text stringlengths 216 39.6k | conversation_id int64 219 108k | embedding list | cluster int64 11 11 |
|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Create a program that inputs the test result data of the visual acuity test and outputs the number of people who apply to each judgment based on the following visual acuity judgment table for each of the left and right eyesight.
Judgment | Sight
--- | ---
A | 1.1 or above
B | 0.6 or more and less than 1.1
C | 0.2 or more and less than 0.6
D | less than 0.2
Input
The input is given in the following format:
l1 r1
l2 r2
l3 r3
::
::
On line i, the real number li, which represents the left visual acuity of the i-th person, and the real number ri, which represents the right visual acuity, are given, separated by blanks. However, visual acuity is 0.1 or more and 2.0 or less, and is given in 0.1 increments.
The number of lines of input does not exceed 40.
Output
Please output the judgment table in the following format.
Line 1 Left eyesight is the number of people A Right eyesight is the number of people (separated by blanks)
2nd line The number of people whose left eyesight is B The number of people whose right eyesight is B (separated by blanks)
3rd line The number of people whose left eyesight is C
4th line Number of people with left eyesight D Number of people with right eyesight D (separated by blanks)
Example
Input
1.0 1.2
0.8 1.5
1.2 0.7
2.0 2.0
Output
2 3
2 1
0 0
0 0
Submitted Solution:
```
# -*- coding: utf-8 -*-
import sys
import os
A = [0, 0]
B = [0, 0]
C = [0, 0]
D = [0, 0]
for s in sys.stdin:
l, r = map(float, s.split())
if l >= 1.1:
A[0] += 1
elif l >= 0.6:
B[0] += 1
elif l >= 0.2:
C[0] += 1
else:
D[0] += 1
if r >= 1.1:
A[1] += 1
elif l >= 0.6:
B[1] += 1
elif l >= 0.2:
C[1] += 1
else:
D[1] += 1
print(*A)
print(*B)
print(*C)
print(*D)
```
No
| 32,414 | [
0.29150390625,
-0.18017578125,
-0.06597900390625,
0.3359375,
-1.2373046875,
-0.513671875,
0.053192138671875,
0.347412109375,
0.12127685546875,
0.7109375,
0.2176513671875,
-0.53515625,
0.3623046875,
-0.27392578125,
-0.09112548828125,
-0.06646728515625,
-0.693359375,
-0.556640625,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Create a program that inputs the test result data of the visual acuity test and outputs the number of people who apply to each judgment based on the following visual acuity judgment table for each of the left and right eyesight.
Judgment | Sight
--- | ---
A | 1.1 or above
B | 0.6 or more and less than 1.1
C | 0.2 or more and less than 0.6
D | less than 0.2
Input
The input is given in the following format:
l1 r1
l2 r2
l3 r3
::
::
On line i, the real number li, which represents the left visual acuity of the i-th person, and the real number ri, which represents the right visual acuity, are given, separated by blanks. However, visual acuity is 0.1 or more and 2.0 or less, and is given in 0.1 increments.
The number of lines of input does not exceed 40.
Output
Please output the judgment table in the following format.
Line 1 Left eyesight is the number of people A Right eyesight is the number of people (separated by blanks)
2nd line The number of people whose left eyesight is B The number of people whose right eyesight is B (separated by blanks)
3rd line The number of people whose left eyesight is C
4th line Number of people with left eyesight D Number of people with right eyesight D (separated by blanks)
Example
Input
1.0 1.2
0.8 1.5
1.2 0.7
2.0 2.0
Output
2 3
2 1
0 0
0 0
Submitted Solution:
```
lvl = [1.1, 0.6, 0.2, 0.0]
rst = [[0]*4 for _ in range(2)]
eyes = []
while True:
try:
l,r = map(float,input().strip().split())
eyes.append([l,r])
except EOFError:
break
for e in eyes:
for lr in range(2):
for l in range(len(lvl)):
if e[lr]>=lvl[l]:
rst[lr][l] += 1
break
else:
raise ValueError('Do not come here')
for lv in range(len(lvl)):
print(rst[0][lv]," ",rst[1][lv])
```
No
| 32,415 | [
0.2459716796875,
-0.1351318359375,
-0.058258056640625,
0.348388671875,
-1.2119140625,
-0.52880859375,
0.1043701171875,
0.302490234375,
0.190673828125,
0.716796875,
0.184814453125,
-0.51953125,
0.368408203125,
-0.3369140625,
-0.0743408203125,
-0.1007080078125,
-0.6943359375,
-0.5820... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Create a program that inputs the test result data of the visual acuity test and outputs the number of people who apply to each judgment based on the following visual acuity judgment table for each of the left and right eyesight.
Judgment | Sight
--- | ---
A | 1.1 or above
B | 0.6 or more and less than 1.1
C | 0.2 or more and less than 0.6
D | less than 0.2
Input
The input is given in the following format:
l1 r1
l2 r2
l3 r3
::
::
On line i, the real number li, which represents the left visual acuity of the i-th person, and the real number ri, which represents the right visual acuity, are given, separated by blanks. However, visual acuity is 0.1 or more and 2.0 or less, and is given in 0.1 increments.
The number of lines of input does not exceed 40.
Output
Please output the judgment table in the following format.
Line 1 Left eyesight is the number of people A Right eyesight is the number of people (separated by blanks)
2nd line The number of people whose left eyesight is B The number of people whose right eyesight is B (separated by blanks)
3rd line The number of people whose left eyesight is C
4th line Number of people with left eyesight D Number of people with right eyesight D (separated by blanks)
Example
Input
1.0 1.2
0.8 1.5
1.2 0.7
2.0 2.0
Output
2 3
2 1
0 0
0 0
Submitted Solution:
```
ld, rd = [0, 0, 0, 0]. [0, 0, 0, 0]
while True:
try:
l, r = map(float, input().split())
if l < 0.2:
ld[3] += 1
elif l < 0.6:
ld[2] += 1
elif l < 1.1:
ld[1] += 1
else:
ld[0] += 1
if r < 0.2:
rd[3] += 1
elif r < 0.6:
rd[2] += 1
elif r < 1.1:
rd[1] += 1
else:
rd[0] += 1
except:
break
for l, r in zip(ld, rd):
print(l, r)
```
No
| 32,416 | [
0.2685546875,
-0.130615234375,
-0.059906005859375,
0.378662109375,
-1.248046875,
-0.5,
0.040496826171875,
0.371826171875,
0.1683349609375,
0.68701171875,
0.2227783203125,
-0.501953125,
0.3662109375,
-0.275634765625,
-0.1417236328125,
-0.031494140625,
-0.68408203125,
-0.5625,
-0.3... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
Currently, people's entertainment is limited to programming contests. The activity of the entertainment club of a junior high school to which she belongs is to plan and run a programming contest. Her job is not to create problems. It's a behind-the-scenes job of soliciting issues from many, organizing referees, and promoting the contest. Unlike charismatic authors and prominent algorithms, people who do such work rarely get the light. She took pride in her work, which had no presence but was indispensable.
The entertainment department is always looking for problems, and these problems are classified into the following six types.
* Math
* Greedy
* Geometry
* DP
* Graph
* Other
Fortunately, there were so many problems that she decided to hold a lot of contests. The contest consists of three questions, but she decided to hold four types of contests to make the contest more educational.
1. Math Game Contest: A set of 3 questions including Math and DP questions
2. Algorithm Game Contest: A set of 3 questions, including Greedy questions and Graph questions.
3. Implementation Game Contest: A set of 3 questions including Geometry questions and Other questions
4. Well-balanced contest: 1 question from Math or DP, 1 question from Greedy or Graph, 1 question from Geometry or Other, a total of 3 question sets
Of course, questions in one contest cannot be asked in another. Her hope is to hold as many contests as possible. I know the stock numbers for the six questions, but how many times can I hold a contest? This is a difficult problem for her, but as a charismatic algorithm, you should be able to solve it.
Input
The input consists of multiple cases. Each case is given in the following format.
nMath nGreedy nGeometry nDP nGraph nOther
The value of each input represents the number of stocks of each type of problem.
The end of input
0 0 0 0 0 0
Given by a line consisting of.
Each value satisfies the following conditions.
nMath + nGreedy + nGeometry + nDP + nGraph + nOther ≤ 100,000,000
The number of test cases does not exceed 20,000.
Output
Output the maximum number of contests that can be held on one line.
Example
Input
1 1 1 1 1 1
1 1 1 0 0 0
1 0 0 0 1 1
3 0 0 3 0 0
3 1 0 1 3 1
1 2 0 2 0 1
0 0 1 1 0 3
1 0 0 1 1 0
0 0 0 0 0 0
Output
2
1
1
2
3
1
1
0
"Correct Solution:
```
while True:
ns = list(map(int, input().split()))
if not any(ns):
break
a = ns[0] + ns[3]
b = ns[1] + ns[4]
c = ns[2] + ns[5]
av, ar = a // 3, a % 3
bv, br = b // 3, b % 3
cv, cr = c // 3, c % 3
if max(ar, br, cr) == 0:
print(av + bv + cv)
elif max(ar, br, cr) == 1:
print(av + bv + cv + min(ar, br, cr))
elif max(ar, br, cr) == 2:
if [ar, br, cr].count(2) == 3:
print(av + bv + cv + 2)
elif [ar, br, cr].count(2) == 2:
for xr, xv in ((ar, av), (br, bv), (cr, cv)):
if xr == 2:
continue
if xv == 0:
print(av + bv + cv + xr)
else:
print(av + bv + cv + 1)
break
else:
print(av + bv + cv + min(ar, br, cr))
```
| 32,443 | [
0.5625,
0.141357421875,
-0.179443359375,
0.08734130859375,
-0.62646484375,
-0.5498046875,
0.0134735107421875,
0.38916015625,
0.2578125,
0.76171875,
0.2083740234375,
-0.272216796875,
0.302978515625,
-0.6357421875,
-0.447998046875,
0.11370849609375,
-0.8564453125,
-1.1044921875,
-0... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
Currently, people's entertainment is limited to programming contests. The activity of the entertainment club of a junior high school to which she belongs is to plan and run a programming contest. Her job is not to create problems. It's a behind-the-scenes job of soliciting issues from many, organizing referees, and promoting the contest. Unlike charismatic authors and prominent algorithms, people who do such work rarely get the light. She took pride in her work, which had no presence but was indispensable.
The entertainment department is always looking for problems, and these problems are classified into the following six types.
* Math
* Greedy
* Geometry
* DP
* Graph
* Other
Fortunately, there were so many problems that she decided to hold a lot of contests. The contest consists of three questions, but she decided to hold four types of contests to make the contest more educational.
1. Math Game Contest: A set of 3 questions including Math and DP questions
2. Algorithm Game Contest: A set of 3 questions, including Greedy questions and Graph questions.
3. Implementation Game Contest: A set of 3 questions including Geometry questions and Other questions
4. Well-balanced contest: 1 question from Math or DP, 1 question from Greedy or Graph, 1 question from Geometry or Other, a total of 3 question sets
Of course, questions in one contest cannot be asked in another. Her hope is to hold as many contests as possible. I know the stock numbers for the six questions, but how many times can I hold a contest? This is a difficult problem for her, but as a charismatic algorithm, you should be able to solve it.
Input
The input consists of multiple cases. Each case is given in the following format.
nMath nGreedy nGeometry nDP nGraph nOther
The value of each input represents the number of stocks of each type of problem.
The end of input
0 0 0 0 0 0
Given by a line consisting of.
Each value satisfies the following conditions.
nMath + nGreedy + nGeometry + nDP + nGraph + nOther ≤ 100,000,000
The number of test cases does not exceed 20,000.
Output
Output the maximum number of contests that can be held on one line.
Example
Input
1 1 1 1 1 1
1 1 1 0 0 0
1 0 0 0 1 1
3 0 0 3 0 0
3 1 0 1 3 1
1 2 0 2 0 1
0 0 1 1 0 3
1 0 0 1 1 0
0 0 0 0 0 0
Output
2
1
1
2
3
1
1
0
"Correct Solution:
```
# AOJ 1077: The Great Summer Contest
# Python3 2018.7.10 bal4u
while True:
n, k, f = [0]*3, [0]*3, [0]*3
a = list(map(int, input().split()))
if sum(a) == 0: break
for i in range(3): n[i] += a[i]+a[3+i]
ans = 0
for i in range(3):
if n[i] >= 3: f[i] = 1
ans += n[i]//3
n[i] %= 3
k[n[i]] += 1
if k[0] > 0:
if k[2] == 2:
i = 0
while n[i]: i += 1
if f[i]: ans += 1
elif k[1] > 0: ans += 1
elif k[2] > 0: ans += 2
print(ans)
```
| 32,444 | [
0.5625,
0.141357421875,
-0.179443359375,
0.08734130859375,
-0.62646484375,
-0.5498046875,
0.0134735107421875,
0.38916015625,
0.2578125,
0.76171875,
0.2083740234375,
-0.272216796875,
0.302978515625,
-0.6357421875,
-0.447998046875,
0.11370849609375,
-0.8564453125,
-1.1044921875,
-0... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Currently, people's entertainment is limited to programming contests. The activity of the entertainment club of a junior high school to which she belongs is to plan and run a programming contest. Her job is not to create problems. It's a behind-the-scenes job of soliciting issues from many, organizing referees, and promoting the contest. Unlike charismatic authors and prominent algorithms, people who do such work rarely get the light. She took pride in her work, which had no presence but was indispensable.
The entertainment department is always looking for problems, and these problems are classified into the following six types.
* Math
* Greedy
* Geometry
* DP
* Graph
* Other
Fortunately, there were so many problems that she decided to hold a lot of contests. The contest consists of three questions, but she decided to hold four types of contests to make the contest more educational.
1. Math Game Contest: A set of 3 questions including Math and DP questions
2. Algorithm Game Contest: A set of 3 questions, including Greedy questions and Graph questions.
3. Implementation Game Contest: A set of 3 questions including Geometry questions and Other questions
4. Well-balanced contest: 1 question from Math or DP, 1 question from Greedy or Graph, 1 question from Geometry or Other, a total of 3 question sets
Of course, questions in one contest cannot be asked in another. Her hope is to hold as many contests as possible. I know the stock numbers for the six questions, but how many times can I hold a contest? This is a difficult problem for her, but as a charismatic algorithm, you should be able to solve it.
Input
The input consists of multiple cases. Each case is given in the following format.
nMath nGreedy nGeometry nDP nGraph nOther
The value of each input represents the number of stocks of each type of problem.
The end of input
0 0 0 0 0 0
Given by a line consisting of.
Each value satisfies the following conditions.
nMath + nGreedy + nGeometry + nDP + nGraph + nOther ≤ 100,000,000
The number of test cases does not exceed 20,000.
Output
Output the maximum number of contests that can be held on one line.
Example
Input
1 1 1 1 1 1
1 1 1 0 0 0
1 0 0 0 1 1
3 0 0 3 0 0
3 1 0 1 3 1
1 2 0 2 0 1
0 0 1 1 0 3
1 0 0 1 1 0
0 0 0 0 0 0
Output
2
1
1
2
3
1
1
0
Submitted Solution:
```
# AOJ 1077: The Great Summer Contest
# Python3 2018.7.10 bal4u
while True:
n, k, f = [0]*3, [0]*3, [0]*3
a = list(map(int, input().split()))
if sum(a) == 0: break
for i in range(3): n[i] += a[i]+a[3+i]
ans = 0
for i in range(3):
if n[i] >= 3: f[i] = 1
ans += n[i]//3
n[i] %= 3
k[n[i]] += 1
if k[0] > 0:
if k[2] == 2:
for i in range(3):
if n[i] == 0: break
if f[i]: ans += 1
elif k[1] > 0: ans += 1
elif k[2] > 0: ans += 2
print(ans)
```
No
| 32,445 | [
0.56591796875,
0.0643310546875,
-0.1746826171875,
0.130615234375,
-0.55078125,
-0.489013671875,
-0.08905029296875,
0.457275390625,
0.1890869140625,
0.7587890625,
0.2435302734375,
-0.2376708984375,
0.257080078125,
-0.6474609375,
-0.466064453125,
-0.034912109375,
-0.86474609375,
-0.9... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Example
Input
2
10 10
Output
40.00000000
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 998244353
dd = [(0,-1),(1,0),(0,1),(-1,0)]
ddn = [(0,-1),(1,-1),(1,0),(1,1),(0,1),(-1,-1),(-1,0),(-1,1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
rr = []
while True:
n = I()
a = LI()
t = [a[0]]
for c in a[1:]:
b = t[-1]
if b <= c:
t.append(c)
continue
tt = b - c
k = ((b+c) ** 2 - tt**2) ** 0.5
if k > b:
t.append(c)
a = t[::-1]
t = [a[0]]
for c in a[1:]:
b = t[-1]
if b <= c:
t.append(c)
continue
tt = b - c
k = ((b+c) ** 2 - tt**2) ** 0.5
if k > b:
t.append(c)
a = t[::-1]
l = len(a)
if l == 1:
rr.append(a[0]*2)
break
r = a[0] + a[-1]
for i in range(l-1):
b = a[i]
c = a[i+1]
if b == c:
r += b + c
continue
t = abs(b-c)
r += ((b+c) ** 2 - t**2) ** 0.5
rr.append('{:00.9f}'.format(r))
break
return '\n'.join(map(str, rr))
print(main())
```
No
| 32,448 | [
0.33642578125,
0.060638427734375,
0.0389404296875,
0.0007686614990234375,
-0.93994140625,
-0.1978759765625,
-0.2578125,
0.1519775390625,
0.141357421875,
1.2314453125,
0.26708984375,
-0.1649169921875,
0.061614990234375,
-0.5126953125,
-0.583984375,
-0.039520263671875,
-0.6982421875,
... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
Problem Statement
Nathan O. Davis is a student at the department of integrated systems.
Today's agenda in the class is audio signal processing. Nathan was given a lot of homework out. One of the homework was to write a program to process an audio signal. He copied the given audio signal to his USB memory and brought it back to his home.
When he started his homework, he unfortunately dropped the USB memory to the floor. He checked the contents of the USB memory and found that the audio signal data got broken.
There are several characteristics in the audio signal that he copied.
* The audio signal is a sequence of $N$ samples.
* Each sample in the audio signal is numbered from $1$ to $N$ and represented as an integer value.
* Each value of the odd-numbered sample(s) is strictly smaller than the value(s) of its neighboring sample(s).
* Each value of the even-numbered sample(s) is strictly larger than the value(s) of its neighboring sample(s).
He got into a panic and asked you for a help. You tried to recover the audio signal from his USB memory but some samples of the audio signal are broken and could not be recovered. Fortunately, you found from the metadata that all the broken samples have the same integer value.
Your task is to write a program, which takes the broken audio signal extracted from his USB memory as its input, to detect whether the audio signal can be recovered uniquely.
Input
The input consists of multiple datasets. The form of each dataset is described below.
> $N$
> $a_{1}$ $a_{2}$ ... $a_{N}$
The first line of each dataset consists of an integer, $N (2 \le N \le 1{,}000)$. $N$ denotes the number of samples in the given audio signal. The second line of each dataset consists of $N$ values separated by spaces. The $i$-th value, $a_{i}$, is either a character `x` or an integer between $-10^9$ and $10^9$, inclusive. It represents the $i$-th sample of the broken audio signal. If $a_{i}$ is a character `x` , it denotes that $i$-th sample in the audio signal is broken. Otherwise it denotes the value of the $i$-th sample.
The end of input is indicated by a single $0$. This is not included in the datasets.
You may assume that the number of the datasets does not exceed $100$.
Output
For each dataset, output the value of the broken samples in one line if the original audio signal can be recovered uniquely. If there are multiple possible values, output `ambiguous`. If there are no possible values, output `none`.
Sample Input
5
1 x 2 4 x
2
x x
2
1 2
2
2 1
2
1000000000 x
4
x 2 1 x
0
Output for the Sample Input
3
none
ambiguous
none
ambiguous
none
Example
Input
5
1 x 2 4 x
2
x x
2
1 2
2
2 1
2
1000000000 x
4
x 2 1 x
0
Output
3
none
ambiguous
none
ambiguous
none
"Correct Solution:
```
INF = float('inf')
def judge(src):
x_max = INF
x_min = -INF
for i,(s1,s2) in enumerate(zip(src,src[1:])):
if s1 == s2:
return 'none'
if i%2:
s1,s2 = s2,s1
if s1 != 'x' and s2 != 'x':
if s1 < s2:
return 'none'
else:
continue
if s1 == 'x':
if x_max <= s2:
return 'none'
x_min = max(x_min, s2 + 1)
elif s2 == 'x':
if s1 <= x_min:
return 'none'
x_max = min(x_max, s1 - 1)
if x_max < x_min:
return 'none'
return x_max if x_max == x_min else 'ambiguous'
while True:
N = int(input())
if N == 0: break
src = [INF] + list(map(lambda x:x if x=='x' else int(x), input().split())) + [INF if N%2 else -INF]
print(judge(src))
```
| 32,454 | [
-0.03472900390625,
0.183349609375,
-0.0302581787109375,
0.26953125,
-0.59033203125,
-0.328369140625,
-0.34716796875,
-0.074462890625,
0.57421875,
0.662109375,
0.5546875,
-0.207763671875,
0.422607421875,
-0.88916015625,
-0.69970703125,
0.0298309326171875,
-0.58544921875,
-0.99316406... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
Problem Statement
Nathan O. Davis is a student at the department of integrated systems.
Today's agenda in the class is audio signal processing. Nathan was given a lot of homework out. One of the homework was to write a program to process an audio signal. He copied the given audio signal to his USB memory and brought it back to his home.
When he started his homework, he unfortunately dropped the USB memory to the floor. He checked the contents of the USB memory and found that the audio signal data got broken.
There are several characteristics in the audio signal that he copied.
* The audio signal is a sequence of $N$ samples.
* Each sample in the audio signal is numbered from $1$ to $N$ and represented as an integer value.
* Each value of the odd-numbered sample(s) is strictly smaller than the value(s) of its neighboring sample(s).
* Each value of the even-numbered sample(s) is strictly larger than the value(s) of its neighboring sample(s).
He got into a panic and asked you for a help. You tried to recover the audio signal from his USB memory but some samples of the audio signal are broken and could not be recovered. Fortunately, you found from the metadata that all the broken samples have the same integer value.
Your task is to write a program, which takes the broken audio signal extracted from his USB memory as its input, to detect whether the audio signal can be recovered uniquely.
Input
The input consists of multiple datasets. The form of each dataset is described below.
> $N$
> $a_{1}$ $a_{2}$ ... $a_{N}$
The first line of each dataset consists of an integer, $N (2 \le N \le 1{,}000)$. $N$ denotes the number of samples in the given audio signal. The second line of each dataset consists of $N$ values separated by spaces. The $i$-th value, $a_{i}$, is either a character `x` or an integer between $-10^9$ and $10^9$, inclusive. It represents the $i$-th sample of the broken audio signal. If $a_{i}$ is a character `x` , it denotes that $i$-th sample in the audio signal is broken. Otherwise it denotes the value of the $i$-th sample.
The end of input is indicated by a single $0$. This is not included in the datasets.
You may assume that the number of the datasets does not exceed $100$.
Output
For each dataset, output the value of the broken samples in one line if the original audio signal can be recovered uniquely. If there are multiple possible values, output `ambiguous`. If there are no possible values, output `none`.
Sample Input
5
1 x 2 4 x
2
x x
2
1 2
2
2 1
2
1000000000 x
4
x 2 1 x
0
Output for the Sample Input
3
none
ambiguous
none
ambiguous
none
Example
Input
5
1 x 2 4 x
2
x x
2
1 2
2
2 1
2
1000000000 x
4
x 2 1 x
0
Output
3
none
ambiguous
none
ambiguous
none
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 998244353
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
rr = []
while True:
n = I()
if n == 0:
break
a = LS()
r = None
for i in range(n):
if a[i] == 'x':
if i > 0 and a[i-1] == 'x':
r = 'none'
break
else:
a[i] = int(a[i])
if i > 0 and a[i-1] != 'x':
if i % 2 == 0:
if a[i-1] <= a[i]:
r = 'none'
break
elif a[i-1] >= a[i]:
r = 'none'
break
if r:
rr.append(r)
continue
ma = -inf
mi = inf
for i in range(n):
if a[i] != 'x':
continue
if i % 2 == 0:
if i > 0:
mi = min(mi, a[i-1]-1)
if i < n-1:
mi = min(mi, a[i+1]-1)
else:
ma = max(ma, a[i-1]+1)
if i < n-1:
ma = max(ma, a[i+1]+1)
if ma == mi:
rr.append(ma)
elif ma == -1 or mi == inf:
rr.append('ambiguous')
elif ma > mi:
rr.append('none')
else:
rr.append('ambiguous')
return '\n'.join(map(str, rr))
print(main())
```
| 32,455 | [
-0.03472900390625,
0.183349609375,
-0.0302581787109375,
0.26953125,
-0.59033203125,
-0.328369140625,
-0.34716796875,
-0.074462890625,
0.57421875,
0.662109375,
0.5546875,
-0.207763671875,
0.422607421875,
-0.88916015625,
-0.69970703125,
0.0298309326171875,
-0.58544921875,
-0.99316406... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem Statement
Nathan O. Davis is a student at the department of integrated systems.
Today's agenda in the class is audio signal processing. Nathan was given a lot of homework out. One of the homework was to write a program to process an audio signal. He copied the given audio signal to his USB memory and brought it back to his home.
When he started his homework, he unfortunately dropped the USB memory to the floor. He checked the contents of the USB memory and found that the audio signal data got broken.
There are several characteristics in the audio signal that he copied.
* The audio signal is a sequence of $N$ samples.
* Each sample in the audio signal is numbered from $1$ to $N$ and represented as an integer value.
* Each value of the odd-numbered sample(s) is strictly smaller than the value(s) of its neighboring sample(s).
* Each value of the even-numbered sample(s) is strictly larger than the value(s) of its neighboring sample(s).
He got into a panic and asked you for a help. You tried to recover the audio signal from his USB memory but some samples of the audio signal are broken and could not be recovered. Fortunately, you found from the metadata that all the broken samples have the same integer value.
Your task is to write a program, which takes the broken audio signal extracted from his USB memory as its input, to detect whether the audio signal can be recovered uniquely.
Input
The input consists of multiple datasets. The form of each dataset is described below.
> $N$
> $a_{1}$ $a_{2}$ ... $a_{N}$
The first line of each dataset consists of an integer, $N (2 \le N \le 1{,}000)$. $N$ denotes the number of samples in the given audio signal. The second line of each dataset consists of $N$ values separated by spaces. The $i$-th value, $a_{i}$, is either a character `x` or an integer between $-10^9$ and $10^9$, inclusive. It represents the $i$-th sample of the broken audio signal. If $a_{i}$ is a character `x` , it denotes that $i$-th sample in the audio signal is broken. Otherwise it denotes the value of the $i$-th sample.
The end of input is indicated by a single $0$. This is not included in the datasets.
You may assume that the number of the datasets does not exceed $100$.
Output
For each dataset, output the value of the broken samples in one line if the original audio signal can be recovered uniquely. If there are multiple possible values, output `ambiguous`. If there are no possible values, output `none`.
Sample Input
5
1 x 2 4 x
2
x x
2
1 2
2
2 1
2
1000000000 x
4
x 2 1 x
0
Output for the Sample Input
3
none
ambiguous
none
ambiguous
none
Example
Input
5
1 x 2 4 x
2
x x
2
1 2
2
2 1
2
1000000000 x
4
x 2 1 x
0
Output
3
none
ambiguous
none
ambiguous
none
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 998244353
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
rr = []
while True:
n = I()
if n == 0:
break
a = LS()
r = None
for i in range(n):
if a[i] == 'x':
if i > 0 and a[i-1] == 'x':
r = 'none'
break
else:
a[i] = int(a[i])
if i > 0 and a[i-1] != 'x':
if i % 2 == 0:
if a[i-1] <= a[i]:
r = 'none'
break
elif a[i-1] >= a[i]:
r = 'none'
break
if r:
rr.append(r)
continue
ma = -1
mi = inf
for i in range(n):
if a[i] != 'x':
continue
if i % 2 == 0:
if i > 0:
mi = min(mi, a[i-1]-1)
if i < n-1:
mi = min(mi, a[i+1]-1)
else:
ma = max(ma, a[i-1]+1)
if i < n-1:
ma = max(ma, a[i+1]+1)
if ma == mi:
rr.append(ma)
elif ma == -1 or mi == inf:
rr.append('ambiguous')
elif ma > mi:
rr.append('none')
else:
rr.append('ambiguous')
return '\n'.join(map(str, rr))
print(main())
```
No
| 32,456 | [
-0.0673828125,
0.2030029296875,
-0.0394287109375,
0.26171875,
-0.6337890625,
-0.2744140625,
-0.377685546875,
-0.0116119384765625,
0.50341796875,
0.744140625,
0.5634765625,
-0.1761474609375,
0.31396484375,
-0.83984375,
-0.6298828125,
0.08843994140625,
-0.55224609375,
-0.900390625,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
Vasya and Vitya play a game. Vasya thought of two integers a and b from 1 to n and Vitya tries to guess them. Each round he tells Vasya two numbers x and y from 1 to n. If both x=a and y=b then Vitya wins. Else Vasya must say one of the three phrases:
1. x is less than a;
2. y is less than b;
3. x is greater than a or y is greater than b.
Vasya can't lie, but if multiple phrases are true, he may choose any of them. For example, if Vasya thought of numbers 2 and 4, then he answers with the phrase 3 to a query (3, 4), and he can answer with the phrase 1 or phrase 3 to a query (1, 5).
Help Vitya win in no more than 600 rounds.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^{18}) — the upper limit of the numbers.
Interaction
First, you need to read the number n, after that you can make queries.
To make a query, print two integers: x and y (1 ≤ x, y ≤ n), then flush the output.
After each query, read a single integer ans (0 ≤ ans ≤ 3).
If ans > 0, then it is the number of the phrase said by Vasya.
If ans = 0, it means that you win and your program should terminate.
If you make more than 600 queries or make an incorrect query, you will get Wrong Answer.
Your solution will get Idleness Limit Exceeded, if you don't print anything or forget to flush the output.
To flush you need to do the following right after printing a query and a line end:
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* For other languages see documentation.
Hacks format
For hacks, use the following format:
In the first line, print a single integer n (1 ≤ n ≤ 10^{18}) — the upper limit of the numbers.
In the second line, print two integers a and b (1 ≤ a, b ≤ n) — the numbers which Vasya thought of.
In the third line, print a single integer m (1 ≤ m ≤ 10^5) — the number of instructions for the interactor.
In each of the next m lines, print five integers: x_i, y_i, r^{12}_i, r^{13}_i, and r^{23}_i (1 ≤ x_i, y_i ≤ n), where r^{ST}_i equals to either number S or number T.
While answering the query x y, the interactor finds a number i from 1 to n with the minimal value |x-x_i| + |y-y_i|. If multiple numbers can be chosen, the least i is preferred. Then the interactor answers to the query, but if there are two phrases S and T that can be given, then r^{ST}_i is chosen.
For example, the sample test data file contains the following:
5
2 4
2
2 5 1 1 2
4 1 2 3 3
Example
Input
5
3
3
2
1
0
Output
4 3
3 4
3 3
1 5
2 4
Note
Let's analyze the sample test. The chosen numbers are 2 and 4. The interactor was given two instructions.
For the query (4, 3), it can return 2 or 3. Out of the two instructions the second one is chosen, so the interactor returns a^{23}_2=3.
For the query (3, 4), it can return only 3.
For the query (3, 3), it can return 2 or 3. Out of the two instructions the first one is chosen (since in case of equal values, the least number is preferred), so the interactor returns a^{23}_1=2.
For the query (1, 5), it can return 1 or 3. Out of the two instructions the first one is chosen, so the interactor returns a^{13}_1=1.
In the fifth query (2, 4), the numbers are guessed correctly, the player wins.
Submitted Solution:
```
n = int(input())
a = [(1, 1, n + 1, n + 1)]
b = [(1, 1, n + 1, n + 1)]
while True:
sumarea = sum((x2 - x1) * (y2 - y1) for x1, y1, x2, y2 in a)
target = sumarea // 3
sumarea = 0
for x1, y1, x2, y2 in a:
x = x1 + (target - sumarea) // (y2 - y1)
if x <= x2:
break
sumarea += (x2 - x1) * (y2 - y1)
sumarea = 0
for x1, y1, x2, y2 in b:
y = y1 + (target - sumarea) // (x2 - x1)
if y <= y2:
break
sumarea += (x2 - x1) * (y2 - y1)
print(x, y, flush=True)
ans = int(input())
if ans == 0:
break
elif ans == 1:
a = [(x + 1, y1, x2, y2) for x1, y1, x2, y2 in a
if x + 1 < x2]
b = [(x + 1, y1, x2, y2) for x1, y1, x2, y2 in b
if x + 1 < x2]
elif ans == 2:
a = [(x1, y + 1, x2, y2) for x1, y1, x2, y2 in a
if y + 1 < y2]
b = [(x1, y + 1, x2, y2) for x1, y1, x2, y2 in b
if y + 1 < y2]
elif ans == 3:
a = [z for x1, y1, x2, y2 in a for z in
(((x1, y1, x2, y2),) if x2 <= x else
((x1, y1, x2, y),) if x <= x1 else
((x1, y1, x, y2), (x, y1, x2, y)))]
b = [z for x1, y1, x2, y2 in b for z in
(((x1, y1, x2, y2),) if y2 <= y else
((x1, y1, x, y2),) if y <= y1 else
((x1, y1, x2, y), (x1, y, x, y2)))]
```
No
| 33,319 | [
0.48583984375,
0.082763671875,
-0.165771484375,
0.060211181640625,
-0.70849609375,
-0.400390625,
-0.0384521484375,
0.12237548828125,
-0.054443359375,
0.87646484375,
0.22119140625,
0.03985595703125,
0.03643798828125,
-0.69677734375,
-0.28076171875,
-0.228759765625,
-0.513671875,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You survived several months of exam wars and were able to enter ICPC University on a sunny day. On the day of admission, there was an enthusiastic recruitment of circles on the campus of the university, and you received a large number of pamphlets and returned. When you return to your room, you find one of the pamphlets you received that interests you. The pamphlet was given by the university's public relations department.
The following problems were described in the pamphlet.
> How many combinations of two or more consecutive positive integers have a sum of N? For example, 9 has two combinations, 2 + 3 + 4 and 4 + 5.
If you were curious about the answer to this question, you decided to write a program to find out the answer. Therefore, your job is to write a program that outputs the answer to the problem for the positive integer N given as the input.
Input
The input is a sequence of datasets. Each dataset is a line of one integer N. Where 1 ≤ N ≤ 1000.
The end of the input is indicated by a single line of zeros.
Output
The output is the order of the input datasets, which is the answer to the question for the positive integers represented by each dataset of the inputs. No other characters should be in the output.
Example
Input
9
500
0
Output
2
3
Submitted Solution:
```
import math
while True:
n = int(input())
if n == 0:break
c = 0
for x in reversed(range(1,math.ceil(n/2)+1)):
sumx = 0
add_c = 0
for y in reversed(range(1,x+1)):
sumx += y
add_c += 1
if sumx >= n:
break
if sumx == n and add_c > 1:
c += 1
print(c)
```
Yes
| 34,215 | [
0.580078125,
-0.0462646484375,
0.01479339599609375,
-0.08734130859375,
-0.1864013671875,
-0.342529296875,
-0.1953125,
0.218505859375,
0.2122802734375,
0.97021484375,
0.5498046875,
-0.00481414794921875,
0.09930419921875,
-0.62109375,
-0.5166015625,
-0.30078125,
-0.5322265625,
-0.994... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You survived several months of exam wars and were able to enter ICPC University on a sunny day. On the day of admission, there was an enthusiastic recruitment of circles on the campus of the university, and you received a large number of pamphlets and returned. When you return to your room, you find one of the pamphlets you received that interests you. The pamphlet was given by the university's public relations department.
The following problems were described in the pamphlet.
> How many combinations of two or more consecutive positive integers have a sum of N? For example, 9 has two combinations, 2 + 3 + 4 and 4 + 5.
If you were curious about the answer to this question, you decided to write a program to find out the answer. Therefore, your job is to write a program that outputs the answer to the problem for the positive integer N given as the input.
Input
The input is a sequence of datasets. Each dataset is a line of one integer N. Where 1 ≤ N ≤ 1000.
The end of the input is indicated by a single line of zeros.
Output
The output is the order of the input datasets, which is the answer to the question for the positive integers represented by each dataset of the inputs. No other characters should be in the output.
Example
Input
9
500
0
Output
2
3
Submitted Solution:
```
while True:
n=int(input())
p=0
if n==0:
break
else:
s=[i for i in range(1,n)]
for i in range(2,n):
for j in range(0,n-i+1):
ss=sum(s[j:i:])
if ss==n:
p+=1
print(p)
```
Yes
| 34,216 | [
0.5380859375,
-0.0275726318359375,
0.0701904296875,
-0.12054443359375,
-0.41748046875,
-0.318115234375,
-0.329345703125,
0.3203125,
0.125732421875,
0.79150390625,
0.48583984375,
-0.10595703125,
0.07073974609375,
-0.5947265625,
-0.496826171875,
-0.2171630859375,
-0.57861328125,
-0.7... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You survived several months of exam wars and were able to enter ICPC University on a sunny day. On the day of admission, there was an enthusiastic recruitment of circles on the campus of the university, and you received a large number of pamphlets and returned. When you return to your room, you find one of the pamphlets you received that interests you. The pamphlet was given by the university's public relations department.
The following problems were described in the pamphlet.
> How many combinations of two or more consecutive positive integers have a sum of N? For example, 9 has two combinations, 2 + 3 + 4 and 4 + 5.
If you were curious about the answer to this question, you decided to write a program to find out the answer. Therefore, your job is to write a program that outputs the answer to the problem for the positive integer N given as the input.
Input
The input is a sequence of datasets. Each dataset is a line of one integer N. Where 1 ≤ N ≤ 1000.
The end of the input is indicated by a single line of zeros.
Output
The output is the order of the input datasets, which is the answer to the question for the positive integers represented by each dataset of the inputs. No other characters should be in the output.
Example
Input
9
500
0
Output
2
3
Submitted Solution:
```
while True:
N=int(input())
if(N==0):
break
ans=0
for i in range((N//2)+1,0,-1):
SUM=i
k=i-1
while SUM<=N and k>0:
SUM+=k
if SUM==N:
ans+=1
k-=1
print(ans)
```
Yes
| 34,217 | [
0.6025390625,
-0.06439208984375,
0.0063018798828125,
-0.14599609375,
-0.348876953125,
-0.402099609375,
-0.2081298828125,
0.267578125,
0.1688232421875,
0.78466796875,
0.473876953125,
-0.10015869140625,
0.05059814453125,
-0.63671875,
-0.55810546875,
-0.1773681640625,
-0.53857421875,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You survived several months of exam wars and were able to enter ICPC University on a sunny day. On the day of admission, there was an enthusiastic recruitment of circles on the campus of the university, and you received a large number of pamphlets and returned. When you return to your room, you find one of the pamphlets you received that interests you. The pamphlet was given by the university's public relations department.
The following problems were described in the pamphlet.
> How many combinations of two or more consecutive positive integers have a sum of N? For example, 9 has two combinations, 2 + 3 + 4 and 4 + 5.
If you were curious about the answer to this question, you decided to write a program to find out the answer. Therefore, your job is to write a program that outputs the answer to the problem for the positive integer N given as the input.
Input
The input is a sequence of datasets. Each dataset is a line of one integer N. Where 1 ≤ N ≤ 1000.
The end of the input is indicated by a single line of zeros.
Output
The output is the order of the input datasets, which is the answer to the question for the positive integers represented by each dataset of the inputs. No other characters should be in the output.
Example
Input
9
500
0
Output
2
3
Submitted Solution:
```
while True:
n = int(input())
ans = 0
if n == 0:
break
count = n
for i in range(1, n):
a = 0
for j in range (i, n):
a += j
if a == n:
ans += 1
break
if a > n:
break
# for i in range(2, n):
# a = int(n / i)
# print("a=", a)
# b = a
# for j in range(0,i-1):
# a += 1
# b += a
# print("b=", b)
# if a == b:
# ans += 1
print(ans)
```
Yes
| 34,218 | [
0.5419921875,
-0.041412353515625,
0.04217529296875,
-0.10986328125,
-0.369140625,
-0.443359375,
-0.1531982421875,
0.27880859375,
0.2421875,
0.84765625,
0.3798828125,
-0.050323486328125,
0.03375244140625,
-0.6416015625,
-0.56396484375,
-0.22705078125,
-0.58544921875,
-0.82373046875,... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You survived several months of exam wars and were able to enter ICPC University on a sunny day. On the day of admission, there was an enthusiastic recruitment of circles on the campus of the university, and you received a large number of pamphlets and returned. When you return to your room, you find one of the pamphlets you received that interests you. The pamphlet was given by the university's public relations department.
The following problems were described in the pamphlet.
> How many combinations of two or more consecutive positive integers have a sum of N? For example, 9 has two combinations, 2 + 3 + 4 and 4 + 5.
If you were curious about the answer to this question, you decided to write a program to find out the answer. Therefore, your job is to write a program that outputs the answer to the problem for the positive integer N given as the input.
Input
The input is a sequence of datasets. Each dataset is a line of one integer N. Where 1 ≤ N ≤ 1000.
The end of the input is indicated by a single line of zeros.
Output
The output is the order of the input datasets, which is the answer to the question for the positive integers represented by each dataset of the inputs. No other characters should be in the output.
Example
Input
9
500
0
Output
2
3
Submitted Solution:
```
while True:
n = int(input())
if n == 0:
break
count = 0
for i in range(int((n+1)/2)):
s = 0
j = i
while s < n:
s += j
j += 1
if s == n:
count += 1
print(count)
```
No
| 34,219 | [
0.64013671875,
-0.053466796875,
0.00557708740234375,
-0.12481689453125,
-0.391845703125,
-0.3984375,
-0.1961669921875,
0.2685546875,
0.173583984375,
0.77978515625,
0.5,
-0.07818603515625,
0.049468994140625,
-0.607421875,
-0.5205078125,
-0.297119140625,
-0.5263671875,
-0.71240234375... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You survived several months of exam wars and were able to enter ICPC University on a sunny day. On the day of admission, there was an enthusiastic recruitment of circles on the campus of the university, and you received a large number of pamphlets and returned. When you return to your room, you find one of the pamphlets you received that interests you. The pamphlet was given by the university's public relations department.
The following problems were described in the pamphlet.
> How many combinations of two or more consecutive positive integers have a sum of N? For example, 9 has two combinations, 2 + 3 + 4 and 4 + 5.
If you were curious about the answer to this question, you decided to write a program to find out the answer. Therefore, your job is to write a program that outputs the answer to the problem for the positive integer N given as the input.
Input
The input is a sequence of datasets. Each dataset is a line of one integer N. Where 1 ≤ N ≤ 1000.
The end of the input is indicated by a single line of zeros.
Output
The output is the order of the input datasets, which is the answer to the question for the positive integers represented by each dataset of the inputs. No other characters should be in the output.
Example
Input
9
500
0
Output
2
3
Submitted Solution:
```
a=0
while True:
n=int(input())
if n==0: break
elif n%2!=0: a+=1
for i in range(3,n//2,2):
if n%i==0: a+=1
print(a)
```
No
| 34,220 | [
0.70458984375,
-0.10748291015625,
0.009246826171875,
-0.17578125,
-0.39453125,
-0.3759765625,
-0.12103271484375,
0.301025390625,
0.1744384765625,
0.8134765625,
0.485107421875,
-0.0775146484375,
-0.0087738037109375,
-0.576171875,
-0.544921875,
-0.29736328125,
-0.51806640625,
-0.7016... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You survived several months of exam wars and were able to enter ICPC University on a sunny day. On the day of admission, there was an enthusiastic recruitment of circles on the campus of the university, and you received a large number of pamphlets and returned. When you return to your room, you find one of the pamphlets you received that interests you. The pamphlet was given by the university's public relations department.
The following problems were described in the pamphlet.
> How many combinations of two or more consecutive positive integers have a sum of N? For example, 9 has two combinations, 2 + 3 + 4 and 4 + 5.
If you were curious about the answer to this question, you decided to write a program to find out the answer. Therefore, your job is to write a program that outputs the answer to the problem for the positive integer N given as the input.
Input
The input is a sequence of datasets. Each dataset is a line of one integer N. Where 1 ≤ N ≤ 1000.
The end of the input is indicated by a single line of zeros.
Output
The output is the order of the input datasets, which is the answer to the question for the positive integers represented by each dataset of the inputs. No other characters should be in the output.
Example
Input
9
500
0
Output
2
3
Submitted Solution:
```
if __name__ == "__main__":
while True:
n = int(input())
if n ==0:break
nline = [i for i in range(n)]
count = 0
for i in range(n):
for j in range(i,n):
if sum(nline[i:j]) > n:break
if sum(nline[i:j]) == n:count +=1
print(count)
```
No
| 34,221 | [
0.59130859375,
-0.0633544921875,
-0.054351806640625,
-0.142578125,
-0.396728515625,
-0.43701171875,
-0.10595703125,
0.31396484375,
0.1904296875,
0.87451171875,
0.442626953125,
-0.11016845703125,
0.033843994140625,
-0.60302734375,
-0.61181640625,
-0.1968994140625,
-0.65283203125,
-0... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You survived several months of exam wars and were able to enter ICPC University on a sunny day. On the day of admission, there was an enthusiastic recruitment of circles on the campus of the university, and you received a large number of pamphlets and returned. When you return to your room, you find one of the pamphlets you received that interests you. The pamphlet was given by the university's public relations department.
The following problems were described in the pamphlet.
> How many combinations of two or more consecutive positive integers have a sum of N? For example, 9 has two combinations, 2 + 3 + 4 and 4 + 5.
If you were curious about the answer to this question, you decided to write a program to find out the answer. Therefore, your job is to write a program that outputs the answer to the problem for the positive integer N given as the input.
Input
The input is a sequence of datasets. Each dataset is a line of one integer N. Where 1 ≤ N ≤ 1000.
The end of the input is indicated by a single line of zeros.
Output
The output is the order of the input datasets, which is the answer to the question for the positive integers represented by each dataset of the inputs. No other characters should be in the output.
Example
Input
9
500
0
Output
2
3
Submitted Solution:
```
#include <bits/stdc++.h>
int main(void) {
int n, i, j, cnt, tmp;
while (scanf("%d", &n)) {
if (n == 0) {
break;
}
cnt = 0;
for (i = 1; i < n / 2 + 1; i++) {
for (j = i; j < n; j++) {
tmp = (i + j) * (j - i + 1) / 2;
if (tmp == n) {
cnt++;
break;
}
if (tmp > n) {
break;
}
}
}
printf("%d\n", cnt);
}
return 0;
}
```
No
| 34,222 | [
0.464111328125,
-0.10418701171875,
0.09246826171875,
-0.1097412109375,
-0.351318359375,
-0.294921875,
-0.16796875,
0.298095703125,
0.10272216796875,
0.6865234375,
0.38427734375,
-0.083740234375,
0.10723876953125,
-0.61279296875,
-0.6298828125,
-0.1982421875,
-0.56591796875,
-0.7924... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
Example
Input
2 2
1 2 0
3 4 1
Output
2
"Correct Solution:
```
from itertools import product
from collections import deque
import sys
readline = sys.stdin.readline
write = sys.stdout.write
def solve():
MOD = 10**9 + 9
N, M = map(int, readline().split())
K = 3*N
fact = [1]*(K+1)
rfact = [1]*(K+1)
r = 1
for i in range(1, K+1):
fact[i] = r = r * i % MOD
rfact[K] = r = pow(fact[K], MOD-2, MOD)
for i in range(K, 0, -1):
rfact[i-1] = r = r * i % MOD
pr6 = [1]*(N+1)
base = pow(6, MOD-2, MOD)
r = 1
for i in range(N):
pr6[i+1] = r = base * r % MOD
mp = {}
G = [[] for i in range(M*2)]
cur = 0
m = 0
E = []
for i in range(M):
a, b, c = map(int, readline().split())
if a not in mp:
mp[a] = cur
cur += 1
ma = mp[a]
if b not in mp:
mp[b] = cur
cur += 1
mb = mp[b]
if c == 0:
G[ma].append(mb)
G[mb].append(ma)
else:
E.append((ma, mb))
m += 1
L = cur
cr = 0
lb = [-1]*L
sz = []
zz = []
cc = [0, 0, 0]
u = [0]*L
que = deque()
for i in range(L):
if u[i]:
continue
que.append(i)
vs = []
u[i] = 1
while que:
v = que.popleft()
vs.append(v)
for w in G[v]:
if u[w]:
continue
u[w] = 1
que.append(w)
s = len(vs)
if s > 3:
write("0\n")
return
for v in vs:
lb[v] = cr
sz.append(s)
zz.append(vs)
cc[s-1] += 1
cr += 1
used = [0]*(1 << m)
ans = 0
def dfs(state, c, lb, cc):
nonlocal ans
if used[state]:
return
used[state] = 1
x = cc[0] + (K - L); y = cc[1]
if x >= y:
k = (x - y) // 3
if c & 1 == 0:
ans += fact[x] * pr6[k] * rfact[k] % MOD
else:
ans -= fact[x] * pr6[k] * rfact[k] % MOD
cc0 = [0]*3
for i in range(m):
if state & (1 << i):
continue
p, q = E[i]
pl = lb[p]; ql = lb[q]
if pl != ql:
s = sz[pl] + sz[ql]
if s > 3:
continue
cc0[:] = cc
cc0[sz[pl]-1] -= 1
cc0[sz[ql]-1] -= 1
cc0[s-1] += 1
nl = len(sz)
vs = zz[pl] + zz[ql]
for v in vs:
lb[v] = nl
sz.append(s)
zz.append(vs)
dfs(state | (1 << i), c+1, lb, cc0)
for v in zz[pl]:
lb[v] = pl
for v in zz[ql]:
lb[v] = ql
else:
dfs(state | (1 << i), c+1, lb, cc)
dfs(0, 0, lb, cc)
ans %= MOD
write("%d\n" % ans)
solve()
```
| 34,223 | [
0.6328125,
-0.11871337890625,
0.050048828125,
0.050262451171875,
-0.404052734375,
-0.153564453125,
-0.20751953125,
-0.186279296875,
0.1588134765625,
0.92724609375,
0.404296875,
-0.09954833984375,
0.283447265625,
-0.62841796875,
-0.170166015625,
0.2288818359375,
-0.095703125,
-0.812... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given array a_1, a_2, ..., a_n. Find the subsegment a_l, a_{l+1}, ..., a_r (1 ≤ l ≤ r ≤ n) with maximum arithmetic mean (1)/(r - l + 1)∑_{i=l}^{r}{a_i} (in floating-point numbers, i.e. without any rounding).
If there are many such subsegments find the longest one.
Input
The first line contains single integer n (1 ≤ n ≤ 10^5) — length of the array a.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the array a.
Output
Print the single integer — the length of the longest subsegment with maximum possible arithmetic mean.
Example
Input
5
6 1 6 6 0
Output
2
Note
The subsegment [3, 4] is the longest among all subsegments with maximum arithmetic mean.
Submitted Solution:
```
def read_input():
n = int(input())
s2 = input()
a = [int(i) for i in s2.split()]
return n, a
def problem_a(n, a):
""" Find largest streak of largest element
"""
largest = max(a)
i = 0
old_streak_begin = 0
old_streak_end = 0
while i < n:
if a[i] == largest:
streak_begin = i
while i < n and a[i] == largest:
i += 1
streak_end = i - 1
i += 1
if (streak_end - streak_begin) < (old_streak_end - old_streak_begin):
streak_begin = old_streak_begin
streak_end = old_streak_end
print(streak_end - streak_begin + 1)
if __name__ == '__main__':
n, a = read_input()
problem_a(n, a)
```
No
| 34,298 | [
0.457763671875,
0.0701904296875,
-0.35986328125,
0.6630859375,
-0.378662109375,
-0.375244140625,
-0.1629638671875,
0.016265869140625,
0.401123046875,
0.89501953125,
0.20263671875,
0.01161956787109375,
0.09674072265625,
-0.83056640625,
-0.7861328125,
0.1243896484375,
-0.57275390625,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is the faculty of Computer Science in Berland. In the social net "TheContact!" for each course of this faculty there is the special group whose name equals the year of university entrance of corresponding course of students at the university.
Each of students joins the group of his course and joins all groups for which the year of student's university entrance differs by no more than x from the year of university entrance of this student, where x — some non-negative integer. A value x is not given, but it can be uniquely determined from the available data. Note that students don't join other groups.
You are given the list of groups which the student Igor joined. According to this information you need to determine the year of Igor's university entrance.
Input
The first line contains the positive odd integer n (1 ≤ n ≤ 5) — the number of groups which Igor joined.
The next line contains n distinct integers a1, a2, ..., an (2010 ≤ ai ≤ 2100) — years of student's university entrance for each group in which Igor is the member.
It is guaranteed that the input data is correct and the answer always exists. Groups are given randomly.
Output
Print the year of Igor's university entrance.
Examples
Input
3
2014 2016 2015
Output
2015
Input
1
2050
Output
2050
Note
In the first test the value x = 1. Igor entered the university in 2015. So he joined groups members of which are students who entered the university in 2014, 2015 and 2016.
In the second test the value x = 0. Igor entered only the group which corresponds to the year of his university entrance.
Submitted Solution:
```
input()
a = list(map(int, input().split()))
mi = ma = a[0]
for i in a:
mi = min(mi, i)
ma = max(ma, i)
print ((mi+ma) // 2)
```
Yes
| 34,809 | [
0.41064453125,
-0.1524658203125,
-0.12445068359375,
-0.047637939453125,
-0.560546875,
-0.07861328125,
0.198974609375,
0.372314453125,
0.03289794921875,
0.7587890625,
0.41943359375,
0.047882080078125,
0.46875,
-0.78857421875,
-0.626953125,
0.06219482421875,
-0.43994140625,
-0.849121... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is the faculty of Computer Science in Berland. In the social net "TheContact!" for each course of this faculty there is the special group whose name equals the year of university entrance of corresponding course of students at the university.
Each of students joins the group of his course and joins all groups for which the year of student's university entrance differs by no more than x from the year of university entrance of this student, where x — some non-negative integer. A value x is not given, but it can be uniquely determined from the available data. Note that students don't join other groups.
You are given the list of groups which the student Igor joined. According to this information you need to determine the year of Igor's university entrance.
Input
The first line contains the positive odd integer n (1 ≤ n ≤ 5) — the number of groups which Igor joined.
The next line contains n distinct integers a1, a2, ..., an (2010 ≤ ai ≤ 2100) — years of student's university entrance for each group in which Igor is the member.
It is guaranteed that the input data is correct and the answer always exists. Groups are given randomly.
Output
Print the year of Igor's university entrance.
Examples
Input
3
2014 2016 2015
Output
2015
Input
1
2050
Output
2050
Note
In the first test the value x = 1. Igor entered the university in 2015. So he joined groups members of which are students who entered the university in 2014, 2015 and 2016.
In the second test the value x = 0. Igor entered only the group which corresponds to the year of his university entrance.
Submitted Solution:
```
n = int(input())
t = [int(i) for i in input().split()]
t1 = []
if n == 1 :
print(t[0])
else :
for i in range(0,len(t)-1):
b = abs(t[i]-t[i+1])
t1.append(b)
c = max(t1)
print(t[0] + c)
```
No
| 34,810 | [
0.350830078125,
-0.17431640625,
-0.07940673828125,
-0.000988006591796875,
-0.498046875,
-0.043243408203125,
0.209716796875,
0.351318359375,
0.0330810546875,
0.8017578125,
0.422119140625,
0.019317626953125,
0.50537109375,
-0.75830078125,
-0.5986328125,
0.0174560546875,
-0.4814453125,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is the faculty of Computer Science in Berland. In the social net "TheContact!" for each course of this faculty there is the special group whose name equals the year of university entrance of corresponding course of students at the university.
Each of students joins the group of his course and joins all groups for which the year of student's university entrance differs by no more than x from the year of university entrance of this student, where x — some non-negative integer. A value x is not given, but it can be uniquely determined from the available data. Note that students don't join other groups.
You are given the list of groups which the student Igor joined. According to this information you need to determine the year of Igor's university entrance.
Input
The first line contains the positive odd integer n (1 ≤ n ≤ 5) — the number of groups which Igor joined.
The next line contains n distinct integers a1, a2, ..., an (2010 ≤ ai ≤ 2100) — years of student's university entrance for each group in which Igor is the member.
It is guaranteed that the input data is correct and the answer always exists. Groups are given randomly.
Output
Print the year of Igor's university entrance.
Examples
Input
3
2014 2016 2015
Output
2015
Input
1
2050
Output
2050
Note
In the first test the value x = 1. Igor entered the university in 2015. So he joined groups members of which are students who entered the university in 2014, 2015 and 2016.
In the second test the value x = 0. Igor entered only the group which corresponds to the year of his university entrance.
Submitted Solution:
```
n = int(input())
t = [int(i) for i in input().split()]
t1 = []
if n == 1 :
print(t[0])
else :
for i in range(0,len(t)-1):
b = abs(t[i]-t[i+1])
t1.append(b)
c = min(t1)
print(t[0] + c)
```
No
| 34,812 | [
0.350341796875,
-0.1781005859375,
-0.0821533203125,
-0.0118865966796875,
-0.497314453125,
-0.03582763671875,
0.20556640625,
0.35009765625,
0.03167724609375,
0.79931640625,
0.426513671875,
0.010528564453125,
0.4990234375,
-0.7578125,
-0.60595703125,
0.0134735107421875,
-0.474365234375... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is the faculty of Computer Science in Berland. In the social net "TheContact!" for each course of this faculty there is the special group whose name equals the year of university entrance of corresponding course of students at the university.
Each of students joins the group of his course and joins all groups for which the year of student's university entrance differs by no more than x from the year of university entrance of this student, where x — some non-negative integer. A value x is not given, but it can be uniquely determined from the available data. Note that students don't join other groups.
You are given the list of groups which the student Igor joined. According to this information you need to determine the year of Igor's university entrance.
Input
The first line contains the positive odd integer n (1 ≤ n ≤ 5) — the number of groups which Igor joined.
The next line contains n distinct integers a1, a2, ..., an (2010 ≤ ai ≤ 2100) — years of student's university entrance for each group in which Igor is the member.
It is guaranteed that the input data is correct and the answer always exists. Groups are given randomly.
Output
Print the year of Igor's university entrance.
Examples
Input
3
2014 2016 2015
Output
2015
Input
1
2050
Output
2050
Note
In the first test the value x = 1. Igor entered the university in 2015. So he joined groups members of which are students who entered the university in 2014, 2015 and 2016.
In the second test the value x = 0. Igor entered only the group which corresponds to the year of his university entrance.
Submitted Solution:
```
def solve(n, arr):
if n == 1:
return '0'
arr = sorted(list(set(arr)))
return arr[1]-arr[0]
n = int(input())
arr = list(map(int,input().strip().split()))[:n]
print(solve(n, arr))
```
No
| 34,813 | [
0.395263671875,
-0.1177978515625,
-0.1534423828125,
0.025299072265625,
-0.448974609375,
-0.08087158203125,
0.12890625,
0.4326171875,
0.0826416015625,
0.7431640625,
0.521484375,
-0.054595947265625,
0.51904296875,
-0.79052734375,
-0.646484375,
0.05908203125,
-0.428955078125,
-0.75683... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's another Start[c]up, and that means there are T-shirts to order. In order to make sure T-shirts are shipped as soon as possible, we've decided that this year we're going to order all of the necessary T-shirts before the actual competition. The top C contestants are going to be awarded T-shirts, but we obviously don't know which contestants that will be. The plan is to get the T-Shirt sizes of all contestants before the actual competition, and then order enough T-shirts so that no matter who is in the top C we'll have T-shirts available in order to award them.
In order to get the T-shirt sizes of the contestants, we will send out a survey. The survey will allow contestants to either specify a single desired T-shirt size, or two adjacent T-shirt sizes. If a contestant specifies two sizes, it means that they can be awarded either size.
As you can probably tell, this plan could require ordering a lot of unnecessary T-shirts. We'd like your help to determine the minimum number of T-shirts we'll need to order to ensure that we'll be able to award T-shirts no matter the outcome of the competition.
Input
Input will begin with two integers N and C (1 ≤ N ≤ 2·105, 1 ≤ C), the number of T-shirt sizes and number of T-shirts to be awarded, respectively.
Following this is a line with 2·N - 1 integers, s1 through s2·N - 1 (0 ≤ si ≤ 108). For odd i, si indicates the number of contestants desiring T-shirt size ((i + 1) / 2). For even i, si indicates the number of contestants okay receiving either of T-shirt sizes (i / 2) and (i / 2 + 1). C will not exceed the total number of contestants.
Output
Print the minimum number of T-shirts we need to buy.
Examples
Input
2 200
100 250 100
Output
200
Input
4 160
88 69 62 29 58 52 44
Output
314
Note
In the first example, we can buy 100 of each size.
Submitted Solution:
```
_, n = [int(x) for x in input().split()]
ar = [int(x) for x in input().split()]
s = 0
for i in range(0, len(ar), 2):
s += min(ar[i], n)
dop = []
for i in range(0, len(ar) - 2, 2):
dop.append(max(min(n - ar[i] - ar[i+2], ar[i+1]), 0))
if len(dop) > 0:
for i in range(1, len(dop) - 1):
a, b = dop[i-1], dop[i]
s += max(a,b)
if dop[i - 1] > dop[i]:
dop[i - 1] = 0
dop[i] = 0
else:
dop[i - 1] = 0
dop[i] = 0
dop[i + 1] -= b - a
s += dop[-1]
print(s)
```
No
| 34,852 | [
0.361083984375,
0.042083740234375,
0.11151123046875,
0.0992431640625,
-0.791015625,
-0.365478515625,
-0.26025390625,
0.359619140625,
-0.248291015625,
0.8447265625,
0.413818359375,
-0.1307373046875,
0.043426513671875,
-0.3359375,
-0.491455078125,
-0.0196533203125,
-0.41943359375,
-0... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's another Start[c]up, and that means there are T-shirts to order. In order to make sure T-shirts are shipped as soon as possible, we've decided that this year we're going to order all of the necessary T-shirts before the actual competition. The top C contestants are going to be awarded T-shirts, but we obviously don't know which contestants that will be. The plan is to get the T-Shirt sizes of all contestants before the actual competition, and then order enough T-shirts so that no matter who is in the top C we'll have T-shirts available in order to award them.
In order to get the T-shirt sizes of the contestants, we will send out a survey. The survey will allow contestants to either specify a single desired T-shirt size, or two adjacent T-shirt sizes. If a contestant specifies two sizes, it means that they can be awarded either size.
As you can probably tell, this plan could require ordering a lot of unnecessary T-shirts. We'd like your help to determine the minimum number of T-shirts we'll need to order to ensure that we'll be able to award T-shirts no matter the outcome of the competition.
Input
Input will begin with two integers N and C (1 ≤ N ≤ 2·105, 1 ≤ C), the number of T-shirt sizes and number of T-shirts to be awarded, respectively.
Following this is a line with 2·N - 1 integers, s1 through s2·N - 1 (0 ≤ si ≤ 108). For odd i, si indicates the number of contestants desiring T-shirt size ((i + 1) / 2). For even i, si indicates the number of contestants okay receiving either of T-shirt sizes (i / 2) and (i / 2 + 1). C will not exceed the total number of contestants.
Output
Print the minimum number of T-shirts we need to buy.
Examples
Input
2 200
100 250 100
Output
200
Input
4 160
88 69 62 29 58 52 44
Output
314
Note
In the first example, we can buy 100 of each size.
Submitted Solution:
```
_, n = [int(x) for x in input().split()]
ar = [int(x) for x in input().split()]
s = 0
for i in range(0, len(ar), 2):
s += min(ar[i], n)
dop = []
for i in range(0, len(ar) - 2, 2):
dop.append(max(min(n - ar[i] - ar[i+2], ar[i+1]), 0))
print(dop)
for i in range(1, len(dop) - 1):
a, b = dop[i-1], dop[i]
s += max(a,b)
if dop[i - 1] > dop[i]:
dop[i - 1] = 0
dop[i] = 0
else:
dop[i - 1] = 0
dop[i] = 0
dop[i + 1] -= b - a
s += dop[-1]
print(s)
```
No
| 34,853 | [
0.361083984375,
0.042083740234375,
0.11151123046875,
0.0992431640625,
-0.791015625,
-0.365478515625,
-0.26025390625,
0.359619140625,
-0.248291015625,
0.8447265625,
0.413818359375,
-0.1307373046875,
0.043426513671875,
-0.3359375,
-0.491455078125,
-0.0196533203125,
-0.41943359375,
-0... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's another Start[c]up, and that means there are T-shirts to order. In order to make sure T-shirts are shipped as soon as possible, we've decided that this year we're going to order all of the necessary T-shirts before the actual competition. The top C contestants are going to be awarded T-shirts, but we obviously don't know which contestants that will be. The plan is to get the T-Shirt sizes of all contestants before the actual competition, and then order enough T-shirts so that no matter who is in the top C we'll have T-shirts available in order to award them.
In order to get the T-shirt sizes of the contestants, we will send out a survey. The survey will allow contestants to either specify a single desired T-shirt size, or two adjacent T-shirt sizes. If a contestant specifies two sizes, it means that they can be awarded either size.
As you can probably tell, this plan could require ordering a lot of unnecessary T-shirts. We'd like your help to determine the minimum number of T-shirts we'll need to order to ensure that we'll be able to award T-shirts no matter the outcome of the competition.
Input
Input will begin with two integers N and C (1 ≤ N ≤ 2·105, 1 ≤ C), the number of T-shirt sizes and number of T-shirts to be awarded, respectively.
Following this is a line with 2·N - 1 integers, s1 through s2·N - 1 (0 ≤ si ≤ 108). For odd i, si indicates the number of contestants desiring T-shirt size ((i + 1) / 2). For even i, si indicates the number of contestants okay receiving either of T-shirt sizes (i / 2) and (i / 2 + 1). C will not exceed the total number of contestants.
Output
Print the minimum number of T-shirts we need to buy.
Examples
Input
2 200
100 250 100
Output
200
Input
4 160
88 69 62 29 58 52 44
Output
314
Note
In the first example, we can buy 100 of each size.
Submitted Solution:
```
_, n = [int(x) for x in input().split()]
ar = [int(x) for x in input().split()]
s = 0
for i in range(0, len(ar), 2):
s += ar[i]
dop = []
for i in range(0, len(ar) - 2, 2):
dop.append(max(min(n - ar[i] - ar[i+2], ar[i+1]), 0))
print(dop)
for i in range(1, len(dop) - 1):
a, b = dop[i-1], dop[i]
s += max(a,b)
if dop[i - 1] > dop[i]:
dop[i - 1] = 0
dop[i] = 0
else:
dop[i - 1] = 0
dop[i] = 0
dop[i + 1] -= b - a
s += dop[-1]
print(s)
```
No
| 34,854 | [
0.361083984375,
0.042083740234375,
0.11151123046875,
0.0992431640625,
-0.791015625,
-0.365478515625,
-0.26025390625,
0.359619140625,
-0.248291015625,
0.8447265625,
0.413818359375,
-0.1307373046875,
0.043426513671875,
-0.3359375,
-0.491455078125,
-0.0196533203125,
-0.41943359375,
-0... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's another Start[c]up, and that means there are T-shirts to order. In order to make sure T-shirts are shipped as soon as possible, we've decided that this year we're going to order all of the necessary T-shirts before the actual competition. The top C contestants are going to be awarded T-shirts, but we obviously don't know which contestants that will be. The plan is to get the T-Shirt sizes of all contestants before the actual competition, and then order enough T-shirts so that no matter who is in the top C we'll have T-shirts available in order to award them.
In order to get the T-shirt sizes of the contestants, we will send out a survey. The survey will allow contestants to either specify a single desired T-shirt size, or two adjacent T-shirt sizes. If a contestant specifies two sizes, it means that they can be awarded either size.
As you can probably tell, this plan could require ordering a lot of unnecessary T-shirts. We'd like your help to determine the minimum number of T-shirts we'll need to order to ensure that we'll be able to award T-shirts no matter the outcome of the competition.
Input
Input will begin with two integers N and C (1 ≤ N ≤ 2·105, 1 ≤ C), the number of T-shirt sizes and number of T-shirts to be awarded, respectively.
Following this is a line with 2·N - 1 integers, s1 through s2·N - 1 (0 ≤ si ≤ 108). For odd i, si indicates the number of contestants desiring T-shirt size ((i + 1) / 2). For even i, si indicates the number of contestants okay receiving either of T-shirt sizes (i / 2) and (i / 2 + 1). C will not exceed the total number of contestants.
Output
Print the minimum number of T-shirts we need to buy.
Examples
Input
2 200
100 250 100
Output
200
Input
4 160
88 69 62 29 58 52 44
Output
314
Note
In the first example, we can buy 100 of each size.
Submitted Solution:
```
_, n = [int(x) for x in input().split()]
ar = [int(x) for x in input().split()]
s = 0
for i in range(0, len(ar), 2):
s += min(ar[i], n)
dop = []
for i in range(0, len(ar) - 2, 2):
dop.append(max(min(n - ar[i] - ar[i+2], ar[i+1]), 0))
for i in range(len(dop)):
min_val = min(dop)
ind = dop.index(min_val)
if ind > 0:
dop[ind - 1] = max(dop[ind - 1] - min_val, 0)
if ind < len(dop) - 1:
dop[ind + 1] = max(dop[ind + 1] - min_val, 0)
dop[ind] = 10**20
s += min_val
print(dop)
print(s)
```
No
| 34,855 | [
0.361083984375,
0.042083740234375,
0.11151123046875,
0.0992431640625,
-0.791015625,
-0.365478515625,
-0.26025390625,
0.359619140625,
-0.248291015625,
0.8447265625,
0.413818359375,
-0.1307373046875,
0.043426513671875,
-0.3359375,
-0.491455078125,
-0.0196533203125,
-0.41943359375,
-0... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi is solving quizzes. He has easily solved all but the last one.
The last quiz has three choices: 1, 2, and 3.
With his supernatural power, Takahashi has found out that the choices A and B are both wrong.
Print the correct choice for this problem.
Constraints
* Each of the numbers A and B is 1, 2, or 3.
* A and B are different.
Input
Input is given from Standard Input in the following format:
A
B
Output
Print the correct choice.
Examples
Input
3
1
Output
2
Input
1
2
Output
3
Submitted Solution:
```
in1 = int(input())
in1 += int(input())
print(6 - in1)
```
Yes
| 34,937 | [
0.6953125,
0.0277252197265625,
-0.2425537109375,
-0.20556640625,
-0.75830078125,
-0.1842041015625,
-0.162109375,
0.15869140625,
-0.0243377685546875,
1.0009765625,
0.49169921875,
0.173828125,
0.301513671875,
-0.61669921875,
-0.76171875,
-0.3115234375,
-0.5693359375,
-0.307373046875,... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi is solving quizzes. He has easily solved all but the last one.
The last quiz has three choices: 1, 2, and 3.
With his supernatural power, Takahashi has found out that the choices A and B are both wrong.
Print the correct choice for this problem.
Constraints
* Each of the numbers A and B is 1, 2, or 3.
* A and B are different.
Input
Input is given from Standard Input in the following format:
A
B
Output
Print the correct choice.
Examples
Input
3
1
Output
2
Input
1
2
Output
3
Submitted Solution:
```
a = ["1","2","3"]
a.remove(input())
a.remove(input())
print(a[0])
```
Yes
| 34,938 | [
0.54345703125,
0.0173797607421875,
-0.0911865234375,
-0.08099365234375,
-0.91796875,
-0.218505859375,
-0.262939453125,
0.1981201171875,
-0.04296875,
0.9990234375,
0.4619140625,
0.0858154296875,
0.244873046875,
-0.6796875,
-0.92822265625,
-0.236572265625,
-0.57666015625,
-0.41015625... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi is solving quizzes. He has easily solved all but the last one.
The last quiz has three choices: 1, 2, and 3.
With his supernatural power, Takahashi has found out that the choices A and B are both wrong.
Print the correct choice for this problem.
Constraints
* Each of the numbers A and B is 1, 2, or 3.
* A and B are different.
Input
Input is given from Standard Input in the following format:
A
B
Output
Print the correct choice.
Examples
Input
3
1
Output
2
Input
1
2
Output
3
Submitted Solution:
```
A = int(input())
B = int(input())
print(int((6/A)/B))
```
Yes
| 34,939 | [
0.6220703125,
-0.0343017578125,
-0.218994140625,
-0.124755859375,
-0.76416015625,
-0.2176513671875,
-0.147705078125,
0.1895751953125,
-0.07916259765625,
1.0283203125,
0.456787109375,
0.114013671875,
0.1932373046875,
-0.56982421875,
-0.74951171875,
-0.2017822265625,
-0.580078125,
-0... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi is solving quizzes. He has easily solved all but the last one.
The last quiz has three choices: 1, 2, and 3.
With his supernatural power, Takahashi has found out that the choices A and B are both wrong.
Print the correct choice for this problem.
Constraints
* Each of the numbers A and B is 1, 2, or 3.
* A and B are different.
Input
Input is given from Standard Input in the following format:
A
B
Output
Print the correct choice.
Examples
Input
3
1
Output
2
Input
1
2
Output
3
Submitted Solution:
```
s=input()
t=input()
print(6-int(s)-int(t))
```
Yes
| 34,940 | [
0.650390625,
0.004863739013671875,
-0.1676025390625,
-0.11798095703125,
-0.7822265625,
-0.120849609375,
-0.1591796875,
0.1549072265625,
-0.097412109375,
1.0263671875,
0.39501953125,
0.11322021484375,
0.26025390625,
-0.63037109375,
-0.75830078125,
-0.30615234375,
-0.529296875,
-0.29... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi is solving quizzes. He has easily solved all but the last one.
The last quiz has three choices: 1, 2, and 3.
With his supernatural power, Takahashi has found out that the choices A and B are both wrong.
Print the correct choice for this problem.
Constraints
* Each of the numbers A and B is 1, 2, or 3.
* A and B are different.
Input
Input is given from Standard Input in the following format:
A
B
Output
Print the correct choice.
Examples
Input
3
1
Output
2
Input
1
2
Output
3
Submitted Solution:
```
icase=0
if icase==0:
a=input()
b=input()
print(6-a-b)
```
No
| 34,941 | [
0.61083984375,
0.08099365234375,
-0.157470703125,
-0.140380859375,
-0.8935546875,
-0.1328125,
-0.2744140625,
0.2093505859375,
-0.1287841796875,
0.9873046875,
0.36376953125,
0.0687255859375,
0.220703125,
-0.6533203125,
-0.72998046875,
-0.256103515625,
-0.61962890625,
-0.330810546875... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi is solving quizzes. He has easily solved all but the last one.
The last quiz has three choices: 1, 2, and 3.
With his supernatural power, Takahashi has found out that the choices A and B are both wrong.
Print the correct choice for this problem.
Constraints
* Each of the numbers A and B is 1, 2, or 3.
* A and B are different.
Input
Input is given from Standard Input in the following format:
A
B
Output
Print the correct choice.
Examples
Input
3
1
Output
2
Input
1
2
Output
3
Submitted Solution:
```
A=map(int,input())
B=map(int,input())
if (A=1 and B=2)or(A=2 and B=1):
print(3)
if (A=2 and B=3)or(A=3 and B=2):
print(1)
else:
print(2)
```
No
| 34,942 | [
0.64208984375,
0.08843994140625,
-0.218994140625,
-0.08636474609375,
-0.74072265625,
-0.1939697265625,
-0.150390625,
0.2052001953125,
-0.0865478515625,
1.0107421875,
0.5517578125,
0.0570068359375,
0.47216796875,
-0.685546875,
-0.65283203125,
-0.224853515625,
-0.57763671875,
-0.3686... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi is solving quizzes. He has easily solved all but the last one.
The last quiz has three choices: 1, 2, and 3.
With his supernatural power, Takahashi has found out that the choices A and B are both wrong.
Print the correct choice for this problem.
Constraints
* Each of the numbers A and B is 1, 2, or 3.
* A and B are different.
Input
Input is given from Standard Input in the following format:
A
B
Output
Print the correct choice.
Examples
Input
3
1
Output
2
Input
1
2
Output
3
Submitted Solution:
```
a = int(input())
b = ont(input())
print(6 - a - b)
```
No
| 34,943 | [
0.681640625,
-0.053070068359375,
-0.16650390625,
-0.1688232421875,
-0.75830078125,
-0.2158203125,
-0.178466796875,
0.1651611328125,
-0.05157470703125,
0.99462890625,
0.453857421875,
0.128662109375,
0.256103515625,
-0.583984375,
-0.7099609375,
-0.28125,
-0.59619140625,
-0.2868652343... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi is solving quizzes. He has easily solved all but the last one.
The last quiz has three choices: 1, 2, and 3.
With his supernatural power, Takahashi has found out that the choices A and B are both wrong.
Print the correct choice for this problem.
Constraints
* Each of the numbers A and B is 1, 2, or 3.
* A and B are different.
Input
Input is given from Standard Input in the following format:
A
B
Output
Print the correct choice.
Examples
Input
3
1
Output
2
Input
1
2
Output
3
Submitted Solution:
```
input1 = int(input())
input2 = int(input())
if input1 == 1 and input2 == 2:
print(3)
elif input1 == 1 and input2 == 3:
print(2)
else:
print(1)
```
No
| 34,944 | [
0.654296875,
0.09283447265625,
-0.19482421875,
-0.1307373046875,
-0.8349609375,
-0.2452392578125,
-0.12432861328125,
0.205322265625,
-0.00262451171875,
0.994140625,
0.44580078125,
0.09954833984375,
0.355224609375,
-0.56787109375,
-0.767578125,
-0.2431640625,
-0.5234375,
-0.37866210... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
Prize
Segtree entered a programming contest with a team of $ N $ and won a $ K $ yen prize! I'm trying to distribute this prize now.
Each $ N $ team member, including Segtree, is numbered from $ 1 $ to $ N $ in order of ability. Segtree is $ 1 $.
If the prize amount of $ i $'s teammate $ (i \ geq 2) $ is less than "$ i-half of the prize amount of $ i's teammate rounded down to an integer", that person Get angry.
When distributing the $ K $ Yen prize so that no one gets angry, find the maximum prize that Segtree can get.
input
Input is given from standard input in the following format.
N K
output
Please output the maximum prize money that Segtree can receive.
However, insert a line break at the end.
Constraint
* $ 1 \ leq N, K \ leq 10 ^ {18} $
* All inputs are integers.
Input example 1
1 1
Output example 1
1
Input example 2
819875141880895728 349993004923078537
Output example 2
174996502461539284
Example
Input
1 1
Output
1
"Correct Solution:
```
n,k = map(int,input().split())
ok,ng = 0,k+1
while ng-ok>1:
mid = (ng+ok)//2
s = 0
m = mid
for i in range(n):
s += m
m //= 2
if m == 0:
break
if s <= k:
ok = mid
else:
ng = mid
print(ok)
```
| 35,108 | [
0.4912109375,
0.15185546875,
-0.2115478515625,
0.490234375,
-0.5009765625,
-0.50830078125,
-0.0034770965576171875,
0.544921875,
0.091064453125,
0.93310546875,
-0.098388671875,
-0.1505126953125,
0.05767822265625,
-0.5146484375,
-0.2841796875,
0.1741943359375,
-0.755859375,
-0.751953... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
Prize
Segtree entered a programming contest with a team of $ N $ and won a $ K $ yen prize! I'm trying to distribute this prize now.
Each $ N $ team member, including Segtree, is numbered from $ 1 $ to $ N $ in order of ability. Segtree is $ 1 $.
If the prize amount of $ i $'s teammate $ (i \ geq 2) $ is less than "$ i-half of the prize amount of $ i's teammate rounded down to an integer", that person Get angry.
When distributing the $ K $ Yen prize so that no one gets angry, find the maximum prize that Segtree can get.
input
Input is given from standard input in the following format.
N K
output
Please output the maximum prize money that Segtree can receive.
However, insert a line break at the end.
Constraint
* $ 1 \ leq N, K \ leq 10 ^ {18} $
* All inputs are integers.
Input example 1
1 1
Output example 1
1
Input example 2
819875141880895728 349993004923078537
Output example 2
174996502461539284
Example
Input
1 1
Output
1
"Correct Solution:
```
from itertools import *
from bisect import *
from math import *
from collections import *
from heapq import *
from random import *
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 MI1(): return map(int1, sys.stdin.readline().split())
def MF(): return map(float, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LI1(): return list(map(int1, sys.stdin.readline().split()))
def LF(): return list(map(float, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
dij = [(1, 0), (0, 1), (-1, 0), (0, -1)]
def main():
def ok(m):
s=0
cnt=0
while m:
s+=m
cnt+=1
if s>k:return False
if cnt==n:break
m>>=1
return True
n,k=MI()
l=0
r=k+1
while l+1<r:
m=(l+r)//2
if ok(m):l=m
else:r=m
print(l)
main()
```
| 35,109 | [
0.43310546875,
0.162841796875,
-0.2371826171875,
0.57275390625,
-0.54541015625,
-0.34130859375,
-0.08209228515625,
0.416015625,
0.05517578125,
0.9189453125,
-0.1446533203125,
-0.2451171875,
0.0269622802734375,
-0.486328125,
-0.294189453125,
0.1251220703125,
-0.6953125,
-0.844238281... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
Prize
Segtree entered a programming contest with a team of $ N $ and won a $ K $ yen prize! I'm trying to distribute this prize now.
Each $ N $ team member, including Segtree, is numbered from $ 1 $ to $ N $ in order of ability. Segtree is $ 1 $.
If the prize amount of $ i $'s teammate $ (i \ geq 2) $ is less than "$ i-half of the prize amount of $ i's teammate rounded down to an integer", that person Get angry.
When distributing the $ K $ Yen prize so that no one gets angry, find the maximum prize that Segtree can get.
input
Input is given from standard input in the following format.
N K
output
Please output the maximum prize money that Segtree can receive.
However, insert a line break at the end.
Constraint
* $ 1 \ leq N, K \ leq 10 ^ {18} $
* All inputs are integers.
Input example 1
1 1
Output example 1
1
Input example 2
819875141880895728 349993004923078537
Output example 2
174996502461539284
Example
Input
1 1
Output
1
"Correct Solution:
```
n,k=map(int,input().split())
l,r=0,k+1
while r-l>1:
m=(l+r)//2
res = m
t = m
for i in range(n-1):
t //= 2
res += t
if t == 0:
break
if res > k:
r = m
else:
l = m
print(l)
```
| 35,110 | [
0.4423828125,
0.162841796875,
-0.259033203125,
0.464599609375,
-0.51953125,
-0.48779296875,
-0.057525634765625,
0.52392578125,
0.05914306640625,
0.904296875,
-0.12396240234375,
-0.1673583984375,
0.03436279296875,
-0.53125,
-0.2666015625,
0.18994140625,
-0.76220703125,
-0.8134765625... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
Prize
Segtree entered a programming contest with a team of $ N $ and won a $ K $ yen prize! I'm trying to distribute this prize now.
Each $ N $ team member, including Segtree, is numbered from $ 1 $ to $ N $ in order of ability. Segtree is $ 1 $.
If the prize amount of $ i $'s teammate $ (i \ geq 2) $ is less than "$ i-half of the prize amount of $ i's teammate rounded down to an integer", that person Get angry.
When distributing the $ K $ Yen prize so that no one gets angry, find the maximum prize that Segtree can get.
input
Input is given from standard input in the following format.
N K
output
Please output the maximum prize money that Segtree can receive.
However, insert a line break at the end.
Constraint
* $ 1 \ leq N, K \ leq 10 ^ {18} $
* All inputs are integers.
Input example 1
1 1
Output example 1
1
Input example 2
819875141880895728 349993004923078537
Output example 2
174996502461539284
Example
Input
1 1
Output
1
"Correct Solution:
```
N,K=map(int,input().split())
OK=0
NG=K+1
while NG-OK>1:
#print(OK,NG)
mid=(OK+NG)//2
ANS=0
money=mid
for i in range(N):
ANS+=money
money//=2
if money==0:
break
if ANS<=K:
OK=mid
else:
NG=mid
print(OK)
```
| 35,111 | [
0.50048828125,
0.1883544921875,
-0.2215576171875,
0.479736328125,
-0.50146484375,
-0.5185546875,
0.017791748046875,
0.54443359375,
0.08233642578125,
0.89404296875,
-0.1207275390625,
-0.158203125,
0.06024169921875,
-0.52001953125,
-0.28857421875,
0.1473388671875,
-0.76123046875,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The average miner Vaganych took refresher courses. As soon as a miner completes the courses, he should take exams. The hardest one is a computer test called "Testing Pants for Sadness".
The test consists of n questions; the questions are to be answered strictly in the order in which they are given, from question 1 to question n. Question i contains ai answer variants, exactly one of them is correct.
A click is regarded as selecting any answer in any question. The goal is to select the correct answer for each of the n questions. If Vaganych selects a wrong answer for some question, then all selected answers become unselected and the test starts from the very beginning, from question 1 again. But Vaganych remembers everything. The order of answers for each question and the order of questions remain unchanged, as well as the question and answers themselves.
Vaganych is very smart and his memory is superb, yet he is unbelievably unlucky and knows nothing whatsoever about the test's theme. How many clicks will he have to perform in the worst case?
Input
The first line contains a positive integer n (1 ≤ n ≤ 100). It is the number of questions in the test. The second line contains space-separated n positive integers ai (1 ≤ ai ≤ 109), the number of answer variants to question i.
Output
Print a single number — the minimal number of clicks needed to pass the test it the worst-case scenario.
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Examples
Input
2
1 1
Output
2
Input
2
2 2
Output
5
Input
1
10
Output
10
Note
Note to the second sample. In the worst-case scenario you will need five clicks:
* the first click selects the first variant to the first question, this answer turns out to be wrong.
* the second click selects the second variant to the first question, it proves correct and we move on to the second question;
* the third click selects the first variant to the second question, it is wrong and we go back to question 1;
* the fourth click selects the second variant to the first question, it proves as correct as it was and we move on to the second question;
* the fifth click selects the second variant to the second question, it proves correct, the test is finished.
Submitted Solution:
```
def trausers(lst):
result = 0
for i in range(len(lst)):
result += (lst[i] - 1) * (i + 1) + 1
return result
n = int(input())
a = [int(j) for j in input().split()]
print(trausers(a))
```
Yes
| 35,147 | [
0.5693359375,
-0.040374755859375,
-0.07159423828125,
0.1485595703125,
-0.84912109375,
-0.417724609375,
-0.29638671875,
0.238037109375,
-0.09515380859375,
0.90234375,
0.50634765625,
-0.204345703125,
0.184814453125,
-0.8251953125,
-0.59716796875,
-0.288330078125,
-0.650390625,
-0.593... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The average miner Vaganych took refresher courses. As soon as a miner completes the courses, he should take exams. The hardest one is a computer test called "Testing Pants for Sadness".
The test consists of n questions; the questions are to be answered strictly in the order in which they are given, from question 1 to question n. Question i contains ai answer variants, exactly one of them is correct.
A click is regarded as selecting any answer in any question. The goal is to select the correct answer for each of the n questions. If Vaganych selects a wrong answer for some question, then all selected answers become unselected and the test starts from the very beginning, from question 1 again. But Vaganych remembers everything. The order of answers for each question and the order of questions remain unchanged, as well as the question and answers themselves.
Vaganych is very smart and his memory is superb, yet he is unbelievably unlucky and knows nothing whatsoever about the test's theme. How many clicks will he have to perform in the worst case?
Input
The first line contains a positive integer n (1 ≤ n ≤ 100). It is the number of questions in the test. The second line contains space-separated n positive integers ai (1 ≤ ai ≤ 109), the number of answer variants to question i.
Output
Print a single number — the minimal number of clicks needed to pass the test it the worst-case scenario.
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Examples
Input
2
1 1
Output
2
Input
2
2 2
Output
5
Input
1
10
Output
10
Note
Note to the second sample. In the worst-case scenario you will need five clicks:
* the first click selects the first variant to the first question, this answer turns out to be wrong.
* the second click selects the second variant to the first question, it proves correct and we move on to the second question;
* the third click selects the first variant to the second question, it is wrong and we go back to question 1;
* the fourth click selects the second variant to the first question, it proves as correct as it was and we move on to the second question;
* the fifth click selects the second variant to the second question, it proves correct, the test is finished.
Submitted Solution:
```
a = int(input())
arr = list(map(int,input().split()))
ans = 0
for i in range(1,a+1):
ans += (arr[i-1] - 1) * i + 1
print(ans)
```
Yes
| 35,148 | [
0.5693359375,
-0.040374755859375,
-0.07159423828125,
0.1485595703125,
-0.84912109375,
-0.417724609375,
-0.29638671875,
0.238037109375,
-0.09515380859375,
0.90234375,
0.50634765625,
-0.204345703125,
0.184814453125,
-0.8251953125,
-0.59716796875,
-0.288330078125,
-0.650390625,
-0.593... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The average miner Vaganych took refresher courses. As soon as a miner completes the courses, he should take exams. The hardest one is a computer test called "Testing Pants for Sadness".
The test consists of n questions; the questions are to be answered strictly in the order in which they are given, from question 1 to question n. Question i contains ai answer variants, exactly one of them is correct.
A click is regarded as selecting any answer in any question. The goal is to select the correct answer for each of the n questions. If Vaganych selects a wrong answer for some question, then all selected answers become unselected and the test starts from the very beginning, from question 1 again. But Vaganych remembers everything. The order of answers for each question and the order of questions remain unchanged, as well as the question and answers themselves.
Vaganych is very smart and his memory is superb, yet he is unbelievably unlucky and knows nothing whatsoever about the test's theme. How many clicks will he have to perform in the worst case?
Input
The first line contains a positive integer n (1 ≤ n ≤ 100). It is the number of questions in the test. The second line contains space-separated n positive integers ai (1 ≤ ai ≤ 109), the number of answer variants to question i.
Output
Print a single number — the minimal number of clicks needed to pass the test it the worst-case scenario.
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Examples
Input
2
1 1
Output
2
Input
2
2 2
Output
5
Input
1
10
Output
10
Note
Note to the second sample. In the worst-case scenario you will need five clicks:
* the first click selects the first variant to the first question, this answer turns out to be wrong.
* the second click selects the second variant to the first question, it proves correct and we move on to the second question;
* the third click selects the first variant to the second question, it is wrong and we go back to question 1;
* the fourth click selects the second variant to the first question, it proves as correct as it was and we move on to the second question;
* the fifth click selects the second variant to the second question, it proves correct, the test is finished.
Submitted Solution:
```
input()
print(sum(i * a - i + 1 for i, a in enumerate(map(int, input().split()), 1)))
```
Yes
| 35,149 | [
0.5693359375,
-0.040374755859375,
-0.07159423828125,
0.1485595703125,
-0.84912109375,
-0.417724609375,
-0.29638671875,
0.238037109375,
-0.09515380859375,
0.90234375,
0.50634765625,
-0.204345703125,
0.184814453125,
-0.8251953125,
-0.59716796875,
-0.288330078125,
-0.650390625,
-0.593... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The average miner Vaganych took refresher courses. As soon as a miner completes the courses, he should take exams. The hardest one is a computer test called "Testing Pants for Sadness".
The test consists of n questions; the questions are to be answered strictly in the order in which they are given, from question 1 to question n. Question i contains ai answer variants, exactly one of them is correct.
A click is regarded as selecting any answer in any question. The goal is to select the correct answer for each of the n questions. If Vaganych selects a wrong answer for some question, then all selected answers become unselected and the test starts from the very beginning, from question 1 again. But Vaganych remembers everything. The order of answers for each question and the order of questions remain unchanged, as well as the question and answers themselves.
Vaganych is very smart and his memory is superb, yet he is unbelievably unlucky and knows nothing whatsoever about the test's theme. How many clicks will he have to perform in the worst case?
Input
The first line contains a positive integer n (1 ≤ n ≤ 100). It is the number of questions in the test. The second line contains space-separated n positive integers ai (1 ≤ ai ≤ 109), the number of answer variants to question i.
Output
Print a single number — the minimal number of clicks needed to pass the test it the worst-case scenario.
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Examples
Input
2
1 1
Output
2
Input
2
2 2
Output
5
Input
1
10
Output
10
Note
Note to the second sample. In the worst-case scenario you will need five clicks:
* the first click selects the first variant to the first question, this answer turns out to be wrong.
* the second click selects the second variant to the first question, it proves correct and we move on to the second question;
* the third click selects the first variant to the second question, it is wrong and we go back to question 1;
* the fourth click selects the second variant to the first question, it proves as correct as it was and we move on to the second question;
* the fifth click selects the second variant to the second question, it proves correct, the test is finished.
Submitted Solution:
```
"""
Template written to be used by Python Programmers.
Use at your own risk!!!!
Owned by adi0311(rating - 5 star at CodeChef and Specialist at Codeforces).
"""
import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush, nlargest, nsmallest, _heapify_max, _heapreplace_max
from math import ceil, floor, gcd, fabs, factorial, fmod, sqrt, inf, log
from collections import defaultdict as dd, deque, Counter as c
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from fractions import Fraction
# sys.setrecursionlimit(2*pow(10, 6))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
mod = pow(10, 9) + 7
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(var): sys.stdout.write(str(var))
def outln(var): sys.stdout.write(str(var)+"\n")
def l(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
n = int(data())
arr = l()
for i in range(n-2, -1, -1):
arr[i] = arr[i+1] + arr[i] - 1
print(sum(arr))
```
Yes
| 35,150 | [
0.5693359375,
-0.040374755859375,
-0.07159423828125,
0.1485595703125,
-0.84912109375,
-0.417724609375,
-0.29638671875,
0.238037109375,
-0.09515380859375,
0.90234375,
0.50634765625,
-0.204345703125,
0.184814453125,
-0.8251953125,
-0.59716796875,
-0.288330078125,
-0.650390625,
-0.593... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The average miner Vaganych took refresher courses. As soon as a miner completes the courses, he should take exams. The hardest one is a computer test called "Testing Pants for Sadness".
The test consists of n questions; the questions are to be answered strictly in the order in which they are given, from question 1 to question n. Question i contains ai answer variants, exactly one of them is correct.
A click is regarded as selecting any answer in any question. The goal is to select the correct answer for each of the n questions. If Vaganych selects a wrong answer for some question, then all selected answers become unselected and the test starts from the very beginning, from question 1 again. But Vaganych remembers everything. The order of answers for each question and the order of questions remain unchanged, as well as the question and answers themselves.
Vaganych is very smart and his memory is superb, yet he is unbelievably unlucky and knows nothing whatsoever about the test's theme. How many clicks will he have to perform in the worst case?
Input
The first line contains a positive integer n (1 ≤ n ≤ 100). It is the number of questions in the test. The second line contains space-separated n positive integers ai (1 ≤ ai ≤ 109), the number of answer variants to question i.
Output
Print a single number — the minimal number of clicks needed to pass the test it the worst-case scenario.
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Examples
Input
2
1 1
Output
2
Input
2
2 2
Output
5
Input
1
10
Output
10
Note
Note to the second sample. In the worst-case scenario you will need five clicks:
* the first click selects the first variant to the first question, this answer turns out to be wrong.
* the second click selects the second variant to the first question, it proves correct and we move on to the second question;
* the third click selects the first variant to the second question, it is wrong and we go back to question 1;
* the fourth click selects the second variant to the first question, it proves as correct as it was and we move on to the second question;
* the fifth click selects the second variant to the second question, it proves correct, the test is finished.
Submitted Solution:
```
n = int(input())
l = list(map(int,input().split()))
if len(set(l)) == 1 and l[0] == 1 or n == 1 :
print(sum(l))
else:
cnt = l[0]
for i in range(1 , n):
cnt += l[i] + i
print(cnt)
```
No
| 35,151 | [
0.5693359375,
-0.040374755859375,
-0.07159423828125,
0.1485595703125,
-0.84912109375,
-0.417724609375,
-0.29638671875,
0.238037109375,
-0.09515380859375,
0.90234375,
0.50634765625,
-0.204345703125,
0.184814453125,
-0.8251953125,
-0.59716796875,
-0.288330078125,
-0.650390625,
-0.593... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The average miner Vaganych took refresher courses. As soon as a miner completes the courses, he should take exams. The hardest one is a computer test called "Testing Pants for Sadness".
The test consists of n questions; the questions are to be answered strictly in the order in which they are given, from question 1 to question n. Question i contains ai answer variants, exactly one of them is correct.
A click is regarded as selecting any answer in any question. The goal is to select the correct answer for each of the n questions. If Vaganych selects a wrong answer for some question, then all selected answers become unselected and the test starts from the very beginning, from question 1 again. But Vaganych remembers everything. The order of answers for each question and the order of questions remain unchanged, as well as the question and answers themselves.
Vaganych is very smart and his memory is superb, yet he is unbelievably unlucky and knows nothing whatsoever about the test's theme. How many clicks will he have to perform in the worst case?
Input
The first line contains a positive integer n (1 ≤ n ≤ 100). It is the number of questions in the test. The second line contains space-separated n positive integers ai (1 ≤ ai ≤ 109), the number of answer variants to question i.
Output
Print a single number — the minimal number of clicks needed to pass the test it the worst-case scenario.
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Examples
Input
2
1 1
Output
2
Input
2
2 2
Output
5
Input
1
10
Output
10
Note
Note to the second sample. In the worst-case scenario you will need five clicks:
* the first click selects the first variant to the first question, this answer turns out to be wrong.
* the second click selects the second variant to the first question, it proves correct and we move on to the second question;
* the third click selects the first variant to the second question, it is wrong and we go back to question 1;
* the fourth click selects the second variant to the first question, it proves as correct as it was and we move on to the second question;
* the fifth click selects the second variant to the second question, it proves correct, the test is finished.
Submitted Solution:
```
import itertools as it
n, a = int(input()), list(map(int, input().split()))
b = list(it.accumulate([int(c > 1) for c in a[::-1]]))[::-1]
print(sum([a[i] + b[i + 1] for i in range(n - 1)]) + a[-1])
```
No
| 35,152 | [
0.5693359375,
-0.040374755859375,
-0.07159423828125,
0.1485595703125,
-0.84912109375,
-0.417724609375,
-0.29638671875,
0.238037109375,
-0.09515380859375,
0.90234375,
0.50634765625,
-0.204345703125,
0.184814453125,
-0.8251953125,
-0.59716796875,
-0.288330078125,
-0.650390625,
-0.593... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The average miner Vaganych took refresher courses. As soon as a miner completes the courses, he should take exams. The hardest one is a computer test called "Testing Pants for Sadness".
The test consists of n questions; the questions are to be answered strictly in the order in which they are given, from question 1 to question n. Question i contains ai answer variants, exactly one of them is correct.
A click is regarded as selecting any answer in any question. The goal is to select the correct answer for each of the n questions. If Vaganych selects a wrong answer for some question, then all selected answers become unselected and the test starts from the very beginning, from question 1 again. But Vaganych remembers everything. The order of answers for each question and the order of questions remain unchanged, as well as the question and answers themselves.
Vaganych is very smart and his memory is superb, yet he is unbelievably unlucky and knows nothing whatsoever about the test's theme. How many clicks will he have to perform in the worst case?
Input
The first line contains a positive integer n (1 ≤ n ≤ 100). It is the number of questions in the test. The second line contains space-separated n positive integers ai (1 ≤ ai ≤ 109), the number of answer variants to question i.
Output
Print a single number — the minimal number of clicks needed to pass the test it the worst-case scenario.
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Examples
Input
2
1 1
Output
2
Input
2
2 2
Output
5
Input
1
10
Output
10
Note
Note to the second sample. In the worst-case scenario you will need five clicks:
* the first click selects the first variant to the first question, this answer turns out to be wrong.
* the second click selects the second variant to the first question, it proves correct and we move on to the second question;
* the third click selects the first variant to the second question, it is wrong and we go back to question 1;
* the fourth click selects the second variant to the first question, it proves as correct as it was and we move on to the second question;
* the fifth click selects the second variant to the second question, it proves correct, the test is finished.
Submitted Solution:
```
n=int(input())
l=list(map(int,input().split()))
i=0
x=0
for j in l:
x=x+j+i
if(j>1):
i=i+1
print(x)
```
No
| 35,153 | [
0.5693359375,
-0.040374755859375,
-0.07159423828125,
0.1485595703125,
-0.84912109375,
-0.417724609375,
-0.29638671875,
0.238037109375,
-0.09515380859375,
0.90234375,
0.50634765625,
-0.204345703125,
0.184814453125,
-0.8251953125,
-0.59716796875,
-0.288330078125,
-0.650390625,
-0.593... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The average miner Vaganych took refresher courses. As soon as a miner completes the courses, he should take exams. The hardest one is a computer test called "Testing Pants for Sadness".
The test consists of n questions; the questions are to be answered strictly in the order in which they are given, from question 1 to question n. Question i contains ai answer variants, exactly one of them is correct.
A click is regarded as selecting any answer in any question. The goal is to select the correct answer for each of the n questions. If Vaganych selects a wrong answer for some question, then all selected answers become unselected and the test starts from the very beginning, from question 1 again. But Vaganych remembers everything. The order of answers for each question and the order of questions remain unchanged, as well as the question and answers themselves.
Vaganych is very smart and his memory is superb, yet he is unbelievably unlucky and knows nothing whatsoever about the test's theme. How many clicks will he have to perform in the worst case?
Input
The first line contains a positive integer n (1 ≤ n ≤ 100). It is the number of questions in the test. The second line contains space-separated n positive integers ai (1 ≤ ai ≤ 109), the number of answer variants to question i.
Output
Print a single number — the minimal number of clicks needed to pass the test it the worst-case scenario.
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Examples
Input
2
1 1
Output
2
Input
2
2 2
Output
5
Input
1
10
Output
10
Note
Note to the second sample. In the worst-case scenario you will need five clicks:
* the first click selects the first variant to the first question, this answer turns out to be wrong.
* the second click selects the second variant to the first question, it proves correct and we move on to the second question;
* the third click selects the first variant to the second question, it is wrong and we go back to question 1;
* the fourth click selects the second variant to the first question, it proves as correct as it was and we move on to the second question;
* the fifth click selects the second variant to the second question, it proves correct, the test is finished.
Submitted Solution:
```
import sys,math
s = int(input())
arr = list(map(int,input().split(' ')))
m = arr[0]
for i in range(1,s):
m = m + (arr[i])-1
arr[i] = m
print(sum(arr))
'''
2 2 2
1+1 2
1+1+1 3
(1+1+1,1) 4
3 3
1+1+1 3
1+1+1+1+1 5
1+1+1,1+1+1,1 7
'''
```
No
| 35,154 | [
0.5693359375,
-0.040374755859375,
-0.07159423828125,
0.1485595703125,
-0.84912109375,
-0.417724609375,
-0.29638671875,
0.238037109375,
-0.09515380859375,
0.90234375,
0.50634765625,
-0.204345703125,
0.184814453125,
-0.8251953125,
-0.59716796875,
-0.288330078125,
-0.650390625,
-0.593... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Input
The input contains two integers a1, a2 (0 ≤ ai ≤ 109), separated by a single space.
Output
Output a single integer.
Examples
Input
3 14
Output
44
Input
27 12
Output
48
Input
100 200
Output
102
Submitted Solution:
```
"""====================================================================================
====================================================================================
___ _______ ___ _______ ___ ___
| /\ | | \ | | / | | | | |\ /|
| / \ | | \ | | / | | | | | \ / |
|___ /____\ | | \ | |/ |___| | | | \/ |
| / \ | | / | |\ |\ | | | |
| / \ | | / | | \ | \ | | | |
___|/ \___|___ |___/ ___|___ | \ | \ |___| | |
====================================================================================
====================================================================================
"""
# ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥
a1,a2 = map(str,input().split())
a = int(a1) + int(a2[::-1])
print(a)
# ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥
"""====================================================================================
====================================================================================
___ _______ ___ _______ ___ ___
| /\ | | \ | | / | | | | |\ /|
| / \ | | \ | | / | | | | | \ / |
|___ /____\ | | \ | |/ |___| | | | \/ |
| / \ | | / | |\ |\ | | | |
| / \ | | / | | \ | \ | | | |
___|/ \___|___ |___/ ___|___ | \ | \ |___| | |
====================================================================================
====================================================================================
"""
```
Yes
| 35,432 | [
0.412353515625,
-0.007762908935546875,
-0.10882568359375,
-0.103515625,
-1.1767578125,
-0.48974609375,
-0.1104736328125,
0.327392578125,
0.041259765625,
0.89794921875,
0.4814453125,
0.2283935546875,
-0.09600830078125,
-0.486328125,
-0.46728515625,
-0.11492919921875,
-0.460205078125,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
SmallR is an archer. SmallR is taking a match of archer with Zanoes. They try to shoot in the target in turns, and SmallR shoots first. The probability of shooting the target each time is <image> for SmallR while <image> for Zanoes. The one who shoots in the target first should be the winner.
Output the probability that SmallR will win the match.
Input
A single line contains four integers <image>.
Output
Print a single real number, the probability that SmallR will win the match.
The answer will be considered correct if the absolute or relative error doesn't exceed 10 - 6.
Examples
Input
1 2 1 2
Output
0.666666666667
Submitted Solution:
```
import sys
input = sys.stdin.readline
a, b, c, d = map(int, input().split())
p = a/b
q = (1-(a/b))*(1-(c/d))
print(p/(1-q))
```
Yes
| 35,489 | [
0.3876953125,
-0.0251617431640625,
0.0889892578125,
-0.07598876953125,
-0.63037109375,
-0.54150390625,
-0.0265655517578125,
0.33349609375,
0.0648193359375,
0.72900390625,
0.84130859375,
-0.016845703125,
-0.1749267578125,
-0.58984375,
-0.5068359375,
0.08807373046875,
-0.458740234375,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
SmallR is an archer. SmallR is taking a match of archer with Zanoes. They try to shoot in the target in turns, and SmallR shoots first. The probability of shooting the target each time is <image> for SmallR while <image> for Zanoes. The one who shoots in the target first should be the winner.
Output the probability that SmallR will win the match.
Input
A single line contains four integers <image>.
Output
Print a single real number, the probability that SmallR will win the match.
The answer will be considered correct if the absolute or relative error doesn't exceed 10 - 6.
Examples
Input
1 2 1 2
Output
0.666666666667
Submitted Solution:
```
a, b, c, d = map(int, input().split())
n = 10**6
s = a/b
prev_term = a/b
for i in range(n):
prev_term *= ((b-a)/b) * ((d-c)/d)
s += prev_term
print(s)
```
Yes
| 35,490 | [
0.4384765625,
-0.0255584716796875,
0.00240325927734375,
-0.0186767578125,
-0.51806640625,
-0.499755859375,
-0.018463134765625,
0.337890625,
0.039215087890625,
0.70556640625,
0.93798828125,
0.0060882568359375,
-0.1866455078125,
-0.6318359375,
-0.435546875,
0.177490234375,
-0.462158203... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
SmallR is an archer. SmallR is taking a match of archer with Zanoes. They try to shoot in the target in turns, and SmallR shoots first. The probability of shooting the target each time is <image> for SmallR while <image> for Zanoes. The one who shoots in the target first should be the winner.
Output the probability that SmallR will win the match.
Input
A single line contains four integers <image>.
Output
Print a single real number, the probability that SmallR will win the match.
The answer will be considered correct if the absolute or relative error doesn't exceed 10 - 6.
Examples
Input
1 2 1 2
Output
0.666666666667
Submitted Solution:
```
######################################################
############Created by Devesh Kumar###################
#############devesh1102@gmail.com####################
##########For CodeForces(Devesh1102)#################
#####################2020#############################
######################################################
import sys
input = sys.stdin.readline
# import sys
import heapq
import copy
import math
import decimal
# import sys.stdout.flush as flush
# from decimal import *
#heapq.heapify(li)
#
#heapq.heappush(li,4)
#
#heapq.heappop(li)
#
# & Bitwise AND Operator 10 & 7 = 2
# | Bitwise OR Operator 10 | 7 = 15
# ^ Bitwise XOR Operator 10 ^ 7 = 13
# << Bitwise Left Shift operator 10<<2 = 40
# >> Bitwise Right Shift Operator
# '''############ ---- Input Functions ---- #######Start#####'''
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def insr2():
s = input()
return((s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
############ ---- Input Functions ---- #######End
# #####
ans = 0
def pr_list(a):
print( *a , sep=" ")
def main():
# tests = inp()
tests = 1
mod = 1000000007
limit = 10**18
ans = 0
stack = []
hashm = {}
arr = []
heapq.heapify(arr)
for test in range(tests):
# l1 = insr()(
[a,b,c,d] = inlt()
print( abs((a/b)*(1/ (1 - (1 - (a/b)) * (1 - (c/d)) ))) )
if __name__== "__main__":
main()
```
Yes
| 35,491 | [
0.372802734375,
-0.1396484375,
0.0205078125,
-0.007450103759765625,
-0.5966796875,
-0.5419921875,
-0.037322998046875,
0.30029296875,
0.09027099609375,
0.5830078125,
0.77783203125,
-0.1622314453125,
-0.11968994140625,
-0.5966796875,
-0.5380859375,
0.1453857421875,
-0.5751953125,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
SmallR is an archer. SmallR is taking a match of archer with Zanoes. They try to shoot in the target in turns, and SmallR shoots first. The probability of shooting the target each time is <image> for SmallR while <image> for Zanoes. The one who shoots in the target first should be the winner.
Output the probability that SmallR will win the match.
Input
A single line contains four integers <image>.
Output
Print a single real number, the probability that SmallR will win the match.
The answer will be considered correct if the absolute or relative error doesn't exceed 10 - 6.
Examples
Input
1 2 1 2
Output
0.666666666667
Submitted Solution:
```
# URL: http://codeforces.com/problemset/problem/312/B
from typing import List
def parse_input() -> List[int]:
return [int(x) for x in input().split()]
def solve(a: int, b: int, c: int, d: int) -> float:
p = a/b
q = c/d
return p / (p + q - q*p)
if __name__ == '__main__':
a, b, c, d = parse_input()
print(solve(a, b, c, d))
```
Yes
| 35,492 | [
0.3955078125,
-0.08099365234375,
0.018585205078125,
-0.058990478515625,
-0.51708984375,
-0.43896484375,
0.11163330078125,
0.357177734375,
0.1302490234375,
0.662109375,
0.83203125,
0.0135498046875,
-0.265869140625,
-0.6884765625,
-0.5419921875,
0.09307861328125,
-0.4228515625,
-0.66... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
SmallR is an archer. SmallR is taking a match of archer with Zanoes. They try to shoot in the target in turns, and SmallR shoots first. The probability of shooting the target each time is <image> for SmallR while <image> for Zanoes. The one who shoots in the target first should be the winner.
Output the probability that SmallR will win the match.
Input
A single line contains four integers <image>.
Output
Print a single real number, the probability that SmallR will win the match.
The answer will be considered correct if the absolute or relative error doesn't exceed 10 - 6.
Examples
Input
1 2 1 2
Output
0.666666666667
Submitted Solution:
```
a, b, c, d=list(map(int, input().split()))
ans=(a/b)/(1-((a/b)*((d-c)/d)))
print(ans)
```
No
| 35,493 | [
0.417724609375,
-0.043609619140625,
-0.0014600753784179688,
-0.07733154296875,
-0.55419921875,
-0.55859375,
0.01389312744140625,
0.334228515625,
0.1256103515625,
0.6962890625,
0.90185546875,
-0.0245513916015625,
-0.1417236328125,
-0.6044921875,
-0.50048828125,
0.1102294921875,
-0.461... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
SmallR is an archer. SmallR is taking a match of archer with Zanoes. They try to shoot in the target in turns, and SmallR shoots first. The probability of shooting the target each time is <image> for SmallR while <image> for Zanoes. The one who shoots in the target first should be the winner.
Output the probability that SmallR will win the match.
Input
A single line contains four integers <image>.
Output
Print a single real number, the probability that SmallR will win the match.
The answer will be considered correct if the absolute or relative error doesn't exceed 10 - 6.
Examples
Input
1 2 1 2
Output
0.666666666667
Submitted Solution:
```
a,b,c,d = map(int,input().split())
print((a/b)/(1-((c*c)/(d*d))))
```
No
| 35,494 | [
0.435791015625,
-0.026153564453125,
-0.047637939453125,
-0.033477783203125,
-0.54443359375,
-0.5517578125,
0.007232666015625,
0.324951171875,
0.11712646484375,
0.69677734375,
0.90771484375,
-0.0195159912109375,
-0.157958984375,
-0.6201171875,
-0.4658203125,
0.076171875,
-0.4514160156... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
SmallR is an archer. SmallR is taking a match of archer with Zanoes. They try to shoot in the target in turns, and SmallR shoots first. The probability of shooting the target each time is <image> for SmallR while <image> for Zanoes. The one who shoots in the target first should be the winner.
Output the probability that SmallR will win the match.
Input
A single line contains four integers <image>.
Output
Print a single real number, the probability that SmallR will win the match.
The answer will be considered correct if the absolute or relative error doesn't exceed 10 - 6.
Examples
Input
1 2 1 2
Output
0.666666666667
Submitted Solution:
```
a,b,c,d=map(int,input().split())
r=a/b
z=c/d
print(r/(1-(r*(1-z))))
```
No
| 35,495 | [
0.405029296875,
-0.0401611328125,
-0.002655029296875,
-0.035888671875,
-0.54541015625,
-0.56787109375,
0.00493621826171875,
0.343505859375,
0.0863037109375,
0.69677734375,
0.916015625,
-0.00893402099609375,
-0.1678466796875,
-0.60205078125,
-0.454345703125,
0.11346435546875,
-0.43701... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
SmallR is an archer. SmallR is taking a match of archer with Zanoes. They try to shoot in the target in turns, and SmallR shoots first. The probability of shooting the target each time is <image> for SmallR while <image> for Zanoes. The one who shoots in the target first should be the winner.
Output the probability that SmallR will win the match.
Input
A single line contains four integers <image>.
Output
Print a single real number, the probability that SmallR will win the match.
The answer will be considered correct if the absolute or relative error doesn't exceed 10 - 6.
Examples
Input
1 2 1 2
Output
0.666666666667
Submitted Solution:
```
a = input().split()
r1 = int(a[0]) / int(a[1])
r2 = int(a[2]) / int(a[3])
total = r1
last = 1-r1
for _ in range(300):
Bnotwin = last-(last*r2)
res = Bnotwin * r1
total += res
last = Bnotwin - res
print(total)
```
No
| 35,496 | [
0.373046875,
-0.08941650390625,
0.017608642578125,
0.0174560546875,
-0.52587890625,
-0.61181640625,
0.06793212890625,
0.351318359375,
0.12744140625,
0.6748046875,
0.90869140625,
-0.0628662109375,
-0.283447265625,
-0.609375,
-0.490966796875,
0.109619140625,
-0.4814453125,
-0.7138671... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Iahub and Sorin are the best competitive programmers in their town. However, they can't both qualify to an important contest. The selection will be made with the help of a single problem. Blatnatalag, a friend of Iahub, managed to get hold of the problem before the contest. Because he wants to make sure Iahub will be the one qualified, he tells Iahub the following task.
You're given an (1-based) array a with n elements. Let's define function f(i, j) (1 ≤ i, j ≤ n) as (i - j)2 + g(i, j)2. Function g is calculated by the following pseudo-code:
int g(int i, int j) {
int sum = 0;
for (int k = min(i, j) + 1; k <= max(i, j); k = k + 1)
sum = sum + a[k];
return sum;
}
Find a value mini ≠ j f(i, j).
Probably by now Iahub already figured out the solution to this problem. Can you?
Input
The first line of input contains a single integer n (2 ≤ n ≤ 100000). Next line contains n integers a[1], a[2], ..., a[n] ( - 104 ≤ a[i] ≤ 104).
Output
Output a single integer — the value of mini ≠ j f(i, j).
Examples
Input
4
1 0 0 -1
Output
1
Input
2
1 -1
Output
2
Tags: data structures, divide and conquer, geometry
Correct Solution:
```
import sys
max_Int = int(1e9)
class splitFeature:
def __init__(self, position, value):
self.position = position
self.value = value
def bruteForce(features, left, right):
min_distance = max_Int
for i in range(left, right):
for j in range(i+1, right):
min_distance = min(min_distance,
(features[i].position - features[j].position)**2 + (features[i].value - features[j].value)**2)
return min_distance
def enhanceData(features, left, right, mid, min_distance):
selected_population = []
for i in range(left, right):
if (features[i].position - features[mid].position) ** 2 <= min_distance:
selected_population.append(features[i])
selected_population.sort(key = lambda x: x.value)
l = len(selected_population)
result = max_Int
for i in range(l):
for j in range(i+1, l):
if (selected_population[i].value - selected_population[j].value) ** 2 >= min_distance:
break
distance = (selected_population[i].position - selected_population[j].position)**2 + (selected_population[i].value - selected_population[j].value)**2
result = min(result, distance)
return result
def analyzeData(features, left, right):
if right - left <= 3:
return bruteForce(features, left, right)
mid = (left + right) // 2
min_distance = min(analyzeData(features, left, mid),
analyzeData(features, mid+1, right))
return min(min_distance,
enhanceData(features, left, right, mid, min_distance))
def main():
n = int(sys.stdin.readline())
A = list(map(int, sys.stdin.readline().split()))
features = []
for i in range(n):
if (i > 0):
A[i] += A[i-1]
features.append(splitFeature(i , A[i]))
sys.stdout.write(str(analyzeData(features, 0, n)))
main() #optimizeCode
```
| 35,514 | [
0.427734375,
0.00458526611328125,
-0.0288238525390625,
0.362548828125,
-0.54833984375,
-0.36328125,
-0.09259033203125,
0.1683349609375,
0.34716796875,
0.60595703125,
0.5693359375,
0.039581298828125,
-0.137939453125,
-0.505859375,
-0.16796875,
-0.059295654296875,
-0.2890625,
-0.6127... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Iahub and Sorin are the best competitive programmers in their town. However, they can't both qualify to an important contest. The selection will be made with the help of a single problem. Blatnatalag, a friend of Iahub, managed to get hold of the problem before the contest. Because he wants to make sure Iahub will be the one qualified, he tells Iahub the following task.
You're given an (1-based) array a with n elements. Let's define function f(i, j) (1 ≤ i, j ≤ n) as (i - j)2 + g(i, j)2. Function g is calculated by the following pseudo-code:
int g(int i, int j) {
int sum = 0;
for (int k = min(i, j) + 1; k <= max(i, j); k = k + 1)
sum = sum + a[k];
return sum;
}
Find a value mini ≠ j f(i, j).
Probably by now Iahub already figured out the solution to this problem. Can you?
Input
The first line of input contains a single integer n (2 ≤ n ≤ 100000). Next line contains n integers a[1], a[2], ..., a[n] ( - 104 ≤ a[i] ≤ 104).
Output
Output a single integer — the value of mini ≠ j f(i, j).
Examples
Input
4
1 0 0 -1
Output
1
Input
2
1 -1
Output
2
Tags: data structures, divide and conquer, geometry
Correct Solution:
```
import sys
max_Int = int(1e9)
class splitFeature:
def __init__(self, position, value):
self.position = position
self.value = value
def bruteForce(features, left, right):
min_distance = max_Int
for i in range(left, right):
for j in range(i+1, right):
min_distance = min(min_distance,
(features[i].position - features[j].position)**2 + (features[i].value - features[j].value)**2)
return min_distance
def enhanceData(features, left, right, mid, min_distance):
selected_population = []
for i in range(left, right):
if (features[i].position - features[mid].position) ** 2 <= min_distance:
selected_population.append(features[i])
selected_population.sort(key = lambda x: x.value)
l = len(selected_population)
result = max_Int
for i in range(l):
for j in range(i+1, l):
if (selected_population[i].value - selected_population[j].value) ** 2 >= min_distance:
break
distance = (selected_population[i].position - selected_population[j].position)**2 + (selected_population[i].value - selected_population[j].value)**2
result = min(result, distance)
return result
def analyzeData(features, left, right):
if right - left <= 3:
return bruteForce(features, left, right)
mid = (left + right) // 2
min_distance = min(analyzeData(features, left, mid),
analyzeData(features, mid+1, right))
return min(min_distance,
enhanceData(features, left, right, mid, min_distance))
def main():
n = int(sys.stdin.readline())
A = list(map(int, sys.stdin.readline().split()))
features = []
for i in range(n):
if (i > 0):
A[i] += A[i-1]
features.append(splitFeature(i , A[i]))
sys.stdout.write(str(analyzeData(features, 0, n)))
main()
```
| 35,515 | [
0.427734375,
0.00458526611328125,
-0.0288238525390625,
0.362548828125,
-0.54833984375,
-0.36328125,
-0.09259033203125,
0.1683349609375,
0.34716796875,
0.60595703125,
0.5693359375,
0.039581298828125,
-0.137939453125,
-0.505859375,
-0.16796875,
-0.059295654296875,
-0.2890625,
-0.6127... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Iahub and Sorin are the best competitive programmers in their town. However, they can't both qualify to an important contest. The selection will be made with the help of a single problem. Blatnatalag, a friend of Iahub, managed to get hold of the problem before the contest. Because he wants to make sure Iahub will be the one qualified, he tells Iahub the following task.
You're given an (1-based) array a with n elements. Let's define function f(i, j) (1 ≤ i, j ≤ n) as (i - j)2 + g(i, j)2. Function g is calculated by the following pseudo-code:
int g(int i, int j) {
int sum = 0;
for (int k = min(i, j) + 1; k <= max(i, j); k = k + 1)
sum = sum + a[k];
return sum;
}
Find a value mini ≠ j f(i, j).
Probably by now Iahub already figured out the solution to this problem. Can you?
Input
The first line of input contains a single integer n (2 ≤ n ≤ 100000). Next line contains n integers a[1], a[2], ..., a[n] ( - 104 ≤ a[i] ≤ 104).
Output
Output a single integer — the value of mini ≠ j f(i, j).
Examples
Input
4
1 0 0 -1
Output
1
Input
2
1 -1
Output
2
Tags: data structures, divide and conquer, geometry
Correct Solution:
```
import sys
sys.setrecursionlimit(100000)
INF = 1e9
class Point:
def __init__(self, x, y):
self.x = x
self.y = y
def distance(p1, p2):
x = p1.x - p2.x
y = p1.y - p2.y
return (x ** 2 + y ** 2) ** 0.5
def brute_force(point_set, left, right):
min_dist = INF
for i in range(left, right):
for j in range(i + 1, right):
min_dist = min(min_dist, distance(point_set[i], point_set[j]))
return min_dist
def strip_closest(point_set, left, right, mid, dist_min):
point_mid = point_set[mid]
splitted_points = []
for i in range(left, right):
if abs(point_set[i].x - point_mid.x) <= dist_min:
splitted_points.append(point_set[i])
splitted_points.sort(key=lambda p: p.y)
smallest = INF
l = len(splitted_points)
for i in range(0, l):
for j in range(i + 1, l):
if not (splitted_points[j].y - splitted_points[i].y) < dist_min:
break
d = distance(splitted_points[i], splitted_points[j])
smallest = min(smallest, d)
return smallest
def closest_util(point_set, left, right):
if right - left <= 3:
return brute_force(point_set, left, right)
mid = (right + left) // 2
dist_left = closest_util(point_set, left, mid)
dist_right = closest_util(point_set, mid + 1, right)
dist_min = min(dist_left, dist_right)
return min(dist_min, strip_closest(point_set, left, right, mid, dist_min))
n = int(input())
a = list(map(int, input().split()))
point_set = []
for i in range(n):
if (i > 0):a[i] += a[i - 1]
point_set.append(Point(i , a[i]))
point_set.sort(key=lambda a: a.x)
ans = closest_util(point_set, 0, n)
print('%.0f' % (ans * ans))
```
| 35,516 | [
0.3828125,
-0.0162200927734375,
0.042205810546875,
0.339599609375,
-0.52880859375,
-0.319580078125,
-0.005466461181640625,
0.2149658203125,
0.369140625,
0.62451171875,
0.5869140625,
-0.0731201171875,
-0.0772705078125,
-0.56396484375,
-0.2374267578125,
-0.1287841796875,
-0.60595703125... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Iahub and Sorin are the best competitive programmers in their town. However, they can't both qualify to an important contest. The selection will be made with the help of a single problem. Blatnatalag, a friend of Iahub, managed to get hold of the problem before the contest. Because he wants to make sure Iahub will be the one qualified, he tells Iahub the following task.
You're given an (1-based) array a with n elements. Let's define function f(i, j) (1 ≤ i, j ≤ n) as (i - j)2 + g(i, j)2. Function g is calculated by the following pseudo-code:
int g(int i, int j) {
int sum = 0;
for (int k = min(i, j) + 1; k <= max(i, j); k = k + 1)
sum = sum + a[k];
return sum;
}
Find a value mini ≠ j f(i, j).
Probably by now Iahub already figured out the solution to this problem. Can you?
Input
The first line of input contains a single integer n (2 ≤ n ≤ 100000). Next line contains n integers a[1], a[2], ..., a[n] ( - 104 ≤ a[i] ≤ 104).
Output
Output a single integer — the value of mini ≠ j f(i, j).
Examples
Input
4
1 0 0 -1
Output
1
Input
2
1 -1
Output
2
Tags: data structures, divide and conquer, geometry
Correct Solution:
```
import sys
max_Int = int(1e9)
class splitFeature:
def __init__(self, position, value):
self.position = position
self.value = value
def bruteForce(features, left, right):
min_distance = max_Int
for i in range(left, right):
for j in range(i+1, right):
min_distance = min(min_distance,
(features[i].position - features[j].position)**2 + (features[i].value - features[j].value)**2)
return min_distance
def enhanceData(features, left, right, mid, min_distance):
selected_population = []
for i in range(left, right):
if (features[i].position - features[mid].position) ** 2 <= min_distance:
selected_population.append(features[i])
selected_population.sort(key = lambda x: x.value)
l = len(selected_population)
result = max_Int
for i in range(l):
for j in range(i+1, l):
if (selected_population[i].value - selected_population[j].value) ** 2 >= min_distance:
break
distance = (selected_population[i].position - selected_population[j].position)**2 + (selected_population[i].value - selected_population[j].value)**2
result = min(result, distance)
return result
def analyzeData(features, left, right):
if right - left <= 3:
return bruteForce(features, left, right)
mid = (left + right) // 2
min_distance = min(analyzeData(features, left, mid),
analyzeData(features, mid+1, right))
return min(min_distance,
enhanceData(features, left, right, mid, min_distance))
def main():
n = int(input())
A = list(map(int, input().split()))
features = []
for i in range(n):
if (i > 0):
A[i] += A[i-1]
features.append(splitFeature(i , A[i]))
print(analyzeData(features, 0, n))
main()
```
| 35,517 | [
0.427734375,
0.00458526611328125,
-0.0288238525390625,
0.362548828125,
-0.54833984375,
-0.36328125,
-0.09259033203125,
0.1683349609375,
0.34716796875,
0.60595703125,
0.5693359375,
0.039581298828125,
-0.137939453125,
-0.505859375,
-0.16796875,
-0.059295654296875,
-0.2890625,
-0.6127... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Iahub and Sorin are the best competitive programmers in their town. However, they can't both qualify to an important contest. The selection will be made with the help of a single problem. Blatnatalag, a friend of Iahub, managed to get hold of the problem before the contest. Because he wants to make sure Iahub will be the one qualified, he tells Iahub the following task.
You're given an (1-based) array a with n elements. Let's define function f(i, j) (1 ≤ i, j ≤ n) as (i - j)2 + g(i, j)2. Function g is calculated by the following pseudo-code:
int g(int i, int j) {
int sum = 0;
for (int k = min(i, j) + 1; k <= max(i, j); k = k + 1)
sum = sum + a[k];
return sum;
}
Find a value mini ≠ j f(i, j).
Probably by now Iahub already figured out the solution to this problem. Can you?
Input
The first line of input contains a single integer n (2 ≤ n ≤ 100000). Next line contains n integers a[1], a[2], ..., a[n] ( - 104 ≤ a[i] ≤ 104).
Output
Output a single integer — the value of mini ≠ j f(i, j).
Examples
Input
4
1 0 0 -1
Output
1
Input
2
1 -1
Output
2
Tags: data structures, divide and conquer, geometry
Correct Solution:
```
from sys import stdin, stdout
INF = int(1e9)
class Point:
def __init__(self, x=0, y=0):
self.x = x
self.y = y
def distance(p1, p2):
x = p1.x - p2.x
y = p1.y - p2.y
return x*x + y*y
def bruteForce(point_set, left, right):
min_dist = INF
for i in range(left, right):
for j in range(i+1, right):
min_dist = min(min_dist, distance(point_set[i], point_set[j]))
return min_dist
def stripClosest(point_set, left, right, mid, min_dist):
point_mid = point_set[mid]
splitted_points = []
for i in range(left, right):
if (point_set[i].x - point_mid.x) ** 2 <= min_dist:
splitted_points.append(point_set[i])
splitted_points.sort(key=lambda point: point.y)
l = len(splitted_points)
smallest = INF
for i in range(l):
for j in range(i+1, l):
if (splitted_points[i].y - splitted_points[j].y) ** 2 >= min_dist:
break
d = distance(splitted_points[i], splitted_points[j])
smallest = min(smallest, d)
return smallest
def closestUtil(point_set, left, right):
if right - left <= 3:
return bruteForce(point_set, left, right)
mid = (left + right) // 2
dist_left = closestUtil(point_set, left, mid)
dist_right = closestUtil(point_set, mid+1, right)
dist_min = min(dist_left, dist_right)
return min(dist_min, stripClosest(point_set, left, right, mid, dist_min))
n = int(stdin.readline())
a = list(map(int, stdin.readline().split()))
pref = [0]
for i in range(n):
pref.append(pref[i] + a[i])
point_set = []
for i in range(n):
point_set.append(Point(i, pref[i+1]))
ans = closestUtil(point_set, 0, n)
stdout.write(str(ans))
```
| 35,519 | [
0.378173828125,
-0.0491943359375,
0.02813720703125,
0.2958984375,
-0.56787109375,
-0.33349609375,
-0.08172607421875,
0.2283935546875,
0.330322265625,
0.6279296875,
0.59423828125,
-0.0286865234375,
-0.1082763671875,
-0.578125,
-0.248291015625,
-0.1695556640625,
-0.591796875,
-0.5336... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub and Sorin are the best competitive programmers in their town. However, they can't both qualify to an important contest. The selection will be made with the help of a single problem. Blatnatalag, a friend of Iahub, managed to get hold of the problem before the contest. Because he wants to make sure Iahub will be the one qualified, he tells Iahub the following task.
You're given an (1-based) array a with n elements. Let's define function f(i, j) (1 ≤ i, j ≤ n) as (i - j)2 + g(i, j)2. Function g is calculated by the following pseudo-code:
int g(int i, int j) {
int sum = 0;
for (int k = min(i, j) + 1; k <= max(i, j); k = k + 1)
sum = sum + a[k];
return sum;
}
Find a value mini ≠ j f(i, j).
Probably by now Iahub already figured out the solution to this problem. Can you?
Input
The first line of input contains a single integer n (2 ≤ n ≤ 100000). Next line contains n integers a[1], a[2], ..., a[n] ( - 104 ≤ a[i] ≤ 104).
Output
Output a single integer — the value of mini ≠ j f(i, j).
Examples
Input
4
1 0 0 -1
Output
1
Input
2
1 -1
Output
2
Submitted Solution:
```
def g(i,j,array):
s = 0
for k in range(min(i,j)+1,max(i,j)+1):
s += array[k]
return s
n = int(input())
array = input().split()
array = [int(i) for i in array]
result = []
for i in range(1,n):
result.append(1+g(i,i-1,array)**2)
print(min(result))
```
No
| 35,520 | [
0.3916015625,
0.09356689453125,
-0.18896484375,
0.189208984375,
-0.62060546875,
-0.361328125,
-0.1380615234375,
0.403564453125,
0.1986083984375,
0.62841796875,
0.63232421875,
-0.2122802734375,
-0.19677734375,
-0.66845703125,
-0.3134765625,
-0.251953125,
-0.4736328125,
-0.609375,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub and Sorin are the best competitive programmers in their town. However, they can't both qualify to an important contest. The selection will be made with the help of a single problem. Blatnatalag, a friend of Iahub, managed to get hold of the problem before the contest. Because he wants to make sure Iahub will be the one qualified, he tells Iahub the following task.
You're given an (1-based) array a with n elements. Let's define function f(i, j) (1 ≤ i, j ≤ n) as (i - j)2 + g(i, j)2. Function g is calculated by the following pseudo-code:
int g(int i, int j) {
int sum = 0;
for (int k = min(i, j) + 1; k <= max(i, j); k = k + 1)
sum = sum + a[k];
return sum;
}
Find a value mini ≠ j f(i, j).
Probably by now Iahub already figured out the solution to this problem. Can you?
Input
The first line of input contains a single integer n (2 ≤ n ≤ 100000). Next line contains n integers a[1], a[2], ..., a[n] ( - 104 ≤ a[i] ≤ 104).
Output
Output a single integer — the value of mini ≠ j f(i, j).
Examples
Input
4
1 0 0 -1
Output
1
Input
2
1 -1
Output
2
Submitted Solution:
```
import sys
sys.setrecursionlimit(1000000)
def classifyData(individual_1, individual_2):
return min(individual_1, individual_2)
def computeFitness(individual_i, individual_j):
return (individual_i[0] - individual_j[0])*(individual_i[0] - individual_j[0]) + (individual_i[1] - individual_j[1])*(individual_i[1] - individual_j[1])
def bruteForce(features, left, right):
min_distance = max_Int
for i in range(left, right):
for j in range(i+1, right):
min_distance = classifyData(min_distance,
computeFitness(features[i], features[j]))
return min_distance
def selectIndividual(features, left, right, mid, min_distance):
selected_population = []
for i in range(left, right):
if (features[i][0] - features[mid][0]) ** 2 <= min_distance:
selected_population.append(features[i])
return selected_population
def enhanceData(features, left, right, mid, min_distance):
selected_population = selectIndividual(features, left, right, mid, min_distance)
l = len(selected_population)
result = max_Int
for i in range(l):
for j in range(i+1, l):
#optimizeLoop
if (selected_population[i][1] - selected_population[j][1]) ** 2 >= min_distance:
break
distance = computeFitness(selected_population[i],
selected_population[j])
result = classifyData(result, distance)
return result
def analyzeData(features, left, right):
if right - left <= 3:
return bruteForce(features, left, right)
mid = (left + right) // 2
min_distance = classifyData(analyzeData(features, left, mid),
analyzeData(features, mid+1, right))
return classifyData(min_distance,
enhanceData(features, left, right, mid, min_distance))
#loadData
n = int(input())
A = list(map(int, input().split()))
#preprocessData
max_Int = int(1e9)
features = []
for i in range(n):
if (i > 0):
A[i] += A[i-1]
features.append([i , A[i]])
print(analyzeData(features, 0, n))
#print(bruteForce(features, 0, n))
```
No
| 35,521 | [
0.375244140625,
0.11798095703125,
-0.2132568359375,
0.178955078125,
-0.64306640625,
-0.07318115234375,
-0.20361328125,
0.29638671875,
0.2222900390625,
0.552734375,
0.60009765625,
-0.1700439453125,
-0.210205078125,
-0.671875,
-0.233642578125,
-0.1328125,
-0.4853515625,
-0.5375976562... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub and Sorin are the best competitive programmers in their town. However, they can't both qualify to an important contest. The selection will be made with the help of a single problem. Blatnatalag, a friend of Iahub, managed to get hold of the problem before the contest. Because he wants to make sure Iahub will be the one qualified, he tells Iahub the following task.
You're given an (1-based) array a with n elements. Let's define function f(i, j) (1 ≤ i, j ≤ n) as (i - j)2 + g(i, j)2. Function g is calculated by the following pseudo-code:
int g(int i, int j) {
int sum = 0;
for (int k = min(i, j) + 1; k <= max(i, j); k = k + 1)
sum = sum + a[k];
return sum;
}
Find a value mini ≠ j f(i, j).
Probably by now Iahub already figured out the solution to this problem. Can you?
Input
The first line of input contains a single integer n (2 ≤ n ≤ 100000). Next line contains n integers a[1], a[2], ..., a[n] ( - 104 ≤ a[i] ≤ 104).
Output
Output a single integer — the value of mini ≠ j f(i, j).
Examples
Input
4
1 0 0 -1
Output
1
Input
2
1 -1
Output
2
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))[1:]
res=[[a[0],1]]
for i in range(1,n-1):
if (res[-1][0]+a[i])**2+(res[-1][1]+1)**2<a[i]**2+1:
res.append([res[-1][0]+a[i],res[-1][1]+1])
else:
res.append([a[i],1])
m=res[0][0]**2+res[0][1]**2
for i in res:
if i[0]**2+i[1]**2<m:
m=i[0]**2+i[1]**2
print(m)
```
No
| 35,522 | [
0.292724609375,
0.15380859375,
-0.2049560546875,
0.0712890625,
-0.646484375,
-0.317138671875,
-0.144775390625,
0.304443359375,
0.20849609375,
0.732421875,
0.5791015625,
-0.193115234375,
-0.165283203125,
-0.69189453125,
-0.224609375,
-0.2734375,
-0.53173828125,
-0.58203125,
-0.185... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub and Sorin are the best competitive programmers in their town. However, they can't both qualify to an important contest. The selection will be made with the help of a single problem. Blatnatalag, a friend of Iahub, managed to get hold of the problem before the contest. Because he wants to make sure Iahub will be the one qualified, he tells Iahub the following task.
You're given an (1-based) array a with n elements. Let's define function f(i, j) (1 ≤ i, j ≤ n) as (i - j)2 + g(i, j)2. Function g is calculated by the following pseudo-code:
int g(int i, int j) {
int sum = 0;
for (int k = min(i, j) + 1; k <= max(i, j); k = k + 1)
sum = sum + a[k];
return sum;
}
Find a value mini ≠ j f(i, j).
Probably by now Iahub already figured out the solution to this problem. Can you?
Input
The first line of input contains a single integer n (2 ≤ n ≤ 100000). Next line contains n integers a[1], a[2], ..., a[n] ( - 104 ≤ a[i] ≤ 104).
Output
Output a single integer — the value of mini ≠ j f(i, j).
Examples
Input
4
1 0 0 -1
Output
1
Input
2
1 -1
Output
2
Submitted Solution:
```
import math
INF = 10**6
def calc(left, right):
min_dist = INF
for i in range(left,right):
for j in range(i+1,right):
min_dist = min(min_dist,distance(i,j))
return min_dist
def divide(left,right):
if right-left<=3:
return calc(left,right)
mid = (left+right)//2
min_right = divide(mid+1,right)
min_left = divide(left,mid)
return combine(min(min_left,min_right),mid,left,right)
def combine(dist,mid,left,right):
global s
new = []
for i in range(left,right):
if (i-mid)**2<=dist:
new.append(i)
new.sort(key=lambda x: s[x])
for i in range(len(new)-1):
for j in range(i+1,len(new)):
if (s[i]-s[j])**2>=dist:
break
dist = min(distance(new[i],new[j]),dist)
return dist
def distance(i,j):
global a,s
return (j-i)**2 + (s[j]-s[i])**2
n = int(input())
a = list(map(int,input().split()))
s = []
cur = 0
for i in a:
cur+=i
s.append(cur)
ans = divide(0,n)
print(ans)
```
No
| 35,523 | [
0.378662109375,
-0.021453857421875,
-0.258544921875,
0.128662109375,
-0.634765625,
-0.26806640625,
0.002986907958984375,
0.36962890625,
0.07806396484375,
0.806640625,
0.4853515625,
-0.1318359375,
-0.2353515625,
-0.72998046875,
-0.466796875,
-0.31640625,
-0.51611328125,
-0.480957031... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Makes solves problems on Decoforces and lots of other different online judges. Each problem is denoted by its difficulty — a positive integer number. Difficulties are measured the same across all the judges (the problem with difficulty d on Decoforces is as hard as the problem with difficulty d on any other judge).
Makes has chosen n problems to solve on Decoforces with difficulties a1, a2, ..., an. He can solve these problems in arbitrary order. Though he can solve problem i with difficulty ai only if he had already solved some problem with difficulty <image> (no matter on what online judge was it).
Before starting this chosen list of problems, Makes has already solved problems with maximum difficulty k.
With given conditions it's easy to see that Makes sometimes can't solve all the chosen problems, no matter what order he chooses. So he wants to solve some problems on other judges to finish solving problems from his list.
For every positive integer y there exist some problem with difficulty y on at least one judge besides Decoforces.
Makes can solve problems on any judge at any time, it isn't necessary to do problems from the chosen list one right after another.
Makes doesn't have too much free time, so he asked you to calculate the minimum number of problems he should solve on other judges in order to solve all the chosen problems from Decoforces.
Input
The first line contains two integer numbers n, k (1 ≤ n ≤ 103, 1 ≤ k ≤ 109).
The second line contains n space-separated integer numbers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
Print minimum number of problems Makes should solve on other judges in order to solve all chosen problems on Decoforces.
Examples
Input
3 3
2 1 9
Output
1
Input
4 20
10 3 6 3
Output
0
Note
In the first example Makes at first solves problems 1 and 2. Then in order to solve the problem with difficulty 9, he should solve problem with difficulty no less than 5. The only available are difficulties 5 and 6 on some other judge. Solving any of these will give Makes opportunity to solve problem 3.
In the second example he can solve every problem right from the start.
Tags: greedy, implementation
Correct Solution:
```
n,k = map(int,input().split())
arr = [int(x) for x in input().split()]
arr.sort()
ans = 0
i = 0
while(i < n):
if(arr[i] <= k):
i+=1
else:
break
diff = k
for i in range(i, n):
if(diff*2 < arr[i]):
ans += 1
diff<<=1
while(diff*2 < arr[i]):
diff<<=1
ans += 1
diff = arr[i]
print(ans)
```
| 35,677 | [
0.156005859375,
-0.11419677734375,
-0.1258544921875,
0.1663818359375,
-0.5400390625,
-0.6220703125,
-0.302001953125,
0.1361083984375,
-0.1370849609375,
1.0146484375,
0.5244140625,
-0.11724853515625,
0.452880859375,
-0.80810546875,
-0.2115478515625,
0.0183258056640625,
-0.603515625,
... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Makes solves problems on Decoforces and lots of other different online judges. Each problem is denoted by its difficulty — a positive integer number. Difficulties are measured the same across all the judges (the problem with difficulty d on Decoforces is as hard as the problem with difficulty d on any other judge).
Makes has chosen n problems to solve on Decoforces with difficulties a1, a2, ..., an. He can solve these problems in arbitrary order. Though he can solve problem i with difficulty ai only if he had already solved some problem with difficulty <image> (no matter on what online judge was it).
Before starting this chosen list of problems, Makes has already solved problems with maximum difficulty k.
With given conditions it's easy to see that Makes sometimes can't solve all the chosen problems, no matter what order he chooses. So he wants to solve some problems on other judges to finish solving problems from his list.
For every positive integer y there exist some problem with difficulty y on at least one judge besides Decoforces.
Makes can solve problems on any judge at any time, it isn't necessary to do problems from the chosen list one right after another.
Makes doesn't have too much free time, so he asked you to calculate the minimum number of problems he should solve on other judges in order to solve all the chosen problems from Decoforces.
Input
The first line contains two integer numbers n, k (1 ≤ n ≤ 103, 1 ≤ k ≤ 109).
The second line contains n space-separated integer numbers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
Print minimum number of problems Makes should solve on other judges in order to solve all chosen problems on Decoforces.
Examples
Input
3 3
2 1 9
Output
1
Input
4 20
10 3 6 3
Output
0
Note
In the first example Makes at first solves problems 1 and 2. Then in order to solve the problem with difficulty 9, he should solve problem with difficulty no less than 5. The only available are difficulties 5 and 6 on some other judge. Solving any of these will give Makes opportunity to solve problem 3.
In the second example he can solve every problem right from the start.
Tags: greedy, implementation
Correct Solution:
```
import sys
input = sys.stdin.readline
n, k = map(int, input().split())
a = sorted(map(int, input().split()))
k *= 2
ans = 0
for x in a:
while k < x:
k *= 2
ans += 1
k = max(k, x*2)
print(ans)
```
| 35,678 | [
0.1112060546875,
-0.132568359375,
-0.06787109375,
0.13232421875,
-0.580078125,
-0.59130859375,
-0.31005859375,
0.1510009765625,
-0.1817626953125,
1.021484375,
0.446044921875,
-0.154296875,
0.49462890625,
-0.75390625,
-0.2059326171875,
0.01119232177734375,
-0.59228515625,
-0.5405273... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Makes solves problems on Decoforces and lots of other different online judges. Each problem is denoted by its difficulty — a positive integer number. Difficulties are measured the same across all the judges (the problem with difficulty d on Decoforces is as hard as the problem with difficulty d on any other judge).
Makes has chosen n problems to solve on Decoforces with difficulties a1, a2, ..., an. He can solve these problems in arbitrary order. Though he can solve problem i with difficulty ai only if he had already solved some problem with difficulty <image> (no matter on what online judge was it).
Before starting this chosen list of problems, Makes has already solved problems with maximum difficulty k.
With given conditions it's easy to see that Makes sometimes can't solve all the chosen problems, no matter what order he chooses. So he wants to solve some problems on other judges to finish solving problems from his list.
For every positive integer y there exist some problem with difficulty y on at least one judge besides Decoforces.
Makes can solve problems on any judge at any time, it isn't necessary to do problems from the chosen list one right after another.
Makes doesn't have too much free time, so he asked you to calculate the minimum number of problems he should solve on other judges in order to solve all the chosen problems from Decoforces.
Input
The first line contains two integer numbers n, k (1 ≤ n ≤ 103, 1 ≤ k ≤ 109).
The second line contains n space-separated integer numbers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
Print minimum number of problems Makes should solve on other judges in order to solve all chosen problems on Decoforces.
Examples
Input
3 3
2 1 9
Output
1
Input
4 20
10 3 6 3
Output
0
Note
In the first example Makes at first solves problems 1 and 2. Then in order to solve the problem with difficulty 9, he should solve problem with difficulty no less than 5. The only available are difficulties 5 and 6 on some other judge. Solving any of these will give Makes opportunity to solve problem 3.
In the second example he can solve every problem right from the start.
Tags: greedy, implementation
Correct Solution:
```
n,k=map(int,input().split())
l=list(map(int,input().split()))
l.sort()
ka=0
for i in range(n):
while(l[i]>(2*k)):
k=k*2
ka=ka+1
k=max(k,l[i])
print(ka)
```
| 35,679 | [
0.1531982421875,
-0.1181640625,
-0.1373291015625,
0.1617431640625,
-0.55224609375,
-0.6005859375,
-0.2890625,
0.1856689453125,
-0.1490478515625,
1.009765625,
0.493408203125,
-0.1099853515625,
0.4677734375,
-0.8095703125,
-0.197265625,
-0.00519561767578125,
-0.607421875,
-0.51904296... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Makes solves problems on Decoforces and lots of other different online judges. Each problem is denoted by its difficulty — a positive integer number. Difficulties are measured the same across all the judges (the problem with difficulty d on Decoforces is as hard as the problem with difficulty d on any other judge).
Makes has chosen n problems to solve on Decoforces with difficulties a1, a2, ..., an. He can solve these problems in arbitrary order. Though he can solve problem i with difficulty ai only if he had already solved some problem with difficulty <image> (no matter on what online judge was it).
Before starting this chosen list of problems, Makes has already solved problems with maximum difficulty k.
With given conditions it's easy to see that Makes sometimes can't solve all the chosen problems, no matter what order he chooses. So he wants to solve some problems on other judges to finish solving problems from his list.
For every positive integer y there exist some problem with difficulty y on at least one judge besides Decoforces.
Makes can solve problems on any judge at any time, it isn't necessary to do problems from the chosen list one right after another.
Makes doesn't have too much free time, so he asked you to calculate the minimum number of problems he should solve on other judges in order to solve all the chosen problems from Decoforces.
Input
The first line contains two integer numbers n, k (1 ≤ n ≤ 103, 1 ≤ k ≤ 109).
The second line contains n space-separated integer numbers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
Print minimum number of problems Makes should solve on other judges in order to solve all chosen problems on Decoforces.
Examples
Input
3 3
2 1 9
Output
1
Input
4 20
10 3 6 3
Output
0
Note
In the first example Makes at first solves problems 1 and 2. Then in order to solve the problem with difficulty 9, he should solve problem with difficulty no less than 5. The only available are difficulties 5 and 6 on some other judge. Solving any of these will give Makes opportunity to solve problem 3.
In the second example he can solve every problem right from the start.
Tags: greedy, implementation
Correct Solution:
```
n, k = map(int, input().split())
dif = list(map(int, input().split()))
dif.sort()
i = 0
ans = 0
while i < n:
while i < n and dif[i] <= 2 * k:
k = max(k, dif[i])
i += 1
if i < n:
while 2 * k < dif[i]:
k *= 2
ans += 1
print(ans)
```
| 35,680 | [
0.1473388671875,
-0.1248779296875,
-0.1348876953125,
0.163818359375,
-0.5458984375,
-0.6044921875,
-0.28076171875,
0.1710205078125,
-0.1456298828125,
1.029296875,
0.479248046875,
-0.10479736328125,
0.471923828125,
-0.80908203125,
-0.1903076171875,
-0.00806427001953125,
-0.6103515625,... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Makes solves problems on Decoforces and lots of other different online judges. Each problem is denoted by its difficulty — a positive integer number. Difficulties are measured the same across all the judges (the problem with difficulty d on Decoforces is as hard as the problem with difficulty d on any other judge).
Makes has chosen n problems to solve on Decoforces with difficulties a1, a2, ..., an. He can solve these problems in arbitrary order. Though he can solve problem i with difficulty ai only if he had already solved some problem with difficulty <image> (no matter on what online judge was it).
Before starting this chosen list of problems, Makes has already solved problems with maximum difficulty k.
With given conditions it's easy to see that Makes sometimes can't solve all the chosen problems, no matter what order he chooses. So he wants to solve some problems on other judges to finish solving problems from his list.
For every positive integer y there exist some problem with difficulty y on at least one judge besides Decoforces.
Makes can solve problems on any judge at any time, it isn't necessary to do problems from the chosen list one right after another.
Makes doesn't have too much free time, so he asked you to calculate the minimum number of problems he should solve on other judges in order to solve all the chosen problems from Decoforces.
Input
The first line contains two integer numbers n, k (1 ≤ n ≤ 103, 1 ≤ k ≤ 109).
The second line contains n space-separated integer numbers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
Print minimum number of problems Makes should solve on other judges in order to solve all chosen problems on Decoforces.
Examples
Input
3 3
2 1 9
Output
1
Input
4 20
10 3 6 3
Output
0
Note
In the first example Makes at first solves problems 1 and 2. Then in order to solve the problem with difficulty 9, he should solve problem with difficulty no less than 5. The only available are difficulties 5 and 6 on some other judge. Solving any of these will give Makes opportunity to solve problem 3.
In the second example he can solve every problem right from the start.
Tags: greedy, implementation
Correct Solution:
```
n, k = map(int, input().split())
a = list(map(int, input().split()))
a = list(sorted(a))
d = k
ans = 0
for ai in a:
while 2 * d < ai:
d *= 2
ans += 1
d = max(d, ai)
print(ans)
```
| 35,681 | [
0.1510009765625,
-0.114013671875,
-0.142822265625,
0.1544189453125,
-0.55078125,
-0.60693359375,
-0.29150390625,
0.1881103515625,
-0.1456298828125,
1.009765625,
0.49560546875,
-0.10791015625,
0.465576171875,
-0.8095703125,
-0.1929931640625,
-0.00402069091796875,
-0.60302734375,
-0.... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Makes solves problems on Decoforces and lots of other different online judges. Each problem is denoted by its difficulty — a positive integer number. Difficulties are measured the same across all the judges (the problem with difficulty d on Decoforces is as hard as the problem with difficulty d on any other judge).
Makes has chosen n problems to solve on Decoforces with difficulties a1, a2, ..., an. He can solve these problems in arbitrary order. Though he can solve problem i with difficulty ai only if he had already solved some problem with difficulty <image> (no matter on what online judge was it).
Before starting this chosen list of problems, Makes has already solved problems with maximum difficulty k.
With given conditions it's easy to see that Makes sometimes can't solve all the chosen problems, no matter what order he chooses. So he wants to solve some problems on other judges to finish solving problems from his list.
For every positive integer y there exist some problem with difficulty y on at least one judge besides Decoforces.
Makes can solve problems on any judge at any time, it isn't necessary to do problems from the chosen list one right after another.
Makes doesn't have too much free time, so he asked you to calculate the minimum number of problems he should solve on other judges in order to solve all the chosen problems from Decoforces.
Input
The first line contains two integer numbers n, k (1 ≤ n ≤ 103, 1 ≤ k ≤ 109).
The second line contains n space-separated integer numbers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
Print minimum number of problems Makes should solve on other judges in order to solve all chosen problems on Decoforces.
Examples
Input
3 3
2 1 9
Output
1
Input
4 20
10 3 6 3
Output
0
Note
In the first example Makes at first solves problems 1 and 2. Then in order to solve the problem with difficulty 9, he should solve problem with difficulty no less than 5. The only available are difficulties 5 and 6 on some other judge. Solving any of these will give Makes opportunity to solve problem 3.
In the second example he can solve every problem right from the start.
Tags: greedy, implementation
Correct Solution:
```
n, k = map(int, input().split())
a = list(map(int, input().split()))
a.sort()
c = 0
for i in range(n):
if (k * 2 >= a[i]):
k = max(k, a[i])
else:
while(k * 2 < a[i]):
k *= 2
c += 1
k = max(k, a[i])
print (c)
```
| 35,682 | [
0.1510009765625,
-0.114013671875,
-0.142822265625,
0.1544189453125,
-0.55078125,
-0.60693359375,
-0.29150390625,
0.1881103515625,
-0.1456298828125,
1.009765625,
0.49560546875,
-0.10791015625,
0.465576171875,
-0.8095703125,
-0.1929931640625,
-0.00402069091796875,
-0.60302734375,
-0.... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Makes solves problems on Decoforces and lots of other different online judges. Each problem is denoted by its difficulty — a positive integer number. Difficulties are measured the same across all the judges (the problem with difficulty d on Decoforces is as hard as the problem with difficulty d on any other judge).
Makes has chosen n problems to solve on Decoforces with difficulties a1, a2, ..., an. He can solve these problems in arbitrary order. Though he can solve problem i with difficulty ai only if he had already solved some problem with difficulty <image> (no matter on what online judge was it).
Before starting this chosen list of problems, Makes has already solved problems with maximum difficulty k.
With given conditions it's easy to see that Makes sometimes can't solve all the chosen problems, no matter what order he chooses. So he wants to solve some problems on other judges to finish solving problems from his list.
For every positive integer y there exist some problem with difficulty y on at least one judge besides Decoforces.
Makes can solve problems on any judge at any time, it isn't necessary to do problems from the chosen list one right after another.
Makes doesn't have too much free time, so he asked you to calculate the minimum number of problems he should solve on other judges in order to solve all the chosen problems from Decoforces.
Input
The first line contains two integer numbers n, k (1 ≤ n ≤ 103, 1 ≤ k ≤ 109).
The second line contains n space-separated integer numbers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
Print minimum number of problems Makes should solve on other judges in order to solve all chosen problems on Decoforces.
Examples
Input
3 3
2 1 9
Output
1
Input
4 20
10 3 6 3
Output
0
Note
In the first example Makes at first solves problems 1 and 2. Then in order to solve the problem with difficulty 9, he should solve problem with difficulty no less than 5. The only available are difficulties 5 and 6 on some other judge. Solving any of these will give Makes opportunity to solve problem 3.
In the second example he can solve every problem right from the start.
Tags: greedy, implementation
Correct Solution:
```
#Bhargey Mehta (Sophomore)
#DA-IICT, Gandhinagar
import sys, math, queue, bisect
#sys.stdin = open("input.txt", "r")
MOD = 10**9+7
sys.setrecursionlimit(1000000)
n, k = map(int, input().split())
a = sorted(map(int, input().split()))
ans = 0
for i in range(n):
while a[i] > 2*k:
k = k*2
ans += 1
k = max(k, a[i])
print(ans)
```
| 35,683 | [
0.11578369140625,
-0.1375732421875,
-0.12017822265625,
0.264404296875,
-0.61669921875,
-0.546875,
-0.271240234375,
0.1259765625,
-0.139892578125,
1.0595703125,
0.53662109375,
-0.14208984375,
0.46337890625,
-0.78564453125,
-0.246337890625,
0.04913330078125,
-0.59423828125,
-0.469482... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Makes solves problems on Decoforces and lots of other different online judges. Each problem is denoted by its difficulty — a positive integer number. Difficulties are measured the same across all the judges (the problem with difficulty d on Decoforces is as hard as the problem with difficulty d on any other judge).
Makes has chosen n problems to solve on Decoforces with difficulties a1, a2, ..., an. He can solve these problems in arbitrary order. Though he can solve problem i with difficulty ai only if he had already solved some problem with difficulty <image> (no matter on what online judge was it).
Before starting this chosen list of problems, Makes has already solved problems with maximum difficulty k.
With given conditions it's easy to see that Makes sometimes can't solve all the chosen problems, no matter what order he chooses. So he wants to solve some problems on other judges to finish solving problems from his list.
For every positive integer y there exist some problem with difficulty y on at least one judge besides Decoforces.
Makes can solve problems on any judge at any time, it isn't necessary to do problems from the chosen list one right after another.
Makes doesn't have too much free time, so he asked you to calculate the minimum number of problems he should solve on other judges in order to solve all the chosen problems from Decoforces.
Input
The first line contains two integer numbers n, k (1 ≤ n ≤ 103, 1 ≤ k ≤ 109).
The second line contains n space-separated integer numbers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
Print minimum number of problems Makes should solve on other judges in order to solve all chosen problems on Decoforces.
Examples
Input
3 3
2 1 9
Output
1
Input
4 20
10 3 6 3
Output
0
Note
In the first example Makes at first solves problems 1 and 2. Then in order to solve the problem with difficulty 9, he should solve problem with difficulty no less than 5. The only available are difficulties 5 and 6 on some other judge. Solving any of these will give Makes opportunity to solve problem 3.
In the second example he can solve every problem right from the start.
Tags: greedy, implementation
Correct Solution:
```
n, k = map(int, input().split())
tasks = sorted(filter(lambda x: x > k, set(map(int, input().split()))))
res = 0
for task in tasks:
while (k << 1) < task:
k <<= 1
res += 1
k = task
print(res)
```
| 35,684 | [
0.1431884765625,
-0.1451416015625,
-0.1268310546875,
0.188720703125,
-0.5556640625,
-0.576171875,
-0.31591796875,
0.1619873046875,
-0.1558837890625,
1.0048828125,
0.47705078125,
-0.1319580078125,
0.466552734375,
-0.7763671875,
-0.21044921875,
0.01149749755859375,
-0.630859375,
-0.5... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Makes solves problems on Decoforces and lots of other different online judges. Each problem is denoted by its difficulty — a positive integer number. Difficulties are measured the same across all the judges (the problem with difficulty d on Decoforces is as hard as the problem with difficulty d on any other judge).
Makes has chosen n problems to solve on Decoforces with difficulties a1, a2, ..., an. He can solve these problems in arbitrary order. Though he can solve problem i with difficulty ai only if he had already solved some problem with difficulty <image> (no matter on what online judge was it).
Before starting this chosen list of problems, Makes has already solved problems with maximum difficulty k.
With given conditions it's easy to see that Makes sometimes can't solve all the chosen problems, no matter what order he chooses. So he wants to solve some problems on other judges to finish solving problems from his list.
For every positive integer y there exist some problem with difficulty y on at least one judge besides Decoforces.
Makes can solve problems on any judge at any time, it isn't necessary to do problems from the chosen list one right after another.
Makes doesn't have too much free time, so he asked you to calculate the minimum number of problems he should solve on other judges in order to solve all the chosen problems from Decoforces.
Input
The first line contains two integer numbers n, k (1 ≤ n ≤ 103, 1 ≤ k ≤ 109).
The second line contains n space-separated integer numbers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
Print minimum number of problems Makes should solve on other judges in order to solve all chosen problems on Decoforces.
Examples
Input
3 3
2 1 9
Output
1
Input
4 20
10 3 6 3
Output
0
Note
In the first example Makes at first solves problems 1 and 2. Then in order to solve the problem with difficulty 9, he should solve problem with difficulty no less than 5. The only available are difficulties 5 and 6 on some other judge. Solving any of these will give Makes opportunity to solve problem 3.
In the second example he can solve every problem right from the start.
Submitted Solution:
```
n,k=map(int,input().split())
a=[int(c) for c in input().split()]
a.sort()
i=0
res=0
while i<n:
if k>=a[i]/2:
k=max(a[i],k)
i+=1
else:
k*=2
res+=1
print(res)
```
Yes
| 35,685 | [
0.1614990234375,
-0.00344085693359375,
-0.09954833984375,
0.138916015625,
-0.60693359375,
-0.5615234375,
-0.30224609375,
0.26220703125,
-0.2435302734375,
1.0703125,
0.404296875,
-0.005382537841796875,
0.423095703125,
-0.80712890625,
-0.192626953125,
-0.05401611328125,
-0.580078125,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Makes solves problems on Decoforces and lots of other different online judges. Each problem is denoted by its difficulty — a positive integer number. Difficulties are measured the same across all the judges (the problem with difficulty d on Decoforces is as hard as the problem with difficulty d on any other judge).
Makes has chosen n problems to solve on Decoforces with difficulties a1, a2, ..., an. He can solve these problems in arbitrary order. Though he can solve problem i with difficulty ai only if he had already solved some problem with difficulty <image> (no matter on what online judge was it).
Before starting this chosen list of problems, Makes has already solved problems with maximum difficulty k.
With given conditions it's easy to see that Makes sometimes can't solve all the chosen problems, no matter what order he chooses. So he wants to solve some problems on other judges to finish solving problems from his list.
For every positive integer y there exist some problem with difficulty y on at least one judge besides Decoforces.
Makes can solve problems on any judge at any time, it isn't necessary to do problems from the chosen list one right after another.
Makes doesn't have too much free time, so he asked you to calculate the minimum number of problems he should solve on other judges in order to solve all the chosen problems from Decoforces.
Input
The first line contains two integer numbers n, k (1 ≤ n ≤ 103, 1 ≤ k ≤ 109).
The second line contains n space-separated integer numbers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
Print minimum number of problems Makes should solve on other judges in order to solve all chosen problems on Decoforces.
Examples
Input
3 3
2 1 9
Output
1
Input
4 20
10 3 6 3
Output
0
Note
In the first example Makes at first solves problems 1 and 2. Then in order to solve the problem with difficulty 9, he should solve problem with difficulty no less than 5. The only available are difficulties 5 and 6 on some other judge. Solving any of these will give Makes opportunity to solve problem 3.
In the second example he can solve every problem right from the start.
Submitted Solution:
```
def judge():
count = 0
S = input()
T = input()
sList = S.split()
sList = [int(i) for i in sList]
tList = T.split()
n = sList[0]
k = sList[1]
tList = [int(i) for i in tList]
uList = sorted(tList)
for num in uList:
if (2*k >= num):
if (k<num):
k = num
pass
else:
while True:
k = 2*k
count += 1
if (2*k >= num):
if (k < num):
k = num
break
print(count)
judge()
```
Yes
| 35,686 | [
0.171142578125,
-0.02520751953125,
-0.0718994140625,
0.128662109375,
-0.599609375,
-0.552734375,
-0.273193359375,
0.26611328125,
-0.2196044921875,
1.0458984375,
0.39453125,
-0.06402587890625,
0.4375,
-0.78369140625,
-0.1727294921875,
-0.0677490234375,
-0.5654296875,
-0.436767578125... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Makes solves problems on Decoforces and lots of other different online judges. Each problem is denoted by its difficulty — a positive integer number. Difficulties are measured the same across all the judges (the problem with difficulty d on Decoforces is as hard as the problem with difficulty d on any other judge).
Makes has chosen n problems to solve on Decoforces with difficulties a1, a2, ..., an. He can solve these problems in arbitrary order. Though he can solve problem i with difficulty ai only if he had already solved some problem with difficulty <image> (no matter on what online judge was it).
Before starting this chosen list of problems, Makes has already solved problems with maximum difficulty k.
With given conditions it's easy to see that Makes sometimes can't solve all the chosen problems, no matter what order he chooses. So he wants to solve some problems on other judges to finish solving problems from his list.
For every positive integer y there exist some problem with difficulty y on at least one judge besides Decoforces.
Makes can solve problems on any judge at any time, it isn't necessary to do problems from the chosen list one right after another.
Makes doesn't have too much free time, so he asked you to calculate the minimum number of problems he should solve on other judges in order to solve all the chosen problems from Decoforces.
Input
The first line contains two integer numbers n, k (1 ≤ n ≤ 103, 1 ≤ k ≤ 109).
The second line contains n space-separated integer numbers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
Print minimum number of problems Makes should solve on other judges in order to solve all chosen problems on Decoforces.
Examples
Input
3 3
2 1 9
Output
1
Input
4 20
10 3 6 3
Output
0
Note
In the first example Makes at first solves problems 1 and 2. Then in order to solve the problem with difficulty 9, he should solve problem with difficulty no less than 5. The only available are difficulties 5 and 6 on some other judge. Solving any of these will give Makes opportunity to solve problem 3.
In the second example he can solve every problem right from the start.
Submitted Solution:
```
n, k = map(int, input().split())
a = list(map(int, input().split()))
a = sorted(a)
num = 0
for i in range(n):
if 2 * k >= a[i]:
if k < a[i]:
k = a[i]
else:
while 2 * k < a[i]:
k = 2 * k
num += 1
if k < a[i]:
k = a[i]
print(num)
```
Yes
| 35,687 | [
0.1617431640625,
-0.00519561767578125,
-0.11126708984375,
0.14208984375,
-0.6103515625,
-0.56298828125,
-0.305908203125,
0.277587890625,
-0.246337890625,
1.072265625,
0.399658203125,
-0.00036072731018066406,
0.44189453125,
-0.8017578125,
-0.182373046875,
-0.05145263671875,
-0.578125,... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Makes solves problems on Decoforces and lots of other different online judges. Each problem is denoted by its difficulty — a positive integer number. Difficulties are measured the same across all the judges (the problem with difficulty d on Decoforces is as hard as the problem with difficulty d on any other judge).
Makes has chosen n problems to solve on Decoforces with difficulties a1, a2, ..., an. He can solve these problems in arbitrary order. Though he can solve problem i with difficulty ai only if he had already solved some problem with difficulty <image> (no matter on what online judge was it).
Before starting this chosen list of problems, Makes has already solved problems with maximum difficulty k.
With given conditions it's easy to see that Makes sometimes can't solve all the chosen problems, no matter what order he chooses. So he wants to solve some problems on other judges to finish solving problems from his list.
For every positive integer y there exist some problem with difficulty y on at least one judge besides Decoforces.
Makes can solve problems on any judge at any time, it isn't necessary to do problems from the chosen list one right after another.
Makes doesn't have too much free time, so he asked you to calculate the minimum number of problems he should solve on other judges in order to solve all the chosen problems from Decoforces.
Input
The first line contains two integer numbers n, k (1 ≤ n ≤ 103, 1 ≤ k ≤ 109).
The second line contains n space-separated integer numbers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
Print minimum number of problems Makes should solve on other judges in order to solve all chosen problems on Decoforces.
Examples
Input
3 3
2 1 9
Output
1
Input
4 20
10 3 6 3
Output
0
Note
In the first example Makes at first solves problems 1 and 2. Then in order to solve the problem with difficulty 9, he should solve problem with difficulty no less than 5. The only available are difficulties 5 and 6 on some other judge. Solving any of these will give Makes opportunity to solve problem 3.
In the second example he can solve every problem right from the start.
Submitted Solution:
```
import sys
a = sys.stdin.readline().strip('\n')
n,m = map(int,a.split(' '))
list1 = list(map(int,input().split(' ')))
sum = 0
list1.sort()
for i in list1:
while 2*m < i:
m *= 2
sum += 1
if i > m:
m = i
print(sum)
```
Yes
| 35,688 | [
0.1461181640625,
-0.037628173828125,
-0.05120849609375,
0.1483154296875,
-0.615234375,
-0.51220703125,
-0.268798828125,
0.265380859375,
-0.240966796875,
1.1083984375,
0.3642578125,
-0.0208892822265625,
0.44482421875,
-0.77001953125,
-0.199951171875,
-0.048828125,
-0.52587890625,
-0... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Makes solves problems on Decoforces and lots of other different online judges. Each problem is denoted by its difficulty — a positive integer number. Difficulties are measured the same across all the judges (the problem with difficulty d on Decoforces is as hard as the problem with difficulty d on any other judge).
Makes has chosen n problems to solve on Decoforces with difficulties a1, a2, ..., an. He can solve these problems in arbitrary order. Though he can solve problem i with difficulty ai only if he had already solved some problem with difficulty <image> (no matter on what online judge was it).
Before starting this chosen list of problems, Makes has already solved problems with maximum difficulty k.
With given conditions it's easy to see that Makes sometimes can't solve all the chosen problems, no matter what order he chooses. So he wants to solve some problems on other judges to finish solving problems from his list.
For every positive integer y there exist some problem with difficulty y on at least one judge besides Decoforces.
Makes can solve problems on any judge at any time, it isn't necessary to do problems from the chosen list one right after another.
Makes doesn't have too much free time, so he asked you to calculate the minimum number of problems he should solve on other judges in order to solve all the chosen problems from Decoforces.
Input
The first line contains two integer numbers n, k (1 ≤ n ≤ 103, 1 ≤ k ≤ 109).
The second line contains n space-separated integer numbers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
Print minimum number of problems Makes should solve on other judges in order to solve all chosen problems on Decoforces.
Examples
Input
3 3
2 1 9
Output
1
Input
4 20
10 3 6 3
Output
0
Note
In the first example Makes at first solves problems 1 and 2. Then in order to solve the problem with difficulty 9, he should solve problem with difficulty no less than 5. The only available are difficulties 5 and 6 on some other judge. Solving any of these will give Makes opportunity to solve problem 3.
In the second example he can solve every problem right from the start.
Submitted Solution:
```
# ===================================
# (c) MidAndFeed aka ASilentVoice
# ===================================
# import math, fractions, collections
# ===================================
n, k = [int(x) for x in input().split()]
q = [int(x) for x in input().split()]
q.sort()
ans = 0
for x in q:
if x <= 2*k:
k = max(x, k)
else:
k *= 2
ans += 1
print(ans)
```
No
| 35,689 | [
0.1748046875,
-0.022186279296875,
-0.10205078125,
0.2027587890625,
-0.638671875,
-0.52099609375,
-0.27880859375,
0.302490234375,
-0.2249755859375,
1.083984375,
0.410400390625,
-0.0241241455078125,
0.402099609375,
-0.78076171875,
-0.199462890625,
-0.047119140625,
-0.5517578125,
-0.4... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Makes solves problems on Decoforces and lots of other different online judges. Each problem is denoted by its difficulty — a positive integer number. Difficulties are measured the same across all the judges (the problem with difficulty d on Decoforces is as hard as the problem with difficulty d on any other judge).
Makes has chosen n problems to solve on Decoforces with difficulties a1, a2, ..., an. He can solve these problems in arbitrary order. Though he can solve problem i with difficulty ai only if he had already solved some problem with difficulty <image> (no matter on what online judge was it).
Before starting this chosen list of problems, Makes has already solved problems with maximum difficulty k.
With given conditions it's easy to see that Makes sometimes can't solve all the chosen problems, no matter what order he chooses. So he wants to solve some problems on other judges to finish solving problems from his list.
For every positive integer y there exist some problem with difficulty y on at least one judge besides Decoforces.
Makes can solve problems on any judge at any time, it isn't necessary to do problems from the chosen list one right after another.
Makes doesn't have too much free time, so he asked you to calculate the minimum number of problems he should solve on other judges in order to solve all the chosen problems from Decoforces.
Input
The first line contains two integer numbers n, k (1 ≤ n ≤ 103, 1 ≤ k ≤ 109).
The second line contains n space-separated integer numbers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
Print minimum number of problems Makes should solve on other judges in order to solve all chosen problems on Decoforces.
Examples
Input
3 3
2 1 9
Output
1
Input
4 20
10 3 6 3
Output
0
Note
In the first example Makes at first solves problems 1 and 2. Then in order to solve the problem with difficulty 9, he should solve problem with difficulty no less than 5. The only available are difficulties 5 and 6 on some other judge. Solving any of these will give Makes opportunity to solve problem 3.
In the second example he can solve every problem right from the start.
Submitted Solution:
```
n, IQ = map(int, input().split())
a = [int(x) for x in input().split()]
ans = 0
for i in a:
if IQ < i:
ans += 1
IQ *= 2
print(ans)
```
No
| 35,690 | [
0.17041015625,
0.0170745849609375,
-0.1002197265625,
0.1546630859375,
-0.61962890625,
-0.57373046875,
-0.31005859375,
0.247802734375,
-0.248291015625,
1.078125,
0.384521484375,
0.017547607421875,
0.3974609375,
-0.82275390625,
-0.179931640625,
-0.050262451171875,
-0.57666015625,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Makes solves problems on Decoforces and lots of other different online judges. Each problem is denoted by its difficulty — a positive integer number. Difficulties are measured the same across all the judges (the problem with difficulty d on Decoforces is as hard as the problem with difficulty d on any other judge).
Makes has chosen n problems to solve on Decoforces with difficulties a1, a2, ..., an. He can solve these problems in arbitrary order. Though he can solve problem i with difficulty ai only if he had already solved some problem with difficulty <image> (no matter on what online judge was it).
Before starting this chosen list of problems, Makes has already solved problems with maximum difficulty k.
With given conditions it's easy to see that Makes sometimes can't solve all the chosen problems, no matter what order he chooses. So he wants to solve some problems on other judges to finish solving problems from his list.
For every positive integer y there exist some problem with difficulty y on at least one judge besides Decoforces.
Makes can solve problems on any judge at any time, it isn't necessary to do problems from the chosen list one right after another.
Makes doesn't have too much free time, so he asked you to calculate the minimum number of problems he should solve on other judges in order to solve all the chosen problems from Decoforces.
Input
The first line contains two integer numbers n, k (1 ≤ n ≤ 103, 1 ≤ k ≤ 109).
The second line contains n space-separated integer numbers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
Print minimum number of problems Makes should solve on other judges in order to solve all chosen problems on Decoforces.
Examples
Input
3 3
2 1 9
Output
1
Input
4 20
10 3 6 3
Output
0
Note
In the first example Makes at first solves problems 1 and 2. Then in order to solve the problem with difficulty 9, he should solve problem with difficulty no less than 5. The only available are difficulties 5 and 6 on some other judge. Solving any of these will give Makes opportunity to solve problem 3.
In the second example he can solve every problem right from the start.
Submitted Solution:
```
N, K = list(map(int, input().split(' ')))
a = list(map(int, input().split(' ')))
m = max(a) / 2
i = K
j = 0
while i < m:
i *= 2
j += 1
print(j)
```
No
| 35,691 | [
0.1519775390625,
-0.0206298828125,
-0.0888671875,
0.14111328125,
-0.61572265625,
-0.5390625,
-0.30126953125,
0.2802734375,
-0.2421875,
1.0546875,
0.3984375,
0.004093170166015625,
0.435302734375,
-0.7939453125,
-0.1905517578125,
-0.04461669921875,
-0.5703125,
-0.441650390625,
-0.6... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Makes solves problems on Decoforces and lots of other different online judges. Each problem is denoted by its difficulty — a positive integer number. Difficulties are measured the same across all the judges (the problem with difficulty d on Decoforces is as hard as the problem with difficulty d on any other judge).
Makes has chosen n problems to solve on Decoforces with difficulties a1, a2, ..., an. He can solve these problems in arbitrary order. Though he can solve problem i with difficulty ai only if he had already solved some problem with difficulty <image> (no matter on what online judge was it).
Before starting this chosen list of problems, Makes has already solved problems with maximum difficulty k.
With given conditions it's easy to see that Makes sometimes can't solve all the chosen problems, no matter what order he chooses. So he wants to solve some problems on other judges to finish solving problems from his list.
For every positive integer y there exist some problem with difficulty y on at least one judge besides Decoforces.
Makes can solve problems on any judge at any time, it isn't necessary to do problems from the chosen list one right after another.
Makes doesn't have too much free time, so he asked you to calculate the minimum number of problems he should solve on other judges in order to solve all the chosen problems from Decoforces.
Input
The first line contains two integer numbers n, k (1 ≤ n ≤ 103, 1 ≤ k ≤ 109).
The second line contains n space-separated integer numbers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
Print minimum number of problems Makes should solve on other judges in order to solve all chosen problems on Decoforces.
Examples
Input
3 3
2 1 9
Output
1
Input
4 20
10 3 6 3
Output
0
Note
In the first example Makes at first solves problems 1 and 2. Then in order to solve the problem with difficulty 9, he should solve problem with difficulty no less than 5. The only available are difficulties 5 and 6 on some other judge. Solving any of these will give Makes opportunity to solve problem 3.
In the second example he can solve every problem right from the start.
Submitted Solution:
```
import math
o=lambda:[int(f)for f in input().split()]
n,k=o()
a=sorted(o())
i=0
for x in a:
if math.ceil(x/2)>k:
k=x
i+=1
print(i)
```
No
| 35,692 | [
0.1766357421875,
0.0067596435546875,
-0.1207275390625,
0.161865234375,
-0.59814453125,
-0.5078125,
-0.26220703125,
0.2861328125,
-0.1943359375,
1.0810546875,
0.405517578125,
-0.0341796875,
0.391845703125,
-0.79736328125,
-0.175537109375,
-0.0869140625,
-0.5556640625,
-0.4365234375,... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting only of characters + and -. You perform some process with this string. This process can be described by the following pseudocode:
res = 0
for init = 0 to inf
cur = init
ok = true
for i = 1 to |s|
res = res + 1
if s[i] == '+'
cur = cur + 1
else
cur = cur - 1
if cur < 0
ok = false
break
if ok
break
Note that the inf denotes infinity, and the characters of the string are numbered from 1 to |s|.
You have to calculate the value of the res after the process ends.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The only lines of each test case contains string s (1 ≤ |s| ≤ 10^6) consisting only of characters + and -.
It's guaranteed that sum of |s| over all test cases doesn't exceed 10^6.
Output
For each test case print one integer — the value of the res after the process ends.
Example
Input
3
--+-
---
++--+-
Output
7
9
6
Submitted Solution:
```
import math
import bisect
import sys
input = sys.stdin.readline
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(s[:len(s)-1])
def invr():
return(map(int,input().split()))
# N, M = inlt()
# A = inlt()
T = inp()
# print(T)
for t in range(T):
s = insr()
res = 0
c = 0
init = 0
for i in range(len(s)):
if s[i] == "-":
c -= 1
else:
c += 1
if c + init < 0:
init += 1
res += i+1
print(res+len(s))
```
Yes
| 36,253 | [
0.1619873046875,
0.1778564453125,
-0.0762939453125,
-0.1849365234375,
-0.77001953125,
-0.351806640625,
0.133056640625,
-0.072265625,
0.08343505859375,
0.83837890625,
0.5537109375,
-0.0723876953125,
0.2939453125,
-0.9736328125,
-0.414306640625,
-0.286865234375,
-0.5751953125,
-0.819... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting only of characters + and -. You perform some process with this string. This process can be described by the following pseudocode:
res = 0
for init = 0 to inf
cur = init
ok = true
for i = 1 to |s|
res = res + 1
if s[i] == '+'
cur = cur + 1
else
cur = cur - 1
if cur < 0
ok = false
break
if ok
break
Note that the inf denotes infinity, and the characters of the string are numbered from 1 to |s|.
You have to calculate the value of the res after the process ends.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The only lines of each test case contains string s (1 ≤ |s| ≤ 10^6) consisting only of characters + and -.
It's guaranteed that sum of |s| over all test cases doesn't exceed 10^6.
Output
For each test case print one integer — the value of the res after the process ends.
Example
Input
3
--+-
---
++--+-
Output
7
9
6
Submitted Solution:
```
for _ in " "*int(input()):
a=input();res=len(a);k,f=0,0
for i in range(len(a)):
k+= -1 if a[i]=="-" else 1
if k<f:res+=(i+1);f=k
print(res)
```
Yes
| 36,254 | [
0.2705078125,
0.1396484375,
-0.022705078125,
-0.11138916015625,
-0.61328125,
-0.35791015625,
0.1533203125,
-0.07354736328125,
0.1318359375,
0.755859375,
0.50732421875,
-0.146728515625,
0.250732421875,
-0.84765625,
-0.368896484375,
-0.26025390625,
-0.57666015625,
-0.658203125,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting only of characters + and -. You perform some process with this string. This process can be described by the following pseudocode:
res = 0
for init = 0 to inf
cur = init
ok = true
for i = 1 to |s|
res = res + 1
if s[i] == '+'
cur = cur + 1
else
cur = cur - 1
if cur < 0
ok = false
break
if ok
break
Note that the inf denotes infinity, and the characters of the string are numbered from 1 to |s|.
You have to calculate the value of the res after the process ends.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The only lines of each test case contains string s (1 ≤ |s| ≤ 10^6) consisting only of characters + and -.
It's guaranteed that sum of |s| over all test cases doesn't exceed 10^6.
Output
For each test case print one integer — the value of the res after the process ends.
Example
Input
3
--+-
---
++--+-
Output
7
9
6
Submitted Solution:
```
for _ in range(int(input())):
s = input()
cur, mn, res = 0, 0, len(s)
for i in range(len(s)):
if s[i] == '+':
cur += 1
else:
cur -= 1
if cur < mn:
mn = cur
res += i + 1
#print("res: ", res)
print(res)
```
Yes
| 36,255 | [
0.2286376953125,
0.0872802734375,
-0.005580902099609375,
-0.1151123046875,
-0.62158203125,
-0.366943359375,
0.208251953125,
-0.089599609375,
0.169677734375,
0.7626953125,
0.54541015625,
-0.12384033203125,
0.2410888671875,
-0.873046875,
-0.3818359375,
-0.285400390625,
-0.56591796875,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting only of characters + and -. You perform some process with this string. This process can be described by the following pseudocode:
res = 0
for init = 0 to inf
cur = init
ok = true
for i = 1 to |s|
res = res + 1
if s[i] == '+'
cur = cur + 1
else
cur = cur - 1
if cur < 0
ok = false
break
if ok
break
Note that the inf denotes infinity, and the characters of the string are numbered from 1 to |s|.
You have to calculate the value of the res after the process ends.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The only lines of each test case contains string s (1 ≤ |s| ≤ 10^6) consisting only of characters + and -.
It's guaranteed that sum of |s| over all test cases doesn't exceed 10^6.
Output
For each test case print one integer — the value of the res after the process ends.
Example
Input
3
--+-
---
++--+-
Output
7
9
6
Submitted Solution:
```
for t in range(int(input())):
s=input()
res=0
for init in range(0,10):
cur=init
ok=True
for i in range(0,len(s)):
res+=1
if s[i] == '+':
cur = cur + 1
else:
cur = cur - 1
if cur < 0:
ok =False
break
if ok:
break
init+=1
print(res)
```
No
| 36,256 | [
0.2413330078125,
0.1302490234375,
-0.032562255859375,
-0.06475830078125,
-0.61962890625,
-0.42529296875,
0.2012939453125,
-0.060516357421875,
0.136962890625,
0.7607421875,
0.5107421875,
-0.1363525390625,
0.2491455078125,
-0.85595703125,
-0.3759765625,
-0.2490234375,
-0.5673828125,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting only of characters + and -. You perform some process with this string. This process can be described by the following pseudocode:
res = 0
for init = 0 to inf
cur = init
ok = true
for i = 1 to |s|
res = res + 1
if s[i] == '+'
cur = cur + 1
else
cur = cur - 1
if cur < 0
ok = false
break
if ok
break
Note that the inf denotes infinity, and the characters of the string are numbered from 1 to |s|.
You have to calculate the value of the res after the process ends.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The only lines of each test case contains string s (1 ≤ |s| ≤ 10^6) consisting only of characters + and -.
It's guaranteed that sum of |s| over all test cases doesn't exceed 10^6.
Output
For each test case print one integer — the value of the res after the process ends.
Example
Input
3
--+-
---
++--+-
Output
7
9
6
Submitted Solution:
```
t = int(input())
for i in range(t):
s = input()
res = 0
for x in range(0,120):
cur = x
ok = True
for i in s:
res+=1
if i == "+":
cur+=1
else:
cur-=1
if cur<0:
ok= False
break
if ok:
break
print(res)
```
No
| 36,257 | [
0.2861328125,
0.123046875,
0.01013946533203125,
-0.09112548828125,
-0.61181640625,
-0.390380859375,
0.17529296875,
-0.0587158203125,
0.1676025390625,
0.744140625,
0.5244140625,
-0.1171875,
0.22021484375,
-0.845703125,
-0.344970703125,
-0.30615234375,
-0.55029296875,
-0.66357421875,... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting only of characters + and -. You perform some process with this string. This process can be described by the following pseudocode:
res = 0
for init = 0 to inf
cur = init
ok = true
for i = 1 to |s|
res = res + 1
if s[i] == '+'
cur = cur + 1
else
cur = cur - 1
if cur < 0
ok = false
break
if ok
break
Note that the inf denotes infinity, and the characters of the string are numbered from 1 to |s|.
You have to calculate the value of the res after the process ends.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The only lines of each test case contains string s (1 ≤ |s| ≤ 10^6) consisting only of characters + and -.
It's guaranteed that sum of |s| over all test cases doesn't exceed 10^6.
Output
For each test case print one integer — the value of the res after the process ends.
Example
Input
3
--+-
---
++--+-
Output
7
9
6
Submitted Solution:
```
for _ in range(int(input())):
s=input()
minus=0
plus=0
count=0
for i in range(len(s)):
if i==0:
if s[i]=='-':
count+=2
minus+=1
else:
count+=1
plus+=1
else:
if s[i]=='-' and s[i-1]=='+':
count+=1
minus+=1
elif s[i]=='-' and s[i-1]=='-':
minus+=1
if minus>plus:
count+=(i+2)
else:
count+=1
else:
count+=1
plus+=1
print(count)
```
No
| 36,258 | [
0.2410888671875,
0.1304931640625,
-0.08782958984375,
-0.05010986328125,
-0.6162109375,
-0.45751953125,
0.1832275390625,
-0.0125274658203125,
0.11865234375,
0.78955078125,
0.476318359375,
-0.197265625,
0.237548828125,
-0.82666015625,
-0.458740234375,
-0.293701171875,
-0.59423828125,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting only of characters + and -. You perform some process with this string. This process can be described by the following pseudocode:
res = 0
for init = 0 to inf
cur = init
ok = true
for i = 1 to |s|
res = res + 1
if s[i] == '+'
cur = cur + 1
else
cur = cur - 1
if cur < 0
ok = false
break
if ok
break
Note that the inf denotes infinity, and the characters of the string are numbered from 1 to |s|.
You have to calculate the value of the res after the process ends.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The only lines of each test case contains string s (1 ≤ |s| ≤ 10^6) consisting only of characters + and -.
It's guaranteed that sum of |s| over all test cases doesn't exceed 10^6.
Output
For each test case print one integer — the value of the res after the process ends.
Example
Input
3
--+-
---
++--+-
Output
7
9
6
Submitted Solution:
```
t = int(input())
for i in range(t):
s = input()
min_v = float('inf')
value = 0
max_index = 0
for i in range(len(s)):
if s[i]=='+':
value += 1
else:
value -= 1
if value < min_v:
min_v = value
# print(min_v)
ans = 0
if min_v<0:
r = abs(min_v)
ans += int(r*(r+1)/2)
print(ans+len(s))
```
No
| 36,259 | [
0.1968994140625,
0.129150390625,
-0.035491943359375,
0.04815673828125,
-0.6513671875,
-0.490478515625,
0.205322265625,
0.0105743408203125,
0.17626953125,
0.7431640625,
0.5546875,
-0.1942138671875,
0.21630859375,
-0.9482421875,
-0.498779296875,
-0.2454833984375,
-0.5888671875,
-0.76... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n (n ≥ 3) positive integers. It is known that in this array, all the numbers except one are the same (for example, in the array [4, 11, 4, 4] all numbers except one are equal to 4).
Print the index of the element that does not equal others. The numbers in the array are numbered from one.
Input
The first line contains a single integer t (1 ≤ t ≤ 100). Then t test cases follow.
The first line of each test case contains a single integer n (3 ≤ n ≤ 100) — the length of the array a.
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 100).
It is guaranteed that all the numbers except one in the a array are the same.
Output
For each test case, output a single integer — the index of the element that is not equal to others.
Example
Input
4
4
11 13 11 11
5
1 4 4 4 4
10
3 3 3 3 10 3 3 3 3 3
3
20 20 10
Output
2
1
5
3
Submitted Solution:
```
def solve(ar, n):
for i in range(n):
flag = 1
for j in range(i+1, n):
if ar[i]==ar[j]:
flag = 0
if flag:
return (i+1)
for _ in range(int(input())):
n = int(input())
ar = list(map(int, input().split()))
print(solve(ar, n))
```
No
| 36,342 | [
0.371826171875,
0.281005859375,
-0.289306640625,
0.051116943359375,
-0.52880859375,
-0.52490234375,
-0.137939453125,
0.117919921875,
0.262939453125,
1.109375,
0.8154296875,
-0.0906982421875,
0.0899658203125,
-1.076171875,
-0.59521484375,
-0.1912841796875,
-0.50244140625,
-0.3383789... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Soon there will be held the world's largest programming contest, but the testing system still has m bugs. The contest organizer, a well-known university, has no choice but to attract university students to fix all the bugs. The university has n students able to perform such work. The students realize that they are the only hope of the organizers, so they don't want to work for free: the i-th student wants to get ci 'passes' in his subjects (regardless of the volume of his work).
Bugs, like students, are not the same: every bug is characterized by complexity aj, and every student has the level of his abilities bi. Student i can fix a bug j only if the level of his abilities is not less than the complexity of the bug: bi ≥ aj, and he does it in one day. Otherwise, the bug will have to be fixed by another student. Of course, no student can work on a few bugs in one day. All bugs are not dependent on each other, so they can be corrected in any order, and different students can work simultaneously.
The university wants to fix all the bugs as quickly as possible, but giving the students the total of not more than s passes. Determine which students to use for that and come up with the schedule of work saying which student should fix which bug.
Input
The first line contains three space-separated integers: n, m and s (1 ≤ n, m ≤ 105, 0 ≤ s ≤ 109) — the number of students, the number of bugs in the system and the maximum number of passes the university is ready to give the students.
The next line contains m space-separated integers a1, a2, ..., am (1 ≤ ai ≤ 109) — the bugs' complexities.
The next line contains n space-separated integers b1, b2, ..., bn (1 ≤ bi ≤ 109) — the levels of the students' abilities.
The next line contains n space-separated integers c1, c2, ..., cn (0 ≤ ci ≤ 109) — the numbers of the passes the students want to get for their help.
Output
If the university can't correct all bugs print "NO".
Otherwise, on the first line print "YES", and on the next line print m space-separated integers: the i-th of these numbers should equal the number of the student who corrects the i-th bug in the optimal answer. The bugs should be corrected as quickly as possible (you must spend the minimum number of days), and the total given passes mustn't exceed s. If there are multiple optimal answers, you can output any of them.
Examples
Input
3 4 9
1 3 1 2
2 1 3
4 3 6
Output
YES
2 3 2 3
Input
3 4 10
2 3 1 2
2 1 3
4 3 6
Output
YES
1 3 1 3
Input
3 4 9
2 3 1 2
2 1 3
4 3 6
Output
YES
3 3 2 3
Input
3 4 5
1 3 1 2
2 1 3
5 3 6
Output
NO
Note
Consider the first sample.
The third student (with level 3) must fix the 2nd and 4th bugs (complexities 3 and 2 correspondingly) and the second student (with level 1) must fix the 1st and 3rd bugs (their complexity also equals 1). Fixing each bug takes one day for each student, so it takes 2 days to fix all bugs (the students can work in parallel).
The second student wants 3 passes for his assistance, the third student wants 6 passes. It meets the university's capabilities as it is ready to give at most 9 passes.
Tags: binary search, data structures, greedy, sortings
Correct Solution:
```
# from sys import stdin
import heapq
def f(x):
budzet=0
ost_wrzucony=-1
kolejka=[]
for i in range(0,il_bugow,x):
bug=bugi[i][0]
for j in range(ost_wrzucony+1,il_stud):
if studenci[j][1]>=bug:
heapq.heappush(kolejka,studenci[j])
ost_wrzucony=j
else:
break
if len(kolejka)<=0:
return False
st=heapq.heappop(kolejka)
budzet+=st[0]
if budzet>il_kasy:
return False
return True
def fw(x):
kolejka=[]
budzet=0
ost_wrzucony=-1
kolejnosc=[-1 for x in range(il_bugow)]
# print("x:",x)
for i in range(0,il_bugow,x):
bug=bugi[i][0]
for j in range(ost_wrzucony+1,il_stud):
if studenci[j][1]>=bug:
heapq.heappush(kolejka,studenci[j])
ost_wrzucony=j
else:
break
if len(kolejka)<=0:
return False,[]
st=heapq.heappop(kolejka)
budzet+=st[0]
for c in range(i,min(i+x,il_bugow)):
# print(f"c {c}, bugi[c][1]={bugi[c][1]}, st[1]={st[1]}, budzet={budzet} ")
kolejnosc[bugi[c][1]]=st[2]
if budzet>il_kasy:
return False, []
# print("x:",x)
return True, kolejnosc
def bs(start,koniec,zmi):
while start<koniec:
srodek=(start+koniec)//2
# print(f" bin_serach{start},{koniec},{srodek}")
if f(srodek):
koniec=srodek
else:
start=srodek+1
# print(f" bin_serach_koniec {start},{koniec},{srodek}")
return fw(koniec)
# return zmi,[]
zmi=0
il_stud, il_bugow, il_kasy=[int(x) for x in input().split()]
b=[int(x) for x in input().split()]
bugi=[]
for i in range(il_bugow):
bugi.append((b[i],i))
u=[int(x) for x in input().split()]
k=[int(x) for x in input().split()]
studenci=list(zip(k,u,range(1,il_stud+1)))
# print(studenci)
# studenci=[]
# for i in range(il_stud):
# studenci.append((k[i],u[i],i+1))
# print(studenci)
studenci.sort(reverse=True, key= lambda a: a[1])
bugi.sort(reverse= True)
# print(studenci, bugi)
zmi,kol=bs(1,il_bugow+1,zmi)
if zmi:
print('YES')
print(*kol)
else:
print('NO')
```
| 36,444 | [
0.0880126953125,
-0.2301025390625,
-0.10675048828125,
0.10076904296875,
-0.451416015625,
-0.377685546875,
0.09149169921875,
0.10888671875,
0.0190887451171875,
0.86865234375,
0.5283203125,
0.1788330078125,
0.2249755859375,
-0.9326171875,
-0.5888671875,
-0.402099609375,
-0.6083984375,
... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Soon there will be held the world's largest programming contest, but the testing system still has m bugs. The contest organizer, a well-known university, has no choice but to attract university students to fix all the bugs. The university has n students able to perform such work. The students realize that they are the only hope of the organizers, so they don't want to work for free: the i-th student wants to get ci 'passes' in his subjects (regardless of the volume of his work).
Bugs, like students, are not the same: every bug is characterized by complexity aj, and every student has the level of his abilities bi. Student i can fix a bug j only if the level of his abilities is not less than the complexity of the bug: bi ≥ aj, and he does it in one day. Otherwise, the bug will have to be fixed by another student. Of course, no student can work on a few bugs in one day. All bugs are not dependent on each other, so they can be corrected in any order, and different students can work simultaneously.
The university wants to fix all the bugs as quickly as possible, but giving the students the total of not more than s passes. Determine which students to use for that and come up with the schedule of work saying which student should fix which bug.
Input
The first line contains three space-separated integers: n, m and s (1 ≤ n, m ≤ 105, 0 ≤ s ≤ 109) — the number of students, the number of bugs in the system and the maximum number of passes the university is ready to give the students.
The next line contains m space-separated integers a1, a2, ..., am (1 ≤ ai ≤ 109) — the bugs' complexities.
The next line contains n space-separated integers b1, b2, ..., bn (1 ≤ bi ≤ 109) — the levels of the students' abilities.
The next line contains n space-separated integers c1, c2, ..., cn (0 ≤ ci ≤ 109) — the numbers of the passes the students want to get for their help.
Output
If the university can't correct all bugs print "NO".
Otherwise, on the first line print "YES", and on the next line print m space-separated integers: the i-th of these numbers should equal the number of the student who corrects the i-th bug in the optimal answer. The bugs should be corrected as quickly as possible (you must spend the minimum number of days), and the total given passes mustn't exceed s. If there are multiple optimal answers, you can output any of them.
Examples
Input
3 4 9
1 3 1 2
2 1 3
4 3 6
Output
YES
2 3 2 3
Input
3 4 10
2 3 1 2
2 1 3
4 3 6
Output
YES
1 3 1 3
Input
3 4 9
2 3 1 2
2 1 3
4 3 6
Output
YES
3 3 2 3
Input
3 4 5
1 3 1 2
2 1 3
5 3 6
Output
NO
Note
Consider the first sample.
The third student (with level 3) must fix the 2nd and 4th bugs (complexities 3 and 2 correspondingly) and the second student (with level 1) must fix the 1st and 3rd bugs (their complexity also equals 1). Fixing each bug takes one day for each student, so it takes 2 days to fix all bugs (the students can work in parallel).
The second student wants 3 passes for his assistance, the third student wants 6 passes. It meets the university's capabilities as it is ready to give at most 9 passes.
Tags: binary search, data structures, greedy, sortings
Correct Solution:
```
from sys import stdin
import heapq
n,m,s = [int(x) for x in stdin.readline().split()]
bugs = [int(x) for x in stdin.readline().split()]
bugs = sorted([(bugs[x],x) for x in range(m)])
order = [x[1] for x in bugs]
bugs = [x[0] for x in bugs]
students = [int(x) for x in stdin.readline().split()]
rate = [int(x) for x in stdin.readline().split()]
valid = False
for x in range(n):
if students[x] >= bugs[-1] and rate[x] <= s:
valid = True
if not valid:
print('NO')
else:
print('YES')
#print(students)
for i,x in enumerate(students):
low = 0
high = m-1
while high >= low:
mid = (high+low)//2
if bugs[mid] > x:
high = mid-1
else:
low = mid+1
#print(x,high)
students[i] = high
students = sorted([(students[x]+1,rate[x], x+1) for x in range(n)],reverse=True)
#print(students)
l1 = 1
high = m
lastValid = []
lastD = 100000
while l1 <= high:
mid = (l1+high)//2
shift = (mid-(m%mid))%mid
segs = m//mid
if shift > 0:
segs += 1
ind = 0
q = []
total = 0
group = []
for x in range(segs,0,-1):
while ind<n:
if (students[ind][0]+shift)//mid >= x:
heapq.heappush(q,(students[ind][1],students[ind][2]))
ind += 1
else:
break
if q:
r,i = heapq.heappop(q)
group.append((x,i))
total += r
else:
break
if len(group) == segs and total <= s:
#print(mid,total)
high = mid-1
lastValid = group
lastD = mid
else:
l1 = mid+1
complete = [0 for x in range(m)]
lastValid.sort()
mid = lastD
shift = (mid-(m%mid))%mid
skill = 1
for bruh,i in lastValid:
end = skill*mid-shift
start = max(0,end-mid)
for x in range(start,end):
complete[x] = i
skill += 1
c2 = [0 for x in range(m)]
for i,x in enumerate(complete):
c2[order[i]] = x
print(' '.join([str(x) for x in c2]))
```
| 36,445 | [
0.0880126953125,
-0.2301025390625,
-0.10675048828125,
0.10076904296875,
-0.451416015625,
-0.377685546875,
0.09149169921875,
0.10888671875,
0.0190887451171875,
0.86865234375,
0.5283203125,
0.1788330078125,
0.2249755859375,
-0.9326171875,
-0.5888671875,
-0.402099609375,
-0.6083984375,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Soon there will be held the world's largest programming contest, but the testing system still has m bugs. The contest organizer, a well-known university, has no choice but to attract university students to fix all the bugs. The university has n students able to perform such work. The students realize that they are the only hope of the organizers, so they don't want to work for free: the i-th student wants to get ci 'passes' in his subjects (regardless of the volume of his work).
Bugs, like students, are not the same: every bug is characterized by complexity aj, and every student has the level of his abilities bi. Student i can fix a bug j only if the level of his abilities is not less than the complexity of the bug: bi ≥ aj, and he does it in one day. Otherwise, the bug will have to be fixed by another student. Of course, no student can work on a few bugs in one day. All bugs are not dependent on each other, so they can be corrected in any order, and different students can work simultaneously.
The university wants to fix all the bugs as quickly as possible, but giving the students the total of not more than s passes. Determine which students to use for that and come up with the schedule of work saying which student should fix which bug.
Input
The first line contains three space-separated integers: n, m and s (1 ≤ n, m ≤ 105, 0 ≤ s ≤ 109) — the number of students, the number of bugs in the system and the maximum number of passes the university is ready to give the students.
The next line contains m space-separated integers a1, a2, ..., am (1 ≤ ai ≤ 109) — the bugs' complexities.
The next line contains n space-separated integers b1, b2, ..., bn (1 ≤ bi ≤ 109) — the levels of the students' abilities.
The next line contains n space-separated integers c1, c2, ..., cn (0 ≤ ci ≤ 109) — the numbers of the passes the students want to get for their help.
Output
If the university can't correct all bugs print "NO".
Otherwise, on the first line print "YES", and on the next line print m space-separated integers: the i-th of these numbers should equal the number of the student who corrects the i-th bug in the optimal answer. The bugs should be corrected as quickly as possible (you must spend the minimum number of days), and the total given passes mustn't exceed s. If there are multiple optimal answers, you can output any of them.
Examples
Input
3 4 9
1 3 1 2
2 1 3
4 3 6
Output
YES
2 3 2 3
Input
3 4 10
2 3 1 2
2 1 3
4 3 6
Output
YES
1 3 1 3
Input
3 4 9
2 3 1 2
2 1 3
4 3 6
Output
YES
3 3 2 3
Input
3 4 5
1 3 1 2
2 1 3
5 3 6
Output
NO
Note
Consider the first sample.
The third student (with level 3) must fix the 2nd and 4th bugs (complexities 3 and 2 correspondingly) and the second student (with level 1) must fix the 1st and 3rd bugs (their complexity also equals 1). Fixing each bug takes one day for each student, so it takes 2 days to fix all bugs (the students can work in parallel).
The second student wants 3 passes for his assistance, the third student wants 6 passes. It meets the university's capabilities as it is ready to give at most 9 passes.
Submitted Solution:
```
from sys import stdin
import heapq
def update(tree,n,i):
i += 1
while i <= n:
tree[i] += 1
i += i&(-i)
def get(tree,i):
s = 0
i += 1
while i > 0:
s += tree[i]
i -= i&(-i)
return s
n,m,s = [int(x) for x in stdin.readline().split()]
bugs = [int(x) for x in stdin.readline().split()]
bugs = sorted([(bugs[x],x) for x in range(m)])
order = [x[1] for x in bugs]
bugs = [x[0] for x in bugs]
students = [int(x) for x in stdin.readline().split()]
rate = [int(x) for x in stdin.readline().split()]
valid = False
for x in range(n):
if students[x] >= bugs[-1] and rate[x] <= s:
valid = True
if not valid:
print('NO')
else:
print('YES')
#print(students)
for i,x in enumerate(students):
low = 0
high = m-1
while high >= low:
mid = (high+low)//2
if bugs[mid] > x:
high = mid-1
else:
low = mid+1
#print(x,high)
students[i] = high
students = sorted([(rate[x],students[x]+1, x+1) for x in range(n)])
#print(students)
l1 = 1
high = m
lastValid = []
lastD = 100000
while l1 <= high:
mid = (l1+high)//2
shift = (mid-(m%mid))%mid
segs = m//mid
if shift > 0:
segs += 1
#print(mid,shift,segs)
sCopy = []
for x in students:
sCopy.append((x[0],(x[1]+shift)//mid,x[2]))
#print(sCopy)
low = 1
ind = 0
q = []
total = 0
bTree = [0 for x in range(segs+2)]
group = []
while get(bTree,segs) < segs and ind < n:
r,skill,i = sCopy[ind]
ind += 1
#print(r,skill,get(bTree,skill))
if get(bTree,skill) < skill:
update(bTree,segs+1,skill)
#print(r,skill)
total += r
group.append((skill,i))
#print(total)
#print(group)
#print(get(bTree,segs))
#print(bTree)
if total <= s and get(bTree,segs) >= segs:
high = mid-1
lastD = mid
lastValid = group
else:
l1 = mid+1
complete = [0 for x in range(m)]
lastValid.sort()
mid = lastD
shift = (mid-(m%mid))%mid
skill = 1
for bruh,i in lastValid:
end = skill*mid-shift
start = max(0,end-mid)
for x in range(start,end):
complete[x] = i
skill += 1
c2 = [0 for x in range(m)]
for i,x in enumerate(complete):
c2[order[i]] = x
print(' '.join([str(x) for x in c2]))
```
No
| 36,446 | [
0.15283203125,
-0.182861328125,
-0.1466064453125,
0.032196044921875,
-0.552734375,
-0.27294921875,
0.084716796875,
0.272216796875,
-0.04901123046875,
0.8935546875,
0.4013671875,
0.1839599609375,
0.146484375,
-0.9033203125,
-0.5966796875,
-0.483642578125,
-0.57275390625,
-0.73339843... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Soon there will be held the world's largest programming contest, but the testing system still has m bugs. The contest organizer, a well-known university, has no choice but to attract university students to fix all the bugs. The university has n students able to perform such work. The students realize that they are the only hope of the organizers, so they don't want to work for free: the i-th student wants to get ci 'passes' in his subjects (regardless of the volume of his work).
Bugs, like students, are not the same: every bug is characterized by complexity aj, and every student has the level of his abilities bi. Student i can fix a bug j only if the level of his abilities is not less than the complexity of the bug: bi ≥ aj, and he does it in one day. Otherwise, the bug will have to be fixed by another student. Of course, no student can work on a few bugs in one day. All bugs are not dependent on each other, so they can be corrected in any order, and different students can work simultaneously.
The university wants to fix all the bugs as quickly as possible, but giving the students the total of not more than s passes. Determine which students to use for that and come up with the schedule of work saying which student should fix which bug.
Input
The first line contains three space-separated integers: n, m and s (1 ≤ n, m ≤ 105, 0 ≤ s ≤ 109) — the number of students, the number of bugs in the system and the maximum number of passes the university is ready to give the students.
The next line contains m space-separated integers a1, a2, ..., am (1 ≤ ai ≤ 109) — the bugs' complexities.
The next line contains n space-separated integers b1, b2, ..., bn (1 ≤ bi ≤ 109) — the levels of the students' abilities.
The next line contains n space-separated integers c1, c2, ..., cn (0 ≤ ci ≤ 109) — the numbers of the passes the students want to get for their help.
Output
If the university can't correct all bugs print "NO".
Otherwise, on the first line print "YES", and on the next line print m space-separated integers: the i-th of these numbers should equal the number of the student who corrects the i-th bug in the optimal answer. The bugs should be corrected as quickly as possible (you must spend the minimum number of days), and the total given passes mustn't exceed s. If there are multiple optimal answers, you can output any of them.
Examples
Input
3 4 9
1 3 1 2
2 1 3
4 3 6
Output
YES
2 3 2 3
Input
3 4 10
2 3 1 2
2 1 3
4 3 6
Output
YES
1 3 1 3
Input
3 4 9
2 3 1 2
2 1 3
4 3 6
Output
YES
3 3 2 3
Input
3 4 5
1 3 1 2
2 1 3
5 3 6
Output
NO
Note
Consider the first sample.
The third student (with level 3) must fix the 2nd and 4th bugs (complexities 3 and 2 correspondingly) and the second student (with level 1) must fix the 1st and 3rd bugs (their complexity also equals 1). Fixing each bug takes one day for each student, so it takes 2 days to fix all bugs (the students can work in parallel).
The second student wants 3 passes for his assistance, the third student wants 6 passes. It meets the university's capabilities as it is ready to give at most 9 passes.
Submitted Solution:
```
import heapq
import sys
n, m, money = input().split()
n = int(n)
m = int(m)
money = int(money)
a = input().split()
a = [int(x) for x in a]
b = input().split()
b = [int(x) for x in b]
c = input().split()
c = [int(x) for x in c]
class st:
def __init__(self, b,c, idx):
self.b = b
self.c = c
self.idx = idx
def __lt__(self, obj):
return self.c < obj.c
students = []
for i in range(len(b)):
students.append(st(b[i], c[i], i))
students = sorted(students, key=lambda s: s.b, reverse=True)
tasks = zip(a, range(len(a)))
tasks = sorted(tasks, key = lambda pair: pair[0], reverse = True)
def min_cost(days, money, tasks, students):
if days == 0:
return -1, []
cur_student = 0
st_heap = []
task_number = 0
st_res = []
while task_number < len(tasks):
while cur_student < len(students) and students[cur_student].b >= tasks[task_number][0]:
#print(tasks[i][0])
heapq.heappush(st_heap, students[cur_student])
cur_student += 1
if len(st_heap) == 0:
return -1, []
s = heapq.heappop(st_heap)
st_res += [s]*days
money -= s.c
task_number += days
#print(money, days)
return money, st_res[:len(tasks)]
m, s = min_cost(len(tasks), money, tasks, students)
if m < 0:
print("NO")
sys.exit(0)
else:
print("YES")
begin = 0
end = money+1
while end - begin > 1:
mid = (end + begin) // 2
m, s = min_cost(mid, money, tasks, students)
print(begin, end, mid)
if m < 0:
begin = mid
else:
end = mid
m,s = min_cost(begin, money, tasks, students)
if m < 0:
m,s = min_cost(begin + 1, money, tasks, students)
numbers = [task[1] for task in tasks]
stud_numbers = [stud.idx for stud in s]
#print(stud_numbers)
both = zip(numbers, stud_numbers)
both = sorted(both, key = lambda x: x[0])
answer = ""
for p in both:
answer += str(p[1] + 1)
answer += ' '
print(answer)
```
No
| 36,447 | [
0.15283203125,
-0.182861328125,
-0.1466064453125,
0.032196044921875,
-0.552734375,
-0.27294921875,
0.084716796875,
0.272216796875,
-0.04901123046875,
0.8935546875,
0.4013671875,
0.1839599609375,
0.146484375,
-0.9033203125,
-0.5966796875,
-0.483642578125,
-0.57275390625,
-0.73339843... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Soon there will be held the world's largest programming contest, but the testing system still has m bugs. The contest organizer, a well-known university, has no choice but to attract university students to fix all the bugs. The university has n students able to perform such work. The students realize that they are the only hope of the organizers, so they don't want to work for free: the i-th student wants to get ci 'passes' in his subjects (regardless of the volume of his work).
Bugs, like students, are not the same: every bug is characterized by complexity aj, and every student has the level of his abilities bi. Student i can fix a bug j only if the level of his abilities is not less than the complexity of the bug: bi ≥ aj, and he does it in one day. Otherwise, the bug will have to be fixed by another student. Of course, no student can work on a few bugs in one day. All bugs are not dependent on each other, so they can be corrected in any order, and different students can work simultaneously.
The university wants to fix all the bugs as quickly as possible, but giving the students the total of not more than s passes. Determine which students to use for that and come up with the schedule of work saying which student should fix which bug.
Input
The first line contains three space-separated integers: n, m and s (1 ≤ n, m ≤ 105, 0 ≤ s ≤ 109) — the number of students, the number of bugs in the system and the maximum number of passes the university is ready to give the students.
The next line contains m space-separated integers a1, a2, ..., am (1 ≤ ai ≤ 109) — the bugs' complexities.
The next line contains n space-separated integers b1, b2, ..., bn (1 ≤ bi ≤ 109) — the levels of the students' abilities.
The next line contains n space-separated integers c1, c2, ..., cn (0 ≤ ci ≤ 109) — the numbers of the passes the students want to get for their help.
Output
If the university can't correct all bugs print "NO".
Otherwise, on the first line print "YES", and on the next line print m space-separated integers: the i-th of these numbers should equal the number of the student who corrects the i-th bug in the optimal answer. The bugs should be corrected as quickly as possible (you must spend the minimum number of days), and the total given passes mustn't exceed s. If there are multiple optimal answers, you can output any of them.
Examples
Input
3 4 9
1 3 1 2
2 1 3
4 3 6
Output
YES
2 3 2 3
Input
3 4 10
2 3 1 2
2 1 3
4 3 6
Output
YES
1 3 1 3
Input
3 4 9
2 3 1 2
2 1 3
4 3 6
Output
YES
3 3 2 3
Input
3 4 5
1 3 1 2
2 1 3
5 3 6
Output
NO
Note
Consider the first sample.
The third student (with level 3) must fix the 2nd and 4th bugs (complexities 3 and 2 correspondingly) and the second student (with level 1) must fix the 1st and 3rd bugs (their complexity also equals 1). Fixing each bug takes one day for each student, so it takes 2 days to fix all bugs (the students can work in parallel).
The second student wants 3 passes for his assistance, the third student wants 6 passes. It meets the university's capabilities as it is ready to give at most 9 passes.
Submitted Solution:
```
from sys import stdin
import heapq
n,m,s = [int(x) for x in stdin.readline().split()]
bugs = [int(x) for x in stdin.readline().split()]
bugs = sorted([(bugs[x],x) for x in range(m)])
order = [x[1] for x in bugs]
bugs = [x[0] for x in bugs]
students = [int(x) for x in stdin.readline().split()]
rate = [int(x) for x in stdin.readline().split()]
valid = False
for x in range(n):
if students[x] >= bugs[-1] and rate[x] <= s:
valid = True
if not valid:
print('NO')
else:
print('YES')
for i,x in enumerate(students):
low = 0
high = m-1
while high >= low:
mid = (high+low)//2
if bugs[mid] > x:
high = mid-1
else:
low = mid+1
students[i] = high
students = sorted([(rate[x],students[x]+1, x+1) for x in range(n)])
l1 = 1
high = m
lastValid = []
lastD = 100000
while l1 <= high:
mid = (l1+high)//2
shift = (mid-(m%mid))%mid
segs = m//mid
if shift > 0:
segs += 1
#print(mid,shift,segs)
sCopy = []
for x in students:
sCopy.append((x[0],(x[1]+shift)//mid,x[2]))
#print(sCopy)
low = 1
ind = 0
q = []
total = 0
group = []
while low+len(q) <= segs and ind < n:
r,skill,i = sCopy[ind]
ind += 1
if skill < low:
continue
if skill-low < len(q):
while q[0] == low:
low += 1
heapq.heappop(q)
if not q:
break
if skill >= low:
heapq.heappush(q,skill)
#print(r,skill)
total += r
group.append((i,skill))
else:
heapq.heappush(q,skill)
#print(r,skill)
total += r
group.append((i,skill))
if total <= s and len(q)+low > segs:
high = mid-1
lastD = mid
lastValid = group
else:
l1 = mid+1
complete = [0 for x in range(m)]
mid = lastD
shift = (mid-(m%mid))%mid
for i,skill in lastValid:
end = skill*mid-shift
start = max(0,end-mid)
for x in range(start,end):
complete[x] = i
c2 = [0 for x in range(m)]
for i,x in enumerate(complete):
c2[order[i]] = x
print(' '.join([str(x) for x in c2]))
```
No
| 36,448 | [
0.15283203125,
-0.182861328125,
-0.1466064453125,
0.032196044921875,
-0.552734375,
-0.27294921875,
0.084716796875,
0.272216796875,
-0.04901123046875,
0.8935546875,
0.4013671875,
0.1839599609375,
0.146484375,
-0.9033203125,
-0.5966796875,
-0.483642578125,
-0.57275390625,
-0.73339843... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Soon there will be held the world's largest programming contest, but the testing system still has m bugs. The contest organizer, a well-known university, has no choice but to attract university students to fix all the bugs. The university has n students able to perform such work. The students realize that they are the only hope of the organizers, so they don't want to work for free: the i-th student wants to get ci 'passes' in his subjects (regardless of the volume of his work).
Bugs, like students, are not the same: every bug is characterized by complexity aj, and every student has the level of his abilities bi. Student i can fix a bug j only if the level of his abilities is not less than the complexity of the bug: bi ≥ aj, and he does it in one day. Otherwise, the bug will have to be fixed by another student. Of course, no student can work on a few bugs in one day. All bugs are not dependent on each other, so they can be corrected in any order, and different students can work simultaneously.
The university wants to fix all the bugs as quickly as possible, but giving the students the total of not more than s passes. Determine which students to use for that and come up with the schedule of work saying which student should fix which bug.
Input
The first line contains three space-separated integers: n, m and s (1 ≤ n, m ≤ 105, 0 ≤ s ≤ 109) — the number of students, the number of bugs in the system and the maximum number of passes the university is ready to give the students.
The next line contains m space-separated integers a1, a2, ..., am (1 ≤ ai ≤ 109) — the bugs' complexities.
The next line contains n space-separated integers b1, b2, ..., bn (1 ≤ bi ≤ 109) — the levels of the students' abilities.
The next line contains n space-separated integers c1, c2, ..., cn (0 ≤ ci ≤ 109) — the numbers of the passes the students want to get for their help.
Output
If the university can't correct all bugs print "NO".
Otherwise, on the first line print "YES", and on the next line print m space-separated integers: the i-th of these numbers should equal the number of the student who corrects the i-th bug in the optimal answer. The bugs should be corrected as quickly as possible (you must spend the minimum number of days), and the total given passes mustn't exceed s. If there are multiple optimal answers, you can output any of them.
Examples
Input
3 4 9
1 3 1 2
2 1 3
4 3 6
Output
YES
2 3 2 3
Input
3 4 10
2 3 1 2
2 1 3
4 3 6
Output
YES
1 3 1 3
Input
3 4 9
2 3 1 2
2 1 3
4 3 6
Output
YES
3 3 2 3
Input
3 4 5
1 3 1 2
2 1 3
5 3 6
Output
NO
Note
Consider the first sample.
The third student (with level 3) must fix the 2nd and 4th bugs (complexities 3 and 2 correspondingly) and the second student (with level 1) must fix the 1st and 3rd bugs (their complexity also equals 1). Fixing each bug takes one day for each student, so it takes 2 days to fix all bugs (the students can work in parallel).
The second student wants 3 passes for his assistance, the third student wants 6 passes. It meets the university's capabilities as it is ready to give at most 9 passes.
Submitted Solution:
```
import sys
def check(a_i0, c_b_i0, s0, k, f):
a_i = a_i0[:]
c_b_i = c_b_i0[:]
#��ǰ���
s = s0
#pָ�����������
p = m-1
while p >= 0:
low = a_i[p][0]
#�ҵ�����>=low������˵���
for i in range(len(c_b_i)):
if c_b_i[i][1] < low:
continue
#i�Ѿ�������˵�
if s < c_b_i[i][0]:
#����
return False
#��ļi��������ؼ�¼
for j in range(min(p+1, k)):
f[a_i[p][1]] = c_b_i[i][2]
p -= 1
s -= c_b_i[i][0]
c_b_i.pop(i)
break
else:
#û�ҵ��ܽ���������
return False
return True
n,m,s = (int(x) for x in sys.stdin.readline().split(' '))
a = [int(x) for x in sys.stdin.readline().split(' ')]
b = [int(x) for x in sys.stdin.readline().split(' ')]
c = [int(x) for x in sys.stdin.readline().split(' ')]
a_i = sorted(zip(a, range(m)), key=lambda x:x[0])
c_b_i = sorted(zip(c, b, range(n)), key=lambda x:x[0])
lo, hi = 1, m
while lo < hi - 1:
k = (lo + hi) // 2
if check(a_i, c_b_i, s, k, {}):
hi = k
else:
lo = k
f = {}
if check(a_i, c_b_i, s, hi, f):
print('YES')
print(' '.join([str(f[x]+1) for x in sorted(f.keys())]))
else:
print('NO')
```
No
| 36,449 | [
0.15283203125,
-0.182861328125,
-0.1466064453125,
0.032196044921875,
-0.552734375,
-0.27294921875,
0.084716796875,
0.272216796875,
-0.04901123046875,
0.8935546875,
0.4013671875,
0.1839599609375,
0.146484375,
-0.9033203125,
-0.5966796875,
-0.483642578125,
-0.57275390625,
-0.73339843... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
A programming competition site AtCode regularly holds programming contests.
The next contest on AtCode is called ABC, which is rated for contestants with ratings less than 1200.
The contest after the ABC is called ARC, which is rated for contestants with ratings less than 2800.
The contest after the ARC is called AGC, which is rated for all contestants.
Takahashi's rating on AtCode is R. What is the next contest rated for him?
Constraints
* 0 ≤ R ≤ 4208
* R is an integer.
Input
Input is given from Standard Input in the following format:
R
Output
Print the name of the next contest rated for Takahashi (`ABC`, `ARC` or `AGC`).
Examples
Input
1199
Output
ABC
Input
1200
Output
ARC
Input
4208
Output
AGC
"Correct Solution:
```
n=int(input())
print("ABC" if n<1200 else("ARC" if n<2800 else "AGC"))
```
| 36,703 | [
0.67919921875,
0.2119140625,
-0.488037109375,
-0.09063720703125,
-0.63134765625,
-0.52294921875,
0.166259765625,
0.03289794921875,
-0.0261383056640625,
0.80322265625,
0.8876953125,
-0.427001953125,
-0.1864013671875,
-1.173828125,
-0.281494140625,
-0.263671875,
-0.50537109375,
-0.50... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
A programming competition site AtCode regularly holds programming contests.
The next contest on AtCode is called ABC, which is rated for contestants with ratings less than 1200.
The contest after the ABC is called ARC, which is rated for contestants with ratings less than 2800.
The contest after the ARC is called AGC, which is rated for all contestants.
Takahashi's rating on AtCode is R. What is the next contest rated for him?
Constraints
* 0 ≤ R ≤ 4208
* R is an integer.
Input
Input is given from Standard Input in the following format:
R
Output
Print the name of the next contest rated for Takahashi (`ABC`, `ARC` or `AGC`).
Examples
Input
1199
Output
ABC
Input
1200
Output
ARC
Input
4208
Output
AGC
"Correct Solution:
```
A = int(input())
result = "ABC" if A<1200 else "AGC" if A >= 2800 else "ARC"
print(result)
```
| 36,704 | [
0.6591796875,
0.1988525390625,
-0.47119140625,
-0.10833740234375,
-0.6689453125,
-0.505859375,
0.154296875,
0.038909912109375,
-0.01338958740234375,
0.76806640625,
0.861328125,
-0.40966796875,
-0.2012939453125,
-1.1650390625,
-0.32080078125,
-0.273681640625,
-0.462158203125,
-0.481... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
A programming competition site AtCode regularly holds programming contests.
The next contest on AtCode is called ABC, which is rated for contestants with ratings less than 1200.
The contest after the ABC is called ARC, which is rated for contestants with ratings less than 2800.
The contest after the ARC is called AGC, which is rated for all contestants.
Takahashi's rating on AtCode is R. What is the next contest rated for him?
Constraints
* 0 ≤ R ≤ 4208
* R is an integer.
Input
Input is given from Standard Input in the following format:
R
Output
Print the name of the next contest rated for Takahashi (`ABC`, `ARC` or `AGC`).
Examples
Input
1199
Output
ABC
Input
1200
Output
ARC
Input
4208
Output
AGC
"Correct Solution:
```
a=int(input())
print("ABC" if a<=1199 else "ARC" if a<=2799 else "AGC")
```
| 36,705 | [
0.66748046875,
0.2200927734375,
-0.474365234375,
-0.09368896484375,
-0.677734375,
-0.50390625,
0.1766357421875,
0.034271240234375,
-0.007534027099609375,
0.77685546875,
0.88330078125,
-0.425048828125,
-0.201416015625,
-1.17578125,
-0.319091796875,
-0.268310546875,
-0.481201171875,
... | 11 |
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