text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have x ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with d ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take d ice cream packs comes to the house, then Kay and Gerda will give him d packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
Input
The first line contains two space-separated integers n and x (1 ≤ n ≤ 1000, 0 ≤ x ≤ 109).
Each of the next n lines contains a character '+' or '-', and an integer di, separated by a space (1 ≤ di ≤ 109). Record "+ di" in i-th line means that a carrier with di ice cream packs occupies i-th place from the start of the queue, and record "- di" means that a child who wants to take di packs stands in i-th place.
Output
Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.
Examples
Input
5 7
+ 5
- 10
- 20
+ 40
- 20
Output
22 1
Input
5 17
- 16
- 2
- 98
+ 100
- 98
Output
3 2
Note
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream.
2. Carrier brings 5 more, so now they have 12 packs.
3. A kid asks for 10 packs and receives them. There are only 2 packs remaining.
4. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed.
5. Carrier bring 40 packs, now Kay and Gerda have 42 packs.
6. Kid asks for 20 packs and receives them. There are 22 packs remaining.
Tags: constructive algorithms, implementation
Correct Solution:
```
n, x = map(int,input().split())
c = 0
for i in range(n):
a, b = input().split()
b = int(b)
if a == '+':
x += b
else:
if x<b:
c+=1
else:
x-=b
print(x,c)
```
| 88,000 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have x ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with d ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take d ice cream packs comes to the house, then Kay and Gerda will give him d packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
Input
The first line contains two space-separated integers n and x (1 ≤ n ≤ 1000, 0 ≤ x ≤ 109).
Each of the next n lines contains a character '+' or '-', and an integer di, separated by a space (1 ≤ di ≤ 109). Record "+ di" in i-th line means that a carrier with di ice cream packs occupies i-th place from the start of the queue, and record "- di" means that a child who wants to take di packs stands in i-th place.
Output
Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.
Examples
Input
5 7
+ 5
- 10
- 20
+ 40
- 20
Output
22 1
Input
5 17
- 16
- 2
- 98
+ 100
- 98
Output
3 2
Note
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream.
2. Carrier brings 5 more, so now they have 12 packs.
3. A kid asks for 10 packs and receives them. There are only 2 packs remaining.
4. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed.
5. Carrier bring 40 packs, now Kay and Gerda have 42 packs.
6. Kid asks for 20 packs and receives them. There are 22 packs remaining.
Tags: constructive algorithms, implementation
Correct Solution:
```
n,x=(int(i) for i in input().split())
dis=0
ice=x
for i in range(n):
a,b=(input().split())
b=int(b)
if (a=='+'):
ice+=b
else:
if (b>ice):
dis+=1
else:
ice-=b
print("{} {}".format(ice,dis))
```
| 88,001 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have x ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with d ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take d ice cream packs comes to the house, then Kay and Gerda will give him d packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
Input
The first line contains two space-separated integers n and x (1 ≤ n ≤ 1000, 0 ≤ x ≤ 109).
Each of the next n lines contains a character '+' or '-', and an integer di, separated by a space (1 ≤ di ≤ 109). Record "+ di" in i-th line means that a carrier with di ice cream packs occupies i-th place from the start of the queue, and record "- di" means that a child who wants to take di packs stands in i-th place.
Output
Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.
Examples
Input
5 7
+ 5
- 10
- 20
+ 40
- 20
Output
22 1
Input
5 17
- 16
- 2
- 98
+ 100
- 98
Output
3 2
Note
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream.
2. Carrier brings 5 more, so now they have 12 packs.
3. A kid asks for 10 packs and receives them. There are only 2 packs remaining.
4. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed.
5. Carrier bring 40 packs, now Kay and Gerda have 42 packs.
6. Kid asks for 20 packs and receives them. There are 22 packs remaining.
Tags: constructive algorithms, implementation
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
def solve() :
n,x = map(int,input().split())
ini = x
dis = 0
for i in range(n) :
q = input()
if q[0]=="+" :
ini = ini + int(q[2:])
elif q[0]=="-" :
if ini>=int(q[2:]) :
ini = ini - int(q[2:])
else :
dis = dis + 1
print(ini,dis)
def main():
solve()
# 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")
# endregion
if __name__ == "__main__":
main()
```
| 88,002 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have x ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with d ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take d ice cream packs comes to the house, then Kay and Gerda will give him d packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
Input
The first line contains two space-separated integers n and x (1 ≤ n ≤ 1000, 0 ≤ x ≤ 109).
Each of the next n lines contains a character '+' or '-', and an integer di, separated by a space (1 ≤ di ≤ 109). Record "+ di" in i-th line means that a carrier with di ice cream packs occupies i-th place from the start of the queue, and record "- di" means that a child who wants to take di packs stands in i-th place.
Output
Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.
Examples
Input
5 7
+ 5
- 10
- 20
+ 40
- 20
Output
22 1
Input
5 17
- 16
- 2
- 98
+ 100
- 98
Output
3 2
Note
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream.
2. Carrier brings 5 more, so now they have 12 packs.
3. A kid asks for 10 packs and receives them. There are only 2 packs remaining.
4. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed.
5. Carrier bring 40 packs, now Kay and Gerda have 42 packs.
6. Kid asks for 20 packs and receives them. There are 22 packs remaining.
Tags: constructive algorithms, implementation
Correct Solution:
```
a, b = input().split(" ")
a = int(a)
b = int(b)
f=0
sum = 0
s=b
e=0
for i in range(a):
c,d = input().split(" ")
d = int(d)
if c == '+':
s += d
else:
if s-d<0:
f += 1
else:
s= s-d
print(s, f)
```
| 88,003 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have x ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with d ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take d ice cream packs comes to the house, then Kay and Gerda will give him d packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
Input
The first line contains two space-separated integers n and x (1 ≤ n ≤ 1000, 0 ≤ x ≤ 109).
Each of the next n lines contains a character '+' or '-', and an integer di, separated by a space (1 ≤ di ≤ 109). Record "+ di" in i-th line means that a carrier with di ice cream packs occupies i-th place from the start of the queue, and record "- di" means that a child who wants to take di packs stands in i-th place.
Output
Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.
Examples
Input
5 7
+ 5
- 10
- 20
+ 40
- 20
Output
22 1
Input
5 17
- 16
- 2
- 98
+ 100
- 98
Output
3 2
Note
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream.
2. Carrier brings 5 more, so now they have 12 packs.
3. A kid asks for 10 packs and receives them. There are only 2 packs remaining.
4. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed.
5. Carrier bring 40 packs, now Kay and Gerda have 42 packs.
6. Kid asks for 20 packs and receives them. There are 22 packs remaining.
Tags: constructive algorithms, implementation
Correct Solution:
```
n,d=list(map(int,input().split()))
c=0
for i in range(n):
r,p=input().split()
if r=='+':
d+=int(p)
elif r=='-':
if int(p)<=d:
d-=int(p)
else:
c+=1
print(d,c)
```
| 88,004 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have x ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with d ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take d ice cream packs comes to the house, then Kay and Gerda will give him d packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
Input
The first line contains two space-separated integers n and x (1 ≤ n ≤ 1000, 0 ≤ x ≤ 109).
Each of the next n lines contains a character '+' or '-', and an integer di, separated by a space (1 ≤ di ≤ 109). Record "+ di" in i-th line means that a carrier with di ice cream packs occupies i-th place from the start of the queue, and record "- di" means that a child who wants to take di packs stands in i-th place.
Output
Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.
Examples
Input
5 7
+ 5
- 10
- 20
+ 40
- 20
Output
22 1
Input
5 17
- 16
- 2
- 98
+ 100
- 98
Output
3 2
Note
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream.
2. Carrier brings 5 more, so now they have 12 packs.
3. A kid asks for 10 packs and receives them. There are only 2 packs remaining.
4. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed.
5. Carrier bring 40 packs, now Kay and Gerda have 42 packs.
6. Kid asks for 20 packs and receives them. There are 22 packs remaining.
Tags: constructive algorithms, implementation
Correct Solution:
```
n,x = map(int,input().split())
cnt = 0
for i in range(n):
s,d = input().split()
d = int(d)
if s == '+':
x += d
elif s == '-':
if x >= d :
x-=d
elif x<d:
cnt +=1
print(str(x) + " " + str(cnt))
```
| 88,005 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have x ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with d ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take d ice cream packs comes to the house, then Kay and Gerda will give him d packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
Input
The first line contains two space-separated integers n and x (1 ≤ n ≤ 1000, 0 ≤ x ≤ 109).
Each of the next n lines contains a character '+' or '-', and an integer di, separated by a space (1 ≤ di ≤ 109). Record "+ di" in i-th line means that a carrier with di ice cream packs occupies i-th place from the start of the queue, and record "- di" means that a child who wants to take di packs stands in i-th place.
Output
Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.
Examples
Input
5 7
+ 5
- 10
- 20
+ 40
- 20
Output
22 1
Input
5 17
- 16
- 2
- 98
+ 100
- 98
Output
3 2
Note
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream.
2. Carrier brings 5 more, so now they have 12 packs.
3. A kid asks for 10 packs and receives them. There are only 2 packs remaining.
4. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed.
5. Carrier bring 40 packs, now Kay and Gerda have 42 packs.
6. Kid asks for 20 packs and receives them. There are 22 packs remaining.
Tags: constructive algorithms, implementation
Correct Solution:
```
n,x = [int(y) for y in input().split(' ')]
d = 0
for nn in range(n):
p,q = [y for y in input().split(' ')]
if p == '+': x += int(q)
else:
q = int(q)
if q > x: d += 1
else: x -= q
print(x,d)
```
| 88,006 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have x ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with d ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take d ice cream packs comes to the house, then Kay and Gerda will give him d packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
Input
The first line contains two space-separated integers n and x (1 ≤ n ≤ 1000, 0 ≤ x ≤ 109).
Each of the next n lines contains a character '+' or '-', and an integer di, separated by a space (1 ≤ di ≤ 109). Record "+ di" in i-th line means that a carrier with di ice cream packs occupies i-th place from the start of the queue, and record "- di" means that a child who wants to take di packs stands in i-th place.
Output
Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.
Examples
Input
5 7
+ 5
- 10
- 20
+ 40
- 20
Output
22 1
Input
5 17
- 16
- 2
- 98
+ 100
- 98
Output
3 2
Note
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream.
2. Carrier brings 5 more, so now they have 12 packs.
3. A kid asks for 10 packs and receives them. There are only 2 packs remaining.
4. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed.
5. Carrier bring 40 packs, now Kay and Gerda have 42 packs.
6. Kid asks for 20 packs and receives them. There are 22 packs remaining.
Tags: constructive algorithms, implementation
Correct Solution:
```
a,Sum = map(int , input().split())
n=0
for i in range(a):
x,y = input().split()
if x == '+': Sum+=int(y)
else:
if Sum >= int(y):Sum-=int(y)
else :n+=1
print(Sum , n)
```
| 88,007 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have x ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with d ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take d ice cream packs comes to the house, then Kay and Gerda will give him d packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
Input
The first line contains two space-separated integers n and x (1 ≤ n ≤ 1000, 0 ≤ x ≤ 109).
Each of the next n lines contains a character '+' or '-', and an integer di, separated by a space (1 ≤ di ≤ 109). Record "+ di" in i-th line means that a carrier with di ice cream packs occupies i-th place from the start of the queue, and record "- di" means that a child who wants to take di packs stands in i-th place.
Output
Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.
Examples
Input
5 7
+ 5
- 10
- 20
+ 40
- 20
Output
22 1
Input
5 17
- 16
- 2
- 98
+ 100
- 98
Output
3 2
Note
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream.
2. Carrier brings 5 more, so now they have 12 packs.
3. A kid asks for 10 packs and receives them. There are only 2 packs remaining.
4. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed.
5. Carrier bring 40 packs, now Kay and Gerda have 42 packs.
6. Kid asks for 20 packs and receives them. There are 22 packs remaining.
Submitted Solution:
```
x = list(map(int,input().split()))
n = x[0]
ice = x[1]
temp = []
dis = 0
for i in range(n):
y=list(input().split())
temp.append(y)
for i in temp:
if i[0] == "-":
if int(i[1]) <= ice:
ice -= int(i[1])
else:
dis += 1
else:
ice += int(i[1])
print(str(ice)+" "+str(dis))
```
Yes
| 88,008 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have x ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with d ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take d ice cream packs comes to the house, then Kay and Gerda will give him d packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
Input
The first line contains two space-separated integers n and x (1 ≤ n ≤ 1000, 0 ≤ x ≤ 109).
Each of the next n lines contains a character '+' or '-', and an integer di, separated by a space (1 ≤ di ≤ 109). Record "+ di" in i-th line means that a carrier with di ice cream packs occupies i-th place from the start of the queue, and record "- di" means that a child who wants to take di packs stands in i-th place.
Output
Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.
Examples
Input
5 7
+ 5
- 10
- 20
+ 40
- 20
Output
22 1
Input
5 17
- 16
- 2
- 98
+ 100
- 98
Output
3 2
Note
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream.
2. Carrier brings 5 more, so now they have 12 packs.
3. A kid asks for 10 packs and receives them. There are only 2 packs remaining.
4. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed.
5. Carrier bring 40 packs, now Kay and Gerda have 42 packs.
6. Kid asks for 20 packs and receives them. There are 22 packs remaining.
Submitted Solution:
```
n, x = (int(v) for v in input().split())
ans = 0
for _ in range(n):
inp = input().split()
val = int(inp[1])
sign = inp[0]
if sign == '+':
x += val
else:
if x < val:
ans += 1
else:
x -= val
print(x, ans)
```
Yes
| 88,009 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have x ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with d ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take d ice cream packs comes to the house, then Kay and Gerda will give him d packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
Input
The first line contains two space-separated integers n and x (1 ≤ n ≤ 1000, 0 ≤ x ≤ 109).
Each of the next n lines contains a character '+' or '-', and an integer di, separated by a space (1 ≤ di ≤ 109). Record "+ di" in i-th line means that a carrier with di ice cream packs occupies i-th place from the start of the queue, and record "- di" means that a child who wants to take di packs stands in i-th place.
Output
Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.
Examples
Input
5 7
+ 5
- 10
- 20
+ 40
- 20
Output
22 1
Input
5 17
- 16
- 2
- 98
+ 100
- 98
Output
3 2
Note
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream.
2. Carrier brings 5 more, so now they have 12 packs.
3. A kid asks for 10 packs and receives them. There are only 2 packs remaining.
4. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed.
5. Carrier bring 40 packs, now Kay and Gerda have 42 packs.
6. Kid asks for 20 packs and receives them. There are 22 packs remaining.
Submitted Solution:
```
n, sum_ = map(int, input().split())
cnt = 0
for i in range(n):
op, d = input().split()
d = int(d)
sum_ += d if op == "+" else -d
if sum_ < 0:
cnt += 1
sum_ += d
print(f"{sum_} {cnt}")
```
Yes
| 88,010 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have x ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with d ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take d ice cream packs comes to the house, then Kay and Gerda will give him d packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
Input
The first line contains two space-separated integers n and x (1 ≤ n ≤ 1000, 0 ≤ x ≤ 109).
Each of the next n lines contains a character '+' or '-', and an integer di, separated by a space (1 ≤ di ≤ 109). Record "+ di" in i-th line means that a carrier with di ice cream packs occupies i-th place from the start of the queue, and record "- di" means that a child who wants to take di packs stands in i-th place.
Output
Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.
Examples
Input
5 7
+ 5
- 10
- 20
+ 40
- 20
Output
22 1
Input
5 17
- 16
- 2
- 98
+ 100
- 98
Output
3 2
Note
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream.
2. Carrier brings 5 more, so now they have 12 packs.
3. A kid asks for 10 packs and receives them. There are only 2 packs remaining.
4. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed.
5. Carrier bring 40 packs, now Kay and Gerda have 42 packs.
6. Kid asks for 20 packs and receives them. There are 22 packs remaining.
Submitted Solution:
```
q, x = map(int, (input().split()))
sad_child = 0
for _ in range(q):
n = int(input().replace(' ', ''))
if n >= 0:
x += n
# print(x)
elif n < 0 and x >= abs(n):
x -= abs(n)
# print(x)
elif n < 0 and x < abs(n):
sad_child += 1
# print(x)
print(x, sad_child)
```
Yes
| 88,011 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have x ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with d ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take d ice cream packs comes to the house, then Kay and Gerda will give him d packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
Input
The first line contains two space-separated integers n and x (1 ≤ n ≤ 1000, 0 ≤ x ≤ 109).
Each of the next n lines contains a character '+' or '-', and an integer di, separated by a space (1 ≤ di ≤ 109). Record "+ di" in i-th line means that a carrier with di ice cream packs occupies i-th place from the start of the queue, and record "- di" means that a child who wants to take di packs stands in i-th place.
Output
Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.
Examples
Input
5 7
+ 5
- 10
- 20
+ 40
- 20
Output
22 1
Input
5 17
- 16
- 2
- 98
+ 100
- 98
Output
3 2
Note
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream.
2. Carrier brings 5 more, so now they have 12 packs.
3. A kid asks for 10 packs and receives them. There are only 2 packs remaining.
4. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed.
5. Carrier bring 40 packs, now Kay and Gerda have 42 packs.
6. Kid asks for 20 packs and receives them. There are 22 packs remaining.
Submitted Solution:
```
n, x = map(int,input().split(' '))
n = int(n)
x = int(x)
icVal = x
dCount = 0
while n > 0:
a, b = input().split(' ')
b = int(b)
if a == '-':
b = b * -1
if (icVal + b) > 0:
icVal = icVal + b
else:
dCount = dCount + 1
n = n - 1
print(str(icVal) + " " + str(dCount))
```
No
| 88,012 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have x ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with d ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take d ice cream packs comes to the house, then Kay and Gerda will give him d packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
Input
The first line contains two space-separated integers n and x (1 ≤ n ≤ 1000, 0 ≤ x ≤ 109).
Each of the next n lines contains a character '+' or '-', and an integer di, separated by a space (1 ≤ di ≤ 109). Record "+ di" in i-th line means that a carrier with di ice cream packs occupies i-th place from the start of the queue, and record "- di" means that a child who wants to take di packs stands in i-th place.
Output
Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.
Examples
Input
5 7
+ 5
- 10
- 20
+ 40
- 20
Output
22 1
Input
5 17
- 16
- 2
- 98
+ 100
- 98
Output
3 2
Note
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream.
2. Carrier brings 5 more, so now they have 12 packs.
3. A kid asks for 10 packs and receives them. There are only 2 packs remaining.
4. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed.
5. Carrier bring 40 packs, now Kay and Gerda have 42 packs.
6. Kid asks for 20 packs and receives them. There are 22 packs remaining.
Submitted Solution:
```
n,d=[int(i) for i in input().split()]
for i in range(n):
y=0
x=input().split()
if x[0]=='+':
p=int(x[1])
d=d+p
else:
p=int(x[1])
if d>=p:
d=d-p
else:
y=y+1
print(d,y)
```
No
| 88,013 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have x ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with d ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take d ice cream packs comes to the house, then Kay and Gerda will give him d packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
Input
The first line contains two space-separated integers n and x (1 ≤ n ≤ 1000, 0 ≤ x ≤ 109).
Each of the next n lines contains a character '+' or '-', and an integer di, separated by a space (1 ≤ di ≤ 109). Record "+ di" in i-th line means that a carrier with di ice cream packs occupies i-th place from the start of the queue, and record "- di" means that a child who wants to take di packs stands in i-th place.
Output
Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.
Examples
Input
5 7
+ 5
- 10
- 20
+ 40
- 20
Output
22 1
Input
5 17
- 16
- 2
- 98
+ 100
- 98
Output
3 2
Note
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream.
2. Carrier brings 5 more, so now they have 12 packs.
3. A kid asks for 10 packs and receives them. There are only 2 packs remaining.
4. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed.
5. Carrier bring 40 packs, now Kay and Gerda have 42 packs.
6. Kid asks for 20 packs and receives them. There are 22 packs remaining.
Submitted Solution:
```
#http://codeforces.com/problemset/problem/686/A
inp = list(map(int,input().split()))
tCases = inp[0]
iceCream = inp[1]
store = []
distress = 0
for x in range(tCases):
temp = list(input().split())
store.append(temp)
for x in store:
if x[0] == "-":
if int(x[1]) <iceCream:
iceCream -= int(x[1])
else:
distress += 1
else:
iceCream += int(x[1])
print(str(iceCream) + " " + str(distress))
```
No
| 88,014 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have x ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with d ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take d ice cream packs comes to the house, then Kay and Gerda will give him d packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
Input
The first line contains two space-separated integers n and x (1 ≤ n ≤ 1000, 0 ≤ x ≤ 109).
Each of the next n lines contains a character '+' or '-', and an integer di, separated by a space (1 ≤ di ≤ 109). Record "+ di" in i-th line means that a carrier with di ice cream packs occupies i-th place from the start of the queue, and record "- di" means that a child who wants to take di packs stands in i-th place.
Output
Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.
Examples
Input
5 7
+ 5
- 10
- 20
+ 40
- 20
Output
22 1
Input
5 17
- 16
- 2
- 98
+ 100
- 98
Output
3 2
Note
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream.
2. Carrier brings 5 more, so now they have 12 packs.
3. A kid asks for 10 packs and receives them. There are only 2 packs remaining.
4. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed.
5. Carrier bring 40 packs, now Kay and Gerda have 42 packs.
6. Kid asks for 20 packs and receives them. There are 22 packs remaining.
Submitted Solution:
```
n,x=[int(i) for i in input().split(" ")]
distress=0
for i in range(n):
query=input()
if query[0]=="+":
x+=int(query[2])
elif query[0]=="-":
if x-int(query[2])<0:
distress+=1
else:
x-=int(query[2])
print(x,distress)
```
No
| 88,015 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem. You have to use flush operation right after printing each line. For example, in C++ you should use function fflush(stdout), in Java — System.out.flush(), in Pascal — flush(output) and in Python — sys.stdout.flush().
In this problem, you need to find maximal and minimal elements of an array. What could be simpler?
You can imagine that the jury has an array, and initially you know the only number n — array's length.
Array's elements are numbered from 1 to n. You are allowed to compare two elements of the array by using their indices i and j. There are three possible responses to this query: '<' (if ai is less than aj), '=' (if ai is equal to aj) and finally '>' (if ai is greater than aj).
It's known that it's always possible to find both maximal and minimal elements of the array by using no more than <image> comparisons, where ⌈ x⌉ is the result of rounding x up.
Write the program that will find positions of the minimum and the maximum in the jury's array of length n, by using no more than f(n) comparisons.
Interaction
Each test for this problem will contain one or more arrays. You have to find positions of minimal and maximal elements for each of these arrays. The first line of the input contains integer T (1 ≤ T ≤ 1000) — number of arrays in the test.
Thus, at the beginning, you program should read number T, and then it should solve the problem for T jury's arrays one by one.
Then input for each array goes. Firstly, your program has to read the number n (1 ≤ n ≤ 50) — the length of the array. It will be provided in the next line of the input.
Further, your program can perform comparisons or report that the answer is found.
* To perform a comparison, you have to output string of the following pattern «? i j» (i and j must be integer numbers from 1 to n) — the indices of the elements to compare in the current query.
* To report the indices of minimal and maximal elements of the hidden array, your program have to output a line in the form «! i j» (i and j must be integer numbers from 1 to n), where i is an index of the minimal element of array, and j is an index of the maximal element of the array. If there are several possible answers to the problem, you can output any of them.
There are several possible responses for a comparison:
* '<' — if ai is less than aj,
* '=' — if ai is equal to aj,
* '>' — if ai is greater than aj.
For an array of length n your program can make at most <image> comparisons. Note that the operation of reporting an answer («! i j» ) is not included into the value of f(n).
After the answer is reported, your program has to solve the problem for the next array or it should terminate if all T arrays are processed.
Example
Input
2
2
>
3
=
=
Output
? 1 2
! 2 1
? 3 1
? 2 1
! 2 3
Tags: constructive algorithms, interactive
Correct Solution:
```
from collections import defaultdict
for ik in range(int(input())):
n=int(input())
d=defaultdict(str)
if n==1:
print("! 1 1",flush=True)
continue
elif n==2:
print("? 1 2",flush=True)
s=input()
if s==">":
print("! 2 1",flush=True)
else:
print("! 1 2",flush=True)
continue
last=[i for i in range(1,n+1)]
res=[]
while(len(last)!=1):
for i in range(len(last)//2):
print("?", last[2*i], last[2*i+1], flush=True)
r=input()
d[(last[2 * i], last[2 * i + 1])] = r
if r=='>':
res.append(last[2*i])
else:
res.append(last[2*i+1])
if len(last)%2==1:
res.append(last[-1])
last=res+[]
res=[]
max=last[0]
#print(max)
last = [i for i in range(1, n+1)]
res = []
while (len(last) != 1):
for i in range(len(last)// 2):
if (last[2*i],last[2*i+1]) in d:
r1=d[(last[2*i],last[2*i+1])]
else:
print("?", last[2 * i], last[2 * i + 1], flush=True)
r1 = input()
if r1 == '<':
res.append(last[2 * i])
else:
res.append(last[2 * i + 1])
if len(last) % 2 == 1:
res.append(last[-1])
last = res + []
res = []
min = last[0]
print('!',min,max,flush=True)
```
| 88,016 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem. You have to use flush operation right after printing each line. For example, in C++ you should use function fflush(stdout), in Java — System.out.flush(), in Pascal — flush(output) and in Python — sys.stdout.flush().
In this problem, you need to find maximal and minimal elements of an array. What could be simpler?
You can imagine that the jury has an array, and initially you know the only number n — array's length.
Array's elements are numbered from 1 to n. You are allowed to compare two elements of the array by using their indices i and j. There are three possible responses to this query: '<' (if ai is less than aj), '=' (if ai is equal to aj) and finally '>' (if ai is greater than aj).
It's known that it's always possible to find both maximal and minimal elements of the array by using no more than <image> comparisons, where ⌈ x⌉ is the result of rounding x up.
Write the program that will find positions of the minimum and the maximum in the jury's array of length n, by using no more than f(n) comparisons.
Interaction
Each test for this problem will contain one or more arrays. You have to find positions of minimal and maximal elements for each of these arrays. The first line of the input contains integer T (1 ≤ T ≤ 1000) — number of arrays in the test.
Thus, at the beginning, you program should read number T, and then it should solve the problem for T jury's arrays one by one.
Then input for each array goes. Firstly, your program has to read the number n (1 ≤ n ≤ 50) — the length of the array. It will be provided in the next line of the input.
Further, your program can perform comparisons or report that the answer is found.
* To perform a comparison, you have to output string of the following pattern «? i j» (i and j must be integer numbers from 1 to n) — the indices of the elements to compare in the current query.
* To report the indices of minimal and maximal elements of the hidden array, your program have to output a line in the form «! i j» (i and j must be integer numbers from 1 to n), where i is an index of the minimal element of array, and j is an index of the maximal element of the array. If there are several possible answers to the problem, you can output any of them.
There are several possible responses for a comparison:
* '<' — if ai is less than aj,
* '=' — if ai is equal to aj,
* '>' — if ai is greater than aj.
For an array of length n your program can make at most <image> comparisons. Note that the operation of reporting an answer («! i j» ) is not included into the value of f(n).
After the answer is reported, your program has to solve the problem for the next array or it should terminate if all T arrays are processed.
Example
Input
2
2
>
3
=
=
Output
? 1 2
! 2 1
? 3 1
? 2 1
! 2 3
Tags: constructive algorithms, interactive
Correct Solution:
```
import sys
T = int(input())
def compare(l,r,arr):
print('?',a[l]+1,a[r]+1)
sys.stdout.flush()
res = input()
if (res == '>'): return 1;
else: return 0;
while (T > 0):
T -= 1
Max_a = []
Min_a = []
n = int(input())
a = range(n)
for i in range(int(n/2)):
if (compare(i*2,i*2+1,a) > 0):
Max_a.append(i*2)
Min_a.append(i*2+1)
else :
Max_a.append(i*2+1)
Min_a.append(i*2)
if (n%2 == 1):
Max_a.append(n-1)
Min_a.append(n-1)
Max_index = Max_a[-1]
Min_index = Min_a[-1]
for i in range(len(Max_a)-1):
if (compare(Max_a[i],Max_index,a) > 0):
Max_index = Max_a[i]
for i in range(len(Min_a)-1):
if (compare(Min_a[i],Min_index,a) == 0):
Min_index = Min_a[i]
print('!',Min_index+1,Max_index+1)
sys.stdout.flush()
```
| 88,017 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem. You have to use flush operation right after printing each line. For example, in C++ you should use function fflush(stdout), in Java — System.out.flush(), in Pascal — flush(output) and in Python — sys.stdout.flush().
In this problem, you need to find maximal and minimal elements of an array. What could be simpler?
You can imagine that the jury has an array, and initially you know the only number n — array's length.
Array's elements are numbered from 1 to n. You are allowed to compare two elements of the array by using their indices i and j. There are three possible responses to this query: '<' (if ai is less than aj), '=' (if ai is equal to aj) and finally '>' (if ai is greater than aj).
It's known that it's always possible to find both maximal and minimal elements of the array by using no more than <image> comparisons, where ⌈ x⌉ is the result of rounding x up.
Write the program that will find positions of the minimum and the maximum in the jury's array of length n, by using no more than f(n) comparisons.
Interaction
Each test for this problem will contain one or more arrays. You have to find positions of minimal and maximal elements for each of these arrays. The first line of the input contains integer T (1 ≤ T ≤ 1000) — number of arrays in the test.
Thus, at the beginning, you program should read number T, and then it should solve the problem for T jury's arrays one by one.
Then input for each array goes. Firstly, your program has to read the number n (1 ≤ n ≤ 50) — the length of the array. It will be provided in the next line of the input.
Further, your program can perform comparisons or report that the answer is found.
* To perform a comparison, you have to output string of the following pattern «? i j» (i and j must be integer numbers from 1 to n) — the indices of the elements to compare in the current query.
* To report the indices of minimal and maximal elements of the hidden array, your program have to output a line in the form «! i j» (i and j must be integer numbers from 1 to n), where i is an index of the minimal element of array, and j is an index of the maximal element of the array. If there are several possible answers to the problem, you can output any of them.
There are several possible responses for a comparison:
* '<' — if ai is less than aj,
* '=' — if ai is equal to aj,
* '>' — if ai is greater than aj.
For an array of length n your program can make at most <image> comparisons. Note that the operation of reporting an answer («! i j» ) is not included into the value of f(n).
After the answer is reported, your program has to solve the problem for the next array or it should terminate if all T arrays are processed.
Example
Input
2
2
>
3
=
=
Output
? 1 2
! 2 1
? 3 1
? 2 1
! 2 3
Tags: constructive algorithms, interactive
Correct Solution:
```
import sys
def f(arr):
mn, mx = [], []
for i in range(len(arr) // 2):
print('?', arr[2*i], arr[2*i + 1])
sys.stdout.flush()
if input() == '>':
mn.append(arr[2*i + 1])
mx.append(arr[2*i])
else:
mn.append(arr[2*i])
mx.append(arr[2*i + 1])
if len(arr) % 2 == 1:
mn.append(arr[-1])
mx.append(arr[-1])
return (mn, mx)
for _ in range(int(input())):
n = int(input())
arr = range(1, n+1)
mn, mx = f(arr)
while len(mn) > 1:
mn = f(mn)[0]
while len(mx) > 1:
mx = f(mx)[1]
print('!',mn[0],mx[0])
sys.stdout.flush()
# Made By Mostafa_Khaled
```
| 88,018 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem. You have to use flush operation right after printing each line. For example, in C++ you should use function fflush(stdout), in Java — System.out.flush(), in Pascal — flush(output) and in Python — sys.stdout.flush().
In this problem, you need to find maximal and minimal elements of an array. What could be simpler?
You can imagine that the jury has an array, and initially you know the only number n — array's length.
Array's elements are numbered from 1 to n. You are allowed to compare two elements of the array by using their indices i and j. There are three possible responses to this query: '<' (if ai is less than aj), '=' (if ai is equal to aj) and finally '>' (if ai is greater than aj).
It's known that it's always possible to find both maximal and minimal elements of the array by using no more than <image> comparisons, where ⌈ x⌉ is the result of rounding x up.
Write the program that will find positions of the minimum and the maximum in the jury's array of length n, by using no more than f(n) comparisons.
Interaction
Each test for this problem will contain one or more arrays. You have to find positions of minimal and maximal elements for each of these arrays. The first line of the input contains integer T (1 ≤ T ≤ 1000) — number of arrays in the test.
Thus, at the beginning, you program should read number T, and then it should solve the problem for T jury's arrays one by one.
Then input for each array goes. Firstly, your program has to read the number n (1 ≤ n ≤ 50) — the length of the array. It will be provided in the next line of the input.
Further, your program can perform comparisons or report that the answer is found.
* To perform a comparison, you have to output string of the following pattern «? i j» (i and j must be integer numbers from 1 to n) — the indices of the elements to compare in the current query.
* To report the indices of minimal and maximal elements of the hidden array, your program have to output a line in the form «! i j» (i and j must be integer numbers from 1 to n), where i is an index of the minimal element of array, and j is an index of the maximal element of the array. If there are several possible answers to the problem, you can output any of them.
There are several possible responses for a comparison:
* '<' — if ai is less than aj,
* '=' — if ai is equal to aj,
* '>' — if ai is greater than aj.
For an array of length n your program can make at most <image> comparisons. Note that the operation of reporting an answer («! i j» ) is not included into the value of f(n).
After the answer is reported, your program has to solve the problem for the next array or it should terminate if all T arrays are processed.
Example
Input
2
2
>
3
=
=
Output
? 1 2
! 2 1
? 3 1
? 2 1
! 2 3
Tags: constructive algorithms, interactive
Correct Solution:
```
from sys import stdin, stdout
from math import sin, tan, cos
def ask(i, j):
stdout.write('? ' + str(i) + ' ' + str(j) + '\n')
stdout.flush()
return stdin.readline().strip()
T = int(stdin.readline())
for t in range(T):
n = int(stdin.readline())
if n == 1:
stdout.write('! 1 1\n')
stdout.flush()
continue
l, r = 1, 2
if ask(l, r) == '>':
l, r = r, l
for i in range(3, n + 1, 2):
if i == n:
s = ask(l, i)
f = ask(r, i)
if s == '>':
l = i
if f == '<':
r = i
continue
lb, rb = i, i + 1
if ask(lb, rb) == '>':
lb, rb = rb, lb
if ask(lb, l) == '<':
l = lb
if ask(rb, r) == '>':
r = rb
stdout.write('! ' + str(l) + ' ' + str(r) + '\n')
stdout.flush()
```
| 88,019 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem. You have to use flush operation right after printing each line. For example, in C++ you should use function fflush(stdout), in Java — System.out.flush(), in Pascal — flush(output) and in Python — sys.stdout.flush().
In this problem, you need to find maximal and minimal elements of an array. What could be simpler?
You can imagine that the jury has an array, and initially you know the only number n — array's length.
Array's elements are numbered from 1 to n. You are allowed to compare two elements of the array by using their indices i and j. There are three possible responses to this query: '<' (if ai is less than aj), '=' (if ai is equal to aj) and finally '>' (if ai is greater than aj).
It's known that it's always possible to find both maximal and minimal elements of the array by using no more than <image> comparisons, where ⌈ x⌉ is the result of rounding x up.
Write the program that will find positions of the minimum and the maximum in the jury's array of length n, by using no more than f(n) comparisons.
Interaction
Each test for this problem will contain one or more arrays. You have to find positions of minimal and maximal elements for each of these arrays. The first line of the input contains integer T (1 ≤ T ≤ 1000) — number of arrays in the test.
Thus, at the beginning, you program should read number T, and then it should solve the problem for T jury's arrays one by one.
Then input for each array goes. Firstly, your program has to read the number n (1 ≤ n ≤ 50) — the length of the array. It will be provided in the next line of the input.
Further, your program can perform comparisons or report that the answer is found.
* To perform a comparison, you have to output string of the following pattern «? i j» (i and j must be integer numbers from 1 to n) — the indices of the elements to compare in the current query.
* To report the indices of minimal and maximal elements of the hidden array, your program have to output a line in the form «! i j» (i and j must be integer numbers from 1 to n), where i is an index of the minimal element of array, and j is an index of the maximal element of the array. If there are several possible answers to the problem, you can output any of them.
There are several possible responses for a comparison:
* '<' — if ai is less than aj,
* '=' — if ai is equal to aj,
* '>' — if ai is greater than aj.
For an array of length n your program can make at most <image> comparisons. Note that the operation of reporting an answer («! i j» ) is not included into the value of f(n).
After the answer is reported, your program has to solve the problem for the next array or it should terminate if all T arrays are processed.
Example
Input
2
2
>
3
=
=
Output
? 1 2
! 2 1
? 3 1
? 2 1
! 2 3
Tags: constructive algorithms, interactive
Correct Solution:
```
# Online Python compiler (interpreter) to run Python online.
# Write Python 3 code in this online editor and run it.
# Online Python compiler (interpreter) to run Python online.
# Write Python 3 code in this online editor and run it.
import sys
def obtain_min_max(left, right):
print("?", left, right)
sys.stdout.flush()
k = input()
if k == "=":
return (left, left)
elif k == "<":
return (left, right)
else:
return (right, left)
def solve_aux(left, right):
if left == right:
return (left, left)
if right == left + 1:
return obtain_min_max(left, right)
mid = (left + right)//2
if (mid - left + 1) % 2 == 1 and (right - mid) % 2 == 1:
mid -= 1
(min1, max1) = solve_aux(left, mid)
(min2, max2) = solve_aux(mid + 1, right)
(min_min, min_max) = obtain_min_max(min1, min2)
(max_min, max_max) = obtain_min_max(max1, max2)
return (min_min, max_max)
def solve(k):
return solve_aux(1, k)
n = int(input())
for i in range(n):
k = int(input())
(min, max) = solve(k)
print("!", min, max)
sys.stdout.flush()
```
| 88,020 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem. You have to use flush operation right after printing each line. For example, in C++ you should use function fflush(stdout), in Java — System.out.flush(), in Pascal — flush(output) and in Python — sys.stdout.flush().
In this problem, you need to find maximal and minimal elements of an array. What could be simpler?
You can imagine that the jury has an array, and initially you know the only number n — array's length.
Array's elements are numbered from 1 to n. You are allowed to compare two elements of the array by using their indices i and j. There are three possible responses to this query: '<' (if ai is less than aj), '=' (if ai is equal to aj) and finally '>' (if ai is greater than aj).
It's known that it's always possible to find both maximal and minimal elements of the array by using no more than <image> comparisons, where ⌈ x⌉ is the result of rounding x up.
Write the program that will find positions of the minimum and the maximum in the jury's array of length n, by using no more than f(n) comparisons.
Interaction
Each test for this problem will contain one or more arrays. You have to find positions of minimal and maximal elements for each of these arrays. The first line of the input contains integer T (1 ≤ T ≤ 1000) — number of arrays in the test.
Thus, at the beginning, you program should read number T, and then it should solve the problem for T jury's arrays one by one.
Then input for each array goes. Firstly, your program has to read the number n (1 ≤ n ≤ 50) — the length of the array. It will be provided in the next line of the input.
Further, your program can perform comparisons or report that the answer is found.
* To perform a comparison, you have to output string of the following pattern «? i j» (i and j must be integer numbers from 1 to n) — the indices of the elements to compare in the current query.
* To report the indices of minimal and maximal elements of the hidden array, your program have to output a line in the form «! i j» (i and j must be integer numbers from 1 to n), where i is an index of the minimal element of array, and j is an index of the maximal element of the array. If there are several possible answers to the problem, you can output any of them.
There are several possible responses for a comparison:
* '<' — if ai is less than aj,
* '=' — if ai is equal to aj,
* '>' — if ai is greater than aj.
For an array of length n your program can make at most <image> comparisons. Note that the operation of reporting an answer («! i j» ) is not included into the value of f(n).
After the answer is reported, your program has to solve the problem for the next array or it should terminate if all T arrays are processed.
Example
Input
2
2
>
3
=
=
Output
? 1 2
! 2 1
? 3 1
? 2 1
! 2 3
Tags: constructive algorithms, interactive
Correct Solution:
```
"""
Author - Satwik Tiwari .
"""
#===============================================================================================
#importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from functools import cmp_to_key
# from itertools import *
from heapq import *
from math import gcd, factorial,floor,ceil,sqrt,log2
from copy import deepcopy
from collections import deque
from bisect import bisect_left as bl
from bisect import bisect_right as br
from bisect import bisect
#==============================================================================================
#fast I/O region
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")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
#===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
#### END ITERATE RECURSION ####
#===============================================================================================
#some shortcuts
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var)) #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
# def graph(vertex): return [[] for i in range(0,vertex+1)]
def testcase(t):
for pp in range(t):
solve(pp)
def google(p):
print('Case #'+str(p)+': ',end='')
def lcm(a,b): return (a*b)//gcd(a,b)
def modInverse(b):
g = gcd(b, mod)
if (g != 1):
# print("Inverse doesn't exist")
return -1
else:
# If b and m are relatively prime,
# then modulo inverse is b^(m-2) mode m
return pow(b, mod - 2, mod)
def power(x, y, p) :
y%=(p-1) #not so sure about this. used when y>p-1. if p is prime.
res = 1 # Initialize result
x = x % p # Update x if it is more , than or equal to p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) : # If y is odd, multiply, x with result
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
def isPrime(n) :
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
inf = pow(10,20)
mod = 10**9+7
#===============================================================================================
# code here ;))
def ask(a,b):
print('?',a+1,b+1)
sys.stdout.flush()
s = inp()
return s
def solve(case):
n = int(inp())
if(n == 1):
print('!',1,1)
sys.stdout.flush()
return
max = []
min = []
for i in range(0,n,2):
if(i == n-1):
max.append(i)
min.append(i)
else:
temp = ask(i,i+1)
if(temp == '<' or temp == '='):
min.append(i)
max.append(i+1)
if(temp == '>'):
min.append(i+1)
max.append(i)
while(len(max) > 1):
new = []
for i in range(0,len(max),2):
if(i == len(max) - 1):
new.append(max[i])
else:
temp = ask(max[i],max[i+1])
if(temp == '<' or temp == '='):
new.append(max[i+1])
else:
new.append(max[i])
max = deepcopy(new)
maxans = max[0]
while (len(min) > 1):
new = []
for i in range(0, len(min), 2):
if (i == len(min) - 1):
new.append(min[i])
else:
temp = ask(min[i], min[i + 1])
if (temp == '<' or temp == '='):
new.append(min[i])
else:
new.append(min[i + 1])
min = deepcopy(new)
minans = min[0]
print('!',minans + 1,maxans + 1)
sys.stdout.flush()
# testcase(1)
testcase(int(inp()))
```
| 88,021 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem. You have to use flush operation right after printing each line. For example, in C++ you should use function fflush(stdout), in Java — System.out.flush(), in Pascal — flush(output) and in Python — sys.stdout.flush().
In this problem, you need to find maximal and minimal elements of an array. What could be simpler?
You can imagine that the jury has an array, and initially you know the only number n — array's length.
Array's elements are numbered from 1 to n. You are allowed to compare two elements of the array by using their indices i and j. There are three possible responses to this query: '<' (if ai is less than aj), '=' (if ai is equal to aj) and finally '>' (if ai is greater than aj).
It's known that it's always possible to find both maximal and minimal elements of the array by using no more than <image> comparisons, where ⌈ x⌉ is the result of rounding x up.
Write the program that will find positions of the minimum and the maximum in the jury's array of length n, by using no more than f(n) comparisons.
Interaction
Each test for this problem will contain one or more arrays. You have to find positions of minimal and maximal elements for each of these arrays. The first line of the input contains integer T (1 ≤ T ≤ 1000) — number of arrays in the test.
Thus, at the beginning, you program should read number T, and then it should solve the problem for T jury's arrays one by one.
Then input for each array goes. Firstly, your program has to read the number n (1 ≤ n ≤ 50) — the length of the array. It will be provided in the next line of the input.
Further, your program can perform comparisons or report that the answer is found.
* To perform a comparison, you have to output string of the following pattern «? i j» (i and j must be integer numbers from 1 to n) — the indices of the elements to compare in the current query.
* To report the indices of minimal and maximal elements of the hidden array, your program have to output a line in the form «! i j» (i and j must be integer numbers from 1 to n), where i is an index of the minimal element of array, and j is an index of the maximal element of the array. If there are several possible answers to the problem, you can output any of them.
There are several possible responses for a comparison:
* '<' — if ai is less than aj,
* '=' — if ai is equal to aj,
* '>' — if ai is greater than aj.
For an array of length n your program can make at most <image> comparisons. Note that the operation of reporting an answer («! i j» ) is not included into the value of f(n).
After the answer is reported, your program has to solve the problem for the next array or it should terminate if all T arrays are processed.
Example
Input
2
2
>
3
=
=
Output
? 1 2
! 2 1
? 3 1
? 2 1
! 2 3
Tags: constructive algorithms, interactive
Correct Solution:
```
import sys
def f(arr):
mn, mx = [], []
for i in range(len(arr) // 2):
print('?', arr[2*i], arr[2*i + 1])
sys.stdout.flush()
if input() == '>':
mn.append(arr[2*i + 1])
mx.append(arr[2*i])
else:
mn.append(arr[2*i])
mx.append(arr[2*i + 1])
if len(arr) % 2 == 1:
mn.append(arr[-1])
mx.append(arr[-1])
return (mn, mx)
for _ in range(int(input())):
n = int(input())
arr = range(1, n+1)
mn, mx = f(arr)
while len(mn) > 1:
mn = f(mn)[0]
while len(mx) > 1:
mx = f(mx)[1]
print('!',mn[0],mx[0])
sys.stdout.flush()
```
| 88,022 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem. You have to use flush operation right after printing each line. For example, in C++ you should use function fflush(stdout), in Java — System.out.flush(), in Pascal — flush(output) and in Python — sys.stdout.flush().
In this problem, you need to find maximal and minimal elements of an array. What could be simpler?
You can imagine that the jury has an array, and initially you know the only number n — array's length.
Array's elements are numbered from 1 to n. You are allowed to compare two elements of the array by using their indices i and j. There are three possible responses to this query: '<' (if ai is less than aj), '=' (if ai is equal to aj) and finally '>' (if ai is greater than aj).
It's known that it's always possible to find both maximal and minimal elements of the array by using no more than <image> comparisons, where ⌈ x⌉ is the result of rounding x up.
Write the program that will find positions of the minimum and the maximum in the jury's array of length n, by using no more than f(n) comparisons.
Interaction
Each test for this problem will contain one or more arrays. You have to find positions of minimal and maximal elements for each of these arrays. The first line of the input contains integer T (1 ≤ T ≤ 1000) — number of arrays in the test.
Thus, at the beginning, you program should read number T, and then it should solve the problem for T jury's arrays one by one.
Then input for each array goes. Firstly, your program has to read the number n (1 ≤ n ≤ 50) — the length of the array. It will be provided in the next line of the input.
Further, your program can perform comparisons or report that the answer is found.
* To perform a comparison, you have to output string of the following pattern «? i j» (i and j must be integer numbers from 1 to n) — the indices of the elements to compare in the current query.
* To report the indices of minimal and maximal elements of the hidden array, your program have to output a line in the form «! i j» (i and j must be integer numbers from 1 to n), where i is an index of the minimal element of array, and j is an index of the maximal element of the array. If there are several possible answers to the problem, you can output any of them.
There are several possible responses for a comparison:
* '<' — if ai is less than aj,
* '=' — if ai is equal to aj,
* '>' — if ai is greater than aj.
For an array of length n your program can make at most <image> comparisons. Note that the operation of reporting an answer («! i j» ) is not included into the value of f(n).
After the answer is reported, your program has to solve the problem for the next array or it should terminate if all T arrays are processed.
Example
Input
2
2
>
3
=
=
Output
? 1 2
! 2 1
? 3 1
? 2 1
! 2 3
Tags: constructive algorithms, interactive
Correct Solution:
```
from sys import stdin, stdout
#input = stdin.buffer.readline
#print = stdout.write
def ask(i, j):
print('?', i + 1, j + 1, flush=True)
s = input()
if s == '<':
return -1
if s == '=':
return 0
return 1
def ans(i, j):
print('!', i + 1, j + 1, flush=True)
def f(i, j, c):
if (ask(i, j) > 0) ^ c:
return j
return i
def g(i, j):
if ask(i, j) < 0:
return [i, j]
return [j, i]
for _ in range(int(input())):
n = int(input())
if n == 1:
ans(0, 0)
continue
a = []
for i in range(n // 2):
a.append(g(2 * i, 2 * i + 1))
x, y = a[0][0], a[0][1]
for i in range(1, n // 2):
x = f(x, a[i][0], 0)
y = f(y, a[i][1], 1)
if n & 1:
x = f(x, n - 1, 0)
y = f(y, n - 1, 1)
ans(x, y)
```
| 88,023 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. You have to use flush operation right after printing each line. For example, in C++ you should use function fflush(stdout), in Java — System.out.flush(), in Pascal — flush(output) and in Python — sys.stdout.flush().
In this problem, you need to find maximal and minimal elements of an array. What could be simpler?
You can imagine that the jury has an array, and initially you know the only number n — array's length.
Array's elements are numbered from 1 to n. You are allowed to compare two elements of the array by using their indices i and j. There are three possible responses to this query: '<' (if ai is less than aj), '=' (if ai is equal to aj) and finally '>' (if ai is greater than aj).
It's known that it's always possible to find both maximal and minimal elements of the array by using no more than <image> comparisons, where ⌈ x⌉ is the result of rounding x up.
Write the program that will find positions of the minimum and the maximum in the jury's array of length n, by using no more than f(n) comparisons.
Interaction
Each test for this problem will contain one or more arrays. You have to find positions of minimal and maximal elements for each of these arrays. The first line of the input contains integer T (1 ≤ T ≤ 1000) — number of arrays in the test.
Thus, at the beginning, you program should read number T, and then it should solve the problem for T jury's arrays one by one.
Then input for each array goes. Firstly, your program has to read the number n (1 ≤ n ≤ 50) — the length of the array. It will be provided in the next line of the input.
Further, your program can perform comparisons or report that the answer is found.
* To perform a comparison, you have to output string of the following pattern «? i j» (i and j must be integer numbers from 1 to n) — the indices of the elements to compare in the current query.
* To report the indices of minimal and maximal elements of the hidden array, your program have to output a line in the form «! i j» (i and j must be integer numbers from 1 to n), where i is an index of the minimal element of array, and j is an index of the maximal element of the array. If there are several possible answers to the problem, you can output any of them.
There are several possible responses for a comparison:
* '<' — if ai is less than aj,
* '=' — if ai is equal to aj,
* '>' — if ai is greater than aj.
For an array of length n your program can make at most <image> comparisons. Note that the operation of reporting an answer («! i j» ) is not included into the value of f(n).
After the answer is reported, your program has to solve the problem for the next array or it should terminate if all T arrays are processed.
Example
Input
2
2
>
3
=
=
Output
? 1 2
! 2 1
? 3 1
? 2 1
! 2 3
Submitted Solution:
```
import random
import sys
def myprint(s):
print(s)
sys.stdout.flush()
t = int(input())
while t > 0:
t -= 1
n = int(input())
A = [i+1 for i in range(n)]
allowed = 3 * n // 2 + n % 2 - 2
#print(allowed)
random.shuffle(A)
#print(A)
def cmp(x, y):
myprint("? " + str(A[x]) + " " + str(A[y]))
return input()
mins = []
maxs = []
for i in range(0, n - 1, 2):
cmpres = cmp(i, i+1)
allowed -= 1
if(cmpres == '<'):
mins.append(i)
maxs.append(i+1)
else:
mins.append(i+1)
maxs.append(i)
if n % 2 != 0:
mins.append(n - 1)
maxs.append(n - 1)
while allowed:
mn = []
mx = []
for i in range(0, len(mins) - 1, 2):
if allowed:
cmpres = cmp(mins[i], mins[i+1])
allowed -=1
if cmpres == '<':
mn.append(mins[i])
else:
mn.append(mins[i+1])
if len(mins) % 2:
mn.append(mins[-1])
for i in range(0, len(maxs) - 1, 2):
if allowed:
cmpres = cmp(maxs[i], maxs[i + 1])
allowed -= 1
if cmpres == '>':
mx.append(maxs[i])
else:
mx.append(maxs[i + 1])
if len(maxs) % 2:
mx.append(maxs[-1])
mins = mn
maxs = mx
myprint("! "+ str(A[mins[0]]) + " " + str(A[maxs[0]]))
```
Yes
| 88,024 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. You have to use flush operation right after printing each line. For example, in C++ you should use function fflush(stdout), in Java — System.out.flush(), in Pascal — flush(output) and in Python — sys.stdout.flush().
In this problem, you need to find maximal and minimal elements of an array. What could be simpler?
You can imagine that the jury has an array, and initially you know the only number n — array's length.
Array's elements are numbered from 1 to n. You are allowed to compare two elements of the array by using their indices i and j. There are three possible responses to this query: '<' (if ai is less than aj), '=' (if ai is equal to aj) and finally '>' (if ai is greater than aj).
It's known that it's always possible to find both maximal and minimal elements of the array by using no more than <image> comparisons, where ⌈ x⌉ is the result of rounding x up.
Write the program that will find positions of the minimum and the maximum in the jury's array of length n, by using no more than f(n) comparisons.
Interaction
Each test for this problem will contain one or more arrays. You have to find positions of minimal and maximal elements for each of these arrays. The first line of the input contains integer T (1 ≤ T ≤ 1000) — number of arrays in the test.
Thus, at the beginning, you program should read number T, and then it should solve the problem for T jury's arrays one by one.
Then input for each array goes. Firstly, your program has to read the number n (1 ≤ n ≤ 50) — the length of the array. It will be provided in the next line of the input.
Further, your program can perform comparisons or report that the answer is found.
* To perform a comparison, you have to output string of the following pattern «? i j» (i and j must be integer numbers from 1 to n) — the indices of the elements to compare in the current query.
* To report the indices of minimal and maximal elements of the hidden array, your program have to output a line in the form «! i j» (i and j must be integer numbers from 1 to n), where i is an index of the minimal element of array, and j is an index of the maximal element of the array. If there are several possible answers to the problem, you can output any of them.
There are several possible responses for a comparison:
* '<' — if ai is less than aj,
* '=' — if ai is equal to aj,
* '>' — if ai is greater than aj.
For an array of length n your program can make at most <image> comparisons. Note that the operation of reporting an answer («! i j» ) is not included into the value of f(n).
After the answer is reported, your program has to solve the problem for the next array or it should terminate if all T arrays are processed.
Example
Input
2
2
>
3
=
=
Output
? 1 2
! 2 1
? 3 1
? 2 1
! 2 3
Submitted Solution:
```
from sys import *
def f(t):
n = len(t)
k = n >> 1
u, v = [], []
if n & 1: u, v = [t[-1]], [t[-1]]
for i in range(k):
print('?', t[i], t[k + i])
stdout.flush()
q = k * (input() == '<')
u.append(t[k - q + i])
v.append(t[q + i])
return u, v
for i in range(int(input())):
u, v = f(range(1, int(input()) + 1))
while len(u) > 1: u = f(u)[0]
while len(v) > 1: v = f(v)[1]
print('!', u[0], v[0])
stdout.flush()
```
Yes
| 88,025 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. You have to use flush operation right after printing each line. For example, in C++ you should use function fflush(stdout), in Java — System.out.flush(), in Pascal — flush(output) and in Python — sys.stdout.flush().
In this problem, you need to find maximal and minimal elements of an array. What could be simpler?
You can imagine that the jury has an array, and initially you know the only number n — array's length.
Array's elements are numbered from 1 to n. You are allowed to compare two elements of the array by using their indices i and j. There are three possible responses to this query: '<' (if ai is less than aj), '=' (if ai is equal to aj) and finally '>' (if ai is greater than aj).
It's known that it's always possible to find both maximal and minimal elements of the array by using no more than <image> comparisons, where ⌈ x⌉ is the result of rounding x up.
Write the program that will find positions of the minimum and the maximum in the jury's array of length n, by using no more than f(n) comparisons.
Interaction
Each test for this problem will contain one or more arrays. You have to find positions of minimal and maximal elements for each of these arrays. The first line of the input contains integer T (1 ≤ T ≤ 1000) — number of arrays in the test.
Thus, at the beginning, you program should read number T, and then it should solve the problem for T jury's arrays one by one.
Then input for each array goes. Firstly, your program has to read the number n (1 ≤ n ≤ 50) — the length of the array. It will be provided in the next line of the input.
Further, your program can perform comparisons or report that the answer is found.
* To perform a comparison, you have to output string of the following pattern «? i j» (i and j must be integer numbers from 1 to n) — the indices of the elements to compare in the current query.
* To report the indices of minimal and maximal elements of the hidden array, your program have to output a line in the form «! i j» (i and j must be integer numbers from 1 to n), where i is an index of the minimal element of array, and j is an index of the maximal element of the array. If there are several possible answers to the problem, you can output any of them.
There are several possible responses for a comparison:
* '<' — if ai is less than aj,
* '=' — if ai is equal to aj,
* '>' — if ai is greater than aj.
For an array of length n your program can make at most <image> comparisons. Note that the operation of reporting an answer («! i j» ) is not included into the value of f(n).
After the answer is reported, your program has to solve the problem for the next array or it should terminate if all T arrays are processed.
Example
Input
2
2
>
3
=
=
Output
? 1 2
! 2 1
? 3 1
? 2 1
! 2 3
Submitted Solution:
```
import sys
def find_min_max(l, d=None):
#print(l)
n = len(l)
if n == 1:
return (l[0], l[0])
lesser = []
greater = []
for i in range(n//2):
first = l[2*i]
second = l[2*i + 1]
print("? {} {}".format(first, second))
sys.stdout.flush()
answer = input()
if answer == '<':
lesser.append(first)
greater.append(second)
else:
lesser.append(second)
greater.append(first)
if n%2 == 1:
lesser.append(l[-1])
greater.append(l[-1])
mn = None
mx = None
if d != 'max':
mn = find_min_max(lesser, 'min')[0]
if d != 'min':
mx = find_min_max(greater, 'max')[1]
return (mn, mx)
t = input()
t = int(t)
for k in range(t):
n = input()
n = int(n)
l = list(range(1, n+1))
mn, mx = find_min_max(l)
print("! {} {}".format(mn, mx))
sys.stdout.flush()
```
Yes
| 88,026 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. You have to use flush operation right after printing each line. For example, in C++ you should use function fflush(stdout), in Java — System.out.flush(), in Pascal — flush(output) and in Python — sys.stdout.flush().
In this problem, you need to find maximal and minimal elements of an array. What could be simpler?
You can imagine that the jury has an array, and initially you know the only number n — array's length.
Array's elements are numbered from 1 to n. You are allowed to compare two elements of the array by using their indices i and j. There are three possible responses to this query: '<' (if ai is less than aj), '=' (if ai is equal to aj) and finally '>' (if ai is greater than aj).
It's known that it's always possible to find both maximal and minimal elements of the array by using no more than <image> comparisons, where ⌈ x⌉ is the result of rounding x up.
Write the program that will find positions of the minimum and the maximum in the jury's array of length n, by using no more than f(n) comparisons.
Interaction
Each test for this problem will contain one or more arrays. You have to find positions of minimal and maximal elements for each of these arrays. The first line of the input contains integer T (1 ≤ T ≤ 1000) — number of arrays in the test.
Thus, at the beginning, you program should read number T, and then it should solve the problem for T jury's arrays one by one.
Then input for each array goes. Firstly, your program has to read the number n (1 ≤ n ≤ 50) — the length of the array. It will be provided in the next line of the input.
Further, your program can perform comparisons or report that the answer is found.
* To perform a comparison, you have to output string of the following pattern «? i j» (i and j must be integer numbers from 1 to n) — the indices of the elements to compare in the current query.
* To report the indices of minimal and maximal elements of the hidden array, your program have to output a line in the form «! i j» (i and j must be integer numbers from 1 to n), where i is an index of the minimal element of array, and j is an index of the maximal element of the array. If there are several possible answers to the problem, you can output any of them.
There are several possible responses for a comparison:
* '<' — if ai is less than aj,
* '=' — if ai is equal to aj,
* '>' — if ai is greater than aj.
For an array of length n your program can make at most <image> comparisons. Note that the operation of reporting an answer («! i j» ) is not included into the value of f(n).
After the answer is reported, your program has to solve the problem for the next array or it should terminate if all T arrays are processed.
Example
Input
2
2
>
3
=
=
Output
? 1 2
! 2 1
? 3 1
? 2 1
! 2 3
Submitted Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
mod = 10 ** 9 + 7
mod1 = 998244353
# sys.setrecursionlimit(300000)
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
# sys.setrecursionlimit(300000)
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
# -----------------------------------------------binary seacrh tree---------------------------------------
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: a + b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] < key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
for ik in range(int(input())):
n=int(input())
if n==1:
print("! 1 1",flush=True)
continue
elif n==2:
print("? 1 2",flush=True)
s=input()
if s==">":
print("! 2 1",flush=True)
else:
print("! 1 2",flush=True)
continue
last=1
f=2
res=[]
d=defaultdict(str)
for i in range(n-1):
#print("?",last,f,flush=True)
if (last,f) in d:
r=d[(last,f)]
elif (f,last) in d:
if d[(f,last)]==">":
r="<"
elif d[(f,last)]=="<":
r=">"
else:
r=d[(f,last)]
d[(f,last)]=r
else:
print("?", last, f, flush=True)
r=input()
d[(last,f)]=r
if r=='<':
res.append(last)
last=f
f+=1
else:
f+=1
res.append(f-1)
#print(res,d)
last1=res[0]
f1=res[1]
t=1
for i in range(len(res)-1):
if (last1,f1) in d:
r=d[(last1,f1)]
elif (f1,last1) in d:
if d[(f1,last1)]==">":
r="<"
elif d[(f1,last1)]=="<":
r=">"
else:
r=d[(f1,last1)]
else:
print("?", last1, f1, flush=True)
r = input()
d[(last1, f1)] = r
if r == '>':
res.append(last)
last = f1
if t+1==len(res):
break
f1= res[t+1]
t+=1
else:
if t+1==len(res):
break
f1 = res[t + 1]
t += 1
res.append(res[t-1])
print('!',last1,last,flush=True)
```
No
| 88,027 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. You have to use flush operation right after printing each line. For example, in C++ you should use function fflush(stdout), in Java — System.out.flush(), in Pascal — flush(output) and in Python — sys.stdout.flush().
In this problem, you need to find maximal and minimal elements of an array. What could be simpler?
You can imagine that the jury has an array, and initially you know the only number n — array's length.
Array's elements are numbered from 1 to n. You are allowed to compare two elements of the array by using their indices i and j. There are three possible responses to this query: '<' (if ai is less than aj), '=' (if ai is equal to aj) and finally '>' (if ai is greater than aj).
It's known that it's always possible to find both maximal and minimal elements of the array by using no more than <image> comparisons, where ⌈ x⌉ is the result of rounding x up.
Write the program that will find positions of the minimum and the maximum in the jury's array of length n, by using no more than f(n) comparisons.
Interaction
Each test for this problem will contain one or more arrays. You have to find positions of minimal and maximal elements for each of these arrays. The first line of the input contains integer T (1 ≤ T ≤ 1000) — number of arrays in the test.
Thus, at the beginning, you program should read number T, and then it should solve the problem for T jury's arrays one by one.
Then input for each array goes. Firstly, your program has to read the number n (1 ≤ n ≤ 50) — the length of the array. It will be provided in the next line of the input.
Further, your program can perform comparisons or report that the answer is found.
* To perform a comparison, you have to output string of the following pattern «? i j» (i and j must be integer numbers from 1 to n) — the indices of the elements to compare in the current query.
* To report the indices of minimal and maximal elements of the hidden array, your program have to output a line in the form «! i j» (i and j must be integer numbers from 1 to n), where i is an index of the minimal element of array, and j is an index of the maximal element of the array. If there are several possible answers to the problem, you can output any of them.
There are several possible responses for a comparison:
* '<' — if ai is less than aj,
* '=' — if ai is equal to aj,
* '>' — if ai is greater than aj.
For an array of length n your program can make at most <image> comparisons. Note that the operation of reporting an answer («! i j» ) is not included into the value of f(n).
After the answer is reported, your program has to solve the problem for the next array or it should terminate if all T arrays are processed.
Example
Input
2
2
>
3
=
=
Output
? 1 2
! 2 1
? 3 1
? 2 1
! 2 3
Submitted Solution:
```
import random
import sys
def myprint(s):
print(s)
sys.stdout.flush()
t = int(input())
while t > 0:
t -= 1
n = int(input())
A = [i+1 for i in range(n)]
allowed = 3 * n // 2 + n % 2 - 2
#print(allowed)
random.shuffle(A)
#print(A)
mx = 0
mn = 0
min_eq_max = True
def cmp(x, y):
myprint("? " + str(A[x]) + " " + str(A[y]))
return input()
for i in range(1, n):
withmin = ' '
if allowed:
withmin = cmp(i, mn)
allowed -= 1
if withmin == '<':
mn = i
min_eq_max = False
elif withmin == '=':
pass
elif allowed:
withmax = ' '
if min_eq_max:
withmax = withmin
else:
withmax = cmp(i, mx)
allowed -= 1
if withmax == '>':
mx = i
#print(A[mn], A[mx])
#myprint(allowed)
myprint("! "+ str(A[mn]) + " " + str(A[mx]))
```
No
| 88,028 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. You have to use flush operation right after printing each line. For example, in C++ you should use function fflush(stdout), in Java — System.out.flush(), in Pascal — flush(output) and in Python — sys.stdout.flush().
In this problem, you need to find maximal and minimal elements of an array. What could be simpler?
You can imagine that the jury has an array, and initially you know the only number n — array's length.
Array's elements are numbered from 1 to n. You are allowed to compare two elements of the array by using their indices i and j. There are three possible responses to this query: '<' (if ai is less than aj), '=' (if ai is equal to aj) and finally '>' (if ai is greater than aj).
It's known that it's always possible to find both maximal and minimal elements of the array by using no more than <image> comparisons, where ⌈ x⌉ is the result of rounding x up.
Write the program that will find positions of the minimum and the maximum in the jury's array of length n, by using no more than f(n) comparisons.
Interaction
Each test for this problem will contain one or more arrays. You have to find positions of minimal and maximal elements for each of these arrays. The first line of the input contains integer T (1 ≤ T ≤ 1000) — number of arrays in the test.
Thus, at the beginning, you program should read number T, and then it should solve the problem for T jury's arrays one by one.
Then input for each array goes. Firstly, your program has to read the number n (1 ≤ n ≤ 50) — the length of the array. It will be provided in the next line of the input.
Further, your program can perform comparisons or report that the answer is found.
* To perform a comparison, you have to output string of the following pattern «? i j» (i and j must be integer numbers from 1 to n) — the indices of the elements to compare in the current query.
* To report the indices of minimal and maximal elements of the hidden array, your program have to output a line in the form «! i j» (i and j must be integer numbers from 1 to n), where i is an index of the minimal element of array, and j is an index of the maximal element of the array. If there are several possible answers to the problem, you can output any of them.
There are several possible responses for a comparison:
* '<' — if ai is less than aj,
* '=' — if ai is equal to aj,
* '>' — if ai is greater than aj.
For an array of length n your program can make at most <image> comparisons. Note that the operation of reporting an answer («! i j» ) is not included into the value of f(n).
After the answer is reported, your program has to solve the problem for the next array or it should terminate if all T arrays are processed.
Example
Input
2
2
>
3
=
=
Output
? 1 2
! 2 1
? 3 1
? 2 1
! 2 3
Submitted Solution:
```
import sys
T = int(input())
def compare(l,r,arr):
print('?',a[l]+1,a[r]+1)
sys.stdout.flush()
res = input()
if (res == '>'): return 1;
else: return 0;
while (T > 0):
T -= 1
Max_a = []
Min_a = []
n = int(input())
a = range(n)
for i in range(int(n/2)):
if (compare(i*2,i*2+1,a) > 0):
Max_a.append(i*2)
Min_a.append(i*2+1)
else :
Max_a.append(i*2+1)
Min_a.append(i*2)
if (n%2 == 1):Max_a.append(n-1)
Max_index = Max_a[-1]
Min_index = Min_a[-1]
for i in range(len(Max_a)-1):
if (compare(Max_a[i],Max_index,a) > 0):
Max_index = Max_a[i]
for i in range(len(Min_a)-1):
if (compare(Min_a[i],Min_index,a) == 0):
Min_index = Min_a[i]
print('!',Min_index+1,Max_index+1)
sys.stdout.flush()
```
No
| 88,029 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. You have to use flush operation right after printing each line. For example, in C++ you should use function fflush(stdout), in Java — System.out.flush(), in Pascal — flush(output) and in Python — sys.stdout.flush().
In this problem, you need to find maximal and minimal elements of an array. What could be simpler?
You can imagine that the jury has an array, and initially you know the only number n — array's length.
Array's elements are numbered from 1 to n. You are allowed to compare two elements of the array by using their indices i and j. There are three possible responses to this query: '<' (if ai is less than aj), '=' (if ai is equal to aj) and finally '>' (if ai is greater than aj).
It's known that it's always possible to find both maximal and minimal elements of the array by using no more than <image> comparisons, where ⌈ x⌉ is the result of rounding x up.
Write the program that will find positions of the minimum and the maximum in the jury's array of length n, by using no more than f(n) comparisons.
Interaction
Each test for this problem will contain one or more arrays. You have to find positions of minimal and maximal elements for each of these arrays. The first line of the input contains integer T (1 ≤ T ≤ 1000) — number of arrays in the test.
Thus, at the beginning, you program should read number T, and then it should solve the problem for T jury's arrays one by one.
Then input for each array goes. Firstly, your program has to read the number n (1 ≤ n ≤ 50) — the length of the array. It will be provided in the next line of the input.
Further, your program can perform comparisons or report that the answer is found.
* To perform a comparison, you have to output string of the following pattern «? i j» (i and j must be integer numbers from 1 to n) — the indices of the elements to compare in the current query.
* To report the indices of minimal and maximal elements of the hidden array, your program have to output a line in the form «! i j» (i and j must be integer numbers from 1 to n), where i is an index of the minimal element of array, and j is an index of the maximal element of the array. If there are several possible answers to the problem, you can output any of them.
There are several possible responses for a comparison:
* '<' — if ai is less than aj,
* '=' — if ai is equal to aj,
* '>' — if ai is greater than aj.
For an array of length n your program can make at most <image> comparisons. Note that the operation of reporting an answer («! i j» ) is not included into the value of f(n).
After the answer is reported, your program has to solve the problem for the next array or it should terminate if all T arrays are processed.
Example
Input
2
2
>
3
=
=
Output
? 1 2
! 2 1
? 3 1
? 2 1
! 2 3
Submitted Solution:
```
from sys import stdin, stdout
from math import sin, tan, cos
def ask(i, j):
stdout.write(str(i) + ' ' + str(j) + '\n')
stdout.flush()
return stdin.readline().strip()
T = int(stdin.readline())
for t in range(T):
n = int(stdin.readline())
if n == 1:
stdout.write('! 1 1\n')
stdout.flush()
continue
l, r = 1, 2
if ask(l, r) == '>':
l, r = r, l
for i in range(3, n + 1, 2):
if i == n:
s = ask(l, i)
f = ask(r, i)
if s == '>':
l = i
if f == '<':
r = i
continue
lb, rb = i, i + 1
if ask(lb, rb) == '>':
lb, rb = rb, lb
if ask(lb, l) == '<':
l = lb
if ask(rb, r) == '>':
r = rb
stdout.write('! ' + str(l) + ' ' + str(r) + '\n')
stdout.flush()
```
No
| 88,030 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Stepan has n pens. Every day he uses them, and on the i-th day he uses the pen number i. On the (n + 1)-th day again he uses the pen number 1, on the (n + 2)-th — he uses the pen number 2 and so on.
On every working day (from Monday to Saturday, inclusive) Stepan spends exactly 1 milliliter of ink of the pen he uses that day. On Sunday Stepan has a day of rest, he does not stend the ink of the pen he uses that day.
Stepan knows the current volume of ink in each of his pens. Now it's the Monday morning and Stepan is going to use the pen number 1 today. Your task is to determine which pen will run out of ink before all the rest (that is, there will be no ink left in it), if Stepan will use the pens according to the conditions described above.
Input
The first line contains the integer n (1 ≤ n ≤ 50 000) — the number of pens Stepan has.
The second line contains the sequence of integers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is equal to the number of milliliters of ink which the pen number i currently has.
Output
Print the index of the pen which will run out of ink before all (it means that there will be no ink left in it), if Stepan will use pens according to the conditions described above.
Pens are numbered in the order they are given in input data. The numeration begins from one.
Note that the answer is always unambiguous, since several pens can not end at the same time.
Examples
Input
3
3 3 3
Output
2
Input
5
5 4 5 4 4
Output
5
Note
In the first test Stepan uses ink of pens as follows:
1. on the day number 1 (Monday) Stepan will use the pen number 1, after that there will be 2 milliliters of ink in it;
2. on the day number 2 (Tuesday) Stepan will use the pen number 2, after that there will be 2 milliliters of ink in it;
3. on the day number 3 (Wednesday) Stepan will use the pen number 3, after that there will be 2 milliliters of ink in it;
4. on the day number 4 (Thursday) Stepan will use the pen number 1, after that there will be 1 milliliters of ink in it;
5. on the day number 5 (Friday) Stepan will use the pen number 2, after that there will be 1 milliliters of ink in it;
6. on the day number 6 (Saturday) Stepan will use the pen number 3, after that there will be 1 milliliters of ink in it;
7. on the day number 7 (Sunday) Stepan will use the pen number 1, but it is a day of rest so he will not waste ink of this pen in it;
8. on the day number 8 (Monday) Stepan will use the pen number 2, after that this pen will run out of ink.
So, the first pen which will not have ink is the pen number 2.
Tags: *special, binary search, number theory
Correct Solution:
```
import sys
def Min(x, y):
if x > y:
return y
else:
return x
def Gcd(x, y):
if x == 0:
return y
else:
return Gcd(y % x, x)
def Lcm(x, y):
return x * y // Gcd(x, y)
n = int(input())
a = [int(i) for i in input().split()]
d = [int(0) for i in range(0, n)]
ok = 0
cur = 0
len = Lcm(7, n)
for i in range(0, 7 * n):
if a[i % n] == 0 :
print(i % n + 1)
ok = 1
break
if cur != 6:
a[i % n] -= 1
d[i % n] += 1
cur = (cur + 1) % 7
if ok == 0:
k = 10**20
for i in range(0, n):
a[i] += d[i]
if d[i] == 0: continue
if a[i] % d[i] > 0:
k = Min(k, a[i] // d[i])
else:
k = Min(k, a[i] // d[i] - 1)
if k == 10**20:
k = 0
for i in range(0, n):
a[i] -= k * d[i]
iter = 0
cur = 0
while True:
if a[iter] == 0:
print(iter % n + 1)
break
else:
if cur != 6:
a[iter] -= 1
cur = (cur + 1) % 7
iter = (iter + 1) % n
```
| 88,031 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Stepan has n pens. Every day he uses them, and on the i-th day he uses the pen number i. On the (n + 1)-th day again he uses the pen number 1, on the (n + 2)-th — he uses the pen number 2 and so on.
On every working day (from Monday to Saturday, inclusive) Stepan spends exactly 1 milliliter of ink of the pen he uses that day. On Sunday Stepan has a day of rest, he does not stend the ink of the pen he uses that day.
Stepan knows the current volume of ink in each of his pens. Now it's the Monday morning and Stepan is going to use the pen number 1 today. Your task is to determine which pen will run out of ink before all the rest (that is, there will be no ink left in it), if Stepan will use the pens according to the conditions described above.
Input
The first line contains the integer n (1 ≤ n ≤ 50 000) — the number of pens Stepan has.
The second line contains the sequence of integers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is equal to the number of milliliters of ink which the pen number i currently has.
Output
Print the index of the pen which will run out of ink before all (it means that there will be no ink left in it), if Stepan will use pens according to the conditions described above.
Pens are numbered in the order they are given in input data. The numeration begins from one.
Note that the answer is always unambiguous, since several pens can not end at the same time.
Examples
Input
3
3 3 3
Output
2
Input
5
5 4 5 4 4
Output
5
Note
In the first test Stepan uses ink of pens as follows:
1. on the day number 1 (Monday) Stepan will use the pen number 1, after that there will be 2 milliliters of ink in it;
2. on the day number 2 (Tuesday) Stepan will use the pen number 2, after that there will be 2 milliliters of ink in it;
3. on the day number 3 (Wednesday) Stepan will use the pen number 3, after that there will be 2 milliliters of ink in it;
4. on the day number 4 (Thursday) Stepan will use the pen number 1, after that there will be 1 milliliters of ink in it;
5. on the day number 5 (Friday) Stepan will use the pen number 2, after that there will be 1 milliliters of ink in it;
6. on the day number 6 (Saturday) Stepan will use the pen number 3, after that there will be 1 milliliters of ink in it;
7. on the day number 7 (Sunday) Stepan will use the pen number 1, but it is a day of rest so he will not waste ink of this pen in it;
8. on the day number 8 (Monday) Stepan will use the pen number 2, after that this pen will run out of ink.
So, the first pen which will not have ink is the pen number 2.
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
ans = 100000
min = 10000000000
if ((n % 7) == 0):
for i in range(6, n, 7):
if a[i] < min:
ans = i
min = a[i]
else:
i = 7 % n
while True:
if a[i] < min:
ans = i
min = a[i]
i = (i + 7) % n
if i == (7 % n):
break
print(ans + 1)
```
No
| 88,032 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Stepan has n pens. Every day he uses them, and on the i-th day he uses the pen number i. On the (n + 1)-th day again he uses the pen number 1, on the (n + 2)-th — he uses the pen number 2 and so on.
On every working day (from Monday to Saturday, inclusive) Stepan spends exactly 1 milliliter of ink of the pen he uses that day. On Sunday Stepan has a day of rest, he does not stend the ink of the pen he uses that day.
Stepan knows the current volume of ink in each of his pens. Now it's the Monday morning and Stepan is going to use the pen number 1 today. Your task is to determine which pen will run out of ink before all the rest (that is, there will be no ink left in it), if Stepan will use the pens according to the conditions described above.
Input
The first line contains the integer n (1 ≤ n ≤ 50 000) — the number of pens Stepan has.
The second line contains the sequence of integers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is equal to the number of milliliters of ink which the pen number i currently has.
Output
Print the index of the pen which will run out of ink before all (it means that there will be no ink left in it), if Stepan will use pens according to the conditions described above.
Pens are numbered in the order they are given in input data. The numeration begins from one.
Note that the answer is always unambiguous, since several pens can not end at the same time.
Examples
Input
3
3 3 3
Output
2
Input
5
5 4 5 4 4
Output
5
Note
In the first test Stepan uses ink of pens as follows:
1. on the day number 1 (Monday) Stepan will use the pen number 1, after that there will be 2 milliliters of ink in it;
2. on the day number 2 (Tuesday) Stepan will use the pen number 2, after that there will be 2 milliliters of ink in it;
3. on the day number 3 (Wednesday) Stepan will use the pen number 3, after that there will be 2 milliliters of ink in it;
4. on the day number 4 (Thursday) Stepan will use the pen number 1, after that there will be 1 milliliters of ink in it;
5. on the day number 5 (Friday) Stepan will use the pen number 2, after that there will be 1 milliliters of ink in it;
6. on the day number 6 (Saturday) Stepan will use the pen number 3, after that there will be 1 milliliters of ink in it;
7. on the day number 7 (Sunday) Stepan will use the pen number 1, but it is a day of rest so he will not waste ink of this pen in it;
8. on the day number 8 (Monday) Stepan will use the pen number 2, after that this pen will run out of ink.
So, the first pen which will not have ink is the pen number 2.
Submitted Solution:
```
import sys
def Min(x, y):
if x > y:
return y
else:
return x
n = int(input())
a = [int(i) for i in input().split()]
d = [int(0) for i in range(0, n)]
ok = 0
cur = 0
for i in range(0, 7 * n):
if a[i % n] == 0 :
print(i % n + 1)
ok = 1
break
if cur != 6:
a[i % n] -= 1
d[i % n] += 1
cur = (cur + 1) % 7
if ok == 0:
k = -1
for i in range(0, n):
if d[i] == 0: continue
if a[i] % d[i] > 0:
k = Min(k, a[i] // d[i])
else:
k = Min(k, a[i] // d[i] - 1)
for i in range(0, n):
a[i] -= k * d[i]
iter = 0
cur = 0
while True:
if a[iter] == 0:
print(iter)
break
else:
if cur != 6:
a[iter] -= 1
cur = (cur + 1) % 7
iter = (iter + 1) % n
```
No
| 88,033 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Stepan has n pens. Every day he uses them, and on the i-th day he uses the pen number i. On the (n + 1)-th day again he uses the pen number 1, on the (n + 2)-th — he uses the pen number 2 and so on.
On every working day (from Monday to Saturday, inclusive) Stepan spends exactly 1 milliliter of ink of the pen he uses that day. On Sunday Stepan has a day of rest, he does not stend the ink of the pen he uses that day.
Stepan knows the current volume of ink in each of his pens. Now it's the Monday morning and Stepan is going to use the pen number 1 today. Your task is to determine which pen will run out of ink before all the rest (that is, there will be no ink left in it), if Stepan will use the pens according to the conditions described above.
Input
The first line contains the integer n (1 ≤ n ≤ 50 000) — the number of pens Stepan has.
The second line contains the sequence of integers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is equal to the number of milliliters of ink which the pen number i currently has.
Output
Print the index of the pen which will run out of ink before all (it means that there will be no ink left in it), if Stepan will use pens according to the conditions described above.
Pens are numbered in the order they are given in input data. The numeration begins from one.
Note that the answer is always unambiguous, since several pens can not end at the same time.
Examples
Input
3
3 3 3
Output
2
Input
5
5 4 5 4 4
Output
5
Note
In the first test Stepan uses ink of pens as follows:
1. on the day number 1 (Monday) Stepan will use the pen number 1, after that there will be 2 milliliters of ink in it;
2. on the day number 2 (Tuesday) Stepan will use the pen number 2, after that there will be 2 milliliters of ink in it;
3. on the day number 3 (Wednesday) Stepan will use the pen number 3, after that there will be 2 milliliters of ink in it;
4. on the day number 4 (Thursday) Stepan will use the pen number 1, after that there will be 1 milliliters of ink in it;
5. on the day number 5 (Friday) Stepan will use the pen number 2, after that there will be 1 milliliters of ink in it;
6. on the day number 6 (Saturday) Stepan will use the pen number 3, after that there will be 1 milliliters of ink in it;
7. on the day number 7 (Sunday) Stepan will use the pen number 1, but it is a day of rest so he will not waste ink of this pen in it;
8. on the day number 8 (Monday) Stepan will use the pen number 2, after that this pen will run out of ink.
So, the first pen which will not have ink is the pen number 2.
Submitted Solution:
```
import sys
def Min(x, y):
if x > y:
return y
else:
return x
def Gcd(x, y):
if x == 0:
return y
else:
return Gcd(y % x, x)
def Lcm(x, y):
return x * y // Gcd(x, y)
n = int(input())
a = [int(i) for i in input().split()]
d = [int(0) for i in range(0, n)]
ok = 0
cur = 0
len = Lcm(7, n)
for i in range(0, 7 * n):
if a[i % n] == 0 :
print(i % n + 1)
ok = 1
break
if cur != 6:
a[i % n] -= 1
d[i % n] += 1
cur = (cur + 1) % 7
if ok == 0:
k = -1
for i in range(0, n):
a[i] += d[i]
if d[i] == 0: continue
if a[i] % d[i] > 0:
k = Min(k, a[i] // d[i])
else:
k = Min(k, a[i] // d[i] - 1)
for i in range(0, n):
a[i] -= k * d[i]
iter = 0
cur = 0
while True:
if a[iter] == 0:
print(iter)
break
else:
if cur != 6:
a[iter] -= 1
cur = (cur + 1) % 7
iter = (iter + 1) % n
```
No
| 88,034 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp thinks about the meaning of life very often. He does this constantly, even when typing in the editor. Every time he starts brooding he can no longer fully concentrate and repeatedly presses the keys that need to be pressed only once. For example, instead of the phrase "how are you" he can type "hhoow aaaare yyoouu".
Polycarp decided to automate the process of correcting such errors. He decided to write a plug-in to the text editor that will remove pairs of identical consecutive letters (if there are any in the text). Of course, this is not exactly what Polycarp needs, but he's got to start from something!
Help Polycarp and write the main plug-in module. Your program should remove from a string all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, the program should remove them as well. Technically, its work should be equivalent to the following: while the string contains a pair of consecutive identical letters, the pair should be deleted. Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.
Input
The input data consists of a single line to be processed. The length of the line is from 1 to 2·105 characters inclusive. The string contains only lowercase Latin letters.
Output
Print the given string after it is processed. It is guaranteed that the result will contain at least one character.
Examples
Input
hhoowaaaareyyoouu
Output
wre
Input
reallazy
Output
rezy
Input
abacabaabacabaa
Output
a
Tags: implementation
Correct Solution:
```
string = input()
stack = [len(string)]
for letter in string:
if letter == stack[-1]:
stack.pop()
else:
stack.append(letter)
print(''.join(stack[1:]))
```
| 88,035 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp thinks about the meaning of life very often. He does this constantly, even when typing in the editor. Every time he starts brooding he can no longer fully concentrate and repeatedly presses the keys that need to be pressed only once. For example, instead of the phrase "how are you" he can type "hhoow aaaare yyoouu".
Polycarp decided to automate the process of correcting such errors. He decided to write a plug-in to the text editor that will remove pairs of identical consecutive letters (if there are any in the text). Of course, this is not exactly what Polycarp needs, but he's got to start from something!
Help Polycarp and write the main plug-in module. Your program should remove from a string all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, the program should remove them as well. Technically, its work should be equivalent to the following: while the string contains a pair of consecutive identical letters, the pair should be deleted. Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.
Input
The input data consists of a single line to be processed. The length of the line is from 1 to 2·105 characters inclusive. The string contains only lowercase Latin letters.
Output
Print the given string after it is processed. It is guaranteed that the result will contain at least one character.
Examples
Input
hhoowaaaareyyoouu
Output
wre
Input
reallazy
Output
rezy
Input
abacabaabacabaa
Output
a
Tags: implementation
Correct Solution:
```
##############--->>>>> Deepcoder Amit Kumar Bhuyan <<<<<---##############
"""
Perfection is achieved not when there is nothing more to add, but rather when there is nothing more to take away.
"""
from __future__ import division, print_function
import os,sys
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
def ii(): return int(input())
def si(): return input()
def mi(): return map(int,input().strip().split(" "))
def msi(): return map(str,input().strip().split(" "))
def li(): return list(mi())
def dmain():
sys.setrecursionlimit(1000000)
threading.stack_size(1024000)
thread = threading.Thread(target=main)
thread.start()
#from collections import deque, Counter, OrderedDict,defaultdict
#from heapq import nsmallest, nlargest, heapify,heappop ,heappush, heapreplace
#from math import log,sqrt,factorial,cos,tan,sin,radians
#from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
#from decimal import *
#import threading
#from itertools import permutations
#Copy 2D list m = [x[:] for x in mark] .. Avoid Using Deepcopy
import sys
input = sys.stdin.readline
scanner = lambda: int(input())
string = lambda: input().rstrip()
get_list = lambda: list(read())
read = lambda: map(int, input().split())
get_float = lambda: map(float, input().split())
# from bisect import bisect_left as lower_bound;
# from bisect import bisect_right as upper_bound;
# from math import ceil, factorial;
def ceil(x):
if x != int(x):
x = int(x) + 1
return x
def factorial(x, m):
val = 1
while x>0:
val = (val * x) % m
x -= 1
return val
def fact(x):
val = 1
while x > 0:
val *= x
x -= 1
return val
# swap_array function
def swaparr(arr, a,b):
temp = arr[a];
arr[a] = arr[b];
arr[b] = temp;
## gcd function
def gcd(a,b):
if b == 0:
return a;
return gcd(b, a % b);
## lcm function
def lcm(a, b):
return (a * b) // math.gcd(a, b)
def is_integer(n):
return math.ceil(n) == math.floor(n)
## nCr function efficient using Binomial Cofficient
def nCr(n, k):
if k > n:
return 0
if(k > n - k):
k = n - k
res = 1
for i in range(k):
res = res * (n - i)
res = res / (i + 1)
return int(res)
## upper bound function code -- such that e in a[:i] e < x;
## prime factorization
def primefs(n):
## if n == 1 ## calculating primes
primes = {}
while(n%2 == 0 and n > 0):
primes[2] = primes.get(2, 0) + 1
n = n//2
for i in range(3, int(n**0.5)+2, 2):
while(n%i == 0 and n > 0):
primes[i] = primes.get(i, 0) + 1
n = n//i
if n > 2:
primes[n] = primes.get(n, 0) + 1
## prime factoriazation of n is stored in dictionary
## primes and can be accesed. O(sqrt n)
return primes
## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p):
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
temp = a
a = b
b = temp
return a,b;
# find function with path compression included (recursive)
# def find(x, link):
# if link[x] == x:
# return x
# link[x] = find(link[x], link);
# return link[x];
# find function with path compression (ITERATIVE)
def find(x, link):
p = x;
while( p != link[p]):
p = link[p];
while( x != p):
nex = link[x];
link[x] = p;
x = nex;
return p;
# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
x = find(x, link)
y = find(y, link)
if size[x] < size[y]:
x,y = swap(x,y)
if x != y:
size[x] += size[y]
link[y] = x
## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n):
prime = [True for i in range(n+1)]
prime[0], prime[1] = False, False
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
# Euler's Toitent Function phi
def phi(n) :
result = n
p = 2
while(p * p<= n) :
if (n % p == 0) :
while (n % p == 0) :
n = n // p
result = result * (1.0 - (1.0 / (float) (p)))
p = p + 1
if (n > 1) :
result = result * (1.0 - (1.0 / (float)(n)))
return (int)(result)
def is_prime(n):
if n == 0:
return False
if n == 1:
return True
for i in range(2, int(n ** (1 / 2)) + 1):
if not n % i:
return False
return True
def next_prime(n, primes):
while primes[n] != True:
n += 1
return n
#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e5 + 5)
def spf_sieve():
spf[1] = 1;
for i in range(2, MAXN):
spf[i] = i;
for i in range(4, MAXN, 2):
spf[i] = 2;
for i in range(3, ceil(MAXN ** 0.5), 2):
if spf[i] == i:
for j in range(i*i, MAXN, i):
if spf[j] == j:
spf[j] = i;
## function for storing smallest prime factors (spf) in the array
################## un-comment below 2 lines when using factorization #################
spf = [0 for i in range(MAXN)]
# spf_sieve();
def factoriazation(x):
res = []
for i in range(2, int(x ** 0.5) + 1):
while x % i == 0:
res.append(i)
x //= i
if x != 1:
res.append(x)
return res
## this function is useful for multiple queries only, o/w use
## primefs function above. complexity O(log n)
def factors(n):
res = []
for i in range(1, int(n ** 0.5) + 1):
if n % i == 0:
res.append(i)
res.append(n // i)
return list(set(res))
## taking integer array input
def int_array():
return list(map(int, input().strip().split()));
def float_array():
return list(map(float, input().strip().split()));
## taking string array input
def str_array():
return input().strip().split();
def binary_search(low, high, w, h, n):
while low < high:
mid = low + (high - low) // 2
# print(low, mid, high)
if check(mid, w, h, n):
low = mid + 1
else:
high = mid
return low
## for checking any conditions
def check(beauty, s, n, count):
pass
#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');
alphs = "abcdefghijklmnopqrstuvwxyz"
################### ---------------- TEMPLATE ENDS HERE ---------------- ###################
from itertools import permutations
import math
import bisect as bis
import random
import sys
import collections as collect
# import numpy as np
def solve():
s = string()
ans = []
for i in s:
if ans and i == ans[-1]:
ans.pop()
else:
ans += [i]
print("".join(ans))
# region fastio
# template taken from https://github.com/cheran-senthil/PyRival/blob/master/templates/template.py
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")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
#read()
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
for i in range(1):
solve()
#dmain()
# Comment Read()
# fin_time = datetime.now()
# print("Execution time (for loop): ", (fin_time-init_time))
```
| 88,036 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp thinks about the meaning of life very often. He does this constantly, even when typing in the editor. Every time he starts brooding he can no longer fully concentrate and repeatedly presses the keys that need to be pressed only once. For example, instead of the phrase "how are you" he can type "hhoow aaaare yyoouu".
Polycarp decided to automate the process of correcting such errors. He decided to write a plug-in to the text editor that will remove pairs of identical consecutive letters (if there are any in the text). Of course, this is not exactly what Polycarp needs, but he's got to start from something!
Help Polycarp and write the main plug-in module. Your program should remove from a string all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, the program should remove them as well. Technically, its work should be equivalent to the following: while the string contains a pair of consecutive identical letters, the pair should be deleted. Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.
Input
The input data consists of a single line to be processed. The length of the line is from 1 to 2·105 characters inclusive. The string contains only lowercase Latin letters.
Output
Print the given string after it is processed. It is guaranteed that the result will contain at least one character.
Examples
Input
hhoowaaaareyyoouu
Output
wre
Input
reallazy
Output
rezy
Input
abacabaabacabaa
Output
a
Tags: implementation
Correct Solution:
```
s = input()
stack = []
for char in s:
# print("Comparing", char, "&", stack)
if len(stack)==0:
stack.append(char)
else:
if(char == stack[-1]):
stack.pop()
else:
stack.append(char)
print("".join(stack))
```
| 88,037 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp thinks about the meaning of life very often. He does this constantly, even when typing in the editor. Every time he starts brooding he can no longer fully concentrate and repeatedly presses the keys that need to be pressed only once. For example, instead of the phrase "how are you" he can type "hhoow aaaare yyoouu".
Polycarp decided to automate the process of correcting such errors. He decided to write a plug-in to the text editor that will remove pairs of identical consecutive letters (if there are any in the text). Of course, this is not exactly what Polycarp needs, but he's got to start from something!
Help Polycarp and write the main plug-in module. Your program should remove from a string all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, the program should remove them as well. Technically, its work should be equivalent to the following: while the string contains a pair of consecutive identical letters, the pair should be deleted. Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.
Input
The input data consists of a single line to be processed. The length of the line is from 1 to 2·105 characters inclusive. The string contains only lowercase Latin letters.
Output
Print the given string after it is processed. It is guaranteed that the result will contain at least one character.
Examples
Input
hhoowaaaareyyoouu
Output
wre
Input
reallazy
Output
rezy
Input
abacabaabacabaa
Output
a
Tags: implementation
Correct Solution:
```
s = []
for i in input():
if s and s[-1] == i:
s.pop()
else:
s.append(i)
print("".join(s))
```
| 88,038 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp thinks about the meaning of life very often. He does this constantly, even when typing in the editor. Every time he starts brooding he can no longer fully concentrate and repeatedly presses the keys that need to be pressed only once. For example, instead of the phrase "how are you" he can type "hhoow aaaare yyoouu".
Polycarp decided to automate the process of correcting such errors. He decided to write a plug-in to the text editor that will remove pairs of identical consecutive letters (if there are any in the text). Of course, this is not exactly what Polycarp needs, but he's got to start from something!
Help Polycarp and write the main plug-in module. Your program should remove from a string all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, the program should remove them as well. Technically, its work should be equivalent to the following: while the string contains a pair of consecutive identical letters, the pair should be deleted. Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.
Input
The input data consists of a single line to be processed. The length of the line is from 1 to 2·105 characters inclusive. The string contains only lowercase Latin letters.
Output
Print the given string after it is processed. It is guaranteed that the result will contain at least one character.
Examples
Input
hhoowaaaareyyoouu
Output
wre
Input
reallazy
Output
rezy
Input
abacabaabacabaa
Output
a
Tags: implementation
Correct Solution:
```
s = input()
l = [""]*200000
li = 0
for c in s:
li += 1
if len(l) > 0 and l[li-1] == c:
l[li-1] = ""
li -= 2
else:
l[li] = c
print(*l[:li+1], sep="")
```
| 88,039 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp thinks about the meaning of life very often. He does this constantly, even when typing in the editor. Every time he starts brooding he can no longer fully concentrate and repeatedly presses the keys that need to be pressed only once. For example, instead of the phrase "how are you" he can type "hhoow aaaare yyoouu".
Polycarp decided to automate the process of correcting such errors. He decided to write a plug-in to the text editor that will remove pairs of identical consecutive letters (if there are any in the text). Of course, this is not exactly what Polycarp needs, but he's got to start from something!
Help Polycarp and write the main plug-in module. Your program should remove from a string all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, the program should remove them as well. Technically, its work should be equivalent to the following: while the string contains a pair of consecutive identical letters, the pair should be deleted. Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.
Input
The input data consists of a single line to be processed. The length of the line is from 1 to 2·105 characters inclusive. The string contains only lowercase Latin letters.
Output
Print the given string after it is processed. It is guaranteed that the result will contain at least one character.
Examples
Input
hhoowaaaareyyoouu
Output
wre
Input
reallazy
Output
rezy
Input
abacabaabacabaa
Output
a
Tags: implementation
Correct Solution:
```
p=list(input())
a=[ ] #Creacion d euna lista vacia
a.append(p[0]) #incluir el primer elemnto de p a la lista a
m=len(p) #tamaño de p
i=1
while i<m:
if (len(a)==0):
a.append(p[i])
else:
if(a[-1]==p[i]):
a.pop()
else:
a.append(p[i])
i=i+1
#Convertir listas a strings
print(''.join(a))
```
| 88,040 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp thinks about the meaning of life very often. He does this constantly, even when typing in the editor. Every time he starts brooding he can no longer fully concentrate and repeatedly presses the keys that need to be pressed only once. For example, instead of the phrase "how are you" he can type "hhoow aaaare yyoouu".
Polycarp decided to automate the process of correcting such errors. He decided to write a plug-in to the text editor that will remove pairs of identical consecutive letters (if there are any in the text). Of course, this is not exactly what Polycarp needs, but he's got to start from something!
Help Polycarp and write the main plug-in module. Your program should remove from a string all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, the program should remove them as well. Technically, its work should be equivalent to the following: while the string contains a pair of consecutive identical letters, the pair should be deleted. Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.
Input
The input data consists of a single line to be processed. The length of the line is from 1 to 2·105 characters inclusive. The string contains only lowercase Latin letters.
Output
Print the given string after it is processed. It is guaranteed that the result will contain at least one character.
Examples
Input
hhoowaaaareyyoouu
Output
wre
Input
reallazy
Output
rezy
Input
abacabaabacabaa
Output
a
Tags: implementation
Correct Solution:
```
s=input()
ans=[s[0]]
for i in range(1,len(s)):
if len(ans) and ans[-1]==s[i]:
ans.pop()
else:
ans.append(s[i])
print("".join(ans))
```
| 88,041 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp thinks about the meaning of life very often. He does this constantly, even when typing in the editor. Every time he starts brooding he can no longer fully concentrate and repeatedly presses the keys that need to be pressed only once. For example, instead of the phrase "how are you" he can type "hhoow aaaare yyoouu".
Polycarp decided to automate the process of correcting such errors. He decided to write a plug-in to the text editor that will remove pairs of identical consecutive letters (if there are any in the text). Of course, this is not exactly what Polycarp needs, but he's got to start from something!
Help Polycarp and write the main plug-in module. Your program should remove from a string all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, the program should remove them as well. Technically, its work should be equivalent to the following: while the string contains a pair of consecutive identical letters, the pair should be deleted. Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.
Input
The input data consists of a single line to be processed. The length of the line is from 1 to 2·105 characters inclusive. The string contains only lowercase Latin letters.
Output
Print the given string after it is processed. It is guaranteed that the result will contain at least one character.
Examples
Input
hhoowaaaareyyoouu
Output
wre
Input
reallazy
Output
rezy
Input
abacabaabacabaa
Output
a
Tags: implementation
Correct Solution:
```
string = input()
def plugInX(string):
stringNew = []
for i in string:
if i not in stringNew:
stringNew.append(i)
elif i == stringNew[-1]:
stringNew.pop()
else:
stringNew.append(i)
if len(stringNew) > 0:
for i in stringNew:
print(i, end="")
else:
print("")
plugInX(string)
```
| 88,042 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp thinks about the meaning of life very often. He does this constantly, even when typing in the editor. Every time he starts brooding he can no longer fully concentrate and repeatedly presses the keys that need to be pressed only once. For example, instead of the phrase "how are you" he can type "hhoow aaaare yyoouu".
Polycarp decided to automate the process of correcting such errors. He decided to write a plug-in to the text editor that will remove pairs of identical consecutive letters (if there are any in the text). Of course, this is not exactly what Polycarp needs, but he's got to start from something!
Help Polycarp and write the main plug-in module. Your program should remove from a string all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, the program should remove them as well. Technically, its work should be equivalent to the following: while the string contains a pair of consecutive identical letters, the pair should be deleted. Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.
Input
The input data consists of a single line to be processed. The length of the line is from 1 to 2·105 characters inclusive. The string contains only lowercase Latin letters.
Output
Print the given string after it is processed. It is guaranteed that the result will contain at least one character.
Examples
Input
hhoowaaaareyyoouu
Output
wre
Input
reallazy
Output
rezy
Input
abacabaabacabaa
Output
a
Submitted Solution:
```
line = input()
s = []
for l in line:
if s and s[-1] == l:
s.pop()
else:
s.append(l)
print("".join(s))
```
Yes
| 88,043 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp thinks about the meaning of life very often. He does this constantly, even when typing in the editor. Every time he starts brooding he can no longer fully concentrate and repeatedly presses the keys that need to be pressed only once. For example, instead of the phrase "how are you" he can type "hhoow aaaare yyoouu".
Polycarp decided to automate the process of correcting such errors. He decided to write a plug-in to the text editor that will remove pairs of identical consecutive letters (if there are any in the text). Of course, this is not exactly what Polycarp needs, but he's got to start from something!
Help Polycarp and write the main plug-in module. Your program should remove from a string all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, the program should remove them as well. Technically, its work should be equivalent to the following: while the string contains a pair of consecutive identical letters, the pair should be deleted. Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.
Input
The input data consists of a single line to be processed. The length of the line is from 1 to 2·105 characters inclusive. The string contains only lowercase Latin letters.
Output
Print the given string after it is processed. It is guaranteed that the result will contain at least one character.
Examples
Input
hhoowaaaareyyoouu
Output
wre
Input
reallazy
Output
rezy
Input
abacabaabacabaa
Output
a
Submitted Solution:
```
def main():
s: tuple = tuple(input().strip())
STACK: list = []
for i in range(len(s)):
if not STACK:
STACK.append(s[i])
elif STACK[-1] == s[i]:
STACK.pop()
else:
STACK.append(s[i])
print(''.join(STACK))
if __name__ == '__main__':
main()
```
Yes
| 88,044 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp thinks about the meaning of life very often. He does this constantly, even when typing in the editor. Every time he starts brooding he can no longer fully concentrate and repeatedly presses the keys that need to be pressed only once. For example, instead of the phrase "how are you" he can type "hhoow aaaare yyoouu".
Polycarp decided to automate the process of correcting such errors. He decided to write a plug-in to the text editor that will remove pairs of identical consecutive letters (if there are any in the text). Of course, this is not exactly what Polycarp needs, but he's got to start from something!
Help Polycarp and write the main plug-in module. Your program should remove from a string all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, the program should remove them as well. Technically, its work should be equivalent to the following: while the string contains a pair of consecutive identical letters, the pair should be deleted. Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.
Input
The input data consists of a single line to be processed. The length of the line is from 1 to 2·105 characters inclusive. The string contains only lowercase Latin letters.
Output
Print the given string after it is processed. It is guaranteed that the result will contain at least one character.
Examples
Input
hhoowaaaareyyoouu
Output
wre
Input
reallazy
Output
rezy
Input
abacabaabacabaa
Output
a
Submitted Solution:
```
s = input()
l = []
for c in s:
if len(l) > 0 and l[-1] == c:
del l[-1]
else:
l += [c]
print(*l, sep="")
```
Yes
| 88,045 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp thinks about the meaning of life very often. He does this constantly, even when typing in the editor. Every time he starts brooding he can no longer fully concentrate and repeatedly presses the keys that need to be pressed only once. For example, instead of the phrase "how are you" he can type "hhoow aaaare yyoouu".
Polycarp decided to automate the process of correcting such errors. He decided to write a plug-in to the text editor that will remove pairs of identical consecutive letters (if there are any in the text). Of course, this is not exactly what Polycarp needs, but he's got to start from something!
Help Polycarp and write the main plug-in module. Your program should remove from a string all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, the program should remove them as well. Technically, its work should be equivalent to the following: while the string contains a pair of consecutive identical letters, the pair should be deleted. Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.
Input
The input data consists of a single line to be processed. The length of the line is from 1 to 2·105 characters inclusive. The string contains only lowercase Latin letters.
Output
Print the given string after it is processed. It is guaranteed that the result will contain at least one character.
Examples
Input
hhoowaaaareyyoouu
Output
wre
Input
reallazy
Output
rezy
Input
abacabaabacabaa
Output
a
Submitted Solution:
```
'''
Fuad Ashraful Mehmet
University of Asia Pacific,Bangladesh
Date:31th March 2020
'''
def JoyBangla(s):
st=[]
for c in s:
if not st:
st.append(c)
elif st[-1]==c:
st.pop()
else:
st.append(c)
print(*st,sep='')
JoyBangla(input())
```
Yes
| 88,046 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp thinks about the meaning of life very often. He does this constantly, even when typing in the editor. Every time he starts brooding he can no longer fully concentrate and repeatedly presses the keys that need to be pressed only once. For example, instead of the phrase "how are you" he can type "hhoow aaaare yyoouu".
Polycarp decided to automate the process of correcting such errors. He decided to write a plug-in to the text editor that will remove pairs of identical consecutive letters (if there are any in the text). Of course, this is not exactly what Polycarp needs, but he's got to start from something!
Help Polycarp and write the main plug-in module. Your program should remove from a string all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, the program should remove them as well. Technically, its work should be equivalent to the following: while the string contains a pair of consecutive identical letters, the pair should be deleted. Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.
Input
The input data consists of a single line to be processed. The length of the line is from 1 to 2·105 characters inclusive. The string contains only lowercase Latin letters.
Output
Print the given string after it is processed. It is guaranteed that the result will contain at least one character.
Examples
Input
hhoowaaaareyyoouu
Output
wre
Input
reallazy
Output
rezy
Input
abacabaabacabaa
Output
a
Submitted Solution:
```
import sys
from collections import OrderedDict
f = sys.stdin
#f = open("input.txt", "r")
a = f.readline().strip()
a1 = list(OrderedDict.fromkeys(a))
ans = ""
for i in a1:
if a.count(i)<=1:
ans += i
print(ans)
```
No
| 88,047 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp thinks about the meaning of life very often. He does this constantly, even when typing in the editor. Every time he starts brooding he can no longer fully concentrate and repeatedly presses the keys that need to be pressed only once. For example, instead of the phrase "how are you" he can type "hhoow aaaare yyoouu".
Polycarp decided to automate the process of correcting such errors. He decided to write a plug-in to the text editor that will remove pairs of identical consecutive letters (if there are any in the text). Of course, this is not exactly what Polycarp needs, but he's got to start from something!
Help Polycarp and write the main plug-in module. Your program should remove from a string all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, the program should remove them as well. Technically, its work should be equivalent to the following: while the string contains a pair of consecutive identical letters, the pair should be deleted. Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.
Input
The input data consists of a single line to be processed. The length of the line is from 1 to 2·105 characters inclusive. The string contains only lowercase Latin letters.
Output
Print the given string after it is processed. It is guaranteed that the result will contain at least one character.
Examples
Input
hhoowaaaareyyoouu
Output
wre
Input
reallazy
Output
rezy
Input
abacabaabacabaa
Output
a
Submitted Solution:
```
from collections import deque
def removeRepeats(letters):
queue = deque(list(letters))
if len(queue) == 1:
return letters
answer = ""
candidate = queue.popleft()
repeat = 0
for x in range(len(letters)):
if queue:
front = queue.popleft()
else:
break
if candidate == front:
repeat = 1
continue
else:
if repeat:
candidate = front
repeat = 0
continue
else:
answer += candidate
candidate = front
if letters[-1] != letters[-2]:
answer += letters[-1]
return answer
s = list(input())
old = removeRepeats(s)
new = removeRepeats(old)
while 1:
if new == old:
break
old = removeRepeats(new)
new = removeRepeats(old)
print(new)
x = 5
```
No
| 88,048 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp thinks about the meaning of life very often. He does this constantly, even when typing in the editor. Every time he starts brooding he can no longer fully concentrate and repeatedly presses the keys that need to be pressed only once. For example, instead of the phrase "how are you" he can type "hhoow aaaare yyoouu".
Polycarp decided to automate the process of correcting such errors. He decided to write a plug-in to the text editor that will remove pairs of identical consecutive letters (if there are any in the text). Of course, this is not exactly what Polycarp needs, but he's got to start from something!
Help Polycarp and write the main plug-in module. Your program should remove from a string all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, the program should remove them as well. Technically, its work should be equivalent to the following: while the string contains a pair of consecutive identical letters, the pair should be deleted. Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.
Input
The input data consists of a single line to be processed. The length of the line is from 1 to 2·105 characters inclusive. The string contains only lowercase Latin letters.
Output
Print the given string after it is processed. It is guaranteed that the result will contain at least one character.
Examples
Input
hhoowaaaareyyoouu
Output
wre
Input
reallazy
Output
rezy
Input
abacabaabacabaa
Output
a
Submitted Solution:
```
mistake = list(input())
if len(mistake) == 1:
print(mistake[0])
elif mistake[0] == mistake[1] and len(mistake) == 2:
print("")
else:
iterator = 2
answer = []
while mistake[0] == mistake[1] and len(mistake) >1:
mistake.pop(0)
mistake.pop(0)
answer.append(mistake[0])
answer.append(mistake[1])
while iterator < len(mistake):
if answer[-1] != mistake[iterator]:
answer.append(mistake[iterator])
else:
answer.pop(-1)
if len(answer) == 0 and iterator+1 < len(mistake):
answer.append(mistake[iterator])
iterator += 1
elif len(answer) == 0 and mistake[-1] == mistake[iterator]:
answer.append(mistake[-1])
iterator += 1
print("")
for x in answer:
print(x, end = "")
```
No
| 88,049 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp thinks about the meaning of life very often. He does this constantly, even when typing in the editor. Every time he starts brooding he can no longer fully concentrate and repeatedly presses the keys that need to be pressed only once. For example, instead of the phrase "how are you" he can type "hhoow aaaare yyoouu".
Polycarp decided to automate the process of correcting such errors. He decided to write a plug-in to the text editor that will remove pairs of identical consecutive letters (if there are any in the text). Of course, this is not exactly what Polycarp needs, but he's got to start from something!
Help Polycarp and write the main plug-in module. Your program should remove from a string all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, the program should remove them as well. Technically, its work should be equivalent to the following: while the string contains a pair of consecutive identical letters, the pair should be deleted. Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.
Input
The input data consists of a single line to be processed. The length of the line is from 1 to 2·105 characters inclusive. The string contains only lowercase Latin letters.
Output
Print the given string after it is processed. It is guaranteed that the result will contain at least one character.
Examples
Input
hhoowaaaareyyoouu
Output
wre
Input
reallazy
Output
rezy
Input
abacabaabacabaa
Output
a
Submitted Solution:
```
letters = input()
lettersList = []
newList = []
for y in letters:
lettersList.append(y)
for x in lettersList:
if x not in newList:
newList.append(x)
elif x in newList:
#index = newList.index(x)
lettersList.pop()
print(''.join(lettersList))
```
No
| 88,050 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has just attempted to pass the driving test. He ran over the straight road with the signs of four types.
* speed limit: this sign comes with a positive integer number — maximal speed of the car after the sign (cancel the action of the previous sign of this type);
* overtake is allowed: this sign means that after some car meets it, it can overtake any other car;
* no speed limit: this sign cancels speed limit if any (car can move with arbitrary speed after this sign);
* no overtake allowed: some car can't overtake any other car after this sign.
Polycarp goes past the signs consequentially, each new sign cancels the action of all the previous signs of it's kind (speed limit/overtake). It is possible that two or more "no overtake allowed" signs go one after another with zero "overtake is allowed" signs between them. It works with "no speed limit" and "overtake is allowed" signs as well.
In the beginning of the ride overtake is allowed and there is no speed limit.
You are given the sequence of events in chronological order — events which happened to Polycarp during the ride. There are events of following types:
1. Polycarp changes the speed of his car to specified (this event comes with a positive integer number);
2. Polycarp's car overtakes the other car;
3. Polycarp's car goes past the "speed limit" sign (this sign comes with a positive integer);
4. Polycarp's car goes past the "overtake is allowed" sign;
5. Polycarp's car goes past the "no speed limit";
6. Polycarp's car goes past the "no overtake allowed";
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
After the exam Polycarp can justify his rule violations by telling the driving instructor that he just didn't notice some of the signs. What is the minimal number of signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view?
Input
The first line contains one integer number n (1 ≤ n ≤ 2·105) — number of events.
Each of the next n lines starts with integer t (1 ≤ t ≤ 6) — the type of the event.
An integer s (1 ≤ s ≤ 300) follows in the query of the first and the third type (if it is the query of first type, then it's new speed of Polycarp's car, if it is the query of third type, then it's new speed limit).
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
Output
Print the minimal number of road signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view.
Examples
Input
11
1 100
3 70
4
2
3 120
5
3 120
6
1 150
4
3 300
Output
2
Input
5
1 100
3 200
2
4
5
Output
0
Input
7
1 20
2
6
4
6
6
2
Output
2
Note
In the first example Polycarp should say he didn't notice the "speed limit" sign with the limit of 70 and the second "speed limit" sign with the limit of 120.
In the second example Polycarp didn't make any rule violation.
In the third example Polycarp should say he didn't notice both "no overtake allowed" that came after "overtake is allowed" sign.
Tags: data structures, dp, greedy
Correct Solution:
```
n=int(input())
s=0
ms=[]
o=1
z=0
q=0
for _ in range(n):
a=list(map(int,input().split()))
if a[0]==1:
s=a[1]
while len(ms)>0 and ms[-1]<s:
z+=1
ms.pop(-1)
elif a[0]==2:
if o!=1:
z+=(1-o)
o=1
elif a[0]==3:
ms.append(a[1])
while len(ms)>0 and ms[-1]<s:
z+=1
ms.pop(-1)
elif a[0]==4:
o=1
elif a[0]==5:
ms=[]
elif a[0]==6:
o-=1
print(z)
```
| 88,051 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has just attempted to pass the driving test. He ran over the straight road with the signs of four types.
* speed limit: this sign comes with a positive integer number — maximal speed of the car after the sign (cancel the action of the previous sign of this type);
* overtake is allowed: this sign means that after some car meets it, it can overtake any other car;
* no speed limit: this sign cancels speed limit if any (car can move with arbitrary speed after this sign);
* no overtake allowed: some car can't overtake any other car after this sign.
Polycarp goes past the signs consequentially, each new sign cancels the action of all the previous signs of it's kind (speed limit/overtake). It is possible that two or more "no overtake allowed" signs go one after another with zero "overtake is allowed" signs between them. It works with "no speed limit" and "overtake is allowed" signs as well.
In the beginning of the ride overtake is allowed and there is no speed limit.
You are given the sequence of events in chronological order — events which happened to Polycarp during the ride. There are events of following types:
1. Polycarp changes the speed of his car to specified (this event comes with a positive integer number);
2. Polycarp's car overtakes the other car;
3. Polycarp's car goes past the "speed limit" sign (this sign comes with a positive integer);
4. Polycarp's car goes past the "overtake is allowed" sign;
5. Polycarp's car goes past the "no speed limit";
6. Polycarp's car goes past the "no overtake allowed";
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
After the exam Polycarp can justify his rule violations by telling the driving instructor that he just didn't notice some of the signs. What is the minimal number of signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view?
Input
The first line contains one integer number n (1 ≤ n ≤ 2·105) — number of events.
Each of the next n lines starts with integer t (1 ≤ t ≤ 6) — the type of the event.
An integer s (1 ≤ s ≤ 300) follows in the query of the first and the third type (if it is the query of first type, then it's new speed of Polycarp's car, if it is the query of third type, then it's new speed limit).
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
Output
Print the minimal number of road signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view.
Examples
Input
11
1 100
3 70
4
2
3 120
5
3 120
6
1 150
4
3 300
Output
2
Input
5
1 100
3 200
2
4
5
Output
0
Input
7
1 20
2
6
4
6
6
2
Output
2
Note
In the first example Polycarp should say he didn't notice the "speed limit" sign with the limit of 70 and the second "speed limit" sign with the limit of 120.
In the second example Polycarp didn't make any rule violation.
In the third example Polycarp should say he didn't notice both "no overtake allowed" that came after "overtake is allowed" sign.
Tags: data structures, dp, greedy
Correct Solution:
```
n=int(input())
st=[0]*n
top=-1
curr_speed=0
ans=0
c=0
while(n>0):
n-=1
a=list(map(int,input().split()))
if(a[0]==1):
curr_speed=a[1]
if(top>=0 and curr_speed>st[top]):
while(curr_speed>st[top] and top>=0):
top-=1
ans+=1
if(a[0]==4):
c=0
elif(a[0]==6):
c+=1
if(a[0]==5):
top=-1
if(a[0]==2):
ans+=c
c=0
if(a[0]==3):
if(curr_speed>a[1]):
ans+=1
else:
st[top+1]=a[1]
top+=1
print(ans)
```
| 88,052 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has just attempted to pass the driving test. He ran over the straight road with the signs of four types.
* speed limit: this sign comes with a positive integer number — maximal speed of the car after the sign (cancel the action of the previous sign of this type);
* overtake is allowed: this sign means that after some car meets it, it can overtake any other car;
* no speed limit: this sign cancels speed limit if any (car can move with arbitrary speed after this sign);
* no overtake allowed: some car can't overtake any other car after this sign.
Polycarp goes past the signs consequentially, each new sign cancels the action of all the previous signs of it's kind (speed limit/overtake). It is possible that two or more "no overtake allowed" signs go one after another with zero "overtake is allowed" signs between them. It works with "no speed limit" and "overtake is allowed" signs as well.
In the beginning of the ride overtake is allowed and there is no speed limit.
You are given the sequence of events in chronological order — events which happened to Polycarp during the ride. There are events of following types:
1. Polycarp changes the speed of his car to specified (this event comes with a positive integer number);
2. Polycarp's car overtakes the other car;
3. Polycarp's car goes past the "speed limit" sign (this sign comes with a positive integer);
4. Polycarp's car goes past the "overtake is allowed" sign;
5. Polycarp's car goes past the "no speed limit";
6. Polycarp's car goes past the "no overtake allowed";
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
After the exam Polycarp can justify his rule violations by telling the driving instructor that he just didn't notice some of the signs. What is the minimal number of signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view?
Input
The first line contains one integer number n (1 ≤ n ≤ 2·105) — number of events.
Each of the next n lines starts with integer t (1 ≤ t ≤ 6) — the type of the event.
An integer s (1 ≤ s ≤ 300) follows in the query of the first and the third type (if it is the query of first type, then it's new speed of Polycarp's car, if it is the query of third type, then it's new speed limit).
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
Output
Print the minimal number of road signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view.
Examples
Input
11
1 100
3 70
4
2
3 120
5
3 120
6
1 150
4
3 300
Output
2
Input
5
1 100
3 200
2
4
5
Output
0
Input
7
1 20
2
6
4
6
6
2
Output
2
Note
In the first example Polycarp should say he didn't notice the "speed limit" sign with the limit of 70 and the second "speed limit" sign with the limit of 120.
In the second example Polycarp didn't make any rule violation.
In the third example Polycarp should say he didn't notice both "no overtake allowed" that came after "overtake is allowed" sign.
Tags: data structures, dp, greedy
Correct Solution:
```
n = int(input())
no_overtake = 0
speedlimit = [float("inf")]
speed = 0
vio = 0
ot_ignored = False
for _ in range(n):
cmd = input().split()
if cmd[0] == '1':
speed = int(cmd[1])
while speed > speedlimit[-1]:
vio += 1
speedlimit.pop(-1)
elif cmd[0] == '2':
if no_overtake and not ot_ignored:
vio += no_overtake
ot_ignored = True
elif cmd[0] == '3':
speedlimit.append(int(cmd[1]))
if speed > speedlimit[-1]:
vio += 1
speedlimit.pop(-1)
elif cmd[0] == '4':
no_overtake = 0
ot_ignored = False
elif cmd[0] == '5':
speedlimit = [float('inf')]
elif cmd[0] == '6':
if ot_ignored == 1:
no_overtake = 0
no_overtake += 1
ot_ignored = False
print(vio)
```
| 88,053 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has just attempted to pass the driving test. He ran over the straight road with the signs of four types.
* speed limit: this sign comes with a positive integer number — maximal speed of the car after the sign (cancel the action of the previous sign of this type);
* overtake is allowed: this sign means that after some car meets it, it can overtake any other car;
* no speed limit: this sign cancels speed limit if any (car can move with arbitrary speed after this sign);
* no overtake allowed: some car can't overtake any other car after this sign.
Polycarp goes past the signs consequentially, each new sign cancels the action of all the previous signs of it's kind (speed limit/overtake). It is possible that two or more "no overtake allowed" signs go one after another with zero "overtake is allowed" signs between them. It works with "no speed limit" and "overtake is allowed" signs as well.
In the beginning of the ride overtake is allowed and there is no speed limit.
You are given the sequence of events in chronological order — events which happened to Polycarp during the ride. There are events of following types:
1. Polycarp changes the speed of his car to specified (this event comes with a positive integer number);
2. Polycarp's car overtakes the other car;
3. Polycarp's car goes past the "speed limit" sign (this sign comes with a positive integer);
4. Polycarp's car goes past the "overtake is allowed" sign;
5. Polycarp's car goes past the "no speed limit";
6. Polycarp's car goes past the "no overtake allowed";
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
After the exam Polycarp can justify his rule violations by telling the driving instructor that he just didn't notice some of the signs. What is the minimal number of signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view?
Input
The first line contains one integer number n (1 ≤ n ≤ 2·105) — number of events.
Each of the next n lines starts with integer t (1 ≤ t ≤ 6) — the type of the event.
An integer s (1 ≤ s ≤ 300) follows in the query of the first and the third type (if it is the query of first type, then it's new speed of Polycarp's car, if it is the query of third type, then it's new speed limit).
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
Output
Print the minimal number of road signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view.
Examples
Input
11
1 100
3 70
4
2
3 120
5
3 120
6
1 150
4
3 300
Output
2
Input
5
1 100
3 200
2
4
5
Output
0
Input
7
1 20
2
6
4
6
6
2
Output
2
Note
In the first example Polycarp should say he didn't notice the "speed limit" sign with the limit of 70 and the second "speed limit" sign with the limit of 120.
In the second example Polycarp didn't make any rule violation.
In the third example Polycarp should say he didn't notice both "no overtake allowed" that came after "overtake is allowed" sign.
Tags: data structures, dp, greedy
Correct Solution:
```
#!/usr/bin/env python3
from sys import stdin, stdout
from math import inf
def rint():
return map(int, stdin.readline().split())
#lines = stdin.readlines()
csp = 0
ov = 1
sign = []
ans = 0
spst = [inf]
ovst = [1]
n = int(input())
for i in range(n):
sign = list(rint())
if sign[0] == 1:
csp = sign[1]
while csp > spst[-1]:
spst.pop()
ans += 1
if sign[0] == 2:
while ovst[-1] == 0:
ovst.pop()
ans += 1
if sign[0] == 3:
if sign[1] >= csp:
spst.append(sign[1])
else:
ans += 1
if sign[0] == 4:
ovst.append(1)
if sign[0] == 5:
spst.append(inf)
if sign[0] == 6:
ovst.append(0)
print(ans)
```
| 88,054 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has just attempted to pass the driving test. He ran over the straight road with the signs of four types.
* speed limit: this sign comes with a positive integer number — maximal speed of the car after the sign (cancel the action of the previous sign of this type);
* overtake is allowed: this sign means that after some car meets it, it can overtake any other car;
* no speed limit: this sign cancels speed limit if any (car can move with arbitrary speed after this sign);
* no overtake allowed: some car can't overtake any other car after this sign.
Polycarp goes past the signs consequentially, each new sign cancels the action of all the previous signs of it's kind (speed limit/overtake). It is possible that two or more "no overtake allowed" signs go one after another with zero "overtake is allowed" signs between them. It works with "no speed limit" and "overtake is allowed" signs as well.
In the beginning of the ride overtake is allowed and there is no speed limit.
You are given the sequence of events in chronological order — events which happened to Polycarp during the ride. There are events of following types:
1. Polycarp changes the speed of his car to specified (this event comes with a positive integer number);
2. Polycarp's car overtakes the other car;
3. Polycarp's car goes past the "speed limit" sign (this sign comes with a positive integer);
4. Polycarp's car goes past the "overtake is allowed" sign;
5. Polycarp's car goes past the "no speed limit";
6. Polycarp's car goes past the "no overtake allowed";
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
After the exam Polycarp can justify his rule violations by telling the driving instructor that he just didn't notice some of the signs. What is the minimal number of signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view?
Input
The first line contains one integer number n (1 ≤ n ≤ 2·105) — number of events.
Each of the next n lines starts with integer t (1 ≤ t ≤ 6) — the type of the event.
An integer s (1 ≤ s ≤ 300) follows in the query of the first and the third type (if it is the query of first type, then it's new speed of Polycarp's car, if it is the query of third type, then it's new speed limit).
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
Output
Print the minimal number of road signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view.
Examples
Input
11
1 100
3 70
4
2
3 120
5
3 120
6
1 150
4
3 300
Output
2
Input
5
1 100
3 200
2
4
5
Output
0
Input
7
1 20
2
6
4
6
6
2
Output
2
Note
In the first example Polycarp should say he didn't notice the "speed limit" sign with the limit of 70 and the second "speed limit" sign with the limit of 120.
In the second example Polycarp didn't make any rule violation.
In the third example Polycarp should say he didn't notice both "no overtake allowed" that came after "overtake is allowed" sign.
Tags: data structures, dp, greedy
Correct Solution:
```
import sys
n = int(input())
a = [list(map(int, input().split())) for i in range(n)]
c2 = 0
speed = 0
c = 0
st = []
for i in a:
if i[0] == 1:
speed = i[1]
while len(st) and st[-1] < speed:
c += 1
st.pop()
elif i[0] == 2 and c2 > 0:
c += c2
c2 = 0
elif i[0] == 3:
st.append(i[1])
while len(st) and st[-1] < speed:
c += 1
st.pop()
elif i[0] == 4:
c2 = 0
elif i[0] == 5:
st = []
elif i[0] == 6:
c2 += 1
print(c)
```
| 88,055 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has just attempted to pass the driving test. He ran over the straight road with the signs of four types.
* speed limit: this sign comes with a positive integer number — maximal speed of the car after the sign (cancel the action of the previous sign of this type);
* overtake is allowed: this sign means that after some car meets it, it can overtake any other car;
* no speed limit: this sign cancels speed limit if any (car can move with arbitrary speed after this sign);
* no overtake allowed: some car can't overtake any other car after this sign.
Polycarp goes past the signs consequentially, each new sign cancels the action of all the previous signs of it's kind (speed limit/overtake). It is possible that two or more "no overtake allowed" signs go one after another with zero "overtake is allowed" signs between them. It works with "no speed limit" and "overtake is allowed" signs as well.
In the beginning of the ride overtake is allowed and there is no speed limit.
You are given the sequence of events in chronological order — events which happened to Polycarp during the ride. There are events of following types:
1. Polycarp changes the speed of his car to specified (this event comes with a positive integer number);
2. Polycarp's car overtakes the other car;
3. Polycarp's car goes past the "speed limit" sign (this sign comes with a positive integer);
4. Polycarp's car goes past the "overtake is allowed" sign;
5. Polycarp's car goes past the "no speed limit";
6. Polycarp's car goes past the "no overtake allowed";
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
After the exam Polycarp can justify his rule violations by telling the driving instructor that he just didn't notice some of the signs. What is the minimal number of signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view?
Input
The first line contains one integer number n (1 ≤ n ≤ 2·105) — number of events.
Each of the next n lines starts with integer t (1 ≤ t ≤ 6) — the type of the event.
An integer s (1 ≤ s ≤ 300) follows in the query of the first and the third type (if it is the query of first type, then it's new speed of Polycarp's car, if it is the query of third type, then it's new speed limit).
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
Output
Print the minimal number of road signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view.
Examples
Input
11
1 100
3 70
4
2
3 120
5
3 120
6
1 150
4
3 300
Output
2
Input
5
1 100
3 200
2
4
5
Output
0
Input
7
1 20
2
6
4
6
6
2
Output
2
Note
In the first example Polycarp should say he didn't notice the "speed limit" sign with the limit of 70 and the second "speed limit" sign with the limit of 120.
In the second example Polycarp didn't make any rule violation.
In the third example Polycarp should say he didn't notice both "no overtake allowed" that came after "overtake is allowed" sign.
Tags: data structures, dp, greedy
Correct Solution:
```
n = int(input())
speed = 0
overtake_allowed = True
just_overtake_violation = False
speed_signs = []
violations = 0
overtake_passed = 0
for i in range(n):
arr = [int(a) for a in input().split(' ')]
if arr[0] == 1:
if arr[1] > speed:
speed = arr[1]
for j in range(len(speed_signs) - 1, -1, -1):
if speed > speed_signs[j]:
violations += 1
del speed_signs[j]
else:
break
else:
speed = arr[1]
if arr[0] == 2:
violations += overtake_passed
overtake_passed = 0
if arr[0] == 3:
if speed > arr[1]:
violations += 1
else:
speed_signs.append(arr[1])
if arr[0] == 4:
overtake_passed = 0
if arr[0] == 5:
speed_signs = []
if arr[0] == 6:
overtake_passed += 1
# print(speed_signs)
# print(violated_speed)
# print(overtake_signs)
# print(violated_overtake)
print(violations)
```
| 88,056 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has just attempted to pass the driving test. He ran over the straight road with the signs of four types.
* speed limit: this sign comes with a positive integer number — maximal speed of the car after the sign (cancel the action of the previous sign of this type);
* overtake is allowed: this sign means that after some car meets it, it can overtake any other car;
* no speed limit: this sign cancels speed limit if any (car can move with arbitrary speed after this sign);
* no overtake allowed: some car can't overtake any other car after this sign.
Polycarp goes past the signs consequentially, each new sign cancels the action of all the previous signs of it's kind (speed limit/overtake). It is possible that two or more "no overtake allowed" signs go one after another with zero "overtake is allowed" signs between them. It works with "no speed limit" and "overtake is allowed" signs as well.
In the beginning of the ride overtake is allowed and there is no speed limit.
You are given the sequence of events in chronological order — events which happened to Polycarp during the ride. There are events of following types:
1. Polycarp changes the speed of his car to specified (this event comes with a positive integer number);
2. Polycarp's car overtakes the other car;
3. Polycarp's car goes past the "speed limit" sign (this sign comes with a positive integer);
4. Polycarp's car goes past the "overtake is allowed" sign;
5. Polycarp's car goes past the "no speed limit";
6. Polycarp's car goes past the "no overtake allowed";
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
After the exam Polycarp can justify his rule violations by telling the driving instructor that he just didn't notice some of the signs. What is the minimal number of signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view?
Input
The first line contains one integer number n (1 ≤ n ≤ 2·105) — number of events.
Each of the next n lines starts with integer t (1 ≤ t ≤ 6) — the type of the event.
An integer s (1 ≤ s ≤ 300) follows in the query of the first and the third type (if it is the query of first type, then it's new speed of Polycarp's car, if it is the query of third type, then it's new speed limit).
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
Output
Print the minimal number of road signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view.
Examples
Input
11
1 100
3 70
4
2
3 120
5
3 120
6
1 150
4
3 300
Output
2
Input
5
1 100
3 200
2
4
5
Output
0
Input
7
1 20
2
6
4
6
6
2
Output
2
Note
In the first example Polycarp should say he didn't notice the "speed limit" sign with the limit of 70 and the second "speed limit" sign with the limit of 120.
In the second example Polycarp didn't make any rule violation.
In the third example Polycarp should say he didn't notice both "no overtake allowed" that came after "overtake is allowed" sign.
Tags: data structures, dp, greedy
Correct Solution:
```
speeds = [1000000]
overtakes = [True]
count = 0
speed = 0
n = int(input())
for e in range(n):
inp = list(map(int, input().split()))
# print(inp)
if inp[0] == 4:
overtakes.append(True)
elif inp[0] == 6:
overtakes.append(False)
elif inp[0] == 5:
speeds.append(1000000)
elif inp[0] == 3:
speeds.append(inp[1])
while speed > speeds[-1]:
count += 1
speeds.pop()
elif inp[0] == 2:
while not overtakes[-1]:
count += 1
overtakes.pop()
else:
while inp[1] > speeds[-1]:
count += 1
speeds.pop()
speed = inp[1]
print(count)
"""
Polycarp changes the speed of his car to specified (this event comes with a positive integer number);
Polycarp's car overtakes the other car;
Polycarp's car goes past the "speed limit" sign (this sign comes with a positive integer);
Polycarp's car goes past the "overtake is allowed" sign;
Polycarp's car goes past the "no speed limit";
Polycarp's car goes past the "no overtake allowed";
"""
```
| 88,057 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has just attempted to pass the driving test. He ran over the straight road with the signs of four types.
* speed limit: this sign comes with a positive integer number — maximal speed of the car after the sign (cancel the action of the previous sign of this type);
* overtake is allowed: this sign means that after some car meets it, it can overtake any other car;
* no speed limit: this sign cancels speed limit if any (car can move with arbitrary speed after this sign);
* no overtake allowed: some car can't overtake any other car after this sign.
Polycarp goes past the signs consequentially, each new sign cancels the action of all the previous signs of it's kind (speed limit/overtake). It is possible that two or more "no overtake allowed" signs go one after another with zero "overtake is allowed" signs between them. It works with "no speed limit" and "overtake is allowed" signs as well.
In the beginning of the ride overtake is allowed and there is no speed limit.
You are given the sequence of events in chronological order — events which happened to Polycarp during the ride. There are events of following types:
1. Polycarp changes the speed of his car to specified (this event comes with a positive integer number);
2. Polycarp's car overtakes the other car;
3. Polycarp's car goes past the "speed limit" sign (this sign comes with a positive integer);
4. Polycarp's car goes past the "overtake is allowed" sign;
5. Polycarp's car goes past the "no speed limit";
6. Polycarp's car goes past the "no overtake allowed";
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
After the exam Polycarp can justify his rule violations by telling the driving instructor that he just didn't notice some of the signs. What is the minimal number of signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view?
Input
The first line contains one integer number n (1 ≤ n ≤ 2·105) — number of events.
Each of the next n lines starts with integer t (1 ≤ t ≤ 6) — the type of the event.
An integer s (1 ≤ s ≤ 300) follows in the query of the first and the third type (if it is the query of first type, then it's new speed of Polycarp's car, if it is the query of third type, then it's new speed limit).
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
Output
Print the minimal number of road signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view.
Examples
Input
11
1 100
3 70
4
2
3 120
5
3 120
6
1 150
4
3 300
Output
2
Input
5
1 100
3 200
2
4
5
Output
0
Input
7
1 20
2
6
4
6
6
2
Output
2
Note
In the first example Polycarp should say he didn't notice the "speed limit" sign with the limit of 70 and the second "speed limit" sign with the limit of 120.
In the second example Polycarp didn't make any rule violation.
In the third example Polycarp should say he didn't notice both "no overtake allowed" that came after "overtake is allowed" sign.
Tags: data structures, dp, greedy
Correct Solution:
```
class DrivingTest:
car_speed = 0
speed_limit = [0] # no speed limit
overtake_allowed = [True]
missed_signs = 0
def __init__(self, test=False):
if not test:
event_count = int(input())
for counter in range(event_count):
event = input()
self.handle_event(event)
# print(f"Speed: {self.car_speed}, Limit: {self.speed_limit}, Overtake allowed: {self.overtake_allowed}, Missed signs:{self.missed_signs}")
print(self.missed_signs)
def handle_event(self, event):
event = list(map(int, event.split()))
event_type = event[0]
if event_type == 1:
self.set_car_speed(event[1])
if event_type == 2:
self.overtake_car()
if event_type == 3:
self.set_speed_limit(event[1])
if event_type == 4:
self.set_overtake_allowed(True)
if event_type == 5:
self.set_speed_limit(0)
if event_type == 6:
self.set_overtake_allowed(False)
def set_car_speed(self, car_speed):
self.car_speed = car_speed
self.check_car_speed()
def set_speed_limit(self, speed_limit):
self.speed_limit.append(speed_limit)
self.check_car_speed()
def check_car_speed(self):
latest_speed_limit = self.speed_limit[-1]
while latest_speed_limit != 0 and latest_speed_limit < self.car_speed:
self.missed_signs += 1
self.speed_limit.pop()
latest_speed_limit = self.speed_limit[-1]
def set_overtake_allowed(self, overtake_allowed):
self.overtake_allowed.append(overtake_allowed)
def overtake_car(self):
while not self.overtake_allowed[-1]:
self.missed_signs += 1
self.overtake_allowed.pop()
def get_missed_signs(self):
return self.missed_signs
if __name__ == "__main__":
DrivingTest()
```
| 88,058 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has just attempted to pass the driving test. He ran over the straight road with the signs of four types.
* speed limit: this sign comes with a positive integer number — maximal speed of the car after the sign (cancel the action of the previous sign of this type);
* overtake is allowed: this sign means that after some car meets it, it can overtake any other car;
* no speed limit: this sign cancels speed limit if any (car can move with arbitrary speed after this sign);
* no overtake allowed: some car can't overtake any other car after this sign.
Polycarp goes past the signs consequentially, each new sign cancels the action of all the previous signs of it's kind (speed limit/overtake). It is possible that two or more "no overtake allowed" signs go one after another with zero "overtake is allowed" signs between them. It works with "no speed limit" and "overtake is allowed" signs as well.
In the beginning of the ride overtake is allowed and there is no speed limit.
You are given the sequence of events in chronological order — events which happened to Polycarp during the ride. There are events of following types:
1. Polycarp changes the speed of his car to specified (this event comes with a positive integer number);
2. Polycarp's car overtakes the other car;
3. Polycarp's car goes past the "speed limit" sign (this sign comes with a positive integer);
4. Polycarp's car goes past the "overtake is allowed" sign;
5. Polycarp's car goes past the "no speed limit";
6. Polycarp's car goes past the "no overtake allowed";
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
After the exam Polycarp can justify his rule violations by telling the driving instructor that he just didn't notice some of the signs. What is the minimal number of signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view?
Input
The first line contains one integer number n (1 ≤ n ≤ 2·105) — number of events.
Each of the next n lines starts with integer t (1 ≤ t ≤ 6) — the type of the event.
An integer s (1 ≤ s ≤ 300) follows in the query of the first and the third type (if it is the query of first type, then it's new speed of Polycarp's car, if it is the query of third type, then it's new speed limit).
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
Output
Print the minimal number of road signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view.
Examples
Input
11
1 100
3 70
4
2
3 120
5
3 120
6
1 150
4
3 300
Output
2
Input
5
1 100
3 200
2
4
5
Output
0
Input
7
1 20
2
6
4
6
6
2
Output
2
Note
In the first example Polycarp should say he didn't notice the "speed limit" sign with the limit of 70 and the second "speed limit" sign with the limit of 120.
In the second example Polycarp didn't make any rule violation.
In the third example Polycarp should say he didn't notice both "no overtake allowed" that came after "overtake is allowed" sign.
Submitted Solution:
```
MAX_SPEED = 300
n = int(input())
actions = [list(map(int, input().split())) for _ in range(n)]
speed = [0 for i in range(n)]
for i in range(n):
if actions[i][0] == 1:
lastSpeed = actions[i][1]
speed[i] = lastSpeed
ignored = 0
overtook = False
maxSpeed = -1
for i in range(n-1,-1,-1):
maxSpeed = max(maxSpeed, speed[i])
if actions[i][0] == 1:
pass
elif actions[i][0] == 2:
overtook = True
elif actions[i][0] == 3:
if maxSpeed > actions[i][1]:
ignored += 1
else:
maxSpeed = speed[i-1]
elif actions[i][0] == 4:
overtook = False
elif actions[i][0] == 5:
maxSpeed = speed[i-1]
elif actions[i][0] == 6:
if overtook:
ignored += 1
print(ignored)
```
Yes
| 88,059 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has just attempted to pass the driving test. He ran over the straight road with the signs of four types.
* speed limit: this sign comes with a positive integer number — maximal speed of the car after the sign (cancel the action of the previous sign of this type);
* overtake is allowed: this sign means that after some car meets it, it can overtake any other car;
* no speed limit: this sign cancels speed limit if any (car can move with arbitrary speed after this sign);
* no overtake allowed: some car can't overtake any other car after this sign.
Polycarp goes past the signs consequentially, each new sign cancels the action of all the previous signs of it's kind (speed limit/overtake). It is possible that two or more "no overtake allowed" signs go one after another with zero "overtake is allowed" signs between them. It works with "no speed limit" and "overtake is allowed" signs as well.
In the beginning of the ride overtake is allowed and there is no speed limit.
You are given the sequence of events in chronological order — events which happened to Polycarp during the ride. There are events of following types:
1. Polycarp changes the speed of his car to specified (this event comes with a positive integer number);
2. Polycarp's car overtakes the other car;
3. Polycarp's car goes past the "speed limit" sign (this sign comes with a positive integer);
4. Polycarp's car goes past the "overtake is allowed" sign;
5. Polycarp's car goes past the "no speed limit";
6. Polycarp's car goes past the "no overtake allowed";
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
After the exam Polycarp can justify his rule violations by telling the driving instructor that he just didn't notice some of the signs. What is the minimal number of signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view?
Input
The first line contains one integer number n (1 ≤ n ≤ 2·105) — number of events.
Each of the next n lines starts with integer t (1 ≤ t ≤ 6) — the type of the event.
An integer s (1 ≤ s ≤ 300) follows in the query of the first and the third type (if it is the query of first type, then it's new speed of Polycarp's car, if it is the query of third type, then it's new speed limit).
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
Output
Print the minimal number of road signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view.
Examples
Input
11
1 100
3 70
4
2
3 120
5
3 120
6
1 150
4
3 300
Output
2
Input
5
1 100
3 200
2
4
5
Output
0
Input
7
1 20
2
6
4
6
6
2
Output
2
Note
In the first example Polycarp should say he didn't notice the "speed limit" sign with the limit of 70 and the second "speed limit" sign with the limit of 120.
In the second example Polycarp didn't make any rule violation.
In the third example Polycarp should say he didn't notice both "no overtake allowed" that came after "overtake is allowed" sign.
Submitted Solution:
```
n = int(input())
v = int(input()[2:])
p = [1e9]
d = k = 0
for i in range(n - 1):
s = input()
t = int(s[0])
if t == 1:
v = int(s[2:])
while p[-1] < v:
p.pop()
k += 1
if t == 2:
k += d
d = 0
if t == 3:
u = int(s[2:])
if v > u: k += 1
else: p.append(u)
if t == 4: d = 0
if t == 5: p = [1e9]
if t == 6: d += 1
print(k)
```
Yes
| 88,060 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has just attempted to pass the driving test. He ran over the straight road with the signs of four types.
* speed limit: this sign comes with a positive integer number — maximal speed of the car after the sign (cancel the action of the previous sign of this type);
* overtake is allowed: this sign means that after some car meets it, it can overtake any other car;
* no speed limit: this sign cancels speed limit if any (car can move with arbitrary speed after this sign);
* no overtake allowed: some car can't overtake any other car after this sign.
Polycarp goes past the signs consequentially, each new sign cancels the action of all the previous signs of it's kind (speed limit/overtake). It is possible that two or more "no overtake allowed" signs go one after another with zero "overtake is allowed" signs between them. It works with "no speed limit" and "overtake is allowed" signs as well.
In the beginning of the ride overtake is allowed and there is no speed limit.
You are given the sequence of events in chronological order — events which happened to Polycarp during the ride. There are events of following types:
1. Polycarp changes the speed of his car to specified (this event comes with a positive integer number);
2. Polycarp's car overtakes the other car;
3. Polycarp's car goes past the "speed limit" sign (this sign comes with a positive integer);
4. Polycarp's car goes past the "overtake is allowed" sign;
5. Polycarp's car goes past the "no speed limit";
6. Polycarp's car goes past the "no overtake allowed";
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
After the exam Polycarp can justify his rule violations by telling the driving instructor that he just didn't notice some of the signs. What is the minimal number of signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view?
Input
The first line contains one integer number n (1 ≤ n ≤ 2·105) — number of events.
Each of the next n lines starts with integer t (1 ≤ t ≤ 6) — the type of the event.
An integer s (1 ≤ s ≤ 300) follows in the query of the first and the third type (if it is the query of first type, then it's new speed of Polycarp's car, if it is the query of third type, then it's new speed limit).
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
Output
Print the minimal number of road signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view.
Examples
Input
11
1 100
3 70
4
2
3 120
5
3 120
6
1 150
4
3 300
Output
2
Input
5
1 100
3 200
2
4
5
Output
0
Input
7
1 20
2
6
4
6
6
2
Output
2
Note
In the first example Polycarp should say he didn't notice the "speed limit" sign with the limit of 70 and the second "speed limit" sign with the limit of 120.
In the second example Polycarp didn't make any rule violation.
In the third example Polycarp should say he didn't notice both "no overtake allowed" that came after "overtake is allowed" sign.
Submitted Solution:
```
# D. Driving Test
n = int(input())
ps = None
seen_sl = [float('inf')]
seen_ot = [True]
answer = 0
for i in range(n):
line = list(map(int, input().split()))
t = line[0]
if t == 1:
ps = line[1]
while ps > seen_sl[-1]:
answer += 1
seen_sl.pop(-1)
elif t == 2:
while not seen_ot[-1]:
answer += 1
seen_ot.pop(-1)
elif t == 3:
seen_sl.append(line[1])
while ps > seen_sl[-1]:
answer += 1
seen_sl.pop(-1)
elif t == 4:
seen_ot.append(True)
elif t == 5:
seen_sl.append(float('inf'))
elif t == 6:
seen_ot.append(False)
print(answer)
```
Yes
| 88,061 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has just attempted to pass the driving test. He ran over the straight road with the signs of four types.
* speed limit: this sign comes with a positive integer number — maximal speed of the car after the sign (cancel the action of the previous sign of this type);
* overtake is allowed: this sign means that after some car meets it, it can overtake any other car;
* no speed limit: this sign cancels speed limit if any (car can move with arbitrary speed after this sign);
* no overtake allowed: some car can't overtake any other car after this sign.
Polycarp goes past the signs consequentially, each new sign cancels the action of all the previous signs of it's kind (speed limit/overtake). It is possible that two or more "no overtake allowed" signs go one after another with zero "overtake is allowed" signs between them. It works with "no speed limit" and "overtake is allowed" signs as well.
In the beginning of the ride overtake is allowed and there is no speed limit.
You are given the sequence of events in chronological order — events which happened to Polycarp during the ride. There are events of following types:
1. Polycarp changes the speed of his car to specified (this event comes with a positive integer number);
2. Polycarp's car overtakes the other car;
3. Polycarp's car goes past the "speed limit" sign (this sign comes with a positive integer);
4. Polycarp's car goes past the "overtake is allowed" sign;
5. Polycarp's car goes past the "no speed limit";
6. Polycarp's car goes past the "no overtake allowed";
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
After the exam Polycarp can justify his rule violations by telling the driving instructor that he just didn't notice some of the signs. What is the minimal number of signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view?
Input
The first line contains one integer number n (1 ≤ n ≤ 2·105) — number of events.
Each of the next n lines starts with integer t (1 ≤ t ≤ 6) — the type of the event.
An integer s (1 ≤ s ≤ 300) follows in the query of the first and the third type (if it is the query of first type, then it's new speed of Polycarp's car, if it is the query of third type, then it's new speed limit).
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
Output
Print the minimal number of road signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view.
Examples
Input
11
1 100
3 70
4
2
3 120
5
3 120
6
1 150
4
3 300
Output
2
Input
5
1 100
3 200
2
4
5
Output
0
Input
7
1 20
2
6
4
6
6
2
Output
2
Note
In the first example Polycarp should say he didn't notice the "speed limit" sign with the limit of 70 and the second "speed limit" sign with the limit of 120.
In the second example Polycarp didn't make any rule violation.
In the third example Polycarp should say he didn't notice both "no overtake allowed" that came after "overtake is allowed" sign.
Submitted Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
avl=AvlTree()
#-----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default='z', func=lambda a, b: min(a ,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: a + b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left)/ 2)
# Check if middle element is
# less than or equal to key
if (arr[mid]<=key):
count = mid+1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def countGreater( arr,n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
spe=400
curspe=0
w=[400]
over=0
ans=0
for ik in range(int(input())):
a=list(map(int,input().split()))
if a[0]==4:
over=0
elif a[0]==6:
over+=1
elif a[0]==2:
if over>=1:
ans+=over
over=0
elif a[0]==1:
if a[1]>spe:
while(True):
e=w.pop()
spe=e
if a[1]<=e:
w.append(e)
break
else:
ans+=1
curspe=a[1]
elif a[0]==5:
spe=400
w=[400]
elif a[0]==3:
w.append(a[1])
while(True):
#print(w)
e=w.pop()
spe=e
if curspe<=e:
w.append(e)
break
else:
ans+=1
#print(ans)
print(ans)
```
Yes
| 88,062 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has just attempted to pass the driving test. He ran over the straight road with the signs of four types.
* speed limit: this sign comes with a positive integer number — maximal speed of the car after the sign (cancel the action of the previous sign of this type);
* overtake is allowed: this sign means that after some car meets it, it can overtake any other car;
* no speed limit: this sign cancels speed limit if any (car can move with arbitrary speed after this sign);
* no overtake allowed: some car can't overtake any other car after this sign.
Polycarp goes past the signs consequentially, each new sign cancels the action of all the previous signs of it's kind (speed limit/overtake). It is possible that two or more "no overtake allowed" signs go one after another with zero "overtake is allowed" signs between them. It works with "no speed limit" and "overtake is allowed" signs as well.
In the beginning of the ride overtake is allowed and there is no speed limit.
You are given the sequence of events in chronological order — events which happened to Polycarp during the ride. There are events of following types:
1. Polycarp changes the speed of his car to specified (this event comes with a positive integer number);
2. Polycarp's car overtakes the other car;
3. Polycarp's car goes past the "speed limit" sign (this sign comes with a positive integer);
4. Polycarp's car goes past the "overtake is allowed" sign;
5. Polycarp's car goes past the "no speed limit";
6. Polycarp's car goes past the "no overtake allowed";
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
After the exam Polycarp can justify his rule violations by telling the driving instructor that he just didn't notice some of the signs. What is the minimal number of signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view?
Input
The first line contains one integer number n (1 ≤ n ≤ 2·105) — number of events.
Each of the next n lines starts with integer t (1 ≤ t ≤ 6) — the type of the event.
An integer s (1 ≤ s ≤ 300) follows in the query of the first and the third type (if it is the query of first type, then it's new speed of Polycarp's car, if it is the query of third type, then it's new speed limit).
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
Output
Print the minimal number of road signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view.
Examples
Input
11
1 100
3 70
4
2
3 120
5
3 120
6
1 150
4
3 300
Output
2
Input
5
1 100
3 200
2
4
5
Output
0
Input
7
1 20
2
6
4
6
6
2
Output
2
Note
In the first example Polycarp should say he didn't notice the "speed limit" sign with the limit of 70 and the second "speed limit" sign with the limit of 120.
In the second example Polycarp didn't make any rule violation.
In the third example Polycarp should say he didn't notice both "no overtake allowed" that came after "overtake is allowed" sign.
Submitted Solution:
```
events = int(input())
speed = -1
limitSigns = []
overtake = True
noOvertakeSigns = 0
signs = 0
for i in range(events):
l = input().split(" ")
e_type = int(l[0])
if e_type == 1:
speed = int(l[1])
newLimSigns = []
for i in limitSigns:
if speed > i:
signs += 1
else:
newLimSigns.append(i)
break
limitSigns = newLimSigns
if e_type == 2:
if not overtake:
signs += noOvertakeSigns
noOvertakeSigns = 0
if e_type == 3:
limit = int(l[1])
if speed > limit:
signs += 1
else:
limitSigns = [limit] + limitSigns
if e_type == 4:
overtake = True
noOvertakeSigns = 0
if e_type == 5:
limitSigns = []
if e_type == 6:
overtake = False
noOvertakeSigns += 1
print(signs)
```
No
| 88,063 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has just attempted to pass the driving test. He ran over the straight road with the signs of four types.
* speed limit: this sign comes with a positive integer number — maximal speed of the car after the sign (cancel the action of the previous sign of this type);
* overtake is allowed: this sign means that after some car meets it, it can overtake any other car;
* no speed limit: this sign cancels speed limit if any (car can move with arbitrary speed after this sign);
* no overtake allowed: some car can't overtake any other car after this sign.
Polycarp goes past the signs consequentially, each new sign cancels the action of all the previous signs of it's kind (speed limit/overtake). It is possible that two or more "no overtake allowed" signs go one after another with zero "overtake is allowed" signs between them. It works with "no speed limit" and "overtake is allowed" signs as well.
In the beginning of the ride overtake is allowed and there is no speed limit.
You are given the sequence of events in chronological order — events which happened to Polycarp during the ride. There are events of following types:
1. Polycarp changes the speed of his car to specified (this event comes with a positive integer number);
2. Polycarp's car overtakes the other car;
3. Polycarp's car goes past the "speed limit" sign (this sign comes with a positive integer);
4. Polycarp's car goes past the "overtake is allowed" sign;
5. Polycarp's car goes past the "no speed limit";
6. Polycarp's car goes past the "no overtake allowed";
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
After the exam Polycarp can justify his rule violations by telling the driving instructor that he just didn't notice some of the signs. What is the minimal number of signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view?
Input
The first line contains one integer number n (1 ≤ n ≤ 2·105) — number of events.
Each of the next n lines starts with integer t (1 ≤ t ≤ 6) — the type of the event.
An integer s (1 ≤ s ≤ 300) follows in the query of the first and the third type (if it is the query of first type, then it's new speed of Polycarp's car, if it is the query of third type, then it's new speed limit).
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
Output
Print the minimal number of road signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view.
Examples
Input
11
1 100
3 70
4
2
3 120
5
3 120
6
1 150
4
3 300
Output
2
Input
5
1 100
3 200
2
4
5
Output
0
Input
7
1 20
2
6
4
6
6
2
Output
2
Note
In the first example Polycarp should say he didn't notice the "speed limit" sign with the limit of 70 and the second "speed limit" sign with the limit of 120.
In the second example Polycarp didn't make any rule violation.
In the third example Polycarp should say he didn't notice both "no overtake allowed" that came after "overtake is allowed" sign.
Submitted Solution:
```
n=int(input())
s=0
ms=-1
o=1
z=0
q=0
for _ in range(n):
a=list(map(int,input().split()))
if a[0]==1:
s=a[1]
if ms>0 and ms<s:
z+=1
ms=-1
elif a[0]==2:
if o!=1:
z+=(1-o)
o=1
elif a[0]==3:
ms=a[1]
if s>ms:
z+=1
ms=-1
elif a[0]==4:
o=1
elif a[0]==5:
ms=-1
elif a[0]==6:
o-=1
#print(z)
print(z)
#print(' '.join([str(i) for i in a]))
```
No
| 88,064 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has just attempted to pass the driving test. He ran over the straight road with the signs of four types.
* speed limit: this sign comes with a positive integer number — maximal speed of the car after the sign (cancel the action of the previous sign of this type);
* overtake is allowed: this sign means that after some car meets it, it can overtake any other car;
* no speed limit: this sign cancels speed limit if any (car can move with arbitrary speed after this sign);
* no overtake allowed: some car can't overtake any other car after this sign.
Polycarp goes past the signs consequentially, each new sign cancels the action of all the previous signs of it's kind (speed limit/overtake). It is possible that two or more "no overtake allowed" signs go one after another with zero "overtake is allowed" signs between them. It works with "no speed limit" and "overtake is allowed" signs as well.
In the beginning of the ride overtake is allowed and there is no speed limit.
You are given the sequence of events in chronological order — events which happened to Polycarp during the ride. There are events of following types:
1. Polycarp changes the speed of his car to specified (this event comes with a positive integer number);
2. Polycarp's car overtakes the other car;
3. Polycarp's car goes past the "speed limit" sign (this sign comes with a positive integer);
4. Polycarp's car goes past the "overtake is allowed" sign;
5. Polycarp's car goes past the "no speed limit";
6. Polycarp's car goes past the "no overtake allowed";
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
After the exam Polycarp can justify his rule violations by telling the driving instructor that he just didn't notice some of the signs. What is the minimal number of signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view?
Input
The first line contains one integer number n (1 ≤ n ≤ 2·105) — number of events.
Each of the next n lines starts with integer t (1 ≤ t ≤ 6) — the type of the event.
An integer s (1 ≤ s ≤ 300) follows in the query of the first and the third type (if it is the query of first type, then it's new speed of Polycarp's car, if it is the query of third type, then it's new speed limit).
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
Output
Print the minimal number of road signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view.
Examples
Input
11
1 100
3 70
4
2
3 120
5
3 120
6
1 150
4
3 300
Output
2
Input
5
1 100
3 200
2
4
5
Output
0
Input
7
1 20
2
6
4
6
6
2
Output
2
Note
In the first example Polycarp should say he didn't notice the "speed limit" sign with the limit of 70 and the second "speed limit" sign with the limit of 120.
In the second example Polycarp didn't make any rule violation.
In the third example Polycarp should say he didn't notice both "no overtake allowed" that came after "overtake is allowed" sign.
Submitted Solution:
```
import sys, os
n = int(input())
k = 0
res = 0
has = [305]
for i in range(n):
x = input()
if x == "4":
k = 0
continue
if x == "2":
res += k
k = 0
continue
if x == "6":
k += 1
continue
if x == "5":
has = [305]
continue
if x[0] == "3":
d = int(x[2:])
has.append(haste)
if d < haste:
res += 1
has.pop()
continue
if x[0] == "1":
d = int(x[2:])
haste = d
if has[-1] < d:
res += 1
has.pop()
print(res)
```
No
| 88,065 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has just attempted to pass the driving test. He ran over the straight road with the signs of four types.
* speed limit: this sign comes with a positive integer number — maximal speed of the car after the sign (cancel the action of the previous sign of this type);
* overtake is allowed: this sign means that after some car meets it, it can overtake any other car;
* no speed limit: this sign cancels speed limit if any (car can move with arbitrary speed after this sign);
* no overtake allowed: some car can't overtake any other car after this sign.
Polycarp goes past the signs consequentially, each new sign cancels the action of all the previous signs of it's kind (speed limit/overtake). It is possible that two or more "no overtake allowed" signs go one after another with zero "overtake is allowed" signs between them. It works with "no speed limit" and "overtake is allowed" signs as well.
In the beginning of the ride overtake is allowed and there is no speed limit.
You are given the sequence of events in chronological order — events which happened to Polycarp during the ride. There are events of following types:
1. Polycarp changes the speed of his car to specified (this event comes with a positive integer number);
2. Polycarp's car overtakes the other car;
3. Polycarp's car goes past the "speed limit" sign (this sign comes with a positive integer);
4. Polycarp's car goes past the "overtake is allowed" sign;
5. Polycarp's car goes past the "no speed limit";
6. Polycarp's car goes past the "no overtake allowed";
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
After the exam Polycarp can justify his rule violations by telling the driving instructor that he just didn't notice some of the signs. What is the minimal number of signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view?
Input
The first line contains one integer number n (1 ≤ n ≤ 2·105) — number of events.
Each of the next n lines starts with integer t (1 ≤ t ≤ 6) — the type of the event.
An integer s (1 ≤ s ≤ 300) follows in the query of the first and the third type (if it is the query of first type, then it's new speed of Polycarp's car, if it is the query of third type, then it's new speed limit).
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
Output
Print the minimal number of road signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view.
Examples
Input
11
1 100
3 70
4
2
3 120
5
3 120
6
1 150
4
3 300
Output
2
Input
5
1 100
3 200
2
4
5
Output
0
Input
7
1 20
2
6
4
6
6
2
Output
2
Note
In the first example Polycarp should say he didn't notice the "speed limit" sign with the limit of 70 and the second "speed limit" sign with the limit of 120.
In the second example Polycarp didn't make any rule violation.
In the third example Polycarp should say he didn't notice both "no overtake allowed" that came after "overtake is allowed" sign.
Submitted Solution:
```
import collections
n=int(input())
x,v=map(int,input().split())
o=v
k,t=0,0
f=False
q=collections.deque()
for i in range(1,n):
s=input()
if s[0]=='1':
x,v=map(int,s.split())
if v>o:
k+=1
if s[0]=='2':
if len(q)!=0 and not f:
k+=len(q)
elif len(q)!=0:
k+=t
f=True
t=0
if s[0]=='3':
x,o=map(int,s.split())
if v>o:
k+=1
if s[0]=='4':
q.clear()
f=False
t=0
if s[0]=='5':
o=v
if s[0]=='6':
q.append(1)
t+=1
print(k)
```
No
| 88,066 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jafar has n cans of cola. Each can is described by two integers: remaining volume of cola ai and can's capacity bi (ai ≤ bi).
Jafar has decided to pour all remaining cola into just 2 cans, determine if he can do this or not!
Input
The first line of the input contains one integer n (2 ≤ n ≤ 100 000) — number of cola cans.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) — volume of remaining cola in cans.
The third line contains n space-separated integers that b1, b2, ..., bn (ai ≤ bi ≤ 109) — capacities of the cans.
Output
Print "YES" (without quotes) if it is possible to pour all remaining cola in 2 cans. Otherwise print "NO" (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
2
3 5
3 6
Output
YES
Input
3
6 8 9
6 10 12
Output
NO
Input
5
0 0 5 0 0
1 1 8 10 5
Output
YES
Input
4
4 1 0 3
5 2 2 3
Output
YES
Note
In the first sample, there are already 2 cans, so the answer is "YES".
Tags: greedy, implementation
Correct Solution:
```
n = int(input())
A = [int(_) for _ in input().split()]
B = [int(_) for _ in input().split()]
a = sum(A)
b1, b2 = 0,0
for b in B:
if b > b1:
b2 = b1
b1 = b
elif b > b2:
b2 = b
if b1+b2 >= a:
print("YES")
else:
print("NO")
```
| 88,067 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jafar has n cans of cola. Each can is described by two integers: remaining volume of cola ai and can's capacity bi (ai ≤ bi).
Jafar has decided to pour all remaining cola into just 2 cans, determine if he can do this or not!
Input
The first line of the input contains one integer n (2 ≤ n ≤ 100 000) — number of cola cans.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) — volume of remaining cola in cans.
The third line contains n space-separated integers that b1, b2, ..., bn (ai ≤ bi ≤ 109) — capacities of the cans.
Output
Print "YES" (without quotes) if it is possible to pour all remaining cola in 2 cans. Otherwise print "NO" (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
2
3 5
3 6
Output
YES
Input
3
6 8 9
6 10 12
Output
NO
Input
5
0 0 5 0 0
1 1 8 10 5
Output
YES
Input
4
4 1 0 3
5 2 2 3
Output
YES
Note
In the first sample, there are already 2 cans, so the answer is "YES".
Tags: greedy, implementation
Correct Solution:
```
a = int(input())
b = list(map(int,input().split()))
c = list(map(int,input().split()))
c.sort()
z = c[-1] +c[-2]
if sum(b)<=z:print("YES")
else:print("NO")
```
| 88,068 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jafar has n cans of cola. Each can is described by two integers: remaining volume of cola ai and can's capacity bi (ai ≤ bi).
Jafar has decided to pour all remaining cola into just 2 cans, determine if he can do this or not!
Input
The first line of the input contains one integer n (2 ≤ n ≤ 100 000) — number of cola cans.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) — volume of remaining cola in cans.
The third line contains n space-separated integers that b1, b2, ..., bn (ai ≤ bi ≤ 109) — capacities of the cans.
Output
Print "YES" (without quotes) if it is possible to pour all remaining cola in 2 cans. Otherwise print "NO" (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
2
3 5
3 6
Output
YES
Input
3
6 8 9
6 10 12
Output
NO
Input
5
0 0 5 0 0
1 1 8 10 5
Output
YES
Input
4
4 1 0 3
5 2 2 3
Output
YES
Note
In the first sample, there are already 2 cans, so the answer is "YES".
Tags: greedy, implementation
Correct Solution:
```
def ii():
return int(input())
def mi():
return map(int, input().split())
def li():
return list(mi())
n = ii()
a = li()
b = li()
b.sort()
print('YES' if sum(a) <= b[-1] + b[-2] else 'NO')
```
| 88,069 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jafar has n cans of cola. Each can is described by two integers: remaining volume of cola ai and can's capacity bi (ai ≤ bi).
Jafar has decided to pour all remaining cola into just 2 cans, determine if he can do this or not!
Input
The first line of the input contains one integer n (2 ≤ n ≤ 100 000) — number of cola cans.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) — volume of remaining cola in cans.
The third line contains n space-separated integers that b1, b2, ..., bn (ai ≤ bi ≤ 109) — capacities of the cans.
Output
Print "YES" (without quotes) if it is possible to pour all remaining cola in 2 cans. Otherwise print "NO" (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
2
3 5
3 6
Output
YES
Input
3
6 8 9
6 10 12
Output
NO
Input
5
0 0 5 0 0
1 1 8 10 5
Output
YES
Input
4
4 1 0 3
5 2 2 3
Output
YES
Note
In the first sample, there are already 2 cans, so the answer is "YES".
Tags: greedy, implementation
Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
b.sort(reverse=True)
c=b[0]+b[1]
if c>=sum(a):
print("YES")
else:
print("NO")
```
| 88,070 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jafar has n cans of cola. Each can is described by two integers: remaining volume of cola ai and can's capacity bi (ai ≤ bi).
Jafar has decided to pour all remaining cola into just 2 cans, determine if he can do this or not!
Input
The first line of the input contains one integer n (2 ≤ n ≤ 100 000) — number of cola cans.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) — volume of remaining cola in cans.
The third line contains n space-separated integers that b1, b2, ..., bn (ai ≤ bi ≤ 109) — capacities of the cans.
Output
Print "YES" (without quotes) if it is possible to pour all remaining cola in 2 cans. Otherwise print "NO" (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
2
3 5
3 6
Output
YES
Input
3
6 8 9
6 10 12
Output
NO
Input
5
0 0 5 0 0
1 1 8 10 5
Output
YES
Input
4
4 1 0 3
5 2 2 3
Output
YES
Note
In the first sample, there are already 2 cans, so the answer is "YES".
Tags: greedy, implementation
Correct Solution:
```
sum = 0
b = []
n = int(input())
line1 = input().split()
line2 = input().split()
for i in range(n):
sum += int(line1[i])
b.append(int(line2[i]))
continue
b.sort()
cap = b[n-1] + b[n-2]
if sum > cap:
print('NO')
else:
print('YES')
```
| 88,071 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jafar has n cans of cola. Each can is described by two integers: remaining volume of cola ai and can's capacity bi (ai ≤ bi).
Jafar has decided to pour all remaining cola into just 2 cans, determine if he can do this or not!
Input
The first line of the input contains one integer n (2 ≤ n ≤ 100 000) — number of cola cans.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) — volume of remaining cola in cans.
The third line contains n space-separated integers that b1, b2, ..., bn (ai ≤ bi ≤ 109) — capacities of the cans.
Output
Print "YES" (without quotes) if it is possible to pour all remaining cola in 2 cans. Otherwise print "NO" (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
2
3 5
3 6
Output
YES
Input
3
6 8 9
6 10 12
Output
NO
Input
5
0 0 5 0 0
1 1 8 10 5
Output
YES
Input
4
4 1 0 3
5 2 2 3
Output
YES
Note
In the first sample, there are already 2 cans, so the answer is "YES".
Tags: greedy, implementation
Correct Solution:
```
def larg2(arr):
first = None
second = None
for number in arr:
if first is None:
first = number
elif number > first:
second = first
first = number
else:
if second is None:
second = number
elif number > second:
second = number
return [first, second]
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
r_a = sum(a)
l = larg2(b)
if sum(l) >= r_a:
print("YES")
else:
print("NO")
```
| 88,072 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jafar has n cans of cola. Each can is described by two integers: remaining volume of cola ai and can's capacity bi (ai ≤ bi).
Jafar has decided to pour all remaining cola into just 2 cans, determine if he can do this or not!
Input
The first line of the input contains one integer n (2 ≤ n ≤ 100 000) — number of cola cans.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) — volume of remaining cola in cans.
The third line contains n space-separated integers that b1, b2, ..., bn (ai ≤ bi ≤ 109) — capacities of the cans.
Output
Print "YES" (without quotes) if it is possible to pour all remaining cola in 2 cans. Otherwise print "NO" (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
2
3 5
3 6
Output
YES
Input
3
6 8 9
6 10 12
Output
NO
Input
5
0 0 5 0 0
1 1 8 10 5
Output
YES
Input
4
4 1 0 3
5 2 2 3
Output
YES
Note
In the first sample, there are already 2 cans, so the answer is "YES".
Tags: greedy, implementation
Correct Solution:
```
n = int(input())
a = [int(x) for x in input().split()]
b = [int(x) for x in input().split()]
b.sort()
s = b[-1] + b[-2]
if sum(a) > s:
print('NO')
else:
print('YES')
```
| 88,073 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jafar has n cans of cola. Each can is described by two integers: remaining volume of cola ai and can's capacity bi (ai ≤ bi).
Jafar has decided to pour all remaining cola into just 2 cans, determine if he can do this or not!
Input
The first line of the input contains one integer n (2 ≤ n ≤ 100 000) — number of cola cans.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) — volume of remaining cola in cans.
The third line contains n space-separated integers that b1, b2, ..., bn (ai ≤ bi ≤ 109) — capacities of the cans.
Output
Print "YES" (without quotes) if it is possible to pour all remaining cola in 2 cans. Otherwise print "NO" (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
2
3 5
3 6
Output
YES
Input
3
6 8 9
6 10 12
Output
NO
Input
5
0 0 5 0 0
1 1 8 10 5
Output
YES
Input
4
4 1 0 3
5 2 2 3
Output
YES
Note
In the first sample, there are already 2 cans, so the answer is "YES".
Tags: greedy, implementation
Correct Solution:
```
n = int(input())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
if n == 2:
print("YES")
else:
tot = sum(a)
m1 = max(b)
b.remove(m1)
m2 = max(b)
if tot > m1+m2:
print("NO")
else:
print("YES")
```
| 88,074 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jafar has n cans of cola. Each can is described by two integers: remaining volume of cola ai and can's capacity bi (ai ≤ bi).
Jafar has decided to pour all remaining cola into just 2 cans, determine if he can do this or not!
Input
The first line of the input contains one integer n (2 ≤ n ≤ 100 000) — number of cola cans.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) — volume of remaining cola in cans.
The third line contains n space-separated integers that b1, b2, ..., bn (ai ≤ bi ≤ 109) — capacities of the cans.
Output
Print "YES" (without quotes) if it is possible to pour all remaining cola in 2 cans. Otherwise print "NO" (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
2
3 5
3 6
Output
YES
Input
3
6 8 9
6 10 12
Output
NO
Input
5
0 0 5 0 0
1 1 8 10 5
Output
YES
Input
4
4 1 0 3
5 2 2 3
Output
YES
Note
In the first sample, there are already 2 cans, so the answer is "YES".
Submitted Solution:
```
n = int(input())
cola = list(map(int, input().split()))
volume = list(map(int, input().split()))
volume.sort()
if volume[-1] + volume[-2] >= sum(cola):
print('YES')
else:
print('NO')
```
Yes
| 88,075 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jafar has n cans of cola. Each can is described by two integers: remaining volume of cola ai and can's capacity bi (ai ≤ bi).
Jafar has decided to pour all remaining cola into just 2 cans, determine if he can do this or not!
Input
The first line of the input contains one integer n (2 ≤ n ≤ 100 000) — number of cola cans.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) — volume of remaining cola in cans.
The third line contains n space-separated integers that b1, b2, ..., bn (ai ≤ bi ≤ 109) — capacities of the cans.
Output
Print "YES" (without quotes) if it is possible to pour all remaining cola in 2 cans. Otherwise print "NO" (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
2
3 5
3 6
Output
YES
Input
3
6 8 9
6 10 12
Output
NO
Input
5
0 0 5 0 0
1 1 8 10 5
Output
YES
Input
4
4 1 0 3
5 2 2 3
Output
YES
Note
In the first sample, there are already 2 cans, so the answer is "YES".
Submitted Solution:
```
def main():
input()
print(("NO", "YES")[sum(map(int, input().split())) <= sum(sorted(map(int, input().split()), reverse=True)[:2])])
if __name__ == '__main__':
main()
```
Yes
| 88,076 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jafar has n cans of cola. Each can is described by two integers: remaining volume of cola ai and can's capacity bi (ai ≤ bi).
Jafar has decided to pour all remaining cola into just 2 cans, determine if he can do this or not!
Input
The first line of the input contains one integer n (2 ≤ n ≤ 100 000) — number of cola cans.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) — volume of remaining cola in cans.
The third line contains n space-separated integers that b1, b2, ..., bn (ai ≤ bi ≤ 109) — capacities of the cans.
Output
Print "YES" (without quotes) if it is possible to pour all remaining cola in 2 cans. Otherwise print "NO" (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
2
3 5
3 6
Output
YES
Input
3
6 8 9
6 10 12
Output
NO
Input
5
0 0 5 0 0
1 1 8 10 5
Output
YES
Input
4
4 1 0 3
5 2 2 3
Output
YES
Note
In the first sample, there are already 2 cans, so the answer is "YES".
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(): return [tuple(map(int, l.split())) for l in sys.stdin]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
n = I()
a = LI()
b = LI()
s = sum(a)
t = sorted(b)
if t[-1] + t[-2] >= s:
return 'YES'
return 'NO'
print(main())
```
Yes
| 88,077 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jafar has n cans of cola. Each can is described by two integers: remaining volume of cola ai and can's capacity bi (ai ≤ bi).
Jafar has decided to pour all remaining cola into just 2 cans, determine if he can do this or not!
Input
The first line of the input contains one integer n (2 ≤ n ≤ 100 000) — number of cola cans.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) — volume of remaining cola in cans.
The third line contains n space-separated integers that b1, b2, ..., bn (ai ≤ bi ≤ 109) — capacities of the cans.
Output
Print "YES" (without quotes) if it is possible to pour all remaining cola in 2 cans. Otherwise print "NO" (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
2
3 5
3 6
Output
YES
Input
3
6 8 9
6 10 12
Output
NO
Input
5
0 0 5 0 0
1 1 8 10 5
Output
YES
Input
4
4 1 0 3
5 2 2 3
Output
YES
Note
In the first sample, there are already 2 cans, so the answer is "YES".
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
x=sum(a)
y=max(b)
b.remove(y)
z=max(b)
if y+z<x:
print('NO')
else:
print('YES')
```
Yes
| 88,078 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jafar has n cans of cola. Each can is described by two integers: remaining volume of cola ai and can's capacity bi (ai ≤ bi).
Jafar has decided to pour all remaining cola into just 2 cans, determine if he can do this or not!
Input
The first line of the input contains one integer n (2 ≤ n ≤ 100 000) — number of cola cans.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) — volume of remaining cola in cans.
The third line contains n space-separated integers that b1, b2, ..., bn (ai ≤ bi ≤ 109) — capacities of the cans.
Output
Print "YES" (without quotes) if it is possible to pour all remaining cola in 2 cans. Otherwise print "NO" (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
2
3 5
3 6
Output
YES
Input
3
6 8 9
6 10 12
Output
NO
Input
5
0 0 5 0 0
1 1 8 10 5
Output
YES
Input
4
4 1 0 3
5 2 2 3
Output
YES
Note
In the first sample, there are already 2 cans, so the answer is "YES".
Submitted Solution:
```
def greedy(lst1, lst2):
s = sum(lst1)
if s <= lst2[len(lst1) - 1] + lst2[len(lst1) - 2]:
return "YES"
return "NO"
n = int(input())
a = [int(i) for i in input().split()]
b = [int(j) for j in input().split()]
print(greedy(a, b))
```
No
| 88,079 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jafar has n cans of cola. Each can is described by two integers: remaining volume of cola ai and can's capacity bi (ai ≤ bi).
Jafar has decided to pour all remaining cola into just 2 cans, determine if he can do this or not!
Input
The first line of the input contains one integer n (2 ≤ n ≤ 100 000) — number of cola cans.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) — volume of remaining cola in cans.
The third line contains n space-separated integers that b1, b2, ..., bn (ai ≤ bi ≤ 109) — capacities of the cans.
Output
Print "YES" (without quotes) if it is possible to pour all remaining cola in 2 cans. Otherwise print "NO" (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
2
3 5
3 6
Output
YES
Input
3
6 8 9
6 10 12
Output
NO
Input
5
0 0 5 0 0
1 1 8 10 5
Output
YES
Input
4
4 1 0 3
5 2 2 3
Output
YES
Note
In the first sample, there are already 2 cans, so the answer is "YES".
Submitted Solution:
```
input()
list1=[*map(int,input().split())]
list2=[*map(int,input().split())]
sum=0
for a in list1:
sum+=a
print(sum)
m2=0
m=0
for a in list2:
if a>=m:
m,m2=a,m
elif a<=m and a>m2:
m2=a
print(m,m2)
if sum<=m+m2:
print('YES')
else:
print('NO')
```
No
| 88,080 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jafar has n cans of cola. Each can is described by two integers: remaining volume of cola ai and can's capacity bi (ai ≤ bi).
Jafar has decided to pour all remaining cola into just 2 cans, determine if he can do this or not!
Input
The first line of the input contains one integer n (2 ≤ n ≤ 100 000) — number of cola cans.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) — volume of remaining cola in cans.
The third line contains n space-separated integers that b1, b2, ..., bn (ai ≤ bi ≤ 109) — capacities of the cans.
Output
Print "YES" (without quotes) if it is possible to pour all remaining cola in 2 cans. Otherwise print "NO" (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
2
3 5
3 6
Output
YES
Input
3
6 8 9
6 10 12
Output
NO
Input
5
0 0 5 0 0
1 1 8 10 5
Output
YES
Input
4
4 1 0 3
5 2 2 3
Output
YES
Note
In the first sample, there are already 2 cans, so the answer is "YES".
Submitted Solution:
```
number_of_cans = int(input())
volume = list(map(int, input().split()))
capacity = list(map(int, input().split()))
volume_of_all = 0
for volume_N in volume:
volume_of_all+=volume_N
max_capacity = capacity[0]
for counter in range(1,len(capacity)):
if capacity[counter] > capacity[counter-1]:
max_capacity = capacity[counter]
capacity.remove(max_capacity)
max_capacity2 = capacity[0]
for counter in range(1,len(capacity)):
if capacity[counter] > capacity[counter-1]:
max_capacity2 = capacity[counter]
if volume_of_all <= (max_capacity + max_capacity2):
print('YES')
else:
print('NO')
```
No
| 88,081 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jafar has n cans of cola. Each can is described by two integers: remaining volume of cola ai and can's capacity bi (ai ≤ bi).
Jafar has decided to pour all remaining cola into just 2 cans, determine if he can do this or not!
Input
The first line of the input contains one integer n (2 ≤ n ≤ 100 000) — number of cola cans.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) — volume of remaining cola in cans.
The third line contains n space-separated integers that b1, b2, ..., bn (ai ≤ bi ≤ 109) — capacities of the cans.
Output
Print "YES" (without quotes) if it is possible to pour all remaining cola in 2 cans. Otherwise print "NO" (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
2
3 5
3 6
Output
YES
Input
3
6 8 9
6 10 12
Output
NO
Input
5
0 0 5 0 0
1 1 8 10 5
Output
YES
Input
4
4 1 0 3
5 2 2 3
Output
YES
Note
In the first sample, there are already 2 cans, so the answer is "YES".
Submitted Solution:
```
n=int(input())
def solve():
a=[int(i) for i in input().split()]
b=[int(i) for i in input().split()]
v=max(b)
b.remove(v)
v+=max(b)
if sum(a) >=v:
print('yes')
else:
print('no')
solve()
```
No
| 88,082 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given an array a1, a2, ..., an of n integers, find the largest number in the array that is not a perfect square.
A number x is said to be a perfect square if there exists an integer y such that x = y2.
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array.
The second line contains n integers a1, a2, ..., an ( - 106 ≤ ai ≤ 106) — the elements of the array.
It is guaranteed that at least one element of the array is not a perfect square.
Output
Print the largest number in the array which is not a perfect square. It is guaranteed that an answer always exists.
Examples
Input
2
4 2
Output
2
Input
8
1 2 4 8 16 32 64 576
Output
32
Note
In the first sample case, 4 is a perfect square, so the largest number in the array that is not a perfect square is 2.
Tags: brute force, implementation, math
Correct Solution:
```
n = int(input())
a = [int(i)for i in input().split()]
a = sorted(a)
max_ = 0
for i in range(n):
if a[i] >= 0:
if a[i]**0.5 != int(a[i]**0.5):
max_ = a[i]
if max_ != 0:
print(max_)
else:
for i in range(n):
if a[i] < 0:
max_ = a[i]
print(max_)
```
| 88,083 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given an array a1, a2, ..., an of n integers, find the largest number in the array that is not a perfect square.
A number x is said to be a perfect square if there exists an integer y such that x = y2.
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array.
The second line contains n integers a1, a2, ..., an ( - 106 ≤ ai ≤ 106) — the elements of the array.
It is guaranteed that at least one element of the array is not a perfect square.
Output
Print the largest number in the array which is not a perfect square. It is guaranteed that an answer always exists.
Examples
Input
2
4 2
Output
2
Input
8
1 2 4 8 16 32 64 576
Output
32
Note
In the first sample case, 4 is a perfect square, so the largest number in the array that is not a perfect square is 2.
Tags: brute force, implementation, math
Correct Solution:
```
import math
n = int(input())
p = list(map(int, input().split()))
def s(n, p):
if n == 1:
print(p[0])
return
elif max(p) < 0:
print(max(p))
return
while math.sqrt(max(p)) == int(math.sqrt(max(p))):
p.remove(max(p))
if max(p) < 0:
print(max(p))
return
print(max(p))
s(n, p)
```
| 88,084 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given an array a1, a2, ..., an of n integers, find the largest number in the array that is not a perfect square.
A number x is said to be a perfect square if there exists an integer y such that x = y2.
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array.
The second line contains n integers a1, a2, ..., an ( - 106 ≤ ai ≤ 106) — the elements of the array.
It is guaranteed that at least one element of the array is not a perfect square.
Output
Print the largest number in the array which is not a perfect square. It is guaranteed that an answer always exists.
Examples
Input
2
4 2
Output
2
Input
8
1 2 4 8 16 32 64 576
Output
32
Note
In the first sample case, 4 is a perfect square, so the largest number in the array that is not a perfect square is 2.
Tags: brute force, implementation, math
Correct Solution:
```
import math
x=int(input())
l=list(map(int,input().split()))
k={}
for i in l:
k[i]=(abs(i)**0.5)
f=[]
for i in l:
if i<0:
f.append(i)
for i in k.keys():
if k[i]%1==0:
pass
else:
f.append(i)
print(max(f))
```
| 88,085 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given an array a1, a2, ..., an of n integers, find the largest number in the array that is not a perfect square.
A number x is said to be a perfect square if there exists an integer y such that x = y2.
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array.
The second line contains n integers a1, a2, ..., an ( - 106 ≤ ai ≤ 106) — the elements of the array.
It is guaranteed that at least one element of the array is not a perfect square.
Output
Print the largest number in the array which is not a perfect square. It is guaranteed that an answer always exists.
Examples
Input
2
4 2
Output
2
Input
8
1 2 4 8 16 32 64 576
Output
32
Note
In the first sample case, 4 is a perfect square, so the largest number in the array that is not a perfect square is 2.
Tags: brute force, implementation, math
Correct Solution:
```
from math import sqrt
n = int(input())
l = list(map(int,input().split()))
max1=-float("inf")
for i in range(n):
if l[i]<0 or (sqrt(l[i]))!=int(sqrt(l[i])):
if l[i]>max1:
max1=l[i]
print(max1)
```
| 88,086 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given an array a1, a2, ..., an of n integers, find the largest number in the array that is not a perfect square.
A number x is said to be a perfect square if there exists an integer y such that x = y2.
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array.
The second line contains n integers a1, a2, ..., an ( - 106 ≤ ai ≤ 106) — the elements of the array.
It is guaranteed that at least one element of the array is not a perfect square.
Output
Print the largest number in the array which is not a perfect square. It is guaranteed that an answer always exists.
Examples
Input
2
4 2
Output
2
Input
8
1 2 4 8 16 32 64 576
Output
32
Note
In the first sample case, 4 is a perfect square, so the largest number in the array that is not a perfect square is 2.
Tags: brute force, implementation, math
Correct Solution:
```
from math import sqrt, floor
int(input())
for n in sorted(map(int, input().split()))[::-1]:
if abs(n) != n:
print(n)
break
elif floor(sqrt(n)) - sqrt(n) != 0:
print(n)
break
```
| 88,087 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given an array a1, a2, ..., an of n integers, find the largest number in the array that is not a perfect square.
A number x is said to be a perfect square if there exists an integer y such that x = y2.
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array.
The second line contains n integers a1, a2, ..., an ( - 106 ≤ ai ≤ 106) — the elements of the array.
It is guaranteed that at least one element of the array is not a perfect square.
Output
Print the largest number in the array which is not a perfect square. It is guaranteed that an answer always exists.
Examples
Input
2
4 2
Output
2
Input
8
1 2 4 8 16 32 64 576
Output
32
Note
In the first sample case, 4 is a perfect square, so the largest number in the array that is not a perfect square is 2.
Tags: brute force, implementation, math
Correct Solution:
```
import math
n=int(input())
x=list(map(int,input().split()))
#arr=sorted(x)
count=-1000001
for i in range(n):
if((x[i]<0) or (math.sqrt(x[i])!=int(math.sqrt(x[i])))):
count=max(count,x[i])
print(count)
```
| 88,088 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given an array a1, a2, ..., an of n integers, find the largest number in the array that is not a perfect square.
A number x is said to be a perfect square if there exists an integer y such that x = y2.
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array.
The second line contains n integers a1, a2, ..., an ( - 106 ≤ ai ≤ 106) — the elements of the array.
It is guaranteed that at least one element of the array is not a perfect square.
Output
Print the largest number in the array which is not a perfect square. It is guaranteed that an answer always exists.
Examples
Input
2
4 2
Output
2
Input
8
1 2 4 8 16 32 64 576
Output
32
Note
In the first sample case, 4 is a perfect square, so the largest number in the array that is not a perfect square is 2.
Tags: brute force, implementation, math
Correct Solution:
```
import math
def perfect_square(x):
if x<0:
return False
else:
y=x**(1/2)
if y-math.floor(y)==0.0:
return True
not_perfect_square_list=[]
n=int(input())
if 1<=n<=1000:
lis=list(map(int,input().split()))
if len(lis)==n:
for i in lis:
if perfect_square(i):
continue
else:
not_perfect_square_list.append(i)
else:
exit()
print(max(not_perfect_square_list))
else:
exit()
```
| 88,089 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given an array a1, a2, ..., an of n integers, find the largest number in the array that is not a perfect square.
A number x is said to be a perfect square if there exists an integer y such that x = y2.
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array.
The second line contains n integers a1, a2, ..., an ( - 106 ≤ ai ≤ 106) — the elements of the array.
It is guaranteed that at least one element of the array is not a perfect square.
Output
Print the largest number in the array which is not a perfect square. It is guaranteed that an answer always exists.
Examples
Input
2
4 2
Output
2
Input
8
1 2 4 8 16 32 64 576
Output
32
Note
In the first sample case, 4 is a perfect square, so the largest number in the array that is not a perfect square is 2.
Tags: brute force, implementation, math
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
s = {i*i for i in range(1001)}
print(max([i for i in a if i not in s]))
```
| 88,090 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given an array a1, a2, ..., an of n integers, find the largest number in the array that is not a perfect square.
A number x is said to be a perfect square if there exists an integer y such that x = y2.
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array.
The second line contains n integers a1, a2, ..., an ( - 106 ≤ ai ≤ 106) — the elements of the array.
It is guaranteed that at least one element of the array is not a perfect square.
Output
Print the largest number in the array which is not a perfect square. It is guaranteed that an answer always exists.
Examples
Input
2
4 2
Output
2
Input
8
1 2 4 8 16 32 64 576
Output
32
Note
In the first sample case, 4 is a perfect square, so the largest number in the array that is not a perfect square is 2.
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
a=sorted(a,reverse=True)
for i in a:
if i>=0:
b=int(i**0.5)
if b*b!=i:
print(i)
break
else:
print(i)
break
```
Yes
| 88,091 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given an array a1, a2, ..., an of n integers, find the largest number in the array that is not a perfect square.
A number x is said to be a perfect square if there exists an integer y such that x = y2.
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array.
The second line contains n integers a1, a2, ..., an ( - 106 ≤ ai ≤ 106) — the elements of the array.
It is guaranteed that at least one element of the array is not a perfect square.
Output
Print the largest number in the array which is not a perfect square. It is guaranteed that an answer always exists.
Examples
Input
2
4 2
Output
2
Input
8
1 2 4 8 16 32 64 576
Output
32
Note
In the first sample case, 4 is a perfect square, so the largest number in the array that is not a perfect square is 2.
Submitted Solution:
```
z=input
mod = 10**9 + 7
from collections import *
from queue import *
from sys import *
from collections import *
from math import *
from heapq import *
from itertools import *
from bisect import *
from collections import Counter as cc
from math import factorial as f
def lcd(xnum1,xnum2):
return (xnum1*xnum2//gcd(xnum1,xnum2))
################################################################################
"""
n=int(z())
for _ in range(int(z())):
x=int(z())
l=list(map(int,z().split()))
n=int(z())
l=sorted(list(map(int,z().split())))[::-1]
a,b=map(int,z().split())
l=set(map(int,z().split()))
led=(6,2,5,5,4,5,6,3,7,6)
vowel={'a':0,'e':0,'i':0,'o':0,'u':0}
color-4=["G", "GB", "YGB", "YGBI", "OYGBI" ,"OYGBIV",'ROYGBIV' ]
"""
###########################---START-CODING---###############################################
n=int(z())
l=sorted(list(map(int,z().split())))[::-1]
for i in l:
t=i
if i<0:
i=i*-1
print(t)
exit()
if int(i**.5)!=i**.5:
print(t)
exit(0)
```
Yes
| 88,092 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given an array a1, a2, ..., an of n integers, find the largest number in the array that is not a perfect square.
A number x is said to be a perfect square if there exists an integer y such that x = y2.
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array.
The second line contains n integers a1, a2, ..., an ( - 106 ≤ ai ≤ 106) — the elements of the array.
It is guaranteed that at least one element of the array is not a perfect square.
Output
Print the largest number in the array which is not a perfect square. It is guaranteed that an answer always exists.
Examples
Input
2
4 2
Output
2
Input
8
1 2 4 8 16 32 64 576
Output
32
Note
In the first sample case, 4 is a perfect square, so the largest number in the array that is not a perfect square is 2.
Submitted Solution:
```
n = list(map(int, input().strip().split()))
arr = list(map(int, input().strip().split()))
answer = -9999999
for i in arr:
if i < 0:
answer = max(answer, i)
continue
temp = int(i ** (1 / 2))
if (temp - 1) ** 2 == i or temp ** 2 == i or (temp + 1) ** 2 == i:
continue
answer = max(answer, i)
print(answer)
```
Yes
| 88,093 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given an array a1, a2, ..., an of n integers, find the largest number in the array that is not a perfect square.
A number x is said to be a perfect square if there exists an integer y such that x = y2.
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array.
The second line contains n integers a1, a2, ..., an ( - 106 ≤ ai ≤ 106) — the elements of the array.
It is guaranteed that at least one element of the array is not a perfect square.
Output
Print the largest number in the array which is not a perfect square. It is guaranteed that an answer always exists.
Examples
Input
2
4 2
Output
2
Input
8
1 2 4 8 16 32 64 576
Output
32
Note
In the first sample case, 4 is a perfect square, so the largest number in the array that is not a perfect square is 2.
Submitted Solution:
```
sq = {i*i for i in range(0,10000)}
n = int(input())
A = [int(x) for x in input().split()]
B = [a for a in A if a not in sq]
print(max(B))
```
Yes
| 88,094 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given an array a1, a2, ..., an of n integers, find the largest number in the array that is not a perfect square.
A number x is said to be a perfect square if there exists an integer y such that x = y2.
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array.
The second line contains n integers a1, a2, ..., an ( - 106 ≤ ai ≤ 106) — the elements of the array.
It is guaranteed that at least one element of the array is not a perfect square.
Output
Print the largest number in the array which is not a perfect square. It is guaranteed that an answer always exists.
Examples
Input
2
4 2
Output
2
Input
8
1 2 4 8 16 32 64 576
Output
32
Note
In the first sample case, 4 is a perfect square, so the largest number in the array that is not a perfect square is 2.
Submitted Solution:
```
from math import sqrt
def dig_sum(num):
while num >= 10:
num = sum(list(map(int,list(str(num)))))
return num
n = int(input())
nums = list(map(int,input().strip().split(' ')))
nums.sort(reverse=True)
for num in nums:
if num % 10 not in [0,1,4,5,6,9] and dig_sum(num) not in [0,1,4,7]:
print(num)
break
```
No
| 88,095 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given an array a1, a2, ..., an of n integers, find the largest number in the array that is not a perfect square.
A number x is said to be a perfect square if there exists an integer y such that x = y2.
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array.
The second line contains n integers a1, a2, ..., an ( - 106 ≤ ai ≤ 106) — the elements of the array.
It is guaranteed that at least one element of the array is not a perfect square.
Output
Print the largest number in the array which is not a perfect square. It is guaranteed that an answer always exists.
Examples
Input
2
4 2
Output
2
Input
8
1 2 4 8 16 32 64 576
Output
32
Note
In the first sample case, 4 is a perfect square, so the largest number in the array that is not a perfect square is 2.
Submitted Solution:
```
input()
prnt=-10000000
for x in list(map(int,input().split())):
print(prnt, x)
if x>0:
prnt=max(prnt, x) if int(abs(x)**0.5)**2 != abs(x) else prnt
else:
prnt = max(prnt,x)
print(prnt)
```
No
| 88,096 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given an array a1, a2, ..., an of n integers, find the largest number in the array that is not a perfect square.
A number x is said to be a perfect square if there exists an integer y such that x = y2.
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array.
The second line contains n integers a1, a2, ..., an ( - 106 ≤ ai ≤ 106) — the elements of the array.
It is guaranteed that at least one element of the array is not a perfect square.
Output
Print the largest number in the array which is not a perfect square. It is guaranteed that an answer always exists.
Examples
Input
2
4 2
Output
2
Input
8
1 2 4 8 16 32 64 576
Output
32
Note
In the first sample case, 4 is a perfect square, so the largest number in the array that is not a perfect square is 2.
Submitted Solution:
```
import math
mp = 0
mo = 0
n = int(input())
a = input().split()
a = [int (i) for i in a]
ot = []
a = sorted(a)
for j in range(len(a)):
if a[j] < 0:
ot.append(a[j])
else:
break
a.reverse()
if len(ot) > 0:
mo = max(ot)
for i in range(len(a)):
if a[i] == 1 or a[i] == -1:
print(a[i])
break
s = math.sqrt(abs(a[i]))
k = int (s)
if s != abs(k):
mp = a[i]
break
if mp != 0:
print(mp)
else:
print(mo)
```
No
| 88,097 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given an array a1, a2, ..., an of n integers, find the largest number in the array that is not a perfect square.
A number x is said to be a perfect square if there exists an integer y such that x = y2.
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array.
The second line contains n integers a1, a2, ..., an ( - 106 ≤ ai ≤ 106) — the elements of the array.
It is guaranteed that at least one element of the array is not a perfect square.
Output
Print the largest number in the array which is not a perfect square. It is guaranteed that an answer always exists.
Examples
Input
2
4 2
Output
2
Input
8
1 2 4 8 16 32 64 576
Output
32
Note
In the first sample case, 4 is a perfect square, so the largest number in the array that is not a perfect square is 2.
Submitted Solution:
```
from math import sqrt
n = int(input())
s = [int(i) for i in input().split()]
c = -10**6
for i in range(n):
if s[i]<0:
pass
elif sqrt(s[i]).is_integer()==False:
c=max(c,s[i])
print(c)
```
No
| 88,098 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Welcome to another task about breaking the code lock! Explorers Whitfield and Martin came across an unusual safe, inside of which, according to rumors, there are untold riches, among which one can find the solution of the problem of discrete logarithm!
Of course, there is a code lock is installed on the safe. The lock has a screen that displays a string of n lowercase Latin letters. Initially, the screen displays string s. Whitfield and Martin found out that the safe will open when string t will be displayed on the screen.
The string on the screen can be changed using the operation «shift x». In order to apply this operation, explorers choose an integer x from 0 to n inclusive. After that, the current string p = αβ changes to βRα, where the length of β is x, and the length of α is n - x. In other words, the suffix of the length x of string p is reversed and moved to the beginning of the string. For example, after the operation «shift 4» the string «abcacb» will be changed with string «bcacab », since α = ab, β = cacb, βR = bcac.
Explorers are afraid that if they apply too many operations «shift», the lock will be locked forever. They ask you to find a way to get the string t on the screen, using no more than 6100 operations.
Input
The first line contains an integer n, the length of the strings s and t (1 ≤ n ≤ 2 000).
After that, there are two strings s and t, consisting of n lowercase Latin letters each.
Output
If it is impossible to get string t from string s using no more than 6100 operations «shift», print a single number - 1.
Otherwise, in the first line output the number of operations k (0 ≤ k ≤ 6100). In the next line output k numbers xi corresponding to the operations «shift xi» (0 ≤ xi ≤ n) in the order in which they should be applied.
Examples
Input
6
abacbb
babcba
Output
4
6 3 2 3
Input
3
aba
bba
Output
-1
Tags: constructive algorithms, implementation, strings
Correct Solution:
```
n, s, t = int(input()), input(), input()
#print n, s, t
if sorted(s)!=sorted(t):
print(-1)
else:
ans = []
def shift(k, cur):
if k == 0:
return cur
return cur[:-k-1:-1] + cur [:-k]
def DO_SWAP(x):
ans.append(x)
return shift(x, s)
#[begin:end:step]
for i in range(n):
curr = t[i]
j = -1
for k in range(n-i):
if s[k] == curr:
j = k
break
s = DO_SWAP(n-j-1)
s = DO_SWAP(1)
s = DO_SWAP(n-1)
assert s == t
print (len(ans))
for x in ans:
print(x, end = " ")
#print(*args, sep=' ', end='\n', file=None)
```
| 88,099 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.