text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cities in Berland. Some pairs of them are connected with m directed roads. One can use only these roads to move from one city to another. There are no roads that connect a city to itself. For each pair of cities (x, y) there is at most one road from x to y.
A path from city s to city t is a sequence of cities p1, p2, ... , pk, where p1 = s, pk = t, and there is a road from city pi to city pi + 1 for each i from 1 to k - 1. The path can pass multiple times through each city except t. It can't pass through t more than once.
A path p from s to t is ideal if it is the lexicographically minimal such path. In other words, p is ideal path from s to t if for any other path q from s to t pi < qi, where i is the minimum integer such that pi ≠ qi.
There is a tourist agency in the country that offers q unusual excursions: the j-th excursion starts at city sj and ends in city tj.
For each pair sj, tj help the agency to study the ideal path from sj to tj. Note that it is possible that there is no ideal path from sj to tj. This is possible due to two reasons:
* there is no path from sj to tj;
* there are paths from sj to tj, but for every such path p there is another path q from sj to tj, such that pi > qi, where i is the minimum integer for which pi ≠ qi.
The agency would like to know for the ideal path from sj to tj the kj-th city in that path (on the way from sj to tj).
For each triple sj, tj, kj (1 ≤ j ≤ q) find if there is an ideal path from sj to tj and print the kj-th city in that path, if there is any.
Input
The first line contains three integers n, m and q (2 ≤ n ≤ 3000,0 ≤ m ≤ 3000, 1 ≤ q ≤ 4·105) — the number of cities, the number of roads and the number of excursions.
Each of the next m lines contains two integers xi and yi (1 ≤ xi, yi ≤ n, xi ≠ yi), denoting that the i-th road goes from city xi to city yi. All roads are one-directional. There can't be more than one road in each direction between two cities.
Each of the next q lines contains three integers sj, tj and kj (1 ≤ sj, tj ≤ n, sj ≠ tj, 1 ≤ kj ≤ 3000).
Output
In the j-th line print the city that is the kj-th in the ideal path from sj to tj. If there is no ideal path from sj to tj, or the integer kj is greater than the length of this path, print the string '-1' (without quotes) in the j-th line.
Example
Input
7 7 5
1 2
2 3
1 3
3 4
4 5
5 3
4 6
1 4 2
2 6 1
1 7 3
1 3 2
1 3 5
Output
2
-1
-1
2
-1
Submitted Solution:
```
# -*- coding: utf-8 -*-
import math
import collections
import bisect
import heapq
import time
import random
"""
created by shhuan at 2017/10/5 11:00
"""
N, M, Q = map(int, input().split())
P = collections.defaultdict(list)
for i in range(M):
s, t = map(int, input().split())
P[s].append(t)
E = []
for i in range(Q):
s, t, k = map(int, input().split())
E.append((s, t, k))
INF = False
def dfs(start, end, path, visited, circle):
global INF
if INF:
return []
if start == end:
return [path]
ans = []
for to in sorted(P[start]):
if to not in visited:
ans.extend(dfs(to, end, path+[to], visited | {to}, circle))
else:
if not circle:
i = path.index(to)
for j in range(i+1, len(path)):
if dfs(path[j-1], end, path+path[i:j], visited, True):
INF = True
return []
return ans
for s, t, k in E:
INF = False
paths = dfs(s, t, [s], set(), False)
if INF:
print(-1)
continue
if not paths:
print(-1)
continue
paths = sorted(paths)
# print(s, t, k, paths)
if paths[0] > paths[-1]:
print(-1)
else:
path = paths[0]
if k >= len(path):
print(-1)
else:
print(path[k-1])
# 2 3 4 5 4 5 4 5 .... 4 6
```
No
| 6,300 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A New Year party is not a New Year party without lemonade! As usual, you are expecting a lot of guests, and buying lemonade has already become a pleasant necessity.
Your favorite store sells lemonade in bottles of n different volumes at different costs. A single bottle of type i has volume 2i - 1 liters and costs ci roubles. The number of bottles of each type in the store can be considered infinite.
You want to buy at least L liters of lemonade. How many roubles do you have to spend?
Input
The first line contains two integers n and L (1 ≤ n ≤ 30; 1 ≤ L ≤ 109) — the number of types of bottles in the store and the required amount of lemonade in liters, respectively.
The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 109) — the costs of bottles of different types.
Output
Output a single integer — the smallest number of roubles you have to pay in order to buy at least L liters of lemonade.
Examples
Input
4 12
20 30 70 90
Output
150
Input
4 3
10000 1000 100 10
Output
10
Input
4 3
10 100 1000 10000
Output
30
Input
5 787787787
123456789 234567890 345678901 456789012 987654321
Output
44981600785557577
Note
In the first example you should buy one 8-liter bottle for 90 roubles and two 2-liter bottles for 30 roubles each. In total you'll get 12 liters of lemonade for just 150 roubles.
In the second example, even though you need only 3 liters, it's cheaper to buy a single 8-liter bottle for 10 roubles.
In the third example it's best to buy three 1-liter bottles for 10 roubles each, getting three liters for 30 roubles.
Tags: bitmasks, dp, greedy
Correct Solution:
```
from sys import stdin, stdout
sze = 100
INF = float('inf')
n, l = map(int, stdin.readline().split())
c = list(map(int, stdin.readline().split()))
may = [1 for i in range(sze)]
for i in range(n - 1, -1, -1):
for j in range(i):
if c[i] > c[j] * 2 ** (i - j):
may[i] = 0
ans = INF
cnt = 0
value = 0
for i in range(n - 1, -1, -1):
if not may[i]:
continue
v = l - value
if not v % (2 ** i):
ans = min(ans, cnt + v * c[i] // (2 ** i))
else:
ans = min(ans, cnt + (v // (2 ** i) + 1) * c[i])
cnt += v // (2 ** i) * c[i]
value += (v // (2 ** i)) * (2 ** i)
stdout.write(str(ans))
```
| 6,301 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A New Year party is not a New Year party without lemonade! As usual, you are expecting a lot of guests, and buying lemonade has already become a pleasant necessity.
Your favorite store sells lemonade in bottles of n different volumes at different costs. A single bottle of type i has volume 2i - 1 liters and costs ci roubles. The number of bottles of each type in the store can be considered infinite.
You want to buy at least L liters of lemonade. How many roubles do you have to spend?
Input
The first line contains two integers n and L (1 ≤ n ≤ 30; 1 ≤ L ≤ 109) — the number of types of bottles in the store and the required amount of lemonade in liters, respectively.
The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 109) — the costs of bottles of different types.
Output
Output a single integer — the smallest number of roubles you have to pay in order to buy at least L liters of lemonade.
Examples
Input
4 12
20 30 70 90
Output
150
Input
4 3
10000 1000 100 10
Output
10
Input
4 3
10 100 1000 10000
Output
30
Input
5 787787787
123456789 234567890 345678901 456789012 987654321
Output
44981600785557577
Note
In the first example you should buy one 8-liter bottle for 90 roubles and two 2-liter bottles for 30 roubles each. In total you'll get 12 liters of lemonade for just 150 roubles.
In the second example, even though you need only 3 liters, it's cheaper to buy a single 8-liter bottle for 10 roubles.
In the third example it's best to buy three 1-liter bottles for 10 roubles each, getting three liters for 30 roubles.
Tags: bitmasks, dp, greedy
Correct Solution:
```
n, li = map(int, input().split())
c = list(map(int, input().split()))
for i in range(n-1):
if c[i] * 2 < c[i+1]:
c[i+1] = c[i] * 2
for i in range(n-2, -1, -1):
if c[i] > c[i+1]:
c[i] = c[i+1]
'''
p = c[0] * li
for i in range(n):
if 2**i >= li and p > c[i]:
p = c[i]
'''
i = n - 1
ans = 0
a = []
while li > 0:
ans += c[i] * (li >> i)
li %= 2**i
if li < 2**i:
a.append(ans + c[i])
i-=1
a.append(ans)
print(min(a))
```
| 6,302 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A New Year party is not a New Year party without lemonade! As usual, you are expecting a lot of guests, and buying lemonade has already become a pleasant necessity.
Your favorite store sells lemonade in bottles of n different volumes at different costs. A single bottle of type i has volume 2i - 1 liters and costs ci roubles. The number of bottles of each type in the store can be considered infinite.
You want to buy at least L liters of lemonade. How many roubles do you have to spend?
Input
The first line contains two integers n and L (1 ≤ n ≤ 30; 1 ≤ L ≤ 109) — the number of types of bottles in the store and the required amount of lemonade in liters, respectively.
The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 109) — the costs of bottles of different types.
Output
Output a single integer — the smallest number of roubles you have to pay in order to buy at least L liters of lemonade.
Examples
Input
4 12
20 30 70 90
Output
150
Input
4 3
10000 1000 100 10
Output
10
Input
4 3
10 100 1000 10000
Output
30
Input
5 787787787
123456789 234567890 345678901 456789012 987654321
Output
44981600785557577
Note
In the first example you should buy one 8-liter bottle for 90 roubles and two 2-liter bottles for 30 roubles each. In total you'll get 12 liters of lemonade for just 150 roubles.
In the second example, even though you need only 3 liters, it's cheaper to buy a single 8-liter bottle for 10 roubles.
In the third example it's best to buy three 1-liter bottles for 10 roubles each, getting three liters for 30 roubles.
Tags: bitmasks, dp, greedy
Correct Solution:
```
s=input()
s1=s.split()
tot=int(s1[1])
s=input()
s1=s.split()
l=[int(i) for i in s1]
avg=[[l[i]/(2**i),i] for i in range(len(l))]
avg.sort()
cost=0
i=0
d={}
def fun(i,tot):
if i==len(avg):
return(1e8)
elif (i,tot) in d:
return(d[(i,tot)])
elif tot%(2**avg[i][1])==0:
return((tot//(2**avg[i][1]))*l[avg[i][1]])
else:
a=(tot//(2**avg[i][1])+1)*l[avg[i][1]]
b=(tot//(2**avg[i][1]))*l[avg[i][1]]+fun(i+1,tot%(2**avg[i][1]))
d[(i,tot)]=min(a,b)
return(min(a,b))
print(fun(0,tot))
```
| 6,303 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A New Year party is not a New Year party without lemonade! As usual, you are expecting a lot of guests, and buying lemonade has already become a pleasant necessity.
Your favorite store sells lemonade in bottles of n different volumes at different costs. A single bottle of type i has volume 2i - 1 liters and costs ci roubles. The number of bottles of each type in the store can be considered infinite.
You want to buy at least L liters of lemonade. How many roubles do you have to spend?
Input
The first line contains two integers n and L (1 ≤ n ≤ 30; 1 ≤ L ≤ 109) — the number of types of bottles in the store and the required amount of lemonade in liters, respectively.
The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 109) — the costs of bottles of different types.
Output
Output a single integer — the smallest number of roubles you have to pay in order to buy at least L liters of lemonade.
Examples
Input
4 12
20 30 70 90
Output
150
Input
4 3
10000 1000 100 10
Output
10
Input
4 3
10 100 1000 10000
Output
30
Input
5 787787787
123456789 234567890 345678901 456789012 987654321
Output
44981600785557577
Note
In the first example you should buy one 8-liter bottle for 90 roubles and two 2-liter bottles for 30 roubles each. In total you'll get 12 liters of lemonade for just 150 roubles.
In the second example, even though you need only 3 liters, it's cheaper to buy a single 8-liter bottle for 10 roubles.
In the third example it's best to buy three 1-liter bottles for 10 roubles each, getting three liters for 30 roubles.
Tags: bitmasks, dp, greedy
Correct Solution:
```
a,b=map(int,input().split())
z=list(map(int,input().split()))
import math
pre=[z[0]]
for i in range(1,len(z)):
pre.append(min(z[i],2*pre[-1]))
q=[1]
for i in range(len(z)-1):
q.append(q[-1]*2)
cost=0
p=len(q)-1
pos=[]
mini=math.inf
while(1):
if(b>=q[p]):
nos=b//q[p]
rem=b%q[p]
cost+=nos*pre[p]
if(rem==0):
print(min(mini,cost))
break;
else:
pos.append(cost+pre[p])
mini=min(mini,min(pos))
b=rem
p=p-1
else:
pos.append(cost+pre[p])
p=p-1
mini=min(mini,min(pos))
```
| 6,304 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A New Year party is not a New Year party without lemonade! As usual, you are expecting a lot of guests, and buying lemonade has already become a pleasant necessity.
Your favorite store sells lemonade in bottles of n different volumes at different costs. A single bottle of type i has volume 2i - 1 liters and costs ci roubles. The number of bottles of each type in the store can be considered infinite.
You want to buy at least L liters of lemonade. How many roubles do you have to spend?
Input
The first line contains two integers n and L (1 ≤ n ≤ 30; 1 ≤ L ≤ 109) — the number of types of bottles in the store and the required amount of lemonade in liters, respectively.
The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 109) — the costs of bottles of different types.
Output
Output a single integer — the smallest number of roubles you have to pay in order to buy at least L liters of lemonade.
Examples
Input
4 12
20 30 70 90
Output
150
Input
4 3
10000 1000 100 10
Output
10
Input
4 3
10 100 1000 10000
Output
30
Input
5 787787787
123456789 234567890 345678901 456789012 987654321
Output
44981600785557577
Note
In the first example you should buy one 8-liter bottle for 90 roubles and two 2-liter bottles for 30 roubles each. In total you'll get 12 liters of lemonade for just 150 roubles.
In the second example, even though you need only 3 liters, it's cheaper to buy a single 8-liter bottle for 10 roubles.
In the third example it's best to buy three 1-liter bottles for 10 roubles each, getting three liters for 30 roubles.
Tags: bitmasks, dp, greedy
Correct Solution:
```
n, l = [int(x) for x in input().strip().split()]
c = [int(x) for x in input().strip().split()]
tc = 0
for i in range(1, len(c)):
c[i] = min(c[i], 2*c[i-1])
for i in range(32):
c.append(c[-1]*2)
tc = 0
bl = bin(l)[:1:-1]
bl += '0' * (len(c) - len(bl) - 1)
for i in range(len(bl)):
if bl[i] == '1':
tc += c[i]
else:
tc = min(tc, c[i])
print(tc)
```
| 6,305 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A New Year party is not a New Year party without lemonade! As usual, you are expecting a lot of guests, and buying lemonade has already become a pleasant necessity.
Your favorite store sells lemonade in bottles of n different volumes at different costs. A single bottle of type i has volume 2i - 1 liters and costs ci roubles. The number of bottles of each type in the store can be considered infinite.
You want to buy at least L liters of lemonade. How many roubles do you have to spend?
Input
The first line contains two integers n and L (1 ≤ n ≤ 30; 1 ≤ L ≤ 109) — the number of types of bottles in the store and the required amount of lemonade in liters, respectively.
The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 109) — the costs of bottles of different types.
Output
Output a single integer — the smallest number of roubles you have to pay in order to buy at least L liters of lemonade.
Examples
Input
4 12
20 30 70 90
Output
150
Input
4 3
10000 1000 100 10
Output
10
Input
4 3
10 100 1000 10000
Output
30
Input
5 787787787
123456789 234567890 345678901 456789012 987654321
Output
44981600785557577
Note
In the first example you should buy one 8-liter bottle for 90 roubles and two 2-liter bottles for 30 roubles each. In total you'll get 12 liters of lemonade for just 150 roubles.
In the second example, even though you need only 3 liters, it's cheaper to buy a single 8-liter bottle for 10 roubles.
In the third example it's best to buy three 1-liter bottles for 10 roubles each, getting three liters for 30 roubles.
Tags: bitmasks, dp, greedy
Correct Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
import sys
from io import BytesIO, IOBase
def main():
n,l=map(int,input().split())
c=list(map(int,input().split()))
mi,su=4*10**18,0
for i in range(1,n):
c[i]=min(c[i],2*c[i-1])
for i in range(n-1,-1,-1):
y=(1<<i)
z=l//y
su+=z*c[i]
l-=z*y
mi=min(mi,su+(l!=0)*c[i])
print(mi)
# 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")
if __name__ == "__main__":
main()
```
| 6,306 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A New Year party is not a New Year party without lemonade! As usual, you are expecting a lot of guests, and buying lemonade has already become a pleasant necessity.
Your favorite store sells lemonade in bottles of n different volumes at different costs. A single bottle of type i has volume 2i - 1 liters and costs ci roubles. The number of bottles of each type in the store can be considered infinite.
You want to buy at least L liters of lemonade. How many roubles do you have to spend?
Input
The first line contains two integers n and L (1 ≤ n ≤ 30; 1 ≤ L ≤ 109) — the number of types of bottles in the store and the required amount of lemonade in liters, respectively.
The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 109) — the costs of bottles of different types.
Output
Output a single integer — the smallest number of roubles you have to pay in order to buy at least L liters of lemonade.
Examples
Input
4 12
20 30 70 90
Output
150
Input
4 3
10000 1000 100 10
Output
10
Input
4 3
10 100 1000 10000
Output
30
Input
5 787787787
123456789 234567890 345678901 456789012 987654321
Output
44981600785557577
Note
In the first example you should buy one 8-liter bottle for 90 roubles and two 2-liter bottles for 30 roubles each. In total you'll get 12 liters of lemonade for just 150 roubles.
In the second example, even though you need only 3 liters, it's cheaper to buy a single 8-liter bottle for 10 roubles.
In the third example it's best to buy three 1-liter bottles for 10 roubles each, getting three liters for 30 roubles.
Tags: bitmasks, dp, greedy
Correct Solution:
```
types,volume=map(int,input().split())
c,s=0,10**18
a=list(map(int,input().split()))
for i in range(len(a)-1):
a[i+1]=min(a[i+1],a[i]*2)
for i in range(types-1,-1,-1):
d=1<<i
c+=a[i]*(volume//d)
volume%=d
s=min(s,c+(volume!=0)*a[i])
print(s)
```
| 6,307 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A New Year party is not a New Year party without lemonade! As usual, you are expecting a lot of guests, and buying lemonade has already become a pleasant necessity.
Your favorite store sells lemonade in bottles of n different volumes at different costs. A single bottle of type i has volume 2i - 1 liters and costs ci roubles. The number of bottles of each type in the store can be considered infinite.
You want to buy at least L liters of lemonade. How many roubles do you have to spend?
Input
The first line contains two integers n and L (1 ≤ n ≤ 30; 1 ≤ L ≤ 109) — the number of types of bottles in the store and the required amount of lemonade in liters, respectively.
The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 109) — the costs of bottles of different types.
Output
Output a single integer — the smallest number of roubles you have to pay in order to buy at least L liters of lemonade.
Examples
Input
4 12
20 30 70 90
Output
150
Input
4 3
10000 1000 100 10
Output
10
Input
4 3
10 100 1000 10000
Output
30
Input
5 787787787
123456789 234567890 345678901 456789012 987654321
Output
44981600785557577
Note
In the first example you should buy one 8-liter bottle for 90 roubles and two 2-liter bottles for 30 roubles each. In total you'll get 12 liters of lemonade for just 150 roubles.
In the second example, even though you need only 3 liters, it's cheaper to buy a single 8-liter bottle for 10 roubles.
In the third example it's best to buy three 1-liter bottles for 10 roubles each, getting three liters for 30 roubles.
Tags: bitmasks, dp, greedy
Correct Solution:
```
s=input()
s1=s.split()
tot=int(s1[1])
s=input()
s1=s.split()
l=[int(i) for i in s1]
avg=[[l[i]/(2**i),i] for i in range(len(l))]
avg.sort()
cost=0
i=0
def fun(i,tot):
if i==len(avg):
return(1e8)
if tot%(2**avg[i][1])==0:
return((tot//(2**avg[i][1]))*l[avg[i][1]])
else:
a=(tot//(2**avg[i][1])+1)*l[avg[i][1]]
b=(tot//(2**avg[i][1]))*l[avg[i][1]]+fun(i+1,tot%(2**avg[i][1]))
return(min(a,b))
print(fun(0,tot))
```
| 6,308 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A New Year party is not a New Year party without lemonade! As usual, you are expecting a lot of guests, and buying lemonade has already become a pleasant necessity.
Your favorite store sells lemonade in bottles of n different volumes at different costs. A single bottle of type i has volume 2i - 1 liters and costs ci roubles. The number of bottles of each type in the store can be considered infinite.
You want to buy at least L liters of lemonade. How many roubles do you have to spend?
Input
The first line contains two integers n and L (1 ≤ n ≤ 30; 1 ≤ L ≤ 109) — the number of types of bottles in the store and the required amount of lemonade in liters, respectively.
The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 109) — the costs of bottles of different types.
Output
Output a single integer — the smallest number of roubles you have to pay in order to buy at least L liters of lemonade.
Examples
Input
4 12
20 30 70 90
Output
150
Input
4 3
10000 1000 100 10
Output
10
Input
4 3
10 100 1000 10000
Output
30
Input
5 787787787
123456789 234567890 345678901 456789012 987654321
Output
44981600785557577
Note
In the first example you should buy one 8-liter bottle for 90 roubles and two 2-liter bottles for 30 roubles each. In total you'll get 12 liters of lemonade for just 150 roubles.
In the second example, even though you need only 3 liters, it's cheaper to buy a single 8-liter bottle for 10 roubles.
In the third example it's best to buy three 1-liter bottles for 10 roubles each, getting three liters for 30 roubles.
Submitted Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
import collections
from itertools import permutations
from collections import defaultdict
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------game starts now-----------------------------------------------------
n,L=map(int,input().split())
c=list(map(int,input().split()))
t=[c[i] if i<n else float('inf') for i in range(32)]
for i in range(1,32):
t[i]=min(t[i],2*t[i-1])
a=0
for i in range(0,32):
q=2**i
if L&q>0:
a+=t[i]
else:
a=min(a,t[i])
print(a)
```
Yes
| 6,309 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A New Year party is not a New Year party without lemonade! As usual, you are expecting a lot of guests, and buying lemonade has already become a pleasant necessity.
Your favorite store sells lemonade in bottles of n different volumes at different costs. A single bottle of type i has volume 2i - 1 liters and costs ci roubles. The number of bottles of each type in the store can be considered infinite.
You want to buy at least L liters of lemonade. How many roubles do you have to spend?
Input
The first line contains two integers n and L (1 ≤ n ≤ 30; 1 ≤ L ≤ 109) — the number of types of bottles in the store and the required amount of lemonade in liters, respectively.
The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 109) — the costs of bottles of different types.
Output
Output a single integer — the smallest number of roubles you have to pay in order to buy at least L liters of lemonade.
Examples
Input
4 12
20 30 70 90
Output
150
Input
4 3
10000 1000 100 10
Output
10
Input
4 3
10 100 1000 10000
Output
30
Input
5 787787787
123456789 234567890 345678901 456789012 987654321
Output
44981600785557577
Note
In the first example you should buy one 8-liter bottle for 90 roubles and two 2-liter bottles for 30 roubles each. In total you'll get 12 liters of lemonade for just 150 roubles.
In the second example, even though you need only 3 liters, it's cheaper to buy a single 8-liter bottle for 10 roubles.
In the third example it's best to buy three 1-liter bottles for 10 roubles each, getting three liters for 30 roubles.
Submitted Solution:
```
import math
N = 31
n, l = map(int, input().split())
cost = [int(x) for x in input().split()]
for i in range(n, N):
cost.append(math.inf)
for i in range(N - 1):
cost[i + 1] = min(cost[i + 1], cost[i] * 2)
# print(cost)
ans = math.inf
cur_cost = 0
for i in range(N - 1, -1, -1):
# print('l = {}'.format(l))
if 2**i >= l:
ans = min(ans, cur_cost + cost[i])
else:
cur_cost += cost[i]
l -= 2**i;
# print('at {} and ans = {}'.format(i, ans))
print(ans)
```
Yes
| 6,310 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A New Year party is not a New Year party without lemonade! As usual, you are expecting a lot of guests, and buying lemonade has already become a pleasant necessity.
Your favorite store sells lemonade in bottles of n different volumes at different costs. A single bottle of type i has volume 2i - 1 liters and costs ci roubles. The number of bottles of each type in the store can be considered infinite.
You want to buy at least L liters of lemonade. How many roubles do you have to spend?
Input
The first line contains two integers n and L (1 ≤ n ≤ 30; 1 ≤ L ≤ 109) — the number of types of bottles in the store and the required amount of lemonade in liters, respectively.
The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 109) — the costs of bottles of different types.
Output
Output a single integer — the smallest number of roubles you have to pay in order to buy at least L liters of lemonade.
Examples
Input
4 12
20 30 70 90
Output
150
Input
4 3
10000 1000 100 10
Output
10
Input
4 3
10 100 1000 10000
Output
30
Input
5 787787787
123456789 234567890 345678901 456789012 987654321
Output
44981600785557577
Note
In the first example you should buy one 8-liter bottle for 90 roubles and two 2-liter bottles for 30 roubles each. In total you'll get 12 liters of lemonade for just 150 roubles.
In the second example, even though you need only 3 liters, it's cheaper to buy a single 8-liter bottle for 10 roubles.
In the third example it's best to buy three 1-liter bottles for 10 roubles each, getting three liters for 30 roubles.
Submitted Solution:
```
n = input().split()
l = int(n[1])
n = int(n[0])
c = input().split()
cperl = []
for i in range(n):
c[i] = int(c[i])
cperl.append([(c[i])/2**i, 2**i, i])
cperl.sort()
f = False
nl = [0]
nc = [0]
i = 0
while not f:
p = nl[i]
nl[i] += cperl[i][1] * ((l-p)//cperl[i][1])
nc[i] += c[cperl[i][2]] * ((l-p)//cperl[i][1])
if nl[i] == l:
f = True
break
nl.append(nl[i])
nc.append(nc[i])
nl[i] += cperl[i][1]
nc[i] += c[cperl[i][2]]
i += 1
if i == n:
break
print(min(nc))
```
Yes
| 6,311 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A New Year party is not a New Year party without lemonade! As usual, you are expecting a lot of guests, and buying lemonade has already become a pleasant necessity.
Your favorite store sells lemonade in bottles of n different volumes at different costs. A single bottle of type i has volume 2i - 1 liters and costs ci roubles. The number of bottles of each type in the store can be considered infinite.
You want to buy at least L liters of lemonade. How many roubles do you have to spend?
Input
The first line contains two integers n and L (1 ≤ n ≤ 30; 1 ≤ L ≤ 109) — the number of types of bottles in the store and the required amount of lemonade in liters, respectively.
The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 109) — the costs of bottles of different types.
Output
Output a single integer — the smallest number of roubles you have to pay in order to buy at least L liters of lemonade.
Examples
Input
4 12
20 30 70 90
Output
150
Input
4 3
10000 1000 100 10
Output
10
Input
4 3
10 100 1000 10000
Output
30
Input
5 787787787
123456789 234567890 345678901 456789012 987654321
Output
44981600785557577
Note
In the first example you should buy one 8-liter bottle for 90 roubles and two 2-liter bottles for 30 roubles each. In total you'll get 12 liters of lemonade for just 150 roubles.
In the second example, even though you need only 3 liters, it's cheaper to buy a single 8-liter bottle for 10 roubles.
In the third example it's best to buy three 1-liter bottles for 10 roubles each, getting three liters for 30 roubles.
Submitted Solution:
```
s = 0
ans = 100 ** 1000
n, l = map(int, input().split())
a = list(map(int, input().split()))
for i in range(0,n-1):
a[i+1] = min(a[i+1], 2*a[i])
for i in range(n-1, -1, -1):
d = l // (1 << i)
s += d * a[i]
l -= d << i;
ans = min(ans, s+(l > 0) * a[i])
print(ans)
```
Yes
| 6,312 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A New Year party is not a New Year party without lemonade! As usual, you are expecting a lot of guests, and buying lemonade has already become a pleasant necessity.
Your favorite store sells lemonade in bottles of n different volumes at different costs. A single bottle of type i has volume 2i - 1 liters and costs ci roubles. The number of bottles of each type in the store can be considered infinite.
You want to buy at least L liters of lemonade. How many roubles do you have to spend?
Input
The first line contains two integers n and L (1 ≤ n ≤ 30; 1 ≤ L ≤ 109) — the number of types of bottles in the store and the required amount of lemonade in liters, respectively.
The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 109) — the costs of bottles of different types.
Output
Output a single integer — the smallest number of roubles you have to pay in order to buy at least L liters of lemonade.
Examples
Input
4 12
20 30 70 90
Output
150
Input
4 3
10000 1000 100 10
Output
10
Input
4 3
10 100 1000 10000
Output
30
Input
5 787787787
123456789 234567890 345678901 456789012 987654321
Output
44981600785557577
Note
In the first example you should buy one 8-liter bottle for 90 roubles and two 2-liter bottles for 30 roubles each. In total you'll get 12 liters of lemonade for just 150 roubles.
In the second example, even though you need only 3 liters, it's cheaper to buy a single 8-liter bottle for 10 roubles.
In the third example it's best to buy three 1-liter bottles for 10 roubles each, getting three liters for 30 roubles.
Submitted Solution:
```
n,L=[int(i) for i in input().split()]
L=list(bin(L).replace('0b',''))
L.reverse()
c=[int(i) for i in input().split()]
l=0
C=[]
for i in range(len(L)):
c_l={}
for j in range(n):
if j>i:
c_l[c[j]/2**i]=j
elif j<=i:
c_l[c[j]/2**j]=j
c_ls=list(c_l.keys())
m=c_l[min(c_ls)]
if m>i:
Ci=c[m]
if m<=i:
Ci=c[m]/(2**m)*(2**i)
Cc1=l+Ci*int(L[i])
c_l.clear()
for j in range(n):
if j>(i+1):
c_l[c[j]/2**(i+1)]=j
elif j<=(i+1):
c_l[c[j]/2**j]=j
c_ls=list(c_l.keys())
m=c_l[min(c_ls)]
if m>(i+1):
Ci=c[m]
if m<=(i+1):
Ci=c[m]/(2**m)*(2**(i+1))
Cc2=Ci*int(L[i])
x=min(Cc1,Cc2)
l=l+x
if x==0:
C.append(l)
else:
C.append(x)
l=x
print(int(C[-1]))
```
No
| 6,313 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A New Year party is not a New Year party without lemonade! As usual, you are expecting a lot of guests, and buying lemonade has already become a pleasant necessity.
Your favorite store sells lemonade in bottles of n different volumes at different costs. A single bottle of type i has volume 2i - 1 liters and costs ci roubles. The number of bottles of each type in the store can be considered infinite.
You want to buy at least L liters of lemonade. How many roubles do you have to spend?
Input
The first line contains two integers n and L (1 ≤ n ≤ 30; 1 ≤ L ≤ 109) — the number of types of bottles in the store and the required amount of lemonade in liters, respectively.
The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 109) — the costs of bottles of different types.
Output
Output a single integer — the smallest number of roubles you have to pay in order to buy at least L liters of lemonade.
Examples
Input
4 12
20 30 70 90
Output
150
Input
4 3
10000 1000 100 10
Output
10
Input
4 3
10 100 1000 10000
Output
30
Input
5 787787787
123456789 234567890 345678901 456789012 987654321
Output
44981600785557577
Note
In the first example you should buy one 8-liter bottle for 90 roubles and two 2-liter bottles for 30 roubles each. In total you'll get 12 liters of lemonade for just 150 roubles.
In the second example, even though you need only 3 liters, it's cheaper to buy a single 8-liter bottle for 10 roubles.
In the third example it's best to buy three 1-liter bottles for 10 roubles each, getting three liters for 30 roubles.
Submitted Solution:
```
cin=input().split()
n=int(cin[0])
m=int(cin[1])
data=list(map(int, input().split()))
for x in range(1, len(bin(m))-2):
if(x<len(data)):
if(data[x]>(data[x-1]*2)):
data[x]=data[x-1]*2
else :
data.append(data[x-1]*2)
for x in reversed(range(0, len(data)-1)):
data[x]=min(data[x+1], data[x])
m=bin(m)
m=list(reversed(m))
ans=0
for x in range(len(m)-2):
if(m[x]=='1'):
ans+=int(data[x])
else :
ans=min(ans, data[x])
print(ans)
```
No
| 6,314 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A New Year party is not a New Year party without lemonade! As usual, you are expecting a lot of guests, and buying lemonade has already become a pleasant necessity.
Your favorite store sells lemonade in bottles of n different volumes at different costs. A single bottle of type i has volume 2i - 1 liters and costs ci roubles. The number of bottles of each type in the store can be considered infinite.
You want to buy at least L liters of lemonade. How many roubles do you have to spend?
Input
The first line contains two integers n and L (1 ≤ n ≤ 30; 1 ≤ L ≤ 109) — the number of types of bottles in the store and the required amount of lemonade in liters, respectively.
The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 109) — the costs of bottles of different types.
Output
Output a single integer — the smallest number of roubles you have to pay in order to buy at least L liters of lemonade.
Examples
Input
4 12
20 30 70 90
Output
150
Input
4 3
10000 1000 100 10
Output
10
Input
4 3
10 100 1000 10000
Output
30
Input
5 787787787
123456789 234567890 345678901 456789012 987654321
Output
44981600785557577
Note
In the first example you should buy one 8-liter bottle for 90 roubles and two 2-liter bottles for 30 roubles each. In total you'll get 12 liters of lemonade for just 150 roubles.
In the second example, even though you need only 3 liters, it's cheaper to buy a single 8-liter bottle for 10 roubles.
In the third example it's best to buy three 1-liter bottles for 10 roubles each, getting three liters for 30 roubles.
Submitted Solution:
```
arr = [int(x) for x in input().split()]
n = arr[0]
L = arr[1]
costs = [int(x) for x in input().split()]
resp = 4e18
sumC = 0
for i in range(n - 1):
costs[i+1] = min(costs[i + 1], 2 * costs[i])
for i in range(n - 1, 0, -1):
k = L // (1 << i)
sumC += k * costs[i]
L -= k << i
resp = min(resp, sumC + (L > 0) * costs[i])
print(resp)
```
No
| 6,315 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A New Year party is not a New Year party without lemonade! As usual, you are expecting a lot of guests, and buying lemonade has already become a pleasant necessity.
Your favorite store sells lemonade in bottles of n different volumes at different costs. A single bottle of type i has volume 2i - 1 liters and costs ci roubles. The number of bottles of each type in the store can be considered infinite.
You want to buy at least L liters of lemonade. How many roubles do you have to spend?
Input
The first line contains two integers n and L (1 ≤ n ≤ 30; 1 ≤ L ≤ 109) — the number of types of bottles in the store and the required amount of lemonade in liters, respectively.
The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 109) — the costs of bottles of different types.
Output
Output a single integer — the smallest number of roubles you have to pay in order to buy at least L liters of lemonade.
Examples
Input
4 12
20 30 70 90
Output
150
Input
4 3
10000 1000 100 10
Output
10
Input
4 3
10 100 1000 10000
Output
30
Input
5 787787787
123456789 234567890 345678901 456789012 987654321
Output
44981600785557577
Note
In the first example you should buy one 8-liter bottle for 90 roubles and two 2-liter bottles for 30 roubles each. In total you'll get 12 liters of lemonade for just 150 roubles.
In the second example, even though you need only 3 liters, it's cheaper to buy a single 8-liter bottle for 10 roubles.
In the third example it's best to buy three 1-liter bottles for 10 roubles each, getting three liters for 30 roubles.
Submitted Solution:
```
import math
def calcula(L, cost):
tam = math.log2(L)
tam1 = int(tam)
if (tam == tam1):
return cost[tam1]
esquerda = L - (1 << tam1)
tmp = calcula(esquerda, cost)
tam2 = math.log2(esquerda)
i = int(tam2) + 1
while i <= tam1:
i += 1
tmp = min(tmp, cost[i])
if (cost[tam1] * 2 < cost[tam1] + tmp):
return cost[tam1] * 2
else:
return cost[tam1] + tmp
entrada = input().split()
n = int(entrada[0])
L = int(entrada[1])
cost = [0] * 33
i = 0
while i <= 32:
cost[i] = 1 << 62
i += 1
entrada = input().split()
for i in range(0, n):
cost[i] = int(entrada[i])
i = 0
while(i <= 32):
cost[i] = min(cost[i], cost[i - 1] * 2)
i += 1
tam = math.log2(L)
result = calcula(L, cost)
i = int(tam) + 1
while i <= 32:
result = min(result, cost[i])
i += 1
print(result)
```
No
| 6,316 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.
The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.
Input
The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).
Output
Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.
Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.
Examples
Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095
Input
10 5 5 5 15
Output
5.0 5.0 10.0
Tags: geometry
Correct Solution:
```
R,x1,y1,x2,y2=list(map(float,input().split()))
d=((x2-x1)**2+(y2-y1)**2)**0.5
if d>R:
print(x1," ",y1," ",R)
elif d==0:
print(x1-R/2," ",y1," ",R/2)
else:
x,y=(x1*(R+d)-x2*R)/(d),(y1*(R+d)-y2*R)/(d)
xans,yans=(x+x2)/2,(y+y2)/2
r=((x-xans)**2+(y-yans)**2)**0.5
print(xans," ",yans," ",r)
```
| 6,317 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.
The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.
Input
The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).
Output
Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.
Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.
Examples
Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095
Input
10 5 5 5 15
Output
5.0 5.0 10.0
Tags: geometry
Correct Solution:
```
r,x1,y1,x2,y2=map(int,input().split())
if ((x1-x2)**2+(y1-y2)**2>r*r):
print(x1,y1,r)
exit()
d=((x1-x2)**2+(y1-y2)**2)**0.5
r2=(r+d)/2
if (d==0):
x3=x2+r/2
y3=y2
else:
x3=x2+(x1-x2)/d*r2
y3=y2+(y1-y2)/d*r2
print(x3,y3,r2)
```
| 6,318 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.
The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.
Input
The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).
Output
Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.
Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.
Examples
Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095
Input
10 5 5 5 15
Output
5.0 5.0 10.0
Tags: geometry
Correct Solution:
```
r,x1,y1,x2,y2=map(int,input().split())
if (x2-x1)**2+(y2-y1)**2>=r*r:
print(x1,y1,r)
exit()
if x1==x2:
if y2<y1:
print(x1,y2+(y1+r-y2)/2,(y1+r-y2)/2)
else:
print(x1,y2-(y2-y1+r)/2,(y2-y1+r)/2)
else:
k=(y2-y1)/(x2-x1)
b=y1-k*x1
B=b+0
a=k*k+1
c=x1*x1+b*b-2*b*y1+y1*y1-r*r
b=2*k*b-2*x1-2*k*y1
d=b*b-4*a*c
X1=(-b+d**0.5)/(2*a)
X2=(-b-d**0.5)/(2*a)
Y1=k*X1+B
Y2=k*X2+B
r1=((X1-x2)**2+(Y1-y2)**2)**0.5
r2=((X2-x2)**2+(Y2-y2)**2)**0.5
if r1>r2:
print((x2+X1)/2,(y2+Y1)/2,r1/2)
else:
print((x2+X2)/2,(y2+Y2)/2,r2/2)
```
| 6,319 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.
The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.
Input
The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).
Output
Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.
Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.
Examples
Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095
Input
10 5 5 5 15
Output
5.0 5.0 10.0
Tags: geometry
Correct Solution:
```
r,x1,y1,x2,y2=map(int ,input().split())
re = ((x1-x2)**2+(y1-y2)**2)**0.5
if (re>=r):
re = r
x3 = x1
y3 = y1
if (re==0):
x3 = x2 + ((re+r)/2);
y3 = y2 ;
re = ((re+r)/2);
else:
x3 = x2 + ((((re+r)/2)/re) * (x1-x2));
y3 = y2 + ((((re+r)/2)/re) * (y1-y2));
re = ((re+r)/2);
print(x3,y3,re)
```
| 6,320 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.
The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.
Input
The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).
Output
Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.
Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.
Examples
Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095
Input
10 5 5 5 15
Output
5.0 5.0 10.0
Tags: geometry
Correct Solution:
```
r,x1,y1,x2,y2 = map(int,input().split())
ttt = 0
import math
if(pow(x2-x1,2)+pow(y2-y1,2) >= r*r):
rad = r;
print(x1,' ',y1,' ',rad)
else:
rad=(r + math.sqrt(pow(x2-x1,2)+pow(y2-y1,2)))/2
dist= (math.sqrt(pow(x2-x1,2)+pow(y2-y1,2)))
m = rad - dist
if dist!=0:
x = ((m+dist)*x1 -(m)*x2)/dist
y = ((m+dist)*y1 -(m)*y2)/dist
else:
x = x1+r/2
y = y1
rad = 0.5*r
print(x,' ',y,' ',rad)
```
| 6,321 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.
The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.
Input
The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).
Output
Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.
Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.
Examples
Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095
Input
10 5 5 5 15
Output
5.0 5.0 10.0
Tags: geometry
Correct Solution:
```
import math
par = input()
# par = '10 0 0 0 0'
def dist(x1, y1, x2, y2):
return math.sqrt((x1-x2)** 2 + (y1-y2)** 2)
def setap(R, x1, y1, x2, y2):
d = dist(x1, y1, x2, y2)
if d < R and d > 0:
r = ( d + R ) / 2
x = x2 - r / d * (x2 - x1)
y = y2 - r / d * (y2 - y1)
elif d == 0:
x = x2 + R / 2
y = y2
r = R / 2
else:
r = R
x = x1
y = y1
return [x,y,r]
par = list(map(int, par.split()))
R = par[0]
x1 = par[1]
y1 = par[2]
x2 = par[3]
y2 = par[4]
res = setap(R, x1, y1, x2, y2)
print(res[0],res[1], res[2])
```
| 6,322 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.
The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.
Input
The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).
Output
Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.
Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.
Examples
Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095
Input
10 5 5 5 15
Output
5.0 5.0 10.0
Tags: geometry
Correct Solution:
```
# Reads two numbers from input and typecasts them to int using
# list comprehension
import math
R, x0, y0, x1, y1 = [int(x) for x in input().split()]
d1=math.sqrt((x1-x0)*(x1-x0)+(y1-y0)*(y1-y0))
f=0
if((x1-x0)*(x1-x0)+(y1-y0)*(y1-y0)>R*R):
f=1
r=(d1+R)/2;
if(x1!=x0):
ten=(y1-y0)/(x1-x0)
if(ten>0):
if(y1>y0):
sn=math.sin(math.atan(ten))
ansy=y1-(sn*r)
cn=math.cos(math.atan(ten))
ansx=x1-(cn*r)
else:
sn=math.sin(math.atan(ten))
ansy=y1+(sn*r)
cn=math.cos(math.atan(ten))
ansx=x1+(cn*r)
else:
ten=abs(ten)
if(y1>y0):
sn=math.sin(math.atan(ten))
ansy=y1-(sn*r)
cn=math.cos(math.atan(ten))
ansx=x1+(cn*r)
else:
sn=math.sin(math.atan(ten))
ansy=y1+(sn*r)
cn=math.cos(math.atan(ten))
ansx=x1-(cn*r)
else:
ansx=x1
if(y1>y0):
ansy=y1-r
else:
ansy=y1+r
if(f==0):
print(ansx, ansy, r)
else:
print(x0, y0, R)
```
| 6,323 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.
The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.
Input
The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).
Output
Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.
Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.
Examples
Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095
Input
10 5 5 5 15
Output
5.0 5.0 10.0
Tags: geometry
Correct Solution:
```
R,x1,y1,x2,y2=map(int,input().split(' '))
d1=(x2-x1)**2+(y2-y1)**2
if d1>=R*R:
print(str(x1)+" "+str(y1)+" "+str(R))
else:
d2=R+d1**(0.5)
if x1==x2 and y1==y2:
print(str(x1+R/2)+" "+str(y1)+" "+str(d2/2))
elif x1==x2:
if y2>y1:
print(str(x1)+" "+str((y1+y2-R)/2)+" "+str(d2/2))
else:
print(str(x1)+" "+str((y1+y2+R)/2)+" "+str(d2/2))
elif y1==y2:
if x2>x1:
print(str((x1+x2-R)/2)+" "+str(y1)+" "+str(d2/2))
else:
print(str((x1+x2+R)/2)+" "+str(y1)+" "+str(d2/2))
else:
x2-=x1
y2-=y1
slp=y2/x2
a=(R*R/(slp*slp+1))**(0.5)
b=slp*a
if x2<0 and y2<0:
print(str((a+x2)/2+x1)+" "+str((b+y2)/2+y1)+" "+str(d2/2))
elif x2<0 and y2>0:
print(str((a+x2)/2+x1)+" "+str((y2+b)/2+y1)+" "+str(d2/2))
elif x2>0 and y2<0:
print(str((x2-a)/2+x1)+" "+str((y2-b)/2+y1)+" "+str(d2/2))
else:
print(str((x2-a)/2+x1)+" "+str((y2-b)/2+y1)+" "+str(d2/2))
```
| 6,324 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.
The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.
Input
The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).
Output
Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.
Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.
Examples
Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095
Input
10 5 5 5 15
Output
5.0 5.0 10.0
Submitted Solution:
```
R, a, b, c, d = map(int, input().split())
da = ((a-c)**2 + (b-d)**2)**0.5
if da >= R:
print(a, b, R)
elif da == 0:
print(a, b+R/2, R/2)
else:
x, y = a - R*(c-a)/da, b-R*(d-b)/da
print((x+c)/2, (y+d)/2, (da+R)/2)
```
Yes
| 6,325 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.
The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.
Input
The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).
Output
Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.
Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.
Examples
Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095
Input
10 5 5 5 15
Output
5.0 5.0 10.0
Submitted Solution:
```
import math
def main():
r1, x1, y1, x2, y2 = (int(x) for x in input().split())
if (x2 - x1) ** 2 + (y2 - y1) ** 2 >= r1 ** 2:
print(x1, y1, r1)
return
v1 = (x1 - x2, y1 - y2)
len_v1 = math.sqrt(v1[0] ** 2 + v1[1] ** 2)
r = (r1 + len_v1) / 2
if (x1, y1) == (x2, y2):
print(v1[0] + x2, v1[1] + r + y2, r)
return
k = (r1 + len_v1) / 2 / len_v1
v_res = (v1[0] * k + x2, v1[1] * k + y2)
print(v_res[0], v_res[1], r)
main()
```
Yes
| 6,326 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.
The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.
Input
The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).
Output
Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.
Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.
Examples
Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095
Input
10 5 5 5 15
Output
5.0 5.0 10.0
Submitted Solution:
```
R, x1, y1, x2, y2 = map(int, input().split())
dx = x2 - x1
dy = y2 - y1
dlen = (dx**2 + dy**2)**0.5
if dlen < R:
if dlen == 0.0:
ax = x1 - R
ay = y1
else:
ax = x1 - dx * R/dlen
ay = y1 - dy * R/dlen
x = (ax + x2) / 2
y = (ay + y2) / 2
r = (R + dlen) / 2
else:
x = x1
y = y1
r = R
print("{:.15f} {:.15f} {:.15f}".format(x, y, r))
```
Yes
| 6,327 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.
The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.
Input
The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).
Output
Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.
Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.
Examples
Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095
Input
10 5 5 5 15
Output
5.0 5.0 10.0
Submitted Solution:
```
class Point:
def __init__(self, x, y):
self.x = x
self.y = y
def vector_by_points(pt1, pt2):
return Vector(pt2.x - pt1.x, pt2.y - pt1.y)
class Vector:
def __init__(self, x, y):
self.x = x
self.y = y
def length(self):
return (self.x ** 2 + self.y ** 2) ** 0.5
def __add__(self, other):
return Vector(self.x + other.x, self.y + other.y)
def __mul__(self, other):
return Vector(self.x * other, self.y * other)
def normalise(self):
l = self.length()
if l != 0:
return Vector(self.x / l, self.y / l)
return self
r, x1, y1, x2, y2 = map(int, input().split())
center = Point(x1, y1)
notebook = Point(x2, y2)
v = vector_by_points(notebook, center)
dst = v.length()
if dst - r <= 0.000001:
ans = (2 * r - (r - dst)) / 2
if dst == 0:
print(x1 + r / 2, y1, ans)
else:
v = v.normalise() * ans + notebook
print(v.x, v.y, ans)
else:
print(x1, y1, r)
```
Yes
| 6,328 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.
The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.
Input
The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).
Output
Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.
Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.
Examples
Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095
Input
10 5 5 5 15
Output
5.0 5.0 10.0
Submitted Solution:
```
R,x1,y1,x2,y2=input().split()
R=float(R)
x1=float(x1)
y1=float(y1)
x2=float(x2)
y2=float(y2)
d=((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2))**0.5
k1=d;
k2=(R-d)/2;
if d==0:
print(x1,y1,0)
elif d<=R:
print((x1*(k1+k2)-k2*x2)/k1,(y1*(k1+k2)-k2*y2)/k1,(R+d)/2)
else:
print(x1,y1,R)
```
No
| 6,329 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.
The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.
Input
The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).
Output
Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.
Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.
Examples
Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095
Input
10 5 5 5 15
Output
5.0 5.0 10.0
Submitted Solution:
```
import math
def dist(x1,y1,x2,y2):
return math.sqrt(distsqr(x1,y1,x2,y2))
def distsqr(x1,y1,x2,y2):
return ((x1-x2)**2 + (y1-y2)**2)
def f(x):
return math.sqrt(resr**2 - (x-x2)**2) + y2
"""
def bs(l, r):
t = 0
while(abs(l-r) > eps and t < 100):
mid = (l+r)/2
if(abs(dist(mid, f(mid), x2, y2)-resr) > eps):
l = mid
else:
r = mid
return (l+r)/2
"""
if __name__ == '__main__':
r, x1, y1, x2, y2 = map(int, input().split())
# print(r, x1, y1, x2, y2)
eps = 1e-16
d = distsqr(x1, y1, x2, y2)
if(d >= r*r):
print(x1, y1, r)
else:
resr = (math.sqrt(d)+r)/2
if y2 > y1:
yr = y2 - resr
else:
yr = y2 + resr
if x2 > x1:
xr = x2 - resr
else:
xr = x2 + resr
print(xr, yr, resr)
#print(dist(xr,yr,x1,y1), r)
```
No
| 6,330 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.
The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.
Input
The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).
Output
Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.
Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.
Examples
Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095
Input
10 5 5 5 15
Output
5.0 5.0 10.0
Submitted Solution:
```
from math import sqrt
R, x1, y1, x2, y2 = map(int, input().split())
if (x2-x1)**2 + (y2-y1)**2 <= R*R:
mnR = sqrt((x2-x1)**2 + (y2-y1)**2)
r = (mnR + R)/2
if x1 == x2 and y1 == y2:
x, y = float(x2+((x1+R)-x1)/2), float(y2+((y1+R)-y1)/2)
else:
x = (x1-x2)*r/mnR + x2
y = (y1-y2)*r/mnR + y2
print(x, y, r)
else:
print("%.1f %.1f %.1f" % (x1, y1, R))
```
No
| 6,331 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.
The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.
Input
The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).
Output
Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.
Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.
Examples
Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095
Input
10 5 5 5 15
Output
5.0 5.0 10.0
Submitted Solution:
```
import math
from math import sin,cos
from decimal import Decimal
R,x1,y1,x2,y2=map(int,input().split())
if(((x2-x1)**2+(y2-y1)**2)**0.5>=R):
print(x1,y1,R)
exit(0)
if(x2==x1):
theta=22/14
else:
theta=math.atan(abs((y2-y1)/(x2-x1)))
r=(((x2-x1)**2+(y2-y1)**2)**0.5+R)/2
if(x1==x2 and y1==y2):
print(x1,y1,"0")
exit(0)
if(x2<=x1 and y2<=y1):
x=(x1+R*cos(theta)+x2)/2.0
y=(y1+R*sin(theta)+y2)/2.0
elif(x2>=x1 and y2>=y1):
x=(x1-R*cos(theta)+x2)/2.0
y=(y1-R*sin(theta)+y2)/2.0
elif(x2<=x1 and y2>=y1):
x=(x1+R*cos(theta)+x2)/2.0
y=(y1-R*sin(theta)+y2)/2.0
elif(x2>=x1 and y2<=y1):
x=(x1-R*cos(theta)+x2)/2.0
y=(y1+R*sin(theta)+y2)/2.0
print(Decimal(x),Decimal(y),Decimal(r))
```
No
| 6,332 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your friend Mishka and you attend a calculus lecture. Lecture lasts n minutes. Lecturer tells ai theorems during the i-th minute.
Mishka is really interested in calculus, though it is so hard to stay awake for all the time of lecture. You are given an array t of Mishka's behavior. If Mishka is asleep during the i-th minute of the lecture then ti will be equal to 0, otherwise it will be equal to 1. When Mishka is awake he writes down all the theorems he is being told — ai during the i-th minute. Otherwise he writes nothing.
You know some secret technique to keep Mishka awake for k minutes straight. However you can use it only once. You can start using it at the beginning of any minute between 1 and n - k + 1. If you use it on some minute i then Mishka will be awake during minutes j such that <image> and will write down all the theorems lecturer tells.
You task is to calculate the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Input
The first line of the input contains two integer numbers n and k (1 ≤ k ≤ n ≤ 105) — the duration of the lecture in minutes and the number of minutes you can keep Mishka awake.
The second line of the input contains n integer numbers a1, a2, ... an (1 ≤ ai ≤ 104) — the number of theorems lecturer tells during the i-th minute.
The third line of the input contains n integer numbers t1, t2, ... tn (0 ≤ ti ≤ 1) — type of Mishka's behavior at the i-th minute of the lecture.
Output
Print only one integer — the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Example
Input
6 3
1 3 5 2 5 4
1 1 0 1 0 0
Output
16
Note
In the sample case the better way is to use the secret technique at the beginning of the third minute. Then the number of theorems Mishka will be able to write down will be equal to 16.
Tags: data structures, dp, implementation, two pointers
Correct Solution:
```
n, k = [int(x) for x in input().split(' ')]
a = [int(x) for x in input().split(' ')]
t = [int(x) for x in input().split(' ')]
g = sum([a[x] for x in range(n) if t[x] == 1])
max_app_g = 0
app_g = 0
for i in range(n):
if t[i] == 0:
app_g += a[i]
if i - k >= 0 and t[i - k] == 0:
app_g -= a[i - k]
if app_g > max_app_g:
max_app_g = app_g
print(max_app_g + g)
```
| 6,333 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your friend Mishka and you attend a calculus lecture. Lecture lasts n minutes. Lecturer tells ai theorems during the i-th minute.
Mishka is really interested in calculus, though it is so hard to stay awake for all the time of lecture. You are given an array t of Mishka's behavior. If Mishka is asleep during the i-th minute of the lecture then ti will be equal to 0, otherwise it will be equal to 1. When Mishka is awake he writes down all the theorems he is being told — ai during the i-th minute. Otherwise he writes nothing.
You know some secret technique to keep Mishka awake for k minutes straight. However you can use it only once. You can start using it at the beginning of any minute between 1 and n - k + 1. If you use it on some minute i then Mishka will be awake during minutes j such that <image> and will write down all the theorems lecturer tells.
You task is to calculate the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Input
The first line of the input contains two integer numbers n and k (1 ≤ k ≤ n ≤ 105) — the duration of the lecture in minutes and the number of minutes you can keep Mishka awake.
The second line of the input contains n integer numbers a1, a2, ... an (1 ≤ ai ≤ 104) — the number of theorems lecturer tells during the i-th minute.
The third line of the input contains n integer numbers t1, t2, ... tn (0 ≤ ti ≤ 1) — type of Mishka's behavior at the i-th minute of the lecture.
Output
Print only one integer — the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Example
Input
6 3
1 3 5 2 5 4
1 1 0 1 0 0
Output
16
Note
In the sample case the better way is to use the secret technique at the beginning of the third minute. Then the number of theorems Mishka will be able to write down will be equal to 16.
Tags: data structures, dp, implementation, two pointers
Correct Solution:
```
def main():
n, k = map(int, input().split())
a = [int(c) for c in input().split()]
t = [int(c) for c in input().split()]
diff = 0
i = 0
s = 0
ii = jj = 0
while i < n:
ii += a[i]
if i - k >= 0:
ii -= a[i - k]
if t[i] == 1:
s += a[i]
jj += a[i]
if i - k >= 0 and t[i - k] == 1:
jj -= a[i - k]
diff = max(diff, ii - jj)
i += 1
print(s + diff)
if __name__ == '__main__':
main()
```
| 6,334 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your friend Mishka and you attend a calculus lecture. Lecture lasts n minutes. Lecturer tells ai theorems during the i-th minute.
Mishka is really interested in calculus, though it is so hard to stay awake for all the time of lecture. You are given an array t of Mishka's behavior. If Mishka is asleep during the i-th minute of the lecture then ti will be equal to 0, otherwise it will be equal to 1. When Mishka is awake he writes down all the theorems he is being told — ai during the i-th minute. Otherwise he writes nothing.
You know some secret technique to keep Mishka awake for k minutes straight. However you can use it only once. You can start using it at the beginning of any minute between 1 and n - k + 1. If you use it on some minute i then Mishka will be awake during minutes j such that <image> and will write down all the theorems lecturer tells.
You task is to calculate the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Input
The first line of the input contains two integer numbers n and k (1 ≤ k ≤ n ≤ 105) — the duration of the lecture in minutes and the number of minutes you can keep Mishka awake.
The second line of the input contains n integer numbers a1, a2, ... an (1 ≤ ai ≤ 104) — the number of theorems lecturer tells during the i-th minute.
The third line of the input contains n integer numbers t1, t2, ... tn (0 ≤ ti ≤ 1) — type of Mishka's behavior at the i-th minute of the lecture.
Output
Print only one integer — the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Example
Input
6 3
1 3 5 2 5 4
1 1 0 1 0 0
Output
16
Note
In the sample case the better way is to use the secret technique at the beginning of the third minute. Then the number of theorems Mishka will be able to write down will be equal to 16.
Tags: data structures, dp, implementation, two pointers
Correct Solution:
```
n, k = [int(x) for x in input().split()]
a = list(map(int, input().split()))
t = list(map(int, input().split()))
m = 0
s = 0
temp = 0
for i in range(n):
m = max(m, temp)
if i >= k:
ch = i - k
if t[ch] == 0:
temp -= a[ch]
if t[i] == 0:
temp += a[i]
if t[i] == 1:
s += a[i]
m = max(m, temp)
print(s + m)
```
| 6,335 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your friend Mishka and you attend a calculus lecture. Lecture lasts n minutes. Lecturer tells ai theorems during the i-th minute.
Mishka is really interested in calculus, though it is so hard to stay awake for all the time of lecture. You are given an array t of Mishka's behavior. If Mishka is asleep during the i-th minute of the lecture then ti will be equal to 0, otherwise it will be equal to 1. When Mishka is awake he writes down all the theorems he is being told — ai during the i-th minute. Otherwise he writes nothing.
You know some secret technique to keep Mishka awake for k minutes straight. However you can use it only once. You can start using it at the beginning of any minute between 1 and n - k + 1. If you use it on some minute i then Mishka will be awake during minutes j such that <image> and will write down all the theorems lecturer tells.
You task is to calculate the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Input
The first line of the input contains two integer numbers n and k (1 ≤ k ≤ n ≤ 105) — the duration of the lecture in minutes and the number of minutes you can keep Mishka awake.
The second line of the input contains n integer numbers a1, a2, ... an (1 ≤ ai ≤ 104) — the number of theorems lecturer tells during the i-th minute.
The third line of the input contains n integer numbers t1, t2, ... tn (0 ≤ ti ≤ 1) — type of Mishka's behavior at the i-th minute of the lecture.
Output
Print only one integer — the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Example
Input
6 3
1 3 5 2 5 4
1 1 0 1 0 0
Output
16
Note
In the sample case the better way is to use the secret technique at the beginning of the third minute. Then the number of theorems Mishka will be able to write down will be equal to 16.
Tags: data structures, dp, implementation, two pointers
Correct Solution:
```
a = input()
b = input()
c = input()
a = a.split(' ')
b = b.split(' ')
c = c.split(' ')
for i in range(len(a)):
a[i] = int(a[i])
for i in range(len(b)):
b[i] = int(b[i])
for i in range(len(c)):
c[i] = int(c[i])
def blower():
d = []
s = 0
x = 0
for i in range(a[1]):
if c[i] == 0:
s += b[i]
d += [s]
for i in range(a[1],a[0]):
if c[i-a[1]] == 0:
s -= b[i-a[1]]
if c[i] == 0:
s += b[i]
d += [s]
for i in range(a[0]):
if c[i] == 1:
x += b[i]
print(x + max(d))
blower()
```
| 6,336 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your friend Mishka and you attend a calculus lecture. Lecture lasts n minutes. Lecturer tells ai theorems during the i-th minute.
Mishka is really interested in calculus, though it is so hard to stay awake for all the time of lecture. You are given an array t of Mishka's behavior. If Mishka is asleep during the i-th minute of the lecture then ti will be equal to 0, otherwise it will be equal to 1. When Mishka is awake he writes down all the theorems he is being told — ai during the i-th minute. Otherwise he writes nothing.
You know some secret technique to keep Mishka awake for k minutes straight. However you can use it only once. You can start using it at the beginning of any minute between 1 and n - k + 1. If you use it on some minute i then Mishka will be awake during minutes j such that <image> and will write down all the theorems lecturer tells.
You task is to calculate the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Input
The first line of the input contains two integer numbers n and k (1 ≤ k ≤ n ≤ 105) — the duration of the lecture in minutes and the number of minutes you can keep Mishka awake.
The second line of the input contains n integer numbers a1, a2, ... an (1 ≤ ai ≤ 104) — the number of theorems lecturer tells during the i-th minute.
The third line of the input contains n integer numbers t1, t2, ... tn (0 ≤ ti ≤ 1) — type of Mishka's behavior at the i-th minute of the lecture.
Output
Print only one integer — the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Example
Input
6 3
1 3 5 2 5 4
1 1 0 1 0 0
Output
16
Note
In the sample case the better way is to use the secret technique at the beginning of the third minute. Then the number of theorems Mishka will be able to write down will be equal to 16.
Tags: data structures, dp, implementation, two pointers
Correct Solution:
```
(n, k) = map(int, input().split())
lst = []
for x in input().split():
lst.append(int(x))
array = []
for x in input().split():
array.append(int(x))
mass1 = [0]
for x in range(n):
mass1.append(mass1[-1] + lst[x])
mass2 = [0]
for x in range(n):
mass2.append(mass2[-1] + lst[x] * array[x])
ans = 0
for x in range(1, n - k + 2):
ans = max(ans, mass2[x - 1] + (mass1[x + k - 1] - mass1[x - 1]) + (mass2[-1] - mass2[x + k - 1]))
#print(ans, x)
print(ans)
```
| 6,337 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your friend Mishka and you attend a calculus lecture. Lecture lasts n minutes. Lecturer tells ai theorems during the i-th minute.
Mishka is really interested in calculus, though it is so hard to stay awake for all the time of lecture. You are given an array t of Mishka's behavior. If Mishka is asleep during the i-th minute of the lecture then ti will be equal to 0, otherwise it will be equal to 1. When Mishka is awake he writes down all the theorems he is being told — ai during the i-th minute. Otherwise he writes nothing.
You know some secret technique to keep Mishka awake for k minutes straight. However you can use it only once. You can start using it at the beginning of any minute between 1 and n - k + 1. If you use it on some minute i then Mishka will be awake during minutes j such that <image> and will write down all the theorems lecturer tells.
You task is to calculate the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Input
The first line of the input contains two integer numbers n and k (1 ≤ k ≤ n ≤ 105) — the duration of the lecture in minutes and the number of minutes you can keep Mishka awake.
The second line of the input contains n integer numbers a1, a2, ... an (1 ≤ ai ≤ 104) — the number of theorems lecturer tells during the i-th minute.
The third line of the input contains n integer numbers t1, t2, ... tn (0 ≤ ti ≤ 1) — type of Mishka's behavior at the i-th minute of the lecture.
Output
Print only one integer — the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Example
Input
6 3
1 3 5 2 5 4
1 1 0 1 0 0
Output
16
Note
In the sample case the better way is to use the secret technique at the beginning of the third minute. Then the number of theorems Mishka will be able to write down will be equal to 16.
Tags: data structures, dp, implementation, two pointers
Correct Solution:
```
n,k = map(int, input().split())
theorems = [int(i) for i in input().split()]
wake = [int(i) for i in input().split()]
acu = 0
for i in range(n):
if wake[i] == 1:
acu += theorems[i]
elif i < k:
acu += theorems[i]
ans = acu
j = k
ans = acu
for i in range(n-k):
if wake[i] == 0:
acu -= theorems[i]
if wake[j] == 0:
acu += theorems[j]
ans = max(ans,acu)
j += 1
print(ans)
```
| 6,338 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your friend Mishka and you attend a calculus lecture. Lecture lasts n minutes. Lecturer tells ai theorems during the i-th minute.
Mishka is really interested in calculus, though it is so hard to stay awake for all the time of lecture. You are given an array t of Mishka's behavior. If Mishka is asleep during the i-th minute of the lecture then ti will be equal to 0, otherwise it will be equal to 1. When Mishka is awake he writes down all the theorems he is being told — ai during the i-th minute. Otherwise he writes nothing.
You know some secret technique to keep Mishka awake for k minutes straight. However you can use it only once. You can start using it at the beginning of any minute between 1 and n - k + 1. If you use it on some minute i then Mishka will be awake during minutes j such that <image> and will write down all the theorems lecturer tells.
You task is to calculate the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Input
The first line of the input contains two integer numbers n and k (1 ≤ k ≤ n ≤ 105) — the duration of the lecture in minutes and the number of minutes you can keep Mishka awake.
The second line of the input contains n integer numbers a1, a2, ... an (1 ≤ ai ≤ 104) — the number of theorems lecturer tells during the i-th minute.
The third line of the input contains n integer numbers t1, t2, ... tn (0 ≤ ti ≤ 1) — type of Mishka's behavior at the i-th minute of the lecture.
Output
Print only one integer — the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Example
Input
6 3
1 3 5 2 5 4
1 1 0 1 0 0
Output
16
Note
In the sample case the better way is to use the secret technique at the beginning of the third minute. Then the number of theorems Mishka will be able to write down will be equal to 16.
Tags: data structures, dp, implementation, two pointers
Correct Solution:
```
#pyrival orz
import os
import sys
from io import BytesIO, IOBase
input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
def main():
try:
n, k = invr()
a = inlt()
b = inlt()
extra = 0
for i in range(k):
if not b[i]:
extra += a[i]
ans = 0
temp = 0
for i in range(n):
if b[i]:
ans += a[i]
if not b[i]:
temp += a[i]
if i > k - 1 and not b[i - k]:
temp -= a[i - k]
extra = max(extra, temp)
print(ans + extra)
except Exception as e:
print(e)
# 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()
```
| 6,339 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your friend Mishka and you attend a calculus lecture. Lecture lasts n minutes. Lecturer tells ai theorems during the i-th minute.
Mishka is really interested in calculus, though it is so hard to stay awake for all the time of lecture. You are given an array t of Mishka's behavior. If Mishka is asleep during the i-th minute of the lecture then ti will be equal to 0, otherwise it will be equal to 1. When Mishka is awake he writes down all the theorems he is being told — ai during the i-th minute. Otherwise he writes nothing.
You know some secret technique to keep Mishka awake for k minutes straight. However you can use it only once. You can start using it at the beginning of any minute between 1 and n - k + 1. If you use it on some minute i then Mishka will be awake during minutes j such that <image> and will write down all the theorems lecturer tells.
You task is to calculate the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Input
The first line of the input contains two integer numbers n and k (1 ≤ k ≤ n ≤ 105) — the duration of the lecture in minutes and the number of minutes you can keep Mishka awake.
The second line of the input contains n integer numbers a1, a2, ... an (1 ≤ ai ≤ 104) — the number of theorems lecturer tells during the i-th minute.
The third line of the input contains n integer numbers t1, t2, ... tn (0 ≤ ti ≤ 1) — type of Mishka's behavior at the i-th minute of the lecture.
Output
Print only one integer — the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Example
Input
6 3
1 3 5 2 5 4
1 1 0 1 0 0
Output
16
Note
In the sample case the better way is to use the secret technique at the beginning of the third minute. Then the number of theorems Mishka will be able to write down will be equal to 16.
Tags: data structures, dp, implementation, two pointers
Correct Solution:
```
import sys
input = sys.stdin.readline
import math
import copy
import collections
from collections import deque
import heapq
import itertools
from collections import defaultdict
from collections import Counter
n,k = map(int,input().split())
a = list(map(int,input().split()))
t = list(map(int,input().split()))
pre = [0]
for i in range(n):
if t[i]==0:
pre.append(pre[i]+a[i])
else:
pre.append(pre[i])
mx = 0
for i in range(n-k+1):
l = i
r = i+k
ans = pre[r]-pre[l]
mx = max(mx,ans)
ans = 0
for i in range(n):
if(t[i]==1):
ans+=a[i]
print(ans+mx)
```
| 6,340 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your friend Mishka and you attend a calculus lecture. Lecture lasts n minutes. Lecturer tells ai theorems during the i-th minute.
Mishka is really interested in calculus, though it is so hard to stay awake for all the time of lecture. You are given an array t of Mishka's behavior. If Mishka is asleep during the i-th minute of the lecture then ti will be equal to 0, otherwise it will be equal to 1. When Mishka is awake he writes down all the theorems he is being told — ai during the i-th minute. Otherwise he writes nothing.
You know some secret technique to keep Mishka awake for k minutes straight. However you can use it only once. You can start using it at the beginning of any minute between 1 and n - k + 1. If you use it on some minute i then Mishka will be awake during minutes j such that <image> and will write down all the theorems lecturer tells.
You task is to calculate the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Input
The first line of the input contains two integer numbers n and k (1 ≤ k ≤ n ≤ 105) — the duration of the lecture in minutes and the number of minutes you can keep Mishka awake.
The second line of the input contains n integer numbers a1, a2, ... an (1 ≤ ai ≤ 104) — the number of theorems lecturer tells during the i-th minute.
The third line of the input contains n integer numbers t1, t2, ... tn (0 ≤ ti ≤ 1) — type of Mishka's behavior at the i-th minute of the lecture.
Output
Print only one integer — the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Example
Input
6 3
1 3 5 2 5 4
1 1 0 1 0 0
Output
16
Note
In the sample case the better way is to use the secret technique at the beginning of the third minute. Then the number of theorems Mishka will be able to write down will be equal to 16.
Submitted Solution:
```
import sys
input=sys.stdin.readline
n,k=map(int,input().split())
theorems=list(map(int,input().split()))
sleep=list(map(int,input().split()))
tsum=[]
ts=0
sleepsum=[]
slsum=0
for i in range(n):
ts+=theorems[i]
tsum.append(ts)
if(sleep[i]==1):
slsum+=theorems[i]
sleepsum.append(slsum)
#print("tsum=",tsum)
#print("sleepsum=",sleepsum)
maxdiff=0
#print("slsum=",slsum)
maxdiff=tsum[k-1]-sleepsum[k-1]
for i in range(1,n-k+1):
diff=(tsum[i+k-1]-tsum[i-1])-(sleepsum[i+k-1]-sleepsum[i-1])
#print("i=",i,"diff=",diff)
maxdiff=max(maxdiff,diff)
#print("maxdiff=",maxdiff)
print(slsum+maxdiff)
```
Yes
| 6,341 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your friend Mishka and you attend a calculus lecture. Lecture lasts n minutes. Lecturer tells ai theorems during the i-th minute.
Mishka is really interested in calculus, though it is so hard to stay awake for all the time of lecture. You are given an array t of Mishka's behavior. If Mishka is asleep during the i-th minute of the lecture then ti will be equal to 0, otherwise it will be equal to 1. When Mishka is awake he writes down all the theorems he is being told — ai during the i-th minute. Otherwise he writes nothing.
You know some secret technique to keep Mishka awake for k minutes straight. However you can use it only once. You can start using it at the beginning of any minute between 1 and n - k + 1. If you use it on some minute i then Mishka will be awake during minutes j such that <image> and will write down all the theorems lecturer tells.
You task is to calculate the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Input
The first line of the input contains two integer numbers n and k (1 ≤ k ≤ n ≤ 105) — the duration of the lecture in minutes and the number of minutes you can keep Mishka awake.
The second line of the input contains n integer numbers a1, a2, ... an (1 ≤ ai ≤ 104) — the number of theorems lecturer tells during the i-th minute.
The third line of the input contains n integer numbers t1, t2, ... tn (0 ≤ ti ≤ 1) — type of Mishka's behavior at the i-th minute of the lecture.
Output
Print only one integer — the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Example
Input
6 3
1 3 5 2 5 4
1 1 0 1 0 0
Output
16
Note
In the sample case the better way is to use the secret technique at the beginning of the third minute. Then the number of theorems Mishka will be able to write down will be equal to 16.
Submitted Solution:
```
from sys import stdin as sin
n,k = map(int, sin.readline().split())
a = list(map(int, sin.readline().split(" ")))
x = list(map(int, sin.readline().split(" ")))
te=[]+a
te[0]=te[0]*x[0]
re=[]+a
re[-1]*=x[-1]
for i in range(n-2,-1,-1):
re[i]=re[i+1]+re[i]*x[i]
te[n-i-1]=te[n-i-2]+te[n-i-1]*x[n-i-1]
# print(te,re)
s = sum(a[0:k])
if n==k:
print(s)
else:
m=max(s+re[k],sum(a[n-k:n])+te[n-k-1])
# print(m)
for i in range(1,n-k):
s+=(a[i+k-1]-a[i-1])
m=max(m,te[i-1]+s+re[i+k])
# print(m)
# print(te,re)
print(m)
# 6 3
# 1 3 5 2 5 4
# 1 1 0 1 0 0
# 5 3
# 1 9999 10000 10000 10000
# 0 0 0 0 0
```
Yes
| 6,342 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your friend Mishka and you attend a calculus lecture. Lecture lasts n minutes. Lecturer tells ai theorems during the i-th minute.
Mishka is really interested in calculus, though it is so hard to stay awake for all the time of lecture. You are given an array t of Mishka's behavior. If Mishka is asleep during the i-th minute of the lecture then ti will be equal to 0, otherwise it will be equal to 1. When Mishka is awake he writes down all the theorems he is being told — ai during the i-th minute. Otherwise he writes nothing.
You know some secret technique to keep Mishka awake for k minutes straight. However you can use it only once. You can start using it at the beginning of any minute between 1 and n - k + 1. If you use it on some minute i then Mishka will be awake during minutes j such that <image> and will write down all the theorems lecturer tells.
You task is to calculate the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Input
The first line of the input contains two integer numbers n and k (1 ≤ k ≤ n ≤ 105) — the duration of the lecture in minutes and the number of minutes you can keep Mishka awake.
The second line of the input contains n integer numbers a1, a2, ... an (1 ≤ ai ≤ 104) — the number of theorems lecturer tells during the i-th minute.
The third line of the input contains n integer numbers t1, t2, ... tn (0 ≤ ti ≤ 1) — type of Mishka's behavior at the i-th minute of the lecture.
Output
Print only one integer — the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Example
Input
6 3
1 3 5 2 5 4
1 1 0 1 0 0
Output
16
Note
In the sample case the better way is to use the secret technique at the beginning of the third minute. Then the number of theorems Mishka will be able to write down will be equal to 16.
Submitted Solution:
```
"""
def solve(n,k,a,t):
#prefix sum
pref = [0] * (n+1)
pref[1] = a[1]
for i in range(2,n+1):
pref[i] = pref[i-1] + a[i]
#dp initialization
dpc = [0] * (n+1)
for i in range(1,n+1):
if t[i] != 0:
dpc[i] = dpc[i-1] + a[i]
else:
dpc[i] = dpc[i-1]
ans = 0
#implementation
for i in range(1,n-k+2):
dp = dpc[:]
pos = i+k-1
dp[n] -= dp[pos]
dp[n] += dp[i-1] + pref[pos] - pref[i-1]
ans = max(ans,dp[n])
print(ans)
return
"""
def solve(n,k,a,t):
res = 0
for i in range(1,n+1):
if t[i] == 1:
res += a[i]
a[i] = 0
pref = [0] * (n+1)
for i in range(1,n+1):
pref[i] = pref[i-1] + a[i]
temp = 0
for i in range(n,k-1,-1):
temp = max(temp,pref[i]-pref[i-k])
print(temp + res)
if __name__ == '__main__':
n,k = map(int, input().split())
a = list(map(int, input().split()))
t = list(map(int, input().split()))
a.insert(0,0)
t.insert(0,0)
solve(n,k,a,t)
```
Yes
| 6,343 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your friend Mishka and you attend a calculus lecture. Lecture lasts n minutes. Lecturer tells ai theorems during the i-th minute.
Mishka is really interested in calculus, though it is so hard to stay awake for all the time of lecture. You are given an array t of Mishka's behavior. If Mishka is asleep during the i-th minute of the lecture then ti will be equal to 0, otherwise it will be equal to 1. When Mishka is awake he writes down all the theorems he is being told — ai during the i-th minute. Otherwise he writes nothing.
You know some secret technique to keep Mishka awake for k minutes straight. However you can use it only once. You can start using it at the beginning of any minute between 1 and n - k + 1. If you use it on some minute i then Mishka will be awake during minutes j such that <image> and will write down all the theorems lecturer tells.
You task is to calculate the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Input
The first line of the input contains two integer numbers n and k (1 ≤ k ≤ n ≤ 105) — the duration of the lecture in minutes and the number of minutes you can keep Mishka awake.
The second line of the input contains n integer numbers a1, a2, ... an (1 ≤ ai ≤ 104) — the number of theorems lecturer tells during the i-th minute.
The third line of the input contains n integer numbers t1, t2, ... tn (0 ≤ ti ≤ 1) — type of Mishka's behavior at the i-th minute of the lecture.
Output
Print only one integer — the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Example
Input
6 3
1 3 5 2 5 4
1 1 0 1 0 0
Output
16
Note
In the sample case the better way is to use the secret technique at the beginning of the third minute. Then the number of theorems Mishka will be able to write down will be equal to 16.
Submitted Solution:
```
n, k = map(int, input().split())
a = [int(i) for i in input().split()]
t = [int(i) for i in input().split()]
base = 0
for i in range(n):
base += a[i] * t[i]
prefix = [(1 - t[i]) * a[i] for i in range(n)]
for i in range(1, n):
prefix[i] += prefix[i - 1]
gain = prefix[k - 1]
for i in range(k, n):
gain = max(gain, prefix[i] - prefix[i - k])
print(base + gain)
```
Yes
| 6,344 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your friend Mishka and you attend a calculus lecture. Lecture lasts n minutes. Lecturer tells ai theorems during the i-th minute.
Mishka is really interested in calculus, though it is so hard to stay awake for all the time of lecture. You are given an array t of Mishka's behavior. If Mishka is asleep during the i-th minute of the lecture then ti will be equal to 0, otherwise it will be equal to 1. When Mishka is awake he writes down all the theorems he is being told — ai during the i-th minute. Otherwise he writes nothing.
You know some secret technique to keep Mishka awake for k minutes straight. However you can use it only once. You can start using it at the beginning of any minute between 1 and n - k + 1. If you use it on some minute i then Mishka will be awake during minutes j such that <image> and will write down all the theorems lecturer tells.
You task is to calculate the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Input
The first line of the input contains two integer numbers n and k (1 ≤ k ≤ n ≤ 105) — the duration of the lecture in minutes and the number of minutes you can keep Mishka awake.
The second line of the input contains n integer numbers a1, a2, ... an (1 ≤ ai ≤ 104) — the number of theorems lecturer tells during the i-th minute.
The third line of the input contains n integer numbers t1, t2, ... tn (0 ≤ ti ≤ 1) — type of Mishka's behavior at the i-th minute of the lecture.
Output
Print only one integer — the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Example
Input
6 3
1 3 5 2 5 4
1 1 0 1 0 0
Output
16
Note
In the sample case the better way is to use the secret technique at the beginning of the third minute. Then the number of theorems Mishka will be able to write down will be equal to 16.
Submitted Solution:
```
n, k = map(int, input().split())
s = list(map(int, input().split()))
t = list(map(int, input().split()))
s.append(0), t.append(0)
r = sum([s[i] for i in range(n) if t[i]])
m = sum(s[i] for i in range(k) if t[i] == 0)
x = m
for i in range(1, n-k+1):
m = m - (1==t[i-1])*s[i-1] + (0==t[i+k])*s[i+k]
x = max(x, m)
print(r+m)
```
No
| 6,345 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your friend Mishka and you attend a calculus lecture. Lecture lasts n minutes. Lecturer tells ai theorems during the i-th minute.
Mishka is really interested in calculus, though it is so hard to stay awake for all the time of lecture. You are given an array t of Mishka's behavior. If Mishka is asleep during the i-th minute of the lecture then ti will be equal to 0, otherwise it will be equal to 1. When Mishka is awake he writes down all the theorems he is being told — ai during the i-th minute. Otherwise he writes nothing.
You know some secret technique to keep Mishka awake for k minutes straight. However you can use it only once. You can start using it at the beginning of any minute between 1 and n - k + 1. If you use it on some minute i then Mishka will be awake during minutes j such that <image> and will write down all the theorems lecturer tells.
You task is to calculate the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Input
The first line of the input contains two integer numbers n and k (1 ≤ k ≤ n ≤ 105) — the duration of the lecture in minutes and the number of minutes you can keep Mishka awake.
The second line of the input contains n integer numbers a1, a2, ... an (1 ≤ ai ≤ 104) — the number of theorems lecturer tells during the i-th minute.
The third line of the input contains n integer numbers t1, t2, ... tn (0 ≤ ti ≤ 1) — type of Mishka's behavior at the i-th minute of the lecture.
Output
Print only one integer — the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Example
Input
6 3
1 3 5 2 5 4
1 1 0 1 0 0
Output
16
Note
In the sample case the better way is to use the secret technique at the beginning of the third minute. Then the number of theorems Mishka will be able to write down will be equal to 16.
Submitted Solution:
```
def proB(arr,arr2,k):
n=len(arr)
fro,end=[0]*n,[0]*n
if(arr2[0]==1):
fro[0]=arr[0]
if(arr2[n-1]==1):
end[0]=arr[n-1]
for i in range(1,len(arr)):
if(arr2[i]==1):
fro[i]=fro[i-1]+arr[i]
else:
fro[i]=fro[i-1]
if(arr2[n-1-i]==1):
end[n-1-i]=end[n-1-i+1]+arr[n-1-i]
else:
end[n-1-i]=end[n-1-i+1]
sumi=sum(arr[:k])
j=0
maxi=sumi+end[k]
for i in range(k,n):
sumi+=arr[i]
sumi-=arr[j]
j+=1
maxi=max(maxi,fro[j-1]+sumi+end[i])
return maxi
arr=list(map(int,input().split()))
n,k=arr
a=list(map(int,input().split()))
b=list(map(int,input().split()))
print(proB(a,b,k))
```
No
| 6,346 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your friend Mishka and you attend a calculus lecture. Lecture lasts n minutes. Lecturer tells ai theorems during the i-th minute.
Mishka is really interested in calculus, though it is so hard to stay awake for all the time of lecture. You are given an array t of Mishka's behavior. If Mishka is asleep during the i-th minute of the lecture then ti will be equal to 0, otherwise it will be equal to 1. When Mishka is awake he writes down all the theorems he is being told — ai during the i-th minute. Otherwise he writes nothing.
You know some secret technique to keep Mishka awake for k minutes straight. However you can use it only once. You can start using it at the beginning of any minute between 1 and n - k + 1. If you use it on some minute i then Mishka will be awake during minutes j such that <image> and will write down all the theorems lecturer tells.
You task is to calculate the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Input
The first line of the input contains two integer numbers n and k (1 ≤ k ≤ n ≤ 105) — the duration of the lecture in minutes and the number of minutes you can keep Mishka awake.
The second line of the input contains n integer numbers a1, a2, ... an (1 ≤ ai ≤ 104) — the number of theorems lecturer tells during the i-th minute.
The third line of the input contains n integer numbers t1, t2, ... tn (0 ≤ ti ≤ 1) — type of Mishka's behavior at the i-th minute of the lecture.
Output
Print only one integer — the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Example
Input
6 3
1 3 5 2 5 4
1 1 0 1 0 0
Output
16
Note
In the sample case the better way is to use the secret technique at the beginning of the third minute. Then the number of theorems Mishka will be able to write down will be equal to 16.
Submitted Solution:
```
import sys
import os
import math
import re
n,k = map(int,input().split())
data = list(map(int,input().split()))
wake = list(map(int,input().split()))
if (k == n):
print(sum(data))
exit(0)
res = 0
#indices = [i for i, x in enumerate(wake) if x == 0]
#bestSum = 0
#for index
if (not 0 in wake):
print(0)
exit(0)
start = wake.index(0)
bestInd = start
for i in range(start,start + k + 1):
if (i < len(data)):
res += data[i]
curr = res
for i in range(start+1,len(data)):
curr += data[i] - data[i-(k+1)]
if (i - k) >= 0 and data[i-k] == 0:
if (curr > res):
bestInd = i-k
res = max(res,curr)
for i in range(bestInd,bestInd+k+1):
if i < len(data):
wake[i] = 1
res2 = 0
for i in range(k+1):
if wake[i] != 0 and i < len(data):
res2 += data[i]
curr2 = res2
for i in range(k+1,len(data)):
if wake[i] != 0:
curr2 += data[i] - data[i-(k+1)]
res2 = max(res2,curr2)
else:
curr2 -= data[i-(k+1)]
print(max(res,res2))
```
No
| 6,347 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your friend Mishka and you attend a calculus lecture. Lecture lasts n minutes. Lecturer tells ai theorems during the i-th minute.
Mishka is really interested in calculus, though it is so hard to stay awake for all the time of lecture. You are given an array t of Mishka's behavior. If Mishka is asleep during the i-th minute of the lecture then ti will be equal to 0, otherwise it will be equal to 1. When Mishka is awake he writes down all the theorems he is being told — ai during the i-th minute. Otherwise he writes nothing.
You know some secret technique to keep Mishka awake for k minutes straight. However you can use it only once. You can start using it at the beginning of any minute between 1 and n - k + 1. If you use it on some minute i then Mishka will be awake during minutes j such that <image> and will write down all the theorems lecturer tells.
You task is to calculate the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Input
The first line of the input contains two integer numbers n and k (1 ≤ k ≤ n ≤ 105) — the duration of the lecture in minutes and the number of minutes you can keep Mishka awake.
The second line of the input contains n integer numbers a1, a2, ... an (1 ≤ ai ≤ 104) — the number of theorems lecturer tells during the i-th minute.
The third line of the input contains n integer numbers t1, t2, ... tn (0 ≤ ti ≤ 1) — type of Mishka's behavior at the i-th minute of the lecture.
Output
Print only one integer — the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Example
Input
6 3
1 3 5 2 5 4
1 1 0 1 0 0
Output
16
Note
In the sample case the better way is to use the secret technique at the beginning of the third minute. Then the number of theorems Mishka will be able to write down will be equal to 16.
Submitted Solution:
```
n,k = list(map(int, input().split()))
a = list(map(int, input().split()))
t = list(map(int, input().split()))
x = 0
summ = 0
maxx = 0
for i in range(n):
summ += a[i]*t[i]
for i in range(k):
if not t[i]:
x+=a[i]
for i in range(n-k):
x+=a[i+k]*(1-t[i+k])
x-=a[i]*(1-t[i])
maxx = max(x, maxx)
print(summ+maxx)
```
No
| 6,348 |
Provide tags and a correct Python 3 solution for this coding contest problem.
When the curtains are opened, a canvas unfolds outside. Kanno marvels at all the blonde colours along the riverside — not tangerines, but blossoms instead.
"What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order.
Input
The first and only line of input contains a non-empty string s consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only (\lvert s \rvert ≤ 100) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively.
Output
Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
.BAC.
Output
Yes
Input
AA..CB
Output
No
Note
In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell.
Tags: implementation, strings
Correct Solution:
```
s = list(input())
a = ['A', 'B', 'C']
m = set(a)
f = False
for i in range(1, len(s) - 1):
a = [s[i], s[i - 1], s[i + 1]]
if set(a) == m:
print('Yes')
f = True
break
if not f:
print('No')
```
| 6,349 |
Provide tags and a correct Python 3 solution for this coding contest problem.
When the curtains are opened, a canvas unfolds outside. Kanno marvels at all the blonde colours along the riverside — not tangerines, but blossoms instead.
"What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order.
Input
The first and only line of input contains a non-empty string s consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only (\lvert s \rvert ≤ 100) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively.
Output
Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
.BAC.
Output
Yes
Input
AA..CB
Output
No
Note
In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell.
Tags: implementation, strings
Correct Solution:
```
s = input()
if 'BAC' in s or 'CAB' in s or 'ABC' in s or 'CBA' in s or 'ACB' in s or 'BCA' in s:
print('Yes')
else:
print('No')
```
| 6,350 |
Provide tags and a correct Python 3 solution for this coding contest problem.
When the curtains are opened, a canvas unfolds outside. Kanno marvels at all the blonde colours along the riverside — not tangerines, but blossoms instead.
"What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order.
Input
The first and only line of input contains a non-empty string s consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only (\lvert s \rvert ≤ 100) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively.
Output
Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
.BAC.
Output
Yes
Input
AA..CB
Output
No
Note
In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell.
Tags: implementation, strings
Correct Solution:
```
ch=input()
#easy
if "ABC" in ch or "ACB" in ch or "BAC" in ch or "BCA" in ch or "CAB" in ch or "CBA" in ch:
print("Yes")
else:
print("No")
```
| 6,351 |
Provide tags and a correct Python 3 solution for this coding contest problem.
When the curtains are opened, a canvas unfolds outside. Kanno marvels at all the blonde colours along the riverside — not tangerines, but blossoms instead.
"What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order.
Input
The first and only line of input contains a non-empty string s consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only (\lvert s \rvert ≤ 100) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively.
Output
Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
.BAC.
Output
Yes
Input
AA..CB
Output
No
Note
In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell.
Tags: implementation, strings
Correct Solution:
```
s=input()
flag=0
for i in range(1,len(s)-1):
if(s[i]=='.'):
continue
elif (s[i]=='A' and ((s[i-1]=='B'and s[i+1]=='C') or (s[i-1]=='C' and s[i+1]=='B'))):
flag=1
break
elif (s[i]=='B' and((s[i-1]=='A' and s[i+1]=='C') or (s[i-1]=='C' and s[i+1]=='A'))):
flag=1
break
elif (s[i]=='C'and ((s[i-1]=='A'and s[i+1]=='B') or (s[i-1]=='B' and s[i+1]=='A'))):
flag=1
break
if(flag==1):
print("Yes")
elif(len(s)<3):
print("No")
elif(flag==0):
print("No")
```
| 6,352 |
Provide tags and a correct Python 3 solution for this coding contest problem.
When the curtains are opened, a canvas unfolds outside. Kanno marvels at all the blonde colours along the riverside — not tangerines, but blossoms instead.
"What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order.
Input
The first and only line of input contains a non-empty string s consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only (\lvert s \rvert ≤ 100) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively.
Output
Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
.BAC.
Output
Yes
Input
AA..CB
Output
No
Note
In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell.
Tags: implementation, strings
Correct Solution:
```
s=input()
f=0
for i in range(len(s)-2):
p=[]
for j in range(i,i+3):
if s[j]!='.':
if s[j] not in p:
p.append(s[j])
#print(p)
if len(p)==3:
f=1
break
if f==1:
print("YES")
else:
print("NO")
```
| 6,353 |
Provide tags and a correct Python 3 solution for this coding contest problem.
When the curtains are opened, a canvas unfolds outside. Kanno marvels at all the blonde colours along the riverside — not tangerines, but blossoms instead.
"What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order.
Input
The first and only line of input contains a non-empty string s consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only (\lvert s \rvert ≤ 100) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively.
Output
Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
.BAC.
Output
Yes
Input
AA..CB
Output
No
Note
In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell.
Tags: implementation, strings
Correct Solution:
```
string = input()
list = ['ABC', 'ACB', 'BAC', 'BCA', 'CAB', 'CBA']
if any(substring in string for substring in list):
print("Yes")
else:
print("No")
```
| 6,354 |
Provide tags and a correct Python 3 solution for this coding contest problem.
When the curtains are opened, a canvas unfolds outside. Kanno marvels at all the blonde colours along the riverside — not tangerines, but blossoms instead.
"What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order.
Input
The first and only line of input contains a non-empty string s consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only (\lvert s \rvert ≤ 100) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively.
Output
Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
.BAC.
Output
Yes
Input
AA..CB
Output
No
Note
In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell.
Tags: implementation, strings
Correct Solution:
```
a = input()
if 'ABC'in a or 'ACB'in a or'BCA'in a or 'BAC'in a or 'CBA'in a or'CAB'in a :
print("Yes")
else:
print("No")
'''
n,k= map(int,input().split())
w = 0
if n == 999983 and k == 1000:
print(999983001)
exit()
elif n ==994009 and k ==997:
print(991026974)
exit()
elif n ==999883 and k ==200:
print(199976601)
exit()
elif n ==199942 and k ==1000:
print(99971002)
exit()
elif n ==999002 and k ==457:
print(228271959)
exit()
elif n ==999995 and k ==1000:
print(199999005)
exit()
while 1:
w+=1
if (w//k)*(w%k)==n:
print(w)
break
'''
'''
'''
'''
#n,k= map(int,input().split())
#w = 0
#while 1:
# w+=1
# if (w//k)*(w%k)==n:
# print(w)
# break
'''
'''
n=int(input())
m=list(map(int,input().split()))
print(m.count(1))
for j in range(n-1):
if m[j+1]==1:
print(m[j],end=' ')
print(m[-1])
'''
'''
a = int(input())
f1 = 1
f2 = 1
if a < 3:
print(1)
exit()
cnt = 2
for i in range(a - 2):
a = f2
f2 += f1
f1 = a
cnt += f2
print(cnt)
'''
```
| 6,355 |
Provide tags and a correct Python 3 solution for this coding contest problem.
When the curtains are opened, a canvas unfolds outside. Kanno marvels at all the blonde colours along the riverside — not tangerines, but blossoms instead.
"What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order.
Input
The first and only line of input contains a non-empty string s consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only (\lvert s \rvert ≤ 100) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively.
Output
Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
.BAC.
Output
Yes
Input
AA..CB
Output
No
Note
In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell.
Tags: implementation, strings
Correct Solution:
```
s=input()
print("Yes"if"ABC"in s or "ACB"in s or "BCA"in s or "BAC"in s or "CBA"in s or "CAB"in s else "No")
```
| 6,356 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
When the curtains are opened, a canvas unfolds outside. Kanno marvels at all the blonde colours along the riverside — not tangerines, but blossoms instead.
"What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order.
Input
The first and only line of input contains a non-empty string s consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only (\lvert s \rvert ≤ 100) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively.
Output
Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
.BAC.
Output
Yes
Input
AA..CB
Output
No
Note
In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell.
Submitted Solution:
```
# def iscommon(a, b):
# for i in range(2, min(a, b) + 1):
# if a % i == b % i == 0:
# return True
# else:
# return False
# for _ in range(int(input())):
# n = int(input())
# arr = [int(i) for i in input().split()]
# cnt = n
# for i in range(n - 1):
# for j in range(i + 1, n):
# if iscommon(arr[i], arr[j]):
# cnt -= 1
# break
# print(cnt)
s = input()
check = {'ABC', 'ACB', 'BAC', 'BCA', 'CAB', 'CBA'}
if any(i in s for i in check):
print('Yes')
else:
print('No')
```
Yes
| 6,357 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
When the curtains are opened, a canvas unfolds outside. Kanno marvels at all the blonde colours along the riverside — not tangerines, but blossoms instead.
"What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order.
Input
The first and only line of input contains a non-empty string s consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only (\lvert s \rvert ≤ 100) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively.
Output
Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
.BAC.
Output
Yes
Input
AA..CB
Output
No
Note
In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell.
Submitted Solution:
```
s = input()
print(('No', 'Yes')[any(set(s[i:i+3]) == set('ABC') for i in range(len(s) - 2))])
```
Yes
| 6,358 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
When the curtains are opened, a canvas unfolds outside. Kanno marvels at all the blonde colours along the riverside — not tangerines, but blossoms instead.
"What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order.
Input
The first and only line of input contains a non-empty string s consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only (\lvert s \rvert ≤ 100) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively.
Output
Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
.BAC.
Output
Yes
Input
AA..CB
Output
No
Note
In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell.
Submitted Solution:
```
n=input()
a=0
for i in range(1,len(n)-1):
if n[i]!="." and n[i-1]!="." and n[i+1]!="." and n[i]!=n[i+1] and n[i]!=n[i-1] and n[i-1]!=n[i+1]:
print("Yes")
a=1
break
if a==0:
print("NO")
```
Yes
| 6,359 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
When the curtains are opened, a canvas unfolds outside. Kanno marvels at all the blonde colours along the riverside — not tangerines, but blossoms instead.
"What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order.
Input
The first and only line of input contains a non-empty string s consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only (\lvert s \rvert ≤ 100) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively.
Output
Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
.BAC.
Output
Yes
Input
AA..CB
Output
No
Note
In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell.
Submitted Solution:
```
s=input()
if 'ABC' in s or 'BAC' in s or 'CBA' in s or 'ACB' in s or 'BCA' in s or 'CAB' in s:
print('Yes')
else:
print('No')
```
Yes
| 6,360 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
When the curtains are opened, a canvas unfolds outside. Kanno marvels at all the blonde colours along the riverside — not tangerines, but blossoms instead.
"What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order.
Input
The first and only line of input contains a non-empty string s consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only (\lvert s \rvert ≤ 100) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively.
Output
Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
.BAC.
Output
Yes
Input
AA..CB
Output
No
Note
In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell.
Submitted Solution:
```
#Winners never quit, Quitters never win............................................................................
from collections import deque as de
import math
import re
from collections import Counter as cnt
from functools import reduce
from typing import MutableMapping
from itertools import groupby as gb
from fractions import Fraction as fr
from bisect import bisect_left as bl, bisect_right as br
def factors(n):
return set(reduce(list.__add__,
([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
class My_stack():
def __init__(self):
self.data = []
def my_push(self, x):
return (self.data.append(x))
def my_pop(self):
return (self.data.pop())
def my_peak(self):
return (self.data[-1])
def my_contains(self, x):
return (self.data.count(x))
def my_show_all(self):
return (self.data)
def isEmpty(self):
return len(self.data)==0
arrStack = My_stack()
def decimalToBinary(n):
return bin(n).replace("0b", "")
def binarytodecimal(n):
return int(n,2)
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
def get_prime_factors(number):
prime_factors = []
while number % 2 == 0:
prime_factors.append(2)
number = number / 2
for i in range(3, int(math.sqrt(number)) + 1, 2):
while number % i == 0:
prime_factors.append(int(i))
number = number / i
if number > 2:
prime_factors.append(int(number))
return prime_factors
def get_frequency(list):
dic={}
for ele in list:
if ele in dic:
dic[ele] += 1
else:
dic[ele] = 1
return dic
def Log2(x):
return (math.log10(x) /
math.log10(2));
# Function to get product of digits
def getProduct(n):
product = 1
while (n != 0):
product = product * (n % 10)
n = n // 10
return product
def isPowerOfTwo(n):
return (math.ceil(Log2(n)) == math.floor(Log2(n)));
def ceildiv(x,y): return (x+y-1)//y #ceil function gives wrong answer after 10^17 so i have to create my own :)
# because i don't want to doubt on my solution of 900-1000 problem set.
def di():return map(int, input().split())
def li():return list(map(int, input().split()))
#Here we go......................
#Winners never quit, Quitters never win
#concentration and mental toughness are margins of victory
s=input()
ch=1
for i in range(len(s)-3):
if len(set(s[i:i+3]))==3:
print("YES")
ch=0
break
if ch:
print("NO")
```
No
| 6,361 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
When the curtains are opened, a canvas unfolds outside. Kanno marvels at all the blonde colours along the riverside — not tangerines, but blossoms instead.
"What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order.
Input
The first and only line of input contains a non-empty string s consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only (\lvert s \rvert ≤ 100) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively.
Output
Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
.BAC.
Output
Yes
Input
AA..CB
Output
No
Note
In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell.
Submitted Solution:
```
s = input()
flag = True
for i in range(len(s)-3):
s2 = s[i:i+3]
l = list(s)
st = set(l)
if(len(l)==len(st)):
print("Yes")
flag=False
if(flag):
print("No")
```
No
| 6,362 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
When the curtains are opened, a canvas unfolds outside. Kanno marvels at all the blonde colours along the riverside — not tangerines, but blossoms instead.
"What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order.
Input
The first and only line of input contains a non-empty string s consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only (\lvert s \rvert ≤ 100) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively.
Output
Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
.BAC.
Output
Yes
Input
AA..CB
Output
No
Note
In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell.
Submitted Solution:
```
s = input()
if len(s)<3:
for i in range(1,len(s)-1):
if sorted([s[i-1],s[i],s[i+1]]) == ['A','B','C']:
b = 'yes'
break
else:
b = 'no'
print(b)
else:
print('no')
```
No
| 6,363 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
When the curtains are opened, a canvas unfolds outside. Kanno marvels at all the blonde colours along the riverside — not tangerines, but blossoms instead.
"What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order.
Input
The first and only line of input contains a non-empty string s consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only (\lvert s \rvert ≤ 100) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively.
Output
Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
.BAC.
Output
Yes
Input
AA..CB
Output
No
Note
In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell.
Submitted Solution:
```
baccab = input()
if 'BAC' in baccab:
print('Yes')
elif 'CAB' in baccab:
print('Yes')
elif 'ABC' in baccab:
print('Yes')
else:
print('No')
```
No
| 6,364 |
Provide a correct Python 3 solution for this coding contest problem.
We have N colored balls arranged in a row from left to right; the color of the i-th ball from the left is c_i.
You are given Q queries. The i-th query is as follows: how many different colors do the l_i-th through r_i-th balls from the left have?
Constraints
* 1\leq N,Q \leq 5 \times 10^5
* 1\leq c_i \leq N
* 1\leq l_i \leq r_i \leq N
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N Q
c_1 c_2 \cdots c_N
l_1 r_1
l_2 r_2
:
l_Q r_Q
Output
Print Q lines. The i-th line should contain the response to the i-th query.
Examples
Input
4 3
1 2 1 3
1 3
2 4
3 3
Output
2
3
1
Input
10 10
2 5 6 5 2 1 7 9 7 2
5 5
2 4
6 7
2 2
7 8
7 9
1 8
6 9
8 10
6 8
Output
1
2
2
1
2
2
6
3
3
3
"Correct Solution:
```
class fenwick_tree:
"""
区間の一点更新と，区間和の取得がO(log n)で可能なデータ構造
1-indexedで実装
"""
def __init__(self, N):
self.size = N
self.tree = [0] * (N+1)
def sum_until(self, i):
s = 0
while i > 0:
s += self.tree[i]
i -= i & (-i)
return s
def sum_acc(self, i, j):
""" [i,j] の和を返す """
return self.sum_until(j) - self.sum_until(i-1)
def add(self, i, x):
if i <= 0:
return
while i <= self.size:
self.tree[i] += x
i += i & (-i)
def main():
N, Q = (int(i) for i in input().split())
C = [int(i) for i in input().split()]
Query = [[] for _ in range(N+1)]
for j in range(Q):
le, ri = (int(i) for i in input().split())
Query[ri].append((le, j))
Query.append((-1, -1, -1))
lastappend = [-1] * (N + 1)
bit = fenwick_tree(N)
ans = [0]*Q
for i, a in enumerate(C, start=1):
if lastappend[a] != -1:
bit.add(lastappend[a], -1)
lastappend[a] = i
bit.add(i, 1)
for (le, j) in Query[i]:
ri = i
ans[j] = bit.sum_acc(le, ri)
print(*ans, sep="\n")
if __name__ == '__main__':
main()
```
| 6,365 |
Provide a correct Python 3 solution for this coding contest problem.
We have N colored balls arranged in a row from left to right; the color of the i-th ball from the left is c_i.
You are given Q queries. The i-th query is as follows: how many different colors do the l_i-th through r_i-th balls from the left have?
Constraints
* 1\leq N,Q \leq 5 \times 10^5
* 1\leq c_i \leq N
* 1\leq l_i \leq r_i \leq N
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N Q
c_1 c_2 \cdots c_N
l_1 r_1
l_2 r_2
:
l_Q r_Q
Output
Print Q lines. The i-th line should contain the response to the i-th query.
Examples
Input
4 3
1 2 1 3
1 3
2 4
3 3
Output
2
3
1
Input
10 10
2 5 6 5 2 1 7 9 7 2
5 5
2 4
6 7
2 2
7 8
7 9
1 8
6 9
8 10
6 8
Output
1
2
2
1
2
2
6
3
3
3
"Correct Solution:
```
import sys
sys.setrecursionlimit(10 ** 6)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
class BitSum:
def __init__(self, n):
self.n = n + 1
self.table = [0] * self.n
def add(self, i, x):
i += 1
while i < self.n:
self.table[i] += x
i += i & -i
def sum(self, i):
i += 1
res = 0
while i > 0:
res += self.table[i]
i -= i & -i
return res
n,q=MI()
cc=LI()
lri=[]
mx=500005
#mx=10
for i in range(q):
l,r=MI()
l-=1
lri.append((l,r-1,i))
lri.sort(key=lambda x:x[1])
#print(lri)
last=[-1]*mx
st=BitSum(mx)
ans=[0]*q
now=0
for l,r,i in lri:
for j in range(now,r+1):
c=cc[j]
pre=last[c]
if pre!=-1:st.add(pre,-1)
last[c]=j
st.add(j,1)
#print(l,r,st.tree,last)
now=r+1
ans[i]=st.sum(r)-st.sum(l-1)
print(*ans,sep="\n")
```
| 6,366 |
Provide a correct Python 3 solution for this coding contest problem.
We have N colored balls arranged in a row from left to right; the color of the i-th ball from the left is c_i.
You are given Q queries. The i-th query is as follows: how many different colors do the l_i-th through r_i-th balls from the left have?
Constraints
* 1\leq N,Q \leq 5 \times 10^5
* 1\leq c_i \leq N
* 1\leq l_i \leq r_i \leq N
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N Q
c_1 c_2 \cdots c_N
l_1 r_1
l_2 r_2
:
l_Q r_Q
Output
Print Q lines. The i-th line should contain the response to the i-th query.
Examples
Input
4 3
1 2 1 3
1 3
2 4
3 3
Output
2
3
1
Input
10 10
2 5 6 5 2 1 7 9 7 2
5 5
2 4
6 7
2 2
7 8
7 9
1 8
6 9
8 10
6 8
Output
1
2
2
1
2
2
6
3
3
3
"Correct Solution:
```
import sys
input = sys.stdin.buffer.readline
N,Q = map(int,input().split())
BIT = [0]*(N+1)
def BIT_query(idx):
res_sum = 0
while idx > 0:
res_sum += BIT[idx]
idx -= idx&(-idx)
return res_sum
def BIT_update(idx,x):
while idx <= N:
BIT[idx] += x
idx += idx&(-idx)
return
c = [0]+list(map(int,input().split()))
lastAppeared = [-1]*(N+1)
ans = [0]*Q
queries = []
for q in range(Q):
l,r = map(int,input().split())
queries.append(r*10**12+l*10**6+q)
queries.sort()
curR = 0
for i in range(Q):
query = queries[i]
r = query//10**12
query %= 10**12
l = query//10**6
q = query % 10**6
while curR < r:
curR += 1
color = c[curR]
last = lastAppeared[color]
if last != -1:
BIT_update(last,-1)
lastAppeared[color] = curR
BIT_update(curR,1)
ans[q] = BIT_query(r)-BIT_query(l-1)
for i in range(Q):
print(ans[i])
```
| 6,367 |
Provide a correct Python 3 solution for this coding contest problem.
We have N colored balls arranged in a row from left to right; the color of the i-th ball from the left is c_i.
You are given Q queries. The i-th query is as follows: how many different colors do the l_i-th through r_i-th balls from the left have?
Constraints
* 1\leq N,Q \leq 5 \times 10^5
* 1\leq c_i \leq N
* 1\leq l_i \leq r_i \leq N
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N Q
c_1 c_2 \cdots c_N
l_1 r_1
l_2 r_2
:
l_Q r_Q
Output
Print Q lines. The i-th line should contain the response to the i-th query.
Examples
Input
4 3
1 2 1 3
1 3
2 4
3 3
Output
2
3
1
Input
10 10
2 5 6 5 2 1 7 9 7 2
5 5
2 4
6 7
2 2
7 8
7 9
1 8
6 9
8 10
6 8
Output
1
2
2
1
2
2
6
3
3
3
"Correct Solution:
```
def solve():
import sys
input = sys.stdin.readline
class FenwickTree:
def __init__(self, size):
self.size = size
self.array = [0]*size
def add(self, index, value):
while index < self.size:
self.array[index] += value
index += index&(-index)
def sum(self, index):
answer = 0
while index > 0:
answer += self.array[index]
index -= index&(-index)
return answer
def rangesum(self, start, end):
return self.sum(end)-self.sum(start-1)
N, Q = map(int, input().split())
*c, = map(int, input().split())
tree = FenwickTree(N+1)
m = [tuple(map(int, input().split()))+(i,) for i in range(Q)]
m.sort(key=lambda x:x[1])
right = 0
pos = [-1] * (N+1)
ans = [-1] * (Q)
for l, r, idx in m:
for i in range(right, r):
x = pos[c[i]]
if x != -1:
tree.add(x, -1)
pos[c[i]] = i+1
tree.add(i+1, 1)
ans[idx] = tree.rangesum(l, r)
right = r
for i in ans:
print(i)
solve()
```
| 6,368 |
Provide a correct Python 3 solution for this coding contest problem.
We have N colored balls arranged in a row from left to right; the color of the i-th ball from the left is c_i.
You are given Q queries. The i-th query is as follows: how many different colors do the l_i-th through r_i-th balls from the left have?
Constraints
* 1\leq N,Q \leq 5 \times 10^5
* 1\leq c_i \leq N
* 1\leq l_i \leq r_i \leq N
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N Q
c_1 c_2 \cdots c_N
l_1 r_1
l_2 r_2
:
l_Q r_Q
Output
Print Q lines. The i-th line should contain the response to the i-th query.
Examples
Input
4 3
1 2 1 3
1 3
2 4
3 3
Output
2
3
1
Input
10 10
2 5 6 5 2 1 7 9 7 2
5 5
2 4
6 7
2 2
7 8
7 9
1 8
6 9
8 10
6 8
Output
1
2
2
1
2
2
6
3
3
3
"Correct Solution:
```
import sys
input = sys.stdin.readline
n, q = map(int, input().split())
C = [0] + list(map(int, input().split()))
D = [-1]*(n+1)
A = [0]*q
U = 10**6
LR = tuple(tuple(map(int, input().split())) for _ in range(q))
W = tuple(sorted(r*U+i for i, (l, r) in enumerate(LR)))
B = [0]*(n+1)
def add(i, a):
while i <= n:
B[i] += a
i += i & -i
def acc(i):
res = 0
while i:
res += B[i]
i -= i & -i
return res
temp = 1
for w in W:
r, i = divmod(w, U)
while temp <= r:
c = C[temp]
add(temp, 1)
if D[c] != -1:
add(D[c], -1)
D[c] = temp
temp += 1
l = LR[i][0]
A[i] = acc(r) - acc(l-1)
print(*A, sep="\n")
```
| 6,369 |
Provide a correct Python 3 solution for this coding contest problem.
We have N colored balls arranged in a row from left to right; the color of the i-th ball from the left is c_i.
You are given Q queries. The i-th query is as follows: how many different colors do the l_i-th through r_i-th balls from the left have?
Constraints
* 1\leq N,Q \leq 5 \times 10^5
* 1\leq c_i \leq N
* 1\leq l_i \leq r_i \leq N
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N Q
c_1 c_2 \cdots c_N
l_1 r_1
l_2 r_2
:
l_Q r_Q
Output
Print Q lines. The i-th line should contain the response to the i-th query.
Examples
Input
4 3
1 2 1 3
1 3
2 4
3 3
Output
2
3
1
Input
10 10
2 5 6 5 2 1 7 9 7 2
5 5
2 4
6 7
2 2
7 8
7 9
1 8
6 9
8 10
6 8
Output
1
2
2
1
2
2
6
3
3
3
"Correct Solution:
```
def main():
import sys
input = sys.stdin.readline
class BIT:
def __init__(self,n):
self.size = n
self.tree = [0]*(n+1)
def add(self,i,x):
while i <= self.size:
self.tree[i] += x
i += i&-i
def sum(self,i):
s = 0
while i > 0:
s += self.tree[i]
i -= i&-i
return s
N,Q = map(int,input().split())
C = list(map(int,input().split()))
LR = []
for i in range(Q):
l,r = map(int,input().split())
LR.append((l,r,i))
bit = BIT(N)
#color[i] => 色iが最後に登場したインデックス(1-indexed)
color = [0]*(N+1)
ANS = [0]*Q
LR.sort(key=lambda x: x[1])
cur = 1
for l,r,j in LR:
for i in range(cur,r+1):
c = C[i-1]
if color[c] != 0:
bit.add(color[c],-1)
color[c] = i
bit.add(i,1)
cur = r+1
ANS[j] = bit.sum(r)-bit.sum(l-1)
for a in ANS:
print(a)
main()
```
| 6,370 |
Provide a correct Python 3 solution for this coding contest problem.
We have N colored balls arranged in a row from left to right; the color of the i-th ball from the left is c_i.
You are given Q queries. The i-th query is as follows: how many different colors do the l_i-th through r_i-th balls from the left have?
Constraints
* 1\leq N,Q \leq 5 \times 10^5
* 1\leq c_i \leq N
* 1\leq l_i \leq r_i \leq N
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N Q
c_1 c_2 \cdots c_N
l_1 r_1
l_2 r_2
:
l_Q r_Q
Output
Print Q lines. The i-th line should contain the response to the i-th query.
Examples
Input
4 3
1 2 1 3
1 3
2 4
3 3
Output
2
3
1
Input
10 10
2 5 6 5 2 1 7 9 7 2
5 5
2 4
6 7
2 2
7 8
7 9
1 8
6 9
8 10
6 8
Output
1
2
2
1
2
2
6
3
3
3
"Correct Solution:
```
class Bit:
def __init__(self, n):
self.size = n
self.tree = [0] * (n+1)
def sum(self, i): # sum in [0, i)
s = 0
while i > 0:
s += self.tree[i]
i -= i & -i
return s
def add(self, i, x): # i > 0
assert i > 0
while i <= self.size:
self.tree[i] += x
i += i & -i
N, Q = map(int, input().split())
c = list(map(int, input().split()))
lr = []
for i in range(Q):
l, r = map(int, input().split())
lr.append([l-1, r-1, i])
lr.sort(key=lambda x: x[1])
rightest = [-1] * (N + 1)
current_q = 0
bit = Bit(N)
ans = [0] * Q
for i in range(N):
if rightest[c[i]] != -1:
bit.add(rightest[c[i]]+1, -1)
rightest[c[i]] = i
bit.add(i+1, 1)
while current_q < Q and lr[current_q][1] == i:
ans[lr[current_q][2]] = bit.sum(lr[current_q][1] + 1) - bit.sum(lr[current_q][0])
current_q += 1
for i in range(Q):
print(ans[i])
```
| 6,371 |
Provide a correct Python 3 solution for this coding contest problem.
We have N colored balls arranged in a row from left to right; the color of the i-th ball from the left is c_i.
You are given Q queries. The i-th query is as follows: how many different colors do the l_i-th through r_i-th balls from the left have?
Constraints
* 1\leq N,Q \leq 5 \times 10^5
* 1\leq c_i \leq N
* 1\leq l_i \leq r_i \leq N
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N Q
c_1 c_2 \cdots c_N
l_1 r_1
l_2 r_2
:
l_Q r_Q
Output
Print Q lines. The i-th line should contain the response to the i-th query.
Examples
Input
4 3
1 2 1 3
1 3
2 4
3 3
Output
2
3
1
Input
10 10
2 5 6 5 2 1 7 9 7 2
5 5
2 4
6 7
2 2
7 8
7 9
1 8
6 9
8 10
6 8
Output
1
2
2
1
2
2
6
3
3
3
"Correct Solution:
```
import sys
input = sys.stdin.buffer.readline
# 足す時はi番目に足し、返すのは累積和
class sumBIT():
def __init__(self, N):
self.N = N
self.bit = [0 for _ in range(self.N+1)]
def __str__(self):
ret = []
for i in range(1, self.N+1):
ret.append(self.__getitem__(i))
return "[" + ", ".join([str(a) for a in ret]) + "]"
def __getitem__(self, i):
s = 0
while i > 0:
s += self.bit[i]
i -= i & -i
return s
def add(self, i, x):
while i <= self.N:
self.bit[i] += x
i += i & -i
N, Q = map(int, input().split())
C = list(map(int, input().split()))
LR = [list(map(int, input().split())) for _ in range(Q)]
Toask = [[] for _ in range(N+1)]
for i, (l, r) in enumerate(LR):
Toask[l].append((r, i))
bit = sumBIT(N+3)
ans = [-1]*Q
Color = [-1]*(N+1)
for l in reversed(range(1, N+1)):
c = C[l-1]
if Color[c] != -1:
last = Color[c]
bit.add(last, -1)
bit.add(l, 1)
for r, ind in Toask[l]:
ans[ind] = str(bit[r])
Color[c] = l
print("\n".join(ans))
```
| 6,372 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N colored balls arranged in a row from left to right; the color of the i-th ball from the left is c_i.
You are given Q queries. The i-th query is as follows: how many different colors do the l_i-th through r_i-th balls from the left have?
Constraints
* 1\leq N,Q \leq 5 \times 10^5
* 1\leq c_i \leq N
* 1\leq l_i \leq r_i \leq N
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N Q
c_1 c_2 \cdots c_N
l_1 r_1
l_2 r_2
:
l_Q r_Q
Output
Print Q lines. The i-th line should contain the response to the i-th query.
Examples
Input
4 3
1 2 1 3
1 3
2 4
3 3
Output
2
3
1
Input
10 10
2 5 6 5 2 1 7 9 7 2
5 5
2 4
6 7
2 2
7 8
7 9
1 8
6 9
8 10
6 8
Output
1
2
2
1
2
2
6
3
3
3
Submitted Solution:
```
import sys
input = sys.stdin.buffer.readline
N,Q = map(int,input().split())
BIT = [0]*(N+1)
def BIT_query(idx):
res_sum = 0
while idx > 0:
res_sum += BIT[idx]
idx -= idx&(-idx)
return res_sum
def BIT_update(idx,x):
while idx <= N:
BIT[idx] += x
idx += idx&(-idx)
return
c = [0]+list(map(int,input().split()))
lastAppeared = [-1]*(N+1)
ans = [0]*Q
queries = []
for q in range(Q):
l,r = map(int,input().split())
queries.append(r*10**12+l*10**6+q)
queries.sort()
curR = 0
for i in range(Q):
query = queries[i]
r = query//10**12
l = (query//10**6)%(10**6)
q = query % 10**6
while curR < r:
curR += 1
color = c[curR]
last = lastAppeared[color]
if last != -1:
BIT_update(last,-1)
lastAppeared[color] = curR
BIT_update(curR,1)
ans[q] = BIT_query(r)-BIT_query(l-1)
for i in range(Q):
print(ans[i])
```
Yes
| 6,373 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N colored balls arranged in a row from left to right; the color of the i-th ball from the left is c_i.
You are given Q queries. The i-th query is as follows: how many different colors do the l_i-th through r_i-th balls from the left have?
Constraints
* 1\leq N,Q \leq 5 \times 10^5
* 1\leq c_i \leq N
* 1\leq l_i \leq r_i \leq N
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N Q
c_1 c_2 \cdots c_N
l_1 r_1
l_2 r_2
:
l_Q r_Q
Output
Print Q lines. The i-th line should contain the response to the i-th query.
Examples
Input
4 3
1 2 1 3
1 3
2 4
3 3
Output
2
3
1
Input
10 10
2 5 6 5 2 1 7 9 7 2
5 5
2 4
6 7
2 2
7 8
7 9
1 8
6 9
8 10
6 8
Output
1
2
2
1
2
2
6
3
3
3
Submitted Solution:
```
import sys
input = sys.stdin.readline
class Bit:
"""1-indexed"""
def __init__(self, n):
self.size = n
self.tree = [0] * (n + 1)
def sum(self, i):
s = 0
while i > 0:
s += self.tree[i]
i -= i & -i
return s
def add(self, i, x):
while i <= self.size:
self.tree[i] += x
i += i & -i
N, Q = map(int, input().split())
C = tuple(map(int, input().split()))
queries = []
for i in range(Q):
l, r = map(int, input().split())
queries.append((l, r, i))
queries.sort(key=lambda x: x[1])
b = Bit(N)
last = [None] * (N + 1)
cur = 1
ans = [0] * Q
for l, r, i in queries:
while cur <= r:
if last[C[cur - 1]] is not None:
b.add(last[C[cur - 1]], -1)
last[C[cur - 1]] = cur
b.add(cur, 1)
cur += 1
ans[i] = b.sum(r) - b.sum(l - 1)
print(*ans, sep='\n')
```
Yes
| 6,374 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N colored balls arranged in a row from left to right; the color of the i-th ball from the left is c_i.
You are given Q queries. The i-th query is as follows: how many different colors do the l_i-th through r_i-th balls from the left have?
Constraints
* 1\leq N,Q \leq 5 \times 10^5
* 1\leq c_i \leq N
* 1\leq l_i \leq r_i \leq N
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N Q
c_1 c_2 \cdots c_N
l_1 r_1
l_2 r_2
:
l_Q r_Q
Output
Print Q lines. The i-th line should contain the response to the i-th query.
Examples
Input
4 3
1 2 1 3
1 3
2 4
3 3
Output
2
3
1
Input
10 10
2 5 6 5 2 1 7 9 7 2
5 5
2 4
6 7
2 2
7 8
7 9
1 8
6 9
8 10
6 8
Output
1
2
2
1
2
2
6
3
3
3
Submitted Solution:
```
#!usr/bin/env python3
import sys
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LIR(n):
return [[i,LI()] for i in range(n)]
def solve():
def add(i,x):
while i < len(bit):
bit[i] += x
i += i&-i
def sum_(i):
res = 0
while i > 0:
res += bit[i]
i -= i&-i
return res
n,Q = LI()
c = LI()
c = [0]+c
q = LIR(Q)
q.sort(key = lambda x:x[1][1])
bit = [0]*(n+2)
i = 1
p = [None]*(n+1)
ans = [0]*Q
s = 0
for ind,(l, r) in q:
l -= 1
while i <= r:
ci = c[i]
j = p[ci]
if j is not None:
add(j,1)
else:
s += 1
p[ci] = i
i += 1
ans[ind] = s-l+sum_(l)
for i in ans:
print(i)
return
#Solve
if __name__ == "__main__":
solve()
```
Yes
| 6,375 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N colored balls arranged in a row from left to right; the color of the i-th ball from the left is c_i.
You are given Q queries. The i-th query is as follows: how many different colors do the l_i-th through r_i-th balls from the left have?
Constraints
* 1\leq N,Q \leq 5 \times 10^5
* 1\leq c_i \leq N
* 1\leq l_i \leq r_i \leq N
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N Q
c_1 c_2 \cdots c_N
l_1 r_1
l_2 r_2
:
l_Q r_Q
Output
Print Q lines. The i-th line should contain the response to the i-th query.
Examples
Input
4 3
1 2 1 3
1 3
2 4
3 3
Output
2
3
1
Input
10 10
2 5 6 5 2 1 7 9 7 2
5 5
2 4
6 7
2 2
7 8
7 9
1 8
6 9
8 10
6 8
Output
1
2
2
1
2
2
6
3
3
3
Submitted Solution:
```
import sys
input=sys.stdin.readline
n,q=map(int,input().split())
c=list(map(int,input().split()))
l=[]
for i in range(q):
L,R=map(int,input().split())
l.append([L,R,i])
l.sort(key=lambda x:x[1])
L=[-1]*(5*10**5+1)
class Bit:
def __init__(self,n):
self.size=n
self.tree=[0]*(n + 1)
self.depth=n.bit_length()
def sum(self,i):
s=0
while i>0:
s+=self.tree[i]
i-=i&-i
return s
def add(self,i,x):
while i<=self.size:
self.tree[i]+=x
i+=i&-i
BIT=Bit(n+1)
ans=[0]*q
ct=0
for i in range(q):
while ct<=l[i][1]:
if L[c[ct-1]-1]!=-1:
BIT.add(L[c[ct-1]-1],-1)
L[c[ct-1]-1]=ct+1
BIT.add(ct+1,1)
ct+=1
ans[l[i][2]]=BIT.sum(l[i][1]+1)-BIT.sum(l[i][0])
for i in range(q):
print(ans[i])
```
Yes
| 6,376 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N colored balls arranged in a row from left to right; the color of the i-th ball from the left is c_i.
You are given Q queries. The i-th query is as follows: how many different colors do the l_i-th through r_i-th balls from the left have?
Constraints
* 1\leq N,Q \leq 5 \times 10^5
* 1\leq c_i \leq N
* 1\leq l_i \leq r_i \leq N
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N Q
c_1 c_2 \cdots c_N
l_1 r_1
l_2 r_2
:
l_Q r_Q
Output
Print Q lines. The i-th line should contain the response to the i-th query.
Examples
Input
4 3
1 2 1 3
1 3
2 4
3 3
Output
2
3
1
Input
10 10
2 5 6 5 2 1 7 9 7 2
5 5
2 4
6 7
2 2
7 8
7 9
1 8
6 9
8 10
6 8
Output
1
2
2
1
2
2
6
3
3
3
Submitted Solution:
```
def segfunc(x,y):
return x|y
ide_ele=set()
class SegTree():
def __init__(self,init_val,segfunc,ide_ele):
n=len(init_val)
self.segfunc=segfunc
self.ide_ele=ide_ele
self.num=1<<(n-1).bit_length()
self.tree=[ide_ele]*2*self.num
for i in range(n):
self.tree[self.num+i]={init_val[i]}
for i in range(self.num-1,0,-1):
self.tree[i]=self.segfunc(self.tree[2*i], self.tree[2*i+1])
def update(self,k,x):
k+=self.num
self.tree[k]=x
while k>1:
self.tree[k>>1]=self.segfunc(self.tree[k],self.tree[k^1])
k>>=1
def query(self,l,r):
res=self.ide_ele
l+=self.num
r+=self.num
while l<r:
if l&1:
res=self.segfunc(res,self.tree[l])
l+=1
if r&1:
res=self.segfunc(res,self.tree[r-1])
l>>=1
r>>=1
return res
n,q=map(int,input().split())
c=list(map(int,input().split()))
st=SegTree(c,segfunc,ide_ele)
for _ in range(q):
l,r=map(int,input().split())
print(len(st.query(l-1,r)))
```
No
| 6,377 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N colored balls arranged in a row from left to right; the color of the i-th ball from the left is c_i.
You are given Q queries. The i-th query is as follows: how many different colors do the l_i-th through r_i-th balls from the left have?
Constraints
* 1\leq N,Q \leq 5 \times 10^5
* 1\leq c_i \leq N
* 1\leq l_i \leq r_i \leq N
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N Q
c_1 c_2 \cdots c_N
l_1 r_1
l_2 r_2
:
l_Q r_Q
Output
Print Q lines. The i-th line should contain the response to the i-th query.
Examples
Input
4 3
1 2 1 3
1 3
2 4
3 3
Output
2
3
1
Input
10 10
2 5 6 5 2 1 7 9 7 2
5 5
2 4
6 7
2 2
7 8
7 9
1 8
6 9
8 10
6 8
Output
1
2
2
1
2
2
6
3
3
3
Submitted Solution:
```
def main():
n,q=map(int,input().split())
lst=[0]*n
clst=tuple(map(int,input().split()))
work={}
for i in range(n):
if clst[i] in work:
work[clst[i]]+=1
else :
work[clst[i]]=1
lst[i]=work.copy()
for _ in range(q):
l,r=map(int,input().split())
rwork=lst[r-1]
if l==1 :
print(len(rwork))
continue
lwork=lst[l-2]
sm=len(rwork)
for x in rwork:
if x not in lwork : continue
if rwork[x]==lwork[x]:
sm-=1
print(sm)
main()
```
No
| 6,378 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N colored balls arranged in a row from left to right; the color of the i-th ball from the left is c_i.
You are given Q queries. The i-th query is as follows: how many different colors do the l_i-th through r_i-th balls from the left have?
Constraints
* 1\leq N,Q \leq 5 \times 10^5
* 1\leq c_i \leq N
* 1\leq l_i \leq r_i \leq N
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N Q
c_1 c_2 \cdots c_N
l_1 r_1
l_2 r_2
:
l_Q r_Q
Output
Print Q lines. The i-th line should contain the response to the i-th query.
Examples
Input
4 3
1 2 1 3
1 3
2 4
3 3
Output
2
3
1
Input
10 10
2 5 6 5 2 1 7 9 7 2
5 5
2 4
6 7
2 2
7 8
7 9
1 8
6 9
8 10
6 8
Output
1
2
2
1
2
2
6
3
3
3
Submitted Solution:
```
import sys
from operator import itemgetter
buf = sys.stdin.buffer
input = buf.readline
class BIT:
def __init__(self, n):
self.n = n
self.tree = [0] * (n + 1)
def add(self, i, x):
n, tree = self.n, self.tree
while i <= n:
tree[i] += x
i += i & -i
def sum(self, i):
tree = self.tree
s = 0
while i:
s += tree[i]
i -= i & -i
return s
def main():
n, q = map(int, input().split())
res = [0] * q
*c, = map(int, input().split())
q = [(i, l, r) for i, (l, r) in enumerate(zip(*[map(int, buf.read().split())] * 2))]
q.sort(key=itemgetter(2))
bit = BIT(n + 2)
add, sum = bit.add, bit.sum
b = [1] * (n + 1)
j = 1
for i, l, r in q:
while j <= r:
d = c[j - 1]
j += 1
k = b[d]
b[d] = j
add(k + 1, 1)
add(j + 1, -1)
res[i] = sum(l + 1)
print(' '.join(map(str, res)))
main()
```
No
| 6,379 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N colored balls arranged in a row from left to right; the color of the i-th ball from the left is c_i.
You are given Q queries. The i-th query is as follows: how many different colors do the l_i-th through r_i-th balls from the left have?
Constraints
* 1\leq N,Q \leq 5 \times 10^5
* 1\leq c_i \leq N
* 1\leq l_i \leq r_i \leq N
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N Q
c_1 c_2 \cdots c_N
l_1 r_1
l_2 r_2
:
l_Q r_Q
Output
Print Q lines. The i-th line should contain the response to the i-th query.
Examples
Input
4 3
1 2 1 3
1 3
2 4
3 3
Output
2
3
1
Input
10 10
2 5 6 5 2 1 7 9 7 2
5 5
2 4
6 7
2 2
7 8
7 9
1 8
6 9
8 10
6 8
Output
1
2
2
1
2
2
6
3
3
3
Submitted Solution:
```
import bisect
def answer(kintama, lri0):
tama = kintama[:]
tama.sort()
num = bisect.bisect_left(tama, lri0)
return 10**5+1 - num
n, q = [int(i) for i in input().split()]
c = [int(i)-1 for i in input().split()]
lr = []
for i in range(q):
lr_ = [int(_)-1 for _ in input().split()] + [i]
lr.append(lr_)
lr.sort(key=lambda x: x[1])
k = 0
ans = []
kintama = [-10 for i in range(10**5+1)]
for i in range(q):
while lr[i][1] > k:
kintama[c[k]] = k
k += 1
kintama[c[k]] = k
ans.append([answer(kintama, lr[i][0]), lr[i][2]])
ans.sort(key=lambda x: x[1])
[print(ans[i][0]) for i in range(q)]
```
No
| 6,380 |
Provide a correct Python 3 solution for this coding contest problem.
A string S of an odd length is said to be a strong palindrome if and only if all of the following conditions are satisfied:
* S is a palindrome.
* Let N be the length of S. The string formed by the 1-st through ((N-1)/2)-th characters of S is a palindrome.
* The string consisting of the (N+3)/2-st through N-th characters of S is a palindrome.
Determine whether S is a strong palindrome.
Constraints
* S consists of lowercase English letters.
* The length of S is an odd number between 3 and 99 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
If S is a strong palindrome, print `Yes`; otherwise, print `No`.
Examples
Input
akasaka
Output
Yes
Input
level
Output
No
Input
atcoder
Output
No
"Correct Solution:
```
S = input()
N = len(S)
A = (len(S) - 1) // 2
B = (len(S) + 3) // 2
if S[0:A] == S[B - 1:N]:
print('Yes')
else:
print('No')
```
| 6,381 |
Provide a correct Python 3 solution for this coding contest problem.
A string S of an odd length is said to be a strong palindrome if and only if all of the following conditions are satisfied:
* S is a palindrome.
* Let N be the length of S. The string formed by the 1-st through ((N-1)/2)-th characters of S is a palindrome.
* The string consisting of the (N+3)/2-st through N-th characters of S is a palindrome.
Determine whether S is a strong palindrome.
Constraints
* S consists of lowercase English letters.
* The length of S is an odd number between 3 and 99 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
If S is a strong palindrome, print `Yes`; otherwise, print `No`.
Examples
Input
akasaka
Output
Yes
Input
level
Output
No
Input
atcoder
Output
No
"Correct Solution:
```
s = input()
n = len(s)
if s[:int(n//2)] == s[int(n//2)+1:]:
print("Yes")
else:
print("No")
```
| 6,382 |
Provide a correct Python 3 solution for this coding contest problem.
A string S of an odd length is said to be a strong palindrome if and only if all of the following conditions are satisfied:
* S is a palindrome.
* Let N be the length of S. The string formed by the 1-st through ((N-1)/2)-th characters of S is a palindrome.
* The string consisting of the (N+3)/2-st through N-th characters of S is a palindrome.
Determine whether S is a strong palindrome.
Constraints
* S consists of lowercase English letters.
* The length of S is an odd number between 3 and 99 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
If S is a strong palindrome, print `Yes`; otherwise, print `No`.
Examples
Input
akasaka
Output
Yes
Input
level
Output
No
Input
atcoder
Output
No
"Correct Solution:
```
S=list(input())
N=len(S)
if S==S[::-1] and S[:(N-1)//2]==S[(N+3)//2-1:]:
print("Yes")
else:print("No")
```
| 6,383 |
Provide a correct Python 3 solution for this coding contest problem.
A string S of an odd length is said to be a strong palindrome if and only if all of the following conditions are satisfied:
* S is a palindrome.
* Let N be the length of S. The string formed by the 1-st through ((N-1)/2)-th characters of S is a palindrome.
* The string consisting of the (N+3)/2-st through N-th characters of S is a palindrome.
Determine whether S is a strong palindrome.
Constraints
* S consists of lowercase English letters.
* The length of S is an odd number between 3 and 99 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
If S is a strong palindrome, print `Yes`; otherwise, print `No`.
Examples
Input
akasaka
Output
Yes
Input
level
Output
No
Input
atcoder
Output
No
"Correct Solution:
```
s = input()
if s[len(s) // 2 + 1:] == s[:len(s) // 2]:
print('Yes')
else:
print('No')
```
| 6,384 |
Provide a correct Python 3 solution for this coding contest problem.
A string S of an odd length is said to be a strong palindrome if and only if all of the following conditions are satisfied:
* S is a palindrome.
* Let N be the length of S. The string formed by the 1-st through ((N-1)/2)-th characters of S is a palindrome.
* The string consisting of the (N+3)/2-st through N-th characters of S is a palindrome.
Determine whether S is a strong palindrome.
Constraints
* S consists of lowercase English letters.
* The length of S is an odd number between 3 and 99 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
If S is a strong palindrome, print `Yes`; otherwise, print `No`.
Examples
Input
akasaka
Output
Yes
Input
level
Output
No
Input
atcoder
Output
No
"Correct Solution:
```
s = input()
print("Yes" if s == s[::-1] and s[:len(s)//2] == s[len(s)//2-1::-1] else "No")
```
| 6,385 |
Provide a correct Python 3 solution for this coding contest problem.
A string S of an odd length is said to be a strong palindrome if and only if all of the following conditions are satisfied:
* S is a palindrome.
* Let N be the length of S. The string formed by the 1-st through ((N-1)/2)-th characters of S is a palindrome.
* The string consisting of the (N+3)/2-st through N-th characters of S is a palindrome.
Determine whether S is a strong palindrome.
Constraints
* S consists of lowercase English letters.
* The length of S is an odd number between 3 and 99 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
If S is a strong palindrome, print `Yes`; otherwise, print `No`.
Examples
Input
akasaka
Output
Yes
Input
level
Output
No
Input
atcoder
Output
No
"Correct Solution:
```
S = str(input())
N = len(S)
if S[:(N-1)//2] == S[(N+3)//2 - 1:] :
print('Yes')
else:
print('No')
```
| 6,386 |
Provide a correct Python 3 solution for this coding contest problem.
A string S of an odd length is said to be a strong palindrome if and only if all of the following conditions are satisfied:
* S is a palindrome.
* Let N be the length of S. The string formed by the 1-st through ((N-1)/2)-th characters of S is a palindrome.
* The string consisting of the (N+3)/2-st through N-th characters of S is a palindrome.
Determine whether S is a strong palindrome.
Constraints
* S consists of lowercase English letters.
* The length of S is an odd number between 3 and 99 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
If S is a strong palindrome, print `Yes`; otherwise, print `No`.
Examples
Input
akasaka
Output
Yes
Input
level
Output
No
Input
atcoder
Output
No
"Correct Solution:
```
S = input()
N = len(S)
N1s = (N-1)//2
N2s = (N+3)//2-1
if S[:N1s] == S[N2s:]:
print("Yes")
else:
print("No")
```
| 6,387 |
Provide a correct Python 3 solution for this coding contest problem.
A string S of an odd length is said to be a strong palindrome if and only if all of the following conditions are satisfied:
* S is a palindrome.
* Let N be the length of S. The string formed by the 1-st through ((N-1)/2)-th characters of S is a palindrome.
* The string consisting of the (N+3)/2-st through N-th characters of S is a palindrome.
Determine whether S is a strong palindrome.
Constraints
* S consists of lowercase English letters.
* The length of S is an odd number between 3 and 99 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
If S is a strong palindrome, print `Yes`; otherwise, print `No`.
Examples
Input
akasaka
Output
Yes
Input
level
Output
No
Input
atcoder
Output
No
"Correct Solution:
```
s=input();h=len(s)//2;print(' YNeos'[s[:h]==s[h-1::-1]==s[-h:]::2])
```
| 6,388 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string S of an odd length is said to be a strong palindrome if and only if all of the following conditions are satisfied:
* S is a palindrome.
* Let N be the length of S. The string formed by the 1-st through ((N-1)/2)-th characters of S is a palindrome.
* The string consisting of the (N+3)/2-st through N-th characters of S is a palindrome.
Determine whether S is a strong palindrome.
Constraints
* S consists of lowercase English letters.
* The length of S is an odd number between 3 and 99 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
If S is a strong palindrome, print `Yes`; otherwise, print `No`.
Examples
Input
akasaka
Output
Yes
Input
level
Output
No
Input
atcoder
Output
No
Submitted Solution:
```
S = input()
H = len(S)//2
if S[:H] == S[-H:] and S[:H] == S[-H:][::-1]:
print("Yes")
else:
print("No")
```
Yes
| 6,389 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string S of an odd length is said to be a strong palindrome if and only if all of the following conditions are satisfied:
* S is a palindrome.
* Let N be the length of S. The string formed by the 1-st through ((N-1)/2)-th characters of S is a palindrome.
* The string consisting of the (N+3)/2-st through N-th characters of S is a palindrome.
Determine whether S is a strong palindrome.
Constraints
* S consists of lowercase English letters.
* The length of S is an odd number between 3 and 99 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
If S is a strong palindrome, print `Yes`; otherwise, print `No`.
Examples
Input
akasaka
Output
Yes
Input
level
Output
No
Input
atcoder
Output
No
Submitted Solution:
```
s = list(input())
if s[0:(len(s) - 1)//2] == s[(len(s)+2)//2:len(s)]:
print('Yes')
else:
print('No')
```
Yes
| 6,390 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string S of an odd length is said to be a strong palindrome if and only if all of the following conditions are satisfied:
* S is a palindrome.
* Let N be the length of S. The string formed by the 1-st through ((N-1)/2)-th characters of S is a palindrome.
* The string consisting of the (N+3)/2-st through N-th characters of S is a palindrome.
Determine whether S is a strong palindrome.
Constraints
* S consists of lowercase English letters.
* The length of S is an odd number between 3 and 99 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
If S is a strong palindrome, print `Yes`; otherwise, print `No`.
Examples
Input
akasaka
Output
Yes
Input
level
Output
No
Input
atcoder
Output
No
Submitted Solution:
```
s = input()
n = len(s)
a = s[:((n-1)//2)]
if a == a[::-1] and s == s[::-1]:
print('Yes')
else:
print('No')
```
Yes
| 6,391 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string S of an odd length is said to be a strong palindrome if and only if all of the following conditions are satisfied:
* S is a palindrome.
* Let N be the length of S. The string formed by the 1-st through ((N-1)/2)-th characters of S is a palindrome.
* The string consisting of the (N+3)/2-st through N-th characters of S is a palindrome.
Determine whether S is a strong palindrome.
Constraints
* S consists of lowercase English letters.
* The length of S is an odd number between 3 and 99 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
If S is a strong palindrome, print `Yes`; otherwise, print `No`.
Examples
Input
akasaka
Output
Yes
Input
level
Output
No
Input
atcoder
Output
No
Submitted Solution:
```
S = input()
N = len(S)
if S[0:(N-1)//2] == S[((N+3)//2)-1:N]:
print("Yes")
else:
print("No")
```
Yes
| 6,392 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string S of an odd length is said to be a strong palindrome if and only if all of the following conditions are satisfied:
* S is a palindrome.
* Let N be the length of S. The string formed by the 1-st through ((N-1)/2)-th characters of S is a palindrome.
* The string consisting of the (N+3)/2-st through N-th characters of S is a palindrome.
Determine whether S is a strong palindrome.
Constraints
* S consists of lowercase English letters.
* The length of S is an odd number between 3 and 99 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
If S is a strong palindrome, print `Yes`; otherwise, print `No`.
Examples
Input
akasaka
Output
Yes
Input
level
Output
No
Input
atcoder
Output
No
Submitted Solution:
```
s = list(input())
n = len(s)
ls = []
l1 = (n-1)//2
l2 = (n+3)//2
h = l1//2
f = True
for i in range(h):
if s[i] == s[l1-i-1]:
pass
else:
f = False
if s[-i-1] == s[l2+i-1]:
pass
else:
f = False
if f:
print("Yes")
else:
print("No")
```
No
| 6,393 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string S of an odd length is said to be a strong palindrome if and only if all of the following conditions are satisfied:
* S is a palindrome.
* Let N be the length of S. The string formed by the 1-st through ((N-1)/2)-th characters of S is a palindrome.
* The string consisting of the (N+3)/2-st through N-th characters of S is a palindrome.
Determine whether S is a strong palindrome.
Constraints
* S consists of lowercase English letters.
* The length of S is an odd number between 3 and 99 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
If S is a strong palindrome, print `Yes`; otherwise, print `No`.
Examples
Input
akasaka
Output
Yes
Input
level
Output
No
Input
atcoder
Output
No
Submitted Solution:
```
S = input()
N = len(S)
i = list(S)
if i[0:((N+3)//2-2)] == [((N+3)//2):N]:
print('Yes')
elif:
print('No')
```
No
| 6,394 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string S of an odd length is said to be a strong palindrome if and only if all of the following conditions are satisfied:
* S is a palindrome.
* Let N be the length of S. The string formed by the 1-st through ((N-1)/2)-th characters of S is a palindrome.
* The string consisting of the (N+3)/2-st through N-th characters of S is a palindrome.
Determine whether S is a strong palindrome.
Constraints
* S consists of lowercase English letters.
* The length of S is an odd number between 3 and 99 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
If S is a strong palindrome, print `Yes`; otherwise, print `No`.
Examples
Input
akasaka
Output
Yes
Input
level
Output
No
Input
atcoder
Output
No
Submitted Solution:
```
def palindrome(str):
n = len(str)
for i in range(n // 2):
if str[i] != str[n - 1 - i]:
return False
return True
str = input()
is_palindrome = palindrome(str)
n = len(str)
if n % 2 == 1:
is_palindrome = palindrome(str[:n // 2]) and palindrome(str[n // 2 + 1:])
print("Yes" if is_palindrome else "No")
```
No
| 6,395 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string S of an odd length is said to be a strong palindrome if and only if all of the following conditions are satisfied:
* S is a palindrome.
* Let N be the length of S. The string formed by the 1-st through ((N-1)/2)-th characters of S is a palindrome.
* The string consisting of the (N+3)/2-st through N-th characters of S is a palindrome.
Determine whether S is a strong palindrome.
Constraints
* S consists of lowercase English letters.
* The length of S is an odd number between 3 and 99 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
If S is a strong palindrome, print `Yes`; otherwise, print `No`.
Examples
Input
akasaka
Output
Yes
Input
level
Output
No
Input
atcoder
Output
No
Submitted Solution:
```
def check(a):
i=0
j=len(a)-1
b=True
while(i<=j):
if(a[i]!=a[j]):
return False
i+=1;
j-=1
return b
a=(input())
b=a[:(len(a)-1)//2:]
c=a[(len(a)+3)//2::]
if(check(a)==True and check(b)==True and check(c)==True):
print("Yes")
else:
print("No")
```
No
| 6,396 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi is at an all-you-can-eat restaurant.
The restaurant offers N kinds of dishes. It takes A_i minutes to eat the i-th dish, whose deliciousness is B_i.
The restaurant has the following rules:
* You can only order one dish at a time. The dish ordered will be immediately served and ready to eat.
* You cannot order the same kind of dish more than once.
* Until you finish eating the dish already served, you cannot order a new dish.
* After T-0.5 minutes from the first order, you can no longer place a new order, but you can continue eating the dish already served.
Let Takahashi's happiness be the sum of the deliciousness of the dishes he eats in this restaurant.
What is the maximum possible happiness achieved by making optimal choices?
Constraints
* 2 \leq N \leq 3000
* 1 \leq T \leq 3000
* 1 \leq A_i \leq 3000
* 1 \leq B_i \leq 3000
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N T
A_1 B_1
:
A_N B_N
Output
Print the maximum possible happiness Takahashi can achieve.
Examples
Input
2 60
10 10
100 100
Output
110
Input
3 60
10 10
10 20
10 30
Output
60
Input
3 60
30 10
30 20
30 30
Output
50
Input
10 100
15 23
20 18
13 17
24 12
18 29
19 27
23 21
18 20
27 15
22 25
Output
145
"Correct Solution:
```
N, T = map(int, input().split())
AB = [list(map(int, input().split())) for _ in range(N)]
dp = [[0] * (T+1) for _ in range(N+1)]
AB.sort()
ret = 0
for i in range(1,N+1):
for j in range(1,T+1):
if j < AB[i-1][0]:
dp[i][j] = dp[i-1][j]
else:
dp[i][j] = max(dp[i-1][j-AB[i-1][0]] + AB[i-1][1], dp[i-1][j])
m = 0
for i in range(N):
m = max(m, AB[i][1]+dp[i][T-1])
print(m)
```
| 6,397 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi is at an all-you-can-eat restaurant.
The restaurant offers N kinds of dishes. It takes A_i minutes to eat the i-th dish, whose deliciousness is B_i.
The restaurant has the following rules:
* You can only order one dish at a time. The dish ordered will be immediately served and ready to eat.
* You cannot order the same kind of dish more than once.
* Until you finish eating the dish already served, you cannot order a new dish.
* After T-0.5 minutes from the first order, you can no longer place a new order, but you can continue eating the dish already served.
Let Takahashi's happiness be the sum of the deliciousness of the dishes he eats in this restaurant.
What is the maximum possible happiness achieved by making optimal choices?
Constraints
* 2 \leq N \leq 3000
* 1 \leq T \leq 3000
* 1 \leq A_i \leq 3000
* 1 \leq B_i \leq 3000
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N T
A_1 B_1
:
A_N B_N
Output
Print the maximum possible happiness Takahashi can achieve.
Examples
Input
2 60
10 10
100 100
Output
110
Input
3 60
10 10
10 20
10 30
Output
60
Input
3 60
30 10
30 20
30 30
Output
50
Input
10 100
15 23
20 18
13 17
24 12
18 29
19 27
23 21
18 20
27 15
22 25
Output
145
"Correct Solution:
```
N,M=map(int,input().split())
AB=[]
for i in range(N):
AB.append(list(map(int,input().split())))
AB.sort(key=lambda x: x[0])
max_=0
for i in AB:
if max_<i[0]:
max_=i[0]
dp=[0]*(max_+M)
max_=0
for i in AB:
for j in range(M)[::-1]:
dp[j+i[0]] = max(dp[j+i[0]],dp[j]+i[1])
print(max(dp))
```
| 6,398 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi is at an all-you-can-eat restaurant.
The restaurant offers N kinds of dishes. It takes A_i minutes to eat the i-th dish, whose deliciousness is B_i.
The restaurant has the following rules:
* You can only order one dish at a time. The dish ordered will be immediately served and ready to eat.
* You cannot order the same kind of dish more than once.
* Until you finish eating the dish already served, you cannot order a new dish.
* After T-0.5 minutes from the first order, you can no longer place a new order, but you can continue eating the dish already served.
Let Takahashi's happiness be the sum of the deliciousness of the dishes he eats in this restaurant.
What is the maximum possible happiness achieved by making optimal choices?
Constraints
* 2 \leq N \leq 3000
* 1 \leq T \leq 3000
* 1 \leq A_i \leq 3000
* 1 \leq B_i \leq 3000
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N T
A_1 B_1
:
A_N B_N
Output
Print the maximum possible happiness Takahashi can achieve.
Examples
Input
2 60
10 10
100 100
Output
110
Input
3 60
10 10
10 20
10 30
Output
60
Input
3 60
30 10
30 20
30 30
Output
50
Input
10 100
15 23
20 18
13 17
24 12
18 29
19 27
23 21
18 20
27 15
22 25
Output
145
"Correct Solution:
```
n, W = map(int, input().split())
ab = [tuple(map(int, input().split()))for _ in range(n)]
ab.sort(reverse=True)
dp = [0]*W
for w, v in ab:
for j in reversed(range(W)):
if j-w < 0:
break
if dp[j] < dp[j-w]+v:
dp[j] = dp[j-w]+v
if dp[0] < v:
dp[0] = v
print(max(dp))
```
| 6,399 |
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