message stringlengths 2 22.7k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 145 109k | cluster float64 9 9 | __index_level_0__ int64 290 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob decided to eat some fruit. In the kitchen they found a large bag of oranges and apples. Alice immediately took an orange for herself, Bob took an apple. To make the process of sharing the remaining fruit more fun, the friends decided to play a game. They put multiple cards and on each one they wrote a letter, either 'A', or the letter 'B'. Then they began to remove the cards one by one from left to right, every time they removed a card with the letter 'A', Alice gave Bob all the fruits she had at that moment and took out of the bag as many apples and as many oranges as she had before. Thus the number of oranges and apples Alice had, did not change. If the card had written letter 'B', then Bob did the same, that is, he gave Alice all the fruit that he had, and took from the bag the same set of fruit. After the last card way removed, all the fruit in the bag were over.
You know how many oranges and apples was in the bag at first. Your task is to find any sequence of cards that Alice and Bob could have played with.
Input
The first line of the input contains two integers, x, y (1 ≤ x, y ≤ 1018, xy > 1) — the number of oranges and apples that were initially in the bag.
Output
Print any sequence of cards that would meet the problem conditions as a compressed string of characters 'A' and 'B. That means that you need to replace the segments of identical consecutive characters by the number of repetitions of the characters and the actual character. For example, string AAABAABBB should be replaced by string 3A1B2A3B, but cannot be replaced by 2A1A1B2A3B or by 3AB2A3B. See the samples for clarifications of the output format. The string that you print should consist of at most 106 characters. It is guaranteed that if the answer exists, its compressed representation exists, consisting of at most 106 characters. If there are several possible answers, you are allowed to print any of them.
If the sequence of cards that meet the problem statement does not not exist, print a single word Impossible.
Examples
Input
1 4
Output
3B
Input
2 2
Output
Impossible
Input
3 2
Output
1A1B
Note
In the first sample, if the row contained three cards with letter 'B', then Bob should give one apple to Alice three times. So, in the end of the game Alice has one orange and three apples, and Bob has one apple, in total it is one orange and four apples.
In second sample, there is no answer since one card is not enough for game to finish, and two cards will produce at least three apples or three oranges.
In the third sample, cards contain letters 'AB', so after removing the first card Bob has one orange and one apple, and after removal of second card Alice has two oranges and one apple. So, in total it is three oranges and two apples.
Submitted Solution:
```
def gcd(a, b):
if b == 0:
return a
return gcd(b, a % b)
class fraction:
n, m = 0, 0
def __init__(self, n, m):
d = gcd(n, m)
self.n = n/d
self.m = m/d
def add(self, tmp):
return fraction(self.n * tmp.m, self.m * tmp.n)
def compareTo(self ,tmp):
a = self.n * tmp.m
b = self.m * tmp.n
if a > b:
return 1
elif a < b:
return -1
return 0
def sternBrocotAdd(self, tmp):
return fraction(self.n + tmp.n, self.m + tmp.m);
def run(left, right, result):
a = left.n
b = left.m
p = right.n
q = right.m
n = result.n
m = result.m
if left.compareTo(result) == 0:
return;
if right.compareTo(result) == 0:
return;
mid = left.sternBrocotAdd(right)
ch = 'Z'
if mid.compareTo(result) <= 0:
x = int((b * n - a * m) / (p * m - q * n))
left = fraction(a + p * x, b + q * x)
if left.compareTo(result) == 0:
x -= 1
ch = 'A'
else:
x = int((p * m - q * n) / (b * n - a * m))
right = fraction(a * x + p, b * x + q)
if right.compareTo(result) == 0:
x -= 1
ch = 'B'
s = str.format("%d%c"%(x, ch))
print(s, end="")
if left.compareTo(result) == 0 or right.compareTo(result) == 0:
return
run(left, right, result)
p, q = map(int, input().split())
d = gcd(p, q)
if d == 1:
result = fraction(p, q)
right = fraction(1, 0)
left = fraction(0, 1)
run(left, right, result)
else:
print('Impossible')
``` | instruction | 0 | 15,486 | 9 | 30,972 |
No | output | 1 | 15,486 | 9 | 30,973 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob decided to eat some fruit. In the kitchen they found a large bag of oranges and apples. Alice immediately took an orange for herself, Bob took an apple. To make the process of sharing the remaining fruit more fun, the friends decided to play a game. They put multiple cards and on each one they wrote a letter, either 'A', or the letter 'B'. Then they began to remove the cards one by one from left to right, every time they removed a card with the letter 'A', Alice gave Bob all the fruits she had at that moment and took out of the bag as many apples and as many oranges as she had before. Thus the number of oranges and apples Alice had, did not change. If the card had written letter 'B', then Bob did the same, that is, he gave Alice all the fruit that he had, and took from the bag the same set of fruit. After the last card way removed, all the fruit in the bag were over.
You know how many oranges and apples was in the bag at first. Your task is to find any sequence of cards that Alice and Bob could have played with.
Input
The first line of the input contains two integers, x, y (1 ≤ x, y ≤ 1018, xy > 1) — the number of oranges and apples that were initially in the bag.
Output
Print any sequence of cards that would meet the problem conditions as a compressed string of characters 'A' and 'B. That means that you need to replace the segments of identical consecutive characters by the number of repetitions of the characters and the actual character. For example, string AAABAABBB should be replaced by string 3A1B2A3B, but cannot be replaced by 2A1A1B2A3B or by 3AB2A3B. See the samples for clarifications of the output format. The string that you print should consist of at most 106 characters. It is guaranteed that if the answer exists, its compressed representation exists, consisting of at most 106 characters. If there are several possible answers, you are allowed to print any of them.
If the sequence of cards that meet the problem statement does not not exist, print a single word Impossible.
Examples
Input
1 4
Output
3B
Input
2 2
Output
Impossible
Input
3 2
Output
1A1B
Note
In the first sample, if the row contained three cards with letter 'B', then Bob should give one apple to Alice three times. So, in the end of the game Alice has one orange and three apples, and Bob has one apple, in total it is one orange and four apples.
In second sample, there is no answer since one card is not enough for game to finish, and two cards will produce at least three apples or three oranges.
In the third sample, cards contain letters 'AB', so after removing the first card Bob has one orange and one apple, and after removal of second card Alice has two oranges and one apple. So, in total it is three oranges and two apples.
Submitted Solution:
```
a, b = input().split()
a = int(a)
b = int(b)
strin=""
countA=0
countB=0
while a + b !=2:
if a < b:
countA = countA*(-1)
countB = countB + 1
b= b-a
elif a>b:
countB = countB*(-1)
countA = countA + 1
a= a-b
elif a==b and a!=1:
strin = "Impossible"
break
if (a==b and a!=1 and strin=="Impossible"):
break
if countB<0:
strin=strin+str(-countB)+'B'
# print(-countB,"B")
countB=0
elif countA <0:
strin=strin+str(-countA)+'A'
# print(-countA,"A")
countA=0
if (a==b and a==1):
countA = countA*(-1)
countB = countB*(-1)
if countB<0:
strin=strin+str(-countB)+'B'
# print(-countB,"B")
countB=0
elif countA <0:
strin=strin+str(-countA)+'B'
# print(-countA,"A")
countA=0
strin.replace(' ', '')
print(strin)
``` | instruction | 0 | 15,487 | 9 | 30,974 |
No | output | 1 | 15,487 | 9 | 30,975 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob decided to eat some fruit. In the kitchen they found a large bag of oranges and apples. Alice immediately took an orange for herself, Bob took an apple. To make the process of sharing the remaining fruit more fun, the friends decided to play a game. They put multiple cards and on each one they wrote a letter, either 'A', or the letter 'B'. Then they began to remove the cards one by one from left to right, every time they removed a card with the letter 'A', Alice gave Bob all the fruits she had at that moment and took out of the bag as many apples and as many oranges as she had before. Thus the number of oranges and apples Alice had, did not change. If the card had written letter 'B', then Bob did the same, that is, he gave Alice all the fruit that he had, and took from the bag the same set of fruit. After the last card way removed, all the fruit in the bag were over.
You know how many oranges and apples was in the bag at first. Your task is to find any sequence of cards that Alice and Bob could have played with.
Input
The first line of the input contains two integers, x, y (1 ≤ x, y ≤ 1018, xy > 1) — the number of oranges and apples that were initially in the bag.
Output
Print any sequence of cards that would meet the problem conditions as a compressed string of characters 'A' and 'B. That means that you need to replace the segments of identical consecutive characters by the number of repetitions of the characters and the actual character. For example, string AAABAABBB should be replaced by string 3A1B2A3B, but cannot be replaced by 2A1A1B2A3B or by 3AB2A3B. See the samples for clarifications of the output format. The string that you print should consist of at most 106 characters. It is guaranteed that if the answer exists, its compressed representation exists, consisting of at most 106 characters. If there are several possible answers, you are allowed to print any of them.
If the sequence of cards that meet the problem statement does not not exist, print a single word Impossible.
Examples
Input
1 4
Output
3B
Input
2 2
Output
Impossible
Input
3 2
Output
1A1B
Note
In the first sample, if the row contained three cards with letter 'B', then Bob should give one apple to Alice three times. So, in the end of the game Alice has one orange and three apples, and Bob has one apple, in total it is one orange and four apples.
In second sample, there is no answer since one card is not enough for game to finish, and two cards will produce at least three apples or three oranges.
In the third sample, cards contain letters 'AB', so after removing the first card Bob has one orange and one apple, and after removal of second card Alice has two oranges and one apple. So, in total it is three oranges and two apples.
Submitted Solution:
```
def gcd(a, b):
if b == 0:
return a
return gcd(b, a % b)
class fraction:
n, m = 0, 0
def __init__(self, n, m):
d = gcd(n, m)
self.n = n/d
self.m = m/d
def add(self, tmp):
return fraction(self.n * tmp.m, self.m * tmp.n)
def compareTo(self ,tmp):
a = self.n * tmp.m
b = self.m * tmp.n
if a > b:
return 1
elif a < b:
return -1
return 0
def sternBrocotAdd(self, tmp):
return fraction(self.n + tmp.n, self.m + tmp.m);
def run(left, right, result):
a = left.n
b = left.m
p = right.n
q = right.m
n = result.n
m = result.m
if left.compareTo(result) == 0:
return;
if right.compareTo(result) == 0:
return;
mid = left.sternBrocotAdd(right)
ch = 'Z'
if mid.compareTo(result) <= 0:
x = int((b * n - a * m) / (p * m - q * n))
left = fraction(a + p * x, b + q * x)
if left.compareTo(result) == 0:
x -= 1
ch = 'A'
else:
x = int((p * m - q * n) / (b * n - a * m))
right = fraction(a * x + p, b * x + q)
if right.compareTo(result) == 0:
x -= 1
ch = 'B'
s = str.format("%d%c"%(x, ch))
print(s, end="")
if left.compareTo(result) == 0 or right.compareTo(result) == 0:
return
run(left, right, result)
p, q = map(int, input().split())
print("%d %d"%(p, q))
result = fraction(p, q)
right = fraction(1, 0)
left = fraction(0, 1)
run(left, right, result)
``` | instruction | 0 | 15,488 | 9 | 30,976 |
No | output | 1 | 15,488 | 9 | 30,977 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob decided to eat some fruit. In the kitchen they found a large bag of oranges and apples. Alice immediately took an orange for herself, Bob took an apple. To make the process of sharing the remaining fruit more fun, the friends decided to play a game. They put multiple cards and on each one they wrote a letter, either 'A', or the letter 'B'. Then they began to remove the cards one by one from left to right, every time they removed a card with the letter 'A', Alice gave Bob all the fruits she had at that moment and took out of the bag as many apples and as many oranges as she had before. Thus the number of oranges and apples Alice had, did not change. If the card had written letter 'B', then Bob did the same, that is, he gave Alice all the fruit that he had, and took from the bag the same set of fruit. After the last card way removed, all the fruit in the bag were over.
You know how many oranges and apples was in the bag at first. Your task is to find any sequence of cards that Alice and Bob could have played with.
Input
The first line of the input contains two integers, x, y (1 ≤ x, y ≤ 1018, xy > 1) — the number of oranges and apples that were initially in the bag.
Output
Print any sequence of cards that would meet the problem conditions as a compressed string of characters 'A' and 'B. That means that you need to replace the segments of identical consecutive characters by the number of repetitions of the characters and the actual character. For example, string AAABAABBB should be replaced by string 3A1B2A3B, but cannot be replaced by 2A1A1B2A3B or by 3AB2A3B. See the samples for clarifications of the output format. The string that you print should consist of at most 106 characters. It is guaranteed that if the answer exists, its compressed representation exists, consisting of at most 106 characters. If there are several possible answers, you are allowed to print any of them.
If the sequence of cards that meet the problem statement does not not exist, print a single word Impossible.
Examples
Input
1 4
Output
3B
Input
2 2
Output
Impossible
Input
3 2
Output
1A1B
Note
In the first sample, if the row contained three cards with letter 'B', then Bob should give one apple to Alice three times. So, in the end of the game Alice has one orange and three apples, and Bob has one apple, in total it is one orange and four apples.
In second sample, there is no answer since one card is not enough for game to finish, and two cards will produce at least three apples or three oranges.
In the third sample, cards contain letters 'AB', so after removing the first card Bob has one orange and one apple, and after removal of second card Alice has two oranges and one apple. So, in total it is three oranges and two apples.
Submitted Solution:
```
def gcd(a, b):
if b == 0:
return a
return gcd(b, a % b)
class fraction:
n, m = 0, 0
def __init__(self, n, m):
d = gcd(n, m)
self.n = n/d
self.m = m/d
def add(self, tmp):
return fraction(self.n * tmp.m, self.m * tmp.n)
def compareTo(self ,tmp):
a = self.n * tmp.m
b = self.m * tmp.n
if a > b:
return 1
elif a < b:
return -1
return 0
def sternBrocotAdd(self, tmp):
return fraction(self.n + tmp.n, self.m + tmp.m);
def run(left, right, result):
a = left.n
b = left.m
p = right.n
q = right.m
n = result.n
m = result.m
mid = left.sternBrocotAdd(right)
if mid.compareTo(result) == 0:
return
ch = 'Z'
x = 0
if mid.compareTo(result) <= 0:
x = int((b * n - a * m) / (p * m - q * n))
left = fraction(a + p * x, b + q * x)
ch = 'A'
if left.compareTo(result) == 0:
x -= 1
else:
x = int((p * m - q * n) / (b * n - a * m))
right = fraction(a * x + p, b * x + q)
ch = 'B'
if right.compareTo(result) == 0:
x -= 1
s = str.format("%d%c"%(x, ch))
print(s, end="")
if left.compareTo(result) == 0 or right.compareTo(result) == 0:
return
run(left, right, result)
p, q = map(int, input().split())
d = gcd(p, q)
if d == 1:
result = fraction(p, q)
right = fraction(1, 0)
left = fraction(0, 1)
run(left, right, result)
else:
print('Impossible')
``` | instruction | 0 | 15,489 | 9 | 30,978 |
No | output | 1 | 15,489 | 9 | 30,979 |
Provide a correct Python 3 solution for this coding contest problem.
Mr. Tanaka died leaving the orchard of HW Earl. The orchard is divided into H x W plots in the north, south, east, and west directions, and apples, oysters, and oranges are planted in each plot. Mr. Tanaka left such a will.
Divide the orchard into as many relatives as possible on a parcel basis. However, if the same kind of fruit is planted in a plot that is located in either the north, south, east, or west direction of a plot, treat them as one large plot because the boundaries of the plot are unknown.
For example, in the following 3 × 10 section ('li' represents an apple,'ka' represents an oyster, and'mi' represents a mandarin orange)
<image>
Eliminating the boundaries between plots with the same tree gives:
<image>
In the end, it will be divided into 10 compartments, or 10 people.
The distribution must be completed before it snows and the boundaries of the plot disappear. Your job is to determine the number of plots to distribute based on the map of the orchard.
Create a program that reads the map of the orchard and outputs the number of relatives who can receive the distribution.
Input
Given multiple datasets. Each dataset is given a string of H lines consisting of the characters H x W, starting with a line containing H, W (H, W ≤ 100) separated by blanks. Only three characters appear in this string:'@' for apples,'#' for oysters, and'*' for oranges.
The input ends with two lines of zeros. The number of datasets does not exceed 20.
Output
For each dataset, output the number of people to be distributed on one line.
Examples
Input
10 10
####*****@
@#@@@@#*#*
@##***@@@*
#****#*@**
##@*#@@*##
*@@@@*@@@#
***#@*@##*
*@@@*@@##@
*@*#*@##**
@****#@@#@
0 0
Output
33
Input
10 10
*****@
@#@@@@#*#*
@##***@@@*
****#*@**
@*#@@*##
*@@@@*@@@#
***#@*@##*
*@@@*@@##@
*@*#*@##**
@****#@@#@
0 0
Output
33 | instruction | 0 | 15,844 | 9 | 31,688 |
"Correct Solution:
```
import sys
sys.setrecursionlimit(100000)
def load_fields():
fields = []
i = 0
H, W = 0, 0
field = []
for line in sys.stdin:
line = line.rstrip()
if line[0] not in ["#", "@", "*"]:
line = line.split(" ")
H = int(line[0])
W = int(line[1])
if (len(field) != 0):
fields.append(field)
field = []
if (H == 0 and W == 0):
break
else:
field.append(list(line))
i += 1
return fields
def dfs(x, y, fruit):
field[x][y] = "0"
for dx in [-1, 0, 1]:
if dx == -1: width = [0]
if dx == 0: width = [-1, 0, 1]
if dx == 1: width = [0]
for dy in width:
nx = x+dx
ny = y+dy
inField = (0 <= nx) and (0 <= ny) and (nx < H) and (ny < W)
if inField and field[nx][ny] == fruit:
dfs(nx, ny, fruit)
fields = load_fields()
count = 0
for field in fields:
H = len(field)
W = len(field[0])
for x in range(H):
for y in range(W):
if (field[x][y] != "0"):
dfs(x, y, field[x][y])
count += 1
print(count)
count = 0
``` | output | 1 | 15,844 | 9 | 31,689 |
Provide a correct Python 3 solution for this coding contest problem.
Mr. Tanaka died leaving the orchard of HW Earl. The orchard is divided into H x W plots in the north, south, east, and west directions, and apples, oysters, and oranges are planted in each plot. Mr. Tanaka left such a will.
Divide the orchard into as many relatives as possible on a parcel basis. However, if the same kind of fruit is planted in a plot that is located in either the north, south, east, or west direction of a plot, treat them as one large plot because the boundaries of the plot are unknown.
For example, in the following 3 × 10 section ('li' represents an apple,'ka' represents an oyster, and'mi' represents a mandarin orange)
<image>
Eliminating the boundaries between plots with the same tree gives:
<image>
In the end, it will be divided into 10 compartments, or 10 people.
The distribution must be completed before it snows and the boundaries of the plot disappear. Your job is to determine the number of plots to distribute based on the map of the orchard.
Create a program that reads the map of the orchard and outputs the number of relatives who can receive the distribution.
Input
Given multiple datasets. Each dataset is given a string of H lines consisting of the characters H x W, starting with a line containing H, W (H, W ≤ 100) separated by blanks. Only three characters appear in this string:'@' for apples,'#' for oysters, and'*' for oranges.
The input ends with two lines of zeros. The number of datasets does not exceed 20.
Output
For each dataset, output the number of people to be distributed on one line.
Examples
Input
10 10
####*****@
@#@@@@#*#*
@##***@@@*
#****#*@**
##@*#@@*##
*@@@@*@@@#
***#@*@##*
*@@@*@@##@
*@*#*@##**
@****#@@#@
0 0
Output
33
Input
10 10
*****@
@#@@@@#*#*
@##***@@@*
****#*@**
@*#@@*##
*@@@@*@@@#
***#@*@##*
*@@@*@@##@
*@*#*@##**
@****#@@#@
0 0
Output
33 | instruction | 0 | 15,845 | 9 | 31,690 |
"Correct Solution:
```
import sys
def setLine(tempH, line):
for i in range(0, len(line)):
geo[tempH][i] = line[i:i+1]
def solve():
person = 0
for i in range(0,H):
for j in range(0,W):
if geo[i][j] is not "_":
search(i,j)
person += 1
print(person)
def search(i,j):
temp = geo[i][j]
geo[i][j] = "_"
for a in range(0,4):
idx, jdy = i + dx[a], j + dy[a]
if(isOnMap(idx, jdy)):
if(isNeededToSolve(temp, idx, jdy)):
search(idx,jdy)
def isOnMap(i,j): return (0<=i and 0<=j and i<H and j<W)
def isNeededToSolve(temp,i,j):
target = geo[i][j]
return (target is not "_" and temp is target)
limit = 10**7
sys.setrecursionlimit(limit)
H, W, tempH = -1, -1, 0
dx = [-1,0,1,0]
dy = [0,-1,0,1]
geo = [[0 for i in range(1)]for j in range(1)]
while True:
line = input()
if H is -1 and W is -1:
p = line.split(" ")
H, W = int(p[0]), int(p[1])
geo = [[0 for i in range(W)] for j in range(H)]
else:
setLine(tempH, line)
tempH+=1
if H is 0 and W is 0:break
if tempH is H:
solve()
H, W, tempH = -1, -1, 0
``` | output | 1 | 15,845 | 9 | 31,691 |
Provide a correct Python 3 solution for this coding contest problem.
Mr. Tanaka died leaving the orchard of HW Earl. The orchard is divided into H x W plots in the north, south, east, and west directions, and apples, oysters, and oranges are planted in each plot. Mr. Tanaka left such a will.
Divide the orchard into as many relatives as possible on a parcel basis. However, if the same kind of fruit is planted in a plot that is located in either the north, south, east, or west direction of a plot, treat them as one large plot because the boundaries of the plot are unknown.
For example, in the following 3 × 10 section ('li' represents an apple,'ka' represents an oyster, and'mi' represents a mandarin orange)
<image>
Eliminating the boundaries between plots with the same tree gives:
<image>
In the end, it will be divided into 10 compartments, or 10 people.
The distribution must be completed before it snows and the boundaries of the plot disappear. Your job is to determine the number of plots to distribute based on the map of the orchard.
Create a program that reads the map of the orchard and outputs the number of relatives who can receive the distribution.
Input
Given multiple datasets. Each dataset is given a string of H lines consisting of the characters H x W, starting with a line containing H, W (H, W ≤ 100) separated by blanks. Only three characters appear in this string:'@' for apples,'#' for oysters, and'*' for oranges.
The input ends with two lines of zeros. The number of datasets does not exceed 20.
Output
For each dataset, output the number of people to be distributed on one line.
Examples
Input
10 10
####*****@
@#@@@@#*#*
@##***@@@*
#****#*@**
##@*#@@*##
*@@@@*@@@#
***#@*@##*
*@@@*@@##@
*@*#*@##**
@****#@@#@
0 0
Output
33
Input
10 10
*****@
@#@@@@#*#*
@##***@@@*
****#*@**
@*#@@*##
*@@@@*@@@#
***#@*@##*
*@@@*@@##@
*@*#*@##**
@****#@@#@
0 0
Output
33 | instruction | 0 | 15,846 | 9 | 31,692 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
import sys
sys.setrecursionlimit(10**6)
def search(values,hp,vp,item):
if not (0<=hp<len(values)): return
if not (0<=vp<len(values[hp])): return
if item!=values[hp][vp]: return
values[hp][vp]=True
# for dh,dv in [[0,1],[0,-1],[1,0],[-1,0]]:
# search(values,hp+dh,vp+dv,item)
search(values,hp-1,vp,item)
search(values,hp+1,vp,item)
search(values,hp,vp-1,item)
search(values,hp,vp+1,item)
def solve(values):
count,valid_items=0,set(['@','#','*'])
for i in range(len(values)):
for j in range(len(values[i])):
if values[i][j] in valid_items:
search(values,i,j,values[i][j])
count+=1
return count
def main():
line, values = input().strip(), list()
while line!='0 0':
H,W = list(map(int,line.split(' ')))
value = list()
for _ in range(H):
# value.append(list(x for x in input().strip()))
value.append(list(input().strip()))
print(solve(value))
# values.append(value)
line = input().strip()
# for value in values:
# print(solve(value))
if __name__=='__main__':
main()
``` | output | 1 | 15,846 | 9 | 31,693 |
Provide a correct Python 3 solution for this coding contest problem.
Mr. Tanaka died leaving the orchard of HW Earl. The orchard is divided into H x W plots in the north, south, east, and west directions, and apples, oysters, and oranges are planted in each plot. Mr. Tanaka left such a will.
Divide the orchard into as many relatives as possible on a parcel basis. However, if the same kind of fruit is planted in a plot that is located in either the north, south, east, or west direction of a plot, treat them as one large plot because the boundaries of the plot are unknown.
For example, in the following 3 × 10 section ('li' represents an apple,'ka' represents an oyster, and'mi' represents a mandarin orange)
<image>
Eliminating the boundaries between plots with the same tree gives:
<image>
In the end, it will be divided into 10 compartments, or 10 people.
The distribution must be completed before it snows and the boundaries of the plot disappear. Your job is to determine the number of plots to distribute based on the map of the orchard.
Create a program that reads the map of the orchard and outputs the number of relatives who can receive the distribution.
Input
Given multiple datasets. Each dataset is given a string of H lines consisting of the characters H x W, starting with a line containing H, W (H, W ≤ 100) separated by blanks. Only three characters appear in this string:'@' for apples,'#' for oysters, and'*' for oranges.
The input ends with two lines of zeros. The number of datasets does not exceed 20.
Output
For each dataset, output the number of people to be distributed on one line.
Examples
Input
10 10
####*****@
@#@@@@#*#*
@##***@@@*
#****#*@**
##@*#@@*##
*@@@@*@@@#
***#@*@##*
*@@@*@@##@
*@*#*@##**
@****#@@#@
0 0
Output
33
Input
10 10
*****@
@#@@@@#*#*
@##***@@@*
****#*@**
@*#@@*##
*@@@@*@@@#
***#@*@##*
*@@@*@@##@
*@*#*@##**
@****#@@#@
0 0
Output
33 | instruction | 0 | 15,847 | 9 | 31,694 |
"Correct Solution:
```
import sys
sys.setrecursionlimit(100000)
def count(h, w, fruits):
if h < 0 or h >= H:
return False
if w < 0 or w >= W:
return False
if fruits == '_':
return False
if fruits != land[h][w]:
return False
land[h][w] = '_'
count(h+1, w, fruits)
count(h-1, w, fruits)
count(h, w+1, fruits)
count(h, w-1, fruits)
while True:
H, W = list(map(int, input().split()))
if H == 0:
break
ans = 0
land = [list(input()) for _ in range(H)]
for i in range(H):
for j in range(W):
if land[i][j] != '_':
ans += 1
count(i, j, land[i][j])
print(ans)
``` | output | 1 | 15,847 | 9 | 31,695 |
Provide a correct Python 3 solution for this coding contest problem.
Mr. Tanaka died leaving the orchard of HW Earl. The orchard is divided into H x W plots in the north, south, east, and west directions, and apples, oysters, and oranges are planted in each plot. Mr. Tanaka left such a will.
Divide the orchard into as many relatives as possible on a parcel basis. However, if the same kind of fruit is planted in a plot that is located in either the north, south, east, or west direction of a plot, treat them as one large plot because the boundaries of the plot are unknown.
For example, in the following 3 × 10 section ('li' represents an apple,'ka' represents an oyster, and'mi' represents a mandarin orange)
<image>
Eliminating the boundaries between plots with the same tree gives:
<image>
In the end, it will be divided into 10 compartments, or 10 people.
The distribution must be completed before it snows and the boundaries of the plot disappear. Your job is to determine the number of plots to distribute based on the map of the orchard.
Create a program that reads the map of the orchard and outputs the number of relatives who can receive the distribution.
Input
Given multiple datasets. Each dataset is given a string of H lines consisting of the characters H x W, starting with a line containing H, W (H, W ≤ 100) separated by blanks. Only three characters appear in this string:'@' for apples,'#' for oysters, and'*' for oranges.
The input ends with two lines of zeros. The number of datasets does not exceed 20.
Output
For each dataset, output the number of people to be distributed on one line.
Examples
Input
10 10
####*****@
@#@@@@#*#*
@##***@@@*
#****#*@**
##@*#@@*##
*@@@@*@@@#
***#@*@##*
*@@@*@@##@
*@*#*@##**
@****#@@#@
0 0
Output
33
Input
10 10
*****@
@#@@@@#*#*
@##***@@@*
****#*@**
@*#@@*##
*@@@@*@@@#
***#@*@##*
*@@@*@@##@
*@*#*@##**
@****#@@#@
0 0
Output
33 | instruction | 0 | 15,848 | 9 | 31,696 |
"Correct Solution:
```
def Labeling(terrain):
label = terrain
offset = [[-1, 0], [0, -1], [1, 0], [0, 1]]
signs = ["@", "#", "*"]
step = 0
for row in range(1, len(label) - 1):
for col in range(1, len(label[row]) - 1):
mark = label[row][col]
if mark in signs:
pointStack = [[row, col]]
label[row][col] = step
while 0 < len(pointStack):
rowIndex, colIndex = pointStack.pop()
for x, y in offset:
offsetRow, offsetCol = rowIndex + x, colIndex + y
if label[offsetRow][offsetCol] == mark:
pointStack.append([offsetRow, offsetCol])
label[offsetRow][offsetCol] = step
step += 1
return step
Sentinel = "1"
while True:
col, row = [int(item) for item in input().split()]
if row == 0 and col == 0:
break
land = []
land.append([item for item in Sentinel * row + Sentinel * 2])
for lp in range(col):
part = [item for item in input()]
land.append([])
land[-1].extend([Sentinel])
land[-1].extend(part)
land[-1].extend([Sentinel])
land.append([item for item in Sentinel * row + Sentinel * 2])
result = Labeling(land)
print(result)
``` | output | 1 | 15,848 | 9 | 31,697 |
Provide a correct Python 3 solution for this coding contest problem.
Mr. Tanaka died leaving the orchard of HW Earl. The orchard is divided into H x W plots in the north, south, east, and west directions, and apples, oysters, and oranges are planted in each plot. Mr. Tanaka left such a will.
Divide the orchard into as many relatives as possible on a parcel basis. However, if the same kind of fruit is planted in a plot that is located in either the north, south, east, or west direction of a plot, treat them as one large plot because the boundaries of the plot are unknown.
For example, in the following 3 × 10 section ('li' represents an apple,'ka' represents an oyster, and'mi' represents a mandarin orange)
<image>
Eliminating the boundaries between plots with the same tree gives:
<image>
In the end, it will be divided into 10 compartments, or 10 people.
The distribution must be completed before it snows and the boundaries of the plot disappear. Your job is to determine the number of plots to distribute based on the map of the orchard.
Create a program that reads the map of the orchard and outputs the number of relatives who can receive the distribution.
Input
Given multiple datasets. Each dataset is given a string of H lines consisting of the characters H x W, starting with a line containing H, W (H, W ≤ 100) separated by blanks. Only three characters appear in this string:'@' for apples,'#' for oysters, and'*' for oranges.
The input ends with two lines of zeros. The number of datasets does not exceed 20.
Output
For each dataset, output the number of people to be distributed on one line.
Examples
Input
10 10
####*****@
@#@@@@#*#*
@##***@@@*
#****#*@**
##@*#@@*##
*@@@@*@@@#
***#@*@##*
*@@@*@@##@
*@*#*@##**
@****#@@#@
0 0
Output
33
Input
10 10
*****@
@#@@@@#*#*
@##***@@@*
****#*@**
@*#@@*##
*@@@@*@@@#
***#@*@##*
*@@@*@@##@
*@*#*@##**
@****#@@#@
0 0
Output
33 | instruction | 0 | 15,849 | 9 | 31,698 |
"Correct Solution:
```
# AOJ 0118 Property Distribution
# Python3 2018.6.23 bal4u
# 隣接する同種ものを集める、Unionセットに帰着
# UNION-FIND library
class UnionSet:
def __init__(self, nmax):
self.size = [1]*nmax
self.id = [i for i in range(nmax+1)]
def root(self, i):
while i != self.id[i]:
self.id[i] = self.id[self.id[i]]
i = self.id[i]
return i
def connected(self, p, q): return self.root(p) == self.root(q)
def unite(self, p, q):
i, j = self.root(p), self.root(q)
if i == j: return
if self.size[i] < self.size[j]:
self.id[i] = j
self.size[j] += self.size[i]
else:
self.id[j] = i
self.size[i] += self.size[j]
# UNION-FIND library
while 1:
H, W = map(int, input().split())
if H == 0 and W == 0: break
tbl = ['']*H
for r in range(H): tbl[r] = list(input())
u = UnionSet(H*W)
for r in range(H):
for c in range(W):
if c+1 < W and tbl[r][c] == tbl[r][c+1]: u.unite(r*W+c, r*W+c+1)
if r+1 < H and tbl[r][c] == tbl[r+1][c]: u.unite(r*W+c, (r+1)*W+c)
ans = 0
for k in range(H*W):
if u.root(k) == k: ans += 1
print(ans)
``` | output | 1 | 15,849 | 9 | 31,699 |
Provide a correct Python 3 solution for this coding contest problem.
Mr. Tanaka died leaving the orchard of HW Earl. The orchard is divided into H x W plots in the north, south, east, and west directions, and apples, oysters, and oranges are planted in each plot. Mr. Tanaka left such a will.
Divide the orchard into as many relatives as possible on a parcel basis. However, if the same kind of fruit is planted in a plot that is located in either the north, south, east, or west direction of a plot, treat them as one large plot because the boundaries of the plot are unknown.
For example, in the following 3 × 10 section ('li' represents an apple,'ka' represents an oyster, and'mi' represents a mandarin orange)
<image>
Eliminating the boundaries between plots with the same tree gives:
<image>
In the end, it will be divided into 10 compartments, or 10 people.
The distribution must be completed before it snows and the boundaries of the plot disappear. Your job is to determine the number of plots to distribute based on the map of the orchard.
Create a program that reads the map of the orchard and outputs the number of relatives who can receive the distribution.
Input
Given multiple datasets. Each dataset is given a string of H lines consisting of the characters H x W, starting with a line containing H, W (H, W ≤ 100) separated by blanks. Only three characters appear in this string:'@' for apples,'#' for oysters, and'*' for oranges.
The input ends with two lines of zeros. The number of datasets does not exceed 20.
Output
For each dataset, output the number of people to be distributed on one line.
Examples
Input
10 10
####*****@
@#@@@@#*#*
@##***@@@*
#****#*@**
##@*#@@*##
*@@@@*@@@#
***#@*@##*
*@@@*@@##@
*@*#*@##**
@****#@@#@
0 0
Output
33
Input
10 10
*****@
@#@@@@#*#*
@##***@@@*
****#*@**
@*#@@*##
*@@@@*@@@#
***#@*@##*
*@@@*@@##@
*@*#*@##**
@****#@@#@
0 0
Output
33 | instruction | 0 | 15,850 | 9 | 31,700 |
"Correct Solution:
```
# AOJ 0118 Property Distribution
# Python3 2018.6.23 bal4u
# 隣接する同種ものを集める、Unionセットに帰着
# UNION-FIND library
class UnionSet:
def __init__(self, nmax):
self.size = [1]*nmax
self.id = [i for i in range(nmax+1)]
def root(self, i):
while i != self.id[i]:
self.id[i] = self.id[self.id[i]]
i = self.id[i]
return i
def connected(self, p, q): return self.root(p) == self.root(q)
def unite(self, p, q):
i, j = self.root(p), self.root(q)
if i == j: return
if self.size[i] < self.size[j]:
self.id[i] = j
self.size[j] += self.size[i]
else:
self.id[j] = i
self.size[i] += self.size[j]
# UNION-FIND library
while 1:
H, W = map(int, input().split())
if H == 0 and W == 0: break
tbl = ['']*H
for r in range(H): tbl[r] = input()
u = UnionSet(H*W)
rr = 0
for r in range(H):
for c in range(W):
if c+1 < W and tbl[r][c] == tbl[r][c+1]: u.unite(rr+c, rr+c+1)
if r+1 < H and tbl[r][c] == tbl[r+1][c]: u.unite(rr+c, rr+W+c)
rr += W
ans = 0
for k in range(H*W):
if u.root(k) == k: ans += 1
print(ans)
``` | output | 1 | 15,850 | 9 | 31,701 |
Provide a correct Python 3 solution for this coding contest problem.
Mr. Tanaka died leaving the orchard of HW Earl. The orchard is divided into H x W plots in the north, south, east, and west directions, and apples, oysters, and oranges are planted in each plot. Mr. Tanaka left such a will.
Divide the orchard into as many relatives as possible on a parcel basis. However, if the same kind of fruit is planted in a plot that is located in either the north, south, east, or west direction of a plot, treat them as one large plot because the boundaries of the plot are unknown.
For example, in the following 3 × 10 section ('li' represents an apple,'ka' represents an oyster, and'mi' represents a mandarin orange)
<image>
Eliminating the boundaries between plots with the same tree gives:
<image>
In the end, it will be divided into 10 compartments, or 10 people.
The distribution must be completed before it snows and the boundaries of the plot disappear. Your job is to determine the number of plots to distribute based on the map of the orchard.
Create a program that reads the map of the orchard and outputs the number of relatives who can receive the distribution.
Input
Given multiple datasets. Each dataset is given a string of H lines consisting of the characters H x W, starting with a line containing H, W (H, W ≤ 100) separated by blanks. Only three characters appear in this string:'@' for apples,'#' for oysters, and'*' for oranges.
The input ends with two lines of zeros. The number of datasets does not exceed 20.
Output
For each dataset, output the number of people to be distributed on one line.
Examples
Input
10 10
####*****@
@#@@@@#*#*
@##***@@@*
#****#*@**
##@*#@@*##
*@@@@*@@@#
***#@*@##*
*@@@*@@##@
*@*#*@##**
@****#@@#@
0 0
Output
33
Input
10 10
*****@
@#@@@@#*#*
@##***@@@*
****#*@**
@*#@@*##
*@@@@*@@@#
***#@*@##*
*@@@*@@##@
*@*#*@##**
@****#@@#@
0 0
Output
33 | instruction | 0 | 15,851 | 9 | 31,702 |
"Correct Solution:
```
import sys
sys.setrecursionlimit(100000)
def load_fields():
fields = []
i = 0
H, W = 0, 0
field = []
for line in sys.stdin:
line = line.rstrip()
if line[0] not in ["#", "@", "*"]:
line = line.split(" ")
H, W = int(line[0]), int(line[1])
if (len(field) != 0):
fields.append(field)
field = []
if (H == 0 and W == 0):
break
else:
field.append(list(line))
i += 1
return fields
def dfs(x, y, fruit):
field[x][y] = "0"
for dx in [-1, 0, 1]:
if dx == -1: width = [0]
if dx == 0: width = [-1, 0, 1]
if dx == 1: width = [0]
for dy in width:
nx = x+dx
ny = y+dy
inField = (0 <= nx < H) and (0 <= ny < W)
if inField and (field[nx][ny] == fruit):
dfs(nx, ny, fruit)
fields = load_fields()
count = 0
for field in fields:
H = len(field)
W = len(field[0])
for x in range(H):
for y in range(W):
if (field[x][y] != "0"):
dfs(x, y, field[x][y])
count += 1
print(count)
count = 0
``` | output | 1 | 15,851 | 9 | 31,703 |
Provide a correct Python 3 solution for this coding contest problem.
A: four tea
problem
Tea is indispensable for programming contests. Tea has the effect of relieving constant tension [citation needed]
There are N players participating in the contest, so I would like to prepare tea for this number of people. There are four types of tea packages, A, B, C, and D, all of which are the same variety but have different contents. For a package X, the price of one package is p_X yen, and it is known that if you buy one, you can make t_X cups of tea.
Find the minimum amount needed to make tea for N people. You may have a package that you don't buy at all, and you don't have to buy a package for just N people (if you can make more than N people).
Input format
The input is given in the following format.
N
p_A p_B p_C p_D
t_A t_B t_C t_D
* The first line gives the number of players participating in the contest.
* In the second line, the prices of tea in packages A, B, C and D are given separated by blanks.
* The third line gives the number of cups of tea that can be made from packages A, B, C, and D, separated by blanks.
Constraint
* 1 \ leq N \ leq 100
* 1 \ leq p_X \ leq 100
* 1 \ leq t_X \ leq 100
* All inputs are given as integers.
Output format
Output the minimum amount required to make tea for N people in one line.
Input example 1
Ten
1 2 3 4
1 2 4 8
Output example 1
6
* It's best to buy one package B and one D.
Input example 2
Five
2 9 9 8
1 4 5 100
Output example 2
8
* You will have 20 times more tea than you need, but buying one Package D is the cheapest way to get more than 5 cups of tea.
Input example 3
twenty four
2 3 4 7
7 9 11 20
Output example 3
8
* It's best to buy two packages A and one C. It may not be possible to make just enough tea for the number of people as in this case.
Example
Input
10
1 2 3 4
1 2 4 8
Output
6 | instruction | 0 | 15,901 | 9 | 31,802 |
"Correct Solution:
```
n = int(input())
lis1 = list(map(int,input().split()))
lis2 = list(map(int,input().split()))
money = [10**7+9]*(n+1)
money[0] = 0
for price,num in zip(lis1,lis2):
i = 0
while(i<n+1):
if i-num < 0:
money[i] = min(price,money[i])
else:
money[i] = min(money[i],price+money[i-num])
i += 1
print(money[n])
``` | output | 1 | 15,901 | 9 | 31,803 |
Provide a correct Python 3 solution for this coding contest problem.
A: four tea
problem
Tea is indispensable for programming contests. Tea has the effect of relieving constant tension [citation needed]
There are N players participating in the contest, so I would like to prepare tea for this number of people. There are four types of tea packages, A, B, C, and D, all of which are the same variety but have different contents. For a package X, the price of one package is p_X yen, and it is known that if you buy one, you can make t_X cups of tea.
Find the minimum amount needed to make tea for N people. You may have a package that you don't buy at all, and you don't have to buy a package for just N people (if you can make more than N people).
Input format
The input is given in the following format.
N
p_A p_B p_C p_D
t_A t_B t_C t_D
* The first line gives the number of players participating in the contest.
* In the second line, the prices of tea in packages A, B, C and D are given separated by blanks.
* The third line gives the number of cups of tea that can be made from packages A, B, C, and D, separated by blanks.
Constraint
* 1 \ leq N \ leq 100
* 1 \ leq p_X \ leq 100
* 1 \ leq t_X \ leq 100
* All inputs are given as integers.
Output format
Output the minimum amount required to make tea for N people in one line.
Input example 1
Ten
1 2 3 4
1 2 4 8
Output example 1
6
* It's best to buy one package B and one D.
Input example 2
Five
2 9 9 8
1 4 5 100
Output example 2
8
* You will have 20 times more tea than you need, but buying one Package D is the cheapest way to get more than 5 cups of tea.
Input example 3
twenty four
2 3 4 7
7 9 11 20
Output example 3
8
* It's best to buy two packages A and one C. It may not be possible to make just enough tea for the number of people as in this case.
Example
Input
10
1 2 3 4
1 2 4 8
Output
6 | instruction | 0 | 15,902 | 9 | 31,804 |
"Correct Solution:
```
N = int(input())
p = list(map(int, input().split()))
t = list(map(int, input().split()))
ans = 10 ** 10
#aの最大個数
A = max(0, (N + t[0] - 1)// t[0])
for a in range(A+1):
#aの個数が決まったときのbの最大個数
B = max(0, (N - a * t[0] + t[1] - 1)//t[1])
for b in range(B+1):
#a,bの個数が決まったときのcの最大個数
C = max(0, (N - a * t[0] - b * t[1] + t[2] - 1)//t[2])
for c in range(C+1):
#a,b,cの個数が決まったときのdの個数(残りは全てdとする)
D = max(0, (N - a * t[0] - b * t[1] - c * t[2] + t[3] - 1)//t[3])
cost = a * p[0] + b *p[1] + c * p[2] + D * p[3]
# print (a, b, c, D, cost)
ans = min(ans, cost)
print (ans)
``` | output | 1 | 15,902 | 9 | 31,805 |
Provide a correct Python 3 solution for this coding contest problem.
A: four tea
problem
Tea is indispensable for programming contests. Tea has the effect of relieving constant tension [citation needed]
There are N players participating in the contest, so I would like to prepare tea for this number of people. There are four types of tea packages, A, B, C, and D, all of which are the same variety but have different contents. For a package X, the price of one package is p_X yen, and it is known that if you buy one, you can make t_X cups of tea.
Find the minimum amount needed to make tea for N people. You may have a package that you don't buy at all, and you don't have to buy a package for just N people (if you can make more than N people).
Input format
The input is given in the following format.
N
p_A p_B p_C p_D
t_A t_B t_C t_D
* The first line gives the number of players participating in the contest.
* In the second line, the prices of tea in packages A, B, C and D are given separated by blanks.
* The third line gives the number of cups of tea that can be made from packages A, B, C, and D, separated by blanks.
Constraint
* 1 \ leq N \ leq 100
* 1 \ leq p_X \ leq 100
* 1 \ leq t_X \ leq 100
* All inputs are given as integers.
Output format
Output the minimum amount required to make tea for N people in one line.
Input example 1
Ten
1 2 3 4
1 2 4 8
Output example 1
6
* It's best to buy one package B and one D.
Input example 2
Five
2 9 9 8
1 4 5 100
Output example 2
8
* You will have 20 times more tea than you need, but buying one Package D is the cheapest way to get more than 5 cups of tea.
Input example 3
twenty four
2 3 4 7
7 9 11 20
Output example 3
8
* It's best to buy two packages A and one C. It may not be possible to make just enough tea for the number of people as in this case.
Example
Input
10
1 2 3 4
1 2 4 8
Output
6 | instruction | 0 | 15,903 | 9 | 31,806 |
"Correct Solution:
```
N = int(input())
p = list(map(int, input().split()))
t = list(map(int, input().split()))
res = 100000
for i in range(N+1):
for j in range(N+1):
for k in range(N+1):
if N < t[0] * i + t[1] * j + t[2] * k:
if res > p[0]*i + p[1]*j + p[2]*k:
res = p[0]*i + p[1]*j + p[2]*k
else:
d = (N-t[0]*i-t[1]*j-t[2]*k + t[3] - 1) // t[3]
if res > p[0]*i + p[1]*j + p[2]*k + p[3]*d:
res = p[0]*i + p[1]*j + p[2]*k + p[3]*d
print(res)
``` | output | 1 | 15,903 | 9 | 31,807 |
Provide a correct Python 3 solution for this coding contest problem.
A: four tea
problem
Tea is indispensable for programming contests. Tea has the effect of relieving constant tension [citation needed]
There are N players participating in the contest, so I would like to prepare tea for this number of people. There are four types of tea packages, A, B, C, and D, all of which are the same variety but have different contents. For a package X, the price of one package is p_X yen, and it is known that if you buy one, you can make t_X cups of tea.
Find the minimum amount needed to make tea for N people. You may have a package that you don't buy at all, and you don't have to buy a package for just N people (if you can make more than N people).
Input format
The input is given in the following format.
N
p_A p_B p_C p_D
t_A t_B t_C t_D
* The first line gives the number of players participating in the contest.
* In the second line, the prices of tea in packages A, B, C and D are given separated by blanks.
* The third line gives the number of cups of tea that can be made from packages A, B, C, and D, separated by blanks.
Constraint
* 1 \ leq N \ leq 100
* 1 \ leq p_X \ leq 100
* 1 \ leq t_X \ leq 100
* All inputs are given as integers.
Output format
Output the minimum amount required to make tea for N people in one line.
Input example 1
Ten
1 2 3 4
1 2 4 8
Output example 1
6
* It's best to buy one package B and one D.
Input example 2
Five
2 9 9 8
1 4 5 100
Output example 2
8
* You will have 20 times more tea than you need, but buying one Package D is the cheapest way to get more than 5 cups of tea.
Input example 3
twenty four
2 3 4 7
7 9 11 20
Output example 3
8
* It's best to buy two packages A and one C. It may not be possible to make just enough tea for the number of people as in this case.
Example
Input
10
1 2 3 4
1 2 4 8
Output
6 | instruction | 0 | 15,904 | 9 | 31,808 |
"Correct Solution:
```
# from sys import exit
N = int(input())
P = [int(n) for n in input().split()]
t = [int(n) for n in input().split()]
Pt = sorted([(p, t) for p, t in zip(P,t)])
memo = [0 for _ in range(N)]
def rec(n):
if n >= N:
return 0
elif memo[n] != 0:
return memo[n]
else:
memo[n] = min([rec(n+t[i]) + P[i] for i in range(4)])
return memo[n]
print(rec(0))
``` | output | 1 | 15,904 | 9 | 31,809 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Today, Yasser and Adel are at the shop buying cupcakes. There are n cupcake types, arranged from 1 to n on the shelf, and there are infinitely many of each type. The tastiness of a cupcake of type i is an integer a_i. There are both tasty and nasty cupcakes, so the tastiness can be positive, zero or negative.
Yasser, of course, wants to try them all, so he will buy exactly one cupcake of each type.
On the other hand, Adel will choose some segment [l, r] (1 ≤ l ≤ r ≤ n) that does not include all of cupcakes (he can't choose [l, r] = [1, n]) and buy exactly one cupcake of each of types l, l + 1, ..., r.
After that they will compare the total tastiness of the cupcakes each of them have bought. Yasser will be happy if the total tastiness of cupcakes he buys is strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice.
For example, let the tastinesses of the cupcakes be [7, 4, -1]. Yasser will buy all of them, the total tastiness will be 7 + 4 - 1 = 10. Adel can choose segments [7], [4], [-1], [7, 4] or [4, -1], their total tastinesses are 7, 4, -1, 11 and 3, respectively. Adel can choose segment with tastiness 11, and as 10 is not strictly greater than 11, Yasser won't be happy :(
Find out if Yasser will be happy after visiting the shop.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). The description of the test cases follows.
The first line of each test case contains n (2 ≤ n ≤ 10^5).
The second line of each test case contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9), where a_i represents the tastiness of the i-th type of cupcake.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print "YES", if the total tastiness of cupcakes Yasser buys will always be strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice. Otherwise, print "NO".
Example
Input
3
4
1 2 3 4
3
7 4 -1
3
5 -5 5
Output
YES
NO
NO
Note
In the first example, the total tastiness of any segment Adel can choose is less than the total tastiness of all cupcakes.
In the second example, Adel will choose the segment [1, 2] with total tastiness 11, which is not less than the total tastiness of all cupcakes, which is 10.
In the third example, Adel can choose the segment [3, 3] with total tastiness of 5. Note that Yasser's cupcakes' total tastiness is also 5, so in that case, the total tastiness of Yasser's cupcakes isn't strictly greater than the total tastiness of Adel's cupcakes. | instruction | 0 | 16,066 | 9 | 32,132 |
Tags: dp, greedy, implementation
Correct Solution:
```
for _ in range(int(input())):
i1=int(input())
ar=list(map(int,input().split()))
ts=sum(ar)
s1=0
s2=0
flag=True
for i in range(i1-1):
s1+=ar[i]
s2+=ar[-(i+1)]
if(s1>=ts or s2>=ts):
flag=False
break
if(flag):
print('YES')
else:
print('NO')
``` | output | 1 | 16,066 | 9 | 32,133 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Today, Yasser and Adel are at the shop buying cupcakes. There are n cupcake types, arranged from 1 to n on the shelf, and there are infinitely many of each type. The tastiness of a cupcake of type i is an integer a_i. There are both tasty and nasty cupcakes, so the tastiness can be positive, zero or negative.
Yasser, of course, wants to try them all, so he will buy exactly one cupcake of each type.
On the other hand, Adel will choose some segment [l, r] (1 ≤ l ≤ r ≤ n) that does not include all of cupcakes (he can't choose [l, r] = [1, n]) and buy exactly one cupcake of each of types l, l + 1, ..., r.
After that they will compare the total tastiness of the cupcakes each of them have bought. Yasser will be happy if the total tastiness of cupcakes he buys is strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice.
For example, let the tastinesses of the cupcakes be [7, 4, -1]. Yasser will buy all of them, the total tastiness will be 7 + 4 - 1 = 10. Adel can choose segments [7], [4], [-1], [7, 4] or [4, -1], their total tastinesses are 7, 4, -1, 11 and 3, respectively. Adel can choose segment with tastiness 11, and as 10 is not strictly greater than 11, Yasser won't be happy :(
Find out if Yasser will be happy after visiting the shop.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). The description of the test cases follows.
The first line of each test case contains n (2 ≤ n ≤ 10^5).
The second line of each test case contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9), where a_i represents the tastiness of the i-th type of cupcake.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print "YES", if the total tastiness of cupcakes Yasser buys will always be strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice. Otherwise, print "NO".
Example
Input
3
4
1 2 3 4
3
7 4 -1
3
5 -5 5
Output
YES
NO
NO
Note
In the first example, the total tastiness of any segment Adel can choose is less than the total tastiness of all cupcakes.
In the second example, Adel will choose the segment [1, 2] with total tastiness 11, which is not less than the total tastiness of all cupcakes, which is 10.
In the third example, Adel can choose the segment [3, 3] with total tastiness of 5. Note that Yasser's cupcakes' total tastiness is also 5, so in that case, the total tastiness of Yasser's cupcakes isn't strictly greater than the total tastiness of Adel's cupcakes. | instruction | 0 | 16,067 | 9 | 32,134 |
Tags: dp, greedy, implementation
Correct Solution:
```
INT_MAX=10**18+7
MOD=10**9+7
def INPUT():return list(int(i) for i in input().split())
def LIST_1D_ARRAY(n):return [0 for _ in range(n)]
def LIST_2D_ARRAY(m,n):return [[0 for _ in range(n)]for _ in range(m)]
#################################################################################
def clac(A,n):
ans1=0
sum1=0
for i in range(1,n):
if sum1+A[i]>0:
sum1+=A[i]
else:
sum1=0
ans1=max(ans1,sum1)
sum2=0
ans2=0
for i in range(n-1):
if sum2+A[i]>0:
sum2+=A[i]
else:
sum2=0
ans2=max(ans2,sum2)
return max(ans2,ans1)
for _ in range(int(input())):
n=int(input())
A=INPUT()
s=sum(A)
if s>clac(A,n):
print("YES")
else:
print("NO")
``` | output | 1 | 16,067 | 9 | 32,135 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Today, Yasser and Adel are at the shop buying cupcakes. There are n cupcake types, arranged from 1 to n on the shelf, and there are infinitely many of each type. The tastiness of a cupcake of type i is an integer a_i. There are both tasty and nasty cupcakes, so the tastiness can be positive, zero or negative.
Yasser, of course, wants to try them all, so he will buy exactly one cupcake of each type.
On the other hand, Adel will choose some segment [l, r] (1 ≤ l ≤ r ≤ n) that does not include all of cupcakes (he can't choose [l, r] = [1, n]) and buy exactly one cupcake of each of types l, l + 1, ..., r.
After that they will compare the total tastiness of the cupcakes each of them have bought. Yasser will be happy if the total tastiness of cupcakes he buys is strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice.
For example, let the tastinesses of the cupcakes be [7, 4, -1]. Yasser will buy all of them, the total tastiness will be 7 + 4 - 1 = 10. Adel can choose segments [7], [4], [-1], [7, 4] or [4, -1], their total tastinesses are 7, 4, -1, 11 and 3, respectively. Adel can choose segment with tastiness 11, and as 10 is not strictly greater than 11, Yasser won't be happy :(
Find out if Yasser will be happy after visiting the shop.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). The description of the test cases follows.
The first line of each test case contains n (2 ≤ n ≤ 10^5).
The second line of each test case contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9), where a_i represents the tastiness of the i-th type of cupcake.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print "YES", if the total tastiness of cupcakes Yasser buys will always be strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice. Otherwise, print "NO".
Example
Input
3
4
1 2 3 4
3
7 4 -1
3
5 -5 5
Output
YES
NO
NO
Note
In the first example, the total tastiness of any segment Adel can choose is less than the total tastiness of all cupcakes.
In the second example, Adel will choose the segment [1, 2] with total tastiness 11, which is not less than the total tastiness of all cupcakes, which is 10.
In the third example, Adel can choose the segment [3, 3] with total tastiness of 5. Note that Yasser's cupcakes' total tastiness is also 5, so in that case, the total tastiness of Yasser's cupcakes isn't strictly greater than the total tastiness of Adel's cupcakes. | instruction | 0 | 16,068 | 9 | 32,136 |
Tags: dp, greedy, implementation
Correct Solution:
```
def solve():
s = 0
for i in range(n):
s += v[i]
if s <= 0:
return 0
s = 0
for i in range(n-1, -1, -1):
s += v[i]
if s <= 0:
return 0
return 1
t = int(input())
while t:
n = int(input())
v = [int(x) for x in input().split()]
if solve() == 0: print("NO")
else: print("YES")
t -= 1
``` | output | 1 | 16,068 | 9 | 32,137 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Today, Yasser and Adel are at the shop buying cupcakes. There are n cupcake types, arranged from 1 to n on the shelf, and there are infinitely many of each type. The tastiness of a cupcake of type i is an integer a_i. There are both tasty and nasty cupcakes, so the tastiness can be positive, zero or negative.
Yasser, of course, wants to try them all, so he will buy exactly one cupcake of each type.
On the other hand, Adel will choose some segment [l, r] (1 ≤ l ≤ r ≤ n) that does not include all of cupcakes (he can't choose [l, r] = [1, n]) and buy exactly one cupcake of each of types l, l + 1, ..., r.
After that they will compare the total tastiness of the cupcakes each of them have bought. Yasser will be happy if the total tastiness of cupcakes he buys is strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice.
For example, let the tastinesses of the cupcakes be [7, 4, -1]. Yasser will buy all of them, the total tastiness will be 7 + 4 - 1 = 10. Adel can choose segments [7], [4], [-1], [7, 4] or [4, -1], their total tastinesses are 7, 4, -1, 11 and 3, respectively. Adel can choose segment with tastiness 11, and as 10 is not strictly greater than 11, Yasser won't be happy :(
Find out if Yasser will be happy after visiting the shop.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). The description of the test cases follows.
The first line of each test case contains n (2 ≤ n ≤ 10^5).
The second line of each test case contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9), where a_i represents the tastiness of the i-th type of cupcake.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print "YES", if the total tastiness of cupcakes Yasser buys will always be strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice. Otherwise, print "NO".
Example
Input
3
4
1 2 3 4
3
7 4 -1
3
5 -5 5
Output
YES
NO
NO
Note
In the first example, the total tastiness of any segment Adel can choose is less than the total tastiness of all cupcakes.
In the second example, Adel will choose the segment [1, 2] with total tastiness 11, which is not less than the total tastiness of all cupcakes, which is 10.
In the third example, Adel can choose the segment [3, 3] with total tastiness of 5. Note that Yasser's cupcakes' total tastiness is also 5, so in that case, the total tastiness of Yasser's cupcakes isn't strictly greater than the total tastiness of Adel's cupcakes. | instruction | 0 | 16,069 | 9 | 32,138 |
Tags: dp, greedy, implementation
Correct Solution:
```
"""
Template written to be used by Python Programmers.
Use at your own risk!!!!
Owned by adi0311(rating - 1989 at CodeChef and 1335 at Codeforces).
"""
import sys
from bisect import bisect_left as bl, bisect_right as br, bisect #Binary Search alternative
import math
from itertools import zip_longest as zl #zl(x, y) return [(x[0], y[0]), ...]
from itertools import groupby as gb #gb(x, y)
from itertools import combinations as comb #comb(x, y) return [all subsets of x with len == y]
from itertools import combinations_with_replacement as cwr
from collections import defaultdict as dd #defaultdict(<datatype>) Free of KeyError.
from collections import deque as dq #deque(list) append(), appendleft(), pop(), popleft() - O(1)
from collections import Counter as c #Counter(list) return a dict with {key: count}
# sys.setrecursionlimit(2*pow(10, 6))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
mod = pow(10, 9)+7
mod2 = 998244353
# def data(): return sys.stdin.readline().strip()
# def out(var): sys.stdout.write(var)
def data(): return input()
def l(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [[val for i in range(n)] for j in range(m)]
def kadane(a, size):
max_so_far = -pow(10, 9) - 1
max_ending_here = 0
for i in range(0, size):
max_ending_here = max_ending_here + a[i]
if (max_so_far < max_ending_here):
max_so_far = max_ending_here
if max_ending_here < 0:
max_ending_here = 0
return max_so_far
for _ in range(int(data())):
n = int(data())
arr = l()
cnt = 0
for i in arr:
if i < 0:
cnt += 1
if kadane(arr[1:], n-1) >= sum(arr) or kadane(arr[:n-1], n-1) >= sum(arr):
print("NO")
else:
print("YES")
``` | output | 1 | 16,069 | 9 | 32,139 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Today, Yasser and Adel are at the shop buying cupcakes. There are n cupcake types, arranged from 1 to n on the shelf, and there are infinitely many of each type. The tastiness of a cupcake of type i is an integer a_i. There are both tasty and nasty cupcakes, so the tastiness can be positive, zero or negative.
Yasser, of course, wants to try them all, so he will buy exactly one cupcake of each type.
On the other hand, Adel will choose some segment [l, r] (1 ≤ l ≤ r ≤ n) that does not include all of cupcakes (he can't choose [l, r] = [1, n]) and buy exactly one cupcake of each of types l, l + 1, ..., r.
After that they will compare the total tastiness of the cupcakes each of them have bought. Yasser will be happy if the total tastiness of cupcakes he buys is strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice.
For example, let the tastinesses of the cupcakes be [7, 4, -1]. Yasser will buy all of them, the total tastiness will be 7 + 4 - 1 = 10. Adel can choose segments [7], [4], [-1], [7, 4] or [4, -1], their total tastinesses are 7, 4, -1, 11 and 3, respectively. Adel can choose segment with tastiness 11, and as 10 is not strictly greater than 11, Yasser won't be happy :(
Find out if Yasser will be happy after visiting the shop.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). The description of the test cases follows.
The first line of each test case contains n (2 ≤ n ≤ 10^5).
The second line of each test case contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9), where a_i represents the tastiness of the i-th type of cupcake.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print "YES", if the total tastiness of cupcakes Yasser buys will always be strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice. Otherwise, print "NO".
Example
Input
3
4
1 2 3 4
3
7 4 -1
3
5 -5 5
Output
YES
NO
NO
Note
In the first example, the total tastiness of any segment Adel can choose is less than the total tastiness of all cupcakes.
In the second example, Adel will choose the segment [1, 2] with total tastiness 11, which is not less than the total tastiness of all cupcakes, which is 10.
In the third example, Adel can choose the segment [3, 3] with total tastiness of 5. Note that Yasser's cupcakes' total tastiness is also 5, so in that case, the total tastiness of Yasser's cupcakes isn't strictly greater than the total tastiness of Adel's cupcakes. | instruction | 0 | 16,070 | 9 | 32,140 |
Tags: dp, greedy, implementation
Correct Solution:
```
t = int(input())
while t:
t -= 1
n = int(input())
a = list(map(int, input().split()))
s1 = sum(a)
curr_sum = 0
flag = 0
maxi = float("-inf")
pos = 0
for i, x in enumerate(a):
curr_sum += x
if curr_sum >= s1 and not(i == n-1 and pos == 0):
flag = 1
break
if curr_sum <= 0:
pos = i+1
if flag:
print("NO")
else:
print("YES")
``` | output | 1 | 16,070 | 9 | 32,141 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Today, Yasser and Adel are at the shop buying cupcakes. There are n cupcake types, arranged from 1 to n on the shelf, and there are infinitely many of each type. The tastiness of a cupcake of type i is an integer a_i. There are both tasty and nasty cupcakes, so the tastiness can be positive, zero or negative.
Yasser, of course, wants to try them all, so he will buy exactly one cupcake of each type.
On the other hand, Adel will choose some segment [l, r] (1 ≤ l ≤ r ≤ n) that does not include all of cupcakes (he can't choose [l, r] = [1, n]) and buy exactly one cupcake of each of types l, l + 1, ..., r.
After that they will compare the total tastiness of the cupcakes each of them have bought. Yasser will be happy if the total tastiness of cupcakes he buys is strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice.
For example, let the tastinesses of the cupcakes be [7, 4, -1]. Yasser will buy all of them, the total tastiness will be 7 + 4 - 1 = 10. Adel can choose segments [7], [4], [-1], [7, 4] or [4, -1], their total tastinesses are 7, 4, -1, 11 and 3, respectively. Adel can choose segment with tastiness 11, and as 10 is not strictly greater than 11, Yasser won't be happy :(
Find out if Yasser will be happy after visiting the shop.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). The description of the test cases follows.
The first line of each test case contains n (2 ≤ n ≤ 10^5).
The second line of each test case contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9), where a_i represents the tastiness of the i-th type of cupcake.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print "YES", if the total tastiness of cupcakes Yasser buys will always be strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice. Otherwise, print "NO".
Example
Input
3
4
1 2 3 4
3
7 4 -1
3
5 -5 5
Output
YES
NO
NO
Note
In the first example, the total tastiness of any segment Adel can choose is less than the total tastiness of all cupcakes.
In the second example, Adel will choose the segment [1, 2] with total tastiness 11, which is not less than the total tastiness of all cupcakes, which is 10.
In the third example, Adel can choose the segment [3, 3] with total tastiness of 5. Note that Yasser's cupcakes' total tastiness is also 5, so in that case, the total tastiness of Yasser's cupcakes isn't strictly greater than the total tastiness of Adel's cupcakes. | instruction | 0 | 16,071 | 9 | 32,142 |
Tags: dp, greedy, implementation
Correct Solution:
```
tests=int(input())
n=[]
numbers=[]
for i in range(0, tests):
n.append(int(input()))
numbers.append(list(map(int, input().split())))
for tastes in numbers:
flag = 0
sum=0
for i in tastes:
sum+=i
partial_sum=0
for i in tastes:
if i<=0:
if partial_sum>=sum or -i>=partial_sum:
print("NO")
flag = 1
break
partial_sum+=i
if flag == 0:
print("YES")
``` | output | 1 | 16,071 | 9 | 32,143 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Today, Yasser and Adel are at the shop buying cupcakes. There are n cupcake types, arranged from 1 to n on the shelf, and there are infinitely many of each type. The tastiness of a cupcake of type i is an integer a_i. There are both tasty and nasty cupcakes, so the tastiness can be positive, zero or negative.
Yasser, of course, wants to try them all, so he will buy exactly one cupcake of each type.
On the other hand, Adel will choose some segment [l, r] (1 ≤ l ≤ r ≤ n) that does not include all of cupcakes (he can't choose [l, r] = [1, n]) and buy exactly one cupcake of each of types l, l + 1, ..., r.
After that they will compare the total tastiness of the cupcakes each of them have bought. Yasser will be happy if the total tastiness of cupcakes he buys is strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice.
For example, let the tastinesses of the cupcakes be [7, 4, -1]. Yasser will buy all of them, the total tastiness will be 7 + 4 - 1 = 10. Adel can choose segments [7], [4], [-1], [7, 4] or [4, -1], their total tastinesses are 7, 4, -1, 11 and 3, respectively. Adel can choose segment with tastiness 11, and as 10 is not strictly greater than 11, Yasser won't be happy :(
Find out if Yasser will be happy after visiting the shop.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). The description of the test cases follows.
The first line of each test case contains n (2 ≤ n ≤ 10^5).
The second line of each test case contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9), where a_i represents the tastiness of the i-th type of cupcake.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print "YES", if the total tastiness of cupcakes Yasser buys will always be strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice. Otherwise, print "NO".
Example
Input
3
4
1 2 3 4
3
7 4 -1
3
5 -5 5
Output
YES
NO
NO
Note
In the first example, the total tastiness of any segment Adel can choose is less than the total tastiness of all cupcakes.
In the second example, Adel will choose the segment [1, 2] with total tastiness 11, which is not less than the total tastiness of all cupcakes, which is 10.
In the third example, Adel can choose the segment [3, 3] with total tastiness of 5. Note that Yasser's cupcakes' total tastiness is also 5, so in that case, the total tastiness of Yasser's cupcakes isn't strictly greater than the total tastiness of Adel's cupcakes. | instruction | 0 | 16,072 | 9 | 32,144 |
Tags: dp, greedy, implementation
Correct Solution:
```
def maxSubArraySum(a,size,initial):
max_so_far =a[initial]
curr_max = a[initial]
for i in range(initial+1,size):
curr_max = max(a[i], curr_max + a[i])
max_so_far = max(max_so_far,curr_max)
return max_so_far
t=int(input())
while(t>0):
n=int(input())
arr = list(map(int, input().split()))
yasser=0
for i in range(0,n):
yasser+=arr[i]
sum1=maxSubArraySum(arr,n-1,0)
sum2=maxSubArraySum(arr,n,1)
if(sum1>sum2):
adel=sum1
else:
adel=sum2
if(yasser>adel):
print("YES")
else:
print("NO")
t=t-1
``` | output | 1 | 16,072 | 9 | 32,145 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Today, Yasser and Adel are at the shop buying cupcakes. There are n cupcake types, arranged from 1 to n on the shelf, and there are infinitely many of each type. The tastiness of a cupcake of type i is an integer a_i. There are both tasty and nasty cupcakes, so the tastiness can be positive, zero or negative.
Yasser, of course, wants to try them all, so he will buy exactly one cupcake of each type.
On the other hand, Adel will choose some segment [l, r] (1 ≤ l ≤ r ≤ n) that does not include all of cupcakes (he can't choose [l, r] = [1, n]) and buy exactly one cupcake of each of types l, l + 1, ..., r.
After that they will compare the total tastiness of the cupcakes each of them have bought. Yasser will be happy if the total tastiness of cupcakes he buys is strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice.
For example, let the tastinesses of the cupcakes be [7, 4, -1]. Yasser will buy all of them, the total tastiness will be 7 + 4 - 1 = 10. Adel can choose segments [7], [4], [-1], [7, 4] or [4, -1], their total tastinesses are 7, 4, -1, 11 and 3, respectively. Adel can choose segment with tastiness 11, and as 10 is not strictly greater than 11, Yasser won't be happy :(
Find out if Yasser will be happy after visiting the shop.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). The description of the test cases follows.
The first line of each test case contains n (2 ≤ n ≤ 10^5).
The second line of each test case contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9), where a_i represents the tastiness of the i-th type of cupcake.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print "YES", if the total tastiness of cupcakes Yasser buys will always be strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice. Otherwise, print "NO".
Example
Input
3
4
1 2 3 4
3
7 4 -1
3
5 -5 5
Output
YES
NO
NO
Note
In the first example, the total tastiness of any segment Adel can choose is less than the total tastiness of all cupcakes.
In the second example, Adel will choose the segment [1, 2] with total tastiness 11, which is not less than the total tastiness of all cupcakes, which is 10.
In the third example, Adel can choose the segment [3, 3] with total tastiness of 5. Note that Yasser's cupcakes' total tastiness is also 5, so in that case, the total tastiness of Yasser's cupcakes isn't strictly greater than the total tastiness of Adel's cupcakes. | instruction | 0 | 16,073 | 9 | 32,146 |
Tags: dp, greedy, implementation
Correct Solution:
```
def maxSubArraySum(a,size):
max_so_far = -1000000009
max_ending_here = 0
for i in range(0, size-1):
max_ending_here = max_ending_here + a[i]
if (max_so_far < max_ending_here):
max_so_far = max_ending_here
if max_ending_here < 0:
max_ending_here = 0
return max_so_far
def maxSubArraySum1(a,size):
max_so_far = -1000000009
max_ending_here = 0
for i in range(1, size):
max_ending_here = max_ending_here + a[i]
if (max_so_far < max_ending_here):
max_so_far = max_ending_here
if max_ending_here < 0:
max_ending_here = 0
return max_so_far
t=int(input())
for _ in range(t):
n=int(input())
arr=list(map(int, input().strip().split()))
ya=sum(arr)
ad1=maxSubArraySum(arr,len(arr))
ad2=maxSubArraySum1(arr,len(arr))
if(ad1>ad2):
ad=ad1
else:
ad=ad2
if(ya>ad):
print("YES")
else:
print("NO")
``` | output | 1 | 16,073 | 9 | 32,147 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today, Yasser and Adel are at the shop buying cupcakes. There are n cupcake types, arranged from 1 to n on the shelf, and there are infinitely many of each type. The tastiness of a cupcake of type i is an integer a_i. There are both tasty and nasty cupcakes, so the tastiness can be positive, zero or negative.
Yasser, of course, wants to try them all, so he will buy exactly one cupcake of each type.
On the other hand, Adel will choose some segment [l, r] (1 ≤ l ≤ r ≤ n) that does not include all of cupcakes (he can't choose [l, r] = [1, n]) and buy exactly one cupcake of each of types l, l + 1, ..., r.
After that they will compare the total tastiness of the cupcakes each of them have bought. Yasser will be happy if the total tastiness of cupcakes he buys is strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice.
For example, let the tastinesses of the cupcakes be [7, 4, -1]. Yasser will buy all of them, the total tastiness will be 7 + 4 - 1 = 10. Adel can choose segments [7], [4], [-1], [7, 4] or [4, -1], their total tastinesses are 7, 4, -1, 11 and 3, respectively. Adel can choose segment with tastiness 11, and as 10 is not strictly greater than 11, Yasser won't be happy :(
Find out if Yasser will be happy after visiting the shop.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). The description of the test cases follows.
The first line of each test case contains n (2 ≤ n ≤ 10^5).
The second line of each test case contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9), where a_i represents the tastiness of the i-th type of cupcake.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print "YES", if the total tastiness of cupcakes Yasser buys will always be strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice. Otherwise, print "NO".
Example
Input
3
4
1 2 3 4
3
7 4 -1
3
5 -5 5
Output
YES
NO
NO
Note
In the first example, the total tastiness of any segment Adel can choose is less than the total tastiness of all cupcakes.
In the second example, Adel will choose the segment [1, 2] with total tastiness 11, which is not less than the total tastiness of all cupcakes, which is 10.
In the third example, Adel can choose the segment [3, 3] with total tastiness of 5. Note that Yasser's cupcakes' total tastiness is also 5, so in that case, the total tastiness of Yasser's cupcakes isn't strictly greater than the total tastiness of Adel's cupcakes.
Submitted Solution:
```
def kadane(a,l,r):
sum=0
ans=0
for x in range(l,r):
sum+=a[x]
ans=max(ans,sum)
if sum<0:
sum=0
return ans
def solve() :
m=int(input())
sum=0
a=[int(a) for a in input().split()]
for x in range(0,len(a)):
sum+=a[x]
k1=kadane(a,0,len(a)-1)
k2=kadane(a,1,len(a))
if max(k1,k2)>=sum:
print("NO")
else:
print("YES")
n=int(input())
while n!=0 :
solve()
n=n-1
``` | instruction | 0 | 16,074 | 9 | 32,148 |
Yes | output | 1 | 16,074 | 9 | 32,149 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today, Yasser and Adel are at the shop buying cupcakes. There are n cupcake types, arranged from 1 to n on the shelf, and there are infinitely many of each type. The tastiness of a cupcake of type i is an integer a_i. There are both tasty and nasty cupcakes, so the tastiness can be positive, zero or negative.
Yasser, of course, wants to try them all, so he will buy exactly one cupcake of each type.
On the other hand, Adel will choose some segment [l, r] (1 ≤ l ≤ r ≤ n) that does not include all of cupcakes (he can't choose [l, r] = [1, n]) and buy exactly one cupcake of each of types l, l + 1, ..., r.
After that they will compare the total tastiness of the cupcakes each of them have bought. Yasser will be happy if the total tastiness of cupcakes he buys is strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice.
For example, let the tastinesses of the cupcakes be [7, 4, -1]. Yasser will buy all of them, the total tastiness will be 7 + 4 - 1 = 10. Adel can choose segments [7], [4], [-1], [7, 4] or [4, -1], their total tastinesses are 7, 4, -1, 11 and 3, respectively. Adel can choose segment with tastiness 11, and as 10 is not strictly greater than 11, Yasser won't be happy :(
Find out if Yasser will be happy after visiting the shop.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). The description of the test cases follows.
The first line of each test case contains n (2 ≤ n ≤ 10^5).
The second line of each test case contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9), where a_i represents the tastiness of the i-th type of cupcake.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print "YES", if the total tastiness of cupcakes Yasser buys will always be strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice. Otherwise, print "NO".
Example
Input
3
4
1 2 3 4
3
7 4 -1
3
5 -5 5
Output
YES
NO
NO
Note
In the first example, the total tastiness of any segment Adel can choose is less than the total tastiness of all cupcakes.
In the second example, Adel will choose the segment [1, 2] with total tastiness 11, which is not less than the total tastiness of all cupcakes, which is 10.
In the third example, Adel can choose the segment [3, 3] with total tastiness of 5. Note that Yasser's cupcakes' total tastiness is also 5, so in that case, the total tastiness of Yasser's cupcakes isn't strictly greater than the total tastiness of Adel's cupcakes.
Submitted Solution:
```
import sys
input=sys.stdin.readline
t=int(input())
while t>0:
t-=1
n=int(input())
a=[int(x) for x in input().split()]
s=0
j=0
maxi=-9999999999999999999999999
p=0;q=0
z=0
for i in range(n):
s+=a[i]
maxi=max(maxi,s)
if s<0:
s=0
p=i
#print(i,p)
if i-p==n-1:
mini1=mini2=9999999999999999999999999
s=0
for i in range(n):
s+=a[i]
mini1=min(mini1,s)
s=0
for i in range(n-1,-1,-1):
s+=a[i]
mini2=min(mini2,s)
maxi-=min(mini1,mini2)
#print(maxi,a,sum(a))
print("YES" if sum(a)>maxi else "NO")
``` | instruction | 0 | 16,075 | 9 | 32,150 |
Yes | output | 1 | 16,075 | 9 | 32,151 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today, Yasser and Adel are at the shop buying cupcakes. There are n cupcake types, arranged from 1 to n on the shelf, and there are infinitely many of each type. The tastiness of a cupcake of type i is an integer a_i. There are both tasty and nasty cupcakes, so the tastiness can be positive, zero or negative.
Yasser, of course, wants to try them all, so he will buy exactly one cupcake of each type.
On the other hand, Adel will choose some segment [l, r] (1 ≤ l ≤ r ≤ n) that does not include all of cupcakes (he can't choose [l, r] = [1, n]) and buy exactly one cupcake of each of types l, l + 1, ..., r.
After that they will compare the total tastiness of the cupcakes each of them have bought. Yasser will be happy if the total tastiness of cupcakes he buys is strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice.
For example, let the tastinesses of the cupcakes be [7, 4, -1]. Yasser will buy all of them, the total tastiness will be 7 + 4 - 1 = 10. Adel can choose segments [7], [4], [-1], [7, 4] or [4, -1], their total tastinesses are 7, 4, -1, 11 and 3, respectively. Adel can choose segment with tastiness 11, and as 10 is not strictly greater than 11, Yasser won't be happy :(
Find out if Yasser will be happy after visiting the shop.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). The description of the test cases follows.
The first line of each test case contains n (2 ≤ n ≤ 10^5).
The second line of each test case contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9), where a_i represents the tastiness of the i-th type of cupcake.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print "YES", if the total tastiness of cupcakes Yasser buys will always be strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice. Otherwise, print "NO".
Example
Input
3
4
1 2 3 4
3
7 4 -1
3
5 -5 5
Output
YES
NO
NO
Note
In the first example, the total tastiness of any segment Adel can choose is less than the total tastiness of all cupcakes.
In the second example, Adel will choose the segment [1, 2] with total tastiness 11, which is not less than the total tastiness of all cupcakes, which is 10.
In the third example, Adel can choose the segment [3, 3] with total tastiness of 5. Note that Yasser's cupcakes' total tastiness is also 5, so in that case, the total tastiness of Yasser's cupcakes isn't strictly greater than the total tastiness of Adel's cupcakes.
Submitted Solution:
```
n = int(input())
for i in range(n):
t = int(input())
pies = list(map(int, input().split(" ")))
flag = True
tmp_sum = 0
for j in range(len(pies)):
tmp_sum += pies[j]
if tmp_sum <= 0:
print("NO")
flag = False
break
if flag != False:
tmp_sum = 0
for j in range(len(pies) - 1, 0, -1):
tmp_sum += pies[j]
if tmp_sum <= 0:
print("NO")
flag = False
break
if flag != False:
print("YES")
``` | instruction | 0 | 16,076 | 9 | 32,152 |
Yes | output | 1 | 16,076 | 9 | 32,153 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today, Yasser and Adel are at the shop buying cupcakes. There are n cupcake types, arranged from 1 to n on the shelf, and there are infinitely many of each type. The tastiness of a cupcake of type i is an integer a_i. There are both tasty and nasty cupcakes, so the tastiness can be positive, zero or negative.
Yasser, of course, wants to try them all, so he will buy exactly one cupcake of each type.
On the other hand, Adel will choose some segment [l, r] (1 ≤ l ≤ r ≤ n) that does not include all of cupcakes (he can't choose [l, r] = [1, n]) and buy exactly one cupcake of each of types l, l + 1, ..., r.
After that they will compare the total tastiness of the cupcakes each of them have bought. Yasser will be happy if the total tastiness of cupcakes he buys is strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice.
For example, let the tastinesses of the cupcakes be [7, 4, -1]. Yasser will buy all of them, the total tastiness will be 7 + 4 - 1 = 10. Adel can choose segments [7], [4], [-1], [7, 4] or [4, -1], their total tastinesses are 7, 4, -1, 11 and 3, respectively. Adel can choose segment with tastiness 11, and as 10 is not strictly greater than 11, Yasser won't be happy :(
Find out if Yasser will be happy after visiting the shop.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). The description of the test cases follows.
The first line of each test case contains n (2 ≤ n ≤ 10^5).
The second line of each test case contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9), where a_i represents the tastiness of the i-th type of cupcake.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print "YES", if the total tastiness of cupcakes Yasser buys will always be strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice. Otherwise, print "NO".
Example
Input
3
4
1 2 3 4
3
7 4 -1
3
5 -5 5
Output
YES
NO
NO
Note
In the first example, the total tastiness of any segment Adel can choose is less than the total tastiness of all cupcakes.
In the second example, Adel will choose the segment [1, 2] with total tastiness 11, which is not less than the total tastiness of all cupcakes, which is 10.
In the third example, Adel can choose the segment [3, 3] with total tastiness of 5. Note that Yasser's cupcakes' total tastiness is also 5, so in that case, the total tastiness of Yasser's cupcakes isn't strictly greater than the total tastiness of Adel's cupcakes.
Submitted Solution:
```
t=(int)(input())
while t>0:
n=(int)(input())
a=list(map(int,input().split()))
yas=sum(a)
b=a[1:]
c=a[0:len(a)-1]
cursum=0
ans1=min(b)
ans2=min(c)
for i in range(len(b)):
cursum=cursum+b[i]
ans1=max(ans1,cursum)
if cursum<0:
cursum=0
cursum=0
for i in range(len(c)):
cursum=cursum+c[i]
ans2=max(ans2,cursum)
if cursum<0:
cursum=0
adel=max(ans1,ans2)
adel=max(adel,a[0])
adel=max(adel,a[-1])
#print(ans1,ans2)
if adel>=yas:
print("NO")
else:
print("YES")
t=t-1
``` | instruction | 0 | 16,077 | 9 | 32,154 |
Yes | output | 1 | 16,077 | 9 | 32,155 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today, Yasser and Adel are at the shop buying cupcakes. There are n cupcake types, arranged from 1 to n on the shelf, and there are infinitely many of each type. The tastiness of a cupcake of type i is an integer a_i. There are both tasty and nasty cupcakes, so the tastiness can be positive, zero or negative.
Yasser, of course, wants to try them all, so he will buy exactly one cupcake of each type.
On the other hand, Adel will choose some segment [l, r] (1 ≤ l ≤ r ≤ n) that does not include all of cupcakes (he can't choose [l, r] = [1, n]) and buy exactly one cupcake of each of types l, l + 1, ..., r.
After that they will compare the total tastiness of the cupcakes each of them have bought. Yasser will be happy if the total tastiness of cupcakes he buys is strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice.
For example, let the tastinesses of the cupcakes be [7, 4, -1]. Yasser will buy all of them, the total tastiness will be 7 + 4 - 1 = 10. Adel can choose segments [7], [4], [-1], [7, 4] or [4, -1], their total tastinesses are 7, 4, -1, 11 and 3, respectively. Adel can choose segment with tastiness 11, and as 10 is not strictly greater than 11, Yasser won't be happy :(
Find out if Yasser will be happy after visiting the shop.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). The description of the test cases follows.
The first line of each test case contains n (2 ≤ n ≤ 10^5).
The second line of each test case contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9), where a_i represents the tastiness of the i-th type of cupcake.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print "YES", if the total tastiness of cupcakes Yasser buys will always be strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice. Otherwise, print "NO".
Example
Input
3
4
1 2 3 4
3
7 4 -1
3
5 -5 5
Output
YES
NO
NO
Note
In the first example, the total tastiness of any segment Adel can choose is less than the total tastiness of all cupcakes.
In the second example, Adel will choose the segment [1, 2] with total tastiness 11, which is not less than the total tastiness of all cupcakes, which is 10.
In the third example, Adel can choose the segment [3, 3] with total tastiness of 5. Note that Yasser's cupcakes' total tastiness is also 5, so in that case, the total tastiness of Yasser's cupcakes isn't strictly greater than the total tastiness of Adel's cupcakes.
Submitted Solution:
```
n = int(input())
sum = 0
flag = 1
for i in range(0, n):
len = int(input())
data = [int(i) for i in filter(None, input().split(" "))]
for j in range(0, len):
sum += data[j]
if sum <= 0:
print("NO")
flag = 0
sum = 0
if flag == 1:
for j in range(len-1, 0, -1):
sum += data[j]
if sum <= 0:
print("NO")
flag = 0
if flag == 1:
print("YES")
``` | instruction | 0 | 16,078 | 9 | 32,156 |
No | output | 1 | 16,078 | 9 | 32,157 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today, Yasser and Adel are at the shop buying cupcakes. There are n cupcake types, arranged from 1 to n on the shelf, and there are infinitely many of each type. The tastiness of a cupcake of type i is an integer a_i. There are both tasty and nasty cupcakes, so the tastiness can be positive, zero or negative.
Yasser, of course, wants to try them all, so he will buy exactly one cupcake of each type.
On the other hand, Adel will choose some segment [l, r] (1 ≤ l ≤ r ≤ n) that does not include all of cupcakes (he can't choose [l, r] = [1, n]) and buy exactly one cupcake of each of types l, l + 1, ..., r.
After that they will compare the total tastiness of the cupcakes each of them have bought. Yasser will be happy if the total tastiness of cupcakes he buys is strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice.
For example, let the tastinesses of the cupcakes be [7, 4, -1]. Yasser will buy all of them, the total tastiness will be 7 + 4 - 1 = 10. Adel can choose segments [7], [4], [-1], [7, 4] or [4, -1], their total tastinesses are 7, 4, -1, 11 and 3, respectively. Adel can choose segment with tastiness 11, and as 10 is not strictly greater than 11, Yasser won't be happy :(
Find out if Yasser will be happy after visiting the shop.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). The description of the test cases follows.
The first line of each test case contains n (2 ≤ n ≤ 10^5).
The second line of each test case contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9), where a_i represents the tastiness of the i-th type of cupcake.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print "YES", if the total tastiness of cupcakes Yasser buys will always be strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice. Otherwise, print "NO".
Example
Input
3
4
1 2 3 4
3
7 4 -1
3
5 -5 5
Output
YES
NO
NO
Note
In the first example, the total tastiness of any segment Adel can choose is less than the total tastiness of all cupcakes.
In the second example, Adel will choose the segment [1, 2] with total tastiness 11, which is not less than the total tastiness of all cupcakes, which is 10.
In the third example, Adel can choose the segment [3, 3] with total tastiness of 5. Note that Yasser's cupcakes' total tastiness is also 5, so in that case, the total tastiness of Yasser's cupcakes isn't strictly greater than the total tastiness of Adel's cupcakes.
Submitted Solution:
```
from math import *
from copy import *
from string import * # alpha = ascii_lowercase
from random import *
from sys import stdin
from sys import maxsize
from operator import * # d = sorted(d.items(), key=itemgetter(1))
from itertools import *
from collections import Counter # d = dict(Counter(l))
import math
def bin1(l,r,k,t,b,val,ans):
if(l>r):
return ans
else:
mid=(l+r)//2
v=k**mid
if(v==val):
return v
elif(v>val):
ans=mid
return bin1(mid+1,r,k,t,b,val,ans)
else:
return bin1(l,mid-1,k,t,b,val,ans)
def bin2(l,r,k,t,b,val,ans):
if(l>r):
return ans
else:
mid=(l+r)//2
v=t*(k**mid)+b*(mid)
if(v==val):
return v
elif(v>val):
ans=mid
return bin2(l,mid-1,k,t,b,val,ans)
else:
return bin2(mid+1,r,k,t,b,val,ans)
def SieveOfEratosthenes(n):
# Create a boolean array "prime[0..n]" and initialize
# all entries it as true. A value in prime[i] will
# finally be false if i is Not a prime, else true.
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
# If prime[p] is not changed, then it is a prime
if (prime[p] == True):
# Update all multiples of p
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
l=[]
for i in range(2,n+1):
if(prime[i]):
l.append(i)
return l
def bin(l,r,ll,val):
if(l>r):
return -1
else:
mid=(l+r)//2
if(val>=ll[mid][0] and val<=ll[mid][1]):
return mid
elif(val<ll[mid][0]):
return bin(l,mid-1,ll,val)
else:
return bin(mid+1,r,ll,val)
def deci(n):
s=""
while(n!=0):
if(n%2==0):
n=n//2
s="0"+s
else:
n=n//2
s="1"+s
return s
def diff(s1,s2):
if(len(s1)<len(s2)):
v=len(s1)
while(v!=len(s2)):
s1="0"+s1
v=v+1
else:
v=len(s2)
while(v!=len(s1)):
s2="0"+s2
v=v+1
c=0
for i in range(len(s1)):
if(s1[i:i+1]!=s2[i:i+1]):
c=c+1
return c
from sys import stdin, stdout
def fac(a,b):
v=a
while(a!=b):
v*=a-1
a=a-1
return v
def bino(l,r,n):
if(l>r):
return -1
else:
mid=(l+r)//2
val1=math.log((n/mid)+1,2)
val2=int(val1)
if(val1==val2):
return val1
elif(val1<1.0):
return bino(l,mid-1,n)
else:
return bino(mid+1,r,n)
def binary(l,r,ll,val,ans):
if(l>r):
return ans
else:
mid=(l+r)//2
if(ll[mid]<=val):
ans=mid+1
return binary(mid+1,r,ll,val,ans)
else:
return binary(l,mid-1,ll,val,ans)
def odd(n,ans,s):
if(n%2!=0 or n in s):
return ans
else:
s.add(n)
return odd(n//2,ans+1,s)
if __name__ == '__main__':
t=int(stdin.readline())
for i in range(t):
n=int(stdin.readline())
l=list(map(int,stdin.readline().split(" ")))
max1=0
max2=0
i1=1
i2=1
for i in range(n):
max2+=l[i]
if(max2<0):
max2=0
i1=i+1
i2=i+1
if(max2>max1):
max1=max2
i2=i+1
max3=sum(l)
if(max3>max1):
print("YES")
else:
if(max3==max1 and i1==1 and i2==n):
print("YES")
else:
print("NO")
``` | instruction | 0 | 16,079 | 9 | 32,158 |
No | output | 1 | 16,079 | 9 | 32,159 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today, Yasser and Adel are at the shop buying cupcakes. There are n cupcake types, arranged from 1 to n on the shelf, and there are infinitely many of each type. The tastiness of a cupcake of type i is an integer a_i. There are both tasty and nasty cupcakes, so the tastiness can be positive, zero or negative.
Yasser, of course, wants to try them all, so he will buy exactly one cupcake of each type.
On the other hand, Adel will choose some segment [l, r] (1 ≤ l ≤ r ≤ n) that does not include all of cupcakes (he can't choose [l, r] = [1, n]) and buy exactly one cupcake of each of types l, l + 1, ..., r.
After that they will compare the total tastiness of the cupcakes each of them have bought. Yasser will be happy if the total tastiness of cupcakes he buys is strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice.
For example, let the tastinesses of the cupcakes be [7, 4, -1]. Yasser will buy all of them, the total tastiness will be 7 + 4 - 1 = 10. Adel can choose segments [7], [4], [-1], [7, 4] or [4, -1], their total tastinesses are 7, 4, -1, 11 and 3, respectively. Adel can choose segment with tastiness 11, and as 10 is not strictly greater than 11, Yasser won't be happy :(
Find out if Yasser will be happy after visiting the shop.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). The description of the test cases follows.
The first line of each test case contains n (2 ≤ n ≤ 10^5).
The second line of each test case contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9), where a_i represents the tastiness of the i-th type of cupcake.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print "YES", if the total tastiness of cupcakes Yasser buys will always be strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice. Otherwise, print "NO".
Example
Input
3
4
1 2 3 4
3
7 4 -1
3
5 -5 5
Output
YES
NO
NO
Note
In the first example, the total tastiness of any segment Adel can choose is less than the total tastiness of all cupcakes.
In the second example, Adel will choose the segment [1, 2] with total tastiness 11, which is not less than the total tastiness of all cupcakes, which is 10.
In the third example, Adel can choose the segment [3, 3] with total tastiness of 5. Note that Yasser's cupcakes' total tastiness is also 5, so in that case, the total tastiness of Yasser's cupcakes isn't strictly greater than the total tastiness of Adel's cupcakes.
Submitted Solution:
```
for case in range(int(input())):
n = int(input())
arr = list(map(int,input().split()))
x = sum(arr)
m = 0
dp = [0]*n
dp[0] = arr[0]
if(dp[0]>=x):
m=1
for i in range(1,n-1):
dp[i] = max(dp[i-1]+arr[i],arr[i])
if(dp[i]>=x):
m=1
break
dp = [0]*n
dp[1] = arr[1]
for i in range(2,n):
dp[i] = max(dp[i-1]+arr[i],arr[i])
if(dp[i]>=x):
m=1
break
if(m==1):
print("NO")
else:
print("YES")
#print(dp)
``` | instruction | 0 | 16,080 | 9 | 32,160 |
No | output | 1 | 16,080 | 9 | 32,161 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today, Yasser and Adel are at the shop buying cupcakes. There are n cupcake types, arranged from 1 to n on the shelf, and there are infinitely many of each type. The tastiness of a cupcake of type i is an integer a_i. There are both tasty and nasty cupcakes, so the tastiness can be positive, zero or negative.
Yasser, of course, wants to try them all, so he will buy exactly one cupcake of each type.
On the other hand, Adel will choose some segment [l, r] (1 ≤ l ≤ r ≤ n) that does not include all of cupcakes (he can't choose [l, r] = [1, n]) and buy exactly one cupcake of each of types l, l + 1, ..., r.
After that they will compare the total tastiness of the cupcakes each of them have bought. Yasser will be happy if the total tastiness of cupcakes he buys is strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice.
For example, let the tastinesses of the cupcakes be [7, 4, -1]. Yasser will buy all of them, the total tastiness will be 7 + 4 - 1 = 10. Adel can choose segments [7], [4], [-1], [7, 4] or [4, -1], their total tastinesses are 7, 4, -1, 11 and 3, respectively. Adel can choose segment with tastiness 11, and as 10 is not strictly greater than 11, Yasser won't be happy :(
Find out if Yasser will be happy after visiting the shop.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). The description of the test cases follows.
The first line of each test case contains n (2 ≤ n ≤ 10^5).
The second line of each test case contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9), where a_i represents the tastiness of the i-th type of cupcake.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print "YES", if the total tastiness of cupcakes Yasser buys will always be strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice. Otherwise, print "NO".
Example
Input
3
4
1 2 3 4
3
7 4 -1
3
5 -5 5
Output
YES
NO
NO
Note
In the first example, the total tastiness of any segment Adel can choose is less than the total tastiness of all cupcakes.
In the second example, Adel will choose the segment [1, 2] with total tastiness 11, which is not less than the total tastiness of all cupcakes, which is 10.
In the third example, Adel can choose the segment [3, 3] with total tastiness of 5. Note that Yasser's cupcakes' total tastiness is also 5, so in that case, the total tastiness of Yasser's cupcakes isn't strictly greater than the total tastiness of Adel's cupcakes.
Submitted Solution:
```
for k in range(int(input())):
n = int(input())
t = list(map(int,input().split()))
v=[]
yasir = 0
u=0
for j in t:
if j not in v:
if j<0:
u+=1
v.append(j)
yasir+=(j)
if u==0:
print('YES')
else:
ans = -999999999
u=0
f=[]
for j in range(n):
if t[j]>=0:
if t[j] not in f:
u +=t[j]
f.append(t[j])
else:
ans = max(u, ans )
u=0
f=[]
ans = max(u,ans)
if ans >= yasir:
print('NO')
else:
print('YES')
``` | instruction | 0 | 16,081 | 9 | 32,162 |
No | output | 1 | 16,081 | 9 | 32,163 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Phoenix is picking berries in his backyard. There are n shrubs, and each shrub has a_i red berries and b_i blue berries.
Each basket can contain k berries. But, Phoenix has decided that each basket may only contain berries from the same shrub or berries of the same color (red or blue). In other words, all berries in a basket must be from the same shrub or/and have the same color.
For example, if there are two shrubs with 5 red and 2 blue berries in the first shrub and 2 red and 1 blue berries in the second shrub then Phoenix can fill 2 baskets of capacity 4 completely:
* the first basket will contain 3 red and 1 blue berries from the first shrub;
* the second basket will contain the 2 remaining red berries from the first shrub and 2 red berries from the second shrub.
Help Phoenix determine the maximum number of baskets he can fill completely!
Input
The first line contains two integers n and k ( 1≤ n, k ≤ 500) — the number of shrubs and the basket capacity, respectively.
The i-th of the next n lines contain two integers a_i and b_i (0 ≤ a_i, b_i ≤ 10^9) — the number of red and blue berries in the i-th shrub, respectively.
Output
Output one integer — the maximum number of baskets that Phoenix can fill completely.
Examples
Input
2 4
5 2
2 1
Output
2
Input
1 5
2 3
Output
1
Input
2 5
2 1
1 3
Output
0
Input
1 2
1000000000 1
Output
500000000
Note
The first example is described above.
In the second example, Phoenix can fill one basket fully using all the berries from the first (and only) shrub.
In the third example, Phoenix cannot fill any basket completely because there are less than 5 berries in each shrub, less than 5 total red berries, and less than 5 total blue berries.
In the fourth example, Phoenix can put all the red berries into baskets, leaving an extra blue berry behind. | instruction | 0 | 16,114 | 9 | 32,228 |
Tags: brute force, dp, greedy, math
Correct Solution:
```
import sys
from array import array # noqa: F401
import typing as Tp # noqa: F401
def input():
return sys.stdin.buffer.readline().decode('utf-8')
def output(*args):
sys.stdout.buffer.write(
('\n'.join(map(str, args)) + '\n').encode('utf-8')
)
def main():
n, k = map(int, input().split())
trees = [tuple(map(int, input().split())) for _ in range(n)]
r_mod, b_mod = 0, 0
dp = array('b', [1] + [0] * (k - 1))
next_dp = array('b', [0]) * k
r_total, b_total = 0, 0
for ri, bi in trees:
r_total += ri
b_total += bi
for r in range(k):
if not dp[r]:
continue
next_dp[(r + ri) % k] = 1
for sub_r, sub_b in zip(range(1, k), range(k - 1, 0, -1)):
if sub_r <= ri and sub_b <= bi:
next_dp[(r + ri - sub_r) % k] = 1
r_mod, b_mod = (r_mod + ri) % k, (b_mod + bi) % k
dp, next_dp = next_dp, dp
for i in range(k):
next_dp[i] = 0
for r, req_b in zip(range(r_mod - 1, -1, -1), range(k - 1, -1, -1)):
if dp[r] and b_mod >= req_b:
print(r_total // k + b_total // k + 1)
break
else:
print(r_total // k + b_total // k)
if __name__ == '__main__':
main()
``` | output | 1 | 16,114 | 9 | 32,229 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Phoenix is picking berries in his backyard. There are n shrubs, and each shrub has a_i red berries and b_i blue berries.
Each basket can contain k berries. But, Phoenix has decided that each basket may only contain berries from the same shrub or berries of the same color (red or blue). In other words, all berries in a basket must be from the same shrub or/and have the same color.
For example, if there are two shrubs with 5 red and 2 blue berries in the first shrub and 2 red and 1 blue berries in the second shrub then Phoenix can fill 2 baskets of capacity 4 completely:
* the first basket will contain 3 red and 1 blue berries from the first shrub;
* the second basket will contain the 2 remaining red berries from the first shrub and 2 red berries from the second shrub.
Help Phoenix determine the maximum number of baskets he can fill completely!
Input
The first line contains two integers n and k ( 1≤ n, k ≤ 500) — the number of shrubs and the basket capacity, respectively.
The i-th of the next n lines contain two integers a_i and b_i (0 ≤ a_i, b_i ≤ 10^9) — the number of red and blue berries in the i-th shrub, respectively.
Output
Output one integer — the maximum number of baskets that Phoenix can fill completely.
Examples
Input
2 4
5 2
2 1
Output
2
Input
1 5
2 3
Output
1
Input
2 5
2 1
1 3
Output
0
Input
1 2
1000000000 1
Output
500000000
Note
The first example is described above.
In the second example, Phoenix can fill one basket fully using all the berries from the first (and only) shrub.
In the third example, Phoenix cannot fill any basket completely because there are less than 5 berries in each shrub, less than 5 total red berries, and less than 5 total blue berries.
In the fourth example, Phoenix can put all the red berries into baskets, leaving an extra blue berry behind. | instruction | 0 | 16,115 | 9 | 32,230 |
Tags: brute force, dp, greedy, math
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------------------
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
def comb(n, m): return factorial(n) / (factorial(m) * factorial(n - m)) if n >= m else 0
def perm(n, m): return factorial(n) // (factorial(n - m)) if n >= m else 0
def mdis(x1, y1, x2, y2): return abs(x1 - x2) + abs(y1 - y2)
mod = 998244353
INF = float('inf')
from math import factorial
from collections import Counter, defaultdict, deque
from heapq import heapify, heappop, heappush
# ------------------------------
# f = open('../input.txt')
# sys.stdin = f
def main():
n, k = RL()
arr = [RLL() for _ in range(n)]
sa = sb = 0
for a, b in arr:
sa+=a
sb+=b
res = sa//k + sb//k
da, db = sa%k, sb%k
if res==(sa+sb)//k:
print(res)
exit()
dp = [0]*k
dp[0] = 1
for i in range(n):
a, b = arr[i]
l, r = max(0, k-a), min(k, b+1)
for ind, v in enumerate(dp[::]):
if v:
for j in range(l, r):
dp[(ind+j)%k] = 1
for i in range(k):
if dp[i]:
res = max(res, (sa+i)//k + (sb-i)//k)
print(res)
if __name__ == "__main__":
main()
``` | output | 1 | 16,115 | 9 | 32,231 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Phoenix is picking berries in his backyard. There are n shrubs, and each shrub has a_i red berries and b_i blue berries.
Each basket can contain k berries. But, Phoenix has decided that each basket may only contain berries from the same shrub or berries of the same color (red or blue). In other words, all berries in a basket must be from the same shrub or/and have the same color.
For example, if there are two shrubs with 5 red and 2 blue berries in the first shrub and 2 red and 1 blue berries in the second shrub then Phoenix can fill 2 baskets of capacity 4 completely:
* the first basket will contain 3 red and 1 blue berries from the first shrub;
* the second basket will contain the 2 remaining red berries from the first shrub and 2 red berries from the second shrub.
Help Phoenix determine the maximum number of baskets he can fill completely!
Input
The first line contains two integers n and k ( 1≤ n, k ≤ 500) — the number of shrubs and the basket capacity, respectively.
The i-th of the next n lines contain two integers a_i and b_i (0 ≤ a_i, b_i ≤ 10^9) — the number of red and blue berries in the i-th shrub, respectively.
Output
Output one integer — the maximum number of baskets that Phoenix can fill completely.
Examples
Input
2 4
5 2
2 1
Output
2
Input
1 5
2 3
Output
1
Input
2 5
2 1
1 3
Output
0
Input
1 2
1000000000 1
Output
500000000
Note
The first example is described above.
In the second example, Phoenix can fill one basket fully using all the berries from the first (and only) shrub.
In the third example, Phoenix cannot fill any basket completely because there are less than 5 berries in each shrub, less than 5 total red berries, and less than 5 total blue berries.
In the fourth example, Phoenix can put all the red berries into baskets, leaving an extra blue berry behind. | instruction | 0 | 16,116 | 9 | 32,232 |
Tags: brute force, dp, greedy, math
Correct Solution:
```
import sys
readline = sys.stdin.readline
def merge(A, B):
res = A[:]
for i in range(len(A)):
if A[i]:
for j in range(len(B)):
if B[j]:
res[(i+j)%K] = 1
return res
N, K = map(int, readline().split())
R = 0
B = 0
flag = False
table = [0]*K
table[0] = 1
for _ in range(N):
r, b = map(int, readline().split())
R += r
B += b
if r >= K and b >= K:
flag = True
elif r+b >= K:
st, en = max(0, K-b), min(K, r)
t2 = [0]*K
for i in range(st, en+1):
t2[i%K] = 1
table = merge(table, t2)
if flag:
print((R+B)//K)
elif (R//K + B//K == (R+B)//K):
print((R+B)//K)
else:
pr = R%K
pb = B%K
ans = R//K + B//K
for i in range(K):
if table[i]:
if (pr-i)%K + (pb-K+i)%K < K:
ans += 1
break
print(ans)
``` | output | 1 | 16,116 | 9 | 32,233 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Phoenix is picking berries in his backyard. There are n shrubs, and each shrub has a_i red berries and b_i blue berries.
Each basket can contain k berries. But, Phoenix has decided that each basket may only contain berries from the same shrub or berries of the same color (red or blue). In other words, all berries in a basket must be from the same shrub or/and have the same color.
For example, if there are two shrubs with 5 red and 2 blue berries in the first shrub and 2 red and 1 blue berries in the second shrub then Phoenix can fill 2 baskets of capacity 4 completely:
* the first basket will contain 3 red and 1 blue berries from the first shrub;
* the second basket will contain the 2 remaining red berries from the first shrub and 2 red berries from the second shrub.
Help Phoenix determine the maximum number of baskets he can fill completely!
Input
The first line contains two integers n and k ( 1≤ n, k ≤ 500) — the number of shrubs and the basket capacity, respectively.
The i-th of the next n lines contain two integers a_i and b_i (0 ≤ a_i, b_i ≤ 10^9) — the number of red and blue berries in the i-th shrub, respectively.
Output
Output one integer — the maximum number of baskets that Phoenix can fill completely.
Examples
Input
2 4
5 2
2 1
Output
2
Input
1 5
2 3
Output
1
Input
2 5
2 1
1 3
Output
0
Input
1 2
1000000000 1
Output
500000000
Note
The first example is described above.
In the second example, Phoenix can fill one basket fully using all the berries from the first (and only) shrub.
In the third example, Phoenix cannot fill any basket completely because there are less than 5 berries in each shrub, less than 5 total red berries, and less than 5 total blue berries.
In the fourth example, Phoenix can put all the red berries into baskets, leaving an extra blue berry behind. | instruction | 0 | 16,117 | 9 | 32,234 |
Tags: brute force, dp, greedy, math
Correct Solution:
```
n,k=map(int,input().split())
a=[0]*n
b=[0]*n
for i in range(n):
a[i],b[i]=map(int,input().split())
base=(sum(a)//k)+(sum(b)//k)
A=sum(a)%k
B=sum(b)%k
data=[False for i in range(k)]
for i in range(n):
ndata=[data[j] for j in range(k)]
for j in range(1,k):
if j<=a[i] and k-j<=b[i]:
ndata[j]=True
for x in range(1,k):
prev=(x-j)%k
if data[prev]:
ndata[x]=True
data=ndata
check=False
for i in range(1,k):
if i<=A and k-i<=B:
if data[i]:
check=True
print(base+check)
``` | output | 1 | 16,117 | 9 | 32,235 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Phoenix is picking berries in his backyard. There are n shrubs, and each shrub has a_i red berries and b_i blue berries.
Each basket can contain k berries. But, Phoenix has decided that each basket may only contain berries from the same shrub or berries of the same color (red or blue). In other words, all berries in a basket must be from the same shrub or/and have the same color.
For example, if there are two shrubs with 5 red and 2 blue berries in the first shrub and 2 red and 1 blue berries in the second shrub then Phoenix can fill 2 baskets of capacity 4 completely:
* the first basket will contain 3 red and 1 blue berries from the first shrub;
* the second basket will contain the 2 remaining red berries from the first shrub and 2 red berries from the second shrub.
Help Phoenix determine the maximum number of baskets he can fill completely!
Input
The first line contains two integers n and k ( 1≤ n, k ≤ 500) — the number of shrubs and the basket capacity, respectively.
The i-th of the next n lines contain two integers a_i and b_i (0 ≤ a_i, b_i ≤ 10^9) — the number of red and blue berries in the i-th shrub, respectively.
Output
Output one integer — the maximum number of baskets that Phoenix can fill completely.
Examples
Input
2 4
5 2
2 1
Output
2
Input
1 5
2 3
Output
1
Input
2 5
2 1
1 3
Output
0
Input
1 2
1000000000 1
Output
500000000
Note
The first example is described above.
In the second example, Phoenix can fill one basket fully using all the berries from the first (and only) shrub.
In the third example, Phoenix cannot fill any basket completely because there are less than 5 berries in each shrub, less than 5 total red berries, and less than 5 total blue berries.
In the fourth example, Phoenix can put all the red berries into baskets, leaving an extra blue berry behind. | instruction | 0 | 16,118 | 9 | 32,236 |
Tags: brute force, dp, greedy, math
Correct Solution:
```
import itertools as it
n, k = map(int, input().split())
bushes = [tuple(map(int, input().split())) for i in range(n)]
red = sum(bush[0] for bush in bushes)
blue = sum(bush[1] for bush in bushes)
r0 = red % k
r1 = blue % k
max_ = (red + blue) // k
if (r0 + r1) < k:
print(max_)
exit()
r0_required = set(range(r0 + r1 - k + 1))
r0_available = {r0}
for red, blue in bushes:
if red + blue < k:
continue
max_red = min(k - 1, red)
min_red = max(0, k - blue)
for ex_r, diff in it.product(list(r0_available), range(min_red, max_red + 1)):
new_r = ex_r - diff
if new_r < 0:
new_r += k
if new_r in r0_required:
print(max_)
exit()
r0_available.add(new_r)
print(max_ - 1)
``` | output | 1 | 16,118 | 9 | 32,237 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Phoenix is picking berries in his backyard. There are n shrubs, and each shrub has a_i red berries and b_i blue berries.
Each basket can contain k berries. But, Phoenix has decided that each basket may only contain berries from the same shrub or berries of the same color (red or blue). In other words, all berries in a basket must be from the same shrub or/and have the same color.
For example, if there are two shrubs with 5 red and 2 blue berries in the first shrub and 2 red and 1 blue berries in the second shrub then Phoenix can fill 2 baskets of capacity 4 completely:
* the first basket will contain 3 red and 1 blue berries from the first shrub;
* the second basket will contain the 2 remaining red berries from the first shrub and 2 red berries from the second shrub.
Help Phoenix determine the maximum number of baskets he can fill completely!
Input
The first line contains two integers n and k ( 1≤ n, k ≤ 500) — the number of shrubs and the basket capacity, respectively.
The i-th of the next n lines contain two integers a_i and b_i (0 ≤ a_i, b_i ≤ 10^9) — the number of red and blue berries in the i-th shrub, respectively.
Output
Output one integer — the maximum number of baskets that Phoenix can fill completely.
Examples
Input
2 4
5 2
2 1
Output
2
Input
1 5
2 3
Output
1
Input
2 5
2 1
1 3
Output
0
Input
1 2
1000000000 1
Output
500000000
Note
The first example is described above.
In the second example, Phoenix can fill one basket fully using all the berries from the first (and only) shrub.
In the third example, Phoenix cannot fill any basket completely because there are less than 5 berries in each shrub, less than 5 total red berries, and less than 5 total blue berries.
In the fourth example, Phoenix can put all the red berries into baskets, leaving an extra blue berry behind. | instruction | 0 | 16,119 | 9 | 32,238 |
Tags: brute force, dp, greedy, math
Correct Solution:
```
#!/usr/bin/env python3
from collections import defaultdict,deque
from heapq import heappush, heappop
from bisect import bisect_left, bisect_right
import sys, random, itertools, math
sys.setrecursionlimit(10**5)
input = sys.stdin.readline
sqrt = math.sqrt
def LI(): return list(map(int, input().split()))
def LF(): return list(map(float, input().split()))
def LI_(): return list(map(lambda x: int(x)-1, input().split()))
def II(): return int(input())
def IF(): return float(input())
def S(): return input().rstrip()
def LS(): return S().split()
def IR(n): return [II() for _ in range(n)]
def LIR(n): return [LI() for _ in range(n)]
def FR(n): return [IF() for _ in range(n)]
def LFR(n): return [LI() for _ in range(n)]
def LIR_(n): return [LI_() for _ in range(n)]
def SR(n): return [S() for _ in range(n)]
def LSR(n): return [LS() for _ in range(n)]
mod = 1000000007
inf = float('INF')
#solve
def solve():
n, k = LI()
ab = LIR(n)
dp = [False] * k
dp[0] = True
A = 0
B = 0
for a, b in ab:
ndp = dp[::1]
A += a
B += b
for i in range(k):
if dp[i]:
for j in range(max((k - a), 1), min(k, (b + 1))):
ndp[(i+j)%k] = True
dp = ndp[::1]
ans = A // k + B // k
for i in range(k):
if dp[i]:
ans = max(ans, (A + i) // k + (B - i) // k)
print(ans)
return
#main
if __name__ == '__main__':
solve()
``` | output | 1 | 16,119 | 9 | 32,239 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Phoenix is picking berries in his backyard. There are n shrubs, and each shrub has a_i red berries and b_i blue berries.
Each basket can contain k berries. But, Phoenix has decided that each basket may only contain berries from the same shrub or berries of the same color (red or blue). In other words, all berries in a basket must be from the same shrub or/and have the same color.
For example, if there are two shrubs with 5 red and 2 blue berries in the first shrub and 2 red and 1 blue berries in the second shrub then Phoenix can fill 2 baskets of capacity 4 completely:
* the first basket will contain 3 red and 1 blue berries from the first shrub;
* the second basket will contain the 2 remaining red berries from the first shrub and 2 red berries from the second shrub.
Help Phoenix determine the maximum number of baskets he can fill completely!
Input
The first line contains two integers n and k ( 1≤ n, k ≤ 500) — the number of shrubs and the basket capacity, respectively.
The i-th of the next n lines contain two integers a_i and b_i (0 ≤ a_i, b_i ≤ 10^9) — the number of red and blue berries in the i-th shrub, respectively.
Output
Output one integer — the maximum number of baskets that Phoenix can fill completely.
Examples
Input
2 4
5 2
2 1
Output
2
Input
1 5
2 3
Output
1
Input
2 5
2 1
1 3
Output
0
Input
1 2
1000000000 1
Output
500000000
Note
The first example is described above.
In the second example, Phoenix can fill one basket fully using all the berries from the first (and only) shrub.
In the third example, Phoenix cannot fill any basket completely because there are less than 5 berries in each shrub, less than 5 total red berries, and less than 5 total blue berries.
In the fourth example, Phoenix can put all the red berries into baskets, leaving an extra blue berry behind. | instruction | 0 | 16,120 | 9 | 32,240 |
Tags: brute force, dp, greedy, math
Correct Solution:
```
import sys
sys.setrecursionlimit(10 ** 6)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
def main():
n,k=MI()
ab=LLI(n)
pre=1
sa=sb=0
mask=(1<<k)-1
for a,b in ab:
sa+=a
sb+=b
if a+b<k:continue
mn=max(k-b,0)
mx=min(a,k-1)
now=pre
for s in range(mn,mx+1):
now|=pre<<s
now|=now>>k
now&=mask
pre=now
#print(bin(pre))
ans=0
for r in range(k):
if pre >> r & 1: ans = max(ans, (sa - r) // k + (sb + r) // k)
print(ans)
main()
``` | output | 1 | 16,120 | 9 | 32,241 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Phoenix is picking berries in his backyard. There are n shrubs, and each shrub has a_i red berries and b_i blue berries.
Each basket can contain k berries. But, Phoenix has decided that each basket may only contain berries from the same shrub or berries of the same color (red or blue). In other words, all berries in a basket must be from the same shrub or/and have the same color.
For example, if there are two shrubs with 5 red and 2 blue berries in the first shrub and 2 red and 1 blue berries in the second shrub then Phoenix can fill 2 baskets of capacity 4 completely:
* the first basket will contain 3 red and 1 blue berries from the first shrub;
* the second basket will contain the 2 remaining red berries from the first shrub and 2 red berries from the second shrub.
Help Phoenix determine the maximum number of baskets he can fill completely!
Input
The first line contains two integers n and k ( 1≤ n, k ≤ 500) — the number of shrubs and the basket capacity, respectively.
The i-th of the next n lines contain two integers a_i and b_i (0 ≤ a_i, b_i ≤ 10^9) — the number of red and blue berries in the i-th shrub, respectively.
Output
Output one integer — the maximum number of baskets that Phoenix can fill completely.
Examples
Input
2 4
5 2
2 1
Output
2
Input
1 5
2 3
Output
1
Input
2 5
2 1
1 3
Output
0
Input
1 2
1000000000 1
Output
500000000
Note
The first example is described above.
In the second example, Phoenix can fill one basket fully using all the berries from the first (and only) shrub.
In the third example, Phoenix cannot fill any basket completely because there are less than 5 berries in each shrub, less than 5 total red berries, and less than 5 total blue berries.
In the fourth example, Phoenix can put all the red berries into baskets, leaving an extra blue berry behind. | instruction | 0 | 16,121 | 9 | 32,242 |
Tags: brute force, dp, greedy, math
Correct Solution:
```
import sys
input = sys.stdin.readline
n, k = map(int, input().split())
a = [None]
b = [None]
for _ in range(n):
x, y = map(int, input().split())
a.append(x)
b.append(y)
dp = [[None] * 505 for _ in range(505)]
totA = sum(a[1:])
totB = sum(b[1:])
dp[0][0] = True
for i in range(1, n + 1):
for j in range(k):
dp[i][j] = dp[i - 1][(j - a[i] % k + k) % k]
for l in range(min(k - 1, a[i]) + 1):
if (a[i] - l) % k + b[i] >= k:
dp[i][j] = dp[i][j] or dp[i - 1][(j - l + k) % k]
ans = 0
for i in range(k):
if dp[n][i]:
ans = max(ans, (totA + totB - i) // k)
print(ans)
``` | output | 1 | 16,121 | 9 | 32,243 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Phoenix is picking berries in his backyard. There are n shrubs, and each shrub has a_i red berries and b_i blue berries.
Each basket can contain k berries. But, Phoenix has decided that each basket may only contain berries from the same shrub or berries of the same color (red or blue). In other words, all berries in a basket must be from the same shrub or/and have the same color.
For example, if there are two shrubs with 5 red and 2 blue berries in the first shrub and 2 red and 1 blue berries in the second shrub then Phoenix can fill 2 baskets of capacity 4 completely:
* the first basket will contain 3 red and 1 blue berries from the first shrub;
* the second basket will contain the 2 remaining red berries from the first shrub and 2 red berries from the second shrub.
Help Phoenix determine the maximum number of baskets he can fill completely!
Input
The first line contains two integers n and k ( 1≤ n, k ≤ 500) — the number of shrubs and the basket capacity, respectively.
The i-th of the next n lines contain two integers a_i and b_i (0 ≤ a_i, b_i ≤ 10^9) — the number of red and blue berries in the i-th shrub, respectively.
Output
Output one integer — the maximum number of baskets that Phoenix can fill completely.
Examples
Input
2 4
5 2
2 1
Output
2
Input
1 5
2 3
Output
1
Input
2 5
2 1
1 3
Output
0
Input
1 2
1000000000 1
Output
500000000
Note
The first example is described above.
In the second example, Phoenix can fill one basket fully using all the berries from the first (and only) shrub.
In the third example, Phoenix cannot fill any basket completely because there are less than 5 berries in each shrub, less than 5 total red berries, and less than 5 total blue berries.
In the fourth example, Phoenix can put all the red berries into baskets, leaving an extra blue berry behind.
Submitted Solution:
```
import copy
shroud = []
aBerrySum = 0
bBerrySum = 0
n,k = map(int,input().split())
isAllRemainderCovered = False
remainderCovered = [False for _ in range(k)]
remainderCovered[0] = True
for _ in range(n):
a,b = map(int,input().split())
shroud.append([a,b])
aBerrySum += a
bBerrySum += b
if a >= k and b >= k:
isAllRemainderCovered = True
elif not isAllRemainderCovered and a + b >= k:
newRemainderCovered = copy.copy(remainderCovered)
for i in range(min(a+1,k+1)):
if k - i <= b:
for j1 in range(k):
if remainderCovered[j1 % k]:
newRemainderCovered[(j1 + i) % k] = True
remainderCovered = newRemainderCovered
if isAllRemainderCovered:
print((aBerrySum + bBerrySum) // k)
else:
aRemainder = aBerrySum % k
bRemainder = bBerrySum % k
if aRemainder + bRemainder < k:
print((aBerrySum + bBerrySum) // k)
else:
isRemainderCovered = False
for i in range(aRemainder + 1):
if (k - i) <= bRemainder and remainderCovered[i]:
isRemainderCovered = True
break
if isRemainderCovered:
print((aBerrySum + bBerrySum) // k)
else:
print((aBerrySum + bBerrySum) // k - 1)
``` | instruction | 0 | 16,122 | 9 | 32,244 |
Yes | output | 1 | 16,122 | 9 | 32,245 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Phoenix is picking berries in his backyard. There are n shrubs, and each shrub has a_i red berries and b_i blue berries.
Each basket can contain k berries. But, Phoenix has decided that each basket may only contain berries from the same shrub or berries of the same color (red or blue). In other words, all berries in a basket must be from the same shrub or/and have the same color.
For example, if there are two shrubs with 5 red and 2 blue berries in the first shrub and 2 red and 1 blue berries in the second shrub then Phoenix can fill 2 baskets of capacity 4 completely:
* the first basket will contain 3 red and 1 blue berries from the first shrub;
* the second basket will contain the 2 remaining red berries from the first shrub and 2 red berries from the second shrub.
Help Phoenix determine the maximum number of baskets he can fill completely!
Input
The first line contains two integers n and k ( 1≤ n, k ≤ 500) — the number of shrubs and the basket capacity, respectively.
The i-th of the next n lines contain two integers a_i and b_i (0 ≤ a_i, b_i ≤ 10^9) — the number of red and blue berries in the i-th shrub, respectively.
Output
Output one integer — the maximum number of baskets that Phoenix can fill completely.
Examples
Input
2 4
5 2
2 1
Output
2
Input
1 5
2 3
Output
1
Input
2 5
2 1
1 3
Output
0
Input
1 2
1000000000 1
Output
500000000
Note
The first example is described above.
In the second example, Phoenix can fill one basket fully using all the berries from the first (and only) shrub.
In the third example, Phoenix cannot fill any basket completely because there are less than 5 berries in each shrub, less than 5 total red berries, and less than 5 total blue berries.
In the fourth example, Phoenix can put all the red berries into baskets, leaving an extra blue berry behind.
Submitted Solution:
```
def check(k, aas, bs, a_rem, b_rem):
if a_rem + b_rem < k:
return False
a_lo = k - b_rem
a_hi = a_rem
rems = set()
rems.add(0)
for a, b in zip(aas, bs):
if a + b < k:
continue
for i in range(max(0, k - b), min(a, k) + 1):
rem = i % k
for j in list(rems):
rems.add((j + rem) % k)
for rem in rems:
if rem >= a_lo and rem <= a_hi:
return True
return False
n, k = [int(x) for x in input().split()]
aas = []
bs = []
a_total = 0
b_total = 0
for i in range(n):
a, b = [int(x) for x in input().split()]
aas.append(a)
bs.append(b)
a_total += a
b_total += b
ans = a_total // k + b_total // k
if check(k, aas, bs, a_total % k, b_total % k):
print(ans + 1)
else:
print(ans)
``` | instruction | 0 | 16,123 | 9 | 32,246 |
Yes | output | 1 | 16,123 | 9 | 32,247 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Phoenix is picking berries in his backyard. There are n shrubs, and each shrub has a_i red berries and b_i blue berries.
Each basket can contain k berries. But, Phoenix has decided that each basket may only contain berries from the same shrub or berries of the same color (red or blue). In other words, all berries in a basket must be from the same shrub or/and have the same color.
For example, if there are two shrubs with 5 red and 2 blue berries in the first shrub and 2 red and 1 blue berries in the second shrub then Phoenix can fill 2 baskets of capacity 4 completely:
* the first basket will contain 3 red and 1 blue berries from the first shrub;
* the second basket will contain the 2 remaining red berries from the first shrub and 2 red berries from the second shrub.
Help Phoenix determine the maximum number of baskets he can fill completely!
Input
The first line contains two integers n and k ( 1≤ n, k ≤ 500) — the number of shrubs and the basket capacity, respectively.
The i-th of the next n lines contain two integers a_i and b_i (0 ≤ a_i, b_i ≤ 10^9) — the number of red and blue berries in the i-th shrub, respectively.
Output
Output one integer — the maximum number of baskets that Phoenix can fill completely.
Examples
Input
2 4
5 2
2 1
Output
2
Input
1 5
2 3
Output
1
Input
2 5
2 1
1 3
Output
0
Input
1 2
1000000000 1
Output
500000000
Note
The first example is described above.
In the second example, Phoenix can fill one basket fully using all the berries from the first (and only) shrub.
In the third example, Phoenix cannot fill any basket completely because there are less than 5 berries in each shrub, less than 5 total red berries, and less than 5 total blue berries.
In the fourth example, Phoenix can put all the red berries into baskets, leaving an extra blue berry behind.
Submitted Solution:
```
MOD = 10 ** 9 + 7
def main():
for _ in inputt(1):
n, k = inputi()
T = [0] * k
T[0] = 1
A, B = 0, 0
for _ in range(n):
a, b = inputi()
A, B = A + a, B + b
mi, ma = max(k - a, 1), min(k, b + 1)
for t, c in list(enumerate(T)):
if c:
for i in range(mi, ma):
T[(t + i) % k] = 1
s = A // k + B // k
for t in range(k):
if T[t]:
s = max((A + t) // k + (B - t) // k, s)
print(s)
# region M
# region import
# 所有import部分
from math import *
from heapq import *
from itertools import *
from functools import reduce, lru_cache, partial
from collections import Counter, defaultdict
import re, copy, operator, cmath
import sys, io, os, builtins
sys.setrecursionlimit(1000)
# endregion
# region fastio
BUFSIZE = 8192
class FastIO(io.IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = io.BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(io.IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
if args:
sys.stdout.write(str(args[0]))
split = kwargs.pop("split", " ")
for arg in args[1:]:
sys.stdout.write(split)
sys.stdout.write(str(arg))
sys.stdout.write(kwargs.pop("end", "\n"))
def debug(*args, **kwargs):
print("debug", *args, **kwargs)
sys.stdout.flush()
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip()
inputt = lambda t = 0: range(t) if t else range(int(input()))
inputs = lambda: input().split()
inputi = lambda k = int: map(k, inputs())
inputl = lambda t = 0, k = lambda: list(inputi()): [k() for _ in range(t)] if t else list(k())
# endregion
# region bisect
def len(a):
if isinstance(a, range):
return -((a.start - a.stop) // a.step)
return builtins.len(a)
def bisect_left(a, x, key = None, lo = 0, hi = None):
if lo < 0: lo = 0
if hi == None: hi = len(a)
if key == None: key = do_nothing
while lo < hi:
mid = (lo + hi) // 2
if key(a[mid]) < x: lo = mid + 1
else: hi = mid
return lo
def bisect_right(a, x, key = None, lo = 0, hi = None):
if lo < 0: lo = 0
if hi == None: hi = len(a)
if key == None: key = do_nothing
while lo < hi:
mid = (lo + hi) // 2
if x < key(a[mid]): hi = mid
else: lo = mid + 1
return lo
def insort_left(a, x, key = None, lo = 0, hi = None):
lo = bisect_left(a, x, key, lo, hi)
a.insert(lo, x)
def insort_right(a, x, key = None, lo = 0, hi = None):
lo = bisect_right(a, x, key, lo, hi)
a.insert(lo, x)
do_nothing = lambda x: x
bisect = bisect_right
insort = insort_right
def index(a, x, default = None, key = None, lo = 0, hi = None):
if lo < 0: lo = 0
if hi == None: hi = len(a)
if key == None: key = do_nothing
i = bisect_left(a, x, key, lo, hi)
if lo <= i < hi and key(a[i]) == x: return a[i]
if default != None: return default
raise ValueError
def find_lt(a, x, default = None, key = None, lo = 0, hi = None):
if lo < 0: lo = 0
if hi == None: hi = len(a)
i = bisect_left(a, x, key, lo, hi)
if lo < i <= hi: return a[i - 1]
if default != None: return default
raise ValueError
def find_le(a, x, default = None, key = None, lo = 0, hi = None):
if lo < 0: lo = 0
if hi == None: hi = len(a)
i = bisect_right(a, x, key, lo, hi)
if lo < i <= hi: return a[i - 1]
if default != None: return default
raise ValueError
def find_gt(a, x, default = None, key = None, lo = 0, hi = None):
if lo < 0: lo = 0
if hi == None: hi = len(a)
i = bisect_right(a, x, key, lo, hi)
if lo <= i < hi: return a[i]
if default != None: return default
raise ValueError
def find_ge(a, x, default = None, key = None, lo = 0, hi = None):
if lo < 0: lo = 0
if hi == None: hi = len(a)
i = bisect_left(a, x, key, lo, hi)
if lo <= i < hi: return a[i]
if default != None: return default
raise ValueError
# endregion
# region csgraph
# TODO
class Tree:
def __init__(n):
self._n = n
self._conn = [[] for _ in range(n)]
self._list = [0] * n
def connect(a, b):
pass
# endregion
# region ntheory
class Sieve:
def __init__(self):
self._n = 6
self._list = [2, 3, 5, 7, 11, 13]
def extend(self, n):
if n <= self._list[-1]: return
maxbase = int(n ** 0.5) + 1
self.extend(maxbase)
begin = self._list[-1] + 1
newsieve = [i for i in range(begin, n + 1)]
for p in self.primerange(2, maxbase):
for i in range(-begin % p, len(newsieve), p):
newsieve[i] = 0
self._list.extend([x for x in newsieve if x])
def extend_to_no(self, i):
while len(self._list) < i:
self.extend(int(self._list[-1] * 1.5))
def primerange(self, a, b):
a = max(2, a)
if a >= b: return
self.extend(b)
i = self.search(a)[1]
maxi = len(self._list) + 1
while i < maxi:
p = self._list[i - 1]
if p < b:
yield p
i += 1
else: return
def search(self, n):
if n < 2: raise ValueError
if n > self._list[-1]: self.extend(n)
b = bisect(self._list, n)
if self._list[b - 1] == n: return b, b
else: return b, b + 1
def __contains__(self, n):
if n < 2: raise ValueError
if not n % 2: return n == 2
a, b = self.search(n)
return a == b
def __getitem__(self, n):
if isinstance(n, slice):
self.extend_to_no(n.stop + 1)
return self._list[n.start: n.stop: n.step]
else:
self.extend_to_no(n + 1)
return self._list[n]
sieve = Sieve()
def isprime(n):
if n <= sieve._list[-1]:
return n in sieve
for i in sieve:
if not n % i: return False
if n < i * i: return True
prime = sieve.__getitem__
primerange = lambda a, b = 0: sieve.primerange(a, b) if b else sieve.primerange(2, a)
def factorint(n):
factors = []
for i in sieve:
if n < i * i: break
while not n % i:
factors.append(i)
n //= i
if n != 1: factors.append(n)
return factors
factordict = lambda n: Counter(factorint(n))
# endregion
# region main
if __name__ == "__main__":
main()
# endregion
# endregion
``` | instruction | 0 | 16,124 | 9 | 32,248 |
Yes | output | 1 | 16,124 | 9 | 32,249 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Phoenix is picking berries in his backyard. There are n shrubs, and each shrub has a_i red berries and b_i blue berries.
Each basket can contain k berries. But, Phoenix has decided that each basket may only contain berries from the same shrub or berries of the same color (red or blue). In other words, all berries in a basket must be from the same shrub or/and have the same color.
For example, if there are two shrubs with 5 red and 2 blue berries in the first shrub and 2 red and 1 blue berries in the second shrub then Phoenix can fill 2 baskets of capacity 4 completely:
* the first basket will contain 3 red and 1 blue berries from the first shrub;
* the second basket will contain the 2 remaining red berries from the first shrub and 2 red berries from the second shrub.
Help Phoenix determine the maximum number of baskets he can fill completely!
Input
The first line contains two integers n and k ( 1≤ n, k ≤ 500) — the number of shrubs and the basket capacity, respectively.
The i-th of the next n lines contain two integers a_i and b_i (0 ≤ a_i, b_i ≤ 10^9) — the number of red and blue berries in the i-th shrub, respectively.
Output
Output one integer — the maximum number of baskets that Phoenix can fill completely.
Examples
Input
2 4
5 2
2 1
Output
2
Input
1 5
2 3
Output
1
Input
2 5
2 1
1 3
Output
0
Input
1 2
1000000000 1
Output
500000000
Note
The first example is described above.
In the second example, Phoenix can fill one basket fully using all the berries from the first (and only) shrub.
In the third example, Phoenix cannot fill any basket completely because there are less than 5 berries in each shrub, less than 5 total red berries, and less than 5 total blue berries.
In the fourth example, Phoenix can put all the red berries into baskets, leaving an extra blue berry behind.
Submitted Solution:
```
n,k=map(int,input().split())
sumr=0
sumb=0
sumtot=0
possr=[0]*k
possr[0]=1
for i in range(n):
a,b=map(int,input().split())
sumr+=a
sumb+=b
sumtot+=a
sumtot+=b
tot=a+b
poss2=[0]*k
for j in range(k):
rest=a-j
rest%=k
if (rest+b>=k or rest==0) and j<=a:
for l in range(k):
if possr[l]==1:
poss2[(l+j)%k]=1
for j in range(k):
possr[j]=poss2[j]
sol1=sumtot//k
sol2=sumr//k+sumb//k
if sol1==sol2:
print(sol1)
else:
i=0
while possr[i]==0:
i+=1
sumr%=k
sumb%=k
sumr-=i
sumb+=sumr
if sumb>=k:
print(sol1)
else:
print(sol2)
``` | instruction | 0 | 16,125 | 9 | 32,250 |
Yes | output | 1 | 16,125 | 9 | 32,251 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Phoenix is picking berries in his backyard. There are n shrubs, and each shrub has a_i red berries and b_i blue berries.
Each basket can contain k berries. But, Phoenix has decided that each basket may only contain berries from the same shrub or berries of the same color (red or blue). In other words, all berries in a basket must be from the same shrub or/and have the same color.
For example, if there are two shrubs with 5 red and 2 blue berries in the first shrub and 2 red and 1 blue berries in the second shrub then Phoenix can fill 2 baskets of capacity 4 completely:
* the first basket will contain 3 red and 1 blue berries from the first shrub;
* the second basket will contain the 2 remaining red berries from the first shrub and 2 red berries from the second shrub.
Help Phoenix determine the maximum number of baskets he can fill completely!
Input
The first line contains two integers n and k ( 1≤ n, k ≤ 500) — the number of shrubs and the basket capacity, respectively.
The i-th of the next n lines contain two integers a_i and b_i (0 ≤ a_i, b_i ≤ 10^9) — the number of red and blue berries in the i-th shrub, respectively.
Output
Output one integer — the maximum number of baskets that Phoenix can fill completely.
Examples
Input
2 4
5 2
2 1
Output
2
Input
1 5
2 3
Output
1
Input
2 5
2 1
1 3
Output
0
Input
1 2
1000000000 1
Output
500000000
Note
The first example is described above.
In the second example, Phoenix can fill one basket fully using all the berries from the first (and only) shrub.
In the third example, Phoenix cannot fill any basket completely because there are less than 5 berries in each shrub, less than 5 total red berries, and less than 5 total blue berries.
In the fourth example, Phoenix can put all the red berries into baskets, leaving an extra blue berry behind.
Submitted Solution:
```
n,k = map(int,input().split())
re,bl=0,0
total = 0
s = []
d = []
for i in range(n):
a,b = map(int,input().split())
if a>=k:
re+=a-(k-1)
a=k-1
if b>=k:
bl+=b-(k-1)
b=k-1
s.append([a,b])
d.append([a,b])
for i in d:
if i[0]+i[1]>=k:
if i[0]>=i[1]:
total+=1
bl+=i[1]
re+=i[0]-(k-i[1])
i[1]=0
else :
total+=1
re+=i[0]
bl+=i[1]-(k-i[0])
i[0]=0
else:
re+=i[0]
bl+=i[1]
i[0],i[1]=0,0
total += (re//k) + (bl//k)
re = re-re%k
bl = bl-bl%k
for i in s:
if re+bl>=k:
if i[0]+i[1]>=k:
if i[0]<=re and i[1]<=bl:
total+=1
break
print(total)
``` | instruction | 0 | 16,126 | 9 | 32,252 |
No | output | 1 | 16,126 | 9 | 32,253 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Phoenix is picking berries in his backyard. There are n shrubs, and each shrub has a_i red berries and b_i blue berries.
Each basket can contain k berries. But, Phoenix has decided that each basket may only contain berries from the same shrub or berries of the same color (red or blue). In other words, all berries in a basket must be from the same shrub or/and have the same color.
For example, if there are two shrubs with 5 red and 2 blue berries in the first shrub and 2 red and 1 blue berries in the second shrub then Phoenix can fill 2 baskets of capacity 4 completely:
* the first basket will contain 3 red and 1 blue berries from the first shrub;
* the second basket will contain the 2 remaining red berries from the first shrub and 2 red berries from the second shrub.
Help Phoenix determine the maximum number of baskets he can fill completely!
Input
The first line contains two integers n and k ( 1≤ n, k ≤ 500) — the number of shrubs and the basket capacity, respectively.
The i-th of the next n lines contain two integers a_i and b_i (0 ≤ a_i, b_i ≤ 10^9) — the number of red and blue berries in the i-th shrub, respectively.
Output
Output one integer — the maximum number of baskets that Phoenix can fill completely.
Examples
Input
2 4
5 2
2 1
Output
2
Input
1 5
2 3
Output
1
Input
2 5
2 1
1 3
Output
0
Input
1 2
1000000000 1
Output
500000000
Note
The first example is described above.
In the second example, Phoenix can fill one basket fully using all the berries from the first (and only) shrub.
In the third example, Phoenix cannot fill any basket completely because there are less than 5 berries in each shrub, less than 5 total red berries, and less than 5 total blue berries.
In the fourth example, Phoenix can put all the red berries into baskets, leaving an extra blue berry behind.
Submitted Solution:
```
import os,io
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
shroud = []
aBerrySum = 0
bBerrySum = 0
n,k = map(int,input().split())
for i in range(n):
a,b = map(int,input().split())
shroud.append([a,b])
aBerrySum += a
bBerrySum += b
basket = 0
aBasket,aRemainder = divmod(aBerrySum,k)
bBasket,bRemainder = divmod(aBerrySum,k)
basket += aBasket + bBasket
if aRemainder + bRemainder < k:
print(basket)
else:
isLastBasketFillable = False
for data in shroud:
if min(data[0],aRemainder) + min(data[1],bRemainder) >= k:
isLastBasketFillable = True
if isLastBasketFillable:
print(basket + 1)
else:
print(basket)
``` | instruction | 0 | 16,127 | 9 | 32,254 |
No | output | 1 | 16,127 | 9 | 32,255 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Phoenix is picking berries in his backyard. There are n shrubs, and each shrub has a_i red berries and b_i blue berries.
Each basket can contain k berries. But, Phoenix has decided that each basket may only contain berries from the same shrub or berries of the same color (red or blue). In other words, all berries in a basket must be from the same shrub or/and have the same color.
For example, if there are two shrubs with 5 red and 2 blue berries in the first shrub and 2 red and 1 blue berries in the second shrub then Phoenix can fill 2 baskets of capacity 4 completely:
* the first basket will contain 3 red and 1 blue berries from the first shrub;
* the second basket will contain the 2 remaining red berries from the first shrub and 2 red berries from the second shrub.
Help Phoenix determine the maximum number of baskets he can fill completely!
Input
The first line contains two integers n and k ( 1≤ n, k ≤ 500) — the number of shrubs and the basket capacity, respectively.
The i-th of the next n lines contain two integers a_i and b_i (0 ≤ a_i, b_i ≤ 10^9) — the number of red and blue berries in the i-th shrub, respectively.
Output
Output one integer — the maximum number of baskets that Phoenix can fill completely.
Examples
Input
2 4
5 2
2 1
Output
2
Input
1 5
2 3
Output
1
Input
2 5
2 1
1 3
Output
0
Input
1 2
1000000000 1
Output
500000000
Note
The first example is described above.
In the second example, Phoenix can fill one basket fully using all the berries from the first (and only) shrub.
In the third example, Phoenix cannot fill any basket completely because there are less than 5 berries in each shrub, less than 5 total red berries, and less than 5 total blue berries.
In the fourth example, Phoenix can put all the red berries into baskets, leaving an extra blue berry behind.
Submitted Solution:
```
n, k = map(int, input().split())
a = []
b = []
sa = 0
sb = 0
for i in range(n):
x, y = map(int, input().split())
a.append(x)
b.append(y)
sa += x
sb += y
# print(a, b)
m = 0
for i in range(n):
if(sa != 0 and sb != 0):
if(a[i] > b[i]):
x = b[i]
if(k - x < 0):
sb -= (k)
m += 1
continue
elif(k - x == 0):
sb -= b[i]
m += 1
continue
else:
if(sa < k - x):
break
if(a[i] < k - x):
continue
sb -= x
sa -= (k - x)
m += 1
else:
x = a[i]
if(k - x <= 0):
sa -= k
m += 1
continue
else:
if(sb < k - x):
break
if(b[i] < k - x):
continue
sa -= x
sb -= (k - x)
m += 1
else:
break
m += (sa // k + sb // k)
print(m)
``` | instruction | 0 | 16,128 | 9 | 32,256 |
No | output | 1 | 16,128 | 9 | 32,257 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Phoenix is picking berries in his backyard. There are n shrubs, and each shrub has a_i red berries and b_i blue berries.
Each basket can contain k berries. But, Phoenix has decided that each basket may only contain berries from the same shrub or berries of the same color (red or blue). In other words, all berries in a basket must be from the same shrub or/and have the same color.
For example, if there are two shrubs with 5 red and 2 blue berries in the first shrub and 2 red and 1 blue berries in the second shrub then Phoenix can fill 2 baskets of capacity 4 completely:
* the first basket will contain 3 red and 1 blue berries from the first shrub;
* the second basket will contain the 2 remaining red berries from the first shrub and 2 red berries from the second shrub.
Help Phoenix determine the maximum number of baskets he can fill completely!
Input
The first line contains two integers n and k ( 1≤ n, k ≤ 500) — the number of shrubs and the basket capacity, respectively.
The i-th of the next n lines contain two integers a_i and b_i (0 ≤ a_i, b_i ≤ 10^9) — the number of red and blue berries in the i-th shrub, respectively.
Output
Output one integer — the maximum number of baskets that Phoenix can fill completely.
Examples
Input
2 4
5 2
2 1
Output
2
Input
1 5
2 3
Output
1
Input
2 5
2 1
1 3
Output
0
Input
1 2
1000000000 1
Output
500000000
Note
The first example is described above.
In the second example, Phoenix can fill one basket fully using all the berries from the first (and only) shrub.
In the third example, Phoenix cannot fill any basket completely because there are less than 5 berries in each shrub, less than 5 total red berries, and less than 5 total blue berries.
In the fourth example, Phoenix can put all the red berries into baskets, leaving an extra blue berry behind.
Submitted Solution:
```
shroud = []
aBerrySum = 0
bBerrySum = 0
n,k = map(int,input().split())
isAllRemainderCovered = False
remainderCovered = [False for _ in range(k)]
remainderCovered[0] = True
for _ in range(n):
a,b = map(int,input().split())
shroud.append([a,b])
aBerrySum += a
bBerrySum += b
if a >= k and b >= k:
isAllRemainderCovered = True
elif not isAllRemainderCovered and a + b >= k:
for i in range(min(a+1,k+1)):
if k - i >= b:
for j1 in range(k):
if remainderCovered[j1 % k]:
remainderCovered[(j1 + i) % k] = True
print(remainderCovered)
if isAllRemainderCovered:
print((aBerrySum + bBerrySum) // k)
else:
aRemainder = aBerrySum % k
bRemainder = bBerrySum % k
if aRemainder + bRemainder < k:
print((aBerrySum + bBerrySum) // k)
else:
isRemainderCovered = False
for i in range(aRemainder):
if (k - i) <= bRemainder and remainderCovered[i]:
isRemainderCovered = True
break
if isRemainderCovered:
print((aBerrySum + bBerrySum) // k)
else:
print((aBerrySum + bBerrySum) // k - 1)
``` | instruction | 0 | 16,129 | 9 | 32,258 |
No | output | 1 | 16,129 | 9 | 32,259 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is an old tradition of keeping 4 boxes of candies in the house in Cyberland. The numbers of candies are special if their arithmetic mean, their median and their range are all equal. By definition, for a set {x1, x2, x3, x4} (x1 ≤ x2 ≤ x3 ≤ x4) arithmetic mean is <image>, median is <image> and range is x4 - x1. The arithmetic mean and median are not necessary integer. It is well-known that if those three numbers are same, boxes will create a "debugging field" and codes in the field will have no bugs.
For example, 1, 1, 3, 3 is the example of 4 numbers meeting the condition because their mean, median and range are all equal to 2.
Jeff has 4 special boxes of candies. However, something bad has happened! Some of the boxes could have been lost and now there are only n (0 ≤ n ≤ 4) boxes remaining. The i-th remaining box contains ai candies.
Now Jeff wants to know: is there a possible way to find the number of candies of the 4 - n missing boxes, meeting the condition above (the mean, median and range are equal)?
Input
The first line of input contains an only integer n (0 ≤ n ≤ 4).
The next n lines contain integers ai, denoting the number of candies in the i-th box (1 ≤ ai ≤ 500).
Output
In the first output line, print "YES" if a solution exists, or print "NO" if there is no solution.
If a solution exists, you should output 4 - n more lines, each line containing an integer b, denoting the number of candies in a missing box.
All your numbers b must satisfy inequality 1 ≤ b ≤ 106. It is guaranteed that if there exists a positive integer solution, you can always find such b's meeting the condition. If there are multiple answers, you are allowed to print any of them.
Given numbers ai may follow in any order in the input, not necessary in non-decreasing.
ai may have stood at any positions in the original set, not necessary on lowest n first positions.
Examples
Input
2
1
1
Output
YES
3
3
Input
3
1
1
1
Output
NO
Input
4
1
2
2
3
Output
YES
Note
For the first sample, the numbers of candies in 4 boxes can be 1, 1, 3, 3. The arithmetic mean, the median and the range of them are all 2.
For the second sample, it's impossible to find the missing number of candies.
In the third example no box has been lost and numbers satisfy the condition.
You may output b in any order. | instruction | 0 | 16,346 | 9 | 32,692 |
Tags: brute force, constructive algorithms, math
Correct Solution:
```
__author__ = 'zhan'
candy = [0]
n = int(input())
for i in range(n):
candy.append(int(input()))
if n == 0:
print("YES\n1\n1\n3\n3")
elif n == 4:
candy.sort()
if candy[4] == 3*candy[1] and (candy[2]+candy[3]) == 4*candy[1]:
print("YES")
else:
print("NO")
elif n == 1:
ans = [str(2*candy[1]), str(2*candy[1]), str(3*candy[1])]
ans.insert(0, "YES")
print("\n".join(ans))
elif n == 3:
candy.sort()
if candy[3] % 3 == 0 and 3*(candy[1]+candy[2]) == 4*candy[3]:
print("YES")
print(candy[3]//3)
elif 3*candy[1] == candy[3]:
print("YES")
print(4*candy[1]-candy[2])
elif 4*candy[1] == candy[2] + candy[3]:
print("YES")
print(3*candy[1])
else:
print("NO")
else:
candy.sort()
if candy[2] == 3*candy[1]:
ans = ["YES", str(2*candy[1]), str(2*candy[1])]
print("\n".join(ans))
elif candy[2] < 3*candy[1]:
ans = ["YES", str(4*candy[1]-candy[2]), str(3*candy[1])]
print("\n".join(ans))
else:
print("NO")
``` | output | 1 | 16,346 | 9 | 32,693 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is an old tradition of keeping 4 boxes of candies in the house in Cyberland. The numbers of candies are special if their arithmetic mean, their median and their range are all equal. By definition, for a set {x1, x2, x3, x4} (x1 ≤ x2 ≤ x3 ≤ x4) arithmetic mean is <image>, median is <image> and range is x4 - x1. The arithmetic mean and median are not necessary integer. It is well-known that if those three numbers are same, boxes will create a "debugging field" and codes in the field will have no bugs.
For example, 1, 1, 3, 3 is the example of 4 numbers meeting the condition because their mean, median and range are all equal to 2.
Jeff has 4 special boxes of candies. However, something bad has happened! Some of the boxes could have been lost and now there are only n (0 ≤ n ≤ 4) boxes remaining. The i-th remaining box contains ai candies.
Now Jeff wants to know: is there a possible way to find the number of candies of the 4 - n missing boxes, meeting the condition above (the mean, median and range are equal)?
Input
The first line of input contains an only integer n (0 ≤ n ≤ 4).
The next n lines contain integers ai, denoting the number of candies in the i-th box (1 ≤ ai ≤ 500).
Output
In the first output line, print "YES" if a solution exists, or print "NO" if there is no solution.
If a solution exists, you should output 4 - n more lines, each line containing an integer b, denoting the number of candies in a missing box.
All your numbers b must satisfy inequality 1 ≤ b ≤ 106. It is guaranteed that if there exists a positive integer solution, you can always find such b's meeting the condition. If there are multiple answers, you are allowed to print any of them.
Given numbers ai may follow in any order in the input, not necessary in non-decreasing.
ai may have stood at any positions in the original set, not necessary on lowest n first positions.
Examples
Input
2
1
1
Output
YES
3
3
Input
3
1
1
1
Output
NO
Input
4
1
2
2
3
Output
YES
Note
For the first sample, the numbers of candies in 4 boxes can be 1, 1, 3, 3. The arithmetic mean, the median and the range of them are all 2.
For the second sample, it's impossible to find the missing number of candies.
In the third example no box has been lost and numbers satisfy the condition.
You may output b in any order. | instruction | 0 | 16,347 | 9 | 32,694 |
Tags: brute force, constructive algorithms, math
Correct Solution:
```
def check(a, b, c, d):
bad = [a, b, c, d]
bad.sort()
x, y, z, w = bad[0], bad[1], bad[2], bad[3]
if w == 3*x and x+w==y+z:
return True
return False
a = int(input())
if a == 0:
print("YES")
print("1")
print("1")
print("3")
print("3")
if a == 1:
x = int(input())
print("YES")
print(x)
print(3*x)
print(3*x)
if a == 2:
x = int(input())
y = int(input())
x, y = min(x, y), max(x, y)
if y > 3 * x:
print("NO")
else:
print("YES")
print(4*x-y)
print(3*x)
if a == 3:
nums = [int(input()) for i in range(3)]
nums.sort()
x, y, z = nums[0], nums[1], nums[2]
for i in range(1, 10**5):
if check(x, y, z, i):
print("YES")
print(i)
quit()
print("NO")
if a == 4:
nums = [int(input()) for i in range(4)]
nums.sort()
if nums[-1] == 3 * nums[0] and nums[0]+nums[3] == nums[1]+nums[2]:
print("YES")
else:
print("NO")
``` | output | 1 | 16,347 | 9 | 32,695 |
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