message stringlengths 2 44.5k | message_type stringclasses 2
values | message_id int64 0 1 | conversation_id int64 42 109k | cluster float64 5 5 | __index_level_0__ int64 84 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are positive integers N and K.
Determine if the 3N integers K, K+1, ..., K+3N-1 can be partitioned into N triples (a_1,b_1,c_1), ..., (a_N,b_N,c_N) so that the condition below is satisfied. Any of the integers K, K+1, ..., K+3N-1 must appear in exactly one of those triples.
* For every integer i from 1 to N, a_i + b_i \leq c_i holds.
If the answer is yes, construct one such partition.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq K \leq 10^9
Input
Input is given from Standard Input in the following format:
N K
Output
If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:
a_1 b_1 c_1
:
a_N b_N c_N
Output
If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:
a_1 b_1 c_1
:
a_N b_N c_N
Examples
Input
1 1
Output
1 2 3
Input
3 3
Output
-1
Submitted Solution:
```
def solve(N,K):
if 2*K < N-1:
print(-1)
return
def it():
for i in range((N+1)//2):
yield (K+i, K+N+i+N//2, K+N*2+i*2)
for i in range(N//2):
yield (K+(N+1)//2 + i, K+N+i, K+N*2+i*2+1)
for a,b,c in it():
print(a,b,c)
N,K = map(int,input().split())
solve(N,K)
``` | instruction | 0 | 36,681 | 5 | 73,362 |
No | output | 1 | 36,681 | 5 | 73,363 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are positive integers N and K.
Determine if the 3N integers K, K+1, ..., K+3N-1 can be partitioned into N triples (a_1,b_1,c_1), ..., (a_N,b_N,c_N) so that the condition below is satisfied. Any of the integers K, K+1, ..., K+3N-1 must appear in exactly one of those triples.
* For every integer i from 1 to N, a_i + b_i \leq c_i holds.
If the answer is yes, construct one such partition.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq K \leq 10^9
Input
Input is given from Standard Input in the following format:
N K
Output
If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:
a_1 b_1 c_1
:
a_N b_N c_N
Output
If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:
a_1 b_1 c_1
:
a_N b_N c_N
Examples
Input
1 1
Output
1 2 3
Input
3 3
Output
-1
Submitted Solution:
```
import sys
readline = sys.stdin.readline
from itertools import accumulate
from collections import Counter
from bisect import bisect as br, bisect_left as bl
class PMS:
#1-indexed
def __init__(self, A, B, issum = False):
#Aに初期状態の要素をすべて入れる,Bは値域のリスト
self.X, self.comp = self.compress(B)
self.size = len(self.X)
self.tree = [0] * (self.size + 1)
self.p = 2**(self.size.bit_length() - 1)
self.dep = self.size.bit_length()
CA = Counter(A)
S = [0] + list(accumulate([CA[self.X[i]] for i in range(self.size)]))
for i in range(1, 1+self.size):
self.tree[i] = S[i] - S[i - (i&-i)]
if issum:
self.sumtree = [0] * (self.size + 1)
Ssum = [0] + list(accumulate([CA[self.X[i]]*self.X[i] for i in range(self.size)]))
for i in range(1, 1+self.size):
self.sumtree[i] = Ssum[i] - Ssum[i - (i&-i)]
def compress(self, L):
#座圧
L2 = list(set(L))
L2.sort()
C = {v : k for k, v in enumerate(L2, 1)}
# 1-indexed
return L2, C
def leng(self):
#今入っている個数を取得
return self.count(self.X[-1])
def count(self, v):
#v(Bの元)以下の個数を取得
i = self.comp[v]
s = 0
while i > 0:
s += self.tree[i]
i -= i & -i
return s
def less(self, v):
#v(Bの元である必要はない)未満の個数を取得
i = bl(self.X, v)
s = 0
while i > 0:
s += self.tree[i]
i -= i & -i
return s
def leq(self, v):
#v(Bの元である必要はない)以下の個数を取得
i = br(self.X, v)
s = 0
while i > 0:
s += self.tree[i]
i -= i & -i
return s
def add(self, v, x):
#vをx個入れる,負のxで取り出す,iの個数以上取り出すとエラーを出さずにバグる
i = self.comp[v]
while i <= self.size:
self.tree[i] += x
i += i & -i
def get(self, i):
# i番目の値を取得
if i <= 0:
return -1
s = 0
k = self.p
for _ in range(self.dep):
if s + k <= self.size and self.tree[s+k] < i:
s += k
i -= self.tree[s]
k //= 2
return self.X[s]
def gets(self, v):
#累積和がv以下となる最大のindexを返す
v1 = v
s = 0
k = self.p
for _ in range(self.dep):
if s + k <= self.size and self.sumtree[s+k] < v:
s += k
v -= self.sumtree[s]
k //= 2
if s == self.size:
return self.leng()
return self.count(self.X[s]) + (v1 - self.countsum(self.X[s]))//self.X[s]
def addsum(self, i, x):
#sumを扱いたいときにaddの代わりに使う
self.add(i, x)
x *= i
i = self.comp[i]
while i <= self.size:
self.sumtree[i] += x
i += i & -i
def countsum(self, v):
#v(Bの元)以下のsumを取得
i = self.comp[v]
s = 0
while i > 0:
s += self.sumtree[i]
i -= i & -i
return s
def getsum(self, i):
#i番目までのsumを取得
x = self.get(i)
return self.countsum(x) - x*(self.count(x) - i)
N, K = map(int, readline().split())
C = list(range(K+2*N, K+3*N))
A = list(range(K, K+N))
B = list(range(K+N, K+2*N))
Ans = []
ans = 1
BB = PMS(B, B)
if sum(A) + sum(B) > sum(C):
ans = -1
else:
Ao = [A[i] for i in range(N) if A[i]%2]
Ae = [A[i] for i in range(N) if not A[i]%2]
if len(Ao) < len(Ae):
AA = Ae + Ao
else:
AA = Ao + Ae
for a, c in zip(AA, C):
bx = c - a
k = BB.leq(bx)
if k == 0:
ans = -1
break
b = BB.get(k)
Ans.append((a, b, c))
BB.add(b, -1)
if ans != -1:
print('\n'.join('{} {} {}'.format(*an) for an in Ans))
else:
print(ans)
for _ in range(test_case):
import sys
readline = sys.stdin.readline
from itertools import accumulate
from collections import Counter
from bisect import bisect as br, bisect_left as bl
class PMS:
#1-indexed
def __init__(self, A, B, issum = False):
#Aに初期状態の要素をすべて入れる,Bは値域のリスト
self.X, self.comp = self.compress(B)
self.size = len(self.X)
self.tree = [0] * (self.size + 1)
self.p = 2**(self.size.bit_length() - 1)
self.dep = self.size.bit_length()
CA = Counter(A)
S = [0] + list(accumulate([CA[self.X[i]] for i in range(self.size)]))
for i in range(1, 1+self.size):
self.tree[i] = S[i] - S[i - (i&-i)]
if issum:
self.sumtree = [0] * (self.size + 1)
Ssum = [0] + list(accumulate([CA[self.X[i]]*self.X[i] for i in range(self.size)]))
for i in range(1, 1+self.size):
self.sumtree[i] = Ssum[i] - Ssum[i - (i&-i)]
def compress(self, L):
#座圧
L2 = list(set(L))
L2.sort()
C = {v : k for k, v in enumerate(L2, 1)}
# 1-indexed
return L2, C
def leng(self):
#今入っている個数を取得
return self.count(self.X[-1])
def count(self, v):
#v(Bの元)以下の個数を取得
i = self.comp[v]
s = 0
while i > 0:
s += self.tree[i]
i -= i & -i
return s
def less(self, v):
#v(Bの元である必要はない)未満の個数を取得
i = bl(self.X, v)
s = 0
while i > 0:
s += self.tree[i]
i -= i & -i
return s
def leq(self, v):
#v(Bの元である必要はない)以下の個数を取得
i = br(self.X, v)
s = 0
while i > 0:
s += self.tree[i]
i -= i & -i
return s
def add(self, v, x):
#vをx個入れる,負のxで取り出す,iの個数以上取り出すとエラーを出さずにバグる
i = self.comp[v]
while i <= self.size:
self.tree[i] += x
i += i & -i
def get(self, i):
# i番目の値を取得
if i <= 0:
return -1
s = 0
k = self.p
for _ in range(self.dep):
if s + k <= self.size and self.tree[s+k] < i:
s += k
i -= self.tree[s]
k //= 2
return self.X[s]
def gets(self, v):
#累積和がv以下となる最大のindexを返す
v1 = v
s = 0
k = self.p
for _ in range(self.dep):
if s + k <= self.size and self.sumtree[s+k] < v:
s += k
v -= self.sumtree[s]
k //= 2
if s == self.size:
return self.leng()
return self.count(self.X[s]) + (v1 - self.countsum(self.X[s]))//self.X[s]
def addsum(self, i, x):
#sumを扱いたいときにaddの代わりに使う
self.add(i, x)
x *= i
i = self.comp[i]
while i <= self.size:
self.sumtree[i] += x
i += i & -i
def countsum(self, v):
#v(Bの元)以下のsumを取得
i = self.comp[v]
s = 0
while i > 0:
s += self.sumtree[i]
i -= i & -i
return s
def getsum(self, i):
#i番目までのsumを取得
x = self.get(i)
return self.countsum(x) - x*(self.count(x) - i)
N, K = map(int, readline().split())
C = list(range(K+2*N, K+3*N))
A = list(range(K, K+N))
B = list(range(K+N, K+2*N))
Ans = []
ans = 1
BB = PMS(B, B)
if sum(A) + sum(B) > sum(C):
ans = -1
else:
Ao = [A[i] for i in range(N) if A[i]%2]
Ae = [A[i] for i in range(N) if not A[i]%2]
if len(Ao) < len(Ae):
AA = Ae + Ao
else:
AA = Ao + Ae
for a, c in zip(AA, C):
bx = c - a
k = BB.leq(bx)
if k == 0:
ans = -1
break
b = BB.get(k)
Ans.append((a, b, c))
BB.add(b, -1)
if ans != -1:
print('\n'.join('{} {} {}'.format(*an) for an in Ans))
else:
print(ans)
``` | instruction | 0 | 36,682 | 5 | 73,364 |
No | output | 1 | 36,682 | 5 | 73,365 |
Provide a correct Python 3 solution for this coding contest problem.
Example
Input
201
Output
701 | instruction | 0 | 36,805 | 5 | 73,610 |
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 998244353
dd = [(0,-1),(1,0),(0,1),(-1,0)]
ddn = [(0,-1),(1,-1),(1,0),(1,1),(0,1),(-1,-1),(-1,0),(-1,1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
s = S()
l = len(s)
n = int(s)
m = inf
r = 0
t = 10**l
for a in itertools.permutations(range(10), l):
b = 0
for c in a:
b = b * 10 + c
tm = max(abs(n-b),t-abs(n-b))
if m > tm:
m = tm
r = ''.join(map(str,a))
return r
print(main())
``` | output | 1 | 36,805 | 5 | 73,611 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads two integers a and b, and calculates the following values:
* a ÷ b: d (in integer)
* remainder of a ÷ b: r (in integer)
* a ÷ b: f (in real number)
Constraints
* 1 ≤ a, b ≤ 109
Input
Two integers a and b are given in a line.
Output
Print d, r and f separated by a space in a line. For f, the output should not contain an absolute error greater than 10-5.
Example
Input
3 2
Output
1 1 1.50000 | instruction | 0 | 36,830 | 5 | 73,660 |
"Correct Solution:
```
a,b=map(int,input().split())
print(a//b,a%b,"{0:10.6f}".format(a/b))
``` | output | 1 | 36,830 | 5 | 73,661 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads two integers a and b, and calculates the following values:
* a ÷ b: d (in integer)
* remainder of a ÷ b: r (in integer)
* a ÷ b: f (in real number)
Constraints
* 1 ≤ a, b ≤ 109
Input
Two integers a and b are given in a line.
Output
Print d, r and f separated by a space in a line. For f, the output should not contain an absolute error greater than 10-5.
Example
Input
3 2
Output
1 1 1.50000 | instruction | 0 | 36,831 | 5 | 73,662 |
"Correct Solution:
```
a,b=map(int,input().split())
d=a//b
r=a%b
f=round(a/b,5)
print(d,r,f)
``` | output | 1 | 36,831 | 5 | 73,663 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads two integers a and b, and calculates the following values:
* a ÷ b: d (in integer)
* remainder of a ÷ b: r (in integer)
* a ÷ b: f (in real number)
Constraints
* 1 ≤ a, b ≤ 109
Input
Two integers a and b are given in a line.
Output
Print d, r and f separated by a space in a line. For f, the output should not contain an absolute error greater than 10-5.
Example
Input
3 2
Output
1 1 1.50000 | instruction | 0 | 36,832 | 5 | 73,664 |
"Correct Solution:
```
a,b=map(int,input().split())
print(f"{a//b} {a%b} {(a/b):.10f}")
``` | output | 1 | 36,832 | 5 | 73,665 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads two integers a and b, and calculates the following values:
* a ÷ b: d (in integer)
* remainder of a ÷ b: r (in integer)
* a ÷ b: f (in real number)
Constraints
* 1 ≤ a, b ≤ 109
Input
Two integers a and b are given in a line.
Output
Print d, r and f separated by a space in a line. For f, the output should not contain an absolute error greater than 10-5.
Example
Input
3 2
Output
1 1 1.50000 | instruction | 0 | 36,833 | 5 | 73,666 |
"Correct Solution:
```
a, b = map(int, input().split())
print(a//b, a%b, "{0:.8f}".format(a/b))
``` | output | 1 | 36,833 | 5 | 73,667 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads two integers a and b, and calculates the following values:
* a ÷ b: d (in integer)
* remainder of a ÷ b: r (in integer)
* a ÷ b: f (in real number)
Constraints
* 1 ≤ a, b ≤ 109
Input
Two integers a and b are given in a line.
Output
Print d, r and f separated by a space in a line. For f, the output should not contain an absolute error greater than 10-5.
Example
Input
3 2
Output
1 1 1.50000 | instruction | 0 | 36,834 | 5 | 73,668 |
"Correct Solution:
```
a,b = map(int, input().split())
print(a//b, a%b, '{:.6f}'.format(a/b))
``` | output | 1 | 36,834 | 5 | 73,669 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads two integers a and b, and calculates the following values:
* a ÷ b: d (in integer)
* remainder of a ÷ b: r (in integer)
* a ÷ b: f (in real number)
Constraints
* 1 ≤ a, b ≤ 109
Input
Two integers a and b are given in a line.
Output
Print d, r and f separated by a space in a line. For f, the output should not contain an absolute error greater than 10-5.
Example
Input
3 2
Output
1 1 1.50000 | instruction | 0 | 36,835 | 5 | 73,670 |
"Correct Solution:
```
a,b=map(int,input().split())
print(a//b, a%b, f'{a/b:.8f}')
``` | output | 1 | 36,835 | 5 | 73,671 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads two integers a and b, and calculates the following values:
* a ÷ b: d (in integer)
* remainder of a ÷ b: r (in integer)
* a ÷ b: f (in real number)
Constraints
* 1 ≤ a, b ≤ 109
Input
Two integers a and b are given in a line.
Output
Print d, r and f separated by a space in a line. For f, the output should not contain an absolute error greater than 10-5.
Example
Input
3 2
Output
1 1 1.50000 | instruction | 0 | 36,836 | 5 | 73,672 |
"Correct Solution:
```
a,b = map(int,input().split())
print('{0} {1} {2:.8f}'.format(a//b,a%b,a/b))
``` | output | 1 | 36,836 | 5 | 73,673 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads two integers a and b, and calculates the following values:
* a ÷ b: d (in integer)
* remainder of a ÷ b: r (in integer)
* a ÷ b: f (in real number)
Constraints
* 1 ≤ a, b ≤ 109
Input
Two integers a and b are given in a line.
Output
Print d, r and f separated by a space in a line. For f, the output should not contain an absolute error greater than 10-5.
Example
Input
3 2
Output
1 1 1.50000 | instruction | 0 | 36,837 | 5 | 73,674 |
"Correct Solution:
```
a,b = map(int,input().split())
print(int(a/b),a%b,round(a/b,6))
``` | output | 1 | 36,837 | 5 | 73,675 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which reads two integers a and b, and calculates the following values:
* a ÷ b: d (in integer)
* remainder of a ÷ b: r (in integer)
* a ÷ b: f (in real number)
Constraints
* 1 ≤ a, b ≤ 109
Input
Two integers a and b are given in a line.
Output
Print d, r and f separated by a space in a line. For f, the output should not contain an absolute error greater than 10-5.
Example
Input
3 2
Output
1 1 1.50000
Submitted Solution:
```
a, b = map(int, input().split())
print(a // b, a % b, f'{a / b:.20f}')
``` | instruction | 0 | 36,838 | 5 | 73,676 |
Yes | output | 1 | 36,838 | 5 | 73,677 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which reads two integers a and b, and calculates the following values:
* a ÷ b: d (in integer)
* remainder of a ÷ b: r (in integer)
* a ÷ b: f (in real number)
Constraints
* 1 ≤ a, b ≤ 109
Input
Two integers a and b are given in a line.
Output
Print d, r and f separated by a space in a line. For f, the output should not contain an absolute error greater than 10-5.
Example
Input
3 2
Output
1 1 1.50000
Submitted Solution:
```
x=input().split()
a=int(x[0])
b=int(x[1])
print("%d %d %f"%(a//b, a%b, a/b))
``` | instruction | 0 | 36,839 | 5 | 73,678 |
Yes | output | 1 | 36,839 | 5 | 73,679 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which reads two integers a and b, and calculates the following values:
* a ÷ b: d (in integer)
* remainder of a ÷ b: r (in integer)
* a ÷ b: f (in real number)
Constraints
* 1 ≤ a, b ≤ 109
Input
Two integers a and b are given in a line.
Output
Print d, r and f separated by a space in a line. For f, the output should not contain an absolute error greater than 10-5.
Example
Input
3 2
Output
1 1 1.50000
Submitted Solution:
```
a, b = map(int, input().split())
print(a//b, a%b, "{0:6f}".format(a/b))
``` | instruction | 0 | 36,840 | 5 | 73,680 |
Yes | output | 1 | 36,840 | 5 | 73,681 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which reads two integers a and b, and calculates the following values:
* a ÷ b: d (in integer)
* remainder of a ÷ b: r (in integer)
* a ÷ b: f (in real number)
Constraints
* 1 ≤ a, b ≤ 109
Input
Two integers a and b are given in a line.
Output
Print d, r and f separated by a space in a line. For f, the output should not contain an absolute error greater than 10-5.
Example
Input
3 2
Output
1 1 1.50000
Submitted Solution:
```
a,b=map(float,input().split())
print(a//b,a%b,round(a/b,6))
``` | instruction | 0 | 36,841 | 5 | 73,682 |
Yes | output | 1 | 36,841 | 5 | 73,683 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which reads two integers a and b, and calculates the following values:
* a ÷ b: d (in integer)
* remainder of a ÷ b: r (in integer)
* a ÷ b: f (in real number)
Constraints
* 1 ≤ a, b ≤ 109
Input
Two integers a and b are given in a line.
Output
Print d, r and f separated by a space in a line. For f, the output should not contain an absolute error greater than 10-5.
Example
Input
3 2
Output
1 1 1.50000
Submitted Solution:
```
a, b = input().split()
a = float(a)
b = float(b)
print(a//b, a%b, a/b)
``` | instruction | 0 | 36,842 | 5 | 73,684 |
No | output | 1 | 36,842 | 5 | 73,685 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which reads two integers a and b, and calculates the following values:
* a ÷ b: d (in integer)
* remainder of a ÷ b: r (in integer)
* a ÷ b: f (in real number)
Constraints
* 1 ≤ a, b ≤ 109
Input
Two integers a and b are given in a line.
Output
Print d, r and f separated by a space in a line. For f, the output should not contain an absolute error greater than 10-5.
Example
Input
3 2
Output
1 1 1.50000
Submitted Solution:
```
a, b = map(int, input().split())
d = a // b
r = a % b
f = a / b
print("{0} {1} {2}".format(d, r, f))
``` | instruction | 0 | 36,843 | 5 | 73,686 |
No | output | 1 | 36,843 | 5 | 73,687 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which reads two integers a and b, and calculates the following values:
* a ÷ b: d (in integer)
* remainder of a ÷ b: r (in integer)
* a ÷ b: f (in real number)
Constraints
* 1 ≤ a, b ≤ 109
Input
Two integers a and b are given in a line.
Output
Print d, r and f separated by a space in a line. For f, the output should not contain an absolute error greater than 10-5.
Example
Input
3 2
Output
1 1 1.50000
Submitted Solution:
```
a,b = (int(x) for x in input().split())
d = a // b
r = a % b
(float) f = a / b
print (str(d) + " " + str(r) + " " + str(f))
``` | instruction | 0 | 36,844 | 5 | 73,688 |
No | output | 1 | 36,844 | 5 | 73,689 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which reads two integers a and b, and calculates the following values:
* a ÷ b: d (in integer)
* remainder of a ÷ b: r (in integer)
* a ÷ b: f (in real number)
Constraints
* 1 ≤ a, b ≤ 109
Input
Two integers a and b are given in a line.
Output
Print d, r and f separated by a space in a line. For f, the output should not contain an absolute error greater than 10-5.
Example
Input
3 2
Output
1 1 1.50000
Submitted Solution:
```
a,b= map(int,input().split())
r = a % b
f = round(a / b)
d = f - r
print(d,r,f)
``` | instruction | 0 | 36,845 | 5 | 73,690 |
No | output | 1 | 36,845 | 5 | 73,691 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given four integers a, b, x and y. Initially, a ≥ x and b ≥ y. You can do the following operation no more than n times:
* Choose either a or b and decrease it by one. However, as a result of this operation, value of a cannot become less than x, and value of b cannot become less than y.
Your task is to find the minimum possible product of a and b (a ⋅ b) you can achieve by applying the given operation no more than n times.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The only line of the test case contains five integers a, b, x, y and n (1 ≤ a, b, x, y, n ≤ 10^9). Additional constraint on the input: a ≥ x and b ≥ y always holds.
Output
For each test case, print one integer: the minimum possible product of a and b (a ⋅ b) you can achieve by applying the given operation no more than n times.
Example
Input
7
10 10 8 5 3
12 8 8 7 2
12343 43 4543 39 123212
1000000000 1000000000 1 1 1
1000000000 1000000000 1 1 1000000000
10 11 2 1 5
10 11 9 1 10
Output
70
77
177177
999999999000000000
999999999
55
10
Note
In the first test case of the example, you need to decrease b three times and obtain 10 ⋅ 7 = 70.
In the second test case of the example, you need to decrease a one time, b one time and obtain 11 ⋅ 7 = 77.
In the sixth test case of the example, you need to decrease a five times and obtain 5 ⋅ 11 = 55.
In the seventh test case of the example, you need to decrease b ten times and obtain 10 ⋅ 1 = 10. | instruction | 0 | 37,012 | 5 | 74,024 |
Tags: brute force, greedy, math
Correct Solution:
```
t = int(input())
for _ in range(t):
a, b , x, y, n = map(int, input().split())
a1, b1, x1, y1, n1 = a, b , x, y, n
t = min(n, a - x)
a -= t
n -= t
if n > 0:
b -= min(n, b - y)
t1 = min(n1, b1 - y1)
b1 -= t1
n1 -= t1
if n1 > 0:
a1 -= min(n1, a1 - x1)
print(min(a1 * b1, a * b))
``` | output | 1 | 37,012 | 5 | 74,025 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given four integers a, b, x and y. Initially, a ≥ x and b ≥ y. You can do the following operation no more than n times:
* Choose either a or b and decrease it by one. However, as a result of this operation, value of a cannot become less than x, and value of b cannot become less than y.
Your task is to find the minimum possible product of a and b (a ⋅ b) you can achieve by applying the given operation no more than n times.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The only line of the test case contains five integers a, b, x, y and n (1 ≤ a, b, x, y, n ≤ 10^9). Additional constraint on the input: a ≥ x and b ≥ y always holds.
Output
For each test case, print one integer: the minimum possible product of a and b (a ⋅ b) you can achieve by applying the given operation no more than n times.
Example
Input
7
10 10 8 5 3
12 8 8 7 2
12343 43 4543 39 123212
1000000000 1000000000 1 1 1
1000000000 1000000000 1 1 1000000000
10 11 2 1 5
10 11 9 1 10
Output
70
77
177177
999999999000000000
999999999
55
10
Note
In the first test case of the example, you need to decrease b three times and obtain 10 ⋅ 7 = 70.
In the second test case of the example, you need to decrease a one time, b one time and obtain 11 ⋅ 7 = 77.
In the sixth test case of the example, you need to decrease a five times and obtain 5 ⋅ 11 = 55.
In the seventh test case of the example, you need to decrease b ten times and obtain 10 ⋅ 1 = 10. | instruction | 0 | 37,013 | 5 | 74,026 |
Tags: brute force, greedy, math
Correct Solution:
```
import sys
def rs(): return sys.stdin.readline().rstrip()
def ri(): return int(sys.stdin.readline())
def ria(): return list(map(int, sys.stdin.readline().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
import math
from collections import defaultdict
for _ in range(ri()):
a,b,x,y,n=ria()
p=a-x
q=b-y
p1,q1=p,q
n1=n
a1,b1=a,b
m=min(n,p)
n-=m
a-=m
m=min(n,q)
n-=m
b-=m
d=a*b
m=min(q,n1)
b1-=m
n1-=m
m=min(p,n1)
n1-=m
a1-=m
wi(min(a*b,a1*b1))
``` | output | 1 | 37,013 | 5 | 74,027 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given four integers a, b, x and y. Initially, a ≥ x and b ≥ y. You can do the following operation no more than n times:
* Choose either a or b and decrease it by one. However, as a result of this operation, value of a cannot become less than x, and value of b cannot become less than y.
Your task is to find the minimum possible product of a and b (a ⋅ b) you can achieve by applying the given operation no more than n times.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The only line of the test case contains five integers a, b, x, y and n (1 ≤ a, b, x, y, n ≤ 10^9). Additional constraint on the input: a ≥ x and b ≥ y always holds.
Output
For each test case, print one integer: the minimum possible product of a and b (a ⋅ b) you can achieve by applying the given operation no more than n times.
Example
Input
7
10 10 8 5 3
12 8 8 7 2
12343 43 4543 39 123212
1000000000 1000000000 1 1 1
1000000000 1000000000 1 1 1000000000
10 11 2 1 5
10 11 9 1 10
Output
70
77
177177
999999999000000000
999999999
55
10
Note
In the first test case of the example, you need to decrease b three times and obtain 10 ⋅ 7 = 70.
In the second test case of the example, you need to decrease a one time, b one time and obtain 11 ⋅ 7 = 77.
In the sixth test case of the example, you need to decrease a five times and obtain 5 ⋅ 11 = 55.
In the seventh test case of the example, you need to decrease b ten times and obtain 10 ⋅ 1 = 10. | instruction | 0 | 37,014 | 5 | 74,028 |
Tags: brute force, greedy, math
Correct Solution:
```
import collections
from functools import lru_cache
import bisect
INF = float("inf")
NEG_INF = float("inf")
ZERO = 0
ONE = 1
def read():
return input().strip()
def readInt():
return int(input().strip())
def readList():
return list(map(int, input().strip().split()))
def solve(a, b, x, y, n):
def go(a, b, x, y, n):
newA = max(x, a-n)
n -= (a - newA)
newB = max(y, b-n)
return newA * newB
return min(go(a, b, x, y, n), go(b, a, y, x, n))
t = readInt()
for i in range(t):
a, b, x, y, n = readList()
print(solve(a, b, x, y, n))
``` | output | 1 | 37,014 | 5 | 74,029 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given four integers a, b, x and y. Initially, a ≥ x and b ≥ y. You can do the following operation no more than n times:
* Choose either a or b and decrease it by one. However, as a result of this operation, value of a cannot become less than x, and value of b cannot become less than y.
Your task is to find the minimum possible product of a and b (a ⋅ b) you can achieve by applying the given operation no more than n times.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The only line of the test case contains five integers a, b, x, y and n (1 ≤ a, b, x, y, n ≤ 10^9). Additional constraint on the input: a ≥ x and b ≥ y always holds.
Output
For each test case, print one integer: the minimum possible product of a and b (a ⋅ b) you can achieve by applying the given operation no more than n times.
Example
Input
7
10 10 8 5 3
12 8 8 7 2
12343 43 4543 39 123212
1000000000 1000000000 1 1 1
1000000000 1000000000 1 1 1000000000
10 11 2 1 5
10 11 9 1 10
Output
70
77
177177
999999999000000000
999999999
55
10
Note
In the first test case of the example, you need to decrease b three times and obtain 10 ⋅ 7 = 70.
In the second test case of the example, you need to decrease a one time, b one time and obtain 11 ⋅ 7 = 77.
In the sixth test case of the example, you need to decrease a five times and obtain 5 ⋅ 11 = 55.
In the seventh test case of the example, you need to decrease b ten times and obtain 10 ⋅ 1 = 10. | instruction | 0 | 37,015 | 5 | 74,030 |
Tags: brute force, greedy, math
Correct Solution:
```
for _ in range(int(input())):
a, b, x, y, n = map(int, input().split())
ans = 10 ** 18 + 1
for i in range(2):
A = min(n, a - x)
B = min((n - A), (b - y))
ans = min(ans, (a - A) * (b - B))
a, b = b, a
x, y = y, x
print(ans)
``` | output | 1 | 37,015 | 5 | 74,031 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given four integers a, b, x and y. Initially, a ≥ x and b ≥ y. You can do the following operation no more than n times:
* Choose either a or b and decrease it by one. However, as a result of this operation, value of a cannot become less than x, and value of b cannot become less than y.
Your task is to find the minimum possible product of a and b (a ⋅ b) you can achieve by applying the given operation no more than n times.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The only line of the test case contains five integers a, b, x, y and n (1 ≤ a, b, x, y, n ≤ 10^9). Additional constraint on the input: a ≥ x and b ≥ y always holds.
Output
For each test case, print one integer: the minimum possible product of a and b (a ⋅ b) you can achieve by applying the given operation no more than n times.
Example
Input
7
10 10 8 5 3
12 8 8 7 2
12343 43 4543 39 123212
1000000000 1000000000 1 1 1
1000000000 1000000000 1 1 1000000000
10 11 2 1 5
10 11 9 1 10
Output
70
77
177177
999999999000000000
999999999
55
10
Note
In the first test case of the example, you need to decrease b three times and obtain 10 ⋅ 7 = 70.
In the second test case of the example, you need to decrease a one time, b one time and obtain 11 ⋅ 7 = 77.
In the sixth test case of the example, you need to decrease a five times and obtain 5 ⋅ 11 = 55.
In the seventh test case of the example, you need to decrease b ten times and obtain 10 ⋅ 1 = 10. | instruction | 0 | 37,016 | 5 | 74,032 |
Tags: brute force, greedy, math
Correct Solution:
```
import sys
si = sys.stdin.readline
def main():
t = int(si())
while t:
t -= 1
a, b, x, y, n = [int(e) for e in si().split()]
smalla, smallb = a-min(a-x, n), b-min(b-y, n)
if smallb < smalla:
a, b = b, a
x, y = y, x
if a-x < n:
n -= a-x
a = x
if b-y < n:
b = y
else:
b -= n
else:
a -= n
print(a*b)
if __name__ == '__main__':
main()
``` | output | 1 | 37,016 | 5 | 74,033 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given four integers a, b, x and y. Initially, a ≥ x and b ≥ y. You can do the following operation no more than n times:
* Choose either a or b and decrease it by one. However, as a result of this operation, value of a cannot become less than x, and value of b cannot become less than y.
Your task is to find the minimum possible product of a and b (a ⋅ b) you can achieve by applying the given operation no more than n times.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The only line of the test case contains five integers a, b, x, y and n (1 ≤ a, b, x, y, n ≤ 10^9). Additional constraint on the input: a ≥ x and b ≥ y always holds.
Output
For each test case, print one integer: the minimum possible product of a and b (a ⋅ b) you can achieve by applying the given operation no more than n times.
Example
Input
7
10 10 8 5 3
12 8 8 7 2
12343 43 4543 39 123212
1000000000 1000000000 1 1 1
1000000000 1000000000 1 1 1000000000
10 11 2 1 5
10 11 9 1 10
Output
70
77
177177
999999999000000000
999999999
55
10
Note
In the first test case of the example, you need to decrease b three times and obtain 10 ⋅ 7 = 70.
In the second test case of the example, you need to decrease a one time, b one time and obtain 11 ⋅ 7 = 77.
In the sixth test case of the example, you need to decrease a five times and obtain 5 ⋅ 11 = 55.
In the seventh test case of the example, you need to decrease b ten times and obtain 10 ⋅ 1 = 10. | instruction | 0 | 37,018 | 5 | 74,036 |
Tags: brute force, greedy, math
Correct Solution:
```
#OM GANESHAY NAMH
#GANPATI BAPPA MORYA
import math,queue,heapq
import sys
sys.setrecursionlimit(10**6)
fastinput=sys.stdin.readline
fastout=sys.stdout.write
t=int(fastinput())
while t:
t-=1
a,b,x,y,n=map(int,fastinput().split())
if max(a-n,x)<max(b-n,y):
req=a-x
if req>n:
print((a-n)*b)
else:
print(x*(max(y,b-(n-req))))
else:
req=b-y
if req>n:
print((b-n)*a)
else:
print(y*(max(x,a-(n-req))))
``` | output | 1 | 37,018 | 5 | 74,037 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given four integers a, b, x and y. Initially, a ≥ x and b ≥ y. You can do the following operation no more than n times:
* Choose either a or b and decrease it by one. However, as a result of this operation, value of a cannot become less than x, and value of b cannot become less than y.
Your task is to find the minimum possible product of a and b (a ⋅ b) you can achieve by applying the given operation no more than n times.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The only line of the test case contains five integers a, b, x, y and n (1 ≤ a, b, x, y, n ≤ 10^9). Additional constraint on the input: a ≥ x and b ≥ y always holds.
Output
For each test case, print one integer: the minimum possible product of a and b (a ⋅ b) you can achieve by applying the given operation no more than n times.
Example
Input
7
10 10 8 5 3
12 8 8 7 2
12343 43 4543 39 123212
1000000000 1000000000 1 1 1
1000000000 1000000000 1 1 1000000000
10 11 2 1 5
10 11 9 1 10
Output
70
77
177177
999999999000000000
999999999
55
10
Note
In the first test case of the example, you need to decrease b three times and obtain 10 ⋅ 7 = 70.
In the second test case of the example, you need to decrease a one time, b one time and obtain 11 ⋅ 7 = 77.
In the sixth test case of the example, you need to decrease a five times and obtain 5 ⋅ 11 = 55.
In the seventh test case of the example, you need to decrease b ten times and obtain 10 ⋅ 1 = 10. | instruction | 0 | 37,019 | 5 | 74,038 |
Tags: brute force, greedy, math
Correct Solution:
```
def solve(a, b, x, y, n):
c, d, m = a, b, n
temp = d - y
d -= min(m, temp)
m -= min(m, temp)
c -= m
if c < x:
c = x
temp = a - x
a -= min(n, temp)
n -= min(n, temp)
b -= n
if b < y:
b = y
print(min(a*b, c*d))
t = int(input())
for _ in range(t):
inp = input().split()
a, b, x, y, n = int(inp[0]), int(inp[1]), int(inp[2]), int(inp[3]), int(inp[4])
solve(a, b, x, y, n)
``` | output | 1 | 37,019 | 5 | 74,039 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given four integers a, b, x and y. Initially, a ≥ x and b ≥ y. You can do the following operation no more than n times:
* Choose either a or b and decrease it by one. However, as a result of this operation, value of a cannot become less than x, and value of b cannot become less than y.
Your task is to find the minimum possible product of a and b (a ⋅ b) you can achieve by applying the given operation no more than n times.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The only line of the test case contains five integers a, b, x, y and n (1 ≤ a, b, x, y, n ≤ 10^9). Additional constraint on the input: a ≥ x and b ≥ y always holds.
Output
For each test case, print one integer: the minimum possible product of a and b (a ⋅ b) you can achieve by applying the given operation no more than n times.
Example
Input
7
10 10 8 5 3
12 8 8 7 2
12343 43 4543 39 123212
1000000000 1000000000 1 1 1
1000000000 1000000000 1 1 1000000000
10 11 2 1 5
10 11 9 1 10
Output
70
77
177177
999999999000000000
999999999
55
10
Note
In the first test case of the example, you need to decrease b three times and obtain 10 ⋅ 7 = 70.
In the second test case of the example, you need to decrease a one time, b one time and obtain 11 ⋅ 7 = 77.
In the sixth test case of the example, you need to decrease a five times and obtain 5 ⋅ 11 = 55.
In the seventh test case of the example, you need to decrease b ten times and obtain 10 ⋅ 1 = 10.
Submitted Solution:
```
import sys
import bisect as bi
import math
from collections import defaultdict as dd
import heapq
import itertools
input=sys.stdin.readline
##from random import randint
##sys.setrecursionlimit(10**7)
mo=10**9+7
def cin():
return map(int,sin().split())
def ain():
return list(map(int,sin().split()))
def sin():
return input()
def inin():
return int(input())
for _ in range(inin()):
a,b,x,y,n=cin()
p,q,r,s,t=a,b,x,y,n
p1,p2=0,0
if(a>b):
a,b=b,a
x,y=y,x
elif(a==b):
if(x>y):
x,y=y,x
d1=a-x
d2=b-y
if(d1>=n):
a=a-n
n=0
else:
a=x
n-=d1
if(n):
if(d2>=n):
b=b-n
n=0
else:
b=y
n=d2
p1=a*b
a,b,x,y,n=p,q,r,s,t
if(a<b):
a,b=b,a
x,y=y,x
elif(a==b):
if(x>y):
x,y=y,x
d1=a-x
d2=b-y
if(d1>=n):
a=a-n
n=0
else:
a=x
n-=d1
if(n):
if(d2>=n):
b=b-n
n=0
else:
b=y
n=d2
p2=a*b
print(min(p1,p2))
## mat=[];l=[0]*n
## for i in range(n):
## mat.append(ain())
## c=1;ans=0
## for i in mat[0]:
## if(i!=c):
## l[c-1]=1
## c+=1
## if(1 in l):
## for i in range(1,n):
## if(l[i]==1 and l[i]!=l[i-1]):
## ans+=1
## print(ans*2)
## else:
## print(0)
##def msb(n):n|=n>>1;n|=n>>2;n|=n>>4;n|=n>>8;n|=n>>16;n|=n>>32;n|=n>>64;return n-(n>>1) #2 ki power
##def pref(a,n,f):
## pre=[0]*n
## if(f==0): ##from beginning
## pre[0]=a[0]
## for i in range(1,n):
## pre[i]=a[i]+pre[i-1]
## else: ##from end
## pre[-1]=a[-1]
## for i in range(n-2,-1,-1):
## pre[i]=pre[i+1]+a[i]
## return pre
##maxint=10**24
##def kadane(a,size):
## max_so_far = -maxint - 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
##def power(x, y):
## if(y == 0):return 1
## temp = power(x, int(y / 2))%mo
## if (y % 2 == 0):return (temp * temp)%mo
## else:
## if(y > 0):return (x * temp * temp)%mo
## else:return ((temp * temp)//x )%mo
``` | instruction | 0 | 37,020 | 5 | 74,040 |
Yes | output | 1 | 37,020 | 5 | 74,041 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given four integers a, b, x and y. Initially, a ≥ x and b ≥ y. You can do the following operation no more than n times:
* Choose either a or b and decrease it by one. However, as a result of this operation, value of a cannot become less than x, and value of b cannot become less than y.
Your task is to find the minimum possible product of a and b (a ⋅ b) you can achieve by applying the given operation no more than n times.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The only line of the test case contains five integers a, b, x, y and n (1 ≤ a, b, x, y, n ≤ 10^9). Additional constraint on the input: a ≥ x and b ≥ y always holds.
Output
For each test case, print one integer: the minimum possible product of a and b (a ⋅ b) you can achieve by applying the given operation no more than n times.
Example
Input
7
10 10 8 5 3
12 8 8 7 2
12343 43 4543 39 123212
1000000000 1000000000 1 1 1
1000000000 1000000000 1 1 1000000000
10 11 2 1 5
10 11 9 1 10
Output
70
77
177177
999999999000000000
999999999
55
10
Note
In the first test case of the example, you need to decrease b three times and obtain 10 ⋅ 7 = 70.
In the second test case of the example, you need to decrease a one time, b one time and obtain 11 ⋅ 7 = 77.
In the sixth test case of the example, you need to decrease a five times and obtain 5 ⋅ 11 = 55.
In the seventh test case of the example, you need to decrease b ten times and obtain 10 ⋅ 1 = 10.
Submitted Solution:
```
from sys import stdin,stdout
from math import gcd,sqrt,factorial,pi
from collections import deque,defaultdict
input=stdin.readline
R=lambda:map(int,input().split())
I=lambda:int(input())
S=lambda:input().rstrip('\n')
L=lambda:list(R())
P=lambda x:stdout.write(x)
lcm=lambda x,y:(x*y)//gcd(x,y)
hg=lambda x,y:((y+x-1)//x)*x
pw=lambda x:1 if x==1 else 1+pw(x//2)
chk=lambda x:chk(x//2) if not x%2 else True if x==1 else False
sm=lambda x:(x**2+x)//2
N=10**9+7
def pro(a,b,x,y,n):
m=max(x,a-n)
n-=a-m
a=m
b=max(y,b-n)
return a*b
for _ in range(I()):
a,b,x,y,n=R()
print(min(pro(a,b,x,y,n),pro(b,a,y,x,n)))
``` | instruction | 0 | 37,021 | 5 | 74,042 |
Yes | output | 1 | 37,021 | 5 | 74,043 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given four integers a, b, x and y. Initially, a ≥ x and b ≥ y. You can do the following operation no more than n times:
* Choose either a or b and decrease it by one. However, as a result of this operation, value of a cannot become less than x, and value of b cannot become less than y.
Your task is to find the minimum possible product of a and b (a ⋅ b) you can achieve by applying the given operation no more than n times.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The only line of the test case contains five integers a, b, x, y and n (1 ≤ a, b, x, y, n ≤ 10^9). Additional constraint on the input: a ≥ x and b ≥ y always holds.
Output
For each test case, print one integer: the minimum possible product of a and b (a ⋅ b) you can achieve by applying the given operation no more than n times.
Example
Input
7
10 10 8 5 3
12 8 8 7 2
12343 43 4543 39 123212
1000000000 1000000000 1 1 1
1000000000 1000000000 1 1 1000000000
10 11 2 1 5
10 11 9 1 10
Output
70
77
177177
999999999000000000
999999999
55
10
Note
In the first test case of the example, you need to decrease b three times and obtain 10 ⋅ 7 = 70.
In the second test case of the example, you need to decrease a one time, b one time and obtain 11 ⋅ 7 = 77.
In the sixth test case of the example, you need to decrease a five times and obtain 5 ⋅ 11 = 55.
In the seventh test case of the example, you need to decrease b ten times and obtain 10 ⋅ 1 = 10.
Submitted Solution:
```
def find(a,b,x,y,n):
t1,t2,t3 = a,b,n
s1 = a-x
et1 = min(s1,n)
a = a-et1
n = n-et1
s2 = b-y
et2 = min(s2,n)
b = b-et2
n = n-et2
answer = a*b
temp = min(t2-y,t3)
t2=t2-temp
t3=t3-temp
temp = min(t1-x,t3)
t1 = t1-temp
t3 = t3-temp
answer = min(t1*t2,answer)
return answer
t = int(input())
for h in range(0,t):
a,b,x,y,n = map(int,input().split())
answer = find(a,b,x,y,n)
print(answer)
``` | instruction | 0 | 37,022 | 5 | 74,044 |
Yes | output | 1 | 37,022 | 5 | 74,045 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given four integers a, b, x and y. Initially, a ≥ x and b ≥ y. You can do the following operation no more than n times:
* Choose either a or b and decrease it by one. However, as a result of this operation, value of a cannot become less than x, and value of b cannot become less than y.
Your task is to find the minimum possible product of a and b (a ⋅ b) you can achieve by applying the given operation no more than n times.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The only line of the test case contains five integers a, b, x, y and n (1 ≤ a, b, x, y, n ≤ 10^9). Additional constraint on the input: a ≥ x and b ≥ y always holds.
Output
For each test case, print one integer: the minimum possible product of a and b (a ⋅ b) you can achieve by applying the given operation no more than n times.
Example
Input
7
10 10 8 5 3
12 8 8 7 2
12343 43 4543 39 123212
1000000000 1000000000 1 1 1
1000000000 1000000000 1 1 1000000000
10 11 2 1 5
10 11 9 1 10
Output
70
77
177177
999999999000000000
999999999
55
10
Note
In the first test case of the example, you need to decrease b three times and obtain 10 ⋅ 7 = 70.
In the second test case of the example, you need to decrease a one time, b one time and obtain 11 ⋅ 7 = 77.
In the sixth test case of the example, you need to decrease a five times and obtain 5 ⋅ 11 = 55.
In the seventh test case of the example, you need to decrease b ten times and obtain 10 ⋅ 1 = 10.
Submitted Solution:
```
for test in range(int(input())):
a, b, x, y, n = map(int, input().split())
if max(x, a - x) + max(y, b - y) >= n:
m1 = max(a-n,x)
m2 = max(b-n,y)
#print(m1, m2)
if m1<=m2:
t = min(a-x,n)
a = a-t
n = n - t
if n>0:
b = b-min(n,b-y)
else:
t = min(b-y,n)
b = b- t
n = n - t
if n>0:
a =a - min(a-x,n)
#print(a,b)
print(a*b)
else:
print(x * y)
``` | instruction | 0 | 37,023 | 5 | 74,046 |
Yes | output | 1 | 37,023 | 5 | 74,047 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given four integers a, b, x and y. Initially, a ≥ x and b ≥ y. You can do the following operation no more than n times:
* Choose either a or b and decrease it by one. However, as a result of this operation, value of a cannot become less than x, and value of b cannot become less than y.
Your task is to find the minimum possible product of a and b (a ⋅ b) you can achieve by applying the given operation no more than n times.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The only line of the test case contains five integers a, b, x, y and n (1 ≤ a, b, x, y, n ≤ 10^9). Additional constraint on the input: a ≥ x and b ≥ y always holds.
Output
For each test case, print one integer: the minimum possible product of a and b (a ⋅ b) you can achieve by applying the given operation no more than n times.
Example
Input
7
10 10 8 5 3
12 8 8 7 2
12343 43 4543 39 123212
1000000000 1000000000 1 1 1
1000000000 1000000000 1 1 1000000000
10 11 2 1 5
10 11 9 1 10
Output
70
77
177177
999999999000000000
999999999
55
10
Note
In the first test case of the example, you need to decrease b three times and obtain 10 ⋅ 7 = 70.
In the second test case of the example, you need to decrease a one time, b one time and obtain 11 ⋅ 7 = 77.
In the sixth test case of the example, you need to decrease a five times and obtain 5 ⋅ 11 = 55.
In the seventh test case of the example, you need to decrease b ten times and obtain 10 ⋅ 1 = 10.
Submitted Solution:
```
t = int(input())
while t:
t -= 1
ans = 1e18
a, b, x, y, n = map(int, input().split())
for i in range(2):
aa = min(n, a - x)
bb = min(n - aa, b - y)
ans = min(ans, (a - aa) * (b - bb))
aa, bb = bb, aa
x, y = y, x
print(ans)
``` | instruction | 0 | 37,024 | 5 | 74,048 |
No | output | 1 | 37,024 | 5 | 74,049 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given four integers a, b, x and y. Initially, a ≥ x and b ≥ y. You can do the following operation no more than n times:
* Choose either a or b and decrease it by one. However, as a result of this operation, value of a cannot become less than x, and value of b cannot become less than y.
Your task is to find the minimum possible product of a and b (a ⋅ b) you can achieve by applying the given operation no more than n times.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The only line of the test case contains five integers a, b, x, y and n (1 ≤ a, b, x, y, n ≤ 10^9). Additional constraint on the input: a ≥ x and b ≥ y always holds.
Output
For each test case, print one integer: the minimum possible product of a and b (a ⋅ b) you can achieve by applying the given operation no more than n times.
Example
Input
7
10 10 8 5 3
12 8 8 7 2
12343 43 4543 39 123212
1000000000 1000000000 1 1 1
1000000000 1000000000 1 1 1000000000
10 11 2 1 5
10 11 9 1 10
Output
70
77
177177
999999999000000000
999999999
55
10
Note
In the first test case of the example, you need to decrease b three times and obtain 10 ⋅ 7 = 70.
In the second test case of the example, you need to decrease a one time, b one time and obtain 11 ⋅ 7 = 77.
In the sixth test case of the example, you need to decrease a five times and obtain 5 ⋅ 11 = 55.
In the seventh test case of the example, you need to decrease b ten times and obtain 10 ⋅ 1 = 10.
Submitted Solution:
```
for _ in range(int(input())):
a,b,x,y,n = map(float,input().split())
if(a<b):
t=a;
a=b;
b=t;
j = (b-y) if n>(b-y) else n
n-=j;
i = (a-x) if n>(a-x) else n
n-=i;
print((a-i)*(b-j))
``` | instruction | 0 | 37,025 | 5 | 74,050 |
No | output | 1 | 37,025 | 5 | 74,051 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given four integers a, b, x and y. Initially, a ≥ x and b ≥ y. You can do the following operation no more than n times:
* Choose either a or b and decrease it by one. However, as a result of this operation, value of a cannot become less than x, and value of b cannot become less than y.
Your task is to find the minimum possible product of a and b (a ⋅ b) you can achieve by applying the given operation no more than n times.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The only line of the test case contains five integers a, b, x, y and n (1 ≤ a, b, x, y, n ≤ 10^9). Additional constraint on the input: a ≥ x and b ≥ y always holds.
Output
For each test case, print one integer: the minimum possible product of a and b (a ⋅ b) you can achieve by applying the given operation no more than n times.
Example
Input
7
10 10 8 5 3
12 8 8 7 2
12343 43 4543 39 123212
1000000000 1000000000 1 1 1
1000000000 1000000000 1 1 1000000000
10 11 2 1 5
10 11 9 1 10
Output
70
77
177177
999999999000000000
999999999
55
10
Note
In the first test case of the example, you need to decrease b three times and obtain 10 ⋅ 7 = 70.
In the second test case of the example, you need to decrease a one time, b one time and obtain 11 ⋅ 7 = 77.
In the sixth test case of the example, you need to decrease a five times and obtain 5 ⋅ 11 = 55.
In the seventh test case of the example, you need to decrease b ten times and obtain 10 ⋅ 1 = 10.
Submitted Solution:
```
from math import ceil, floor, gcd, fabs, factorial, fmod, sqrt, inf
def sp(): return map(int, input().split())
t=int(input())
# t=1
for _ in range(t):
a,b,x,y,n=sp()
get_max_ans=a*b
get_max_ans_2=get_max_ans
if(a<b):
a,b=b,a
x,y=y,x
a1=a;a2=b;
first_dif = b-y
get_max_ans-= a*min(n,b-y)
n-=(b-y)
if(n>0):
b=y;
get_max_ans-= b*(min(n, a-x))
# print("YES", end=" ")
a=a1;b=a2;
get_max_ans_2-= b*min(n,a-x)
n-=(a-x)
if(n>0):
a=x;
get_max_ans_2-= a*(min(n, b-y))
print(min(get_max_ans, get_max_ans_2))
``` | instruction | 0 | 37,026 | 5 | 74,052 |
No | output | 1 | 37,026 | 5 | 74,053 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given four integers a, b, x and y. Initially, a ≥ x and b ≥ y. You can do the following operation no more than n times:
* Choose either a or b and decrease it by one. However, as a result of this operation, value of a cannot become less than x, and value of b cannot become less than y.
Your task is to find the minimum possible product of a and b (a ⋅ b) you can achieve by applying the given operation no more than n times.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The only line of the test case contains five integers a, b, x, y and n (1 ≤ a, b, x, y, n ≤ 10^9). Additional constraint on the input: a ≥ x and b ≥ y always holds.
Output
For each test case, print one integer: the minimum possible product of a and b (a ⋅ b) you can achieve by applying the given operation no more than n times.
Example
Input
7
10 10 8 5 3
12 8 8 7 2
12343 43 4543 39 123212
1000000000 1000000000 1 1 1
1000000000 1000000000 1 1 1000000000
10 11 2 1 5
10 11 9 1 10
Output
70
77
177177
999999999000000000
999999999
55
10
Note
In the first test case of the example, you need to decrease b three times and obtain 10 ⋅ 7 = 70.
In the second test case of the example, you need to decrease a one time, b one time and obtain 11 ⋅ 7 = 77.
In the sixth test case of the example, you need to decrease a five times and obtain 5 ⋅ 11 = 55.
In the seventh test case of the example, you need to decrease b ten times and obtain 10 ⋅ 1 = 10.
Submitted Solution:
```
def manipluateA(a,n,x):
lowest_value_a = a- n
n = 0
if lowest_value_a <0:
n += -lowest_value_a
a = 0
lowest_value_a = a
if lowest_value_a < x:
n += x - lowest_value_a
a = x
else:
a = lowest_value_a
return (a,n,x)
def manipulateB(b,n,y):
lowest_value_b = b - n
n = 0
if lowest_value_b < 0:
n += lowest_value_b
b = 0
lowest_value_b = 0
if lowest_value_b < y:
#print('true')
#print(lowest_value_b,y)
n += y - lowest_value_b
b = y
else:
b = lowest_value_b
return (b,n,y)
t= int(input())
for _ in range(t):
(a,b,x,y,n) = map(int,input().strip().split())# a >b
(tempa,tempb,tempx,tempy,tempn) = (a,b,x,y,n)
(a,n,x) = manipluateA(a,n,x)
#print(n,'n left')
(b,n,x) = manipulateB(b,n,y)
#print(n,'n left')
#print(a,b,'a b case1')
case1 = a * b
(a,b,x,y,n) = (tempa,tempb,tempx,tempy,tempn)
#print((a,b,x,y,n),'debug')
#print(n,'n before pass')
(b,n,x) = manipulateB(b,n,y)
#print(n,'n left')
(a,n,x) = manipluateA(a,n,x)
#print(n,'n left')
case2 = a*b
#print(a,b,'a b case2')
#print(case1,case2)
print(min(case1,case2))
``` | instruction | 0 | 37,027 | 5 | 74,054 |
No | output | 1 | 37,027 | 5 | 74,055 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In this problem at each moment you have a set of intervals. You can move from interval (a, b) from our set to interval (c, d) from our set if and only if c < a < d or c < b < d. Also there is a path from interval I1 from our set to interval I2 from our set if there is a sequence of successive moves starting from I1 so that we can reach I2.
Your program should handle the queries of the following two types:
1. "1 x y" (x < y) — add the new interval (x, y) to the set of intervals. The length of the new interval is guaranteed to be strictly greater than all the previous intervals.
2. "2 a b" (a ≠ b) — answer the question: is there a path from a-th (one-based) added interval to b-th (one-based) added interval?
Answer all the queries. Note, that initially you have an empty set of intervals.
Input
The first line of the input contains integer n denoting the number of queries, (1 ≤ n ≤ 100). Each of the following lines contains a query as described above. All numbers in the input are integers and don't exceed 109 by their absolute value.
It's guaranteed that all queries are correct.
Output
For each query of the second type print "YES" or "NO" on a separate line depending on the answer.
Examples
Input
5
1 1 5
1 5 11
2 1 2
1 2 9
2 1 2
Output
NO
YES
Submitted Solution:
```
read_line = lambda : map(int,input().split())
n, = read_line()
intervals = []
def dfs(start, end):
visited = []
stack = [intervals[start]]
while stack:
v = stack.pop()
if v not in visited:
visited.append(v)
for i in intervals:
if i[0] < v[0] < i[1] or i[0] < v[1] < i[1]:
stack.extend([i])
if intervals.index(v) == end:
return True
return False
for i in range(n):
option, x, y = read_line()
if option == 1:
intervals.append([x,y])
else:
print ("YES" if dfs(x-1, y-1) else "NO")
``` | instruction | 0 | 37,158 | 5 | 74,316 |
Yes | output | 1 | 37,158 | 5 | 74,317 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In this problem at each moment you have a set of intervals. You can move from interval (a, b) from our set to interval (c, d) from our set if and only if c < a < d or c < b < d. Also there is a path from interval I1 from our set to interval I2 from our set if there is a sequence of successive moves starting from I1 so that we can reach I2.
Your program should handle the queries of the following two types:
1. "1 x y" (x < y) — add the new interval (x, y) to the set of intervals. The length of the new interval is guaranteed to be strictly greater than all the previous intervals.
2. "2 a b" (a ≠ b) — answer the question: is there a path from a-th (one-based) added interval to b-th (one-based) added interval?
Answer all the queries. Note, that initially you have an empty set of intervals.
Input
The first line of the input contains integer n denoting the number of queries, (1 ≤ n ≤ 100). Each of the following lines contains a query as described above. All numbers in the input are integers and don't exceed 109 by their absolute value.
It's guaranteed that all queries are correct.
Output
For each query of the second type print "YES" or "NO" on a separate line depending on the answer.
Examples
Input
5
1 1 5
1 5 11
2 1 2
1 2 9
2 1 2
Output
NO
YES
Submitted Solution:
```
def get_interval():
return tuple(map(int, input().split()))
counter = 1
queries = int(input())
visited_intervals = [0] * queries
intervals = {}
for _ in range (queries):
query = get_interval()
if query[0] == 1:
intervals[counter] = query
counter += 1
elif query[0] == 2:
q = []
q.append(query[0])
while len(q) > 1:
q_interval = q.pop(0)
for index in intervals:
c = intervals[index][0]
d = intervals[index][1]
a = q_interval[0]
b = q_interval[1]
if c < a < d or a < b < d:
if visited_intervals[index] == 0:
visited_intervals[index] = 1
q.append(index)
if visited_intervals[query[2]] == 1:
print ("YES")
else:
print ("NO")
``` | instruction | 0 | 37,160 | 5 | 74,320 |
No | output | 1 | 37,160 | 5 | 74,321 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In this problem at each moment you have a set of intervals. You can move from interval (a, b) from our set to interval (c, d) from our set if and only if c < a < d or c < b < d. Also there is a path from interval I1 from our set to interval I2 from our set if there is a sequence of successive moves starting from I1 so that we can reach I2.
Your program should handle the queries of the following two types:
1. "1 x y" (x < y) — add the new interval (x, y) to the set of intervals. The length of the new interval is guaranteed to be strictly greater than all the previous intervals.
2. "2 a b" (a ≠ b) — answer the question: is there a path from a-th (one-based) added interval to b-th (one-based) added interval?
Answer all the queries. Note, that initially you have an empty set of intervals.
Input
The first line of the input contains integer n denoting the number of queries, (1 ≤ n ≤ 100). Each of the following lines contains a query as described above. All numbers in the input are integers and don't exceed 109 by their absolute value.
It's guaranteed that all queries are correct.
Output
For each query of the second type print "YES" or "NO" on a separate line depending on the answer.
Examples
Input
5
1 1 5
1 5 11
2 1 2
1 2 9
2 1 2
Output
NO
YES
Submitted Solution:
```
n = int(input())
g = []
for i in range(0, n):
t, x, y = map(int,input().split())
if(x > y):
x, y = y, x
if(t == 1):
g.insert(0,[x,y])
elif(t == 2):
a = g[x-1][0]
b = g[x-1][1]
c = g[y-1][0]
d = g[y-1][1]
if(c < a < d or c < b < d):
print("YES")
else:
print("NO")
``` | instruction | 0 | 37,161 | 5 | 74,322 |
No | output | 1 | 37,161 | 5 | 74,323 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sereja has two sequences a and b and number p. Sequence a consists of n integers a1, a2, ..., an. Similarly, sequence b consists of m integers b1, b2, ..., bm. As usual, Sereja studies the sequences he has. Today he wants to find the number of positions q (q + (m - 1)·p ≤ n; q ≥ 1), such that sequence b can be obtained from sequence aq, aq + p, aq + 2p, ..., aq + (m - 1)p by rearranging elements.
Sereja needs to rush to the gym, so he asked to find all the described positions of q.
Input
The first line contains three integers n, m and p (1 ≤ n, m ≤ 2·105, 1 ≤ p ≤ 2·105). The next line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109). The next line contains m integers b1, b2, ..., bm (1 ≤ bi ≤ 109).
Output
In the first line print the number of valid qs. In the second line, print the valid values in the increasing order.
Examples
Input
5 3 1
1 2 3 2 1
1 2 3
Output
2
1 3
Input
6 3 2
1 3 2 2 3 1
1 2 3
Output
2
1 2
Submitted Solution:
```
from sys import stdin, stdout
from math import log, sqrt
lines = stdin.readlines()
n = int(lines[0].split()[0])
m = int(lines[0].split()[1])
p = int(lines[0].split()[2])
a = [int(x) for x in lines[1].split()]
b = [int(x) for x in lines[2].split()]
hash_map = {}
def hash_elem(elem):
if hash_map.get(elem, -1) == -1:
elem = int(elem * 1662634645 + 32544235)
elem = int((elem >> 13) + (elem << 19))
hash_map[elem] = int(elem * 361352451)
# hash_map[elem] = int(342153534 + elem + (elem >> 5) + (elem >> 13) + (elem << 17))
# if elem < 1000:
# hash_map[elem] = elem//2 + elem * 1134234546677 + int(elem/3) + int(elem**2) + elem<<2 + elem>>7 + int(log(elem)) + int(sqrt(elem)) + elem&213213 + elem^324234211323 + elem|21319423094023
# else:
# hash_map[elem] = 3 + elem^34
return elem + hash_map[elem]
c = [hash_elem(elem) for elem in a]
c_new = [sum([c[q + p * i] for i in range(m)]) for q in range(min(p, max(0, n - (m - 1) * p)))]
for q in range(p, n - (m - 1) * p):
prev = c_new[q - p]
# print(len(c_new)-1, q, q + p*(m-1))
c_new.append(prev - c[q - p] + c[q + p * (m - 1)])
b_check = sum([hash_elem(elem) for elem in b])
ans1 = 0
ans = []
for q in range(n - (m - 1) * p):
c_check = c_new[q]
if b_check != c_check:
continue
else:
ans1 += 1
ans.append(q + 1)
print(ans1)
stdout.write(' '.join([str(x) for x in ans]))
``` | instruction | 0 | 37,165 | 5 | 74,330 |
No | output | 1 | 37,165 | 5 | 74,331 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sereja has two sequences a and b and number p. Sequence a consists of n integers a1, a2, ..., an. Similarly, sequence b consists of m integers b1, b2, ..., bm. As usual, Sereja studies the sequences he has. Today he wants to find the number of positions q (q + (m - 1)·p ≤ n; q ≥ 1), such that sequence b can be obtained from sequence aq, aq + p, aq + 2p, ..., aq + (m - 1)p by rearranging elements.
Sereja needs to rush to the gym, so he asked to find all the described positions of q.
Input
The first line contains three integers n, m and p (1 ≤ n, m ≤ 2·105, 1 ≤ p ≤ 2·105). The next line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109). The next line contains m integers b1, b2, ..., bm (1 ≤ bi ≤ 109).
Output
In the first line print the number of valid qs. In the second line, print the valid values in the increasing order.
Examples
Input
5 3 1
1 2 3 2 1
1 2 3
Output
2
1 3
Input
6 3 2
1 3 2 2 3 1
1 2 3
Output
2
1 2
Submitted Solution:
```
from sys import stdin, stdout
from math import log, sqrt
lines = stdin.readlines()
n = int(lines[0].split()[0])
m = int(lines[0].split()[1])
p = int(lines[0].split()[2])
a = [int(x) for x in lines[1].split()]
b = [int(x) for x in lines[2].split()]
hash_map = {}
def hash_elem(elem):
if hash_map.get(elem, -1) == -1:
# elem = int(elem * 1662634645)
# elem = int((elem >> 13) + (elem << 19))
# hash_map[elem] = int(elem * 361352451)
hash_map[elem] = int(342153534 + elem + (elem >> 5) + (elem >> 13) + (elem << 17))
# if elem < 1000:
# hash_map[elem] = elem//2 + elem * 1134234546677 + int(elem/3) + int(elem**2) + elem<<2 + elem>>7 + int(log(elem)) + int(sqrt(elem)) + elem&213213 + elem^324234211323 + elem|21319423094023
# else:
# hash_map[elem] = 3 + elem^34
return elem + hash_map[elem]
c = [hash_elem(elem) for elem in a]
c_new = [sum([c[q + p * i] for i in range(m)]) for q in range(min(p, max(0, n - (m - 1) * p)))]
for q in range(p, n - (m - 1) * p):
prev = c_new[q - p]
# print(len(c_new)-1, q, q + p*(m-1))
c_new.append(prev - c[q - p] + c[q + p * (m - 1)])
b_check = sum([hash_elem(elem) for elem in b])
ans1 = 0
ans = []
for q in range(n - (m - 1) * p):
c_check = c_new[q]
if b_check != c_check:
continue
else:
ans1 += 1
ans.append(q + 1)
print(ans1)
stdout.write(' '.join([str(x) for x in ans]))
``` | instruction | 0 | 37,167 | 5 | 74,334 |
No | output | 1 | 37,167 | 5 | 74,335 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sereja has two sequences a and b and number p. Sequence a consists of n integers a1, a2, ..., an. Similarly, sequence b consists of m integers b1, b2, ..., bm. As usual, Sereja studies the sequences he has. Today he wants to find the number of positions q (q + (m - 1)·p ≤ n; q ≥ 1), such that sequence b can be obtained from sequence aq, aq + p, aq + 2p, ..., aq + (m - 1)p by rearranging elements.
Sereja needs to rush to the gym, so he asked to find all the described positions of q.
Input
The first line contains three integers n, m and p (1 ≤ n, m ≤ 2·105, 1 ≤ p ≤ 2·105). The next line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109). The next line contains m integers b1, b2, ..., bm (1 ≤ bi ≤ 109).
Output
In the first line print the number of valid qs. In the second line, print the valid values in the increasing order.
Examples
Input
5 3 1
1 2 3 2 1
1 2 3
Output
2
1 3
Input
6 3 2
1 3 2 2 3 1
1 2 3
Output
2
1 2
Submitted Solution:
```
from sys import stdin, stdout
lines = stdin.readlines()
n = int(lines[0].split()[0])
m = int(lines[0].split()[1])
p = int(lines[0].split()[2])
a = [int(x) for x in lines[1].split()]
b = [int(x) for x in lines[2].split()]
def hash_sum(a):
ret = 0
for elem in a:
ret += elem + elem//2 + elem * 117
return ret
b_check = hash_sum(b)
ans1 = 0
ans = []
for q in range(n - (m-1)*p):
cand = [a[q + p*i] for i in range(m)]
c_check = hash_sum(cand)
if b_check != c_check:
continue
else:
ans1 += 1
ans.append(q+1)
print(ans1)
stdout.write(' '.join([str(x) for x in ans]))
``` | instruction | 0 | 37,168 | 5 | 74,336 |
No | output | 1 | 37,168 | 5 | 74,337 |
Provide a correct Python 3 solution for this coding contest problem.
Print all the integers that satisfies the following in ascending order:
* Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
Constraints
* 1 \leq A \leq B \leq 10^9
* 1 \leq K \leq 100
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B K
Output
Print all the integers that satisfies the condition above in ascending order.
Examples
Input
3 8 2
Output
3
4
7
8
Input
4 8 3
Output
4
5
6
7
8
Input
2 9 100
Output
2
3
4
5
6
7
8
9 | instruction | 0 | 37,572 | 5 | 75,144 |
"Correct Solution:
```
a,b,k=map(int,input().split())
for i in range(a,min(b+1,a+k)):
print(i)
for j in range(max(a+k,b-k+1),b+1):
print(j)
``` | output | 1 | 37,572 | 5 | 75,145 |
Provide a correct Python 3 solution for this coding contest problem.
Print all the integers that satisfies the following in ascending order:
* Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
Constraints
* 1 \leq A \leq B \leq 10^9
* 1 \leq K \leq 100
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B K
Output
Print all the integers that satisfies the condition above in ascending order.
Examples
Input
3 8 2
Output
3
4
7
8
Input
4 8 3
Output
4
5
6
7
8
Input
2 9 100
Output
2
3
4
5
6
7
8
9 | instruction | 0 | 37,573 | 5 | 75,146 |
"Correct Solution:
```
a,b,k=map(int,input().split())
c,d=set(range(a,a+k)),set(range(b-k+1,b+1))
if b-a>=k:
e=c|d
else:
e=c&d
for i in sorted(e):
print(i)
``` | output | 1 | 37,573 | 5 | 75,147 |
Provide a correct Python 3 solution for this coding contest problem.
Print all the integers that satisfies the following in ascending order:
* Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
Constraints
* 1 \leq A \leq B \leq 10^9
* 1 \leq K \leq 100
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B K
Output
Print all the integers that satisfies the condition above in ascending order.
Examples
Input
3 8 2
Output
3
4
7
8
Input
4 8 3
Output
4
5
6
7
8
Input
2 9 100
Output
2
3
4
5
6
7
8
9 | instruction | 0 | 37,574 | 5 | 75,148 |
"Correct Solution:
```
A, B, K = map(int, input().split())
S = sorted(set(list(range(A, B+1)[:K]) + list(range(A, B+1)[-K:])))
print(*S, sep="\n")
``` | output | 1 | 37,574 | 5 | 75,149 |
Provide a correct Python 3 solution for this coding contest problem.
Print all the integers that satisfies the following in ascending order:
* Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
Constraints
* 1 \leq A \leq B \leq 10^9
* 1 \leq K \leq 100
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B K
Output
Print all the integers that satisfies the condition above in ascending order.
Examples
Input
3 8 2
Output
3
4
7
8
Input
4 8 3
Output
4
5
6
7
8
Input
2 9 100
Output
2
3
4
5
6
7
8
9 | instruction | 0 | 37,575 | 5 | 75,150 |
"Correct Solution:
```
A, B, K = map(int, input().split())
A = range(A, B+1)
A = (set(A[:K]) | set(A[-K:]))
[print(i) for i in sorted(A)]
``` | output | 1 | 37,575 | 5 | 75,151 |
Provide a correct Python 3 solution for this coding contest problem.
Print all the integers that satisfies the following in ascending order:
* Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
Constraints
* 1 \leq A \leq B \leq 10^9
* 1 \leq K \leq 100
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B K
Output
Print all the integers that satisfies the condition above in ascending order.
Examples
Input
3 8 2
Output
3
4
7
8
Input
4 8 3
Output
4
5
6
7
8
Input
2 9 100
Output
2
3
4
5
6
7
8
9 | instruction | 0 | 37,576 | 5 | 75,152 |
"Correct Solution:
```
a,b,k = map(int, input().split())
for i in range(a,min(a+k,b+1)):
print(i)
for i in range(max(a,b-k+1,a+k),b+1):
print(i)
``` | output | 1 | 37,576 | 5 | 75,153 |
Provide a correct Python 3 solution for this coding contest problem.
Print all the integers that satisfies the following in ascending order:
* Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
Constraints
* 1 \leq A \leq B \leq 10^9
* 1 \leq K \leq 100
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B K
Output
Print all the integers that satisfies the condition above in ascending order.
Examples
Input
3 8 2
Output
3
4
7
8
Input
4 8 3
Output
4
5
6
7
8
Input
2 9 100
Output
2
3
4
5
6
7
8
9 | instruction | 0 | 37,577 | 5 | 75,154 |
"Correct Solution:
```
a,b,k=map(int,input().split())
if k > b-a:
k = b-a+1
print('\n'.join(map(str,sorted(set(range(a,a+k)) | set(range(b-k+1,b+1))))))
``` | output | 1 | 37,577 | 5 | 75,155 |
Provide a correct Python 3 solution for this coding contest problem.
Print all the integers that satisfies the following in ascending order:
* Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
Constraints
* 1 \leq A \leq B \leq 10^9
* 1 \leq K \leq 100
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B K
Output
Print all the integers that satisfies the condition above in ascending order.
Examples
Input
3 8 2
Output
3
4
7
8
Input
4 8 3
Output
4
5
6
7
8
Input
2 9 100
Output
2
3
4
5
6
7
8
9 | instruction | 0 | 37,578 | 5 | 75,156 |
"Correct Solution:
```
a,b,k=map(int,input().split())
A=set([i for i in range(a,min(a+k,b))])
B=set([i for i in range(max(a,b-k+1),b+1)])
print(*sorted(A|B),sep="\n")
``` | output | 1 | 37,578 | 5 | 75,157 |
Provide a correct Python 3 solution for this coding contest problem.
Print all the integers that satisfies the following in ascending order:
* Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
Constraints
* 1 \leq A \leq B \leq 10^9
* 1 \leq K \leq 100
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B K
Output
Print all the integers that satisfies the condition above in ascending order.
Examples
Input
3 8 2
Output
3
4
7
8
Input
4 8 3
Output
4
5
6
7
8
Input
2 9 100
Output
2
3
4
5
6
7
8
9 | instruction | 0 | 37,579 | 5 | 75,158 |
"Correct Solution:
```
a,b,k=map(int,input().split())
l=min((a+b+1)//2,a+k)
r=max((a+b+1)//2,b-k+1)
for i in range(a,l):
print(i)
for i in range(r,b+1):
print(i)
``` | output | 1 | 37,579 | 5 | 75,159 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Print all the integers that satisfies the following in ascending order:
* Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
Constraints
* 1 \leq A \leq B \leq 10^9
* 1 \leq K \leq 100
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B K
Output
Print all the integers that satisfies the condition above in ascending order.
Examples
Input
3 8 2
Output
3
4
7
8
Input
4 8 3
Output
4
5
6
7
8
Input
2 9 100
Output
2
3
4
5
6
7
8
9
Submitted Solution:
```
a,b,k=map(int,input().split())
for i in range(k):
if a+i<=b:
print(a+i)
for j in range(k,0,-1):
if b-j>=a+k-1:
print(b-j+1)
``` | instruction | 0 | 37,580 | 5 | 75,160 |
Yes | output | 1 | 37,580 | 5 | 75,161 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Print all the integers that satisfies the following in ascending order:
* Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
Constraints
* 1 \leq A \leq B \leq 10^9
* 1 \leq K \leq 100
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B K
Output
Print all the integers that satisfies the condition above in ascending order.
Examples
Input
3 8 2
Output
3
4
7
8
Input
4 8 3
Output
4
5
6
7
8
Input
2 9 100
Output
2
3
4
5
6
7
8
9
Submitted Solution:
```
A,B,K=map(int,input().split())
i=A
while i<=min(B,A+K-1):
print(i)
i+=1
j=B-K+1
while j<=B:
if i<=j:
print(j)
j+=1
``` | instruction | 0 | 37,581 | 5 | 75,162 |
Yes | output | 1 | 37,581 | 5 | 75,163 |
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