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Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya and his friend Vova play a computer game where they need to destroy n monsters to pass a level. Vanya's character performs attack with frequency x hits per second and Vova's character performs attack with frequency y hits per second. Each character spends fixed time to raise a weapon and then he hits (the time to raise the weapon is 1 / x seconds for the first character and 1 / y seconds for the second one). The i-th monster dies after he receives ai hits.
Vanya and Vova wonder who makes the last hit on each monster. If Vanya and Vova make the last hit at the same time, we assume that both of them have made the last hit.
Input
The first line contains three integers n,x,y (1 ≤ n ≤ 105, 1 ≤ x, y ≤ 106) — the number of monsters, the frequency of Vanya's and Vova's attack, correspondingly.
Next n lines contain integers ai (1 ≤ ai ≤ 109) — the number of hits needed do destroy the i-th monster.
Output
Print n lines. In the i-th line print word "Vanya", if the last hit on the i-th monster was performed by Vanya, "Vova", if Vova performed the last hit, or "Both", if both boys performed it at the same time.
Examples
Input
4 3 2
1
2
3
4
Output
Vanya
Vova
Vanya
Both
Input
2 1 1
1
2
Output
Both
Both
Note
In the first sample Vanya makes the first hit at time 1 / 3, Vova makes the second hit at time 1 / 2, Vanya makes the third hit at time 2 / 3, and both boys make the fourth and fifth hit simultaneously at the time 1.
In the second sample Vanya and Vova make the first and second hit simultaneously at time 1.
|
instruction
| 0
| 27,856
| 2
| 55,712
|
Tags: binary search, implementation, math, sortings
Correct Solution:
```
__author__ = 'PrimuS'
n, a, b = (int(x) for x in input().split())
mobs = [0] * n
for i in range(n):
mobs[i] = int(input())
total = a + b
ans = [0] * (total + 1)
aa = 0
bb = 0
ans[0] = 2
i = 1
while i <= total:
t1 = aa + b
t2 = bb + a
if t1 == t2:
aa += b
bb += a
ans[i] = 2
ans[i + 1] = 2
i += 2
elif t1 < t2:
aa = t1
ans[i] = 0
i += 1
else:
bb = t2
ans[i] = 1
i += 1
res = [""] * len(mobs)
i = 0
for m in mobs:
k = m % total
if ans[k] == 0:
res[i] = "Vanya"
elif ans[k] == 1:
res[i] = "Vova"
else:
res[i] = "Both"
i += 1
print("\n".join(res))
```
|
output
| 1
| 27,856
| 2
| 55,713
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya and his friend Vova play a computer game where they need to destroy n monsters to pass a level. Vanya's character performs attack with frequency x hits per second and Vova's character performs attack with frequency y hits per second. Each character spends fixed time to raise a weapon and then he hits (the time to raise the weapon is 1 / x seconds for the first character and 1 / y seconds for the second one). The i-th monster dies after he receives ai hits.
Vanya and Vova wonder who makes the last hit on each monster. If Vanya and Vova make the last hit at the same time, we assume that both of them have made the last hit.
Input
The first line contains three integers n,x,y (1 ≤ n ≤ 105, 1 ≤ x, y ≤ 106) — the number of monsters, the frequency of Vanya's and Vova's attack, correspondingly.
Next n lines contain integers ai (1 ≤ ai ≤ 109) — the number of hits needed do destroy the i-th monster.
Output
Print n lines. In the i-th line print word "Vanya", if the last hit on the i-th monster was performed by Vanya, "Vova", if Vova performed the last hit, or "Both", if both boys performed it at the same time.
Examples
Input
4 3 2
1
2
3
4
Output
Vanya
Vova
Vanya
Both
Input
2 1 1
1
2
Output
Both
Both
Note
In the first sample Vanya makes the first hit at time 1 / 3, Vova makes the second hit at time 1 / 2, Vanya makes the third hit at time 2 / 3, and both boys make the fourth and fifth hit simultaneously at the time 1.
In the second sample Vanya and Vova make the first and second hit simultaneously at time 1.
|
instruction
| 0
| 27,857
| 2
| 55,714
|
Tags: binary search, implementation, math, sortings
Correct Solution:
```
n, x, y = map(int, input().split())
for _ in range(n):
a = int(input())
c1, c2 = ((a + 1) * x // (x + y)) / x, ((a + 1) * y // (x + y)) / y
if c1 == c2:
print('Both')
elif c1 > c2:
print('Vanya')
else:
print('Vova')
```
|
output
| 1
| 27,857
| 2
| 55,715
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya and his friend Vova play a computer game where they need to destroy n monsters to pass a level. Vanya's character performs attack with frequency x hits per second and Vova's character performs attack with frequency y hits per second. Each character spends fixed time to raise a weapon and then he hits (the time to raise the weapon is 1 / x seconds for the first character and 1 / y seconds for the second one). The i-th monster dies after he receives ai hits.
Vanya and Vova wonder who makes the last hit on each monster. If Vanya and Vova make the last hit at the same time, we assume that both of them have made the last hit.
Input
The first line contains three integers n,x,y (1 ≤ n ≤ 105, 1 ≤ x, y ≤ 106) — the number of monsters, the frequency of Vanya's and Vova's attack, correspondingly.
Next n lines contain integers ai (1 ≤ ai ≤ 109) — the number of hits needed do destroy the i-th monster.
Output
Print n lines. In the i-th line print word "Vanya", if the last hit on the i-th monster was performed by Vanya, "Vova", if Vova performed the last hit, or "Both", if both boys performed it at the same time.
Examples
Input
4 3 2
1
2
3
4
Output
Vanya
Vova
Vanya
Both
Input
2 1 1
1
2
Output
Both
Both
Note
In the first sample Vanya makes the first hit at time 1 / 3, Vova makes the second hit at time 1 / 2, Vanya makes the third hit at time 2 / 3, and both boys make the fourth and fifth hit simultaneously at the time 1.
In the second sample Vanya and Vova make the first and second hit simultaneously at time 1.
|
instruction
| 0
| 27,858
| 2
| 55,716
|
Tags: binary search, implementation, math, sortings
Correct Solution:
```
read = lambda: map(int, input().split())
f = lambda a, x, y: (a * x + x + y - 1) // (x + y)
n, x, y = read()
for i in range(n):
a = int(input())
d = f(a, x, y) * y - f(a, y, x) * x
if d < 0: res = 'Vanya'
elif d > 0: res = 'Vova'
elif d == 0: res = 'Both'
print(res)
```
|
output
| 1
| 27,858
| 2
| 55,717
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya and his friend Vova play a computer game where they need to destroy n monsters to pass a level. Vanya's character performs attack with frequency x hits per second and Vova's character performs attack with frequency y hits per second. Each character spends fixed time to raise a weapon and then he hits (the time to raise the weapon is 1 / x seconds for the first character and 1 / y seconds for the second one). The i-th monster dies after he receives ai hits.
Vanya and Vova wonder who makes the last hit on each monster. If Vanya and Vova make the last hit at the same time, we assume that both of them have made the last hit.
Input
The first line contains three integers n,x,y (1 ≤ n ≤ 105, 1 ≤ x, y ≤ 106) — the number of monsters, the frequency of Vanya's and Vova's attack, correspondingly.
Next n lines contain integers ai (1 ≤ ai ≤ 109) — the number of hits needed do destroy the i-th monster.
Output
Print n lines. In the i-th line print word "Vanya", if the last hit on the i-th monster was performed by Vanya, "Vova", if Vova performed the last hit, or "Both", if both boys performed it at the same time.
Examples
Input
4 3 2
1
2
3
4
Output
Vanya
Vova
Vanya
Both
Input
2 1 1
1
2
Output
Both
Both
Note
In the first sample Vanya makes the first hit at time 1 / 3, Vova makes the second hit at time 1 / 2, Vanya makes the third hit at time 2 / 3, and both boys make the fourth and fifth hit simultaneously at the time 1.
In the second sample Vanya and Vova make the first and second hit simultaneously at time 1.
Submitted Solution:
```
input=__import__('sys').stdin.readline
n,x,y=map(int,input().split())
ans=[]
cx=cy=1
for i in range(x+y-1):
if cx/x>cy/y:ans.append(1);cy+=1
elif cx/x<cy/y:ans.append(0);cx+=1
else:ans.append(2);ans.append(2);cx+=1;cy+=1
f=['Vanya','Vova','Both']
for _ in range(n):
n=int(input())-1
print(f[ans[n%(x+y)]])
```
|
instruction
| 0
| 27,859
| 2
| 55,718
|
Yes
|
output
| 1
| 27,859
| 2
| 55,719
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya and his friend Vova play a computer game where they need to destroy n monsters to pass a level. Vanya's character performs attack with frequency x hits per second and Vova's character performs attack with frequency y hits per second. Each character spends fixed time to raise a weapon and then he hits (the time to raise the weapon is 1 / x seconds for the first character and 1 / y seconds for the second one). The i-th monster dies after he receives ai hits.
Vanya and Vova wonder who makes the last hit on each monster. If Vanya and Vova make the last hit at the same time, we assume that both of them have made the last hit.
Input
The first line contains three integers n,x,y (1 ≤ n ≤ 105, 1 ≤ x, y ≤ 106) — the number of monsters, the frequency of Vanya's and Vova's attack, correspondingly.
Next n lines contain integers ai (1 ≤ ai ≤ 109) — the number of hits needed do destroy the i-th monster.
Output
Print n lines. In the i-th line print word "Vanya", if the last hit on the i-th monster was performed by Vanya, "Vova", if Vova performed the last hit, or "Both", if both boys performed it at the same time.
Examples
Input
4 3 2
1
2
3
4
Output
Vanya
Vova
Vanya
Both
Input
2 1 1
1
2
Output
Both
Both
Note
In the first sample Vanya makes the first hit at time 1 / 3, Vova makes the second hit at time 1 / 2, Vanya makes the third hit at time 2 / 3, and both boys make the fourth and fifth hit simultaneously at the time 1.
In the second sample Vanya and Vova make the first and second hit simultaneously at time 1.
Submitted Solution:
```
n, x, y = [int(x) for x in input().split()]
dp = []
cx, cy = 0, 0
while(cx < x and cy < y):
if (cx + 1) / x > (cy + 1) / y:
dp.append("Vova")
cy += 1
elif (cx + 1) / x < (cy + 1) / y:
dp.append("Vanya")
cx += 1
else:
dp.append("Both")
dp.append("Both")
cx += 1
cy += 1
for i in range(n):
print(dp[(int(input()) - 1) % (x + y)])
```
|
instruction
| 0
| 27,860
| 2
| 55,720
|
Yes
|
output
| 1
| 27,860
| 2
| 55,721
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya and his friend Vova play a computer game where they need to destroy n monsters to pass a level. Vanya's character performs attack with frequency x hits per second and Vova's character performs attack with frequency y hits per second. Each character spends fixed time to raise a weapon and then he hits (the time to raise the weapon is 1 / x seconds for the first character and 1 / y seconds for the second one). The i-th monster dies after he receives ai hits.
Vanya and Vova wonder who makes the last hit on each monster. If Vanya and Vova make the last hit at the same time, we assume that both of them have made the last hit.
Input
The first line contains three integers n,x,y (1 ≤ n ≤ 105, 1 ≤ x, y ≤ 106) — the number of monsters, the frequency of Vanya's and Vova's attack, correspondingly.
Next n lines contain integers ai (1 ≤ ai ≤ 109) — the number of hits needed do destroy the i-th monster.
Output
Print n lines. In the i-th line print word "Vanya", if the last hit on the i-th monster was performed by Vanya, "Vova", if Vova performed the last hit, or "Both", if both boys performed it at the same time.
Examples
Input
4 3 2
1
2
3
4
Output
Vanya
Vova
Vanya
Both
Input
2 1 1
1
2
Output
Both
Both
Note
In the first sample Vanya makes the first hit at time 1 / 3, Vova makes the second hit at time 1 / 2, Vanya makes the third hit at time 2 / 3, and both boys make the fourth and fifth hit simultaneously at the time 1.
In the second sample Vanya and Vova make the first and second hit simultaneously at time 1.
Submitted Solution:
```
from sys import stdin
def main():
n, x, y = map(int, stdin.readline().strip().split())
xy = x + y
for health in map(int,(stdin.readline().strip()for _ in range(n))):
health %= xy
hitsx = health * x // xy
hitsy = health * y // xy
health -= hitsx + hitsy
tx = hitsx * y
ty = hitsy * x
while health > 0:
tx1, ty1 = tx + y, ty + x
if tx1 < ty1:
tx = tx1
health -= 1
elif tx1 > ty1:
ty = ty1
health -= 1
else:
tx, ty = tx1, ty1
health -= 1
print('Vova' if tx < ty else 'Vanya' if tx > ty else 'Both')
main()
```
|
instruction
| 0
| 27,861
| 2
| 55,722
|
Yes
|
output
| 1
| 27,861
| 2
| 55,723
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya and his friend Vova play a computer game where they need to destroy n monsters to pass a level. Vanya's character performs attack with frequency x hits per second and Vova's character performs attack with frequency y hits per second. Each character spends fixed time to raise a weapon and then he hits (the time to raise the weapon is 1 / x seconds for the first character and 1 / y seconds for the second one). The i-th monster dies after he receives ai hits.
Vanya and Vova wonder who makes the last hit on each monster. If Vanya and Vova make the last hit at the same time, we assume that both of them have made the last hit.
Input
The first line contains three integers n,x,y (1 ≤ n ≤ 105, 1 ≤ x, y ≤ 106) — the number of monsters, the frequency of Vanya's and Vova's attack, correspondingly.
Next n lines contain integers ai (1 ≤ ai ≤ 109) — the number of hits needed do destroy the i-th monster.
Output
Print n lines. In the i-th line print word "Vanya", if the last hit on the i-th monster was performed by Vanya, "Vova", if Vova performed the last hit, or "Both", if both boys performed it at the same time.
Examples
Input
4 3 2
1
2
3
4
Output
Vanya
Vova
Vanya
Both
Input
2 1 1
1
2
Output
Both
Both
Note
In the first sample Vanya makes the first hit at time 1 / 3, Vova makes the second hit at time 1 / 2, Vanya makes the third hit at time 2 / 3, and both boys make the fourth and fifth hit simultaneously at the time 1.
In the second sample Vanya and Vova make the first and second hit simultaneously at time 1.
Submitted Solution:
```
# n = int(input())
# w, h = map(int, input().split())
# to_zero_based = lambda x: int(x) - 1
from fractions import gcd
def main():
n, x, y = map(int, input().split())
monsters = [int(input()) for _ in range(n)]
if x == y:
results = ['Both'] * n
else:
results = []
g = gcd(x, y)
x //= g
y //= g
turns = [i * x for i in range(1, y)] + [i * y for i in range(1, x)]
turns.sort()
freq_sum = x + y
for monster in monsters:
monster %= freq_sum
if monster == 0 or monster == freq_sum - 1:
results.append('Both')
else:
results.append('Vanya' if turns[monster - 1] % y == 0 else 'Vova')
print('\n'.join(results))
main()
```
|
instruction
| 0
| 27,862
| 2
| 55,724
|
Yes
|
output
| 1
| 27,862
| 2
| 55,725
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya and his friend Vova play a computer game where they need to destroy n monsters to pass a level. Vanya's character performs attack with frequency x hits per second and Vova's character performs attack with frequency y hits per second. Each character spends fixed time to raise a weapon and then he hits (the time to raise the weapon is 1 / x seconds for the first character and 1 / y seconds for the second one). The i-th monster dies after he receives ai hits.
Vanya and Vova wonder who makes the last hit on each monster. If Vanya and Vova make the last hit at the same time, we assume that both of them have made the last hit.
Input
The first line contains three integers n,x,y (1 ≤ n ≤ 105, 1 ≤ x, y ≤ 106) — the number of monsters, the frequency of Vanya's and Vova's attack, correspondingly.
Next n lines contain integers ai (1 ≤ ai ≤ 109) — the number of hits needed do destroy the i-th monster.
Output
Print n lines. In the i-th line print word "Vanya", if the last hit on the i-th monster was performed by Vanya, "Vova", if Vova performed the last hit, or "Both", if both boys performed it at the same time.
Examples
Input
4 3 2
1
2
3
4
Output
Vanya
Vova
Vanya
Both
Input
2 1 1
1
2
Output
Both
Both
Note
In the first sample Vanya makes the first hit at time 1 / 3, Vova makes the second hit at time 1 / 2, Vanya makes the third hit at time 2 / 3, and both boys make the fourth and fifth hit simultaneously at the time 1.
In the second sample Vanya and Vova make the first and second hit simultaneously at time 1.
Submitted Solution:
```
__author__ = 'PrimuS'
n, a, b = (int(x) for x in input().split())
mobs = [0] * n
for i in range(n):
mobs[i] = int(input())
total = a + b
ans = [0] * (total + 1)
a1 = 1.0 / a
b1 = 1.0 / b
aa = 0.0
bb = 0.0
ans[0] = 2
i = 1
while i <= total:
if aa + a1 < bb + b1:
aa += a1
ans[i] = 0
i += 1
elif aa + a1 > bb + b1:
bb += b1
ans[i] = 1
i += 1
else:
aa += a1
bb += b1
ans[i] = 2
ans[i + 1] = 2
i += 2
print (ans)
for m in mobs:
k = m % total
if ans[k] == 0:
print("Vanya")
elif ans[k] == 1:
print("Vova")
else:
print("Both")
```
|
instruction
| 0
| 27,863
| 2
| 55,726
|
No
|
output
| 1
| 27,863
| 2
| 55,727
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya and his friend Vova play a computer game where they need to destroy n monsters to pass a level. Vanya's character performs attack with frequency x hits per second and Vova's character performs attack with frequency y hits per second. Each character spends fixed time to raise a weapon and then he hits (the time to raise the weapon is 1 / x seconds for the first character and 1 / y seconds for the second one). The i-th monster dies after he receives ai hits.
Vanya and Vova wonder who makes the last hit on each monster. If Vanya and Vova make the last hit at the same time, we assume that both of them have made the last hit.
Input
The first line contains three integers n,x,y (1 ≤ n ≤ 105, 1 ≤ x, y ≤ 106) — the number of monsters, the frequency of Vanya's and Vova's attack, correspondingly.
Next n lines contain integers ai (1 ≤ ai ≤ 109) — the number of hits needed do destroy the i-th monster.
Output
Print n lines. In the i-th line print word "Vanya", if the last hit on the i-th monster was performed by Vanya, "Vova", if Vova performed the last hit, or "Both", if both boys performed it at the same time.
Examples
Input
4 3 2
1
2
3
4
Output
Vanya
Vova
Vanya
Both
Input
2 1 1
1
2
Output
Both
Both
Note
In the first sample Vanya makes the first hit at time 1 / 3, Vova makes the second hit at time 1 / 2, Vanya makes the third hit at time 2 / 3, and both boys make the fourth and fifth hit simultaneously at the time 1.
In the second sample Vanya and Vova make the first and second hit simultaneously at time 1.
Submitted Solution:
```
n, x, y = map(int, input().split(' '))
z = x*y
x, y = y, x
def check(m, r):
a = m//x
b = m//y ;
c = m//(x*y)
if a+b-c >= r:
return True
else:
return False
for i in range(n) :
s = int(input())
s = s*z
l = 0
r = 1000000000000000000000000000000
ans = r
while l<=r:
m=(l+r)//2
if check(m, s):
r = m-1
ans = min(ans, m)
else:
l = m+1
if ans%x == 0 and ans%y == 0:
print("Both")
elif ans%x == 0:
print("Vanya")
else:
print("Vova")
```
|
instruction
| 0
| 27,864
| 2
| 55,728
|
No
|
output
| 1
| 27,864
| 2
| 55,729
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya and his friend Vova play a computer game where they need to destroy n monsters to pass a level. Vanya's character performs attack with frequency x hits per second and Vova's character performs attack with frequency y hits per second. Each character spends fixed time to raise a weapon and then he hits (the time to raise the weapon is 1 / x seconds for the first character and 1 / y seconds for the second one). The i-th monster dies after he receives ai hits.
Vanya and Vova wonder who makes the last hit on each monster. If Vanya and Vova make the last hit at the same time, we assume that both of them have made the last hit.
Input
The first line contains three integers n,x,y (1 ≤ n ≤ 105, 1 ≤ x, y ≤ 106) — the number of monsters, the frequency of Vanya's and Vova's attack, correspondingly.
Next n lines contain integers ai (1 ≤ ai ≤ 109) — the number of hits needed do destroy the i-th monster.
Output
Print n lines. In the i-th line print word "Vanya", if the last hit on the i-th monster was performed by Vanya, "Vova", if Vova performed the last hit, or "Both", if both boys performed it at the same time.
Examples
Input
4 3 2
1
2
3
4
Output
Vanya
Vova
Vanya
Both
Input
2 1 1
1
2
Output
Both
Both
Note
In the first sample Vanya makes the first hit at time 1 / 3, Vova makes the second hit at time 1 / 2, Vanya makes the third hit at time 2 / 3, and both boys make the fourth and fifth hit simultaneously at the time 1.
In the second sample Vanya and Vova make the first and second hit simultaneously at time 1.
Submitted Solution:
```
__author__ = 'zhan'
from fractions import Fraction
[n, x, y] = [int(i) for i in input().split()]
ox = fx = Fraction(1, x)
oy = fy = Fraction(1, y)
loop = Fraction(x, y).numerator + Fraction(x, y).denominator
hits = ["null"]
while len(hits) < loop + 1:
if fx < fy:
hits.append("Vanya")
fx += ox
elif fy < fx:
hits.append("Vova")
fy += oy
else:
hits.append("Both")
hits.append("Both")
fx += ox
fy += oy
hits[0] = hits[-1]
print(hits)
for i in range(n):
print(hits[int(input()) % loop])
```
|
instruction
| 0
| 27,865
| 2
| 55,730
|
No
|
output
| 1
| 27,865
| 2
| 55,731
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya and his friend Vova play a computer game where they need to destroy n monsters to pass a level. Vanya's character performs attack with frequency x hits per second and Vova's character performs attack with frequency y hits per second. Each character spends fixed time to raise a weapon and then he hits (the time to raise the weapon is 1 / x seconds for the first character and 1 / y seconds for the second one). The i-th monster dies after he receives ai hits.
Vanya and Vova wonder who makes the last hit on each monster. If Vanya and Vova make the last hit at the same time, we assume that both of them have made the last hit.
Input
The first line contains three integers n,x,y (1 ≤ n ≤ 105, 1 ≤ x, y ≤ 106) — the number of monsters, the frequency of Vanya's and Vova's attack, correspondingly.
Next n lines contain integers ai (1 ≤ ai ≤ 109) — the number of hits needed do destroy the i-th monster.
Output
Print n lines. In the i-th line print word "Vanya", if the last hit on the i-th monster was performed by Vanya, "Vova", if Vova performed the last hit, or "Both", if both boys performed it at the same time.
Examples
Input
4 3 2
1
2
3
4
Output
Vanya
Vova
Vanya
Both
Input
2 1 1
1
2
Output
Both
Both
Note
In the first sample Vanya makes the first hit at time 1 / 3, Vova makes the second hit at time 1 / 2, Vanya makes the third hit at time 2 / 3, and both boys make the fourth and fifth hit simultaneously at the time 1.
In the second sample Vanya and Vova make the first and second hit simultaneously at time 1.
Submitted Solution:
```
n, x, y = map(int, input().split())
EPS = 10 ** (-10)
for i in range(n):
a = int(input())
t = a / (x + y)
dx = (t * x - int (t * x)) / x
dy = (t * y - int (t * y)) / y
if dx < EPS and dy < EPS or abs((1 / x - dx) - (1 / y - dy)) < EPS:
print("Both")
else:
if dx < EPS or 1 / x - dx < 1 / y - dy:
print("Vanya")
else:
print("Vova")
```
|
instruction
| 0
| 27,866
| 2
| 55,732
|
No
|
output
| 1
| 27,866
| 2
| 55,733
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Dante is engaged in a fight with "The Savior". Before he can fight it with his sword, he needs to break its shields. He has two guns, Ebony and Ivory, each of them is able to perform any non-negative number of shots.
For every bullet that hits the shield, Ebony deals a units of damage while Ivory deals b units of damage. In order to break the shield Dante has to deal exactly c units of damage. Find out if this is possible.
Input
The first line of the input contains three integers a, b, c (1 ≤ a, b ≤ 100, 1 ≤ c ≤ 10 000) — the number of units of damage dealt by Ebony gun and Ivory gun, and the total number of damage required to break the shield, respectively.
Output
Print "Yes" (without quotes) if Dante can deal exactly c damage to the shield and "No" (without quotes) otherwise.
Examples
Input
4 6 15
Output
No
Input
3 2 7
Output
Yes
Input
6 11 6
Output
Yes
Note
In the second sample, Dante can fire 1 bullet from Ebony and 2 from Ivory to deal exactly 1·3 + 2·2 = 7 damage. In the third sample, Dante can fire 1 bullet from ebony and no bullets from ivory to do 1·6 + 0·11 = 6 damage.
|
instruction
| 0
| 27,909
| 2
| 55,818
|
Tags: brute force, math, number theory
Correct Solution:
```
__author__ = 'aste'
def main():
a, b, c = [int(x) for x in input().split()]
res = False
for i in range(0, c + 1):
r = c - a*i
if r >= 0 and r % b == 0:
res = True
break
print(res and "Yes" or "No")
main()
```
|
output
| 1
| 27,909
| 2
| 55,819
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Dante is engaged in a fight with "The Savior". Before he can fight it with his sword, he needs to break its shields. He has two guns, Ebony and Ivory, each of them is able to perform any non-negative number of shots.
For every bullet that hits the shield, Ebony deals a units of damage while Ivory deals b units of damage. In order to break the shield Dante has to deal exactly c units of damage. Find out if this is possible.
Input
The first line of the input contains three integers a, b, c (1 ≤ a, b ≤ 100, 1 ≤ c ≤ 10 000) — the number of units of damage dealt by Ebony gun and Ivory gun, and the total number of damage required to break the shield, respectively.
Output
Print "Yes" (without quotes) if Dante can deal exactly c damage to the shield and "No" (without quotes) otherwise.
Examples
Input
4 6 15
Output
No
Input
3 2 7
Output
Yes
Input
6 11 6
Output
Yes
Note
In the second sample, Dante can fire 1 bullet from Ebony and 2 from Ivory to deal exactly 1·3 + 2·2 = 7 damage. In the third sample, Dante can fire 1 bullet from ebony and no bullets from ivory to do 1·6 + 0·11 = 6 damage.
|
instruction
| 0
| 27,910
| 2
| 55,820
|
Tags: brute force, math, number theory
Correct Solution:
```
a, b, c = map(int, input().split())
for i in range(101):
t = (c - a * i) / b
if t >= 0 and int(t) == t:
print('Yes')
exit()
print('No')
```
|
output
| 1
| 27,910
| 2
| 55,821
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Dante is engaged in a fight with "The Savior". Before he can fight it with his sword, he needs to break its shields. He has two guns, Ebony and Ivory, each of them is able to perform any non-negative number of shots.
For every bullet that hits the shield, Ebony deals a units of damage while Ivory deals b units of damage. In order to break the shield Dante has to deal exactly c units of damage. Find out if this is possible.
Input
The first line of the input contains three integers a, b, c (1 ≤ a, b ≤ 100, 1 ≤ c ≤ 10 000) — the number of units of damage dealt by Ebony gun and Ivory gun, and the total number of damage required to break the shield, respectively.
Output
Print "Yes" (without quotes) if Dante can deal exactly c damage to the shield and "No" (without quotes) otherwise.
Examples
Input
4 6 15
Output
No
Input
3 2 7
Output
Yes
Input
6 11 6
Output
Yes
Note
In the second sample, Dante can fire 1 bullet from Ebony and 2 from Ivory to deal exactly 1·3 + 2·2 = 7 damage. In the third sample, Dante can fire 1 bullet from ebony and no bullets from ivory to do 1·6 + 0·11 = 6 damage.
|
instruction
| 0
| 27,911
| 2
| 55,822
|
Tags: brute force, math, number theory
Correct Solution:
```
a, b, c = map(int, input().split())
for i in range(0, 10000):
if (i * a > c):
break
if (c - i * a) % b == 0:
print("Yes")
exit()
print("No")
```
|
output
| 1
| 27,911
| 2
| 55,823
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Dante is engaged in a fight with "The Savior". Before he can fight it with his sword, he needs to break its shields. He has two guns, Ebony and Ivory, each of them is able to perform any non-negative number of shots.
For every bullet that hits the shield, Ebony deals a units of damage while Ivory deals b units of damage. In order to break the shield Dante has to deal exactly c units of damage. Find out if this is possible.
Input
The first line of the input contains three integers a, b, c (1 ≤ a, b ≤ 100, 1 ≤ c ≤ 10 000) — the number of units of damage dealt by Ebony gun and Ivory gun, and the total number of damage required to break the shield, respectively.
Output
Print "Yes" (without quotes) if Dante can deal exactly c damage to the shield and "No" (without quotes) otherwise.
Examples
Input
4 6 15
Output
No
Input
3 2 7
Output
Yes
Input
6 11 6
Output
Yes
Note
In the second sample, Dante can fire 1 bullet from Ebony and 2 from Ivory to deal exactly 1·3 + 2·2 = 7 damage. In the third sample, Dante can fire 1 bullet from ebony and no bullets from ivory to do 1·6 + 0·11 = 6 damage.
|
instruction
| 0
| 27,912
| 2
| 55,824
|
Tags: brute force, math, number theory
Correct Solution:
```
import math
a1,b1,c=map(int,input().split())
if c%math.gcd(a1,b1)!=0:print('No')
else:
a=max(a1,b1)
b=min(a1,b1)
x=0
if c%a==0:
print('Yes')
else:
while a*x<=c:
if (c-(a*x))%b==0:
print('Yes')
break
x+=1
else:
print('No')
```
|
output
| 1
| 27,912
| 2
| 55,825
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Dante is engaged in a fight with "The Savior". Before he can fight it with his sword, he needs to break its shields. He has two guns, Ebony and Ivory, each of them is able to perform any non-negative number of shots.
For every bullet that hits the shield, Ebony deals a units of damage while Ivory deals b units of damage. In order to break the shield Dante has to deal exactly c units of damage. Find out if this is possible.
Input
The first line of the input contains three integers a, b, c (1 ≤ a, b ≤ 100, 1 ≤ c ≤ 10 000) — the number of units of damage dealt by Ebony gun and Ivory gun, and the total number of damage required to break the shield, respectively.
Output
Print "Yes" (without quotes) if Dante can deal exactly c damage to the shield and "No" (without quotes) otherwise.
Examples
Input
4 6 15
Output
No
Input
3 2 7
Output
Yes
Input
6 11 6
Output
Yes
Note
In the second sample, Dante can fire 1 bullet from Ebony and 2 from Ivory to deal exactly 1·3 + 2·2 = 7 damage. In the third sample, Dante can fire 1 bullet from ebony and no bullets from ivory to do 1·6 + 0·11 = 6 damage.
|
instruction
| 0
| 27,913
| 2
| 55,826
|
Tags: brute force, math, number theory
Correct Solution:
```
a,b,c=map(int,input().split())
for i in range((c//a)+1):
for j in range((c//b)+1):
if (i*a)+(j*b)==c:
print('Yes')
exit(0)
if (i*a)+(j*b)>10000:
break
print('No')
```
|
output
| 1
| 27,913
| 2
| 55,827
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Dante is engaged in a fight with "The Savior". Before he can fight it with his sword, he needs to break its shields. He has two guns, Ebony and Ivory, each of them is able to perform any non-negative number of shots.
For every bullet that hits the shield, Ebony deals a units of damage while Ivory deals b units of damage. In order to break the shield Dante has to deal exactly c units of damage. Find out if this is possible.
Input
The first line of the input contains three integers a, b, c (1 ≤ a, b ≤ 100, 1 ≤ c ≤ 10 000) — the number of units of damage dealt by Ebony gun and Ivory gun, and the total number of damage required to break the shield, respectively.
Output
Print "Yes" (without quotes) if Dante can deal exactly c damage to the shield and "No" (without quotes) otherwise.
Examples
Input
4 6 15
Output
No
Input
3 2 7
Output
Yes
Input
6 11 6
Output
Yes
Note
In the second sample, Dante can fire 1 bullet from Ebony and 2 from Ivory to deal exactly 1·3 + 2·2 = 7 damage. In the third sample, Dante can fire 1 bullet from ebony and no bullets from ivory to do 1·6 + 0·11 = 6 damage.
|
instruction
| 0
| 27,914
| 2
| 55,828
|
Tags: brute force, math, number theory
Correct Solution:
```
n,m,o=map(int,input().split())
for i in range(o//n+1):
if (o - i*n) % m == 0:
print("Yes")
break
else:
print("No")
```
|
output
| 1
| 27,914
| 2
| 55,829
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Dante is engaged in a fight with "The Savior". Before he can fight it with his sword, he needs to break its shields. He has two guns, Ebony and Ivory, each of them is able to perform any non-negative number of shots.
For every bullet that hits the shield, Ebony deals a units of damage while Ivory deals b units of damage. In order to break the shield Dante has to deal exactly c units of damage. Find out if this is possible.
Input
The first line of the input contains three integers a, b, c (1 ≤ a, b ≤ 100, 1 ≤ c ≤ 10 000) — the number of units of damage dealt by Ebony gun and Ivory gun, and the total number of damage required to break the shield, respectively.
Output
Print "Yes" (without quotes) if Dante can deal exactly c damage to the shield and "No" (without quotes) otherwise.
Examples
Input
4 6 15
Output
No
Input
3 2 7
Output
Yes
Input
6 11 6
Output
Yes
Note
In the second sample, Dante can fire 1 bullet from Ebony and 2 from Ivory to deal exactly 1·3 + 2·2 = 7 damage. In the third sample, Dante can fire 1 bullet from ebony and no bullets from ivory to do 1·6 + 0·11 = 6 damage.
|
instruction
| 0
| 27,915
| 2
| 55,830
|
Tags: brute force, math, number theory
Correct Solution:
```
'''input
6 11 6
'''
from math import gcd
a, b, c = map(int, input().split())
for x in range(c//a+1):
if (c - x*a) % b == 0:
print("Yes")
break
else:
print("No")
```
|
output
| 1
| 27,915
| 2
| 55,831
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Dante is engaged in a fight with "The Savior". Before he can fight it with his sword, he needs to break its shields. He has two guns, Ebony and Ivory, each of them is able to perform any non-negative number of shots.
For every bullet that hits the shield, Ebony deals a units of damage while Ivory deals b units of damage. In order to break the shield Dante has to deal exactly c units of damage. Find out if this is possible.
Input
The first line of the input contains three integers a, b, c (1 ≤ a, b ≤ 100, 1 ≤ c ≤ 10 000) — the number of units of damage dealt by Ebony gun and Ivory gun, and the total number of damage required to break the shield, respectively.
Output
Print "Yes" (without quotes) if Dante can deal exactly c damage to the shield and "No" (without quotes) otherwise.
Examples
Input
4 6 15
Output
No
Input
3 2 7
Output
Yes
Input
6 11 6
Output
Yes
Note
In the second sample, Dante can fire 1 bullet from Ebony and 2 from Ivory to deal exactly 1·3 + 2·2 = 7 damage. In the third sample, Dante can fire 1 bullet from ebony and no bullets from ivory to do 1·6 + 0·11 = 6 damage.
|
instruction
| 0
| 27,916
| 2
| 55,832
|
Tags: brute force, math, number theory
Correct Solution:
```
a, b, c = map(int, input().split())
for x in range(max(c//b,c//a) + 3):
if (-a * x + c >= 0) and (-a * x + c) % b == 0:
print('Yes')
break
else:
print('No')
```
|
output
| 1
| 27,916
| 2
| 55,833
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dante is engaged in a fight with "The Savior". Before he can fight it with his sword, he needs to break its shields. He has two guns, Ebony and Ivory, each of them is able to perform any non-negative number of shots.
For every bullet that hits the shield, Ebony deals a units of damage while Ivory deals b units of damage. In order to break the shield Dante has to deal exactly c units of damage. Find out if this is possible.
Input
The first line of the input contains three integers a, b, c (1 ≤ a, b ≤ 100, 1 ≤ c ≤ 10 000) — the number of units of damage dealt by Ebony gun and Ivory gun, and the total number of damage required to break the shield, respectively.
Output
Print "Yes" (without quotes) if Dante can deal exactly c damage to the shield and "No" (without quotes) otherwise.
Examples
Input
4 6 15
Output
No
Input
3 2 7
Output
Yes
Input
6 11 6
Output
Yes
Note
In the second sample, Dante can fire 1 bullet from Ebony and 2 from Ivory to deal exactly 1·3 + 2·2 = 7 damage. In the third sample, Dante can fire 1 bullet from ebony and no bullets from ivory to do 1·6 + 0·11 = 6 damage.
Submitted Solution:
```
a,b,c = map(int,input().split(' '))
n = max(c//a,c//b)
f = 0
for i in range(n+1):
for j in range(n+1):
if i*a+j*b == c:
f = 1
break
if f==1:
break
if f==1:
print('YES')
else:
print('NO')
```
|
instruction
| 0
| 27,917
| 2
| 55,834
|
Yes
|
output
| 1
| 27,917
| 2
| 55,835
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dante is engaged in a fight with "The Savior". Before he can fight it with his sword, he needs to break its shields. He has two guns, Ebony and Ivory, each of them is able to perform any non-negative number of shots.
For every bullet that hits the shield, Ebony deals a units of damage while Ivory deals b units of damage. In order to break the shield Dante has to deal exactly c units of damage. Find out if this is possible.
Input
The first line of the input contains three integers a, b, c (1 ≤ a, b ≤ 100, 1 ≤ c ≤ 10 000) — the number of units of damage dealt by Ebony gun and Ivory gun, and the total number of damage required to break the shield, respectively.
Output
Print "Yes" (without quotes) if Dante can deal exactly c damage to the shield and "No" (without quotes) otherwise.
Examples
Input
4 6 15
Output
No
Input
3 2 7
Output
Yes
Input
6 11 6
Output
Yes
Note
In the second sample, Dante can fire 1 bullet from Ebony and 2 from Ivory to deal exactly 1·3 + 2·2 = 7 damage. In the third sample, Dante can fire 1 bullet from ebony and no bullets from ivory to do 1·6 + 0·11 = 6 damage.
Submitted Solution:
```
a,b,c = map(int,input().split())
d = False
r = c//(min(a,b))
for i in range(r+1):
for j in range(r+1):
if i*a+j*b==c:
d=True
break
if d:
print("YES")
else:
print("NO")
```
|
instruction
| 0
| 27,918
| 2
| 55,836
|
Yes
|
output
| 1
| 27,918
| 2
| 55,837
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dante is engaged in a fight with "The Savior". Before he can fight it with his sword, he needs to break its shields. He has two guns, Ebony and Ivory, each of them is able to perform any non-negative number of shots.
For every bullet that hits the shield, Ebony deals a units of damage while Ivory deals b units of damage. In order to break the shield Dante has to deal exactly c units of damage. Find out if this is possible.
Input
The first line of the input contains three integers a, b, c (1 ≤ a, b ≤ 100, 1 ≤ c ≤ 10 000) — the number of units of damage dealt by Ebony gun and Ivory gun, and the total number of damage required to break the shield, respectively.
Output
Print "Yes" (without quotes) if Dante can deal exactly c damage to the shield and "No" (without quotes) otherwise.
Examples
Input
4 6 15
Output
No
Input
3 2 7
Output
Yes
Input
6 11 6
Output
Yes
Note
In the second sample, Dante can fire 1 bullet from Ebony and 2 from Ivory to deal exactly 1·3 + 2·2 = 7 damage. In the third sample, Dante can fire 1 bullet from ebony and no bullets from ivory to do 1·6 + 0·11 = 6 damage.
Submitted Solution:
```
a, b, c = map(int, input().split())
h = 0
if c % a == 0:
k = c // a
else:
k = c // a + 1
if c % b == 0:
m = c // b
else:
m = c // b + 1
if c - a*k < 0 and c - b*m < 0 and ((c < 2 * a) and c < 2*b):
print('No')
else:
for i in range(k+1):
if (c - a*i) % b == 0 and (h == 0):
print('Yes')
h = 1
if h == 0:
print('No')
```
|
instruction
| 0
| 27,919
| 2
| 55,838
|
Yes
|
output
| 1
| 27,919
| 2
| 55,839
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dante is engaged in a fight with "The Savior". Before he can fight it with his sword, he needs to break its shields. He has two guns, Ebony and Ivory, each of them is able to perform any non-negative number of shots.
For every bullet that hits the shield, Ebony deals a units of damage while Ivory deals b units of damage. In order to break the shield Dante has to deal exactly c units of damage. Find out if this is possible.
Input
The first line of the input contains three integers a, b, c (1 ≤ a, b ≤ 100, 1 ≤ c ≤ 10 000) — the number of units of damage dealt by Ebony gun and Ivory gun, and the total number of damage required to break the shield, respectively.
Output
Print "Yes" (without quotes) if Dante can deal exactly c damage to the shield and "No" (without quotes) otherwise.
Examples
Input
4 6 15
Output
No
Input
3 2 7
Output
Yes
Input
6 11 6
Output
Yes
Note
In the second sample, Dante can fire 1 bullet from Ebony and 2 from Ivory to deal exactly 1·3 + 2·2 = 7 damage. In the third sample, Dante can fire 1 bullet from ebony and no bullets from ivory to do 1·6 + 0·11 = 6 damage.
Submitted Solution:
```
a,b,c = list(map(int,input().split()))
for i in range(10004):
x = a*i
if c-x>=0:
d = c-x
if d%b==0:
print("Yes")
exit()
print("No")
exit()
```
|
instruction
| 0
| 27,920
| 2
| 55,840
|
Yes
|
output
| 1
| 27,920
| 2
| 55,841
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dante is engaged in a fight with "The Savior". Before he can fight it with his sword, he needs to break its shields. He has two guns, Ebony and Ivory, each of them is able to perform any non-negative number of shots.
For every bullet that hits the shield, Ebony deals a units of damage while Ivory deals b units of damage. In order to break the shield Dante has to deal exactly c units of damage. Find out if this is possible.
Input
The first line of the input contains three integers a, b, c (1 ≤ a, b ≤ 100, 1 ≤ c ≤ 10 000) — the number of units of damage dealt by Ebony gun and Ivory gun, and the total number of damage required to break the shield, respectively.
Output
Print "Yes" (without quotes) if Dante can deal exactly c damage to the shield and "No" (without quotes) otherwise.
Examples
Input
4 6 15
Output
No
Input
3 2 7
Output
Yes
Input
6 11 6
Output
Yes
Note
In the second sample, Dante can fire 1 bullet from Ebony and 2 from Ivory to deal exactly 1·3 + 2·2 = 7 damage. In the third sample, Dante can fire 1 bullet from ebony and no bullets from ivory to do 1·6 + 0·11 = 6 damage.
Submitted Solution:
```
def MCD_Extend(a,b):
if (b == 0):
x = 1
y = 0
return a,x,y
else:
d,x1,y1= MCD_Extend(b,a%b)
x = y1
y = x1-((a//b)*y1)
return d,x,y
def find_any_solution(a,b,c):
g,X,Y=MCD_Extend(abs(a),abs(b))
if(c%g):
return False
x0=c//g
y0=c//g
if(a<0):
x0=-x0
if(b<0):
y0=-y0
return True
a,b,c= [int(i) for i in input().split()]
if(find_any_solution(a,b,c)):
print('Si')
else:
print('No')
```
|
instruction
| 0
| 27,921
| 2
| 55,842
|
No
|
output
| 1
| 27,921
| 2
| 55,843
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dante is engaged in a fight with "The Savior". Before he can fight it with his sword, he needs to break its shields. He has two guns, Ebony and Ivory, each of them is able to perform any non-negative number of shots.
For every bullet that hits the shield, Ebony deals a units of damage while Ivory deals b units of damage. In order to break the shield Dante has to deal exactly c units of damage. Find out if this is possible.
Input
The first line of the input contains three integers a, b, c (1 ≤ a, b ≤ 100, 1 ≤ c ≤ 10 000) — the number of units of damage dealt by Ebony gun and Ivory gun, and the total number of damage required to break the shield, respectively.
Output
Print "Yes" (without quotes) if Dante can deal exactly c damage to the shield and "No" (without quotes) otherwise.
Examples
Input
4 6 15
Output
No
Input
3 2 7
Output
Yes
Input
6 11 6
Output
Yes
Note
In the second sample, Dante can fire 1 bullet from Ebony and 2 from Ivory to deal exactly 1·3 + 2·2 = 7 damage. In the third sample, Dante can fire 1 bullet from ebony and no bullets from ivory to do 1·6 + 0·11 = 6 damage.
Submitted Solution:
```
nums = list(map(int, input().split()))
e = nums[0]
i = nums[1]
dmg = nums[2]
def check(a, b, dmg, index):
if(a*index + b > dmg):
return False
if((dmg-a*index)%b is 0):
return True
return check(a, b, dmg, index+1)
if(dmg is e or dmg is i or dmg is 0):
print("Yes")
elif(check(e, i, dmg, 1) or check(i, e, dmg, 1)):
print("Yes")
else:
print("No")
```
|
instruction
| 0
| 27,922
| 2
| 55,844
|
No
|
output
| 1
| 27,922
| 2
| 55,845
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dante is engaged in a fight with "The Savior". Before he can fight it with his sword, he needs to break its shields. He has two guns, Ebony and Ivory, each of them is able to perform any non-negative number of shots.
For every bullet that hits the shield, Ebony deals a units of damage while Ivory deals b units of damage. In order to break the shield Dante has to deal exactly c units of damage. Find out if this is possible.
Input
The first line of the input contains three integers a, b, c (1 ≤ a, b ≤ 100, 1 ≤ c ≤ 10 000) — the number of units of damage dealt by Ebony gun and Ivory gun, and the total number of damage required to break the shield, respectively.
Output
Print "Yes" (without quotes) if Dante can deal exactly c damage to the shield and "No" (without quotes) otherwise.
Examples
Input
4 6 15
Output
No
Input
3 2 7
Output
Yes
Input
6 11 6
Output
Yes
Note
In the second sample, Dante can fire 1 bullet from Ebony and 2 from Ivory to deal exactly 1·3 + 2·2 = 7 damage. In the third sample, Dante can fire 1 bullet from ebony and no bullets from ivory to do 1·6 + 0·11 = 6 damage.
Submitted Solution:
```
def MCD_Extend(a,b):
if (b == 0):
x = 1
y = 0
return a,x,y
else:
d,x1,y1= MCD_Extend(b,a%b)
x = y1
y = x1-((a//b)*y1)
return d,x,y
def find_any_solution(a,b,c):
g,X,Y=MCD_Extend(abs(a),abs(b))
if(c%g):
return "NO"
x0=c//g
y0=c//g
if(a<0):
x0=-x0
if(b<0):
y0=-y0
return "SI"
a,b,c= [int(i) for i in input().split()]
find_any_solution(a,b,c)
```
|
instruction
| 0
| 27,923
| 2
| 55,846
|
No
|
output
| 1
| 27,923
| 2
| 55,847
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dante is engaged in a fight with "The Savior". Before he can fight it with his sword, he needs to break its shields. He has two guns, Ebony and Ivory, each of them is able to perform any non-negative number of shots.
For every bullet that hits the shield, Ebony deals a units of damage while Ivory deals b units of damage. In order to break the shield Dante has to deal exactly c units of damage. Find out if this is possible.
Input
The first line of the input contains three integers a, b, c (1 ≤ a, b ≤ 100, 1 ≤ c ≤ 10 000) — the number of units of damage dealt by Ebony gun and Ivory gun, and the total number of damage required to break the shield, respectively.
Output
Print "Yes" (without quotes) if Dante can deal exactly c damage to the shield and "No" (without quotes) otherwise.
Examples
Input
4 6 15
Output
No
Input
3 2 7
Output
Yes
Input
6 11 6
Output
Yes
Note
In the second sample, Dante can fire 1 bullet from Ebony and 2 from Ivory to deal exactly 1·3 + 2·2 = 7 damage. In the third sample, Dante can fire 1 bullet from ebony and no bullets from ivory to do 1·6 + 0·11 = 6 damage.
Submitted Solution:
```
def gcd(a,b):
if a == 0:
return b
else:
return gcd(b%a,a)
a,b,c =map(int,input().split())
if c/gcd(a,b) == 0:
print("Yes")
else:
print("No")
```
|
instruction
| 0
| 27,924
| 2
| 55,848
|
No
|
output
| 1
| 27,924
| 2
| 55,849
|
Provide tags and a correct Python 3 solution for this coding contest problem.
In the official contest this problem has a different statement, for which jury's solution was working incorrectly, and for this reason it was excluded from the contest. This mistake have been fixed and the current given problem statement and model solution corresponds to what jury wanted it to be during the contest.
Vova and Lesha are friends. They often meet at Vova's place and compete against each other in a computer game named The Ancient Papyri: Swordsink. Vova always chooses a warrior as his fighter and Leshac chooses an archer. After that they should choose initial positions for their characters and start the fight. A warrior is good at melee combat, so Vova will try to make the distance between fighters as small as possible. An archer prefers to keep the enemy at a distance, so Lesha will try to make the initial distance as large as possible.
There are n (n is always even) possible starting positions for characters marked along the Ox axis. The positions are given by their distinct coordinates x1, x2, ..., xn, two characters cannot end up at the same position.
Vova and Lesha take turns banning available positions, Vova moves first. During each turn one of the guys bans exactly one of the remaining positions. Banned positions cannot be used by both Vova and Lesha. They continue to make moves until there are only two possible positions remaining (thus, the total number of moves will be n - 2). After that Vova's character takes the position with the lesser coordinate and Lesha's character takes the position with the bigger coordinate and the guys start fighting.
Vova and Lesha are already tired by the game of choosing positions, as they need to play it before every fight, so they asked you (the developer of the The Ancient Papyri: Swordsink) to write a module that would automatically determine the distance at which the warrior and the archer will start fighting if both Vova and Lesha play optimally.
Input
The first line on the input contains a single integer n (2 ≤ n ≤ 200 000, n is even) — the number of positions available initially. The second line contains n distinct integers x1, x2, ..., xn (0 ≤ xi ≤ 109), giving the coordinates of the corresponding positions.
Output
Print the distance between the warrior and the archer at the beginning of the fight, provided that both Vova and Lesha play optimally.
Examples
Input
6
0 1 3 7 15 31
Output
7
Input
2
73 37
Output
36
Note
In the first sample one of the optimum behavior of the players looks like that:
1. Vova bans the position at coordinate 15;
2. Lesha bans the position at coordinate 3;
3. Vova bans the position at coordinate 31;
4. Lesha bans the position at coordinate 1.
After these actions only positions 0 and 7 will remain, and the distance between them is equal to 7.
In the second sample there are only two possible positions, so there will be no bans.
|
instruction
| 0
| 28,659
| 2
| 57,318
|
Tags: games
Correct Solution:
```
n = int(input())
xs = list(map(int, input().split()))
xs.sort()
m = n//2
print(min(a - b for a, b in zip(xs[m:], xs)))
```
|
output
| 1
| 28,659
| 2
| 57,319
|
Provide tags and a correct Python 3 solution for this coding contest problem.
In the official contest this problem has a different statement, for which jury's solution was working incorrectly, and for this reason it was excluded from the contest. This mistake have been fixed and the current given problem statement and model solution corresponds to what jury wanted it to be during the contest.
Vova and Lesha are friends. They often meet at Vova's place and compete against each other in a computer game named The Ancient Papyri: Swordsink. Vova always chooses a warrior as his fighter and Leshac chooses an archer. After that they should choose initial positions for their characters and start the fight. A warrior is good at melee combat, so Vova will try to make the distance between fighters as small as possible. An archer prefers to keep the enemy at a distance, so Lesha will try to make the initial distance as large as possible.
There are n (n is always even) possible starting positions for characters marked along the Ox axis. The positions are given by their distinct coordinates x1, x2, ..., xn, two characters cannot end up at the same position.
Vova and Lesha take turns banning available positions, Vova moves first. During each turn one of the guys bans exactly one of the remaining positions. Banned positions cannot be used by both Vova and Lesha. They continue to make moves until there are only two possible positions remaining (thus, the total number of moves will be n - 2). After that Vova's character takes the position with the lesser coordinate and Lesha's character takes the position with the bigger coordinate and the guys start fighting.
Vova and Lesha are already tired by the game of choosing positions, as they need to play it before every fight, so they asked you (the developer of the The Ancient Papyri: Swordsink) to write a module that would automatically determine the distance at which the warrior and the archer will start fighting if both Vova and Lesha play optimally.
Input
The first line on the input contains a single integer n (2 ≤ n ≤ 200 000, n is even) — the number of positions available initially. The second line contains n distinct integers x1, x2, ..., xn (0 ≤ xi ≤ 109), giving the coordinates of the corresponding positions.
Output
Print the distance between the warrior and the archer at the beginning of the fight, provided that both Vova and Lesha play optimally.
Examples
Input
6
0 1 3 7 15 31
Output
7
Input
2
73 37
Output
36
Note
In the first sample one of the optimum behavior of the players looks like that:
1. Vova bans the position at coordinate 15;
2. Lesha bans the position at coordinate 3;
3. Vova bans the position at coordinate 31;
4. Lesha bans the position at coordinate 1.
After these actions only positions 0 and 7 will remain, and the distance between them is equal to 7.
In the second sample there are only two possible positions, so there will be no bans.
|
instruction
| 0
| 28,660
| 2
| 57,320
|
Tags: games
Correct Solution:
```
def main():
n = int(input())
l = sorted(map(int, input().split()))
print(min(b - a for a, b in zip(l, l[n // 2:])))
if __name__ == '__main__':
main()
```
|
output
| 1
| 28,660
| 2
| 57,321
|
Provide tags and a correct Python 3 solution for this coding contest problem.
In the official contest this problem has a different statement, for which jury's solution was working incorrectly, and for this reason it was excluded from the contest. This mistake have been fixed and the current given problem statement and model solution corresponds to what jury wanted it to be during the contest.
Vova and Lesha are friends. They often meet at Vova's place and compete against each other in a computer game named The Ancient Papyri: Swordsink. Vova always chooses a warrior as his fighter and Leshac chooses an archer. After that they should choose initial positions for their characters and start the fight. A warrior is good at melee combat, so Vova will try to make the distance between fighters as small as possible. An archer prefers to keep the enemy at a distance, so Lesha will try to make the initial distance as large as possible.
There are n (n is always even) possible starting positions for characters marked along the Ox axis. The positions are given by their distinct coordinates x1, x2, ..., xn, two characters cannot end up at the same position.
Vova and Lesha take turns banning available positions, Vova moves first. During each turn one of the guys bans exactly one of the remaining positions. Banned positions cannot be used by both Vova and Lesha. They continue to make moves until there are only two possible positions remaining (thus, the total number of moves will be n - 2). After that Vova's character takes the position with the lesser coordinate and Lesha's character takes the position with the bigger coordinate and the guys start fighting.
Vova and Lesha are already tired by the game of choosing positions, as they need to play it before every fight, so they asked you (the developer of the The Ancient Papyri: Swordsink) to write a module that would automatically determine the distance at which the warrior and the archer will start fighting if both Vova and Lesha play optimally.
Input
The first line on the input contains a single integer n (2 ≤ n ≤ 200 000, n is even) — the number of positions available initially. The second line contains n distinct integers x1, x2, ..., xn (0 ≤ xi ≤ 109), giving the coordinates of the corresponding positions.
Output
Print the distance between the warrior and the archer at the beginning of the fight, provided that both Vova and Lesha play optimally.
Examples
Input
6
0 1 3 7 15 31
Output
7
Input
2
73 37
Output
36
Note
In the first sample one of the optimum behavior of the players looks like that:
1. Vova bans the position at coordinate 15;
2. Lesha bans the position at coordinate 3;
3. Vova bans the position at coordinate 31;
4. Lesha bans the position at coordinate 1.
After these actions only positions 0 and 7 will remain, and the distance between them is equal to 7.
In the second sample there are only two possible positions, so there will be no bans.
|
instruction
| 0
| 28,661
| 2
| 57,322
|
Tags: games
Correct Solution:
```
n = int(input())
v = sorted([int(i) for i in input().split()])
ans = 2 ** 40
for i in range(n//2):
ans = min(ans, v[i + n//2] - v[i])
print(ans)
```
|
output
| 1
| 28,661
| 2
| 57,323
|
Provide tags and a correct Python 3 solution for this coding contest problem.
In the official contest this problem has a different statement, for which jury's solution was working incorrectly, and for this reason it was excluded from the contest. This mistake have been fixed and the current given problem statement and model solution corresponds to what jury wanted it to be during the contest.
Vova and Lesha are friends. They often meet at Vova's place and compete against each other in a computer game named The Ancient Papyri: Swordsink. Vova always chooses a warrior as his fighter and Leshac chooses an archer. After that they should choose initial positions for their characters and start the fight. A warrior is good at melee combat, so Vova will try to make the distance between fighters as small as possible. An archer prefers to keep the enemy at a distance, so Lesha will try to make the initial distance as large as possible.
There are n (n is always even) possible starting positions for characters marked along the Ox axis. The positions are given by their distinct coordinates x1, x2, ..., xn, two characters cannot end up at the same position.
Vova and Lesha take turns banning available positions, Vova moves first. During each turn one of the guys bans exactly one of the remaining positions. Banned positions cannot be used by both Vova and Lesha. They continue to make moves until there are only two possible positions remaining (thus, the total number of moves will be n - 2). After that Vova's character takes the position with the lesser coordinate and Lesha's character takes the position with the bigger coordinate and the guys start fighting.
Vova and Lesha are already tired by the game of choosing positions, as they need to play it before every fight, so they asked you (the developer of the The Ancient Papyri: Swordsink) to write a module that would automatically determine the distance at which the warrior and the archer will start fighting if both Vova and Lesha play optimally.
Input
The first line on the input contains a single integer n (2 ≤ n ≤ 200 000, n is even) — the number of positions available initially. The second line contains n distinct integers x1, x2, ..., xn (0 ≤ xi ≤ 109), giving the coordinates of the corresponding positions.
Output
Print the distance between the warrior and the archer at the beginning of the fight, provided that both Vova and Lesha play optimally.
Examples
Input
6
0 1 3 7 15 31
Output
7
Input
2
73 37
Output
36
Note
In the first sample one of the optimum behavior of the players looks like that:
1. Vova bans the position at coordinate 15;
2. Lesha bans the position at coordinate 3;
3. Vova bans the position at coordinate 31;
4. Lesha bans the position at coordinate 1.
After these actions only positions 0 and 7 will remain, and the distance between them is equal to 7.
In the second sample there are only two possible positions, so there will be no bans.
|
instruction
| 0
| 28,662
| 2
| 57,324
|
Tags: games
Correct Solution:
```
import sys
n=int(sys.stdin.readline())
x=list(map(int,sys.stdin.readline().split()))
z=[]
x.sort()
for i in range(n//2):
z.append(x[i+n//2]-x[i])
sys.stdout.write(str(min(z)))
```
|
output
| 1
| 28,662
| 2
| 57,325
|
Provide tags and a correct Python 3 solution for this coding contest problem.
In the official contest this problem has a different statement, for which jury's solution was working incorrectly, and for this reason it was excluded from the contest. This mistake have been fixed and the current given problem statement and model solution corresponds to what jury wanted it to be during the contest.
Vova and Lesha are friends. They often meet at Vova's place and compete against each other in a computer game named The Ancient Papyri: Swordsink. Vova always chooses a warrior as his fighter and Leshac chooses an archer. After that they should choose initial positions for their characters and start the fight. A warrior is good at melee combat, so Vova will try to make the distance between fighters as small as possible. An archer prefers to keep the enemy at a distance, so Lesha will try to make the initial distance as large as possible.
There are n (n is always even) possible starting positions for characters marked along the Ox axis. The positions are given by their distinct coordinates x1, x2, ..., xn, two characters cannot end up at the same position.
Vova and Lesha take turns banning available positions, Vova moves first. During each turn one of the guys bans exactly one of the remaining positions. Banned positions cannot be used by both Vova and Lesha. They continue to make moves until there are only two possible positions remaining (thus, the total number of moves will be n - 2). After that Vova's character takes the position with the lesser coordinate and Lesha's character takes the position with the bigger coordinate and the guys start fighting.
Vova and Lesha are already tired by the game of choosing positions, as they need to play it before every fight, so they asked you (the developer of the The Ancient Papyri: Swordsink) to write a module that would automatically determine the distance at which the warrior and the archer will start fighting if both Vova and Lesha play optimally.
Input
The first line on the input contains a single integer n (2 ≤ n ≤ 200 000, n is even) — the number of positions available initially. The second line contains n distinct integers x1, x2, ..., xn (0 ≤ xi ≤ 109), giving the coordinates of the corresponding positions.
Output
Print the distance between the warrior and the archer at the beginning of the fight, provided that both Vova and Lesha play optimally.
Examples
Input
6
0 1 3 7 15 31
Output
7
Input
2
73 37
Output
36
Note
In the first sample one of the optimum behavior of the players looks like that:
1. Vova bans the position at coordinate 15;
2. Lesha bans the position at coordinate 3;
3. Vova bans the position at coordinate 31;
4. Lesha bans the position at coordinate 1.
After these actions only positions 0 and 7 will remain, and the distance between them is equal to 7.
In the second sample there are only two possible positions, so there will be no bans.
|
instruction
| 0
| 28,663
| 2
| 57,326
|
Tags: games
Correct Solution:
```
n = int(input())
x = list(map(int, input().split()))
x.sort()
mini = int(1e9)
for i in range(n // 2):
if x[i + n // 2] - x[i] < mini: mini = x[i + n // 2] - x[i]
print(mini)
```
|
output
| 1
| 28,663
| 2
| 57,327
|
Provide tags and a correct Python 3 solution for this coding contest problem.
In the official contest this problem has a different statement, for which jury's solution was working incorrectly, and for this reason it was excluded from the contest. This mistake have been fixed and the current given problem statement and model solution corresponds to what jury wanted it to be during the contest.
Vova and Lesha are friends. They often meet at Vova's place and compete against each other in a computer game named The Ancient Papyri: Swordsink. Vova always chooses a warrior as his fighter and Leshac chooses an archer. After that they should choose initial positions for their characters and start the fight. A warrior is good at melee combat, so Vova will try to make the distance between fighters as small as possible. An archer prefers to keep the enemy at a distance, so Lesha will try to make the initial distance as large as possible.
There are n (n is always even) possible starting positions for characters marked along the Ox axis. The positions are given by their distinct coordinates x1, x2, ..., xn, two characters cannot end up at the same position.
Vova and Lesha take turns banning available positions, Vova moves first. During each turn one of the guys bans exactly one of the remaining positions. Banned positions cannot be used by both Vova and Lesha. They continue to make moves until there are only two possible positions remaining (thus, the total number of moves will be n - 2). After that Vova's character takes the position with the lesser coordinate and Lesha's character takes the position with the bigger coordinate and the guys start fighting.
Vova and Lesha are already tired by the game of choosing positions, as they need to play it before every fight, so they asked you (the developer of the The Ancient Papyri: Swordsink) to write a module that would automatically determine the distance at which the warrior and the archer will start fighting if both Vova and Lesha play optimally.
Input
The first line on the input contains a single integer n (2 ≤ n ≤ 200 000, n is even) — the number of positions available initially. The second line contains n distinct integers x1, x2, ..., xn (0 ≤ xi ≤ 109), giving the coordinates of the corresponding positions.
Output
Print the distance between the warrior and the archer at the beginning of the fight, provided that both Vova and Lesha play optimally.
Examples
Input
6
0 1 3 7 15 31
Output
7
Input
2
73 37
Output
36
Note
In the first sample one of the optimum behavior of the players looks like that:
1. Vova bans the position at coordinate 15;
2. Lesha bans the position at coordinate 3;
3. Vova bans the position at coordinate 31;
4. Lesha bans the position at coordinate 1.
After these actions only positions 0 and 7 will remain, and the distance between them is equal to 7.
In the second sample there are only two possible positions, so there will be no bans.
|
instruction
| 0
| 28,664
| 2
| 57,328
|
Tags: games
Correct Solution:
```
n = int(input())
arr = [*map(int, input().split())]
arr = sorted(arr)
print(min([b - a for a, b in zip(arr[:n//2], arr[n//2:])]))
```
|
output
| 1
| 28,664
| 2
| 57,329
|
Provide tags and a correct Python 3 solution for this coding contest problem.
In the official contest this problem has a different statement, for which jury's solution was working incorrectly, and for this reason it was excluded from the contest. This mistake have been fixed and the current given problem statement and model solution corresponds to what jury wanted it to be during the contest.
Vova and Lesha are friends. They often meet at Vova's place and compete against each other in a computer game named The Ancient Papyri: Swordsink. Vova always chooses a warrior as his fighter and Leshac chooses an archer. After that they should choose initial positions for their characters and start the fight. A warrior is good at melee combat, so Vova will try to make the distance between fighters as small as possible. An archer prefers to keep the enemy at a distance, so Lesha will try to make the initial distance as large as possible.
There are n (n is always even) possible starting positions for characters marked along the Ox axis. The positions are given by their distinct coordinates x1, x2, ..., xn, two characters cannot end up at the same position.
Vova and Lesha take turns banning available positions, Vova moves first. During each turn one of the guys bans exactly one of the remaining positions. Banned positions cannot be used by both Vova and Lesha. They continue to make moves until there are only two possible positions remaining (thus, the total number of moves will be n - 2). After that Vova's character takes the position with the lesser coordinate and Lesha's character takes the position with the bigger coordinate and the guys start fighting.
Vova and Lesha are already tired by the game of choosing positions, as they need to play it before every fight, so they asked you (the developer of the The Ancient Papyri: Swordsink) to write a module that would automatically determine the distance at which the warrior and the archer will start fighting if both Vova and Lesha play optimally.
Input
The first line on the input contains a single integer n (2 ≤ n ≤ 200 000, n is even) — the number of positions available initially. The second line contains n distinct integers x1, x2, ..., xn (0 ≤ xi ≤ 109), giving the coordinates of the corresponding positions.
Output
Print the distance between the warrior and the archer at the beginning of the fight, provided that both Vova and Lesha play optimally.
Examples
Input
6
0 1 3 7 15 31
Output
7
Input
2
73 37
Output
36
Note
In the first sample one of the optimum behavior of the players looks like that:
1. Vova bans the position at coordinate 15;
2. Lesha bans the position at coordinate 3;
3. Vova bans the position at coordinate 31;
4. Lesha bans the position at coordinate 1.
After these actions only positions 0 and 7 will remain, and the distance between them is equal to 7.
In the second sample there are only two possible positions, so there will be no bans.
|
instruction
| 0
| 28,665
| 2
| 57,330
|
Tags: games
Correct Solution:
```
import os
import random
import sys
from typing import List, Dict
class Int:
def __init__(self, val):
self.val = val
def get(self):
return self.val + 111
class Unique:
def __init__(self):
self.s = set()
def add(self, val : int):
self.s.add(val)
def __contains__(self, item : int) -> bool:
return self.s.__contains__(item)
def ceil(top : int, bottom : int) -> int:
return (top + bottom - 1) // bottom
def concat(l : List[int]):
return "".join(map(str, l))
def get(d : Dict[int, str], val : int) -> Dict[int, str]:
return d[val]
#guy who wants small moves first
#then guy who wants large moves
#so lets say we have 4 positions
# 1, 2, 3, 4
#small wants to ban edges, because if he bans 2 or 3 he is fucked
#so he bans 1
# and we have 2, 3, 4
# then large bans middle so we have 2, 4 and the ans is 2
# 0, 1, 2, 3, 4, 5, 6, 7
# 0, 1, 2, 3, 4, 5, 6
# 0, 1, 2, 3, 5, 6
# 0, 1, 2, 3, 5
# 0, 1, 3, 5
# 0, 1, 3
# 0, 3
# 0, 1, 2, 3, 4, 5, 6, 7
# 0, 4
# # 0, 3
#1 5 9 19 21 22
# 5 9 19 21 22
# 5 19 21 22
# 19 21 22
# 0, 1, 3, 7, 15
# 0, 1, 7, 15
# 0, 1, 7
# 0, 7
def slowsolve(a):
a.sort()
small = True
while len(a) > 2:
if small:
if a[1] - a[0] > a[-1] - a[-2]:
a.pop(0)
else:
a.pop()
small = False
else:
a.pop(len(a) // 2)
small = True
return a[1] - a[0]
def solve(a):
a.sort()
candelete = len(a) // 2 - 1
res = 10 ** 18
for leftdelete in range(0, candelete + 1):
leftrem = leftdelete
rightrem = leftdelete + candelete + 1
res = min(res, a[rightrem] - a[leftrem])
return res
def prt(l): return print(' '.join(l))
def rv(): return map(int, input().split())
def rl(n): return [list(map(int, input().split())) for _ in range(n)]
if os.path.exists("test.txt"): sys.stdin = open("test.txt")
n, = rv()
a, = rl(1)
# a = sorted([random.randrange(10**2) for _ in range(6)])
# print(a)
# print(solve(a), slowsolve(a))
print(solve(a))
```
|
output
| 1
| 28,665
| 2
| 57,331
|
Provide tags and a correct Python 3 solution for this coding contest problem.
In the official contest this problem has a different statement, for which jury's solution was working incorrectly, and for this reason it was excluded from the contest. This mistake have been fixed and the current given problem statement and model solution corresponds to what jury wanted it to be during the contest.
Vova and Lesha are friends. They often meet at Vova's place and compete against each other in a computer game named The Ancient Papyri: Swordsink. Vova always chooses a warrior as his fighter and Leshac chooses an archer. After that they should choose initial positions for their characters and start the fight. A warrior is good at melee combat, so Vova will try to make the distance between fighters as small as possible. An archer prefers to keep the enemy at a distance, so Lesha will try to make the initial distance as large as possible.
There are n (n is always even) possible starting positions for characters marked along the Ox axis. The positions are given by their distinct coordinates x1, x2, ..., xn, two characters cannot end up at the same position.
Vova and Lesha take turns banning available positions, Vova moves first. During each turn one of the guys bans exactly one of the remaining positions. Banned positions cannot be used by both Vova and Lesha. They continue to make moves until there are only two possible positions remaining (thus, the total number of moves will be n - 2). After that Vova's character takes the position with the lesser coordinate and Lesha's character takes the position with the bigger coordinate and the guys start fighting.
Vova and Lesha are already tired by the game of choosing positions, as they need to play it before every fight, so they asked you (the developer of the The Ancient Papyri: Swordsink) to write a module that would automatically determine the distance at which the warrior and the archer will start fighting if both Vova and Lesha play optimally.
Input
The first line on the input contains a single integer n (2 ≤ n ≤ 200 000, n is even) — the number of positions available initially. The second line contains n distinct integers x1, x2, ..., xn (0 ≤ xi ≤ 109), giving the coordinates of the corresponding positions.
Output
Print the distance between the warrior and the archer at the beginning of the fight, provided that both Vova and Lesha play optimally.
Examples
Input
6
0 1 3 7 15 31
Output
7
Input
2
73 37
Output
36
Note
In the first sample one of the optimum behavior of the players looks like that:
1. Vova bans the position at coordinate 15;
2. Lesha bans the position at coordinate 3;
3. Vova bans the position at coordinate 31;
4. Lesha bans the position at coordinate 1.
After these actions only positions 0 and 7 will remain, and the distance between them is equal to 7.
In the second sample there are only two possible positions, so there will be no bans.
|
instruction
| 0
| 28,666
| 2
| 57,332
|
Tags: games
Correct Solution:
```
def distance(xi):
xi = sorted(xi)
k = len(xi) // 2
return min(xi[i+k]-xi[i] for i in range(k))
if __name__ == '__main__':
n = int(input())
xi = map(int, input().split())
print(distance(xi))
```
|
output
| 1
| 28,666
| 2
| 57,333
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the official contest this problem has a different statement, for which jury's solution was working incorrectly, and for this reason it was excluded from the contest. This mistake have been fixed and the current given problem statement and model solution corresponds to what jury wanted it to be during the contest.
Vova and Lesha are friends. They often meet at Vova's place and compete against each other in a computer game named The Ancient Papyri: Swordsink. Vova always chooses a warrior as his fighter and Leshac chooses an archer. After that they should choose initial positions for their characters and start the fight. A warrior is good at melee combat, so Vova will try to make the distance between fighters as small as possible. An archer prefers to keep the enemy at a distance, so Lesha will try to make the initial distance as large as possible.
There are n (n is always even) possible starting positions for characters marked along the Ox axis. The positions are given by their distinct coordinates x1, x2, ..., xn, two characters cannot end up at the same position.
Vova and Lesha take turns banning available positions, Vova moves first. During each turn one of the guys bans exactly one of the remaining positions. Banned positions cannot be used by both Vova and Lesha. They continue to make moves until there are only two possible positions remaining (thus, the total number of moves will be n - 2). After that Vova's character takes the position with the lesser coordinate and Lesha's character takes the position with the bigger coordinate and the guys start fighting.
Vova and Lesha are already tired by the game of choosing positions, as they need to play it before every fight, so they asked you (the developer of the The Ancient Papyri: Swordsink) to write a module that would automatically determine the distance at which the warrior and the archer will start fighting if both Vova and Lesha play optimally.
Input
The first line on the input contains a single integer n (2 ≤ n ≤ 200 000, n is even) — the number of positions available initially. The second line contains n distinct integers x1, x2, ..., xn (0 ≤ xi ≤ 109), giving the coordinates of the corresponding positions.
Output
Print the distance between the warrior and the archer at the beginning of the fight, provided that both Vova and Lesha play optimally.
Examples
Input
6
0 1 3 7 15 31
Output
7
Input
2
73 37
Output
36
Note
In the first sample one of the optimum behavior of the players looks like that:
1. Vova bans the position at coordinate 15;
2. Lesha bans the position at coordinate 3;
3. Vova bans the position at coordinate 31;
4. Lesha bans the position at coordinate 1.
After these actions only positions 0 and 7 will remain, and the distance between them is equal to 7.
In the second sample there are only two possible positions, so there will be no bans.
Submitted Solution:
```
R = lambda : map(int,input().split())
n = int(input())
p = sorted(list(R()))
m = 10**19
for i in range(n//2):
m = min(m, p[i+n//2]-p[i])
print(m)
```
|
instruction
| 0
| 28,667
| 2
| 57,334
|
Yes
|
output
| 1
| 28,667
| 2
| 57,335
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the official contest this problem has a different statement, for which jury's solution was working incorrectly, and for this reason it was excluded from the contest. This mistake have been fixed and the current given problem statement and model solution corresponds to what jury wanted it to be during the contest.
Vova and Lesha are friends. They often meet at Vova's place and compete against each other in a computer game named The Ancient Papyri: Swordsink. Vova always chooses a warrior as his fighter and Leshac chooses an archer. After that they should choose initial positions for their characters and start the fight. A warrior is good at melee combat, so Vova will try to make the distance between fighters as small as possible. An archer prefers to keep the enemy at a distance, so Lesha will try to make the initial distance as large as possible.
There are n (n is always even) possible starting positions for characters marked along the Ox axis. The positions are given by their distinct coordinates x1, x2, ..., xn, two characters cannot end up at the same position.
Vova and Lesha take turns banning available positions, Vova moves first. During each turn one of the guys bans exactly one of the remaining positions. Banned positions cannot be used by both Vova and Lesha. They continue to make moves until there are only two possible positions remaining (thus, the total number of moves will be n - 2). After that Vova's character takes the position with the lesser coordinate and Lesha's character takes the position with the bigger coordinate and the guys start fighting.
Vova and Lesha are already tired by the game of choosing positions, as they need to play it before every fight, so they asked you (the developer of the The Ancient Papyri: Swordsink) to write a module that would automatically determine the distance at which the warrior and the archer will start fighting if both Vova and Lesha play optimally.
Input
The first line on the input contains a single integer n (2 ≤ n ≤ 200 000, n is even) — the number of positions available initially. The second line contains n distinct integers x1, x2, ..., xn (0 ≤ xi ≤ 109), giving the coordinates of the corresponding positions.
Output
Print the distance between the warrior and the archer at the beginning of the fight, provided that both Vova and Lesha play optimally.
Examples
Input
6
0 1 3 7 15 31
Output
7
Input
2
73 37
Output
36
Note
In the first sample one of the optimum behavior of the players looks like that:
1. Vova bans the position at coordinate 15;
2. Lesha bans the position at coordinate 3;
3. Vova bans the position at coordinate 31;
4. Lesha bans the position at coordinate 1.
After these actions only positions 0 and 7 will remain, and the distance between them is equal to 7.
In the second sample there are only two possible positions, so there will be no bans.
Submitted Solution:
```
#!/usr/bin/env python3
n = int(input())
n_2 = n // 2
x = sorted([int(tok) for tok in input().split()])
res = min((x[j] - x[j-n_2] for j in range(n_2, n)))
print(res)
```
|
instruction
| 0
| 28,668
| 2
| 57,336
|
Yes
|
output
| 1
| 28,668
| 2
| 57,337
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the official contest this problem has a different statement, for which jury's solution was working incorrectly, and for this reason it was excluded from the contest. This mistake have been fixed and the current given problem statement and model solution corresponds to what jury wanted it to be during the contest.
Vova and Lesha are friends. They often meet at Vova's place and compete against each other in a computer game named The Ancient Papyri: Swordsink. Vova always chooses a warrior as his fighter and Leshac chooses an archer. After that they should choose initial positions for their characters and start the fight. A warrior is good at melee combat, so Vova will try to make the distance between fighters as small as possible. An archer prefers to keep the enemy at a distance, so Lesha will try to make the initial distance as large as possible.
There are n (n is always even) possible starting positions for characters marked along the Ox axis. The positions are given by their distinct coordinates x1, x2, ..., xn, two characters cannot end up at the same position.
Vova and Lesha take turns banning available positions, Vova moves first. During each turn one of the guys bans exactly one of the remaining positions. Banned positions cannot be used by both Vova and Lesha. They continue to make moves until there are only two possible positions remaining (thus, the total number of moves will be n - 2). After that Vova's character takes the position with the lesser coordinate and Lesha's character takes the position with the bigger coordinate and the guys start fighting.
Vova and Lesha are already tired by the game of choosing positions, as they need to play it before every fight, so they asked you (the developer of the The Ancient Papyri: Swordsink) to write a module that would automatically determine the distance at which the warrior and the archer will start fighting if both Vova and Lesha play optimally.
Input
The first line on the input contains a single integer n (2 ≤ n ≤ 200 000, n is even) — the number of positions available initially. The second line contains n distinct integers x1, x2, ..., xn (0 ≤ xi ≤ 109), giving the coordinates of the corresponding positions.
Output
Print the distance between the warrior and the archer at the beginning of the fight, provided that both Vova and Lesha play optimally.
Examples
Input
6
0 1 3 7 15 31
Output
7
Input
2
73 37
Output
36
Note
In the first sample one of the optimum behavior of the players looks like that:
1. Vova bans the position at coordinate 15;
2. Lesha bans the position at coordinate 3;
3. Vova bans the position at coordinate 31;
4. Lesha bans the position at coordinate 1.
After these actions only positions 0 and 7 will remain, and the distance between them is equal to 7.
In the second sample there are only two possible positions, so there will be no bans.
Submitted Solution:
```
n = int(input())
x = sorted(list(map(int, input().split())))
print(min([x[i + n // 2] - x[i] for i in range(n // 2)]))
# Made By Mostafa_Khaled
```
|
instruction
| 0
| 28,669
| 2
| 57,338
|
Yes
|
output
| 1
| 28,669
| 2
| 57,339
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the official contest this problem has a different statement, for which jury's solution was working incorrectly, and for this reason it was excluded from the contest. This mistake have been fixed and the current given problem statement and model solution corresponds to what jury wanted it to be during the contest.
Vova and Lesha are friends. They often meet at Vova's place and compete against each other in a computer game named The Ancient Papyri: Swordsink. Vova always chooses a warrior as his fighter and Leshac chooses an archer. After that they should choose initial positions for their characters and start the fight. A warrior is good at melee combat, so Vova will try to make the distance between fighters as small as possible. An archer prefers to keep the enemy at a distance, so Lesha will try to make the initial distance as large as possible.
There are n (n is always even) possible starting positions for characters marked along the Ox axis. The positions are given by their distinct coordinates x1, x2, ..., xn, two characters cannot end up at the same position.
Vova and Lesha take turns banning available positions, Vova moves first. During each turn one of the guys bans exactly one of the remaining positions. Banned positions cannot be used by both Vova and Lesha. They continue to make moves until there are only two possible positions remaining (thus, the total number of moves will be n - 2). After that Vova's character takes the position with the lesser coordinate and Lesha's character takes the position with the bigger coordinate and the guys start fighting.
Vova and Lesha are already tired by the game of choosing positions, as they need to play it before every fight, so they asked you (the developer of the The Ancient Papyri: Swordsink) to write a module that would automatically determine the distance at which the warrior and the archer will start fighting if both Vova and Lesha play optimally.
Input
The first line on the input contains a single integer n (2 ≤ n ≤ 200 000, n is even) — the number of positions available initially. The second line contains n distinct integers x1, x2, ..., xn (0 ≤ xi ≤ 109), giving the coordinates of the corresponding positions.
Output
Print the distance between the warrior and the archer at the beginning of the fight, provided that both Vova and Lesha play optimally.
Examples
Input
6
0 1 3 7 15 31
Output
7
Input
2
73 37
Output
36
Note
In the first sample one of the optimum behavior of the players looks like that:
1. Vova bans the position at coordinate 15;
2. Lesha bans the position at coordinate 3;
3. Vova bans the position at coordinate 31;
4. Lesha bans the position at coordinate 1.
After these actions only positions 0 and 7 will remain, and the distance between them is equal to 7.
In the second sample there are only two possible positions, so there will be no bans.
Submitted Solution:
```
n = int(input())
x = sorted(list(map(int, input().split())))
print(min([x[i + n // 2] - x[i] for i in range(n // 2)]))
```
|
instruction
| 0
| 28,670
| 2
| 57,340
|
Yes
|
output
| 1
| 28,670
| 2
| 57,341
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the official contest this problem has a different statement, for which jury's solution was working incorrectly, and for this reason it was excluded from the contest. This mistake have been fixed and the current given problem statement and model solution corresponds to what jury wanted it to be during the contest.
Vova and Lesha are friends. They often meet at Vova's place and compete against each other in a computer game named The Ancient Papyri: Swordsink. Vova always chooses a warrior as his fighter and Leshac chooses an archer. After that they should choose initial positions for their characters and start the fight. A warrior is good at melee combat, so Vova will try to make the distance between fighters as small as possible. An archer prefers to keep the enemy at a distance, so Lesha will try to make the initial distance as large as possible.
There are n (n is always even) possible starting positions for characters marked along the Ox axis. The positions are given by their distinct coordinates x1, x2, ..., xn, two characters cannot end up at the same position.
Vova and Lesha take turns banning available positions, Vova moves first. During each turn one of the guys bans exactly one of the remaining positions. Banned positions cannot be used by both Vova and Lesha. They continue to make moves until there are only two possible positions remaining (thus, the total number of moves will be n - 2). After that Vova's character takes the position with the lesser coordinate and Lesha's character takes the position with the bigger coordinate and the guys start fighting.
Vova and Lesha are already tired by the game of choosing positions, as they need to play it before every fight, so they asked you (the developer of the The Ancient Papyri: Swordsink) to write a module that would automatically determine the distance at which the warrior and the archer will start fighting if both Vova and Lesha play optimally.
Input
The first line on the input contains a single integer n (2 ≤ n ≤ 200 000, n is even) — the number of positions available initially. The second line contains n distinct integers x1, x2, ..., xn (0 ≤ xi ≤ 109), giving the coordinates of the corresponding positions.
Output
Print the distance between the warrior and the archer at the beginning of the fight, provided that both Vova and Lesha play optimally.
Examples
Input
6
0 1 3 7 15 31
Output
7
Input
2
73 37
Output
36
Note
In the first sample one of the optimum behavior of the players looks like that:
1. Vova bans the position at coordinate 15;
2. Lesha bans the position at coordinate 3;
3. Vova bans the position at coordinate 31;
4. Lesha bans the position at coordinate 1.
After these actions only positions 0 and 7 will remain, and the distance between them is equal to 7.
In the second sample there are only two possible positions, so there will be no bans.
Submitted Solution:
```
n = int(input())
X = list(map(int, input().split()))
X.sort()
w = (n-3)//2+1
a = (n-2)//2
if n>2:
z = 1
for k in range(n-2):
z = 1-z
if z==0:
del(X[n-w+k//2])
else:
del(X[a-k//2])
"""
if z==0:
del(X[n-1])
else:
Y = [X[i+1]-X[i] for i in range(n-1)]
m = Y[0]
j = 0
for i in range(1,n-1):
if Y[i] <= m:
m = Y[i]
j = i
del(X[j+1])
"""
n = n - 1
print(X[1]-X[0])
```
|
instruction
| 0
| 28,671
| 2
| 57,342
|
No
|
output
| 1
| 28,671
| 2
| 57,343
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the official contest this problem has a different statement, for which jury's solution was working incorrectly, and for this reason it was excluded from the contest. This mistake have been fixed and the current given problem statement and model solution corresponds to what jury wanted it to be during the contest.
Vova and Lesha are friends. They often meet at Vova's place and compete against each other in a computer game named The Ancient Papyri: Swordsink. Vova always chooses a warrior as his fighter and Leshac chooses an archer. After that they should choose initial positions for their characters and start the fight. A warrior is good at melee combat, so Vova will try to make the distance between fighters as small as possible. An archer prefers to keep the enemy at a distance, so Lesha will try to make the initial distance as large as possible.
There are n (n is always even) possible starting positions for characters marked along the Ox axis. The positions are given by their distinct coordinates x1, x2, ..., xn, two characters cannot end up at the same position.
Vova and Lesha take turns banning available positions, Vova moves first. During each turn one of the guys bans exactly one of the remaining positions. Banned positions cannot be used by both Vova and Lesha. They continue to make moves until there are only two possible positions remaining (thus, the total number of moves will be n - 2). After that Vova's character takes the position with the lesser coordinate and Lesha's character takes the position with the bigger coordinate and the guys start fighting.
Vova and Lesha are already tired by the game of choosing positions, as they need to play it before every fight, so they asked you (the developer of the The Ancient Papyri: Swordsink) to write a module that would automatically determine the distance at which the warrior and the archer will start fighting if both Vova and Lesha play optimally.
Input
The first line on the input contains a single integer n (2 ≤ n ≤ 200 000, n is even) — the number of positions available initially. The second line contains n distinct integers x1, x2, ..., xn (0 ≤ xi ≤ 109), giving the coordinates of the corresponding positions.
Output
Print the distance between the warrior and the archer at the beginning of the fight, provided that both Vova and Lesha play optimally.
Examples
Input
6
0 1 3 7 15 31
Output
7
Input
2
73 37
Output
36
Note
In the first sample one of the optimum behavior of the players looks like that:
1. Vova bans the position at coordinate 15;
2. Lesha bans the position at coordinate 3;
3. Vova bans the position at coordinate 31;
4. Lesha bans the position at coordinate 1.
After these actions only positions 0 and 7 will remain, and the distance between them is equal to 7.
In the second sample there are only two possible positions, so there will be no bans.
Submitted Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
import time
n = int(input())
x = sorted([int(i) for i in input().split()])
start = time.time()
dv = divmod(n - 2, 2)
v = dv[0] + dv[1]
for i in range(v):
if x[1] - x[0] < x[-1] - x[-2]:
x.remove(x[-1])
else:
x.remove(x[0])
print(x)
print(x[-1] - x[0])
finish = time.time()
#print(finish - start)
```
|
instruction
| 0
| 28,672
| 2
| 57,344
|
No
|
output
| 1
| 28,672
| 2
| 57,345
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the official contest this problem has a different statement, for which jury's solution was working incorrectly, and for this reason it was excluded from the contest. This mistake have been fixed and the current given problem statement and model solution corresponds to what jury wanted it to be during the contest.
Vova and Lesha are friends. They often meet at Vova's place and compete against each other in a computer game named The Ancient Papyri: Swordsink. Vova always chooses a warrior as his fighter and Leshac chooses an archer. After that they should choose initial positions for their characters and start the fight. A warrior is good at melee combat, so Vova will try to make the distance between fighters as small as possible. An archer prefers to keep the enemy at a distance, so Lesha will try to make the initial distance as large as possible.
There are n (n is always even) possible starting positions for characters marked along the Ox axis. The positions are given by their distinct coordinates x1, x2, ..., xn, two characters cannot end up at the same position.
Vova and Lesha take turns banning available positions, Vova moves first. During each turn one of the guys bans exactly one of the remaining positions. Banned positions cannot be used by both Vova and Lesha. They continue to make moves until there are only two possible positions remaining (thus, the total number of moves will be n - 2). After that Vova's character takes the position with the lesser coordinate and Lesha's character takes the position with the bigger coordinate and the guys start fighting.
Vova and Lesha are already tired by the game of choosing positions, as they need to play it before every fight, so they asked you (the developer of the The Ancient Papyri: Swordsink) to write a module that would automatically determine the distance at which the warrior and the archer will start fighting if both Vova and Lesha play optimally.
Input
The first line on the input contains a single integer n (2 ≤ n ≤ 200 000, n is even) — the number of positions available initially. The second line contains n distinct integers x1, x2, ..., xn (0 ≤ xi ≤ 109), giving the coordinates of the corresponding positions.
Output
Print the distance between the warrior and the archer at the beginning of the fight, provided that both Vova and Lesha play optimally.
Examples
Input
6
0 1 3 7 15 31
Output
7
Input
2
73 37
Output
36
Note
In the first sample one of the optimum behavior of the players looks like that:
1. Vova bans the position at coordinate 15;
2. Lesha bans the position at coordinate 3;
3. Vova bans the position at coordinate 31;
4. Lesha bans the position at coordinate 1.
After these actions only positions 0 and 7 will remain, and the distance between them is equal to 7.
In the second sample there are only two possible positions, so there will be no bans.
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
def main(a, n):
a.sort()
while n != 2:
col1 = a[1] - a[0]
col2 = a[-1] - a[-2]
if col1 >= col2:
a.pop(0)
else:
a.pop()
n -= 1
if n == 2:
break
if len(a) % 2 == 1:
a.pop(n // 2)
else:
if a[n // 2 + 1] - a[n // 2] >= a[n // 2 - 1] - a[n // 2 - 2]:
a.pop(n // 2 - 1)
else:
a.pop(n // 2)
n -= 1
print(a[1] - a[0])
main(a, n)
```
|
instruction
| 0
| 28,673
| 2
| 57,346
|
No
|
output
| 1
| 28,673
| 2
| 57,347
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the official contest this problem has a different statement, for which jury's solution was working incorrectly, and for this reason it was excluded from the contest. This mistake have been fixed and the current given problem statement and model solution corresponds to what jury wanted it to be during the contest.
Vova and Lesha are friends. They often meet at Vova's place and compete against each other in a computer game named The Ancient Papyri: Swordsink. Vova always chooses a warrior as his fighter and Leshac chooses an archer. After that they should choose initial positions for their characters and start the fight. A warrior is good at melee combat, so Vova will try to make the distance between fighters as small as possible. An archer prefers to keep the enemy at a distance, so Lesha will try to make the initial distance as large as possible.
There are n (n is always even) possible starting positions for characters marked along the Ox axis. The positions are given by their distinct coordinates x1, x2, ..., xn, two characters cannot end up at the same position.
Vova and Lesha take turns banning available positions, Vova moves first. During each turn one of the guys bans exactly one of the remaining positions. Banned positions cannot be used by both Vova and Lesha. They continue to make moves until there are only two possible positions remaining (thus, the total number of moves will be n - 2). After that Vova's character takes the position with the lesser coordinate and Lesha's character takes the position with the bigger coordinate and the guys start fighting.
Vova and Lesha are already tired by the game of choosing positions, as they need to play it before every fight, so they asked you (the developer of the The Ancient Papyri: Swordsink) to write a module that would automatically determine the distance at which the warrior and the archer will start fighting if both Vova and Lesha play optimally.
Input
The first line on the input contains a single integer n (2 ≤ n ≤ 200 000, n is even) — the number of positions available initially. The second line contains n distinct integers x1, x2, ..., xn (0 ≤ xi ≤ 109), giving the coordinates of the corresponding positions.
Output
Print the distance between the warrior and the archer at the beginning of the fight, provided that both Vova and Lesha play optimally.
Examples
Input
6
0 1 3 7 15 31
Output
7
Input
2
73 37
Output
36
Note
In the first sample one of the optimum behavior of the players looks like that:
1. Vova bans the position at coordinate 15;
2. Lesha bans the position at coordinate 3;
3. Vova bans the position at coordinate 31;
4. Lesha bans the position at coordinate 1.
After these actions only positions 0 and 7 will remain, and the distance between them is equal to 7.
In the second sample there are only two possible positions, so there will be no bans.
Submitted Solution:
```
def a2():
global mas
j=int((len(mas)-1)/2)
del mas[j]
a1()
def a1():
global mas
if len(mas)!=2:
del mas[-1]
a2()
input()
mas=input().strip().split()
mas=[int(i) for i in mas]
mas.sort()
a1()
print(int(mas[1])-int(mas[0]))
```
|
instruction
| 0
| 28,674
| 2
| 57,348
|
No
|
output
| 1
| 28,674
| 2
| 57,349
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya plays The Elder Trolls IV: Oblivon. Oh, those creators of computer games! What they do not come up with! Absolutely unique monsters have been added to the The Elder Trolls IV: Oblivon. One of these monsters is Unkillable Slug. Why it is "Unkillable"? Firstly, because it can be killed with cutting weapon only, so lovers of two-handed amber hammers should find suitable knife themselves. Secondly, it is necessary to make so many cutting strokes to Unkillable Slug. Extremely many. Too many!
Vasya has already promoted his character to 80-th level and in order to gain level 81 he was asked to kill Unkillable Slug. The monster has a very interesting shape. It looks like a rectangular parallelepiped with size x × y × z, consisting of undestructable cells 1 × 1 × 1. At one stroke Vasya can cut the Slug along an imaginary grid, i.e. cut with a plane parallel to one of the parallelepiped side. Monster dies when amount of parts it is divided reaches some critical value.
All parts of monster do not fall after each cut, they remains exactly on its places. I. e. Vasya can cut several parts with one cut.
Vasya wants to know what the maximum number of pieces he can cut the Unkillable Slug into striking him at most k times.
Vasya's character uses absolutely thin sword with infinite length.
Input
The first line of input contains four integer numbers x, y, z, k (1 ≤ x, y, z ≤ 106, 0 ≤ k ≤ 109).
Output
Output the only number — the answer for the problem.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Examples
Input
2 2 2 3
Output
8
Input
2 2 2 1
Output
2
Note
In the first sample Vasya make 3 pairwise perpendicular cuts. He cuts monster on two parts with the first cut, then he divides each part on two with the second cut, and finally he divides each of the 4 parts on two.
|
instruction
| 0
| 30,577
| 2
| 61,154
|
Tags: greedy, math
Correct Solution:
```
I=lambda: map(int,input().split())
x,y,z,k=I()
a1,a2,a3,q=0,0,0,0
while q<3:
q=0
if a1+a2+a3==k: break
if a1<x-1:
a1+=1
if a1+a2+a3==k: break
else:
q+=1
if a2<y-1:
a2+=1
if a1+a2+a3==k: break
else:
q+=1
if a3<z-1:
a3+=1
if a1+a2+a3==k: break
else:
q+=1
print((a1+1)*(a2+1)*(a3+1))
```
|
output
| 1
| 30,577
| 2
| 61,155
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya plays The Elder Trolls IV: Oblivon. Oh, those creators of computer games! What they do not come up with! Absolutely unique monsters have been added to the The Elder Trolls IV: Oblivon. One of these monsters is Unkillable Slug. Why it is "Unkillable"? Firstly, because it can be killed with cutting weapon only, so lovers of two-handed amber hammers should find suitable knife themselves. Secondly, it is necessary to make so many cutting strokes to Unkillable Slug. Extremely many. Too many!
Vasya has already promoted his character to 80-th level and in order to gain level 81 he was asked to kill Unkillable Slug. The monster has a very interesting shape. It looks like a rectangular parallelepiped with size x × y × z, consisting of undestructable cells 1 × 1 × 1. At one stroke Vasya can cut the Slug along an imaginary grid, i.e. cut with a plane parallel to one of the parallelepiped side. Monster dies when amount of parts it is divided reaches some critical value.
All parts of monster do not fall after each cut, they remains exactly on its places. I. e. Vasya can cut several parts with one cut.
Vasya wants to know what the maximum number of pieces he can cut the Unkillable Slug into striking him at most k times.
Vasya's character uses absolutely thin sword with infinite length.
Input
The first line of input contains four integer numbers x, y, z, k (1 ≤ x, y, z ≤ 106, 0 ≤ k ≤ 109).
Output
Output the only number — the answer for the problem.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Examples
Input
2 2 2 3
Output
8
Input
2 2 2 1
Output
2
Note
In the first sample Vasya make 3 pairwise perpendicular cuts. He cuts monster on two parts with the first cut, then he divides each part on two with the second cut, and finally he divides each of the 4 parts on two.
|
instruction
| 0
| 30,578
| 2
| 61,156
|
Tags: greedy, math
Correct Solution:
```
x, y, z, k = map(int, input().split())
sides = sorted([x, y, z])
cuts = 3 * [ None ]
product = 1
for i in range(3):
a = min(sides[i] - 1, k // (3 - i))
cuts[i] = a
k -= a
product *= a + 1
#print(cuts)
print(product)
```
|
output
| 1
| 30,578
| 2
| 61,157
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya plays The Elder Trolls IV: Oblivon. Oh, those creators of computer games! What they do not come up with! Absolutely unique monsters have been added to the The Elder Trolls IV: Oblivon. One of these monsters is Unkillable Slug. Why it is "Unkillable"? Firstly, because it can be killed with cutting weapon only, so lovers of two-handed amber hammers should find suitable knife themselves. Secondly, it is necessary to make so many cutting strokes to Unkillable Slug. Extremely many. Too many!
Vasya has already promoted his character to 80-th level and in order to gain level 81 he was asked to kill Unkillable Slug. The monster has a very interesting shape. It looks like a rectangular parallelepiped with size x × y × z, consisting of undestructable cells 1 × 1 × 1. At one stroke Vasya can cut the Slug along an imaginary grid, i.e. cut with a plane parallel to one of the parallelepiped side. Monster dies when amount of parts it is divided reaches some critical value.
All parts of monster do not fall after each cut, they remains exactly on its places. I. e. Vasya can cut several parts with one cut.
Vasya wants to know what the maximum number of pieces he can cut the Unkillable Slug into striking him at most k times.
Vasya's character uses absolutely thin sword with infinite length.
Input
The first line of input contains four integer numbers x, y, z, k (1 ≤ x, y, z ≤ 106, 0 ≤ k ≤ 109).
Output
Output the only number — the answer for the problem.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Examples
Input
2 2 2 3
Output
8
Input
2 2 2 1
Output
2
Note
In the first sample Vasya make 3 pairwise perpendicular cuts. He cuts monster on two parts with the first cut, then he divides each part on two with the second cut, and finally he divides each of the 4 parts on two.
|
instruction
| 0
| 30,579
| 2
| 61,158
|
Tags: greedy, math
Correct Solution:
```
x, y, z, k = [int(value) for value in input().split()]
x, y, z = sorted([x, y, z])
a = min(k // 3, x - 1)
k -= a
b = min(k // 2, y - 1)
k -= b
c = min(k, z - 1)
print((a + 1) * (b + 1) * (c + 1))
```
|
output
| 1
| 30,579
| 2
| 61,159
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya plays The Elder Trolls IV: Oblivon. Oh, those creators of computer games! What they do not come up with! Absolutely unique monsters have been added to the The Elder Trolls IV: Oblivon. One of these monsters is Unkillable Slug. Why it is "Unkillable"? Firstly, because it can be killed with cutting weapon only, so lovers of two-handed amber hammers should find suitable knife themselves. Secondly, it is necessary to make so many cutting strokes to Unkillable Slug. Extremely many. Too many!
Vasya has already promoted his character to 80-th level and in order to gain level 81 he was asked to kill Unkillable Slug. The monster has a very interesting shape. It looks like a rectangular parallelepiped with size x × y × z, consisting of undestructable cells 1 × 1 × 1. At one stroke Vasya can cut the Slug along an imaginary grid, i.e. cut with a plane parallel to one of the parallelepiped side. Monster dies when amount of parts it is divided reaches some critical value.
All parts of monster do not fall after each cut, they remains exactly on its places. I. e. Vasya can cut several parts with one cut.
Vasya wants to know what the maximum number of pieces he can cut the Unkillable Slug into striking him at most k times.
Vasya's character uses absolutely thin sword with infinite length.
Input
The first line of input contains four integer numbers x, y, z, k (1 ≤ x, y, z ≤ 106, 0 ≤ k ≤ 109).
Output
Output the only number — the answer for the problem.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Examples
Input
2 2 2 3
Output
8
Input
2 2 2 1
Output
2
Note
In the first sample Vasya make 3 pairwise perpendicular cuts. He cuts monster on two parts with the first cut, then he divides each part on two with the second cut, and finally he divides each of the 4 parts on two.
|
instruction
| 0
| 30,580
| 2
| 61,160
|
Tags: greedy, math
Correct Solution:
```
x,y,z,k=map(int,input().split())
a,b,c=1,1,1
while k//3 and (a<x or b<y or c<z):p=k//((a<x)+(b<y)+(c<z));A=min(x-a,p);B=min(y-b,p);C=min(z-c,p);a+=A;b+=B;c+=C;k-=A+B+C
while k and (a<x or b<y or c<z):
if a<x:a+=1;k-=1
if k and b<y:b+=1;k-=1
if k and c<z:c+=1;k-=1
print(a*b*c)
```
|
output
| 1
| 30,580
| 2
| 61,161
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya plays The Elder Trolls IV: Oblivon. Oh, those creators of computer games! What they do not come up with! Absolutely unique monsters have been added to the The Elder Trolls IV: Oblivon. One of these monsters is Unkillable Slug. Why it is "Unkillable"? Firstly, because it can be killed with cutting weapon only, so lovers of two-handed amber hammers should find suitable knife themselves. Secondly, it is necessary to make so many cutting strokes to Unkillable Slug. Extremely many. Too many!
Vasya has already promoted his character to 80-th level and in order to gain level 81 he was asked to kill Unkillable Slug. The monster has a very interesting shape. It looks like a rectangular parallelepiped with size x × y × z, consisting of undestructable cells 1 × 1 × 1. At one stroke Vasya can cut the Slug along an imaginary grid, i.e. cut with a plane parallel to one of the parallelepiped side. Monster dies when amount of parts it is divided reaches some critical value.
All parts of monster do not fall after each cut, they remains exactly on its places. I. e. Vasya can cut several parts with one cut.
Vasya wants to know what the maximum number of pieces he can cut the Unkillable Slug into striking him at most k times.
Vasya's character uses absolutely thin sword with infinite length.
Input
The first line of input contains four integer numbers x, y, z, k (1 ≤ x, y, z ≤ 106, 0 ≤ k ≤ 109).
Output
Output the only number — the answer for the problem.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Examples
Input
2 2 2 3
Output
8
Input
2 2 2 1
Output
2
Note
In the first sample Vasya make 3 pairwise perpendicular cuts. He cuts monster on two parts with the first cut, then he divides each part on two with the second cut, and finally he divides each of the 4 parts on two.
|
instruction
| 0
| 30,581
| 2
| 61,162
|
Tags: greedy, math
Correct Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
avl=AvlTree()
#-----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default='z', func=lambda a, b: min(a ,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: a + b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left)/ 2)
# Check if middle element is
# less than or equal to key
if (arr[mid]<=key):
count = mid+1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def countGreater( arr,n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
a,b,c,k=map(int,input().split())
ans=1
k1=k//3
l=[a,b,c]
l.sort()
a=l[0]
b=l[1]
c=l[2]
l.sort()
r=min(a-1,k1)
k-=r
ans*=(r+1)
k1=k//2
r=min(b-1,k1)
k-=r
ans*=(r+1)
r=min(c-1,k)
k-=r
ans*=(r+1)
print(ans)
```
|
output
| 1
| 30,581
| 2
| 61,163
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya plays The Elder Trolls IV: Oblivon. Oh, those creators of computer games! What they do not come up with! Absolutely unique monsters have been added to the The Elder Trolls IV: Oblivon. One of these monsters is Unkillable Slug. Why it is "Unkillable"? Firstly, because it can be killed with cutting weapon only, so lovers of two-handed amber hammers should find suitable knife themselves. Secondly, it is necessary to make so many cutting strokes to Unkillable Slug. Extremely many. Too many!
Vasya has already promoted his character to 80-th level and in order to gain level 81 he was asked to kill Unkillable Slug. The monster has a very interesting shape. It looks like a rectangular parallelepiped with size x × y × z, consisting of undestructable cells 1 × 1 × 1. At one stroke Vasya can cut the Slug along an imaginary grid, i.e. cut with a plane parallel to one of the parallelepiped side. Monster dies when amount of parts it is divided reaches some critical value.
All parts of monster do not fall after each cut, they remains exactly on its places. I. e. Vasya can cut several parts with one cut.
Vasya wants to know what the maximum number of pieces he can cut the Unkillable Slug into striking him at most k times.
Vasya's character uses absolutely thin sword with infinite length.
Input
The first line of input contains four integer numbers x, y, z, k (1 ≤ x, y, z ≤ 106, 0 ≤ k ≤ 109).
Output
Output the only number — the answer for the problem.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Examples
Input
2 2 2 3
Output
8
Input
2 2 2 1
Output
2
Note
In the first sample Vasya make 3 pairwise perpendicular cuts. He cuts monster on two parts with the first cut, then he divides each part on two with the second cut, and finally he divides each of the 4 parts on two.
|
instruction
| 0
| 30,582
| 2
| 61,164
|
Tags: greedy, math
Correct Solution:
```
x, y, z, k = map(int, input().split())
x, y, z = sorted((x, y, z))
a = min(k // 3, x - 1)
b = min((k - a) // 2, y - 1)
c = min(k - a - b, z - 1)
print((a + 1) * (b + 1) * (c + 1))
```
|
output
| 1
| 30,582
| 2
| 61,165
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya plays The Elder Trolls IV: Oblivon. Oh, those creators of computer games! What they do not come up with! Absolutely unique monsters have been added to the The Elder Trolls IV: Oblivon. One of these monsters is Unkillable Slug. Why it is "Unkillable"? Firstly, because it can be killed with cutting weapon only, so lovers of two-handed amber hammers should find suitable knife themselves. Secondly, it is necessary to make so many cutting strokes to Unkillable Slug. Extremely many. Too many!
Vasya has already promoted his character to 80-th level and in order to gain level 81 he was asked to kill Unkillable Slug. The monster has a very interesting shape. It looks like a rectangular parallelepiped with size x × y × z, consisting of undestructable cells 1 × 1 × 1. At one stroke Vasya can cut the Slug along an imaginary grid, i.e. cut with a plane parallel to one of the parallelepiped side. Monster dies when amount of parts it is divided reaches some critical value.
All parts of monster do not fall after each cut, they remains exactly on its places. I. e. Vasya can cut several parts with one cut.
Vasya wants to know what the maximum number of pieces he can cut the Unkillable Slug into striking him at most k times.
Vasya's character uses absolutely thin sword with infinite length.
Input
The first line of input contains four integer numbers x, y, z, k (1 ≤ x, y, z ≤ 106, 0 ≤ k ≤ 109).
Output
Output the only number — the answer for the problem.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Examples
Input
2 2 2 3
Output
8
Input
2 2 2 1
Output
2
Note
In the first sample Vasya make 3 pairwise perpendicular cuts. He cuts monster on two parts with the first cut, then he divides each part on two with the second cut, and finally he divides each of the 4 parts on two.
|
instruction
| 0
| 30,583
| 2
| 61,166
|
Tags: greedy, math
Correct Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
from collections import defaultdict
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------game starts now-----------------------------------------------------
x,y,z,k=map(int,input().split())
if x+y+z-3<=k:
print(x*y*z)
else:
r=1
a=[x,y,z]
a.sort()
x=a[0]
y=a[1]
z=a[2]
c=k//3
if x-1<c:
r*=x
k-=x-1
else:
r*=c+1
k-=c
c=k//2
if y-1<c:
r*=y
k-=y-1
else:
r*=c+1
k-=c
if z-1<k:
r*=z
k-=z-1
else:
r*=k+1
k=0
print(r)
```
|
output
| 1
| 30,583
| 2
| 61,167
|
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