message stringlengths 2 39.6k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 450 109k | cluster float64 2 2 | __index_level_0__ int64 900 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are playing a game where your character should overcome different obstacles. The current problem is to come down from a cliff. The cliff has height h, and there is a moving platform on each height x from 1 to h.
Each platform is either hidden inside the cliff or moved out. At first, there are n moved out platforms on heights p_1, p_2, ..., p_n. The platform on height h is moved out (and the character is initially standing there).
If you character is standing on some moved out platform on height x, then he can pull a special lever, which switches the state of two platforms: on height x and x - 1. In other words, the platform you are currently standing on will hide in the cliff and the platform one unit below will change it state: it will hide if it was moved out or move out if it was hidden. In the second case, you will safely land on it. Note that this is the only way to move from one platform to another.
Your character is quite fragile, so it can safely fall from the height no more than 2. In other words falling from the platform x to platform x - 2 is okay, but falling from x to x - 3 (or lower) is certain death.
Sometimes it's not possible to come down from the cliff, but you can always buy (for donate currency) several magic crystals. Each magic crystal can be used to change the state of any single platform (except platform on height h, which is unaffected by the crystals). After being used, the crystal disappears.
What is the minimum number of magic crystal you need to buy to safely land on the 0 ground level?
Input
The first line contains one integer q (1 ≤ q ≤ 100) — the number of queries. Each query contains two lines and is independent of all other queries.
The first line of each query contains two integers h and n (1 ≤ h ≤ 10^9, 1 ≤ n ≤ min(h, 2 ⋅ 10^5)) — the height of the cliff and the number of moved out platforms.
The second line contains n integers p_1, p_2, ..., p_n (h = p_1 > p_2 > ... > p_n ≥ 1) — the corresponding moved out platforms in the descending order of their heights.
The sum of n over all queries does not exceed 2 ⋅ 10^5.
Output
For each query print one integer — the minimum number of magic crystals you have to spend to safely come down on the ground level (with height 0).
Example
Input
4
3 2
3 1
8 6
8 7 6 5 3 2
9 6
9 8 5 4 3 1
1 1
1
Output
0
1
2
0 | instruction | 0 | 43,042 | 2 | 86,084 |
Tags: dp, greedy, math
Correct Solution:
```
q = int(input())
for _ in range(q):
h, n = list(map(int, input().split()))
arr = list(map(int, input().split())) + [0]
ans = 0
cur = h
for i in range(1, n):
if arr[i] == cur:
continue
else:
if arr[i + 1] == arr[i] - 1:
cur = arr[i] - 1
else:
ans += 1
print(ans)
``` | output | 1 | 43,042 | 2 | 86,085 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are playing a game where your character should overcome different obstacles. The current problem is to come down from a cliff. The cliff has height h, and there is a moving platform on each height x from 1 to h.
Each platform is either hidden inside the cliff or moved out. At first, there are n moved out platforms on heights p_1, p_2, ..., p_n. The platform on height h is moved out (and the character is initially standing there).
If you character is standing on some moved out platform on height x, then he can pull a special lever, which switches the state of two platforms: on height x and x - 1. In other words, the platform you are currently standing on will hide in the cliff and the platform one unit below will change it state: it will hide if it was moved out or move out if it was hidden. In the second case, you will safely land on it. Note that this is the only way to move from one platform to another.
Your character is quite fragile, so it can safely fall from the height no more than 2. In other words falling from the platform x to platform x - 2 is okay, but falling from x to x - 3 (or lower) is certain death.
Sometimes it's not possible to come down from the cliff, but you can always buy (for donate currency) several magic crystals. Each magic crystal can be used to change the state of any single platform (except platform on height h, which is unaffected by the crystals). After being used, the crystal disappears.
What is the minimum number of magic crystal you need to buy to safely land on the 0 ground level?
Input
The first line contains one integer q (1 ≤ q ≤ 100) — the number of queries. Each query contains two lines and is independent of all other queries.
The first line of each query contains two integers h and n (1 ≤ h ≤ 10^9, 1 ≤ n ≤ min(h, 2 ⋅ 10^5)) — the height of the cliff and the number of moved out platforms.
The second line contains n integers p_1, p_2, ..., p_n (h = p_1 > p_2 > ... > p_n ≥ 1) — the corresponding moved out platforms in the descending order of their heights.
The sum of n over all queries does not exceed 2 ⋅ 10^5.
Output
For each query print one integer — the minimum number of magic crystals you have to spend to safely come down on the ground level (with height 0).
Example
Input
4
3 2
3 1
8 6
8 7 6 5 3 2
9 6
9 8 5 4 3 1
1 1
1
Output
0
1
2
0 | instruction | 0 | 43,043 | 2 | 86,086 |
Tags: dp, greedy, math
Correct Solution:
```
for q in range(int(input())):
h, n = map(int, input().split())
p = list(map(int, input().split()))
# s = [0] * h
# for plat in p:
# s[h-plat] = 1
p.append(0)
ans = 0
i = 1
# print(s)
while i < n:
if p[i] - 1 == p[i+1]:
i += 2
else:
ans += 1
i += 1
print(ans)
``` | output | 1 | 43,043 | 2 | 86,087 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are playing a game where your character should overcome different obstacles. The current problem is to come down from a cliff. The cliff has height h, and there is a moving platform on each height x from 1 to h.
Each platform is either hidden inside the cliff or moved out. At first, there are n moved out platforms on heights p_1, p_2, ..., p_n. The platform on height h is moved out (and the character is initially standing there).
If you character is standing on some moved out platform on height x, then he can pull a special lever, which switches the state of two platforms: on height x and x - 1. In other words, the platform you are currently standing on will hide in the cliff and the platform one unit below will change it state: it will hide if it was moved out or move out if it was hidden. In the second case, you will safely land on it. Note that this is the only way to move from one platform to another.
Your character is quite fragile, so it can safely fall from the height no more than 2. In other words falling from the platform x to platform x - 2 is okay, but falling from x to x - 3 (or lower) is certain death.
Sometimes it's not possible to come down from the cliff, but you can always buy (for donate currency) several magic crystals. Each magic crystal can be used to change the state of any single platform (except platform on height h, which is unaffected by the crystals). After being used, the crystal disappears.
What is the minimum number of magic crystal you need to buy to safely land on the 0 ground level?
Input
The first line contains one integer q (1 ≤ q ≤ 100) — the number of queries. Each query contains two lines and is independent of all other queries.
The first line of each query contains two integers h and n (1 ≤ h ≤ 10^9, 1 ≤ n ≤ min(h, 2 ⋅ 10^5)) — the height of the cliff and the number of moved out platforms.
The second line contains n integers p_1, p_2, ..., p_n (h = p_1 > p_2 > ... > p_n ≥ 1) — the corresponding moved out platforms in the descending order of their heights.
The sum of n over all queries does not exceed 2 ⋅ 10^5.
Output
For each query print one integer — the minimum number of magic crystals you have to spend to safely come down on the ground level (with height 0).
Example
Input
4
3 2
3 1
8 6
8 7 6 5 3 2
9 6
9 8 5 4 3 1
1 1
1
Output
0
1
2
0 | instruction | 0 | 43,044 | 2 | 86,088 |
Tags: dp, greedy, math
Correct Solution:
```
for _ in range(int(input())):
h, n = map(int, input().split())
P = list(map(int, input().split()))[::-1]
P.pop()
ans = 0
while h > 2 and P:
w = P.pop()
if (h - w) == 1:
if P:
q = P.pop()
if (h - q) != 2:
ans += 1
P.append(q)
else:
ans += 1
h -= 2
else:
h = w + 1
P.append(w)
print(ans)
``` | output | 1 | 43,044 | 2 | 86,089 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are playing a game where your character should overcome different obstacles. The current problem is to come down from a cliff. The cliff has height h, and there is a moving platform on each height x from 1 to h.
Each platform is either hidden inside the cliff or moved out. At first, there are n moved out platforms on heights p_1, p_2, ..., p_n. The platform on height h is moved out (and the character is initially standing there).
If you character is standing on some moved out platform on height x, then he can pull a special lever, which switches the state of two platforms: on height x and x - 1. In other words, the platform you are currently standing on will hide in the cliff and the platform one unit below will change it state: it will hide if it was moved out or move out if it was hidden. In the second case, you will safely land on it. Note that this is the only way to move from one platform to another.
Your character is quite fragile, so it can safely fall from the height no more than 2. In other words falling from the platform x to platform x - 2 is okay, but falling from x to x - 3 (or lower) is certain death.
Sometimes it's not possible to come down from the cliff, but you can always buy (for donate currency) several magic crystals. Each magic crystal can be used to change the state of any single platform (except platform on height h, which is unaffected by the crystals). After being used, the crystal disappears.
What is the minimum number of magic crystal you need to buy to safely land on the 0 ground level?
Input
The first line contains one integer q (1 ≤ q ≤ 100) — the number of queries. Each query contains two lines and is independent of all other queries.
The first line of each query contains two integers h and n (1 ≤ h ≤ 10^9, 1 ≤ n ≤ min(h, 2 ⋅ 10^5)) — the height of the cliff and the number of moved out platforms.
The second line contains n integers p_1, p_2, ..., p_n (h = p_1 > p_2 > ... > p_n ≥ 1) — the corresponding moved out platforms in the descending order of their heights.
The sum of n over all queries does not exceed 2 ⋅ 10^5.
Output
For each query print one integer — the minimum number of magic crystals you have to spend to safely come down on the ground level (with height 0).
Example
Input
4
3 2
3 1
8 6
8 7 6 5 3 2
9 6
9 8 5 4 3 1
1 1
1
Output
0
1
2
0
Submitted Solution:
```
from collections import defaultdict
for _ in range(int(input())):
h, n = map(int, input().split())
mov = list(map(int, input().split()))
# rec = {}
# rec = defaultdict(int)
# for i in mov:
# rec[i] = 1
i = h
ans = 0
# while i > 0:
# if i <= 2:
# break
# if rec.get(i - 1, 0) == 0:
# i -= 1
# else:
# if rec.get(i - 2, 0) == 0:
# ans += 1
# i -= 2
# curr = h
# next = 1
# while next < n:
# if mov[next] == h - 2:
# h -= 2
# next += 1
# else:
#
dp = [0 for i in range(n)]
# if mov[n - 1] % 2 == 0:
# dp[n - 1] = (mov[n - 1] - 1) // 2
# else:
# dp[n - 1] = (mov[n - 1] - 2) // 2
for i in range(n - 2, -1, -1):
if mov[i] <= 2 or mov[i + 1] <= 1:
dp[i] = 0
elif i + 2 < n and mov[i + 1] - mov[i + 2] == 1:
dp[i] = min(dp[i + 2], 1 + dp[i + 1])
else:
dp[i] = 1 + dp[i + 1]
print(dp[0])
``` | instruction | 0 | 43,045 | 2 | 86,090 |
Yes | output | 1 | 43,045 | 2 | 86,091 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are playing a game where your character should overcome different obstacles. The current problem is to come down from a cliff. The cliff has height h, and there is a moving platform on each height x from 1 to h.
Each platform is either hidden inside the cliff or moved out. At first, there are n moved out platforms on heights p_1, p_2, ..., p_n. The platform on height h is moved out (and the character is initially standing there).
If you character is standing on some moved out platform on height x, then he can pull a special lever, which switches the state of two platforms: on height x and x - 1. In other words, the platform you are currently standing on will hide in the cliff and the platform one unit below will change it state: it will hide if it was moved out or move out if it was hidden. In the second case, you will safely land on it. Note that this is the only way to move from one platform to another.
Your character is quite fragile, so it can safely fall from the height no more than 2. In other words falling from the platform x to platform x - 2 is okay, but falling from x to x - 3 (or lower) is certain death.
Sometimes it's not possible to come down from the cliff, but you can always buy (for donate currency) several magic crystals. Each magic crystal can be used to change the state of any single platform (except platform on height h, which is unaffected by the crystals). After being used, the crystal disappears.
What is the minimum number of magic crystal you need to buy to safely land on the 0 ground level?
Input
The first line contains one integer q (1 ≤ q ≤ 100) — the number of queries. Each query contains two lines and is independent of all other queries.
The first line of each query contains two integers h and n (1 ≤ h ≤ 10^9, 1 ≤ n ≤ min(h, 2 ⋅ 10^5)) — the height of the cliff and the number of moved out platforms.
The second line contains n integers p_1, p_2, ..., p_n (h = p_1 > p_2 > ... > p_n ≥ 1) — the corresponding moved out platforms in the descending order of their heights.
The sum of n over all queries does not exceed 2 ⋅ 10^5.
Output
For each query print one integer — the minimum number of magic crystals you have to spend to safely come down on the ground level (with height 0).
Example
Input
4
3 2
3 1
8 6
8 7 6 5 3 2
9 6
9 8 5 4 3 1
1 1
1
Output
0
1
2
0
Submitted Solution:
```
import sys
input = sys.stdin.readline
q=int(input())
for testcases in range(q):
h,n=map(int,input().split())
P=list(map(int,input().split()))
P.append(-10)
NOW=h
ANS=0
ind=1
while NOW>=3:
#print(NOW,ind,ANS)
if P[ind]<NOW-1:
NOW=P[ind]+1
continue
if P[ind]==NOW-1 and P[ind+1]==NOW-2:
NOW-=2
ind+=2
continue
if P[ind]==NOW-1 and P[ind+1]!=NOW-2:
NOW-=2
ind+=1
ANS+=1
print(ANS)
``` | instruction | 0 | 43,046 | 2 | 86,092 |
Yes | output | 1 | 43,046 | 2 | 86,093 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are playing a game where your character should overcome different obstacles. The current problem is to come down from a cliff. The cliff has height h, and there is a moving platform on each height x from 1 to h.
Each platform is either hidden inside the cliff or moved out. At first, there are n moved out platforms on heights p_1, p_2, ..., p_n. The platform on height h is moved out (and the character is initially standing there).
If you character is standing on some moved out platform on height x, then he can pull a special lever, which switches the state of two platforms: on height x and x - 1. In other words, the platform you are currently standing on will hide in the cliff and the platform one unit below will change it state: it will hide if it was moved out or move out if it was hidden. In the second case, you will safely land on it. Note that this is the only way to move from one platform to another.
Your character is quite fragile, so it can safely fall from the height no more than 2. In other words falling from the platform x to platform x - 2 is okay, but falling from x to x - 3 (or lower) is certain death.
Sometimes it's not possible to come down from the cliff, but you can always buy (for donate currency) several magic crystals. Each magic crystal can be used to change the state of any single platform (except platform on height h, which is unaffected by the crystals). After being used, the crystal disappears.
What is the minimum number of magic crystal you need to buy to safely land on the 0 ground level?
Input
The first line contains one integer q (1 ≤ q ≤ 100) — the number of queries. Each query contains two lines and is independent of all other queries.
The first line of each query contains two integers h and n (1 ≤ h ≤ 10^9, 1 ≤ n ≤ min(h, 2 ⋅ 10^5)) — the height of the cliff and the number of moved out platforms.
The second line contains n integers p_1, p_2, ..., p_n (h = p_1 > p_2 > ... > p_n ≥ 1) — the corresponding moved out platforms in the descending order of their heights.
The sum of n over all queries does not exceed 2 ⋅ 10^5.
Output
For each query print one integer — the minimum number of magic crystals you have to spend to safely come down on the ground level (with height 0).
Example
Input
4
3 2
3 1
8 6
8 7 6 5 3 2
9 6
9 8 5 4 3 1
1 1
1
Output
0
1
2
0
Submitted Solution:
```
from sys import stdin
a=lambda:stdin.readline().split()
for _ in range(int(stdin.readline())):
h,n=map(int,a())
lst=[*map(int,a())]
if n==1:print(0);continue
cout,item,res=1,1,0
for i,x in enumerate(lst[1:]):
if x+1==lst[i]:cout+=1
else:
if item==1:
if cout%2==0:res+=1
item=2
else:
if cout%2==1:res+=1
cout=1
if lst[-1]!=1:
if item==1:
if cout%2==0:res+=1
else:
if cout%2==1:res+=1
print(res)
``` | instruction | 0 | 43,047 | 2 | 86,094 |
Yes | output | 1 | 43,047 | 2 | 86,095 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are playing a game where your character should overcome different obstacles. The current problem is to come down from a cliff. The cliff has height h, and there is a moving platform on each height x from 1 to h.
Each platform is either hidden inside the cliff or moved out. At first, there are n moved out platforms on heights p_1, p_2, ..., p_n. The platform on height h is moved out (and the character is initially standing there).
If you character is standing on some moved out platform on height x, then he can pull a special lever, which switches the state of two platforms: on height x and x - 1. In other words, the platform you are currently standing on will hide in the cliff and the platform one unit below will change it state: it will hide if it was moved out or move out if it was hidden. In the second case, you will safely land on it. Note that this is the only way to move from one platform to another.
Your character is quite fragile, so it can safely fall from the height no more than 2. In other words falling from the platform x to platform x - 2 is okay, but falling from x to x - 3 (or lower) is certain death.
Sometimes it's not possible to come down from the cliff, but you can always buy (for donate currency) several magic crystals. Each magic crystal can be used to change the state of any single platform (except platform on height h, which is unaffected by the crystals). After being used, the crystal disappears.
What is the minimum number of magic crystal you need to buy to safely land on the 0 ground level?
Input
The first line contains one integer q (1 ≤ q ≤ 100) — the number of queries. Each query contains two lines and is independent of all other queries.
The first line of each query contains two integers h and n (1 ≤ h ≤ 10^9, 1 ≤ n ≤ min(h, 2 ⋅ 10^5)) — the height of the cliff and the number of moved out platforms.
The second line contains n integers p_1, p_2, ..., p_n (h = p_1 > p_2 > ... > p_n ≥ 1) — the corresponding moved out platforms in the descending order of their heights.
The sum of n over all queries does not exceed 2 ⋅ 10^5.
Output
For each query print one integer — the minimum number of magic crystals you have to spend to safely come down on the ground level (with height 0).
Example
Input
4
3 2
3 1
8 6
8 7 6 5 3 2
9 6
9 8 5 4 3 1
1 1
1
Output
0
1
2
0
Submitted Solution:
```
q = int(input())
for _ in range(q):
h, n = map(int, input().split())
ps = list(map(int, input().split())) + [0, 0]
ps.reverse()
cur = n-1
crystals = 0
while ps and ps[-1] > 1:
if ps[-2] == h - 1:
if ps[-3] == h - 2:
ps.pop()
ps.pop()
else:
ps.pop()
ps[-1] = h - 2
crystals += 1
h -= 2
else:
ps[-1] = ps[-2] + 1
h = ps[-1]
print(crystals)
``` | instruction | 0 | 43,048 | 2 | 86,096 |
Yes | output | 1 | 43,048 | 2 | 86,097 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are playing a game where your character should overcome different obstacles. The current problem is to come down from a cliff. The cliff has height h, and there is a moving platform on each height x from 1 to h.
Each platform is either hidden inside the cliff or moved out. At first, there are n moved out platforms on heights p_1, p_2, ..., p_n. The platform on height h is moved out (and the character is initially standing there).
If you character is standing on some moved out platform on height x, then he can pull a special lever, which switches the state of two platforms: on height x and x - 1. In other words, the platform you are currently standing on will hide in the cliff and the platform one unit below will change it state: it will hide if it was moved out or move out if it was hidden. In the second case, you will safely land on it. Note that this is the only way to move from one platform to another.
Your character is quite fragile, so it can safely fall from the height no more than 2. In other words falling from the platform x to platform x - 2 is okay, but falling from x to x - 3 (or lower) is certain death.
Sometimes it's not possible to come down from the cliff, but you can always buy (for donate currency) several magic crystals. Each magic crystal can be used to change the state of any single platform (except platform on height h, which is unaffected by the crystals). After being used, the crystal disappears.
What is the minimum number of magic crystal you need to buy to safely land on the 0 ground level?
Input
The first line contains one integer q (1 ≤ q ≤ 100) — the number of queries. Each query contains two lines and is independent of all other queries.
The first line of each query contains two integers h and n (1 ≤ h ≤ 10^9, 1 ≤ n ≤ min(h, 2 ⋅ 10^5)) — the height of the cliff and the number of moved out platforms.
The second line contains n integers p_1, p_2, ..., p_n (h = p_1 > p_2 > ... > p_n ≥ 1) — the corresponding moved out platforms in the descending order of their heights.
The sum of n over all queries does not exceed 2 ⋅ 10^5.
Output
For each query print one integer — the minimum number of magic crystals you have to spend to safely come down on the ground level (with height 0).
Example
Input
4
3 2
3 1
8 6
8 7 6 5 3 2
9 6
9 8 5 4 3 1
1 1
1
Output
0
1
2
0
Submitted Solution:
```
"""
NTC here
"""
from sys import stdin, setrecursionlimit
setrecursionlimit(10**7)
def iin(): return int(stdin.readline())
def lin(): return list(map(int, stdin.readline().split()))
# range = xrange
# input = raw_input
def main():
for _ in range(iin()):
h,n=lin()
a=lin()
a=a+[0,0]
sol=0
i=0
while i<n:
df=0
#print('C',i,a[i])
if a[i+1]<=1 or i==n-1:
break
if a[i]>a[i+1]>=a[i]-2:
if a[i+2]+1==a[i+1]:
pass
elif a[i+2]+2==a[i+1] or a[i+2]==0:
sol+=1
i-=1
else:
sol+=1
df=1
i+=1
else:
df=1
sol+=1 if df else 0
i+=1
#print('S',sol,i)
print(sol)
main()
``` | instruction | 0 | 43,049 | 2 | 86,098 |
No | output | 1 | 43,049 | 2 | 86,099 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are playing a game where your character should overcome different obstacles. The current problem is to come down from a cliff. The cliff has height h, and there is a moving platform on each height x from 1 to h.
Each platform is either hidden inside the cliff or moved out. At first, there are n moved out platforms on heights p_1, p_2, ..., p_n. The platform on height h is moved out (and the character is initially standing there).
If you character is standing on some moved out platform on height x, then he can pull a special lever, which switches the state of two platforms: on height x and x - 1. In other words, the platform you are currently standing on will hide in the cliff and the platform one unit below will change it state: it will hide if it was moved out or move out if it was hidden. In the second case, you will safely land on it. Note that this is the only way to move from one platform to another.
Your character is quite fragile, so it can safely fall from the height no more than 2. In other words falling from the platform x to platform x - 2 is okay, but falling from x to x - 3 (or lower) is certain death.
Sometimes it's not possible to come down from the cliff, but you can always buy (for donate currency) several magic crystals. Each magic crystal can be used to change the state of any single platform (except platform on height h, which is unaffected by the crystals). After being used, the crystal disappears.
What is the minimum number of magic crystal you need to buy to safely land on the 0 ground level?
Input
The first line contains one integer q (1 ≤ q ≤ 100) — the number of queries. Each query contains two lines and is independent of all other queries.
The first line of each query contains two integers h and n (1 ≤ h ≤ 10^9, 1 ≤ n ≤ min(h, 2 ⋅ 10^5)) — the height of the cliff and the number of moved out platforms.
The second line contains n integers p_1, p_2, ..., p_n (h = p_1 > p_2 > ... > p_n ≥ 1) — the corresponding moved out platforms in the descending order of their heights.
The sum of n over all queries does not exceed 2 ⋅ 10^5.
Output
For each query print one integer — the minimum number of magic crystals you have to spend to safely come down on the ground level (with height 0).
Example
Input
4
3 2
3 1
8 6
8 7 6 5 3 2
9 6
9 8 5 4 3 1
1 1
1
Output
0
1
2
0
Submitted Solution:
```
for q in range(int(input())):
h,n=map(int,input().split())
a=[int(x) for x in input().split()]
c=0
i=0
while((i+1)<n):
if a[i+1]<2:
break
if (a[i]-a[i+1])>1:
c+=1
i+=1
if n>2 and 3 in a and 4 not in a:
c=c+1
print(c)
``` | instruction | 0 | 43,050 | 2 | 86,100 |
No | output | 1 | 43,050 | 2 | 86,101 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are playing a game where your character should overcome different obstacles. The current problem is to come down from a cliff. The cliff has height h, and there is a moving platform on each height x from 1 to h.
Each platform is either hidden inside the cliff or moved out. At first, there are n moved out platforms on heights p_1, p_2, ..., p_n. The platform on height h is moved out (and the character is initially standing there).
If you character is standing on some moved out platform on height x, then he can pull a special lever, which switches the state of two platforms: on height x and x - 1. In other words, the platform you are currently standing on will hide in the cliff and the platform one unit below will change it state: it will hide if it was moved out or move out if it was hidden. In the second case, you will safely land on it. Note that this is the only way to move from one platform to another.
Your character is quite fragile, so it can safely fall from the height no more than 2. In other words falling from the platform x to platform x - 2 is okay, but falling from x to x - 3 (or lower) is certain death.
Sometimes it's not possible to come down from the cliff, but you can always buy (for donate currency) several magic crystals. Each magic crystal can be used to change the state of any single platform (except platform on height h, which is unaffected by the crystals). After being used, the crystal disappears.
What is the minimum number of magic crystal you need to buy to safely land on the 0 ground level?
Input
The first line contains one integer q (1 ≤ q ≤ 100) — the number of queries. Each query contains two lines and is independent of all other queries.
The first line of each query contains two integers h and n (1 ≤ h ≤ 10^9, 1 ≤ n ≤ min(h, 2 ⋅ 10^5)) — the height of the cliff and the number of moved out platforms.
The second line contains n integers p_1, p_2, ..., p_n (h = p_1 > p_2 > ... > p_n ≥ 1) — the corresponding moved out platforms in the descending order of their heights.
The sum of n over all queries does not exceed 2 ⋅ 10^5.
Output
For each query print one integer — the minimum number of magic crystals you have to spend to safely come down on the ground level (with height 0).
Example
Input
4
3 2
3 1
8 6
8 7 6 5 3 2
9 6
9 8 5 4 3 1
1 1
1
Output
0
1
2
0
Submitted Solution:
```
import sys
import bisect as b
import math
from collections import defaultdict as dd
input=sys.stdin.readline
mo=10**9+7
def cin():
return map(int,sin().split())
def ain():
return list(map(int,sin().split()))
def sin():
return input()
def inin():
return int(input())
def computeGCD(x, y):
while(y):
x, y = y, x % y
return x
for i in range(inin()):
h,n=cin()
l=ain()
## d=dd(bool)
## for i in l:
## d[i]=True
x=h
c=0
if(n<=2):
print(0)
else:
i=0
l+=[0,0]
while(i<n):
## print(x,l[i],i)
if(x<=2 or l[i]<=2):
break
if(l[i+1]==x-1):
if(l[i+2]==x-2):
x=x-2
i+=2
else:
c+=1
x=l[i+2]+1
i+=1
else:
x=l[i+1]+1
print(c)
## if(x>2):
## if(x-1==l[i]):
## if(x-2==l[i+1]):
## x=x-2
## i+=1
## else:
## c+=1
## x=x-2
## i+=1
## while(x>2):
#### print(x,"k")
#### if(d[x]==1):
## if(d[x-1]==True):
## if(d[x-2]==True):
## x=x-2
## else:
## c+=1
## x=x-2
## d[x]=True
#### print('hello')
## else:
## x=x-1
## else:
## c+=1
## print(c)
``` | instruction | 0 | 43,051 | 2 | 86,102 |
No | output | 1 | 43,051 | 2 | 86,103 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are playing a game where your character should overcome different obstacles. The current problem is to come down from a cliff. The cliff has height h, and there is a moving platform on each height x from 1 to h.
Each platform is either hidden inside the cliff or moved out. At first, there are n moved out platforms on heights p_1, p_2, ..., p_n. The platform on height h is moved out (and the character is initially standing there).
If you character is standing on some moved out platform on height x, then he can pull a special lever, which switches the state of two platforms: on height x and x - 1. In other words, the platform you are currently standing on will hide in the cliff and the platform one unit below will change it state: it will hide if it was moved out or move out if it was hidden. In the second case, you will safely land on it. Note that this is the only way to move from one platform to another.
Your character is quite fragile, so it can safely fall from the height no more than 2. In other words falling from the platform x to platform x - 2 is okay, but falling from x to x - 3 (or lower) is certain death.
Sometimes it's not possible to come down from the cliff, but you can always buy (for donate currency) several magic crystals. Each magic crystal can be used to change the state of any single platform (except platform on height h, which is unaffected by the crystals). After being used, the crystal disappears.
What is the minimum number of magic crystal you need to buy to safely land on the 0 ground level?
Input
The first line contains one integer q (1 ≤ q ≤ 100) — the number of queries. Each query contains two lines and is independent of all other queries.
The first line of each query contains two integers h and n (1 ≤ h ≤ 10^9, 1 ≤ n ≤ min(h, 2 ⋅ 10^5)) — the height of the cliff and the number of moved out platforms.
The second line contains n integers p_1, p_2, ..., p_n (h = p_1 > p_2 > ... > p_n ≥ 1) — the corresponding moved out platforms in the descending order of their heights.
The sum of n over all queries does not exceed 2 ⋅ 10^5.
Output
For each query print one integer — the minimum number of magic crystals you have to spend to safely come down on the ground level (with height 0).
Example
Input
4
3 2
3 1
8 6
8 7 6 5 3 2
9 6
9 8 5 4 3 1
1 1
1
Output
0
1
2
0
Submitted Solution:
```
tc=int(input())
while tc>0:
h,n=[int(x) for x in input().split()]
a=[int(x) for x in input().split()]
a.sort(reverse=True)
i=0
s=0
arr=[]
while i<n:
e=a[i]
if i>0 and a[i]!=a[i-1]-1:
arr.append(False)
s+=1
arr.append(True)
i+=1
s+=1
h=s
dp=[10**10]*h
dp[0]=0
for i in range(h-1):
a=0
if not arr[i+1]:
a=0
else:
a=1
dp[i+1]=min(dp[i+1],a+dp[i])
if i+2<h and not arr[i+2]:
if not arr[i+1]:
a=2
else:
a=1
elif i+2<h:
if not arr[i+1]:
a=1
else:
a=0
if i+2<h:
dp[i+2]=min(dp[i+2],a+dp[i])
res=0
if h-2>=0:
a=0
if not arr[h-1]:
a=1
res=dp[h-2]+a
print(min(res,dp[h-1]))
tc-=1
``` | instruction | 0 | 43,052 | 2 | 86,104 |
No | output | 1 | 43,052 | 2 | 86,105 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Emuskald is a well-known illusionist. One of his trademark tricks involves a set of magical boxes. The essence of the trick is in packing the boxes inside other boxes.
From the top view each magical box looks like a square with side length equal to 2k (k is an integer, k ≥ 0) units. A magical box v can be put inside a magical box u, if side length of v is strictly less than the side length of u. In particular, Emuskald can put 4 boxes of side length 2k - 1 into one box of side length 2k, or as in the following figure:
<image>
Emuskald is about to go on tour performing around the world, and needs to pack his magical boxes for the trip. He has decided that the best way to pack them would be inside another magical box, but magical boxes are quite expensive to make. Help him find the smallest magical box that can fit all his boxes.
Input
The first line of input contains an integer n (1 ≤ n ≤ 105), the number of different sizes of boxes Emuskald has. Each of following n lines contains two integers ki and ai (0 ≤ ki ≤ 109, 1 ≤ ai ≤ 109), which means that Emuskald has ai boxes with side length 2ki. It is guaranteed that all of ki are distinct.
Output
Output a single integer p, such that the smallest magical box that can contain all of Emuskald’s boxes has side length 2p.
Examples
Input
2
0 3
1 5
Output
3
Input
1
0 4
Output
1
Input
2
1 10
2 2
Output
3
Note
Picture explanation. If we have 3 boxes with side length 2 and 5 boxes with side length 1, then we can put all these boxes inside a box with side length 4, for example, as shown in the picture.
In the second test case, we can put all four small boxes into a box with side length 2. | instruction | 0 | 43,285 | 2 | 86,570 |
Tags: greedy, math
Correct Solution:
```
from sys import stdin,stdout
nmbr = lambda: int(input())
lst = lambda: list(map(int, input().split()))
for _ in range(1):#nmbr()):
n=nmbr()
a=[lst() for _ in range(n)]
ans=max(v[0] for v in a)+1
for p,b in a:
cnt=0
power=1
while power<b:
power*=4
cnt+=1
ans=max(cnt+p,ans)
print(ans)
``` | output | 1 | 43,285 | 2 | 86,571 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Emuskald is a well-known illusionist. One of his trademark tricks involves a set of magical boxes. The essence of the trick is in packing the boxes inside other boxes.
From the top view each magical box looks like a square with side length equal to 2k (k is an integer, k ≥ 0) units. A magical box v can be put inside a magical box u, if side length of v is strictly less than the side length of u. In particular, Emuskald can put 4 boxes of side length 2k - 1 into one box of side length 2k, or as in the following figure:
<image>
Emuskald is about to go on tour performing around the world, and needs to pack his magical boxes for the trip. He has decided that the best way to pack them would be inside another magical box, but magical boxes are quite expensive to make. Help him find the smallest magical box that can fit all his boxes.
Input
The first line of input contains an integer n (1 ≤ n ≤ 105), the number of different sizes of boxes Emuskald has. Each of following n lines contains two integers ki and ai (0 ≤ ki ≤ 109, 1 ≤ ai ≤ 109), which means that Emuskald has ai boxes with side length 2ki. It is guaranteed that all of ki are distinct.
Output
Output a single integer p, such that the smallest magical box that can contain all of Emuskald’s boxes has side length 2p.
Examples
Input
2
0 3
1 5
Output
3
Input
1
0 4
Output
1
Input
2
1 10
2 2
Output
3
Note
Picture explanation. If we have 3 boxes with side length 2 and 5 boxes with side length 1, then we can put all these boxes inside a box with side length 4, for example, as shown in the picture.
In the second test case, we can put all four small boxes into a box with side length 2. | instruction | 0 | 43,288 | 2 | 86,576 |
Tags: greedy, math
Correct Solution:
```
from math import *
n = int(input())
v = 0
for i in range(n):
a = [int(x) for x in input().split(' ')]
k = ceil(log(a[1],4))
if k == 0: k=1
if k+a[0]>v:
v = k+a[0]
print(v)
``` | output | 1 | 43,288 | 2 | 86,577 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Emuskald is a well-known illusionist. One of his trademark tricks involves a set of magical boxes. The essence of the trick is in packing the boxes inside other boxes.
From the top view each magical box looks like a square with side length equal to 2k (k is an integer, k ≥ 0) units. A magical box v can be put inside a magical box u, if side length of v is strictly less than the side length of u. In particular, Emuskald can put 4 boxes of side length 2k - 1 into one box of side length 2k, or as in the following figure:
<image>
Emuskald is about to go on tour performing around the world, and needs to pack his magical boxes for the trip. He has decided that the best way to pack them would be inside another magical box, but magical boxes are quite expensive to make. Help him find the smallest magical box that can fit all his boxes.
Input
The first line of input contains an integer n (1 ≤ n ≤ 105), the number of different sizes of boxes Emuskald has. Each of following n lines contains two integers ki and ai (0 ≤ ki ≤ 109, 1 ≤ ai ≤ 109), which means that Emuskald has ai boxes with side length 2ki. It is guaranteed that all of ki are distinct.
Output
Output a single integer p, such that the smallest magical box that can contain all of Emuskald’s boxes has side length 2p.
Examples
Input
2
0 3
1 5
Output
3
Input
1
0 4
Output
1
Input
2
1 10
2 2
Output
3
Note
Picture explanation. If we have 3 boxes with side length 2 and 5 boxes with side length 1, then we can put all these boxes inside a box with side length 4, for example, as shown in the picture.
In the second test case, we can put all four small boxes into a box with side length 2. | instruction | 0 | 43,289 | 2 | 86,578 |
Tags: greedy, math
Correct Solution:
```
# Solution to CodeForces 269A Magic Boxes
from math import ceil, log
max_size = -1 # Largest square, answer
for _ in range(int(input().strip())): # Number of cases
size, count = map(int, input().strip().split())
container = ceil(log(count, 4))
max_size = max(max_size, size + max(container, 1))
print(max_size)
``` | output | 1 | 43,289 | 2 | 86,579 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Digital collectible card games have become very popular recently. So Vova decided to try one of these.
Vova has n cards in his collection. Each of these cards is characterised by its power pi, magic number ci and level li. Vova wants to build a deck with total power not less than k, but magic numbers may not allow him to do so — Vova can't place two cards in a deck if the sum of the magic numbers written on these cards is a prime number. Also Vova cannot use a card if its level is greater than the level of Vova's character.
At the moment Vova's character's level is 1. Help Vova to determine the minimum level he needs to reach in order to build a deck with the required total power.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 100000).
Then n lines follow, each of these lines contains three numbers that represent the corresponding card: pi, ci and li (1 ≤ pi ≤ 1000, 1 ≤ ci ≤ 100000, 1 ≤ li ≤ n).
Output
If Vova won't be able to build a deck with required power, print - 1. Otherwise print the minimum level Vova has to reach in order to build a deck.
Examples
Input
5 8
5 5 1
1 5 4
4 6 3
1 12 4
3 12 1
Output
4
Input
3 7
4 4 1
5 8 2
5 3 3
Output
2
Submitted Solution:
```
import sys
import math
def calculate():
str1 = sys.stdin.readline().rstrip().split()
n = int(str1[0])
k = int(str1[1])
maxlevel = 0
i = 0
arr = [] #power, magic number, and level
while (i < n): #read data
str2 = sys.stdin.readline().rstrip().split()
arr.append([int(str2[0]), int(str2[1]), int(str2[2])])
i = i + 1
#Find max level
for i in arr:
if (i[2] > maxlevel):
maxlevel = i[2]
i = 1
success = 0
while (i <= maxlevel):
possible = [] #possible cards to have in deck
#build possible list
for a in arr:
if (a[2] <= i):
conflict = 0
conflictarr = []
for b in possible:
#Magic number conflict
if (prime(b[1] + a[1])):
conflict = 1
conflictarr.append(b)
#There was a conflict
if (conflict == 1):
conflicttotalpower = 0
#Add up power of nodes a conflicts with
for q in conflictarr:
conflicttotalpower = conflicttotalpower + q[0]
#if a has more power than conflict nodes
if (a[0] > conflicttotalpower):
for q in conflictarr:
possible.remove(q)
possible.append(a)
#There wasnt a conflict
if (conflict == 0):
possible.append(a)
#Check to see if possible list matches power requirements
totalpower = 0
for b in possible:
totalpower = totalpower + b[0]
if (totalpower >= k):
success = 1
sys.stdout.write(str(i))
break;
i = i + 1
if (success == 0):
sys.stdout.write("-1")
def prime(num):
#Check if num is a prime number
i = 1
numFactors = 0
if ((num % 2) == 0):
#evens are not primes
return False
while (i < math.sqrt(num)):
if ((num % i) == 0):
numFactors = numFactors + 1;
i = i + 2
if (numFactors > 1):
#Not prime
return False
else:
return True
if __name__ == "__main__":
calculate()
``` | instruction | 0 | 43,489 | 2 | 86,978 |
No | output | 1 | 43,489 | 2 | 86,979 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Digital collectible card games have become very popular recently. So Vova decided to try one of these.
Vova has n cards in his collection. Each of these cards is characterised by its power pi, magic number ci and level li. Vova wants to build a deck with total power not less than k, but magic numbers may not allow him to do so — Vova can't place two cards in a deck if the sum of the magic numbers written on these cards is a prime number. Also Vova cannot use a card if its level is greater than the level of Vova's character.
At the moment Vova's character's level is 1. Help Vova to determine the minimum level he needs to reach in order to build a deck with the required total power.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 100000).
Then n lines follow, each of these lines contains three numbers that represent the corresponding card: pi, ci and li (1 ≤ pi ≤ 1000, 1 ≤ ci ≤ 100000, 1 ≤ li ≤ n).
Output
If Vova won't be able to build a deck with required power, print - 1. Otherwise print the minimum level Vova has to reach in order to build a deck.
Examples
Input
5 8
5 5 1
1 5 4
4 6 3
1 12 4
3 12 1
Output
4
Input
3 7
4 4 1
5 8 2
5 3 3
Output
2
Submitted Solution:
```
import sys
import math
k = 0
def calculate():
str1 = sys.stdin.readline().rstrip().split()
n = int(str1[0])
k = int(str1[1])
arr = [] #power, magic number, and level
possible = []
maxlevel = 0
i = 0
while (i < n): #read data
str2 = sys.stdin.readline().rstrip().split()
arr.append([int(str2[0]), int(str2[1]), int(str2[2])])
i = i + 1
#Find max level
for i in arr:
if (i[2] > maxlevel):
maxlevel = i[2]
sys.stdout.write(str(recursiveCalc(arr, possible, maxlevel)))
def recursiveCalc(arr, possible, maxlevel):
i = 0
found = 0
minlevel = float("inf")
while (i <= maxlevel):
for a in arr:
if (a[2] == i):
valid = 1
#checks for conflicts with possible
for b in possible:
if (prime(b[1] + a[1])):
valid = 0
break
#recursively calls method
if (valid == 1):
found = 1
possible1 = possible
possible1.append(a)
arr1 = arr
arr1.remove(a)
d = recursiveCalc(arr1, possible1, maxlevel)
if (d < minlevel):
minlevel = d
i = i + 1
if (found == 0):
t = 0
#1 complete possible list. Check to see if list meets the requirements
power = 0
for u in possible:
power = u[0] + power
if (power >= k):
#return max level in that list
for i in possible:
if (i[2] > t):
t = i[2]
return t
else:
return -1
else:
return minlevel
def prime(num):
#Check if num is a prime number
i = 1
numFactors = 0
if ((num % 2) == 0):
#evens are not primes
return False
while (i < math.sqrt(num)):
if ((num % i) == 0):
numFactors = numFactors + 1;
i = i + 2
if (numFactors > 1):
#Not prime
return False
else:
return True
if __name__ == "__main__":
calculate()
``` | instruction | 0 | 43,490 | 2 | 86,980 |
No | output | 1 | 43,490 | 2 | 86,981 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Digital collectible card games have become very popular recently. So Vova decided to try one of these.
Vova has n cards in his collection. Each of these cards is characterised by its power pi, magic number ci and level li. Vova wants to build a deck with total power not less than k, but magic numbers may not allow him to do so — Vova can't place two cards in a deck if the sum of the magic numbers written on these cards is a prime number. Also Vova cannot use a card if its level is greater than the level of Vova's character.
At the moment Vova's character's level is 1. Help Vova to determine the minimum level he needs to reach in order to build a deck with the required total power.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 100000).
Then n lines follow, each of these lines contains three numbers that represent the corresponding card: pi, ci and li (1 ≤ pi ≤ 1000, 1 ≤ ci ≤ 100000, 1 ≤ li ≤ n).
Output
If Vova won't be able to build a deck with required power, print - 1. Otherwise print the minimum level Vova has to reach in order to build a deck.
Examples
Input
5 8
5 5 1
1 5 4
4 6 3
1 12 4
3 12 1
Output
4
Input
3 7
4 4 1
5 8 2
5 3 3
Output
2
Submitted Solution:
```
import sys
import math
def calculate():
str1 = sys.stdin.readline().rstrip().split()
n = int(str1[0])
k = int(str1[1])
arr = [] #power, magic number, and level
possible = []
maxlevel = 0
i = 0
while (i < n): #read data
str2 = sys.stdin.readline().rstrip().split()
arr.append([int(str2[0]), int(str2[1]), int(str2[2])])
i = i + 1
#Find max level
for i in arr:
if (i[2] > maxlevel):
maxlevel = i[2]
sys.stdout.write(str(recursiveCalc(arr, possible, maxlevel)))
def recursiveCalc(arr, possible, maxlevel):
i = 0
found = 0
minlevel = float("inf")
while (i <= maxlevel):
for a in arr:
if (a[2] == i):
valid = 1
#checks for conflicts with possible
for b in possible:
if (prime(b[1] + a[1])):
valid = 0
break
#recursively calls method
if (valid == 1):
found = 1
possible1 = possible
possible1.append(a)
arr1 = arr
arr1.remove(a)
d = recursiveCalc(arr1, possible1, maxlevel)
if (d < minlevel):
minlevel = d
i = i + 1
if (found == 0):
k = 0
#1 complete possible list
#return max level in that list
for i in possible:
if (i[2] > k):
k = i[2]
return k
else:
return minlevel
def prime(num):
#Check if num is a prime number
i = 1
numFactors = 0
if ((num % 2) == 0):
#evens are not primes
return False
while (i < math.sqrt(num)):
if ((num % i) == 0):
numFactors = numFactors + 1;
i = i + 2
if (numFactors > 1):
#Not prime
return False
else:
return True
if __name__ == "__main__":
calculate()
``` | instruction | 0 | 43,491 | 2 | 86,982 |
No | output | 1 | 43,491 | 2 | 86,983 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Digital collectible card games have become very popular recently. So Vova decided to try one of these.
Vova has n cards in his collection. Each of these cards is characterised by its power pi, magic number ci and level li. Vova wants to build a deck with total power not less than k, but magic numbers may not allow him to do so — Vova can't place two cards in a deck if the sum of the magic numbers written on these cards is a prime number. Also Vova cannot use a card if its level is greater than the level of Vova's character.
At the moment Vova's character's level is 1. Help Vova to determine the minimum level he needs to reach in order to build a deck with the required total power.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 100000).
Then n lines follow, each of these lines contains three numbers that represent the corresponding card: pi, ci and li (1 ≤ pi ≤ 1000, 1 ≤ ci ≤ 100000, 1 ≤ li ≤ n).
Output
If Vova won't be able to build a deck with required power, print - 1. Otherwise print the minimum level Vova has to reach in order to build a deck.
Examples
Input
5 8
5 5 1
1 5 4
4 6 3
1 12 4
3 12 1
Output
4
Input
3 7
4 4 1
5 8 2
5 3 3
Output
2
Submitted Solution:
```
import sys
import math
def calculate():
str1 = sys.stdin.readline().rstrip().split()
n = int(str1[0])
k = int(str1[1])
maxlevel = 0
i = 0
arr = [] #power, magic number, and level
while (i < n): #read data
str2 = sys.stdin.readline().rstrip().split()
arr.append([int(str2[0]), int(str2[1]), int(str2[2])])
i = i + 1
#Find max level
for i in arr:
if (i[2] > maxlevel):
maxlevel = i[2]
i = 1
success = 0
while (i <= maxlevel):
possible = [] #possible cards to have in deck
#build possible list
for a in arr:
if (a[2] <= i):
conflict = 0
for b in possible:
#Magic number conflict
if (prime(b[1] + a[1])):
conflict = 1
if (a[0] > b[0]):
possible.remove(b)
possible.append(a)
#There wasnt a conflict
if (conflict == 0):
possible.append(a)
#Check to see if possible list matches power requirements
totalpower = 0
for b in possible:
totalpower = totalpower + b[0]
if (totalpower >= k):
print(totalpower)
success = 1
sys.stdout.write(str(i))
break;
i = i + 1
if (success == 0):
sys.stdout.write("-1")
def prime(num):
#Check if num is a prime number
i = 1
numFactors = 0
if ((num % 2) == 0):
#evens are not primes
return False
while (i < math.sqrt(num)):
if ((num % i) == 0):
numFactors = numFactors + 1;
i = i + 2
if (numFactors > 1):
#Not prime
return False
else:
return True
if __name__ == "__main__":
calculate()
``` | instruction | 0 | 43,492 | 2 | 86,984 |
No | output | 1 | 43,492 | 2 | 86,985 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You and your friend are playing the game Mortal Kombat XI. You are trying to pass a challenge tower. There are n bosses in this tower, numbered from 1 to n. The type of the i-th boss is a_i. If the i-th boss is easy then its type is a_i = 0, otherwise this boss is hard and its type is a_i = 1.
During one session, either you or your friend can kill one or two bosses (neither you nor your friend can skip the session, so the minimum number of bosses killed during one session is at least one). After your friend session, your session begins, then again your friend session begins, your session begins, and so on. The first session is your friend's session.
Your friend needs to get good because he can't actually kill hard bosses. To kill them, he uses skip points. One skip point can be used to kill one hard boss.
Your task is to find the minimum number of skip points your friend needs to use so you and your friend kill all n bosses in the given order.
For example: suppose n = 8, a = [1, 0, 1, 1, 0, 1, 1, 1]. Then the best course of action is the following:
* your friend kills two first bosses, using one skip point for the first boss;
* you kill the third and the fourth bosses;
* your friend kills the fifth boss;
* you kill the sixth and the seventh bosses;
* your friend kills the last boss, using one skip point, so the tower is completed using two skip points.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of bosses. The second line of the test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1), where a_i is the type of the i-th boss.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer: the minimum number of skip points your friend needs to use so you and your friend kill all n bosses in the given order.
Example
Input
6
8
1 0 1 1 0 1 1 1
5
1 1 1 1 0
7
1 1 1 1 0 0 1
6
1 1 1 1 1 1
1
1
1
0
Output
2
2
2
2
1
0 | instruction | 0 | 44,788 | 2 | 89,576 |
Tags: dp, graphs, greedy, shortest paths
Correct Solution:
```
"""
NTC here
"""
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
def main():
T = iin()
while T:
T-=1
n = iin()
a = lin()
if n==1:
print(a[0])
else:
dp = [[0, 0] for i in range(n)]
dp[-1][1] = a[-1]
dp[-2][1] = a[-2]
for i in range(n-3, -1, -1):
dp[i][0] = min(dp[i+1][1], dp[i+2][1])
dp[i][1] = a[i]+min(dp[i+1][0], a[i+1]+dp[i+2][0])
# print(i, min(dp[i+1][1], dp[i+1][0]+dp[i+2][1]), min(dp[i+1][0], dp[i+1][1]+dp[i+2][0])+a[i])
print(dp[0][1])
# region fastio
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")
def iin(): return int(input())
def lin(): return list(map(int, input().split()))
# endregion
if __name__ == "__main__":
main()
``` | output | 1 | 44,788 | 2 | 89,577 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You and your friend are playing the game Mortal Kombat XI. You are trying to pass a challenge tower. There are n bosses in this tower, numbered from 1 to n. The type of the i-th boss is a_i. If the i-th boss is easy then its type is a_i = 0, otherwise this boss is hard and its type is a_i = 1.
During one session, either you or your friend can kill one or two bosses (neither you nor your friend can skip the session, so the minimum number of bosses killed during one session is at least one). After your friend session, your session begins, then again your friend session begins, your session begins, and so on. The first session is your friend's session.
Your friend needs to get good because he can't actually kill hard bosses. To kill them, he uses skip points. One skip point can be used to kill one hard boss.
Your task is to find the minimum number of skip points your friend needs to use so you and your friend kill all n bosses in the given order.
For example: suppose n = 8, a = [1, 0, 1, 1, 0, 1, 1, 1]. Then the best course of action is the following:
* your friend kills two first bosses, using one skip point for the first boss;
* you kill the third and the fourth bosses;
* your friend kills the fifth boss;
* you kill the sixth and the seventh bosses;
* your friend kills the last boss, using one skip point, so the tower is completed using two skip points.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of bosses. The second line of the test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1), where a_i is the type of the i-th boss.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer: the minimum number of skip points your friend needs to use so you and your friend kill all n bosses in the given order.
Example
Input
6
8
1 0 1 1 0 1 1 1
5
1 1 1 1 0
7
1 1 1 1 0 0 1
6
1 1 1 1 1 1
1
1
1
0
Output
2
2
2
2
1
0 | instruction | 0 | 44,789 | 2 | 89,578 |
Tags: dp, graphs, greedy, shortest paths
Correct Solution:
```
import sys
sys.setrecursionlimit(10**5)
int1 = lambda x: int(x)-1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
inf=10**16
for _ in range(II()):
n=II()
aa=LI()
def chmin(i,j,val):
if j>n:return
if val<dp[i][j]:dp[i][j]=val
# dp[i][j]...turn i,boss j
dp=[[inf]*(n+1) for _ in range(2)]
dp[0][0]=0
for j in range(n):
for i in range(2):
pre=dp[i][j]
if pre==inf:continue
if i==0:
chmin(1,j+1,pre+aa[j])
if j+1<n:chmin(1,j+2,pre+aa[j]+aa[j+1])
else:
chmin(0,j+1,pre)
chmin(0,j+2,pre)
# print(dp)
ans=min(dp[0][n],dp[1][n])
print(ans)
``` | output | 1 | 44,789 | 2 | 89,579 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You and your friend are playing the game Mortal Kombat XI. You are trying to pass a challenge tower. There are n bosses in this tower, numbered from 1 to n. The type of the i-th boss is a_i. If the i-th boss is easy then its type is a_i = 0, otherwise this boss is hard and its type is a_i = 1.
During one session, either you or your friend can kill one or two bosses (neither you nor your friend can skip the session, so the minimum number of bosses killed during one session is at least one). After your friend session, your session begins, then again your friend session begins, your session begins, and so on. The first session is your friend's session.
Your friend needs to get good because he can't actually kill hard bosses. To kill them, he uses skip points. One skip point can be used to kill one hard boss.
Your task is to find the minimum number of skip points your friend needs to use so you and your friend kill all n bosses in the given order.
For example: suppose n = 8, a = [1, 0, 1, 1, 0, 1, 1, 1]. Then the best course of action is the following:
* your friend kills two first bosses, using one skip point for the first boss;
* you kill the third and the fourth bosses;
* your friend kills the fifth boss;
* you kill the sixth and the seventh bosses;
* your friend kills the last boss, using one skip point, so the tower is completed using two skip points.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of bosses. The second line of the test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1), where a_i is the type of the i-th boss.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer: the minimum number of skip points your friend needs to use so you and your friend kill all n bosses in the given order.
Example
Input
6
8
1 0 1 1 0 1 1 1
5
1 1 1 1 0
7
1 1 1 1 0 0 1
6
1 1 1 1 1 1
1
1
1
0
Output
2
2
2
2
1
0 | instruction | 0 | 44,790 | 2 | 89,580 |
Tags: dp, graphs, greedy, shortest paths
Correct Solution:
```
for tc in range(int(input())):
n = int(input())
arr = '1 1 ' + input()
skips = arr.count('1 1 1')
print(skips)
``` | output | 1 | 44,790 | 2 | 89,581 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You and your friend are playing the game Mortal Kombat XI. You are trying to pass a challenge tower. There are n bosses in this tower, numbered from 1 to n. The type of the i-th boss is a_i. If the i-th boss is easy then its type is a_i = 0, otherwise this boss is hard and its type is a_i = 1.
During one session, either you or your friend can kill one or two bosses (neither you nor your friend can skip the session, so the minimum number of bosses killed during one session is at least one). After your friend session, your session begins, then again your friend session begins, your session begins, and so on. The first session is your friend's session.
Your friend needs to get good because he can't actually kill hard bosses. To kill them, he uses skip points. One skip point can be used to kill one hard boss.
Your task is to find the minimum number of skip points your friend needs to use so you and your friend kill all n bosses in the given order.
For example: suppose n = 8, a = [1, 0, 1, 1, 0, 1, 1, 1]. Then the best course of action is the following:
* your friend kills two first bosses, using one skip point for the first boss;
* you kill the third and the fourth bosses;
* your friend kills the fifth boss;
* you kill the sixth and the seventh bosses;
* your friend kills the last boss, using one skip point, so the tower is completed using two skip points.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of bosses. The second line of the test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1), where a_i is the type of the i-th boss.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer: the minimum number of skip points your friend needs to use so you and your friend kill all n bosses in the given order.
Example
Input
6
8
1 0 1 1 0 1 1 1
5
1 1 1 1 0
7
1 1 1 1 0 0 1
6
1 1 1 1 1 1
1
1
1
0
Output
2
2
2
2
1
0 | instruction | 0 | 44,791 | 2 | 89,582 |
Tags: dp, graphs, greedy, shortest paths
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
arr=list(map(int,input().split()))
i=1
ans=arr[0]
while i<n-2:
if arr[i]+arr[i+1]+arr[i+2]==3:
ans+=1
i+=3
continue
i+=1
print(ans)
``` | output | 1 | 44,791 | 2 | 89,583 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You and your friend are playing the game Mortal Kombat XI. You are trying to pass a challenge tower. There are n bosses in this tower, numbered from 1 to n. The type of the i-th boss is a_i. If the i-th boss is easy then its type is a_i = 0, otherwise this boss is hard and its type is a_i = 1.
During one session, either you or your friend can kill one or two bosses (neither you nor your friend can skip the session, so the minimum number of bosses killed during one session is at least one). After your friend session, your session begins, then again your friend session begins, your session begins, and so on. The first session is your friend's session.
Your friend needs to get good because he can't actually kill hard bosses. To kill them, he uses skip points. One skip point can be used to kill one hard boss.
Your task is to find the minimum number of skip points your friend needs to use so you and your friend kill all n bosses in the given order.
For example: suppose n = 8, a = [1, 0, 1, 1, 0, 1, 1, 1]. Then the best course of action is the following:
* your friend kills two first bosses, using one skip point for the first boss;
* you kill the third and the fourth bosses;
* your friend kills the fifth boss;
* you kill the sixth and the seventh bosses;
* your friend kills the last boss, using one skip point, so the tower is completed using two skip points.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of bosses. The second line of the test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1), where a_i is the type of the i-th boss.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer: the minimum number of skip points your friend needs to use so you and your friend kill all n bosses in the given order.
Example
Input
6
8
1 0 1 1 0 1 1 1
5
1 1 1 1 0
7
1 1 1 1 0 0 1
6
1 1 1 1 1 1
1
1
1
0
Output
2
2
2
2
1
0 | instruction | 0 | 44,792 | 2 | 89,584 |
Tags: dp, graphs, greedy, shortest paths
Correct Solution:
```
for test in range(int(input())):
pocet = int(input())
a = list(map(int, input().split(" ")))
k = 0
if a[0]==1:
k += 1
a.remove(a[0])
ind1_kon = 0
este = len(a)>0
while este:
try:
ind1_zac = a.index(1, ind1_kon)
except:
este = False
if este:
try:
ind1_kon = a.index(0, ind1_zac)
except:
ind1_kon = len(a)
este = False
skip = (ind1_kon - ind1_zac) // 3
k += skip
#print('ind1_zac je '+ str(ind1_zac))
#print('ind1_kon je '+ str(ind1_kon))
print(k)
``` | output | 1 | 44,792 | 2 | 89,585 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You and your friend are playing the game Mortal Kombat XI. You are trying to pass a challenge tower. There are n bosses in this tower, numbered from 1 to n. The type of the i-th boss is a_i. If the i-th boss is easy then its type is a_i = 0, otherwise this boss is hard and its type is a_i = 1.
During one session, either you or your friend can kill one or two bosses (neither you nor your friend can skip the session, so the minimum number of bosses killed during one session is at least one). After your friend session, your session begins, then again your friend session begins, your session begins, and so on. The first session is your friend's session.
Your friend needs to get good because he can't actually kill hard bosses. To kill them, he uses skip points. One skip point can be used to kill one hard boss.
Your task is to find the minimum number of skip points your friend needs to use so you and your friend kill all n bosses in the given order.
For example: suppose n = 8, a = [1, 0, 1, 1, 0, 1, 1, 1]. Then the best course of action is the following:
* your friend kills two first bosses, using one skip point for the first boss;
* you kill the third and the fourth bosses;
* your friend kills the fifth boss;
* you kill the sixth and the seventh bosses;
* your friend kills the last boss, using one skip point, so the tower is completed using two skip points.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of bosses. The second line of the test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1), where a_i is the type of the i-th boss.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer: the minimum number of skip points your friend needs to use so you and your friend kill all n bosses in the given order.
Example
Input
6
8
1 0 1 1 0 1 1 1
5
1 1 1 1 0
7
1 1 1 1 0 0 1
6
1 1 1 1 1 1
1
1
1
0
Output
2
2
2
2
1
0 | instruction | 0 | 44,793 | 2 | 89,586 |
Tags: dp, graphs, greedy, shortest paths
Correct Solution:
```
import math
for _ in range(int(input())):
n=int(input())
l=list(map(int,input().split()))
ans=0
temp=0
if(l[0]==1):
ans+=1
for i in range(1,n):
if(l[i]==1):
temp+=1
else:
ans+=temp//3
temp=0
ans+=temp//3
print(ans)
``` | output | 1 | 44,793 | 2 | 89,587 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You and your friend are playing the game Mortal Kombat XI. You are trying to pass a challenge tower. There are n bosses in this tower, numbered from 1 to n. The type of the i-th boss is a_i. If the i-th boss is easy then its type is a_i = 0, otherwise this boss is hard and its type is a_i = 1.
During one session, either you or your friend can kill one or two bosses (neither you nor your friend can skip the session, so the minimum number of bosses killed during one session is at least one). After your friend session, your session begins, then again your friend session begins, your session begins, and so on. The first session is your friend's session.
Your friend needs to get good because he can't actually kill hard bosses. To kill them, he uses skip points. One skip point can be used to kill one hard boss.
Your task is to find the minimum number of skip points your friend needs to use so you and your friend kill all n bosses in the given order.
For example: suppose n = 8, a = [1, 0, 1, 1, 0, 1, 1, 1]. Then the best course of action is the following:
* your friend kills two first bosses, using one skip point for the first boss;
* you kill the third and the fourth bosses;
* your friend kills the fifth boss;
* you kill the sixth and the seventh bosses;
* your friend kills the last boss, using one skip point, so the tower is completed using two skip points.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of bosses. The second line of the test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1), where a_i is the type of the i-th boss.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer: the minimum number of skip points your friend needs to use so you and your friend kill all n bosses in the given order.
Example
Input
6
8
1 0 1 1 0 1 1 1
5
1 1 1 1 0
7
1 1 1 1 0 0 1
6
1 1 1 1 1 1
1
1
1
0
Output
2
2
2
2
1
0 | instruction | 0 | 44,794 | 2 | 89,588 |
Tags: dp, graphs, greedy, shortest paths
Correct Solution:
```
T = int(input())
ret = []
for t in range(T):
n = int(input())
a = [0]+list(map(int,input().split()))
dpM = [10**7]*(n+1)
dpF = [10**7]*(n+1)
#そこを倒すのに必要な最小のスキップポイント(1-index)
dpM[0]=0
for i in range(1,n+1):
dpF[i]=min(dpF[i],dpM[i-1]+a[i])
dpF[i]=min(dpF[i],dpM[i-2]+a[i]+a[i-1])
dpM[i]=min(dpM[i],dpF[i-1])
dpM[i]=min(dpM[i],dpF[i-2])
ret.append(min(dpF[n],dpM[n]))
for t in range(T):
print(ret[t])
``` | output | 1 | 44,794 | 2 | 89,589 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You and your friend are playing the game Mortal Kombat XI. You are trying to pass a challenge tower. There are n bosses in this tower, numbered from 1 to n. The type of the i-th boss is a_i. If the i-th boss is easy then its type is a_i = 0, otherwise this boss is hard and its type is a_i = 1.
During one session, either you or your friend can kill one or two bosses (neither you nor your friend can skip the session, so the minimum number of bosses killed during one session is at least one). After your friend session, your session begins, then again your friend session begins, your session begins, and so on. The first session is your friend's session.
Your friend needs to get good because he can't actually kill hard bosses. To kill them, he uses skip points. One skip point can be used to kill one hard boss.
Your task is to find the minimum number of skip points your friend needs to use so you and your friend kill all n bosses in the given order.
For example: suppose n = 8, a = [1, 0, 1, 1, 0, 1, 1, 1]. Then the best course of action is the following:
* your friend kills two first bosses, using one skip point for the first boss;
* you kill the third and the fourth bosses;
* your friend kills the fifth boss;
* you kill the sixth and the seventh bosses;
* your friend kills the last boss, using one skip point, so the tower is completed using two skip points.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of bosses. The second line of the test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1), where a_i is the type of the i-th boss.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer: the minimum number of skip points your friend needs to use so you and your friend kill all n bosses in the given order.
Example
Input
6
8
1 0 1 1 0 1 1 1
5
1 1 1 1 0
7
1 1 1 1 0 0 1
6
1 1 1 1 1 1
1
1
1
0
Output
2
2
2
2
1
0 | instruction | 0 | 44,795 | 2 | 89,590 |
Tags: dp, graphs, greedy, shortest paths
Correct Solution:
```
def run():
n = int(input())
nums = "".join(input().split())
ans = 0
ans += int(nums[0])
ans += nums[1:].count('111')
print(ans)
T = 1
if T:
for i in range(int(input())):
run()
else:
run()
#@Time: 2020/09/15 08:14:33
``` | output | 1 | 44,795 | 2 | 89,591 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You and your friend are playing the game Mortal Kombat XI. You are trying to pass a challenge tower. There are n bosses in this tower, numbered from 1 to n. The type of the i-th boss is a_i. If the i-th boss is easy then its type is a_i = 0, otherwise this boss is hard and its type is a_i = 1.
During one session, either you or your friend can kill one or two bosses (neither you nor your friend can skip the session, so the minimum number of bosses killed during one session is at least one). After your friend session, your session begins, then again your friend session begins, your session begins, and so on. The first session is your friend's session.
Your friend needs to get good because he can't actually kill hard bosses. To kill them, he uses skip points. One skip point can be used to kill one hard boss.
Your task is to find the minimum number of skip points your friend needs to use so you and your friend kill all n bosses in the given order.
For example: suppose n = 8, a = [1, 0, 1, 1, 0, 1, 1, 1]. Then the best course of action is the following:
* your friend kills two first bosses, using one skip point for the first boss;
* you kill the third and the fourth bosses;
* your friend kills the fifth boss;
* you kill the sixth and the seventh bosses;
* your friend kills the last boss, using one skip point, so the tower is completed using two skip points.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of bosses. The second line of the test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1), where a_i is the type of the i-th boss.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer: the minimum number of skip points your friend needs to use so you and your friend kill all n bosses in the given order.
Example
Input
6
8
1 0 1 1 0 1 1 1
5
1 1 1 1 0
7
1 1 1 1 0 0 1
6
1 1 1 1 1 1
1
1
1
0
Output
2
2
2
2
1
0
Submitted Solution:
```
T = int(input())
while T > 0:
n = int(input())
arr = list(map(int, input().split()))
dp = [[-1 for i in range(2)] for j in range(n)]
if len(arr) == 1:
if arr[0] == 1:
print(1)
else:
print(0)
else:
dp = [[n for i in range(2)] for j in range(n)]
dp[n-1][0] = 0
dp[n-2][0] = 0
if arr[n-1] == 1:
dp[n-1][1] = 1
else:
dp[n-1][1] = 0
if arr[n-2] == 1:
dp[n-2][1] = 1
else:
dp[n-2][1] = 0
for index in range(n-3,-1,-1):
dp[index][0] = min(dp[index][0],dp[index+1][1])
dp[index][0] = min(dp[index][0],dp[index+2][1])
temp = 0
if arr[index] == 1:
temp += 1
dp[index][1] = min(dp[index][1], temp+dp[index+1][0])
if arr[index+1] == 1:
temp += 1
dp[index][1] = min(dp[index][1], temp+dp[index+2][0])
print(dp[0][1])
T -= 1
``` | instruction | 0 | 44,796 | 2 | 89,592 |
Yes | output | 1 | 44,796 | 2 | 89,593 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You and your friend are playing the game Mortal Kombat XI. You are trying to pass a challenge tower. There are n bosses in this tower, numbered from 1 to n. The type of the i-th boss is a_i. If the i-th boss is easy then its type is a_i = 0, otherwise this boss is hard and its type is a_i = 1.
During one session, either you or your friend can kill one or two bosses (neither you nor your friend can skip the session, so the minimum number of bosses killed during one session is at least one). After your friend session, your session begins, then again your friend session begins, your session begins, and so on. The first session is your friend's session.
Your friend needs to get good because he can't actually kill hard bosses. To kill them, he uses skip points. One skip point can be used to kill one hard boss.
Your task is to find the minimum number of skip points your friend needs to use so you and your friend kill all n bosses in the given order.
For example: suppose n = 8, a = [1, 0, 1, 1, 0, 1, 1, 1]. Then the best course of action is the following:
* your friend kills two first bosses, using one skip point for the first boss;
* you kill the third and the fourth bosses;
* your friend kills the fifth boss;
* you kill the sixth and the seventh bosses;
* your friend kills the last boss, using one skip point, so the tower is completed using two skip points.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of bosses. The second line of the test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1), where a_i is the type of the i-th boss.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer: the minimum number of skip points your friend needs to use so you and your friend kill all n bosses in the given order.
Example
Input
6
8
1 0 1 1 0 1 1 1
5
1 1 1 1 0
7
1 1 1 1 0 0 1
6
1 1 1 1 1 1
1
1
1
0
Output
2
2
2
2
1
0
Submitted Solution:
```
t=int(input())
for i in range(t):
n=int(input())
a=list(map(int,input().split()))
c=0
p=0
if a[0]==1:
c=c+1
for j in range(1,n):
if a[j]==1:
p=p+1
else:
c=c+p//3
p=0
c=c+p//3
print(c)
``` | instruction | 0 | 44,797 | 2 | 89,594 |
Yes | output | 1 | 44,797 | 2 | 89,595 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You and your friend are playing the game Mortal Kombat XI. You are trying to pass a challenge tower. There are n bosses in this tower, numbered from 1 to n. The type of the i-th boss is a_i. If the i-th boss is easy then its type is a_i = 0, otherwise this boss is hard and its type is a_i = 1.
During one session, either you or your friend can kill one or two bosses (neither you nor your friend can skip the session, so the minimum number of bosses killed during one session is at least one). After your friend session, your session begins, then again your friend session begins, your session begins, and so on. The first session is your friend's session.
Your friend needs to get good because he can't actually kill hard bosses. To kill them, he uses skip points. One skip point can be used to kill one hard boss.
Your task is to find the minimum number of skip points your friend needs to use so you and your friend kill all n bosses in the given order.
For example: suppose n = 8, a = [1, 0, 1, 1, 0, 1, 1, 1]. Then the best course of action is the following:
* your friend kills two first bosses, using one skip point for the first boss;
* you kill the third and the fourth bosses;
* your friend kills the fifth boss;
* you kill the sixth and the seventh bosses;
* your friend kills the last boss, using one skip point, so the tower is completed using two skip points.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of bosses. The second line of the test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1), where a_i is the type of the i-th boss.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer: the minimum number of skip points your friend needs to use so you and your friend kill all n bosses in the given order.
Example
Input
6
8
1 0 1 1 0 1 1 1
5
1 1 1 1 0
7
1 1 1 1 0 0 1
6
1 1 1 1 1 1
1
1
1
0
Output
2
2
2
2
1
0
Submitted Solution:
```
for i in range(int(input())):
n = int(input())
A = list(map(int,input().split()))
Data = [[0,0,0,0] for i in range(n+1)]
Data[0] = [100000000, 100000000,100000000,0]
for i in range(1,n+1):
Data[i][0] = Data[i-1][1]+A[i-1]
Data[i][1] = min(Data[i-1][2],Data[i-1][3])+A[i-1]
Data[i][2] = min(Data[i-1][0],Data[i-1][1])
Data[i][3] = Data[i-1][2]
print(min(Data[n]))
``` | instruction | 0 | 44,798 | 2 | 89,596 |
Yes | output | 1 | 44,798 | 2 | 89,597 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You and your friend are playing the game Mortal Kombat XI. You are trying to pass a challenge tower. There are n bosses in this tower, numbered from 1 to n. The type of the i-th boss is a_i. If the i-th boss is easy then its type is a_i = 0, otherwise this boss is hard and its type is a_i = 1.
During one session, either you or your friend can kill one or two bosses (neither you nor your friend can skip the session, so the minimum number of bosses killed during one session is at least one). After your friend session, your session begins, then again your friend session begins, your session begins, and so on. The first session is your friend's session.
Your friend needs to get good because he can't actually kill hard bosses. To kill them, he uses skip points. One skip point can be used to kill one hard boss.
Your task is to find the minimum number of skip points your friend needs to use so you and your friend kill all n bosses in the given order.
For example: suppose n = 8, a = [1, 0, 1, 1, 0, 1, 1, 1]. Then the best course of action is the following:
* your friend kills two first bosses, using one skip point for the first boss;
* you kill the third and the fourth bosses;
* your friend kills the fifth boss;
* you kill the sixth and the seventh bosses;
* your friend kills the last boss, using one skip point, so the tower is completed using two skip points.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of bosses. The second line of the test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1), where a_i is the type of the i-th boss.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer: the minimum number of skip points your friend needs to use so you and your friend kill all n bosses in the given order.
Example
Input
6
8
1 0 1 1 0 1 1 1
5
1 1 1 1 0
7
1 1 1 1 0 0 1
6
1 1 1 1 1 1
1
1
1
0
Output
2
2
2
2
1
0
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
_str = str
BUFSIZE = 8192
def str(x=b''):
return x if type(x) is bytes else _str(x).encode()
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)
def inp():
return sys.stdin.readline()
def mpint():
return map(int, sys.stdin.readline().split(' '))
def itg():
return int(sys.stdin.readline())
# ############################## import
# ############################## main
INF = int(1e6)
for __ in range(itg()):
n = itg()
if n == 1:
print(inp())
continue
arr = tuple(mpint())
# current boss is arr[i]
dp1 = [INF] * n # current session is friend
dp2 = [INF] * n # current session is you
dp1[-1] = arr[-1]
dp1[-2] = arr[-2]
dp2[-2] = dp2[-1] = 0
for i in range(n-3, -1, -1):
# kill one or two
dp1[i] = min(arr[i] + dp2[i + 1], arr[i]+arr[i+1] + dp2[i+2])
dp2[i] = min(dp1[i + 1], dp1[i + 2])
print(dp1[0])
# Please check!
``` | instruction | 0 | 44,799 | 2 | 89,598 |
Yes | output | 1 | 44,799 | 2 | 89,599 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You and your friend are playing the game Mortal Kombat XI. You are trying to pass a challenge tower. There are n bosses in this tower, numbered from 1 to n. The type of the i-th boss is a_i. If the i-th boss is easy then its type is a_i = 0, otherwise this boss is hard and its type is a_i = 1.
During one session, either you or your friend can kill one or two bosses (neither you nor your friend can skip the session, so the minimum number of bosses killed during one session is at least one). After your friend session, your session begins, then again your friend session begins, your session begins, and so on. The first session is your friend's session.
Your friend needs to get good because he can't actually kill hard bosses. To kill them, he uses skip points. One skip point can be used to kill one hard boss.
Your task is to find the minimum number of skip points your friend needs to use so you and your friend kill all n bosses in the given order.
For example: suppose n = 8, a = [1, 0, 1, 1, 0, 1, 1, 1]. Then the best course of action is the following:
* your friend kills two first bosses, using one skip point for the first boss;
* you kill the third and the fourth bosses;
* your friend kills the fifth boss;
* you kill the sixth and the seventh bosses;
* your friend kills the last boss, using one skip point, so the tower is completed using two skip points.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of bosses. The second line of the test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1), where a_i is the type of the i-th boss.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer: the minimum number of skip points your friend needs to use so you and your friend kill all n bosses in the given order.
Example
Input
6
8
1 0 1 1 0 1 1 1
5
1 1 1 1 0
7
1 1 1 1 0 0 1
6
1 1 1 1 1 1
1
1
1
0
Output
2
2
2
2
1
0
Submitted Solution:
```
from sys import stdin,stdout
import math
from bisect import bisect_right
from collections import Counter,deque
L=lambda:list(map(int, stdin.readline().strip().split()))
M=lambda:map(int, stdin.readline().strip().split())
I=lambda:int(stdin.readline().strip())
S=lambda:stdin.readline().strip()
C=lambda:stdin.readline().strip().split()
mod=1000000007
def pr(a):return("".join(list(map(str,a))))
#______________________-------------------------------_____________________#
def solve():
n = I()
a = L()
b = pr(a).split('0')
c=[]
for i in b:
if i:
c+=[len(i)]
ans = 0
if a[0]==1:
ans += max(1,(c[0]+2)//3)
for i in range(1,len(c)):
ans += c[i]//3
print(ans)
for _ in range(I()):
solve()
``` | instruction | 0 | 44,800 | 2 | 89,600 |
No | output | 1 | 44,800 | 2 | 89,601 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You and your friend are playing the game Mortal Kombat XI. You are trying to pass a challenge tower. There are n bosses in this tower, numbered from 1 to n. The type of the i-th boss is a_i. If the i-th boss is easy then its type is a_i = 0, otherwise this boss is hard and its type is a_i = 1.
During one session, either you or your friend can kill one or two bosses (neither you nor your friend can skip the session, so the minimum number of bosses killed during one session is at least one). After your friend session, your session begins, then again your friend session begins, your session begins, and so on. The first session is your friend's session.
Your friend needs to get good because he can't actually kill hard bosses. To kill them, he uses skip points. One skip point can be used to kill one hard boss.
Your task is to find the minimum number of skip points your friend needs to use so you and your friend kill all n bosses in the given order.
For example: suppose n = 8, a = [1, 0, 1, 1, 0, 1, 1, 1]. Then the best course of action is the following:
* your friend kills two first bosses, using one skip point for the first boss;
* you kill the third and the fourth bosses;
* your friend kills the fifth boss;
* you kill the sixth and the seventh bosses;
* your friend kills the last boss, using one skip point, so the tower is completed using two skip points.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of bosses. The second line of the test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1), where a_i is the type of the i-th boss.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer: the minimum number of skip points your friend needs to use so you and your friend kill all n bosses in the given order.
Example
Input
6
8
1 0 1 1 0 1 1 1
5
1 1 1 1 0
7
1 1 1 1 0 0 1
6
1 1 1 1 1 1
1
1
1
0
Output
2
2
2
2
1
0
Submitted Solution:
```
for o in range(int(input())):
n=int(input())
s=list(map(int,input().split()))
i=0;p=s[0];c=0
while(i<n-1):
if s[i+1]==0 and c==0:
p=p+s[i]
i=i+2
c=1
elif s[i+1]==1 and c==0:
p=p+s[i]
i=i+1
c=1
elif s[i+1]==1 and c==1:
i=i+2
elif s[i+1]==0 and c==1:
i=i+1
print(p)
``` | instruction | 0 | 44,801 | 2 | 89,602 |
No | output | 1 | 44,801 | 2 | 89,603 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You and your friend are playing the game Mortal Kombat XI. You are trying to pass a challenge tower. There are n bosses in this tower, numbered from 1 to n. The type of the i-th boss is a_i. If the i-th boss is easy then its type is a_i = 0, otherwise this boss is hard and its type is a_i = 1.
During one session, either you or your friend can kill one or two bosses (neither you nor your friend can skip the session, so the minimum number of bosses killed during one session is at least one). After your friend session, your session begins, then again your friend session begins, your session begins, and so on. The first session is your friend's session.
Your friend needs to get good because he can't actually kill hard bosses. To kill them, he uses skip points. One skip point can be used to kill one hard boss.
Your task is to find the minimum number of skip points your friend needs to use so you and your friend kill all n bosses in the given order.
For example: suppose n = 8, a = [1, 0, 1, 1, 0, 1, 1, 1]. Then the best course of action is the following:
* your friend kills two first bosses, using one skip point for the first boss;
* you kill the third and the fourth bosses;
* your friend kills the fifth boss;
* you kill the sixth and the seventh bosses;
* your friend kills the last boss, using one skip point, so the tower is completed using two skip points.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of bosses. The second line of the test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1), where a_i is the type of the i-th boss.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer: the minimum number of skip points your friend needs to use so you and your friend kill all n bosses in the given order.
Example
Input
6
8
1 0 1 1 0 1 1 1
5
1 1 1 1 0
7
1 1 1 1 0 0 1
6
1 1 1 1 1 1
1
1
1
0
Output
2
2
2
2
1
0
Submitted Solution:
```
t = int(input())
for i in range(t):
n = int(input())
a = input().split()
session = 1
killed = 0
skips = 0
i = 0
while i < n:
if a[i] == '1':
if session == 1 and killed == 0:
skips += 1
if i < n-1 and a[i+1] == '0':
if i < n-3 and a[i+2] == '0' and a[i+3] == '1':
session = 0
else:
killed += 1
else:
session = 0
elif session == 0:
killed += 1
if i < n-3 and a[i+1] == '0' and a[i+2] == '0' and a[i+3] == '1':
session = 1
killed = 0
elif killed > 2:
session = 1
killed = 0
else:
killed += 1
elif a[i] == '0':
killed += 1
if killed > 2:
session = 1 - session
killed = 0
i += 1
print(skips)
``` | instruction | 0 | 44,802 | 2 | 89,604 |
No | output | 1 | 44,802 | 2 | 89,605 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You and your friend are playing the game Mortal Kombat XI. You are trying to pass a challenge tower. There are n bosses in this tower, numbered from 1 to n. The type of the i-th boss is a_i. If the i-th boss is easy then its type is a_i = 0, otherwise this boss is hard and its type is a_i = 1.
During one session, either you or your friend can kill one or two bosses (neither you nor your friend can skip the session, so the minimum number of bosses killed during one session is at least one). After your friend session, your session begins, then again your friend session begins, your session begins, and so on. The first session is your friend's session.
Your friend needs to get good because he can't actually kill hard bosses. To kill them, he uses skip points. One skip point can be used to kill one hard boss.
Your task is to find the minimum number of skip points your friend needs to use so you and your friend kill all n bosses in the given order.
For example: suppose n = 8, a = [1, 0, 1, 1, 0, 1, 1, 1]. Then the best course of action is the following:
* your friend kills two first bosses, using one skip point for the first boss;
* you kill the third and the fourth bosses;
* your friend kills the fifth boss;
* you kill the sixth and the seventh bosses;
* your friend kills the last boss, using one skip point, so the tower is completed using two skip points.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of bosses. The second line of the test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1), where a_i is the type of the i-th boss.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer: the minimum number of skip points your friend needs to use so you and your friend kill all n bosses in the given order.
Example
Input
6
8
1 0 1 1 0 1 1 1
5
1 1 1 1 0
7
1 1 1 1 0 0 1
6
1 1 1 1 1 1
1
1
1
0
Output
2
2
2
2
1
0
Submitted Solution:
```
import sys
input = sys.stdin.readline
t = int(input())
for _ in range(t):
n = int(input())
arr = list(map(int, input().split()))
dp = [999999999999] * (n+10)
dp[-1] = 0
dp[-2] = 0
dp[-3] = 0
dp[-4] = 0
dp[-5] = 0
for i in range(n):
if i>0:
l2 = arr[i-1]
dp[i] = min(dp[i-2] + l2, dp[i])
if i>1:
l3 = arr[i-2]
dp[i] = min(dp[i-3] + l3, dp[i])
if i>2:
l4 = arr[i-2] + arr[i-3]
dp[i] = min(dp[i-4] + l4, dp[i])
if i==n-1:
dp[i] = min(dp[i], dp[i-1] + arr[0])
if n<4:
print(arr[0])
else:
print(dp[n-1])
``` | instruction | 0 | 44,803 | 2 | 89,606 |
No | output | 1 | 44,803 | 2 | 89,607 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A long time ago (probably even in the first book), Nicholas Flamel, a great alchemist and the creator of the Philosopher's Stone, taught Harry Potter three useful spells. The first one allows you to convert a grams of sand into b grams of lead, the second one allows you to convert c grams of lead into d grams of gold and third one allows you to convert e grams of gold into f grams of sand. When Harry told his friends about these spells, Ron Weasley was amazed. After all, if they succeed in turning sand into lead, lead into gold, and then turning part of the gold into sand again and so on, then it will be possible to start with a small amount of sand and get huge amounts of gold! Even an infinite amount of gold! Hermione Granger, by contrast, was skeptical about that idea. She argues that according to the law of conservation of matter getting an infinite amount of matter, even using magic, is impossible. On the contrary, the amount of matter may even decrease during transformation, being converted to magical energy. Though Hermione's theory seems convincing, Ron won't believe her. As far as Ron is concerned, Hermione made up her law of conservation of matter to stop Harry and Ron wasting their time with this nonsense, and to make them go and do homework instead. That's why Ron has already collected a certain amount of sand for the experiments. A quarrel between the friends seems unavoidable...
Help Harry to determine which one of his friends is right, and avoid the quarrel after all. To do this you have to figure out whether it is possible to get the amount of gold greater than any preassigned number from some finite amount of sand.
Input
The first line contains 6 integers a, b, c, d, e, f (0 ≤ a, b, c, d, e, f ≤ 1000).
Output
Print "Ron", if it is possible to get an infinitely large amount of gold having a certain finite amount of sand (and not having any gold and lead at all), i.e., Ron is right. Otherwise, print "Hermione".
Examples
Input
100 200 250 150 200 250
Output
Ron
Input
100 50 50 200 200 100
Output
Hermione
Input
100 10 200 20 300 30
Output
Hermione
Input
0 0 0 0 0 0
Output
Hermione
Input
1 1 0 1 1 1
Output
Ron
Input
1 0 1 2 1 2
Output
Hermione
Input
100 1 100 1 0 1
Output
Ron
Note
Consider the first sample. Let's start with the 500 grams of sand. Apply the first spell 5 times and turn the sand into 1000 grams of lead. Then apply the second spell 4 times to get 600 grams of gold. Let’s take 400 grams from the resulting amount of gold turn them back into sand. We get 500 grams of sand and 200 grams of gold. If we apply the same operations to 500 grams of sand again, we can get extra 200 grams of gold every time. Thus, you can get 200, 400, 600 etc. grams of gold, i.e., starting with a finite amount of sand (500 grams), you can get the amount of gold which is greater than any preassigned number.
In the forth sample it is impossible to get sand, or lead, or gold, applying the spells.
In the fifth sample an infinitely large amount of gold can be obtained by using only the second spell, which allows you to receive 1 gram of gold out of nothing. Note that if such a second spell is available, then the first and the third one do not affect the answer at all.
The seventh sample is more interesting. We can also start with a zero amount of sand there. With the aid of the third spell you can get sand out of nothing. We get 10000 grams of sand in this manner. Let's get 100 grams of lead using the first spell 100 times. Then make 1 gram of gold from them. We managed to receive 1 gram of gold, starting with a zero amount of sand! Clearly, in this manner you can get an infinitely large amount of gold. | instruction | 0 | 44,993 | 2 | 89,986 |
Tags: implementation, math
Correct Solution:
```
def solve(c1: int, c2: int, c3: int, c4: int, c5: int, c6: int):
if (c3 == 0 and c4 != 0) or (c1 == 0 and c2 != 0 and c4 != 0):
return 'Ron'
return "Ron" if c2 * c4 * c6 > c1 * c3 * c5 else 'Hermione'
def main():
inp_str = input()
# inp_str = "100 200 250 150 200 250"
inp = list(map(int, inp_str.split(" ")))
res = solve(*inp)
print(res)
if __name__ == '__main__':
main()
# 100 50 50 200 200 100 = Hermione
# 100 10 200 20 300 30 = Hermione
# 100 200 250 150 200 250 = Ron
``` | output | 1 | 44,993 | 2 | 89,987 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A long time ago (probably even in the first book), Nicholas Flamel, a great alchemist and the creator of the Philosopher's Stone, taught Harry Potter three useful spells. The first one allows you to convert a grams of sand into b grams of lead, the second one allows you to convert c grams of lead into d grams of gold and third one allows you to convert e grams of gold into f grams of sand. When Harry told his friends about these spells, Ron Weasley was amazed. After all, if they succeed in turning sand into lead, lead into gold, and then turning part of the gold into sand again and so on, then it will be possible to start with a small amount of sand and get huge amounts of gold! Even an infinite amount of gold! Hermione Granger, by contrast, was skeptical about that idea. She argues that according to the law of conservation of matter getting an infinite amount of matter, even using magic, is impossible. On the contrary, the amount of matter may even decrease during transformation, being converted to magical energy. Though Hermione's theory seems convincing, Ron won't believe her. As far as Ron is concerned, Hermione made up her law of conservation of matter to stop Harry and Ron wasting their time with this nonsense, and to make them go and do homework instead. That's why Ron has already collected a certain amount of sand for the experiments. A quarrel between the friends seems unavoidable...
Help Harry to determine which one of his friends is right, and avoid the quarrel after all. To do this you have to figure out whether it is possible to get the amount of gold greater than any preassigned number from some finite amount of sand.
Input
The first line contains 6 integers a, b, c, d, e, f (0 ≤ a, b, c, d, e, f ≤ 1000).
Output
Print "Ron", if it is possible to get an infinitely large amount of gold having a certain finite amount of sand (and not having any gold and lead at all), i.e., Ron is right. Otherwise, print "Hermione".
Examples
Input
100 200 250 150 200 250
Output
Ron
Input
100 50 50 200 200 100
Output
Hermione
Input
100 10 200 20 300 30
Output
Hermione
Input
0 0 0 0 0 0
Output
Hermione
Input
1 1 0 1 1 1
Output
Ron
Input
1 0 1 2 1 2
Output
Hermione
Input
100 1 100 1 0 1
Output
Ron
Note
Consider the first sample. Let's start with the 500 grams of sand. Apply the first spell 5 times and turn the sand into 1000 grams of lead. Then apply the second spell 4 times to get 600 grams of gold. Let’s take 400 grams from the resulting amount of gold turn them back into sand. We get 500 grams of sand and 200 grams of gold. If we apply the same operations to 500 grams of sand again, we can get extra 200 grams of gold every time. Thus, you can get 200, 400, 600 etc. grams of gold, i.e., starting with a finite amount of sand (500 grams), you can get the amount of gold which is greater than any preassigned number.
In the forth sample it is impossible to get sand, or lead, or gold, applying the spells.
In the fifth sample an infinitely large amount of gold can be obtained by using only the second spell, which allows you to receive 1 gram of gold out of nothing. Note that if such a second spell is available, then the first and the third one do not affect the answer at all.
The seventh sample is more interesting. We can also start with a zero amount of sand there. With the aid of the third spell you can get sand out of nothing. We get 10000 grams of sand in this manner. Let's get 100 grams of lead using the first spell 100 times. Then make 1 gram of gold from them. We managed to receive 1 gram of gold, starting with a zero amount of sand! Clearly, in this manner you can get an infinitely large amount of gold. | instruction | 0 | 44,994 | 2 | 89,988 |
Tags: implementation, math
Correct Solution:
```
from functools import reduce
from operator import mul
c = list(map(int, input().split()))
r, h = reduce(mul, c[1::2], 1), reduce(mul, c[0::2], 1)
rw = r > h or (c[2] == 0 and c[3] != 0) or (c[0] == 0 and c[1] * c[3] != 0)
print("Ron" if rw else "Hermione")
``` | output | 1 | 44,994 | 2 | 89,989 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A long time ago (probably even in the first book), Nicholas Flamel, a great alchemist and the creator of the Philosopher's Stone, taught Harry Potter three useful spells. The first one allows you to convert a grams of sand into b grams of lead, the second one allows you to convert c grams of lead into d grams of gold and third one allows you to convert e grams of gold into f grams of sand. When Harry told his friends about these spells, Ron Weasley was amazed. After all, if they succeed in turning sand into lead, lead into gold, and then turning part of the gold into sand again and so on, then it will be possible to start with a small amount of sand and get huge amounts of gold! Even an infinite amount of gold! Hermione Granger, by contrast, was skeptical about that idea. She argues that according to the law of conservation of matter getting an infinite amount of matter, even using magic, is impossible. On the contrary, the amount of matter may even decrease during transformation, being converted to magical energy. Though Hermione's theory seems convincing, Ron won't believe her. As far as Ron is concerned, Hermione made up her law of conservation of matter to stop Harry and Ron wasting their time with this nonsense, and to make them go and do homework instead. That's why Ron has already collected a certain amount of sand for the experiments. A quarrel between the friends seems unavoidable...
Help Harry to determine which one of his friends is right, and avoid the quarrel after all. To do this you have to figure out whether it is possible to get the amount of gold greater than any preassigned number from some finite amount of sand.
Input
The first line contains 6 integers a, b, c, d, e, f (0 ≤ a, b, c, d, e, f ≤ 1000).
Output
Print "Ron", if it is possible to get an infinitely large amount of gold having a certain finite amount of sand (and not having any gold and lead at all), i.e., Ron is right. Otherwise, print "Hermione".
Examples
Input
100 200 250 150 200 250
Output
Ron
Input
100 50 50 200 200 100
Output
Hermione
Input
100 10 200 20 300 30
Output
Hermione
Input
0 0 0 0 0 0
Output
Hermione
Input
1 1 0 1 1 1
Output
Ron
Input
1 0 1 2 1 2
Output
Hermione
Input
100 1 100 1 0 1
Output
Ron
Note
Consider the first sample. Let's start with the 500 grams of sand. Apply the first spell 5 times and turn the sand into 1000 grams of lead. Then apply the second spell 4 times to get 600 grams of gold. Let’s take 400 grams from the resulting amount of gold turn them back into sand. We get 500 grams of sand and 200 grams of gold. If we apply the same operations to 500 grams of sand again, we can get extra 200 grams of gold every time. Thus, you can get 200, 400, 600 etc. grams of gold, i.e., starting with a finite amount of sand (500 grams), you can get the amount of gold which is greater than any preassigned number.
In the forth sample it is impossible to get sand, or lead, or gold, applying the spells.
In the fifth sample an infinitely large amount of gold can be obtained by using only the second spell, which allows you to receive 1 gram of gold out of nothing. Note that if such a second spell is available, then the first and the third one do not affect the answer at all.
The seventh sample is more interesting. We can also start with a zero amount of sand there. With the aid of the third spell you can get sand out of nothing. We get 10000 grams of sand in this manner. Let's get 100 grams of lead using the first spell 100 times. Then make 1 gram of gold from them. We managed to receive 1 gram of gold, starting with a zero amount of sand! Clearly, in this manner you can get an infinitely large amount of gold. | instruction | 0 | 44,995 | 2 | 89,990 |
Tags: implementation, math
Correct Solution:
```
t = list(map(int, input().split()))
f0 = 3
while t[f0] != 0:
f0 = (f0 + 5) % 6
if f0 == 3:
break
if f0 % 2 == 0:
print('Ron')
exit()
num = t[1] * t[3] * t[5]
den = t[0] * t[2] * t[4]
if num > den:
print('Ron')
else:
print('Hermione')
``` | output | 1 | 44,995 | 2 | 89,991 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A long time ago (probably even in the first book), Nicholas Flamel, a great alchemist and the creator of the Philosopher's Stone, taught Harry Potter three useful spells. The first one allows you to convert a grams of sand into b grams of lead, the second one allows you to convert c grams of lead into d grams of gold and third one allows you to convert e grams of gold into f grams of sand. When Harry told his friends about these spells, Ron Weasley was amazed. After all, if they succeed in turning sand into lead, lead into gold, and then turning part of the gold into sand again and so on, then it will be possible to start with a small amount of sand and get huge amounts of gold! Even an infinite amount of gold! Hermione Granger, by contrast, was skeptical about that idea. She argues that according to the law of conservation of matter getting an infinite amount of matter, even using magic, is impossible. On the contrary, the amount of matter may even decrease during transformation, being converted to magical energy. Though Hermione's theory seems convincing, Ron won't believe her. As far as Ron is concerned, Hermione made up her law of conservation of matter to stop Harry and Ron wasting their time with this nonsense, and to make them go and do homework instead. That's why Ron has already collected a certain amount of sand for the experiments. A quarrel between the friends seems unavoidable...
Help Harry to determine which one of his friends is right, and avoid the quarrel after all. To do this you have to figure out whether it is possible to get the amount of gold greater than any preassigned number from some finite amount of sand.
Input
The first line contains 6 integers a, b, c, d, e, f (0 ≤ a, b, c, d, e, f ≤ 1000).
Output
Print "Ron", if it is possible to get an infinitely large amount of gold having a certain finite amount of sand (and not having any gold and lead at all), i.e., Ron is right. Otherwise, print "Hermione".
Examples
Input
100 200 250 150 200 250
Output
Ron
Input
100 50 50 200 200 100
Output
Hermione
Input
100 10 200 20 300 30
Output
Hermione
Input
0 0 0 0 0 0
Output
Hermione
Input
1 1 0 1 1 1
Output
Ron
Input
1 0 1 2 1 2
Output
Hermione
Input
100 1 100 1 0 1
Output
Ron
Note
Consider the first sample. Let's start with the 500 grams of sand. Apply the first spell 5 times and turn the sand into 1000 grams of lead. Then apply the second spell 4 times to get 600 grams of gold. Let’s take 400 grams from the resulting amount of gold turn them back into sand. We get 500 grams of sand and 200 grams of gold. If we apply the same operations to 500 grams of sand again, we can get extra 200 grams of gold every time. Thus, you can get 200, 400, 600 etc. grams of gold, i.e., starting with a finite amount of sand (500 grams), you can get the amount of gold which is greater than any preassigned number.
In the forth sample it is impossible to get sand, or lead, or gold, applying the spells.
In the fifth sample an infinitely large amount of gold can be obtained by using only the second spell, which allows you to receive 1 gram of gold out of nothing. Note that if such a second spell is available, then the first and the third one do not affect the answer at all.
The seventh sample is more interesting. We can also start with a zero amount of sand there. With the aid of the third spell you can get sand out of nothing. We get 10000 grams of sand in this manner. Let's get 100 grams of lead using the first spell 100 times. Then make 1 gram of gold from them. We managed to receive 1 gram of gold, starting with a zero amount of sand! Clearly, in this manner you can get an infinitely large amount of gold. | instruction | 0 | 44,996 | 2 | 89,992 |
Tags: implementation, math
Correct Solution:
```
a = list(map(int, input().split()))
if a[0] * a[2] * a[4] < a[1] * a[3] * a[5]:
print('Ron')
else:
if (0 == a[0] * a[2] * a[4]) ^ (0 == a[1] * a[3] * a[5]):
print('Hermione')
elif 0 == a[0] * a[2] * a[4] == a[1] * a[3] * a[5]:
if (a[2] == 0 and a[3] != 0) or (a[0] == 0 and a[1] > 0 \
and a[3] > 0):
print('Ron')
else:
print('Hermione')
else:
print('Hermione')
``` | output | 1 | 44,996 | 2 | 89,993 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A long time ago (probably even in the first book), Nicholas Flamel, a great alchemist and the creator of the Philosopher's Stone, taught Harry Potter three useful spells. The first one allows you to convert a grams of sand into b grams of lead, the second one allows you to convert c grams of lead into d grams of gold and third one allows you to convert e grams of gold into f grams of sand. When Harry told his friends about these spells, Ron Weasley was amazed. After all, if they succeed in turning sand into lead, lead into gold, and then turning part of the gold into sand again and so on, then it will be possible to start with a small amount of sand and get huge amounts of gold! Even an infinite amount of gold! Hermione Granger, by contrast, was skeptical about that idea. She argues that according to the law of conservation of matter getting an infinite amount of matter, even using magic, is impossible. On the contrary, the amount of matter may even decrease during transformation, being converted to magical energy. Though Hermione's theory seems convincing, Ron won't believe her. As far as Ron is concerned, Hermione made up her law of conservation of matter to stop Harry and Ron wasting their time with this nonsense, and to make them go and do homework instead. That's why Ron has already collected a certain amount of sand for the experiments. A quarrel between the friends seems unavoidable...
Help Harry to determine which one of his friends is right, and avoid the quarrel after all. To do this you have to figure out whether it is possible to get the amount of gold greater than any preassigned number from some finite amount of sand.
Input
The first line contains 6 integers a, b, c, d, e, f (0 ≤ a, b, c, d, e, f ≤ 1000).
Output
Print "Ron", if it is possible to get an infinitely large amount of gold having a certain finite amount of sand (and not having any gold and lead at all), i.e., Ron is right. Otherwise, print "Hermione".
Examples
Input
100 200 250 150 200 250
Output
Ron
Input
100 50 50 200 200 100
Output
Hermione
Input
100 10 200 20 300 30
Output
Hermione
Input
0 0 0 0 0 0
Output
Hermione
Input
1 1 0 1 1 1
Output
Ron
Input
1 0 1 2 1 2
Output
Hermione
Input
100 1 100 1 0 1
Output
Ron
Note
Consider the first sample. Let's start with the 500 grams of sand. Apply the first spell 5 times and turn the sand into 1000 grams of lead. Then apply the second spell 4 times to get 600 grams of gold. Let’s take 400 grams from the resulting amount of gold turn them back into sand. We get 500 grams of sand and 200 grams of gold. If we apply the same operations to 500 grams of sand again, we can get extra 200 grams of gold every time. Thus, you can get 200, 400, 600 etc. grams of gold, i.e., starting with a finite amount of sand (500 grams), you can get the amount of gold which is greater than any preassigned number.
In the forth sample it is impossible to get sand, or lead, or gold, applying the spells.
In the fifth sample an infinitely large amount of gold can be obtained by using only the second spell, which allows you to receive 1 gram of gold out of nothing. Note that if such a second spell is available, then the first and the third one do not affect the answer at all.
The seventh sample is more interesting. We can also start with a zero amount of sand there. With the aid of the third spell you can get sand out of nothing. We get 10000 grams of sand in this manner. Let's get 100 grams of lead using the first spell 100 times. Then make 1 gram of gold from them. We managed to receive 1 gram of gold, starting with a zero amount of sand! Clearly, in this manner you can get an infinitely large amount of gold. | instruction | 0 | 44,997 | 2 | 89,994 |
Tags: implementation, math
Correct Solution:
```
a, b, c, d, e, f = map(int, input().split())
print('Ron' if a * e * c < f * d * b or (c == 0 and d) or (a == 0 and b and d) else 'Hermione')
``` | output | 1 | 44,997 | 2 | 89,995 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A long time ago (probably even in the first book), Nicholas Flamel, a great alchemist and the creator of the Philosopher's Stone, taught Harry Potter three useful spells. The first one allows you to convert a grams of sand into b grams of lead, the second one allows you to convert c grams of lead into d grams of gold and third one allows you to convert e grams of gold into f grams of sand. When Harry told his friends about these spells, Ron Weasley was amazed. After all, if they succeed in turning sand into lead, lead into gold, and then turning part of the gold into sand again and so on, then it will be possible to start with a small amount of sand and get huge amounts of gold! Even an infinite amount of gold! Hermione Granger, by contrast, was skeptical about that idea. She argues that according to the law of conservation of matter getting an infinite amount of matter, even using magic, is impossible. On the contrary, the amount of matter may even decrease during transformation, being converted to magical energy. Though Hermione's theory seems convincing, Ron won't believe her. As far as Ron is concerned, Hermione made up her law of conservation of matter to stop Harry and Ron wasting their time with this nonsense, and to make them go and do homework instead. That's why Ron has already collected a certain amount of sand for the experiments. A quarrel between the friends seems unavoidable...
Help Harry to determine which one of his friends is right, and avoid the quarrel after all. To do this you have to figure out whether it is possible to get the amount of gold greater than any preassigned number from some finite amount of sand.
Input
The first line contains 6 integers a, b, c, d, e, f (0 ≤ a, b, c, d, e, f ≤ 1000).
Output
Print "Ron", if it is possible to get an infinitely large amount of gold having a certain finite amount of sand (and not having any gold and lead at all), i.e., Ron is right. Otherwise, print "Hermione".
Examples
Input
100 200 250 150 200 250
Output
Ron
Input
100 50 50 200 200 100
Output
Hermione
Input
100 10 200 20 300 30
Output
Hermione
Input
0 0 0 0 0 0
Output
Hermione
Input
1 1 0 1 1 1
Output
Ron
Input
1 0 1 2 1 2
Output
Hermione
Input
100 1 100 1 0 1
Output
Ron
Note
Consider the first sample. Let's start with the 500 grams of sand. Apply the first spell 5 times and turn the sand into 1000 grams of lead. Then apply the second spell 4 times to get 600 grams of gold. Let’s take 400 grams from the resulting amount of gold turn them back into sand. We get 500 grams of sand and 200 grams of gold. If we apply the same operations to 500 grams of sand again, we can get extra 200 grams of gold every time. Thus, you can get 200, 400, 600 etc. grams of gold, i.e., starting with a finite amount of sand (500 grams), you can get the amount of gold which is greater than any preassigned number.
In the forth sample it is impossible to get sand, or lead, or gold, applying the spells.
In the fifth sample an infinitely large amount of gold can be obtained by using only the second spell, which allows you to receive 1 gram of gold out of nothing. Note that if such a second spell is available, then the first and the third one do not affect the answer at all.
The seventh sample is more interesting. We can also start with a zero amount of sand there. With the aid of the third spell you can get sand out of nothing. We get 10000 grams of sand in this manner. Let's get 100 grams of lead using the first spell 100 times. Then make 1 gram of gold from them. We managed to receive 1 gram of gold, starting with a zero amount of sand! Clearly, in this manner you can get an infinitely large amount of gold. | instruction | 0 | 44,998 | 2 | 89,996 |
Tags: implementation, math
Correct Solution:
```
sand_in, lead_out, lead_in, gold_out, gold_in, sand_out = map(int,
input().split())
def ron():
print('Ron')
import sys; sys.exit()
if lead_out * gold_out * sand_out > sand_in * lead_in * gold_in:
ron()
if gold_out > 0:
if lead_in == 0:
ron()
if lead_out > 0:
if sand_in == 0:
ron()
if sand_out > 0:
if gold_in == 0:
ron()
print('Hermione')
``` | output | 1 | 44,998 | 2 | 89,997 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A long time ago (probably even in the first book), Nicholas Flamel, a great alchemist and the creator of the Philosopher's Stone, taught Harry Potter three useful spells. The first one allows you to convert a grams of sand into b grams of lead, the second one allows you to convert c grams of lead into d grams of gold and third one allows you to convert e grams of gold into f grams of sand. When Harry told his friends about these spells, Ron Weasley was amazed. After all, if they succeed in turning sand into lead, lead into gold, and then turning part of the gold into sand again and so on, then it will be possible to start with a small amount of sand and get huge amounts of gold! Even an infinite amount of gold! Hermione Granger, by contrast, was skeptical about that idea. She argues that according to the law of conservation of matter getting an infinite amount of matter, even using magic, is impossible. On the contrary, the amount of matter may even decrease during transformation, being converted to magical energy. Though Hermione's theory seems convincing, Ron won't believe her. As far as Ron is concerned, Hermione made up her law of conservation of matter to stop Harry and Ron wasting their time with this nonsense, and to make them go and do homework instead. That's why Ron has already collected a certain amount of sand for the experiments. A quarrel between the friends seems unavoidable...
Help Harry to determine which one of his friends is right, and avoid the quarrel after all. To do this you have to figure out whether it is possible to get the amount of gold greater than any preassigned number from some finite amount of sand.
Input
The first line contains 6 integers a, b, c, d, e, f (0 ≤ a, b, c, d, e, f ≤ 1000).
Output
Print "Ron", if it is possible to get an infinitely large amount of gold having a certain finite amount of sand (and not having any gold and lead at all), i.e., Ron is right. Otherwise, print "Hermione".
Examples
Input
100 200 250 150 200 250
Output
Ron
Input
100 50 50 200 200 100
Output
Hermione
Input
100 10 200 20 300 30
Output
Hermione
Input
0 0 0 0 0 0
Output
Hermione
Input
1 1 0 1 1 1
Output
Ron
Input
1 0 1 2 1 2
Output
Hermione
Input
100 1 100 1 0 1
Output
Ron
Note
Consider the first sample. Let's start with the 500 grams of sand. Apply the first spell 5 times and turn the sand into 1000 grams of lead. Then apply the second spell 4 times to get 600 grams of gold. Let’s take 400 grams from the resulting amount of gold turn them back into sand. We get 500 grams of sand and 200 grams of gold. If we apply the same operations to 500 grams of sand again, we can get extra 200 grams of gold every time. Thus, you can get 200, 400, 600 etc. grams of gold, i.e., starting with a finite amount of sand (500 grams), you can get the amount of gold which is greater than any preassigned number.
In the forth sample it is impossible to get sand, or lead, or gold, applying the spells.
In the fifth sample an infinitely large amount of gold can be obtained by using only the second spell, which allows you to receive 1 gram of gold out of nothing. Note that if such a second spell is available, then the first and the third one do not affect the answer at all.
The seventh sample is more interesting. We can also start with a zero amount of sand there. With the aid of the third spell you can get sand out of nothing. We get 10000 grams of sand in this manner. Let's get 100 grams of lead using the first spell 100 times. Then make 1 gram of gold from them. We managed to receive 1 gram of gold, starting with a zero amount of sand! Clearly, in this manner you can get an infinitely large amount of gold. | instruction | 0 | 44,999 | 2 | 89,998 |
Tags: implementation, math
Correct Solution:
```
a, b, c, d, e, f = map(int, input().split())
if c == 0 and d > 0:
print("Ron")
exit(0)
if a == 0 and b > 0 and d > 0:
print("Ron")
exit(0)
if b == 0 or d == 0 or f == 0:
print("Hermione")
exit(0)
if a == 0 or c == 0 or e == 0:
print("Ron")
exit(0)
g = a * b * c * d * e * f
h = g
g = g * b // a
g = g * d // c
g = g * f // e
if g > h:
print("Ron")
else:
print("Hermione")
``` | output | 1 | 44,999 | 2 | 89,999 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A long time ago (probably even in the first book), Nicholas Flamel, a great alchemist and the creator of the Philosopher's Stone, taught Harry Potter three useful spells. The first one allows you to convert a grams of sand into b grams of lead, the second one allows you to convert c grams of lead into d grams of gold and third one allows you to convert e grams of gold into f grams of sand. When Harry told his friends about these spells, Ron Weasley was amazed. After all, if they succeed in turning sand into lead, lead into gold, and then turning part of the gold into sand again and so on, then it will be possible to start with a small amount of sand and get huge amounts of gold! Even an infinite amount of gold! Hermione Granger, by contrast, was skeptical about that idea. She argues that according to the law of conservation of matter getting an infinite amount of matter, even using magic, is impossible. On the contrary, the amount of matter may even decrease during transformation, being converted to magical energy. Though Hermione's theory seems convincing, Ron won't believe her. As far as Ron is concerned, Hermione made up her law of conservation of matter to stop Harry and Ron wasting their time with this nonsense, and to make them go and do homework instead. That's why Ron has already collected a certain amount of sand for the experiments. A quarrel between the friends seems unavoidable...
Help Harry to determine which one of his friends is right, and avoid the quarrel after all. To do this you have to figure out whether it is possible to get the amount of gold greater than any preassigned number from some finite amount of sand.
Input
The first line contains 6 integers a, b, c, d, e, f (0 ≤ a, b, c, d, e, f ≤ 1000).
Output
Print "Ron", if it is possible to get an infinitely large amount of gold having a certain finite amount of sand (and not having any gold and lead at all), i.e., Ron is right. Otherwise, print "Hermione".
Examples
Input
100 200 250 150 200 250
Output
Ron
Input
100 50 50 200 200 100
Output
Hermione
Input
100 10 200 20 300 30
Output
Hermione
Input
0 0 0 0 0 0
Output
Hermione
Input
1 1 0 1 1 1
Output
Ron
Input
1 0 1 2 1 2
Output
Hermione
Input
100 1 100 1 0 1
Output
Ron
Note
Consider the first sample. Let's start with the 500 grams of sand. Apply the first spell 5 times and turn the sand into 1000 grams of lead. Then apply the second spell 4 times to get 600 grams of gold. Let’s take 400 grams from the resulting amount of gold turn them back into sand. We get 500 grams of sand and 200 grams of gold. If we apply the same operations to 500 grams of sand again, we can get extra 200 grams of gold every time. Thus, you can get 200, 400, 600 etc. grams of gold, i.e., starting with a finite amount of sand (500 grams), you can get the amount of gold which is greater than any preassigned number.
In the forth sample it is impossible to get sand, or lead, or gold, applying the spells.
In the fifth sample an infinitely large amount of gold can be obtained by using only the second spell, which allows you to receive 1 gram of gold out of nothing. Note that if such a second spell is available, then the first and the third one do not affect the answer at all.
The seventh sample is more interesting. We can also start with a zero amount of sand there. With the aid of the third spell you can get sand out of nothing. We get 10000 grams of sand in this manner. Let's get 100 grams of lead using the first spell 100 times. Then make 1 gram of gold from them. We managed to receive 1 gram of gold, starting with a zero amount of sand! Clearly, in this manner you can get an infinitely large amount of gold. | instruction | 0 | 45,000 | 2 | 90,000 |
Tags: implementation, math
Correct Solution:
```
def answer(a, b, c, d, e, f):
if c == 0 and d > 0:
return "Ron"
if a == 0 and b > 0 and d > 0:
return "Ron"
if e == 0 and f > 0 and b > 0 and d > 0:
return "Ron"
if b * d * f > a * c * e:
return "Ron"
return "Hermione"
A, B, C, D, E, F = [int(i) for i in input().split()]
print(answer(A, B, C, D, E, F))
``` | output | 1 | 45,000 | 2 | 90,001 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A long time ago (probably even in the first book), Nicholas Flamel, a great alchemist and the creator of the Philosopher's Stone, taught Harry Potter three useful spells. The first one allows you to convert a grams of sand into b grams of lead, the second one allows you to convert c grams of lead into d grams of gold and third one allows you to convert e grams of gold into f grams of sand. When Harry told his friends about these spells, Ron Weasley was amazed. After all, if they succeed in turning sand into lead, lead into gold, and then turning part of the gold into sand again and so on, then it will be possible to start with a small amount of sand and get huge amounts of gold! Even an infinite amount of gold! Hermione Granger, by contrast, was skeptical about that idea. She argues that according to the law of conservation of matter getting an infinite amount of matter, even using magic, is impossible. On the contrary, the amount of matter may even decrease during transformation, being converted to magical energy. Though Hermione's theory seems convincing, Ron won't believe her. As far as Ron is concerned, Hermione made up her law of conservation of matter to stop Harry and Ron wasting their time with this nonsense, and to make them go and do homework instead. That's why Ron has already collected a certain amount of sand for the experiments. A quarrel between the friends seems unavoidable...
Help Harry to determine which one of his friends is right, and avoid the quarrel after all. To do this you have to figure out whether it is possible to get the amount of gold greater than any preassigned number from some finite amount of sand.
Input
The first line contains 6 integers a, b, c, d, e, f (0 ≤ a, b, c, d, e, f ≤ 1000).
Output
Print "Ron", if it is possible to get an infinitely large amount of gold having a certain finite amount of sand (and not having any gold and lead at all), i.e., Ron is right. Otherwise, print "Hermione".
Examples
Input
100 200 250 150 200 250
Output
Ron
Input
100 50 50 200 200 100
Output
Hermione
Input
100 10 200 20 300 30
Output
Hermione
Input
0 0 0 0 0 0
Output
Hermione
Input
1 1 0 1 1 1
Output
Ron
Input
1 0 1 2 1 2
Output
Hermione
Input
100 1 100 1 0 1
Output
Ron
Note
Consider the first sample. Let's start with the 500 grams of sand. Apply the first spell 5 times and turn the sand into 1000 grams of lead. Then apply the second spell 4 times to get 600 grams of gold. Let’s take 400 grams from the resulting amount of gold turn them back into sand. We get 500 grams of sand and 200 grams of gold. If we apply the same operations to 500 grams of sand again, we can get extra 200 grams of gold every time. Thus, you can get 200, 400, 600 etc. grams of gold, i.e., starting with a finite amount of sand (500 grams), you can get the amount of gold which is greater than any preassigned number.
In the forth sample it is impossible to get sand, or lead, or gold, applying the spells.
In the fifth sample an infinitely large amount of gold can be obtained by using only the second spell, which allows you to receive 1 gram of gold out of nothing. Note that if such a second spell is available, then the first and the third one do not affect the answer at all.
The seventh sample is more interesting. We can also start with a zero amount of sand there. With the aid of the third spell you can get sand out of nothing. We get 10000 grams of sand in this manner. Let's get 100 grams of lead using the first spell 100 times. Then make 1 gram of gold from them. We managed to receive 1 gram of gold, starting with a zero amount of sand! Clearly, in this manner you can get an infinitely large amount of gold.
Submitted Solution:
```
a,b,c,d,e,f=map(int,input().split())
if a==0 and b*d>0 or c==0 and d>0:
print("Ron")
else:
if a*c*e<b*d*f:
print("Ron")
else:
print("Hermione")
``` | instruction | 0 | 45,001 | 2 | 90,002 |
Yes | output | 1 | 45,001 | 2 | 90,003 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A long time ago (probably even in the first book), Nicholas Flamel, a great alchemist and the creator of the Philosopher's Stone, taught Harry Potter three useful spells. The first one allows you to convert a grams of sand into b grams of lead, the second one allows you to convert c grams of lead into d grams of gold and third one allows you to convert e grams of gold into f grams of sand. When Harry told his friends about these spells, Ron Weasley was amazed. After all, if they succeed in turning sand into lead, lead into gold, and then turning part of the gold into sand again and so on, then it will be possible to start with a small amount of sand and get huge amounts of gold! Even an infinite amount of gold! Hermione Granger, by contrast, was skeptical about that idea. She argues that according to the law of conservation of matter getting an infinite amount of matter, even using magic, is impossible. On the contrary, the amount of matter may even decrease during transformation, being converted to magical energy. Though Hermione's theory seems convincing, Ron won't believe her. As far as Ron is concerned, Hermione made up her law of conservation of matter to stop Harry and Ron wasting their time with this nonsense, and to make them go and do homework instead. That's why Ron has already collected a certain amount of sand for the experiments. A quarrel between the friends seems unavoidable...
Help Harry to determine which one of his friends is right, and avoid the quarrel after all. To do this you have to figure out whether it is possible to get the amount of gold greater than any preassigned number from some finite amount of sand.
Input
The first line contains 6 integers a, b, c, d, e, f (0 ≤ a, b, c, d, e, f ≤ 1000).
Output
Print "Ron", if it is possible to get an infinitely large amount of gold having a certain finite amount of sand (and not having any gold and lead at all), i.e., Ron is right. Otherwise, print "Hermione".
Examples
Input
100 200 250 150 200 250
Output
Ron
Input
100 50 50 200 200 100
Output
Hermione
Input
100 10 200 20 300 30
Output
Hermione
Input
0 0 0 0 0 0
Output
Hermione
Input
1 1 0 1 1 1
Output
Ron
Input
1 0 1 2 1 2
Output
Hermione
Input
100 1 100 1 0 1
Output
Ron
Note
Consider the first sample. Let's start with the 500 grams of sand. Apply the first spell 5 times and turn the sand into 1000 grams of lead. Then apply the second spell 4 times to get 600 grams of gold. Let’s take 400 grams from the resulting amount of gold turn them back into sand. We get 500 grams of sand and 200 grams of gold. If we apply the same operations to 500 grams of sand again, we can get extra 200 grams of gold every time. Thus, you can get 200, 400, 600 etc. grams of gold, i.e., starting with a finite amount of sand (500 grams), you can get the amount of gold which is greater than any preassigned number.
In the forth sample it is impossible to get sand, or lead, or gold, applying the spells.
In the fifth sample an infinitely large amount of gold can be obtained by using only the second spell, which allows you to receive 1 gram of gold out of nothing. Note that if such a second spell is available, then the first and the third one do not affect the answer at all.
The seventh sample is more interesting. We can also start with a zero amount of sand there. With the aid of the third spell you can get sand out of nothing. We get 10000 grams of sand in this manner. Let's get 100 grams of lead using the first spell 100 times. Then make 1 gram of gold from them. We managed to receive 1 gram of gold, starting with a zero amount of sand! Clearly, in this manner you can get an infinitely large amount of gold.
Submitted Solution:
```
def main():
a, b, c, d, e, f = map(int, input().split())
if c:
c *= a
d *= b
if c:
c *= e
d *= f
print(("Hermione", "Ron")[c < d])
if __name__ == '__main__':
main()
``` | instruction | 0 | 45,002 | 2 | 90,004 |
Yes | output | 1 | 45,002 | 2 | 90,005 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A long time ago (probably even in the first book), Nicholas Flamel, a great alchemist and the creator of the Philosopher's Stone, taught Harry Potter three useful spells. The first one allows you to convert a grams of sand into b grams of lead, the second one allows you to convert c grams of lead into d grams of gold and third one allows you to convert e grams of gold into f grams of sand. When Harry told his friends about these spells, Ron Weasley was amazed. After all, if they succeed in turning sand into lead, lead into gold, and then turning part of the gold into sand again and so on, then it will be possible to start with a small amount of sand and get huge amounts of gold! Even an infinite amount of gold! Hermione Granger, by contrast, was skeptical about that idea. She argues that according to the law of conservation of matter getting an infinite amount of matter, even using magic, is impossible. On the contrary, the amount of matter may even decrease during transformation, being converted to magical energy. Though Hermione's theory seems convincing, Ron won't believe her. As far as Ron is concerned, Hermione made up her law of conservation of matter to stop Harry and Ron wasting their time with this nonsense, and to make them go and do homework instead. That's why Ron has already collected a certain amount of sand for the experiments. A quarrel between the friends seems unavoidable...
Help Harry to determine which one of his friends is right, and avoid the quarrel after all. To do this you have to figure out whether it is possible to get the amount of gold greater than any preassigned number from some finite amount of sand.
Input
The first line contains 6 integers a, b, c, d, e, f (0 ≤ a, b, c, d, e, f ≤ 1000).
Output
Print "Ron", if it is possible to get an infinitely large amount of gold having a certain finite amount of sand (and not having any gold and lead at all), i.e., Ron is right. Otherwise, print "Hermione".
Examples
Input
100 200 250 150 200 250
Output
Ron
Input
100 50 50 200 200 100
Output
Hermione
Input
100 10 200 20 300 30
Output
Hermione
Input
0 0 0 0 0 0
Output
Hermione
Input
1 1 0 1 1 1
Output
Ron
Input
1 0 1 2 1 2
Output
Hermione
Input
100 1 100 1 0 1
Output
Ron
Note
Consider the first sample. Let's start with the 500 grams of sand. Apply the first spell 5 times and turn the sand into 1000 grams of lead. Then apply the second spell 4 times to get 600 grams of gold. Let’s take 400 grams from the resulting amount of gold turn them back into sand. We get 500 grams of sand and 200 grams of gold. If we apply the same operations to 500 grams of sand again, we can get extra 200 grams of gold every time. Thus, you can get 200, 400, 600 etc. grams of gold, i.e., starting with a finite amount of sand (500 grams), you can get the amount of gold which is greater than any preassigned number.
In the forth sample it is impossible to get sand, or lead, or gold, applying the spells.
In the fifth sample an infinitely large amount of gold can be obtained by using only the second spell, which allows you to receive 1 gram of gold out of nothing. Note that if such a second spell is available, then the first and the third one do not affect the answer at all.
The seventh sample is more interesting. We can also start with a zero amount of sand there. With the aid of the third spell you can get sand out of nothing. We get 10000 grams of sand in this manner. Let's get 100 grams of lead using the first spell 100 times. Then make 1 gram of gold from them. We managed to receive 1 gram of gold, starting with a zero amount of sand! Clearly, in this manner you can get an infinitely large amount of gold.
Submitted Solution:
```
a, b, c, d, e, f = map(int, input().split())
if ((b * d * f > a * c * e) or (c == 0 and d != 0) or (a == 0 and b != 0 and d != 0) ) and d != 0 :
print("Ron")
else :
print("Hermione")
``` | instruction | 0 | 45,003 | 2 | 90,006 |
Yes | output | 1 | 45,003 | 2 | 90,007 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A long time ago (probably even in the first book), Nicholas Flamel, a great alchemist and the creator of the Philosopher's Stone, taught Harry Potter three useful spells. The first one allows you to convert a grams of sand into b grams of lead, the second one allows you to convert c grams of lead into d grams of gold and third one allows you to convert e grams of gold into f grams of sand. When Harry told his friends about these spells, Ron Weasley was amazed. After all, if they succeed in turning sand into lead, lead into gold, and then turning part of the gold into sand again and so on, then it will be possible to start with a small amount of sand and get huge amounts of gold! Even an infinite amount of gold! Hermione Granger, by contrast, was skeptical about that idea. She argues that according to the law of conservation of matter getting an infinite amount of matter, even using magic, is impossible. On the contrary, the amount of matter may even decrease during transformation, being converted to magical energy. Though Hermione's theory seems convincing, Ron won't believe her. As far as Ron is concerned, Hermione made up her law of conservation of matter to stop Harry and Ron wasting their time with this nonsense, and to make them go and do homework instead. That's why Ron has already collected a certain amount of sand for the experiments. A quarrel between the friends seems unavoidable...
Help Harry to determine which one of his friends is right, and avoid the quarrel after all. To do this you have to figure out whether it is possible to get the amount of gold greater than any preassigned number from some finite amount of sand.
Input
The first line contains 6 integers a, b, c, d, e, f (0 ≤ a, b, c, d, e, f ≤ 1000).
Output
Print "Ron", if it is possible to get an infinitely large amount of gold having a certain finite amount of sand (and not having any gold and lead at all), i.e., Ron is right. Otherwise, print "Hermione".
Examples
Input
100 200 250 150 200 250
Output
Ron
Input
100 50 50 200 200 100
Output
Hermione
Input
100 10 200 20 300 30
Output
Hermione
Input
0 0 0 0 0 0
Output
Hermione
Input
1 1 0 1 1 1
Output
Ron
Input
1 0 1 2 1 2
Output
Hermione
Input
100 1 100 1 0 1
Output
Ron
Note
Consider the first sample. Let's start with the 500 grams of sand. Apply the first spell 5 times and turn the sand into 1000 grams of lead. Then apply the second spell 4 times to get 600 grams of gold. Let’s take 400 grams from the resulting amount of gold turn them back into sand. We get 500 grams of sand and 200 grams of gold. If we apply the same operations to 500 grams of sand again, we can get extra 200 grams of gold every time. Thus, you can get 200, 400, 600 etc. grams of gold, i.e., starting with a finite amount of sand (500 grams), you can get the amount of gold which is greater than any preassigned number.
In the forth sample it is impossible to get sand, or lead, or gold, applying the spells.
In the fifth sample an infinitely large amount of gold can be obtained by using only the second spell, which allows you to receive 1 gram of gold out of nothing. Note that if such a second spell is available, then the first and the third one do not affect the answer at all.
The seventh sample is more interesting. We can also start with a zero amount of sand there. With the aid of the third spell you can get sand out of nothing. We get 10000 grams of sand in this manner. Let's get 100 grams of lead using the first spell 100 times. Then make 1 gram of gold from them. We managed to receive 1 gram of gold, starting with a zero amount of sand! Clearly, in this manner you can get an infinitely large amount of gold.
Submitted Solution:
```
a, b, c, d, e, f = list(map(int, input().split(" ")))
if((e * c * a < f * d * b) or (a == 0 and b * d > 0) or (c == 0 and d > 0)):
print("Ron")
else:
print("Hermione")
``` | instruction | 0 | 45,004 | 2 | 90,008 |
Yes | output | 1 | 45,004 | 2 | 90,009 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A long time ago (probably even in the first book), Nicholas Flamel, a great alchemist and the creator of the Philosopher's Stone, taught Harry Potter three useful spells. The first one allows you to convert a grams of sand into b grams of lead, the second one allows you to convert c grams of lead into d grams of gold and third one allows you to convert e grams of gold into f grams of sand. When Harry told his friends about these spells, Ron Weasley was amazed. After all, if they succeed in turning sand into lead, lead into gold, and then turning part of the gold into sand again and so on, then it will be possible to start with a small amount of sand and get huge amounts of gold! Even an infinite amount of gold! Hermione Granger, by contrast, was skeptical about that idea. She argues that according to the law of conservation of matter getting an infinite amount of matter, even using magic, is impossible. On the contrary, the amount of matter may even decrease during transformation, being converted to magical energy. Though Hermione's theory seems convincing, Ron won't believe her. As far as Ron is concerned, Hermione made up her law of conservation of matter to stop Harry and Ron wasting their time with this nonsense, and to make them go and do homework instead. That's why Ron has already collected a certain amount of sand for the experiments. A quarrel between the friends seems unavoidable...
Help Harry to determine which one of his friends is right, and avoid the quarrel after all. To do this you have to figure out whether it is possible to get the amount of gold greater than any preassigned number from some finite amount of sand.
Input
The first line contains 6 integers a, b, c, d, e, f (0 ≤ a, b, c, d, e, f ≤ 1000).
Output
Print "Ron", if it is possible to get an infinitely large amount of gold having a certain finite amount of sand (and not having any gold and lead at all), i.e., Ron is right. Otherwise, print "Hermione".
Examples
Input
100 200 250 150 200 250
Output
Ron
Input
100 50 50 200 200 100
Output
Hermione
Input
100 10 200 20 300 30
Output
Hermione
Input
0 0 0 0 0 0
Output
Hermione
Input
1 1 0 1 1 1
Output
Ron
Input
1 0 1 2 1 2
Output
Hermione
Input
100 1 100 1 0 1
Output
Ron
Note
Consider the first sample. Let's start with the 500 grams of sand. Apply the first spell 5 times and turn the sand into 1000 grams of lead. Then apply the second spell 4 times to get 600 grams of gold. Let’s take 400 grams from the resulting amount of gold turn them back into sand. We get 500 grams of sand and 200 grams of gold. If we apply the same operations to 500 grams of sand again, we can get extra 200 grams of gold every time. Thus, you can get 200, 400, 600 etc. grams of gold, i.e., starting with a finite amount of sand (500 grams), you can get the amount of gold which is greater than any preassigned number.
In the forth sample it is impossible to get sand, or lead, or gold, applying the spells.
In the fifth sample an infinitely large amount of gold can be obtained by using only the second spell, which allows you to receive 1 gram of gold out of nothing. Note that if such a second spell is available, then the first and the third one do not affect the answer at all.
The seventh sample is more interesting. We can also start with a zero amount of sand there. With the aid of the third spell you can get sand out of nothing. We get 10000 grams of sand in this manner. Let's get 100 grams of lead using the first spell 100 times. Then make 1 gram of gold from them. We managed to receive 1 gram of gold, starting with a zero amount of sand! Clearly, in this manner you can get an infinitely large amount of gold.
Submitted Solution:
```
nums = [int(r) for r in input().split()]
if nums[3] > 0 and nums[2] == 0:
print("Ron")
exit()
if nums[1]*nums[3]*nums[5] == 0:
print("Hermione")
exit()
elif nums[0]*nums[2]*nums[4] == 0:
print("Ron")
exit()
val = nums[1]*nums[3]*nums[5]/(nums[0]*nums[2]*nums[4])
if val > 1:
print("Ron")
else:
print("Hermione")
``` | instruction | 0 | 45,005 | 2 | 90,010 |
No | output | 1 | 45,005 | 2 | 90,011 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A long time ago (probably even in the first book), Nicholas Flamel, a great alchemist and the creator of the Philosopher's Stone, taught Harry Potter three useful spells. The first one allows you to convert a grams of sand into b grams of lead, the second one allows you to convert c grams of lead into d grams of gold and third one allows you to convert e grams of gold into f grams of sand. When Harry told his friends about these spells, Ron Weasley was amazed. After all, if they succeed in turning sand into lead, lead into gold, and then turning part of the gold into sand again and so on, then it will be possible to start with a small amount of sand and get huge amounts of gold! Even an infinite amount of gold! Hermione Granger, by contrast, was skeptical about that idea. She argues that according to the law of conservation of matter getting an infinite amount of matter, even using magic, is impossible. On the contrary, the amount of matter may even decrease during transformation, being converted to magical energy. Though Hermione's theory seems convincing, Ron won't believe her. As far as Ron is concerned, Hermione made up her law of conservation of matter to stop Harry and Ron wasting their time with this nonsense, and to make them go and do homework instead. That's why Ron has already collected a certain amount of sand for the experiments. A quarrel between the friends seems unavoidable...
Help Harry to determine which one of his friends is right, and avoid the quarrel after all. To do this you have to figure out whether it is possible to get the amount of gold greater than any preassigned number from some finite amount of sand.
Input
The first line contains 6 integers a, b, c, d, e, f (0 ≤ a, b, c, d, e, f ≤ 1000).
Output
Print "Ron", if it is possible to get an infinitely large amount of gold having a certain finite amount of sand (and not having any gold and lead at all), i.e., Ron is right. Otherwise, print "Hermione".
Examples
Input
100 200 250 150 200 250
Output
Ron
Input
100 50 50 200 200 100
Output
Hermione
Input
100 10 200 20 300 30
Output
Hermione
Input
0 0 0 0 0 0
Output
Hermione
Input
1 1 0 1 1 1
Output
Ron
Input
1 0 1 2 1 2
Output
Hermione
Input
100 1 100 1 0 1
Output
Ron
Note
Consider the first sample. Let's start with the 500 grams of sand. Apply the first spell 5 times and turn the sand into 1000 grams of lead. Then apply the second spell 4 times to get 600 grams of gold. Let’s take 400 grams from the resulting amount of gold turn them back into sand. We get 500 grams of sand and 200 grams of gold. If we apply the same operations to 500 grams of sand again, we can get extra 200 grams of gold every time. Thus, you can get 200, 400, 600 etc. grams of gold, i.e., starting with a finite amount of sand (500 grams), you can get the amount of gold which is greater than any preassigned number.
In the forth sample it is impossible to get sand, or lead, or gold, applying the spells.
In the fifth sample an infinitely large amount of gold can be obtained by using only the second spell, which allows you to receive 1 gram of gold out of nothing. Note that if such a second spell is available, then the first and the third one do not affect the answer at all.
The seventh sample is more interesting. We can also start with a zero amount of sand there. With the aid of the third spell you can get sand out of nothing. We get 10000 grams of sand in this manner. Let's get 100 grams of lead using the first spell 100 times. Then make 1 gram of gold from them. We managed to receive 1 gram of gold, starting with a zero amount of sand! Clearly, in this manner you can get an infinitely large amount of gold.
Submitted Solution:
```
a,b,c,d,e,f=map(int,input().split());
if (b*d==0):
print ("Hermione")
exit(0)
if (a*c*e==0):
print ("Ron")
exit(0)
if ((b/a)*(d/c)*(f/e)>1):
print ("Ron")
else:
print ("Hermione")
``` | instruction | 0 | 45,006 | 2 | 90,012 |
No | output | 1 | 45,006 | 2 | 90,013 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A long time ago (probably even in the first book), Nicholas Flamel, a great alchemist and the creator of the Philosopher's Stone, taught Harry Potter three useful spells. The first one allows you to convert a grams of sand into b grams of lead, the second one allows you to convert c grams of lead into d grams of gold and third one allows you to convert e grams of gold into f grams of sand. When Harry told his friends about these spells, Ron Weasley was amazed. After all, if they succeed in turning sand into lead, lead into gold, and then turning part of the gold into sand again and so on, then it will be possible to start with a small amount of sand and get huge amounts of gold! Even an infinite amount of gold! Hermione Granger, by contrast, was skeptical about that idea. She argues that according to the law of conservation of matter getting an infinite amount of matter, even using magic, is impossible. On the contrary, the amount of matter may even decrease during transformation, being converted to magical energy. Though Hermione's theory seems convincing, Ron won't believe her. As far as Ron is concerned, Hermione made up her law of conservation of matter to stop Harry and Ron wasting their time with this nonsense, and to make them go and do homework instead. That's why Ron has already collected a certain amount of sand for the experiments. A quarrel between the friends seems unavoidable...
Help Harry to determine which one of his friends is right, and avoid the quarrel after all. To do this you have to figure out whether it is possible to get the amount of gold greater than any preassigned number from some finite amount of sand.
Input
The first line contains 6 integers a, b, c, d, e, f (0 ≤ a, b, c, d, e, f ≤ 1000).
Output
Print "Ron", if it is possible to get an infinitely large amount of gold having a certain finite amount of sand (and not having any gold and lead at all), i.e., Ron is right. Otherwise, print "Hermione".
Examples
Input
100 200 250 150 200 250
Output
Ron
Input
100 50 50 200 200 100
Output
Hermione
Input
100 10 200 20 300 30
Output
Hermione
Input
0 0 0 0 0 0
Output
Hermione
Input
1 1 0 1 1 1
Output
Ron
Input
1 0 1 2 1 2
Output
Hermione
Input
100 1 100 1 0 1
Output
Ron
Note
Consider the first sample. Let's start with the 500 grams of sand. Apply the first spell 5 times and turn the sand into 1000 grams of lead. Then apply the second spell 4 times to get 600 grams of gold. Let’s take 400 grams from the resulting amount of gold turn them back into sand. We get 500 grams of sand and 200 grams of gold. If we apply the same operations to 500 grams of sand again, we can get extra 200 grams of gold every time. Thus, you can get 200, 400, 600 etc. grams of gold, i.e., starting with a finite amount of sand (500 grams), you can get the amount of gold which is greater than any preassigned number.
In the forth sample it is impossible to get sand, or lead, or gold, applying the spells.
In the fifth sample an infinitely large amount of gold can be obtained by using only the second spell, which allows you to receive 1 gram of gold out of nothing. Note that if such a second spell is available, then the first and the third one do not affect the answer at all.
The seventh sample is more interesting. We can also start with a zero amount of sand there. With the aid of the third spell you can get sand out of nothing. We get 10000 grams of sand in this manner. Let's get 100 grams of lead using the first spell 100 times. Then make 1 gram of gold from them. We managed to receive 1 gram of gold, starting with a zero amount of sand! Clearly, in this manner you can get an infinitely large amount of gold.
Submitted Solution:
```
a,b,c,d,e,f = map(int, input().split())
if (b==0)or(d==0)or(f==0):
print('Hermione')
elif (a==0)or(c==0)or(e==0):
print('Ron')
else :
x = (d*b)/c
y = (x*f)/e
if y>a :
print('Ron')
else :
print('Hermione')
``` | instruction | 0 | 45,007 | 2 | 90,014 |
No | output | 1 | 45,007 | 2 | 90,015 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A long time ago (probably even in the first book), Nicholas Flamel, a great alchemist and the creator of the Philosopher's Stone, taught Harry Potter three useful spells. The first one allows you to convert a grams of sand into b grams of lead, the second one allows you to convert c grams of lead into d grams of gold and third one allows you to convert e grams of gold into f grams of sand. When Harry told his friends about these spells, Ron Weasley was amazed. After all, if they succeed in turning sand into lead, lead into gold, and then turning part of the gold into sand again and so on, then it will be possible to start with a small amount of sand and get huge amounts of gold! Even an infinite amount of gold! Hermione Granger, by contrast, was skeptical about that idea. She argues that according to the law of conservation of matter getting an infinite amount of matter, even using magic, is impossible. On the contrary, the amount of matter may even decrease during transformation, being converted to magical energy. Though Hermione's theory seems convincing, Ron won't believe her. As far as Ron is concerned, Hermione made up her law of conservation of matter to stop Harry and Ron wasting their time with this nonsense, and to make them go and do homework instead. That's why Ron has already collected a certain amount of sand for the experiments. A quarrel between the friends seems unavoidable...
Help Harry to determine which one of his friends is right, and avoid the quarrel after all. To do this you have to figure out whether it is possible to get the amount of gold greater than any preassigned number from some finite amount of sand.
Input
The first line contains 6 integers a, b, c, d, e, f (0 ≤ a, b, c, d, e, f ≤ 1000).
Output
Print "Ron", if it is possible to get an infinitely large amount of gold having a certain finite amount of sand (and not having any gold and lead at all), i.e., Ron is right. Otherwise, print "Hermione".
Examples
Input
100 200 250 150 200 250
Output
Ron
Input
100 50 50 200 200 100
Output
Hermione
Input
100 10 200 20 300 30
Output
Hermione
Input
0 0 0 0 0 0
Output
Hermione
Input
1 1 0 1 1 1
Output
Ron
Input
1 0 1 2 1 2
Output
Hermione
Input
100 1 100 1 0 1
Output
Ron
Note
Consider the first sample. Let's start with the 500 grams of sand. Apply the first spell 5 times and turn the sand into 1000 grams of lead. Then apply the second spell 4 times to get 600 grams of gold. Let’s take 400 grams from the resulting amount of gold turn them back into sand. We get 500 grams of sand and 200 grams of gold. If we apply the same operations to 500 grams of sand again, we can get extra 200 grams of gold every time. Thus, you can get 200, 400, 600 etc. grams of gold, i.e., starting with a finite amount of sand (500 grams), you can get the amount of gold which is greater than any preassigned number.
In the forth sample it is impossible to get sand, or lead, or gold, applying the spells.
In the fifth sample an infinitely large amount of gold can be obtained by using only the second spell, which allows you to receive 1 gram of gold out of nothing. Note that if such a second spell is available, then the first and the third one do not affect the answer at all.
The seventh sample is more interesting. We can also start with a zero amount of sand there. With the aid of the third spell you can get sand out of nothing. We get 10000 grams of sand in this manner. Let's get 100 grams of lead using the first spell 100 times. Then make 1 gram of gold from them. We managed to receive 1 gram of gold, starting with a zero amount of sand! Clearly, in this manner you can get an infinitely large amount of gold.
Submitted Solution:
```
a,b,c,d,e,f = map(int,input().strip().split())
if c != 0 and e != 0:
if (b*d/c)*f/e > a : print('Ron')
else : print('Hermione')
elif a == b == c == d == e == f == 0 : print('Hermione')
elif (a !=0 and b == 0) : print('Hermione')
else : print('Ron')
``` | instruction | 0 | 45,008 | 2 | 90,016 |
No | output | 1 | 45,008 | 2 | 90,017 |
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