Output
21
22
12
-----Note-----
For the first example, you can't do any operation so the optimal string is $s$ itself. $f(s) = f(1010) = 10 + 01 + 10 = 21$.
For the second example, one of the optimal strings you can obtain is "0011000". The string has an $f$ value of $22$.
For the third example, one of the optimal strings you can obtain is "00011". The string has an $f$ value of $12$. | for _ in range(int(input())):
n, k = map(int, input().split())
s = list(map(int, input()))
fs = 0
for i in range(n - 1):
fs += 10 * s[i] + s[i + 1]
if k == 0:
print(fs)
elif s.count(1) == 0:
print(0)
elif s.count(1) == 1:
zf = s.index(1)
s1 = s[:]
s1.reverse()
ze = s1.index(1)
if ze <= k:
print(1)
elif zf <= k:
print(10)
else:
print(11)
else:
zf = s.index(1)
s1 = s[:]
s1.reverse()
ze = s1.index(1)
if 0 < zf + ze <= k:
if zf > 0 and ze > 0:
print(fs - 11)
elif ze > 0:
print(fs - 10)
elif zf > 0:
print(fs - 1)
elif 0 < ze <= k:
print(fs - 10)
elif 0 < zf <= k:
print(fs - 1)
else:
print(fs) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER IF FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER ASSIGN VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER ASSIGN VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR NUMBER IF NUMBER BIN_OP VAR VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER IF NUMBER VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER IF NUMBER VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR |
You are given a binary string $s$ of length $n$.
Let's define $d_i$ as the number whose decimal representation is $s_i s_{i+1}$ (possibly, with a leading zero). We define $f(s)$ to be the sum of all the valid $d_i$. In other words, $f(s) = \sum\limits_{i=1}^{n-1} d_i$.
For example, for the string $s = 1011$:
$d_1 = 10$ (ten);
$d_2 = 01$ (one)
$d_3 = 11$ (eleven);
$f(s) = 10 + 01 + 11 = 22$.
In one operation you can swap any two adjacent elements of the string. Find the minimum value of $f(s)$ that can be achieved if at most $k$ operations are allowed.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). Description of the test cases follows.
First line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^5$, $0 \le k \le 10^9$) — the length of the string and the maximum number of operations allowed.
The second line of each test case contains the binary string $s$ of length $n$, consisting of only zeros and ones.
It is also given that sum of $n$ over all the test cases doesn't exceed $10^5$.
-----Output-----
For each test case, print the minimum value of $f(s)$ you can obtain with at most $k$ operations.
-----Examples-----
Input
3
4 0
1010
7 1
0010100
5 2
00110
Output
21
22
12
-----Note-----
For the first example, you can't do any operation so the optimal string is $s$ itself. $f(s) = f(1010) = 10 + 01 + 10 = 21$.
For the second example, one of the optimal strings you can obtain is "0011000". The string has an $f$ value of $22$.
For the third example, one of the optimal strings you can obtain is "00011". The string has an $f$ value of $12$. | for _ in range(int(input())):
n, k = map(int, input().split())
s = input().strip()
z = sum(11 * int(c) for c in s)
k = min(k, n - 1)
a, b = 0, 0
while a <= k:
if s[-a - 1] == "1":
break
a += 1
while b <= k:
if s[b] == "1":
break
b += 1
if b + a == n - 1 == k:
print(z - 10)
elif b + a <= k:
print(z - 11)
elif a <= k:
print(z - 10)
elif b <= k:
print(z - 1)
else:
print(z) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR BIN_OP NUMBER FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER WHILE VAR VAR IF VAR BIN_OP VAR NUMBER STRING VAR NUMBER WHILE VAR VAR IF VAR VAR STRING VAR NUMBER IF BIN_OP VAR VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER IF BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR |
You are given a binary string $s$ of length $n$.
Let's define $d_i$ as the number whose decimal representation is $s_i s_{i+1}$ (possibly, with a leading zero). We define $f(s)$ to be the sum of all the valid $d_i$. In other words, $f(s) = \sum\limits_{i=1}^{n-1} d_i$.
For example, for the string $s = 1011$:
$d_1 = 10$ (ten);
$d_2 = 01$ (one)
$d_3 = 11$ (eleven);
$f(s) = 10 + 01 + 11 = 22$.
In one operation you can swap any two adjacent elements of the string. Find the minimum value of $f(s)$ that can be achieved if at most $k$ operations are allowed.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). Description of the test cases follows.
First line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^5$, $0 \le k \le 10^9$) — the length of the string and the maximum number of operations allowed.
The second line of each test case contains the binary string $s$ of length $n$, consisting of only zeros and ones.
It is also given that sum of $n$ over all the test cases doesn't exceed $10^5$.
-----Output-----
For each test case, print the minimum value of $f(s)$ you can obtain with at most $k$ operations.
-----Examples-----
Input
3
4 0
1010
7 1
0010100
5 2
00110
Output
21
22
12
-----Note-----
For the first example, you can't do any operation so the optimal string is $s$ itself. $f(s) = f(1010) = 10 + 01 + 10 = 21$.
For the second example, one of the optimal strings you can obtain is "0011000". The string has an $f$ value of $22$.
For the third example, one of the optimal strings you can obtain is "00011". The string has an $f$ value of $12$. | t = int(input())
for e in range(t):
n, k = map(int, input().split())
a = input()
l, r = -1, -1
s = 0
for i in range(n):
if a[i] == "1":
s += 11
r = i
if l == -1:
l = i
if r == -1:
print(s)
continue
f = k
if k >= n - r - 1:
s -= 10
f -= n - r - 1
if l != r or k < n - r - 1:
if f >= l:
s -= 1
print(s) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR IF VAR BIN_OP BIN_OP VAR VAR NUMBER VAR NUMBER VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
You are given a binary string $s$ of length $n$.
Let's define $d_i$ as the number whose decimal representation is $s_i s_{i+1}$ (possibly, with a leading zero). We define $f(s)$ to be the sum of all the valid $d_i$. In other words, $f(s) = \sum\limits_{i=1}^{n-1} d_i$.
For example, for the string $s = 1011$:
$d_1 = 10$ (ten);
$d_2 = 01$ (one)
$d_3 = 11$ (eleven);
$f(s) = 10 + 01 + 11 = 22$.
In one operation you can swap any two adjacent elements of the string. Find the minimum value of $f(s)$ that can be achieved if at most $k$ operations are allowed.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). Description of the test cases follows.
First line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^5$, $0 \le k \le 10^9$) — the length of the string and the maximum number of operations allowed.
The second line of each test case contains the binary string $s$ of length $n$, consisting of only zeros and ones.
It is also given that sum of $n$ over all the test cases doesn't exceed $10^5$.
-----Output-----
For each test case, print the minimum value of $f(s)$ you can obtain with at most $k$ operations.
-----Examples-----
Input
3
4 0
1010
7 1
0010100
5 2
00110
Output
21
22
12
-----Note-----
For the first example, you can't do any operation so the optimal string is $s$ itself. $f(s) = f(1010) = 10 + 01 + 10 = 21$.
For the second example, one of the optimal strings you can obtain is "0011000". The string has an $f$ value of $22$.
For the third example, one of the optimal strings you can obtain is "00011". The string has an $f$ value of $12$. | t = int(input())
for _ in range(t):
n, k = map(int, input().split())
L = list(input())
L = [int(x) for x in L]
if not 1 in L:
print(0)
continue
if L.count(1) == 1:
a = L.index(1)
if k >= n - 1 - a:
print(1)
elif k >= a:
print(10)
else:
print(11)
continue
a = L.index(1)
L.reverse()
b = L.index(1)
b = n - b - 1
c = L.count(1)
if k >= n - 1 - b + a:
ans = c * 11 - 11
elif k >= n - 1 - b:
ans = c * 11 - 10
elif k >= a:
ans = c * 11 - 1
else:
ans = c * 11
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR IF NUMBER VAR EXPR FUNC_CALL VAR NUMBER IF FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER IF VAR BIN_OP BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER IF VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR |
You are given a binary string $s$ of length $n$.
Let's define $d_i$ as the number whose decimal representation is $s_i s_{i+1}$ (possibly, with a leading zero). We define $f(s)$ to be the sum of all the valid $d_i$. In other words, $f(s) = \sum\limits_{i=1}^{n-1} d_i$.
For example, for the string $s = 1011$:
$d_1 = 10$ (ten);
$d_2 = 01$ (one)
$d_3 = 11$ (eleven);
$f(s) = 10 + 01 + 11 = 22$.
In one operation you can swap any two adjacent elements of the string. Find the minimum value of $f(s)$ that can be achieved if at most $k$ operations are allowed.
-----Input-----
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). Description of the test cases follows.
First line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^5$, $0 \le k \le 10^9$) — the length of the string and the maximum number of operations allowed.
The second line of each test case contains the binary string $s$ of length $n$, consisting of only zeros and ones.
It is also given that sum of $n$ over all the test cases doesn't exceed $10^5$.
-----Output-----
For each test case, print the minimum value of $f(s)$ you can obtain with at most $k$ operations.
-----Examples-----
Input
3
4 0
1010
7 1
0010100
5 2
00110
Output
21
22
12
-----Note-----
For the first example, you can't do any operation so the optimal string is $s$ itself. $f(s) = f(1010) = 10 + 01 + 10 = 21$.
For the second example, one of the optimal strings you can obtain is "0011000". The string has an $f$ value of $22$.
For the third example, one of the optimal strings you can obtain is "00011". The string has an $f$ value of $12$. | t = int(input())
for k in range(t):
n, k = map(int, input().split(" "))
s1 = input()
s = []
ones = 0
for i in s1:
if i == "1":
ones += 1
s.append(int(i))
fone = 0
for i in range(n):
if s[i] == 1:
fone = i
break
lone = 0
for i in range(n - 1, -1, -1):
if s[i] == 1:
lone = i
break
add = 0
if ones > 0 and k >= n - lone - 1:
add += 1
ones -= 1
k -= n - lone - 1
if ones > 0 and k >= fone:
add += 10
ones -= 1
k -= fone
temp = 0
print(11 * ones + add) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR VAR IF VAR STRING VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER VAR BIN_OP BIN_OP VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR NUMBER VAR VAR VAR NUMBER VAR NUMBER VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR VAR |
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