message stringlengths 2 39.6k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 219 108k | cluster float64 11 11 | __index_level_0__ int64 438 217k |
|---|---|---|---|---|---|
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))
``` | instruction | 0 | 35,489 | 11 | 70,978 |
Yes | output | 1 | 35,489 | 11 | 70,979 |
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)
``` | instruction | 0 | 35,490 | 11 | 70,980 |
Yes | output | 1 | 35,490 | 11 | 70,981 |
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()
``` | instruction | 0 | 35,491 | 11 | 70,982 |
Yes | output | 1 | 35,491 | 11 | 70,983 |
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))
``` | instruction | 0 | 35,492 | 11 | 70,984 |
Yes | output | 1 | 35,492 | 11 | 70,985 |
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)
``` | instruction | 0 | 35,493 | 11 | 70,986 |
No | output | 1 | 35,493 | 11 | 70,987 |
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))))
``` | instruction | 0 | 35,494 | 11 | 70,988 |
No | output | 1 | 35,494 | 11 | 70,989 |
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))))
``` | instruction | 0 | 35,495 | 11 | 70,990 |
No | output | 1 | 35,495 | 11 | 70,991 |
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)
``` | instruction | 0 | 35,496 | 11 | 70,992 |
No | output | 1 | 35,496 | 11 | 70,993 |
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 | instruction | 0 | 35,514 | 11 | 71,028 |
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
``` | output | 1 | 35,514 | 11 | 71,029 |
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 | instruction | 0 | 35,515 | 11 | 71,030 |
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()
``` | output | 1 | 35,515 | 11 | 71,031 |
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 | instruction | 0 | 35,516 | 11 | 71,032 |
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))
``` | output | 1 | 35,516 | 11 | 71,033 |
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 | instruction | 0 | 35,517 | 11 | 71,034 |
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()
``` | output | 1 | 35,517 | 11 | 71,035 |
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 | instruction | 0 | 35,519 | 11 | 71,038 |
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))
``` | output | 1 | 35,519 | 11 | 71,039 |
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))
``` | instruction | 0 | 35,520 | 11 | 71,040 |
No | output | 1 | 35,520 | 11 | 71,041 |
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))
``` | instruction | 0 | 35,521 | 11 | 71,042 |
No | output | 1 | 35,521 | 11 | 71,043 |
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)
``` | instruction | 0 | 35,522 | 11 | 71,044 |
No | output | 1 | 35,522 | 11 | 71,045 |
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)
``` | instruction | 0 | 35,523 | 11 | 71,046 |
No | output | 1 | 35,523 | 11 | 71,047 |
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. | instruction | 0 | 35,677 | 11 | 71,354 |
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)
``` | output | 1 | 35,677 | 11 | 71,355 |
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. | instruction | 0 | 35,678 | 11 | 71,356 |
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)
``` | output | 1 | 35,678 | 11 | 71,357 |
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. | instruction | 0 | 35,679 | 11 | 71,358 |
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)
``` | output | 1 | 35,679 | 11 | 71,359 |
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. | instruction | 0 | 35,680 | 11 | 71,360 |
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)
``` | output | 1 | 35,680 | 11 | 71,361 |
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. | instruction | 0 | 35,681 | 11 | 71,362 |
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)
``` | output | 1 | 35,681 | 11 | 71,363 |
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. | instruction | 0 | 35,682 | 11 | 71,364 |
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)
``` | output | 1 | 35,682 | 11 | 71,365 |
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. | instruction | 0 | 35,683 | 11 | 71,366 |
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)
``` | output | 1 | 35,683 | 11 | 71,367 |
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. | instruction | 0 | 35,684 | 11 | 71,368 |
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)
``` | output | 1 | 35,684 | 11 | 71,369 |
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)
``` | instruction | 0 | 35,685 | 11 | 71,370 |
Yes | output | 1 | 35,685 | 11 | 71,371 |
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()
``` | instruction | 0 | 35,686 | 11 | 71,372 |
Yes | output | 1 | 35,686 | 11 | 71,373 |
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)
``` | instruction | 0 | 35,687 | 11 | 71,374 |
Yes | output | 1 | 35,687 | 11 | 71,375 |
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)
``` | instruction | 0 | 35,688 | 11 | 71,376 |
Yes | output | 1 | 35,688 | 11 | 71,377 |
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)
``` | instruction | 0 | 35,689 | 11 | 71,378 |
No | output | 1 | 35,689 | 11 | 71,379 |
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)
``` | instruction | 0 | 35,690 | 11 | 71,380 |
No | output | 1 | 35,690 | 11 | 71,381 |
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)
``` | instruction | 0 | 35,691 | 11 | 71,382 |
No | output | 1 | 35,691 | 11 | 71,383 |
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)
``` | instruction | 0 | 35,692 | 11 | 71,384 |
No | output | 1 | 35,692 | 11 | 71,385 |
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))
``` | instruction | 0 | 36,253 | 11 | 72,506 |
Yes | output | 1 | 36,253 | 11 | 72,507 |
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)
``` | instruction | 0 | 36,254 | 11 | 72,508 |
Yes | output | 1 | 36,254 | 11 | 72,509 |
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)
``` | instruction | 0 | 36,255 | 11 | 72,510 |
Yes | output | 1 | 36,255 | 11 | 72,511 |
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)
``` | instruction | 0 | 36,256 | 11 | 72,512 |
No | output | 1 | 36,256 | 11 | 72,513 |
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)
``` | instruction | 0 | 36,257 | 11 | 72,514 |
No | output | 1 | 36,257 | 11 | 72,515 |
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)
``` | instruction | 0 | 36,258 | 11 | 72,516 |
No | output | 1 | 36,258 | 11 | 72,517 |
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))
``` | instruction | 0 | 36,259 | 11 | 72,518 |
No | output | 1 | 36,259 | 11 | 72,519 |
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))
``` | instruction | 0 | 36,342 | 11 | 72,684 |
No | output | 1 | 36,342 | 11 | 72,685 |
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. | instruction | 0 | 36,444 | 11 | 72,888 |
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')
``` | output | 1 | 36,444 | 11 | 72,889 |
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. | instruction | 0 | 36,445 | 11 | 72,890 |
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]))
``` | output | 1 | 36,445 | 11 | 72,891 |
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]))
``` | instruction | 0 | 36,446 | 11 | 72,892 |
No | output | 1 | 36,446 | 11 | 72,893 |
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)
``` | instruction | 0 | 36,447 | 11 | 72,894 |
No | output | 1 | 36,447 | 11 | 72,895 |
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]))
``` | instruction | 0 | 36,448 | 11 | 72,896 |
No | output | 1 | 36,448 | 11 | 72,897 |
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')
``` | instruction | 0 | 36,449 | 11 | 72,898 |
No | output | 1 | 36,449 | 11 | 72,899 |
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 | instruction | 0 | 36,703 | 11 | 73,406 |
"Correct Solution:
```
n=int(input())
print("ABC" if n<1200 else("ARC" if n<2800 else "AGC"))
``` | output | 1 | 36,703 | 11 | 73,407 |
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 | instruction | 0 | 36,704 | 11 | 73,408 |
"Correct Solution:
```
A = int(input())
result = "ABC" if A<1200 else "AGC" if A >= 2800 else "ARC"
print(result)
``` | output | 1 | 36,704 | 11 | 73,409 |
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 | instruction | 0 | 36,705 | 11 | 73,410 |
"Correct Solution:
```
a=int(input())
print("ABC" if a<=1199 else "ARC" if a<=2799 else "AGC")
``` | output | 1 | 36,705 | 11 | 73,411 |
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