text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a binary string s (recall that a string is binary if each character is either 0 or 1).
Let f(t) be the decimal representation of integer t written in binary form (possibly with leading zeroes). For example f(011) = 3, f(00101) = 5, f(00001) = 1, f(10) = 2, f(000) = 0 and f(000100) = 4.
The substring s_{l}, s_{l+1}, ... , s_{r} is good if r - l + 1 = f(s_l ... s_r).
For example string s = 1011 has 5 good substrings: s_1 ... s_1 = 1, s_3 ... s_3 = 1, s_4 ... s_4 = 1, s_1 ... s_2 = 10 and s_2 ... s_4 = 011.
Your task is to calculate the number of good substrings of string s.
You have to answer t independent queries.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of queries.
The only line of each query contains string s (1 β€ |s| β€ 2 β
10^5), consisting of only digits 0 and 1.
It is guaranteed that β_{i=1}^{t} |s_i| β€ 2 β
10^5.
Output
For each query print one integer β the number of good substrings of string s.
Example
Input
4
0110
0101
00001000
0001000
Output
4
3
4
3
Tags: binary search, bitmasks, brute force
Correct Solution:
```
for _ in range(int(input())):
l = input()
n = len(l)
zeros,cnt=0,0
for i in range(n):
if l[i]=="0":zeros+=1
else:
for j in range(i,n):
if zeros+(j-i+1)>=int(l[i:j+1],2):
cnt+=1
else:
break
zeros=0
print(cnt)
```
| 93,600 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a binary string s (recall that a string is binary if each character is either 0 or 1).
Let f(t) be the decimal representation of integer t written in binary form (possibly with leading zeroes). For example f(011) = 3, f(00101) = 5, f(00001) = 1, f(10) = 2, f(000) = 0 and f(000100) = 4.
The substring s_{l}, s_{l+1}, ... , s_{r} is good if r - l + 1 = f(s_l ... s_r).
For example string s = 1011 has 5 good substrings: s_1 ... s_1 = 1, s_3 ... s_3 = 1, s_4 ... s_4 = 1, s_1 ... s_2 = 10 and s_2 ... s_4 = 011.
Your task is to calculate the number of good substrings of string s.
You have to answer t independent queries.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of queries.
The only line of each query contains string s (1 β€ |s| β€ 2 β
10^5), consisting of only digits 0 and 1.
It is guaranteed that β_{i=1}^{t} |s_i| β€ 2 β
10^5.
Output
For each query print one integer β the number of good substrings of string s.
Example
Input
4
0110
0101
00001000
0001000
Output
4
3
4
3
Tags: binary search, bitmasks, brute force
Correct Solution:
```
t = int(input())
a = []
for i in range(t):
a.append(input())
for i in a:
p=-1
l=len(i)
d=0
for j in range(l):
if i[j]=="1":
q=j
s=j-p-1
while(q<l):
if q==(j+18):
break
f=int(i[j:q+1],2)
r=q-j+1
if (r+s)>=f:
d+=1
q+=1
p=j
print(d)
```
| 93,601 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a binary string s (recall that a string is binary if each character is either 0 or 1).
Let f(t) be the decimal representation of integer t written in binary form (possibly with leading zeroes). For example f(011) = 3, f(00101) = 5, f(00001) = 1, f(10) = 2, f(000) = 0 and f(000100) = 4.
The substring s_{l}, s_{l+1}, ... , s_{r} is good if r - l + 1 = f(s_l ... s_r).
For example string s = 1011 has 5 good substrings: s_1 ... s_1 = 1, s_3 ... s_3 = 1, s_4 ... s_4 = 1, s_1 ... s_2 = 10 and s_2 ... s_4 = 011.
Your task is to calculate the number of good substrings of string s.
You have to answer t independent queries.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of queries.
The only line of each query contains string s (1 β€ |s| β€ 2 β
10^5), consisting of only digits 0 and 1.
It is guaranteed that β_{i=1}^{t} |s_i| β€ 2 β
10^5.
Output
For each query print one integer β the number of good substrings of string s.
Example
Input
4
0110
0101
00001000
0001000
Output
4
3
4
3
Tags: binary search, bitmasks, brute force
Correct Solution:
```
for _ in range(int(input())):
s=[int(x) for x in list(input().strip())]
n=len(s)
an=0
c=0
for i in range(n):
if s[i]==1:
p=0
for j in range(18):
if i+j==n:
break
else:
p=p<<1
p+=s[i+j]
if p>=j+1 and p<=j+1+c:
an+=1
c=0
else:
c+=1
print(an)
```
| 93,602 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a binary string s (recall that a string is binary if each character is either 0 or 1).
Let f(t) be the decimal representation of integer t written in binary form (possibly with leading zeroes). For example f(011) = 3, f(00101) = 5, f(00001) = 1, f(10) = 2, f(000) = 0 and f(000100) = 4.
The substring s_{l}, s_{l+1}, ... , s_{r} is good if r - l + 1 = f(s_l ... s_r).
For example string s = 1011 has 5 good substrings: s_1 ... s_1 = 1, s_3 ... s_3 = 1, s_4 ... s_4 = 1, s_1 ... s_2 = 10 and s_2 ... s_4 = 011.
Your task is to calculate the number of good substrings of string s.
You have to answer t independent queries.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of queries.
The only line of each query contains string s (1 β€ |s| β€ 2 β
10^5), consisting of only digits 0 and 1.
It is guaranteed that β_{i=1}^{t} |s_i| β€ 2 β
10^5.
Output
For each query print one integer β the number of good substrings of string s.
Example
Input
4
0110
0101
00001000
0001000
Output
4
3
4
3
Tags: binary search, bitmasks, brute force
Correct Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
from io import BytesIO, IOBase
import sys
po=[1]
for i in range(17):
po.append(po[-1]*2)
def solve(s,x,y,a):
ans,su,j=0,0,x
while j<y:
su+=s[j]*po[j-x]
if su>=j-x+1 and a[j+1]-x>=su:
ans+=1
j=a[j+1]-1
j+=1
return ans
def main():
for _ in range(int(input())):
s=list(map(int,input().rstrip()))
s.reverse()
n,ans=len(s),0
a=[n]*(n+1)
for i in range(n-1,-1,-1):
a[i]=a[i+1]
if s[i]:
a[i]=i
for i in range(n):
z=solve(s,i,min(i+18,n),a)
ans+=z
print(ans)
# FAST INPUT OUTPUT REGION
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
```
| 93,603 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a binary string s (recall that a string is binary if each character is either 0 or 1).
Let f(t) be the decimal representation of integer t written in binary form (possibly with leading zeroes). For example f(011) = 3, f(00101) = 5, f(00001) = 1, f(10) = 2, f(000) = 0 and f(000100) = 4.
The substring s_{l}, s_{l+1}, ... , s_{r} is good if r - l + 1 = f(s_l ... s_r).
For example string s = 1011 has 5 good substrings: s_1 ... s_1 = 1, s_3 ... s_3 = 1, s_4 ... s_4 = 1, s_1 ... s_2 = 10 and s_2 ... s_4 = 011.
Your task is to calculate the number of good substrings of string s.
You have to answer t independent queries.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of queries.
The only line of each query contains string s (1 β€ |s| β€ 2 β
10^5), consisting of only digits 0 and 1.
It is guaranteed that β_{i=1}^{t} |s_i| β€ 2 β
10^5.
Output
For each query print one integer β the number of good substrings of string s.
Example
Input
4
0110
0101
00001000
0001000
Output
4
3
4
3
Submitted Solution:
```
if __name__ == '__main__':
t = int(input())
nm = {}
for i in range(1, 200001):
nm[format(i, 'b')] = i
while t > 0:
s = input()
sm = [0]
for c in s:
sm.append(sm[-1] + (1 if c == '1' else 0))
ans = 0
for sz in range(1, 20):
for i in range(0, len(s) - sz + 1):
if s[i: i + sz] in nm:
num = nm[s[i: i + sz]]
if num == sz:
ans += 1
else:
li = i + sz - num
if li >= 0 == sm[i] - sm[li]:
ans += 1
print(ans)
t -= 1
```
Yes
| 93,604 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a binary string s (recall that a string is binary if each character is either 0 or 1).
Let f(t) be the decimal representation of integer t written in binary form (possibly with leading zeroes). For example f(011) = 3, f(00101) = 5, f(00001) = 1, f(10) = 2, f(000) = 0 and f(000100) = 4.
The substring s_{l}, s_{l+1}, ... , s_{r} is good if r - l + 1 = f(s_l ... s_r).
For example string s = 1011 has 5 good substrings: s_1 ... s_1 = 1, s_3 ... s_3 = 1, s_4 ... s_4 = 1, s_1 ... s_2 = 10 and s_2 ... s_4 = 011.
Your task is to calculate the number of good substrings of string s.
You have to answer t independent queries.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of queries.
The only line of each query contains string s (1 β€ |s| β€ 2 β
10^5), consisting of only digits 0 and 1.
It is guaranteed that β_{i=1}^{t} |s_i| β€ 2 β
10^5.
Output
For each query print one integer β the number of good substrings of string s.
Example
Input
4
0110
0101
00001000
0001000
Output
4
3
4
3
Submitted Solution:
```
# cook your dish here
def binaryToDecimal(n):
return int(n,2)
t=int(input())
for _ in range(t):
s=input()
ct=0
zr=0
for i in range(len(s)):
if s[i]=='0':
zr+=1
else:
for j in range(i,len(s)):
#print(zr)
if binaryToDecimal(s[i:j+1])<=zr+j-i+1:
#print(binaryToDecimal(s[i:j+1]))
#print(j-i+1)
ct+=1
else:
break
zr=0
print(ct)
```
Yes
| 93,605 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a binary string s (recall that a string is binary if each character is either 0 or 1).
Let f(t) be the decimal representation of integer t written in binary form (possibly with leading zeroes). For example f(011) = 3, f(00101) = 5, f(00001) = 1, f(10) = 2, f(000) = 0 and f(000100) = 4.
The substring s_{l}, s_{l+1}, ... , s_{r} is good if r - l + 1 = f(s_l ... s_r).
For example string s = 1011 has 5 good substrings: s_1 ... s_1 = 1, s_3 ... s_3 = 1, s_4 ... s_4 = 1, s_1 ... s_2 = 10 and s_2 ... s_4 = 011.
Your task is to calculate the number of good substrings of string s.
You have to answer t independent queries.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of queries.
The only line of each query contains string s (1 β€ |s| β€ 2 β
10^5), consisting of only digits 0 and 1.
It is guaranteed that β_{i=1}^{t} |s_i| β€ 2 β
10^5.
Output
For each query print one integer β the number of good substrings of string s.
Example
Input
4
0110
0101
00001000
0001000
Output
4
3
4
3
Submitted Solution:
```
#########################################################################################################\
#########################################################################################################
###################################The_Apurv_Rathore#####################################################
#########################################################################################################
#########################################################################################################
import sys,os,io
from sys import stdin
from math import log, gcd, ceil
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop
from bisect import bisect_left , bisect_right
import math
alphabets = list('abcdefghijklmnopqrstuvwxyz')
def isPrime(x):
for i in range(2,x):
if i*i>x:
break
if (x%i==0):
return False
return True
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den,
p - 2, p)) % p
def primeFactors(n):
l = []
while n % 2 == 0:
l.append(2)
n = n / 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
l.append(int(i))
n = n / i
if n > 2:
l.append(n)
return list(set(l))
def power(x, y, p) :
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
def SieveOfEratosthenes(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def si():
return input()
def prefix_sum(arr):
r = [0] * (len(arr)+1)
for i, el in enumerate(arr):
r[i+1] = r[i] + el
return r
def divideCeil(n,x):
if (n%x==0):
return n//x
return n//x+1
def ii():
return int(input())
def li():
return list(map(int,input().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')
def power_set(L):
"""
L is a list.
The function returns the power set, but as a list of lists.
"""
cardinality=len(L)
n=2 ** cardinality
powerset = []
for i in range(n):
a=bin(i)[2:]
subset=[]
for j in range(len(a)):
if a[-j-1]=='1':
subset.append(L[j])
powerset.append(subset)
#the function could stop here closing with
#return powerset
powerset_orderred=[]
for k in range(cardinality+1):
for w in powerset:
if len(w)==k:
powerset_orderred.append(w)
return powerset_orderred
def fastPlrintNextLines(a):
# 12
# 3
# 1
#like this
#a is list of strings
print('\n'.join(map(str,a)))
#__________________________TEMPLATE__________________OVER_______________________________________________________
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r") ; sys.stdout = open("output.txt","w")
# else:
# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def fun(x):
# print(x)
tt1 = x[:]
tt = len(x)
x = ''.join(x)
x = int(x,2)
# print(x,tt,cnt)
if x-tt<=cnt:
# print(tt1)
return True
return False
t = 1
t = int(input())
for _ in range(t):
s = list(si())
n = len(s)
cnt = 0
res = 0
for i in range(n):
if (s[i]=='0'):
cnt+=1
else:
x = []
for j in range(i,n):
x+=s[j]
if fun(x):
res+=1
else:
break
cnt = 0
print(res)
```
Yes
| 93,606 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a binary string s (recall that a string is binary if each character is either 0 or 1).
Let f(t) be the decimal representation of integer t written in binary form (possibly with leading zeroes). For example f(011) = 3, f(00101) = 5, f(00001) = 1, f(10) = 2, f(000) = 0 and f(000100) = 4.
The substring s_{l}, s_{l+1}, ... , s_{r} is good if r - l + 1 = f(s_l ... s_r).
For example string s = 1011 has 5 good substrings: s_1 ... s_1 = 1, s_3 ... s_3 = 1, s_4 ... s_4 = 1, s_1 ... s_2 = 10 and s_2 ... s_4 = 011.
Your task is to calculate the number of good substrings of string s.
You have to answer t independent queries.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of queries.
The only line of each query contains string s (1 β€ |s| β€ 2 β
10^5), consisting of only digits 0 and 1.
It is guaranteed that β_{i=1}^{t} |s_i| β€ 2 β
10^5.
Output
For each query print one integer β the number of good substrings of string s.
Example
Input
4
0110
0101
00001000
0001000
Output
4
3
4
3
Submitted Solution:
```
from sys import stdin
from sys import setrecursionlimit as SRL;
SRL(10 ** 7)
rd = stdin.readline
rrd = lambda: map(int, rd().strip().split())
T = int(input())
while T:
s = "".join(map(str, rd().strip()))
ans = 0
max0 = 0
for i in range(len(s)):
if s[i] == '0':
max0 += 1
else:
ans += 1
tot = 1
for j in range(1, 20):
tot = tot << 1
if i + j < len(s):
if s[i + j] == '1':
tot += 1
if max0 >= tot-j-1:
ans += 1
max0 = 0
print(ans)
T -= 1
```
Yes
| 93,607 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a binary string s (recall that a string is binary if each character is either 0 or 1).
Let f(t) be the decimal representation of integer t written in binary form (possibly with leading zeroes). For example f(011) = 3, f(00101) = 5, f(00001) = 1, f(10) = 2, f(000) = 0 and f(000100) = 4.
The substring s_{l}, s_{l+1}, ... , s_{r} is good if r - l + 1 = f(s_l ... s_r).
For example string s = 1011 has 5 good substrings: s_1 ... s_1 = 1, s_3 ... s_3 = 1, s_4 ... s_4 = 1, s_1 ... s_2 = 10 and s_2 ... s_4 = 011.
Your task is to calculate the number of good substrings of string s.
You have to answer t independent queries.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of queries.
The only line of each query contains string s (1 β€ |s| β€ 2 β
10^5), consisting of only digits 0 and 1.
It is guaranteed that β_{i=1}^{t} |s_i| β€ 2 β
10^5.
Output
For each query print one integer β the number of good substrings of string s.
Example
Input
4
0110
0101
00001000
0001000
Output
4
3
4
3
Submitted Solution:
```
import sys
t = int(input())
for _ in range(t):
s: str = sys.stdin.readline().rstrip() + "*"
n = len(s)
zero = [0]*n
p = []
for i in range(1, n-1):
if s[i-1] == '0':
zero[i] = zero[i-1] + 1
else:
zero[i] = 0
zero[i+1] = zero[i]
if s[i] == '1':
p.append(i)
ans = 0
for i in range(1, 100):
target = bin(i)[2:]
leading_zero = i - len(target)
suf_len = len(target)
if leading_zero + suf_len > n:
break
for j in p:
if (zero[j] >= leading_zero and target == s[j:j+suf_len]):
ans += 1
print(ans)
```
No
| 93,608 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a binary string s (recall that a string is binary if each character is either 0 or 1).
Let f(t) be the decimal representation of integer t written in binary form (possibly with leading zeroes). For example f(011) = 3, f(00101) = 5, f(00001) = 1, f(10) = 2, f(000) = 0 and f(000100) = 4.
The substring s_{l}, s_{l+1}, ... , s_{r} is good if r - l + 1 = f(s_l ... s_r).
For example string s = 1011 has 5 good substrings: s_1 ... s_1 = 1, s_3 ... s_3 = 1, s_4 ... s_4 = 1, s_1 ... s_2 = 10 and s_2 ... s_4 = 011.
Your task is to calculate the number of good substrings of string s.
You have to answer t independent queries.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of queries.
The only line of each query contains string s (1 β€ |s| β€ 2 β
10^5), consisting of only digits 0 and 1.
It is guaranteed that β_{i=1}^{t} |s_i| β€ 2 β
10^5.
Output
For each query print one integer β the number of good substrings of string s.
Example
Input
4
0110
0101
00001000
0001000
Output
4
3
4
3
Submitted Solution:
```
#########################################################################################################\
#########################################################################################################
###################################The_Apurv_Rathore#####################################################
#########################################################################################################
#########################################################################################################
import sys,os,io
from sys import stdin
from math import log, gcd, ceil
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop
from bisect import bisect_left , bisect_right
import math
alphabets = list('abcdefghijklmnopqrstuvwxyz')
def isPrime(x):
for i in range(2,x):
if i*i>x:
break
if (x%i==0):
return False
return True
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den,
p - 2, p)) % p
def primeFactors(n):
l = []
while n % 2 == 0:
l.append(2)
n = n / 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
l.append(int(i))
n = n / i
if n > 2:
l.append(n)
return list(set(l))
def power(x, y, p) :
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
def SieveOfEratosthenes(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def si():
return input()
def prefix_sum(arr):
r = [0] * (len(arr)+1)
for i, el in enumerate(arr):
r[i+1] = r[i] + el
return r
def divideCeil(n,x):
if (n%x==0):
return n//x
return n//x+1
def ii():
return int(input())
def li():
return list(map(int,input().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')
def power_set(L):
"""
L is a list.
The function returns the power set, but as a list of lists.
"""
cardinality=len(L)
n=2 ** cardinality
powerset = []
for i in range(n):
a=bin(i)[2:]
subset=[]
for j in range(len(a)):
if a[-j-1]=='1':
subset.append(L[j])
powerset.append(subset)
#the function could stop here closing with
#return powerset
powerset_orderred=[]
for k in range(cardinality+1):
for w in powerset:
if len(w)==k:
powerset_orderred.append(w)
return powerset_orderred
def fastPlrintNextLines(a):
# 12
# 3
# 1
#like this
#a is list of strings
print('\n'.join(map(str,a)))
#__________________________TEMPLATE__________________OVER_______________________________________________________
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r") ; sys.stdout = open("output.txt","w")
# else:
# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def fun(x):
# print(x)
tt1 = x[:]
tt = len(x)
x = ''.join(x)
x = int(x,2)
# print(x,tt,cnt)
if x-tt<=cnt:
# print(tt1)
return True
return False
t = 1
t = int(input())
for _ in range(t):
s = list(si())
n = len(s)
cnt = 0
res = 0
for i in range(n):
if (s[i]=='0'):
cnt+=1
else:
x = []
for j in range(i,n):
x+=s[j]
if fun(x):
res+=1
else:
break
print(res)
```
No
| 93,609 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a binary string s (recall that a string is binary if each character is either 0 or 1).
Let f(t) be the decimal representation of integer t written in binary form (possibly with leading zeroes). For example f(011) = 3, f(00101) = 5, f(00001) = 1, f(10) = 2, f(000) = 0 and f(000100) = 4.
The substring s_{l}, s_{l+1}, ... , s_{r} is good if r - l + 1 = f(s_l ... s_r).
For example string s = 1011 has 5 good substrings: s_1 ... s_1 = 1, s_3 ... s_3 = 1, s_4 ... s_4 = 1, s_1 ... s_2 = 10 and s_2 ... s_4 = 011.
Your task is to calculate the number of good substrings of string s.
You have to answer t independent queries.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of queries.
The only line of each query contains string s (1 β€ |s| β€ 2 β
10^5), consisting of only digits 0 and 1.
It is guaranteed that β_{i=1}^{t} |s_i| β€ 2 β
10^5.
Output
For each query print one integer β the number of good substrings of string s.
Example
Input
4
0110
0101
00001000
0001000
Output
4
3
4
3
Submitted Solution:
```
if __name__ == '__main__':
t = int(input())
for _ in range(t):
ln = [int(x) for x in input()]
sz, zc, ans = (len(ln), 0, 0)
for i in range(sz):
if ln[i] == 1:
num, j, rl = (0, i, min(i + 20, sz))
while j < rl:
num, j = (num * 2 + ln[j], j + 1)
if j - i + 1 + zc >= num:
ans += 1
zc = 0
else:
zc += 1
print(ans)
```
No
| 93,610 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a binary string s (recall that a string is binary if each character is either 0 or 1).
Let f(t) be the decimal representation of integer t written in binary form (possibly with leading zeroes). For example f(011) = 3, f(00101) = 5, f(00001) = 1, f(10) = 2, f(000) = 0 and f(000100) = 4.
The substring s_{l}, s_{l+1}, ... , s_{r} is good if r - l + 1 = f(s_l ... s_r).
For example string s = 1011 has 5 good substrings: s_1 ... s_1 = 1, s_3 ... s_3 = 1, s_4 ... s_4 = 1, s_1 ... s_2 = 10 and s_2 ... s_4 = 011.
Your task is to calculate the number of good substrings of string s.
You have to answer t independent queries.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of queries.
The only line of each query contains string s (1 β€ |s| β€ 2 β
10^5), consisting of only digits 0 and 1.
It is guaranteed that β_{i=1}^{t} |s_i| β€ 2 β
10^5.
Output
For each query print one integer β the number of good substrings of string s.
Example
Input
4
0110
0101
00001000
0001000
Output
4
3
4
3
Submitted Solution:
```
import sys
def input():
return sys.stdin.readline().rstrip()
def slv():
s = list(input()[::-1])
s = list(map(int, s))
ans = 0
n = len(s)
RPOS = [n]*(n)
for i in range(n):
if s[i] == 1:
RPOS[i] = i
for i in range(n - 1, -1, -1):
if i + 1 < n:
RPOS[i] = min(RPOS[i], RPOS[i + 1])
for l in range(n):
left_comp, right_comp = 0, 0
for r in range(l, min(l + 20, n)):
left_comp += pow(2, r - l) * s[r]
right_comp += 1
if left_comp == right_comp:
ans += 1
m = min(l + 20,n) - 1
if right_comp <= left_comp < RPOS[m] - l:
ans += 1
print(ans)
return
def main():
t = int(input())
for i in range(t):
slv()
return
def main():
t = int(input())
for i in range(t):
slv()
return
if __name__ == "__main__":
main()
```
No
| 93,611 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house.
Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles.
Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room.
For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i.
Input
The first line contains single integer n (1 β€ n β€ 1000) β the number of rooms.
Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 β€ c_i, sum_i β€ 10^4) β the maximum number of radiators and the minimum total number of sections in the i-th room, respectively.
Output
For each room print one integer β the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i.
Example
Input
4
1 10000
10000 1
2 6
4 6
Output
100000000
1
18
10
Note
In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8.
In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section.
In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18.
Tags: math
Correct Solution:
```
n=int(input())
for _ in range(n):
c, s = map(int,input().split())
q = s//c
r = s%c
print(r*(q+1)**2 + (c-r)*q*q)
```
| 93,612 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house.
Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles.
Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room.
For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i.
Input
The first line contains single integer n (1 β€ n β€ 1000) β the number of rooms.
Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 β€ c_i, sum_i β€ 10^4) β the maximum number of radiators and the minimum total number of sections in the i-th room, respectively.
Output
For each room print one integer β the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i.
Example
Input
4
1 10000
10000 1
2 6
4 6
Output
100000000
1
18
10
Note
In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8.
In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section.
In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18.
Tags: math
Correct Solution:
```
t = int(input())
for c in range (t):
ss = 0
a,b= map(int, input().split())
if b<=a:
print(b)
else :
de = int(b/a)
ss = de**2 *(a-b%a) +(de+1)**2 *(b%a)
print(ss)
```
| 93,613 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house.
Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles.
Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room.
For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i.
Input
The first line contains single integer n (1 β€ n β€ 1000) β the number of rooms.
Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 β€ c_i, sum_i β€ 10^4) β the maximum number of radiators and the minimum total number of sections in the i-th room, respectively.
Output
For each room print one integer β the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i.
Example
Input
4
1 10000
10000 1
2 6
4 6
Output
100000000
1
18
10
Note
In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8.
In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section.
In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18.
Tags: math
Correct Solution:
```
''' Hey stalker :) '''
INF = 10**10
def main():
print = out.append
''' Cook your dish here! '''
c, sum = get_list()
x = 0
while (x)*c<=sum: x+=1
res = c*(x-1)**2
res += (sum - (x-1)*c)*(2*x-1)
print(res)
''' Pythonista fLite 1.1 '''
import sys
#from collections import defaultdict, Counter
#from functools import reduce
#import math
input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
out = []
get_int = lambda: int(input())
get_list = lambda: list(map(int, input().split()))
#main()
[main() for _ in range(int(input()))]
print(*out, sep='\n')
```
| 93,614 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house.
Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles.
Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room.
For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i.
Input
The first line contains single integer n (1 β€ n β€ 1000) β the number of rooms.
Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 β€ c_i, sum_i β€ 10^4) β the maximum number of radiators and the minimum total number of sections in the i-th room, respectively.
Output
For each room print one integer β the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i.
Example
Input
4
1 10000
10000 1
2 6
4 6
Output
100000000
1
18
10
Note
In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8.
In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section.
In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18.
Tags: math
Correct Solution:
```
for i in range(int(input())):
c, sumi = [int(x) for x in input().split()]
deg = sumi//c
res = ((deg + 1)**2) * (sumi%c) + (c - sumi%c)*(deg**2)
print(res)
```
| 93,615 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house.
Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles.
Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room.
For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i.
Input
The first line contains single integer n (1 β€ n β€ 1000) β the number of rooms.
Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 β€ c_i, sum_i β€ 10^4) β the maximum number of radiators and the minimum total number of sections in the i-th room, respectively.
Output
For each room print one integer β the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i.
Example
Input
4
1 10000
10000 1
2 6
4 6
Output
100000000
1
18
10
Note
In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8.
In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section.
In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18.
Tags: math
Correct Solution:
```
n = int(input())
for i in range(n):
c, s = map(int, input().strip().split())
left = s % c
v = s // c
o = 0
for i in range(left):
o += (v + 1) ** 2
for i in range(left, c):
o += v ** 2
print(o)
```
| 93,616 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house.
Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles.
Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room.
For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i.
Input
The first line contains single integer n (1 β€ n β€ 1000) β the number of rooms.
Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 β€ c_i, sum_i β€ 10^4) β the maximum number of radiators and the minimum total number of sections in the i-th room, respectively.
Output
For each room print one integer β the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i.
Example
Input
4
1 10000
10000 1
2 6
4 6
Output
100000000
1
18
10
Note
In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8.
In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section.
In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18.
Tags: math
Correct Solution:
```
n = int(input())
for _ in range(n):
c, su = map(int, input().split())
ci = [0] * c
while su > 0:
for i in range(c):
if su > 0:
ci[i] += 1
su -= 1
else:
break
to = 0
for i in range(c):
to += ci[i] ** 2
print(to)
```
| 93,617 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house.
Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles.
Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room.
For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i.
Input
The first line contains single integer n (1 β€ n β€ 1000) β the number of rooms.
Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 β€ c_i, sum_i β€ 10^4) β the maximum number of radiators and the minimum total number of sections in the i-th room, respectively.
Output
For each room print one integer β the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i.
Example
Input
4
1 10000
10000 1
2 6
4 6
Output
100000000
1
18
10
Note
In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8.
In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section.
In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18.
Tags: math
Correct Solution:
```
from collections import defaultdict
t = int(input())
while(t>0):
a,b= [int(i) for i in input().split()]
if a==1 or b==1:
ans = b**2
else:
curr = b//a
rem = b%a
temp = curr-rem
s = a-rem
ans = s*(curr**2)+rem*(curr+1)**2
print(ans)
t-=1
```
| 93,618 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house.
Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles.
Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room.
For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i.
Input
The first line contains single integer n (1 β€ n β€ 1000) β the number of rooms.
Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 β€ c_i, sum_i β€ 10^4) β the maximum number of radiators and the minimum total number of sections in the i-th room, respectively.
Output
For each room print one integer β the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i.
Example
Input
4
1 10000
10000 1
2 6
4 6
Output
100000000
1
18
10
Note
In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8.
In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section.
In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18.
Tags: math
Correct Solution:
```
'''
Created on 2019. 9. 21.
@author: kkhh88
'''
#q = int(input())
#x, y = map(int,input().split(' '))
#print (' '.join(list(map(str, s))))
q = int(input())
for i in range(q):
x, y = map(int,input().split(' '))
c = y // x
d = y % x
sum = (c + 1) * (c + 1) * d + c * c * (x - d)
print (sum)
```
| 93,619 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house.
Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles.
Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room.
For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i.
Input
The first line contains single integer n (1 β€ n β€ 1000) β the number of rooms.
Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 β€ c_i, sum_i β€ 10^4) β the maximum number of radiators and the minimum total number of sections in the i-th room, respectively.
Output
For each room print one integer β the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i.
Example
Input
4
1 10000
10000 1
2 6
4 6
Output
100000000
1
18
10
Note
In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8.
In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section.
In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18.
Submitted Solution:
```
T = int(input())
anwer = ""
for t in range(T):
text = input().split(' ')
maximum = int(text[0])
need = int(text[1])
one = need // maximum
if one == 0:
anwer += str(need) + '\n'
continue
ost = need - maximum * one
if ost == 0:
n = need / one
anwer += str(int(one**2 * n)) + '\n'
continue
boost = ost // maximum
boost2 = ost - maximum * boost
one += boost
cost = 0
k = maximum - boost2
for i in range(maximum):
if boost2:
cost += (one + 1)**2
boost2 -= 1
else:
cost += one**2 * k
break
anwer += str(int(cost)) + '\n'
print(anwer)
```
Yes
| 93,620 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house.
Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles.
Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room.
For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i.
Input
The first line contains single integer n (1 β€ n β€ 1000) β the number of rooms.
Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 β€ c_i, sum_i β€ 10^4) β the maximum number of radiators and the minimum total number of sections in the i-th room, respectively.
Output
For each room print one integer β the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i.
Example
Input
4
1 10000
10000 1
2 6
4 6
Output
100000000
1
18
10
Note
In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8.
In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section.
In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18.
Submitted Solution:
```
n = int(input())
for _ in range(n):
c,k = map(int,input().strip().split())
x = k//c
y = k%c
ans = y*((x+1)**2)+(c-y)*(x**2)
print(ans)
```
Yes
| 93,621 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house.
Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles.
Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room.
For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i.
Input
The first line contains single integer n (1 β€ n β€ 1000) β the number of rooms.
Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 β€ c_i, sum_i β€ 10^4) β the maximum number of radiators and the minimum total number of sections in the i-th room, respectively.
Output
For each room print one integer β the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i.
Example
Input
4
1 10000
10000 1
2 6
4 6
Output
100000000
1
18
10
Note
In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8.
In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section.
In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18.
Submitted Solution:
```
for i in ' '*int(input()):
c,sum=map(int,input().split())
k=sum//c
r=sum%c
print((k+1)**2*r+k**2*(c-r))
```
Yes
| 93,622 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house.
Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles.
Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room.
For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i.
Input
The first line contains single integer n (1 β€ n β€ 1000) β the number of rooms.
Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 β€ c_i, sum_i β€ 10^4) β the maximum number of radiators and the minimum total number of sections in the i-th room, respectively.
Output
For each room print one integer β the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i.
Example
Input
4
1 10000
10000 1
2 6
4 6
Output
100000000
1
18
10
Note
In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8.
In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section.
In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18.
Submitted Solution:
```
for i in range(int(input())):
a, b = map(int, input().split())
if a == 1 or b == 1:
print(pow(b, 2))
else:
ans = [0]*a
while(b != 0):
for i in range(a):
if b == 0:
break
else:
ans[i] += 1
b -= 1
count = 0
for i in range(len(ans)):
count += pow(ans[i], 2)
print(count)
```
Yes
| 93,623 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house.
Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles.
Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room.
For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i.
Input
The first line contains single integer n (1 β€ n β€ 1000) β the number of rooms.
Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 β€ c_i, sum_i β€ 10^4) β the maximum number of radiators and the minimum total number of sections in the i-th room, respectively.
Output
For each room print one integer β the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i.
Example
Input
4
1 10000
10000 1
2 6
4 6
Output
100000000
1
18
10
Note
In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8.
In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section.
In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18.
Submitted Solution:
```
n = int(input())
while(n):
sum = 0
a,b = map(int,input().split())
if a == 1:
print(b^2)
elif a>b:
print(b)
else:
z = (b//a) +1
x = z-1
m = b%a
r = a - m
sum += (m * (z^2))
sum += (r * (x^2))
print(sum)
n-=1
```
No
| 93,624 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house.
Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles.
Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room.
For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i.
Input
The first line contains single integer n (1 β€ n β€ 1000) β the number of rooms.
Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 β€ c_i, sum_i β€ 10^4) β the maximum number of radiators and the minimum total number of sections in the i-th room, respectively.
Output
For each room print one integer β the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i.
Example
Input
4
1 10000
10000 1
2 6
4 6
Output
100000000
1
18
10
Note
In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8.
In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section.
In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18.
Submitted Solution:
```
n = int(input())
for i in range(n):
a, b = map(int,input().split())
count = 0
while (b%a!=0 and b!=0):
count+=1
b-=1
a-=1
count+=((b//a)**2)*a
print(count)
```
No
| 93,625 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house.
Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles.
Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room.
For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i.
Input
The first line contains single integer n (1 β€ n β€ 1000) β the number of rooms.
Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 β€ c_i, sum_i β€ 10^4) β the maximum number of radiators and the minimum total number of sections in the i-th room, respectively.
Output
For each room print one integer β the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i.
Example
Input
4
1 10000
10000 1
2 6
4 6
Output
100000000
1
18
10
Note
In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8.
In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section.
In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18.
Submitted Solution:
```
from math import gcd
for _ in range(int(input())):
ci,si=map(int,input().split())
if si<ci:
print(si)
else:
ans=0
t=si//2
rem = si%2
if rem:
ans+=t**2*(ci-1)
ans+=(rem+t)**2
else:
ans+=t**2*ci
print(ans)
```
No
| 93,626 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house.
Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles.
Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room.
For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i.
Input
The first line contains single integer n (1 β€ n β€ 1000) β the number of rooms.
Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 β€ c_i, sum_i β€ 10^4) β the maximum number of radiators and the minimum total number of sections in the i-th room, respectively.
Output
For each room print one integer β the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i.
Example
Input
4
1 10000
10000 1
2 6
4 6
Output
100000000
1
18
10
Note
In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8.
In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section.
In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18.
Submitted Solution:
```
def process():
a,b = input().split()
a,b = int(a), int(b)
if b == 3 and a == 2:
print(5)
return
if a > b:
print(1)
else:
if a == 1:
print(int(b**2))
return
if b % a == 0:
res = 0
ave = int(b/a)
while b != 0:
res += ave**2
b -= ave
print(res)
else:
res = 0
ave = int(b/a)
tmp = b/a - ave
cnt = 0
for i in range(1,a+1):
if i*tmp == int(i*tmp):
cnt = i
break
for i in range(cnt):
res += int(ave)**2
for i in range(a-cnt):
res += (int(ave)+1)**2
print(res)
tests = int(input())
for test in range(tests):
process()
```
No
| 93,627 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of this problem. The only difference is the constraint on k β the number of gifts in the offer. In this version: k=2.
Vasya came to the store to buy goods for his friends for the New Year. It turned out that he was very lucky β today the offer "k of goods for the price of one" is held in store. Remember, that in this problem k=2.
Using this offer, Vasya can buy exactly k of any goods, paying only for the most expensive of them. Vasya decided to take this opportunity and buy as many goods as possible for his friends with the money he has.
More formally, for each good, its price is determined by a_i β the number of coins it costs. Initially, Vasya has p coins. He wants to buy the maximum number of goods. Vasya can perform one of the following operations as many times as necessary:
* Vasya can buy one good with the index i if he currently has enough coins (i.e p β₯ a_i). After buying this good, the number of Vasya's coins will decrease by a_i, (i.e it becomes p := p - a_i).
* Vasya can buy a good with the index i, and also choose exactly k-1 goods, the price of which does not exceed a_i, if he currently has enough coins (i.e p β₯ a_i). Thus, he buys all these k goods, and his number of coins decreases by a_i (i.e it becomes p := p - a_i).
Please note that each good can be bought no more than once.
For example, if the store now has n=5 goods worth a_1=2, a_2=4, a_3=3, a_4=5, a_5=7, respectively, k=2, and Vasya has 6 coins, then he can buy 3 goods. A good with the index 1 will be bought by Vasya without using the offer and he will pay 2 coins. Goods with the indices 2 and 3 Vasya will buy using the offer and he will pay 4 coins. It can be proved that Vasya can not buy more goods with six coins.
Help Vasya to find out the maximum number of goods he can buy.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases in the test.
The next lines contain a description of t test cases.
The first line of each test case contains three integers n, p, k (2 β€ n β€ 2 β
10^5, 1 β€ p β€ 2β
10^9, k=2) β the number of goods in the store, the number of coins Vasya has and the number of goods that can be bought by the price of the most expensive of them.
The second line of each test case contains n integers a_i (1 β€ a_i β€ 10^4) β the prices of goods.
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5. It is guaranteed that in this version of the problem k=2 for all test cases.
Output
For each test case in a separate line print one integer m β the maximum number of goods that Vasya can buy.
Example
Input
6
5 6 2
2 4 3 5 7
5 11 2
2 4 3 5 7
2 10000 2
10000 10000
2 9999 2
10000 10000
5 13 2
8 2 8 2 5
3 18 2
1 2 3
Output
3
4
2
0
4
3
Tags: dp, greedy, sortings
Correct Solution:
```
t = int(input())
for _ in range(t) :
n,p,k = map(int,input().split())
arr = list(map(int,input().split()))
arr.sort()
m = 0
idx = 0
count = 0
z = p
while idx < len(arr) :
if idx == 0 and p >= arr[idx]:
count +=1
p = p - arr[idx]
idx +=2
elif p >= arr[idx] :
count += 2
p = p -arr[idx]
idx +=2
elif p < arr[idx] and idx > 0 and p >= arr[idx-1] :
count +=1
p = p - arr[idx-1]
break
else :
break
m = max(m,count)
idx = 1
count = 0
p = z
while idx < len(arr) :
if idx == 1 and p >= arr[idx] :
count +=2
p = p - arr[idx]
idx +=2
elif p >= arr[idx] and idx > 1 :
count +=2
p = p - arr[idx]
idx +=2
elif p < arr[idx] and idx >1 and p >= arr[idx-1] :
count +=1
p = p - arr[idx-1]
break
else :
break
m = max(m ,count)
print(m)
```
| 93,628 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of this problem. The only difference is the constraint on k β the number of gifts in the offer. In this version: k=2.
Vasya came to the store to buy goods for his friends for the New Year. It turned out that he was very lucky β today the offer "k of goods for the price of one" is held in store. Remember, that in this problem k=2.
Using this offer, Vasya can buy exactly k of any goods, paying only for the most expensive of them. Vasya decided to take this opportunity and buy as many goods as possible for his friends with the money he has.
More formally, for each good, its price is determined by a_i β the number of coins it costs. Initially, Vasya has p coins. He wants to buy the maximum number of goods. Vasya can perform one of the following operations as many times as necessary:
* Vasya can buy one good with the index i if he currently has enough coins (i.e p β₯ a_i). After buying this good, the number of Vasya's coins will decrease by a_i, (i.e it becomes p := p - a_i).
* Vasya can buy a good with the index i, and also choose exactly k-1 goods, the price of which does not exceed a_i, if he currently has enough coins (i.e p β₯ a_i). Thus, he buys all these k goods, and his number of coins decreases by a_i (i.e it becomes p := p - a_i).
Please note that each good can be bought no more than once.
For example, if the store now has n=5 goods worth a_1=2, a_2=4, a_3=3, a_4=5, a_5=7, respectively, k=2, and Vasya has 6 coins, then he can buy 3 goods. A good with the index 1 will be bought by Vasya without using the offer and he will pay 2 coins. Goods with the indices 2 and 3 Vasya will buy using the offer and he will pay 4 coins. It can be proved that Vasya can not buy more goods with six coins.
Help Vasya to find out the maximum number of goods he can buy.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases in the test.
The next lines contain a description of t test cases.
The first line of each test case contains three integers n, p, k (2 β€ n β€ 2 β
10^5, 1 β€ p β€ 2β
10^9, k=2) β the number of goods in the store, the number of coins Vasya has and the number of goods that can be bought by the price of the most expensive of them.
The second line of each test case contains n integers a_i (1 β€ a_i β€ 10^4) β the prices of goods.
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5. It is guaranteed that in this version of the problem k=2 for all test cases.
Output
For each test case in a separate line print one integer m β the maximum number of goods that Vasya can buy.
Example
Input
6
5 6 2
2 4 3 5 7
5 11 2
2 4 3 5 7
2 10000 2
10000 10000
2 9999 2
10000 10000
5 13 2
8 2 8 2 5
3 18 2
1 2 3
Output
3
4
2
0
4
3
Tags: dp, greedy, sortings
Correct Solution:
```
t = int(input())
for case in range(t):
n, p, k = map(int, input().split())
# print("n: {}, p: {}, k: {}".format(n, p, k))
a = list(map(int, input().split()))
a.sort()
# print("a: {}".format(a))
i = k - 1
to_spend = 0
while i < n and to_spend + a[i] <= p:
to_spend += a[i]
i += k
right = min(n+1, i + 1)
left = i + 1 - k
while right - left > 1:
# print("left: {}, right: {}".format(left, right))
pivot = (right + left) // 2
# print("pivot: {}".format(pivot))
if pivot % k == 0:
to_spend = sum([a[i] for i in range(pivot-1, -1, -k)])
else:
to_spend = sum([a[i] for i in range(pivot-1, k-2, -k)])
to_spend += sum([a[i] for i in range(pivot % k -1, -1, -1)])
# to_spend2 = pivot
# print("to_spend: {}".format(to_spend))
if to_spend <= p:
left = pivot
else:
right = pivot
print(left)
# print()
```
| 93,629 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of this problem. The only difference is the constraint on k β the number of gifts in the offer. In this version: k=2.
Vasya came to the store to buy goods for his friends for the New Year. It turned out that he was very lucky β today the offer "k of goods for the price of one" is held in store. Remember, that in this problem k=2.
Using this offer, Vasya can buy exactly k of any goods, paying only for the most expensive of them. Vasya decided to take this opportunity and buy as many goods as possible for his friends with the money he has.
More formally, for each good, its price is determined by a_i β the number of coins it costs. Initially, Vasya has p coins. He wants to buy the maximum number of goods. Vasya can perform one of the following operations as many times as necessary:
* Vasya can buy one good with the index i if he currently has enough coins (i.e p β₯ a_i). After buying this good, the number of Vasya's coins will decrease by a_i, (i.e it becomes p := p - a_i).
* Vasya can buy a good with the index i, and also choose exactly k-1 goods, the price of which does not exceed a_i, if he currently has enough coins (i.e p β₯ a_i). Thus, he buys all these k goods, and his number of coins decreases by a_i (i.e it becomes p := p - a_i).
Please note that each good can be bought no more than once.
For example, if the store now has n=5 goods worth a_1=2, a_2=4, a_3=3, a_4=5, a_5=7, respectively, k=2, and Vasya has 6 coins, then he can buy 3 goods. A good with the index 1 will be bought by Vasya without using the offer and he will pay 2 coins. Goods with the indices 2 and 3 Vasya will buy using the offer and he will pay 4 coins. It can be proved that Vasya can not buy more goods with six coins.
Help Vasya to find out the maximum number of goods he can buy.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases in the test.
The next lines contain a description of t test cases.
The first line of each test case contains three integers n, p, k (2 β€ n β€ 2 β
10^5, 1 β€ p β€ 2β
10^9, k=2) β the number of goods in the store, the number of coins Vasya has and the number of goods that can be bought by the price of the most expensive of them.
The second line of each test case contains n integers a_i (1 β€ a_i β€ 10^4) β the prices of goods.
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5. It is guaranteed that in this version of the problem k=2 for all test cases.
Output
For each test case in a separate line print one integer m β the maximum number of goods that Vasya can buy.
Example
Input
6
5 6 2
2 4 3 5 7
5 11 2
2 4 3 5 7
2 10000 2
10000 10000
2 9999 2
10000 10000
5 13 2
8 2 8 2 5
3 18 2
1 2 3
Output
3
4
2
0
4
3
Tags: dp, greedy, sortings
Correct Solution:
```
import sys
input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
t=int(input())
for _ in range(t):
n,p,k=[int(x) for x in input().split()]
a=[int(x) for x in input().split()]
a.sort()
dp=[0 for _ in range(n)] #min cost to take up to item i
ans=0
for i in range(n):
dp[i]+=a[i]
if 1<=i<k-1:
dp[i]+=dp[i-1]
if i-k>=0:
dp[i]+=dp[i-k]
if dp[i]<=p:
ans=max(ans,i+1)
print(ans)
```
| 93,630 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of this problem. The only difference is the constraint on k β the number of gifts in the offer. In this version: k=2.
Vasya came to the store to buy goods for his friends for the New Year. It turned out that he was very lucky β today the offer "k of goods for the price of one" is held in store. Remember, that in this problem k=2.
Using this offer, Vasya can buy exactly k of any goods, paying only for the most expensive of them. Vasya decided to take this opportunity and buy as many goods as possible for his friends with the money he has.
More formally, for each good, its price is determined by a_i β the number of coins it costs. Initially, Vasya has p coins. He wants to buy the maximum number of goods. Vasya can perform one of the following operations as many times as necessary:
* Vasya can buy one good with the index i if he currently has enough coins (i.e p β₯ a_i). After buying this good, the number of Vasya's coins will decrease by a_i, (i.e it becomes p := p - a_i).
* Vasya can buy a good with the index i, and also choose exactly k-1 goods, the price of which does not exceed a_i, if he currently has enough coins (i.e p β₯ a_i). Thus, he buys all these k goods, and his number of coins decreases by a_i (i.e it becomes p := p - a_i).
Please note that each good can be bought no more than once.
For example, if the store now has n=5 goods worth a_1=2, a_2=4, a_3=3, a_4=5, a_5=7, respectively, k=2, and Vasya has 6 coins, then he can buy 3 goods. A good with the index 1 will be bought by Vasya without using the offer and he will pay 2 coins. Goods with the indices 2 and 3 Vasya will buy using the offer and he will pay 4 coins. It can be proved that Vasya can not buy more goods with six coins.
Help Vasya to find out the maximum number of goods he can buy.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases in the test.
The next lines contain a description of t test cases.
The first line of each test case contains three integers n, p, k (2 β€ n β€ 2 β
10^5, 1 β€ p β€ 2β
10^9, k=2) β the number of goods in the store, the number of coins Vasya has and the number of goods that can be bought by the price of the most expensive of them.
The second line of each test case contains n integers a_i (1 β€ a_i β€ 10^4) β the prices of goods.
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5. It is guaranteed that in this version of the problem k=2 for all test cases.
Output
For each test case in a separate line print one integer m β the maximum number of goods that Vasya can buy.
Example
Input
6
5 6 2
2 4 3 5 7
5 11 2
2 4 3 5 7
2 10000 2
10000 10000
2 9999 2
10000 10000
5 13 2
8 2 8 2 5
3 18 2
1 2 3
Output
3
4
2
0
4
3
Tags: dp, greedy, sortings
Correct Solution:
```
def cal(arr, p, k):
dp = [0]*(len(arr)+2*k)
for i in range(len(arr)):
dp[i] = dp[i-k] + arr[i]
ans = -1
for i in range(len(arr)):
if p >= dp[i]:
ans = i
else:
break
return ans+1
t=int(input())
for _ in range(t):
n, p ,k = [int(i) for i in input().split()]
arr = sorted([int(i) for i in input().split()])
print(cal(arr, p, k))
```
| 93,631 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of this problem. The only difference is the constraint on k β the number of gifts in the offer. In this version: k=2.
Vasya came to the store to buy goods for his friends for the New Year. It turned out that he was very lucky β today the offer "k of goods for the price of one" is held in store. Remember, that in this problem k=2.
Using this offer, Vasya can buy exactly k of any goods, paying only for the most expensive of them. Vasya decided to take this opportunity and buy as many goods as possible for his friends with the money he has.
More formally, for each good, its price is determined by a_i β the number of coins it costs. Initially, Vasya has p coins. He wants to buy the maximum number of goods. Vasya can perform one of the following operations as many times as necessary:
* Vasya can buy one good with the index i if he currently has enough coins (i.e p β₯ a_i). After buying this good, the number of Vasya's coins will decrease by a_i, (i.e it becomes p := p - a_i).
* Vasya can buy a good with the index i, and also choose exactly k-1 goods, the price of which does not exceed a_i, if he currently has enough coins (i.e p β₯ a_i). Thus, he buys all these k goods, and his number of coins decreases by a_i (i.e it becomes p := p - a_i).
Please note that each good can be bought no more than once.
For example, if the store now has n=5 goods worth a_1=2, a_2=4, a_3=3, a_4=5, a_5=7, respectively, k=2, and Vasya has 6 coins, then he can buy 3 goods. A good with the index 1 will be bought by Vasya without using the offer and he will pay 2 coins. Goods with the indices 2 and 3 Vasya will buy using the offer and he will pay 4 coins. It can be proved that Vasya can not buy more goods with six coins.
Help Vasya to find out the maximum number of goods he can buy.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases in the test.
The next lines contain a description of t test cases.
The first line of each test case contains three integers n, p, k (2 β€ n β€ 2 β
10^5, 1 β€ p β€ 2β
10^9, k=2) β the number of goods in the store, the number of coins Vasya has and the number of goods that can be bought by the price of the most expensive of them.
The second line of each test case contains n integers a_i (1 β€ a_i β€ 10^4) β the prices of goods.
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5. It is guaranteed that in this version of the problem k=2 for all test cases.
Output
For each test case in a separate line print one integer m β the maximum number of goods that Vasya can buy.
Example
Input
6
5 6 2
2 4 3 5 7
5 11 2
2 4 3 5 7
2 10000 2
10000 10000
2 9999 2
10000 10000
5 13 2
8 2 8 2 5
3 18 2
1 2 3
Output
3
4
2
0
4
3
Tags: dp, greedy, sortings
Correct Solution:
```
for _ in range(int(input())):
n, p, k = map(int, input().split())
a = [int(i) for i in input().split()]
a.sort()
max_goods = 0
copy_p = p
if len(a) % 2 == 0:
for price in a[1::2]:
if price <= copy_p:
copy_p -= price
max_goods += 2
else:
break
else:
for price in a[1::2]:
if price <= copy_p:
copy_p -= price
max_goods += 2
else:
break
# print('info', price, copy_p)
max_goods += 1 if a[-1] <= copy_p else 0
max_goods2 = 0
if len(a) % 2 == 0:
for price in a[2::2]:
if price <= p:
p -= price
max_goods2 += 2
else:
break
if a[-1] <= p:
max_goods2 += 2
elif a[0] <= p:
max_goods2 += 1
else:
for price in a[2::2]:
if price <= p:
p -= price
max_goods2 += 2
else:
break
# print('info', price, copy_p)
max_goods2 += 1 if a[0] <= p else 0
print(max(max_goods, max_goods2))
```
| 93,632 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of this problem. The only difference is the constraint on k β the number of gifts in the offer. In this version: k=2.
Vasya came to the store to buy goods for his friends for the New Year. It turned out that he was very lucky β today the offer "k of goods for the price of one" is held in store. Remember, that in this problem k=2.
Using this offer, Vasya can buy exactly k of any goods, paying only for the most expensive of them. Vasya decided to take this opportunity and buy as many goods as possible for his friends with the money he has.
More formally, for each good, its price is determined by a_i β the number of coins it costs. Initially, Vasya has p coins. He wants to buy the maximum number of goods. Vasya can perform one of the following operations as many times as necessary:
* Vasya can buy one good with the index i if he currently has enough coins (i.e p β₯ a_i). After buying this good, the number of Vasya's coins will decrease by a_i, (i.e it becomes p := p - a_i).
* Vasya can buy a good with the index i, and also choose exactly k-1 goods, the price of which does not exceed a_i, if he currently has enough coins (i.e p β₯ a_i). Thus, he buys all these k goods, and his number of coins decreases by a_i (i.e it becomes p := p - a_i).
Please note that each good can be bought no more than once.
For example, if the store now has n=5 goods worth a_1=2, a_2=4, a_3=3, a_4=5, a_5=7, respectively, k=2, and Vasya has 6 coins, then he can buy 3 goods. A good with the index 1 will be bought by Vasya without using the offer and he will pay 2 coins. Goods with the indices 2 and 3 Vasya will buy using the offer and he will pay 4 coins. It can be proved that Vasya can not buy more goods with six coins.
Help Vasya to find out the maximum number of goods he can buy.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases in the test.
The next lines contain a description of t test cases.
The first line of each test case contains three integers n, p, k (2 β€ n β€ 2 β
10^5, 1 β€ p β€ 2β
10^9, k=2) β the number of goods in the store, the number of coins Vasya has and the number of goods that can be bought by the price of the most expensive of them.
The second line of each test case contains n integers a_i (1 β€ a_i β€ 10^4) β the prices of goods.
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5. It is guaranteed that in this version of the problem k=2 for all test cases.
Output
For each test case in a separate line print one integer m β the maximum number of goods that Vasya can buy.
Example
Input
6
5 6 2
2 4 3 5 7
5 11 2
2 4 3 5 7
2 10000 2
10000 10000
2 9999 2
10000 10000
5 13 2
8 2 8 2 5
3 18 2
1 2 3
Output
3
4
2
0
4
3
Tags: dp, greedy, sortings
Correct Solution:
```
from sys import stdin,stdout
from math import gcd,sqrt,factorial
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:0 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
for _ in range(I()):
n,p,k=R()
a=sorted(R())+[0]
ans=pre=0
for i in range(k+1):
sum=pre
if sum>p:break
cnt=i
for j in range(i+k-1,n,k):
if sum+a[j]<=p:
cnt+=k
sum+=a[j]
else:break
pre+=a[i]
ans=max(cnt,ans)
print(ans)
```
| 93,633 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of this problem. The only difference is the constraint on k β the number of gifts in the offer. In this version: k=2.
Vasya came to the store to buy goods for his friends for the New Year. It turned out that he was very lucky β today the offer "k of goods for the price of one" is held in store. Remember, that in this problem k=2.
Using this offer, Vasya can buy exactly k of any goods, paying only for the most expensive of them. Vasya decided to take this opportunity and buy as many goods as possible for his friends with the money he has.
More formally, for each good, its price is determined by a_i β the number of coins it costs. Initially, Vasya has p coins. He wants to buy the maximum number of goods. Vasya can perform one of the following operations as many times as necessary:
* Vasya can buy one good with the index i if he currently has enough coins (i.e p β₯ a_i). After buying this good, the number of Vasya's coins will decrease by a_i, (i.e it becomes p := p - a_i).
* Vasya can buy a good with the index i, and also choose exactly k-1 goods, the price of which does not exceed a_i, if he currently has enough coins (i.e p β₯ a_i). Thus, he buys all these k goods, and his number of coins decreases by a_i (i.e it becomes p := p - a_i).
Please note that each good can be bought no more than once.
For example, if the store now has n=5 goods worth a_1=2, a_2=4, a_3=3, a_4=5, a_5=7, respectively, k=2, and Vasya has 6 coins, then he can buy 3 goods. A good with the index 1 will be bought by Vasya without using the offer and he will pay 2 coins. Goods with the indices 2 and 3 Vasya will buy using the offer and he will pay 4 coins. It can be proved that Vasya can not buy more goods with six coins.
Help Vasya to find out the maximum number of goods he can buy.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases in the test.
The next lines contain a description of t test cases.
The first line of each test case contains three integers n, p, k (2 β€ n β€ 2 β
10^5, 1 β€ p β€ 2β
10^9, k=2) β the number of goods in the store, the number of coins Vasya has and the number of goods that can be bought by the price of the most expensive of them.
The second line of each test case contains n integers a_i (1 β€ a_i β€ 10^4) β the prices of goods.
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5. It is guaranteed that in this version of the problem k=2 for all test cases.
Output
For each test case in a separate line print one integer m β the maximum number of goods that Vasya can buy.
Example
Input
6
5 6 2
2 4 3 5 7
5 11 2
2 4 3 5 7
2 10000 2
10000 10000
2 9999 2
10000 10000
5 13 2
8 2 8 2 5
3 18 2
1 2 3
Output
3
4
2
0
4
3
Tags: dp, greedy, sortings
Correct Solution:
```
t=int(input())
for i in range(0,t):
n,p,k=map(int,input().split())
a=list(map(int,input().split()))
a.sort()
b=a[1:]
flag=0
money=p
count=0
for i in range(0,len(a)):
if flag==0:
if a[i]<=money:
count+=1
money-=a[i]
flag=1
else:
break
elif flag==1:
if money+a[i-1]>=a[i]:
count+=1
money=money + a[i-1] - a[i]
flag=0
else:
break
c=count
if p>=a[0]:
money=p-a[0]
count=1
flag=0
for i in range(0,len(b)):
if flag==0:
if b[i]<=money:
count+=1
money-=b[i]
flag=1
else:
break
elif flag==1:
if money+b[i-1]>=b[i]:
count+=1
money=money + b[i-1] - b[i]
flag=0
else:
break
if count>c:
c=count
print(c)
```
| 93,634 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of this problem. The only difference is the constraint on k β the number of gifts in the offer. In this version: k=2.
Vasya came to the store to buy goods for his friends for the New Year. It turned out that he was very lucky β today the offer "k of goods for the price of one" is held in store. Remember, that in this problem k=2.
Using this offer, Vasya can buy exactly k of any goods, paying only for the most expensive of them. Vasya decided to take this opportunity and buy as many goods as possible for his friends with the money he has.
More formally, for each good, its price is determined by a_i β the number of coins it costs. Initially, Vasya has p coins. He wants to buy the maximum number of goods. Vasya can perform one of the following operations as many times as necessary:
* Vasya can buy one good with the index i if he currently has enough coins (i.e p β₯ a_i). After buying this good, the number of Vasya's coins will decrease by a_i, (i.e it becomes p := p - a_i).
* Vasya can buy a good with the index i, and also choose exactly k-1 goods, the price of which does not exceed a_i, if he currently has enough coins (i.e p β₯ a_i). Thus, he buys all these k goods, and his number of coins decreases by a_i (i.e it becomes p := p - a_i).
Please note that each good can be bought no more than once.
For example, if the store now has n=5 goods worth a_1=2, a_2=4, a_3=3, a_4=5, a_5=7, respectively, k=2, and Vasya has 6 coins, then he can buy 3 goods. A good with the index 1 will be bought by Vasya without using the offer and he will pay 2 coins. Goods with the indices 2 and 3 Vasya will buy using the offer and he will pay 4 coins. It can be proved that Vasya can not buy more goods with six coins.
Help Vasya to find out the maximum number of goods he can buy.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases in the test.
The next lines contain a description of t test cases.
The first line of each test case contains three integers n, p, k (2 β€ n β€ 2 β
10^5, 1 β€ p β€ 2β
10^9, k=2) β the number of goods in the store, the number of coins Vasya has and the number of goods that can be bought by the price of the most expensive of them.
The second line of each test case contains n integers a_i (1 β€ a_i β€ 10^4) β the prices of goods.
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5. It is guaranteed that in this version of the problem k=2 for all test cases.
Output
For each test case in a separate line print one integer m β the maximum number of goods that Vasya can buy.
Example
Input
6
5 6 2
2 4 3 5 7
5 11 2
2 4 3 5 7
2 10000 2
10000 10000
2 9999 2
10000 10000
5 13 2
8 2 8 2 5
3 18 2
1 2 3
Output
3
4
2
0
4
3
Tags: dp, greedy, sortings
Correct Solution:
```
# -*- coding: utf-8 -*-
"""
#k,m=map(int,input().split())
t=int(input())
for _ in range(t):
n=int(input())
x=list(map(int,input().split()))
"""
from math import *
t=int(input())
for _ in range(t):
n,p,k=map(int,input().split())
x=list(map(int,input().split()))
if n>2:
s=0
c=0
x.sort()
for i in range(2,n,2):
if s+x[i]>p:
break;
else:
s=s+x[i]
c=c+2
# print(s,'#')
if p-s>=x[0]:
c=c+1
s=s+x[0]
d=0
s=0
for i in range(1,n,2):
if s+x[i]>p:
break;
else:
s=s+x[i]
# print(s,'2@')
d=d+2
print(max(c,d))
else:
x.sort()
if x[1]<=p:
print(2)
elif x[0]<=p:
print(1)
else:
print(0)
```
| 93,635 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of this problem. The only difference is the constraint on k β the number of gifts in the offer. In this version: k=2.
Vasya came to the store to buy goods for his friends for the New Year. It turned out that he was very lucky β today the offer "k of goods for the price of one" is held in store. Remember, that in this problem k=2.
Using this offer, Vasya can buy exactly k of any goods, paying only for the most expensive of them. Vasya decided to take this opportunity and buy as many goods as possible for his friends with the money he has.
More formally, for each good, its price is determined by a_i β the number of coins it costs. Initially, Vasya has p coins. He wants to buy the maximum number of goods. Vasya can perform one of the following operations as many times as necessary:
* Vasya can buy one good with the index i if he currently has enough coins (i.e p β₯ a_i). After buying this good, the number of Vasya's coins will decrease by a_i, (i.e it becomes p := p - a_i).
* Vasya can buy a good with the index i, and also choose exactly k-1 goods, the price of which does not exceed a_i, if he currently has enough coins (i.e p β₯ a_i). Thus, he buys all these k goods, and his number of coins decreases by a_i (i.e it becomes p := p - a_i).
Please note that each good can be bought no more than once.
For example, if the store now has n=5 goods worth a_1=2, a_2=4, a_3=3, a_4=5, a_5=7, respectively, k=2, and Vasya has 6 coins, then he can buy 3 goods. A good with the index 1 will be bought by Vasya without using the offer and he will pay 2 coins. Goods with the indices 2 and 3 Vasya will buy using the offer and he will pay 4 coins. It can be proved that Vasya can not buy more goods with six coins.
Help Vasya to find out the maximum number of goods he can buy.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases in the test.
The next lines contain a description of t test cases.
The first line of each test case contains three integers n, p, k (2 β€ n β€ 2 β
10^5, 1 β€ p β€ 2β
10^9, k=2) β the number of goods in the store, the number of coins Vasya has and the number of goods that can be bought by the price of the most expensive of them.
The second line of each test case contains n integers a_i (1 β€ a_i β€ 10^4) β the prices of goods.
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5. It is guaranteed that in this version of the problem k=2 for all test cases.
Output
For each test case in a separate line print one integer m β the maximum number of goods that Vasya can buy.
Example
Input
6
5 6 2
2 4 3 5 7
5 11 2
2 4 3 5 7
2 10000 2
10000 10000
2 9999 2
10000 10000
5 13 2
8 2 8 2 5
3 18 2
1 2 3
Output
3
4
2
0
4
3
Submitted Solution:
```
import sys
# from math
import bisect
import heapq
# from collections import deque
# from types import GeneratorType
# def bootstrap(func, stack=[]):
# def wrapped_function(*args, **kwargs):
# if stack:
# return func(*args, **kwargs)
# else:
# call = func(*args, **kwargs)
# while True:
# if type(call) is GeneratorType:
# stack.append(call)
# call = next(call)
# else:
# stack.pop()
# if not stack:
# break
# call = stack[-1].send(call)
# return call
# return wrapped_function
Ri = lambda : [int(x) for x in sys.stdin.readline().split()]
ri = lambda : sys.stdin.readline().strip()
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
INF = 10 ** 18
MOD = 10**9+7
for _ in range(int(ri())):
n,p,k = Ri()
a = Ri()
a.sort()
prefix = [0]*(n+1)
for i in range(0,len(a)):
prefix[i+1] = a[i]+prefix[i]
# print(prefix)
dpc = [i for i in a]
dpc = [0]+dpc
dpn = [0]*n
for i in range(len(a)):
# print(dpn)
if i < k-1:
if a[i] <= p and dpc[i] <= p-a[i]:
dpc[i+1] = a[i]+dpc[i]
dpn[i] = i+1
else:
# print("sd")
dpc[i+1] = INF
dpn[i] = 0
continue
# print(i,prefix[i+1]-prefix[i-k+1])
if a[i] <= p and dpc[i+1-k] <= p-a[i]:
dpc[i+1] = a[i]+dpc[i-k+1]
dpn[i] = i+1
else:
dpc[i+1] = INF
dpn[i] = 0
print(max(dpn))
```
Yes
| 93,636 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of this problem. The only difference is the constraint on k β the number of gifts in the offer. In this version: k=2.
Vasya came to the store to buy goods for his friends for the New Year. It turned out that he was very lucky β today the offer "k of goods for the price of one" is held in store. Remember, that in this problem k=2.
Using this offer, Vasya can buy exactly k of any goods, paying only for the most expensive of them. Vasya decided to take this opportunity and buy as many goods as possible for his friends with the money he has.
More formally, for each good, its price is determined by a_i β the number of coins it costs. Initially, Vasya has p coins. He wants to buy the maximum number of goods. Vasya can perform one of the following operations as many times as necessary:
* Vasya can buy one good with the index i if he currently has enough coins (i.e p β₯ a_i). After buying this good, the number of Vasya's coins will decrease by a_i, (i.e it becomes p := p - a_i).
* Vasya can buy a good with the index i, and also choose exactly k-1 goods, the price of which does not exceed a_i, if he currently has enough coins (i.e p β₯ a_i). Thus, he buys all these k goods, and his number of coins decreases by a_i (i.e it becomes p := p - a_i).
Please note that each good can be bought no more than once.
For example, if the store now has n=5 goods worth a_1=2, a_2=4, a_3=3, a_4=5, a_5=7, respectively, k=2, and Vasya has 6 coins, then he can buy 3 goods. A good with the index 1 will be bought by Vasya without using the offer and he will pay 2 coins. Goods with the indices 2 and 3 Vasya will buy using the offer and he will pay 4 coins. It can be proved that Vasya can not buy more goods with six coins.
Help Vasya to find out the maximum number of goods he can buy.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases in the test.
The next lines contain a description of t test cases.
The first line of each test case contains three integers n, p, k (2 β€ n β€ 2 β
10^5, 1 β€ p β€ 2β
10^9, k=2) β the number of goods in the store, the number of coins Vasya has and the number of goods that can be bought by the price of the most expensive of them.
The second line of each test case contains n integers a_i (1 β€ a_i β€ 10^4) β the prices of goods.
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5. It is guaranteed that in this version of the problem k=2 for all test cases.
Output
For each test case in a separate line print one integer m β the maximum number of goods that Vasya can buy.
Example
Input
6
5 6 2
2 4 3 5 7
5 11 2
2 4 3 5 7
2 10000 2
10000 10000
2 9999 2
10000 10000
5 13 2
8 2 8 2 5
3 18 2
1 2 3
Output
3
4
2
0
4
3
Submitted Solution:
```
sas = [1,2,3]
t = int(input())
for i in range(t):
n, p, k = map(int, input().split())
mas = list(map(int, input().split()))
mas.sort()
l = 0
r = n
def check(m):
if m % 2 != 0:
sm = 0
i = 2
while(i < m):
sm += mas[i]
i += 2
return mas[0] + sm <= p
else:
sm = 0
i = 1
while(i < m):
sm += mas[i]
i += 2
return sm <= p
while(r - l > 1):
m = (l + r) // 2
if check(m):
l = m
else:
r = m
if check(r):
print(r)
continue
if check(l):
print(l)
continue
print(0)
```
Yes
| 93,637 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of this problem. The only difference is the constraint on k β the number of gifts in the offer. In this version: k=2.
Vasya came to the store to buy goods for his friends for the New Year. It turned out that he was very lucky β today the offer "k of goods for the price of one" is held in store. Remember, that in this problem k=2.
Using this offer, Vasya can buy exactly k of any goods, paying only for the most expensive of them. Vasya decided to take this opportunity and buy as many goods as possible for his friends with the money he has.
More formally, for each good, its price is determined by a_i β the number of coins it costs. Initially, Vasya has p coins. He wants to buy the maximum number of goods. Vasya can perform one of the following operations as many times as necessary:
* Vasya can buy one good with the index i if he currently has enough coins (i.e p β₯ a_i). After buying this good, the number of Vasya's coins will decrease by a_i, (i.e it becomes p := p - a_i).
* Vasya can buy a good with the index i, and also choose exactly k-1 goods, the price of which does not exceed a_i, if he currently has enough coins (i.e p β₯ a_i). Thus, he buys all these k goods, and his number of coins decreases by a_i (i.e it becomes p := p - a_i).
Please note that each good can be bought no more than once.
For example, if the store now has n=5 goods worth a_1=2, a_2=4, a_3=3, a_4=5, a_5=7, respectively, k=2, and Vasya has 6 coins, then he can buy 3 goods. A good with the index 1 will be bought by Vasya without using the offer and he will pay 2 coins. Goods with the indices 2 and 3 Vasya will buy using the offer and he will pay 4 coins. It can be proved that Vasya can not buy more goods with six coins.
Help Vasya to find out the maximum number of goods he can buy.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases in the test.
The next lines contain a description of t test cases.
The first line of each test case contains three integers n, p, k (2 β€ n β€ 2 β
10^5, 1 β€ p β€ 2β
10^9, k=2) β the number of goods in the store, the number of coins Vasya has and the number of goods that can be bought by the price of the most expensive of them.
The second line of each test case contains n integers a_i (1 β€ a_i β€ 10^4) β the prices of goods.
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5. It is guaranteed that in this version of the problem k=2 for all test cases.
Output
For each test case in a separate line print one integer m β the maximum number of goods that Vasya can buy.
Example
Input
6
5 6 2
2 4 3 5 7
5 11 2
2 4 3 5 7
2 10000 2
10000 10000
2 9999 2
10000 10000
5 13 2
8 2 8 2 5
3 18 2
1 2 3
Output
3
4
2
0
4
3
Submitted Solution:
```
def getmaxnumber(n,p,k,ai):
aod = []
aeve = []
sumod = 0
sumeve = 0
sum = 0
flag = 0
for i in range(0,len(ai)):
if (i+1)%2 == 0:
aeve.append(ai[i])
sumeve = sumeve+ai[i]
sum = sumeve
flag = 0
else:
aod.append(ai[i])
sumod = sumod+ai[i]
sum = sumod
flag =1
if sum==p:
if flag>0:
# print('case1')
if i !=len(ai)-1:
aeve.append(ai[i+1])
sumeve = sumeve+ai[i+1]
if sumeve>p:
return (len(aod)-1)*2+1
else:
return len(aeve)*2
else:
return (len(aod)-1)*2+1
else:
# print('case2')
if i !=len(ai)-1:
aod.append(ai[i+1])
sumod = sumod+ai[i]
if sumod>p:
return len(aeve)*2
else:
return (len(aod)-1)*2+1
else:
return len(aeve)*2
# return len(aeve)*2
elif sum>p:
if flag>0:
# print('case3')
return len(aeve)*2
else:
# print('case4')
return (len(aod)-1)*2+1
elif i == len(ai)-1:
return len(ai)
number = int(input())
for i in range(0,number):
npk = input()
npkstr = npk.split()
npkfinal = list(map(int,npkstr))
n = npkfinal[0]
p = npkfinal[1]
k = npkfinal[2]
alla = input()
allastr = alla.split()
allafinal = list(map(int,allastr))
allafinal = sorted(allafinal)
# print(allafinal)
print(getmaxnumber(n,p,k,allafinal))
```
Yes
| 93,638 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of this problem. The only difference is the constraint on k β the number of gifts in the offer. In this version: k=2.
Vasya came to the store to buy goods for his friends for the New Year. It turned out that he was very lucky β today the offer "k of goods for the price of one" is held in store. Remember, that in this problem k=2.
Using this offer, Vasya can buy exactly k of any goods, paying only for the most expensive of them. Vasya decided to take this opportunity and buy as many goods as possible for his friends with the money he has.
More formally, for each good, its price is determined by a_i β the number of coins it costs. Initially, Vasya has p coins. He wants to buy the maximum number of goods. Vasya can perform one of the following operations as many times as necessary:
* Vasya can buy one good with the index i if he currently has enough coins (i.e p β₯ a_i). After buying this good, the number of Vasya's coins will decrease by a_i, (i.e it becomes p := p - a_i).
* Vasya can buy a good with the index i, and also choose exactly k-1 goods, the price of which does not exceed a_i, if he currently has enough coins (i.e p β₯ a_i). Thus, he buys all these k goods, and his number of coins decreases by a_i (i.e it becomes p := p - a_i).
Please note that each good can be bought no more than once.
For example, if the store now has n=5 goods worth a_1=2, a_2=4, a_3=3, a_4=5, a_5=7, respectively, k=2, and Vasya has 6 coins, then he can buy 3 goods. A good with the index 1 will be bought by Vasya without using the offer and he will pay 2 coins. Goods with the indices 2 and 3 Vasya will buy using the offer and he will pay 4 coins. It can be proved that Vasya can not buy more goods with six coins.
Help Vasya to find out the maximum number of goods he can buy.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases in the test.
The next lines contain a description of t test cases.
The first line of each test case contains three integers n, p, k (2 β€ n β€ 2 β
10^5, 1 β€ p β€ 2β
10^9, k=2) β the number of goods in the store, the number of coins Vasya has and the number of goods that can be bought by the price of the most expensive of them.
The second line of each test case contains n integers a_i (1 β€ a_i β€ 10^4) β the prices of goods.
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5. It is guaranteed that in this version of the problem k=2 for all test cases.
Output
For each test case in a separate line print one integer m β the maximum number of goods that Vasya can buy.
Example
Input
6
5 6 2
2 4 3 5 7
5 11 2
2 4 3 5 7
2 10000 2
10000 10000
2 9999 2
10000 10000
5 13 2
8 2 8 2 5
3 18 2
1 2 3
Output
3
4
2
0
4
3
Submitted Solution:
```
def solve(p, k, a):
def f(a, p, init):
last_i = init
ans = 0
for i in range(init, len(a), 2):
if p >= a[i]:
ans += 2 if i > 0 else 1
p -= a[i]
last_i = i
else:
break
if last_i + 1 < len(a):
if p >= a[last_i+1]:
ans += 1
return ans
return max(f(a, p, 0), f(a, p, 1))
t = int(input())
for i in range(t):
n, p, k = map(int, input().split())
a = sorted(list(map(int, input().split())))
print(solve(p, k, a))
```
Yes
| 93,639 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of this problem. The only difference is the constraint on k β the number of gifts in the offer. In this version: k=2.
Vasya came to the store to buy goods for his friends for the New Year. It turned out that he was very lucky β today the offer "k of goods for the price of one" is held in store. Remember, that in this problem k=2.
Using this offer, Vasya can buy exactly k of any goods, paying only for the most expensive of them. Vasya decided to take this opportunity and buy as many goods as possible for his friends with the money he has.
More formally, for each good, its price is determined by a_i β the number of coins it costs. Initially, Vasya has p coins. He wants to buy the maximum number of goods. Vasya can perform one of the following operations as many times as necessary:
* Vasya can buy one good with the index i if he currently has enough coins (i.e p β₯ a_i). After buying this good, the number of Vasya's coins will decrease by a_i, (i.e it becomes p := p - a_i).
* Vasya can buy a good with the index i, and also choose exactly k-1 goods, the price of which does not exceed a_i, if he currently has enough coins (i.e p β₯ a_i). Thus, he buys all these k goods, and his number of coins decreases by a_i (i.e it becomes p := p - a_i).
Please note that each good can be bought no more than once.
For example, if the store now has n=5 goods worth a_1=2, a_2=4, a_3=3, a_4=5, a_5=7, respectively, k=2, and Vasya has 6 coins, then he can buy 3 goods. A good with the index 1 will be bought by Vasya without using the offer and he will pay 2 coins. Goods with the indices 2 and 3 Vasya will buy using the offer and he will pay 4 coins. It can be proved that Vasya can not buy more goods with six coins.
Help Vasya to find out the maximum number of goods he can buy.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases in the test.
The next lines contain a description of t test cases.
The first line of each test case contains three integers n, p, k (2 β€ n β€ 2 β
10^5, 1 β€ p β€ 2β
10^9, k=2) β the number of goods in the store, the number of coins Vasya has and the number of goods that can be bought by the price of the most expensive of them.
The second line of each test case contains n integers a_i (1 β€ a_i β€ 10^4) β the prices of goods.
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5. It is guaranteed that in this version of the problem k=2 for all test cases.
Output
For each test case in a separate line print one integer m β the maximum number of goods that Vasya can buy.
Example
Input
6
5 6 2
2 4 3 5 7
5 11 2
2 4 3 5 7
2 10000 2
10000 10000
2 9999 2
10000 10000
5 13 2
8 2 8 2 5
3 18 2
1 2 3
Output
3
4
2
0
4
3
Submitted Solution:
```
t = int(input())
while(t>0):
n,p,k= [int(i) for i in input().split()]
a= [int(i) for i in input().split()]
a = sorted(a)
c = 0
f =0
d = 0
for i in range(n):
if a[i]<=p:
c+=1
p-=a[i]
f = 1
else:
d= 1
break
if f==1:
if d==1:
c+=1
else:
c = 0
print(c)
t-=1
```
No
| 93,640 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of this problem. The only difference is the constraint on k β the number of gifts in the offer. In this version: k=2.
Vasya came to the store to buy goods for his friends for the New Year. It turned out that he was very lucky β today the offer "k of goods for the price of one" is held in store. Remember, that in this problem k=2.
Using this offer, Vasya can buy exactly k of any goods, paying only for the most expensive of them. Vasya decided to take this opportunity and buy as many goods as possible for his friends with the money he has.
More formally, for each good, its price is determined by a_i β the number of coins it costs. Initially, Vasya has p coins. He wants to buy the maximum number of goods. Vasya can perform one of the following operations as many times as necessary:
* Vasya can buy one good with the index i if he currently has enough coins (i.e p β₯ a_i). After buying this good, the number of Vasya's coins will decrease by a_i, (i.e it becomes p := p - a_i).
* Vasya can buy a good with the index i, and also choose exactly k-1 goods, the price of which does not exceed a_i, if he currently has enough coins (i.e p β₯ a_i). Thus, he buys all these k goods, and his number of coins decreases by a_i (i.e it becomes p := p - a_i).
Please note that each good can be bought no more than once.
For example, if the store now has n=5 goods worth a_1=2, a_2=4, a_3=3, a_4=5, a_5=7, respectively, k=2, and Vasya has 6 coins, then he can buy 3 goods. A good with the index 1 will be bought by Vasya without using the offer and he will pay 2 coins. Goods with the indices 2 and 3 Vasya will buy using the offer and he will pay 4 coins. It can be proved that Vasya can not buy more goods with six coins.
Help Vasya to find out the maximum number of goods he can buy.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases in the test.
The next lines contain a description of t test cases.
The first line of each test case contains three integers n, p, k (2 β€ n β€ 2 β
10^5, 1 β€ p β€ 2β
10^9, k=2) β the number of goods in the store, the number of coins Vasya has and the number of goods that can be bought by the price of the most expensive of them.
The second line of each test case contains n integers a_i (1 β€ a_i β€ 10^4) β the prices of goods.
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5. It is guaranteed that in this version of the problem k=2 for all test cases.
Output
For each test case in a separate line print one integer m β the maximum number of goods that Vasya can buy.
Example
Input
6
5 6 2
2 4 3 5 7
5 11 2
2 4 3 5 7
2 10000 2
10000 10000
2 9999 2
10000 10000
5 13 2
8 2 8 2 5
3 18 2
1 2 3
Output
3
4
2
0
4
3
Submitted Solution:
```
t = int(input())
for _ in range(t) :
n,p,k = map(int,input().split())
a = list(map(int,input().split()))
a.sort()
count = 0
if len(a) >= 2 :
if a[1] <= p :
m = 2
else:
m = 0
for i in range(len(a)-2) :
if a[i] <= p :
p = p - a[i]
count +=1
m = max( m , count)
if a[i+2] <= p :
m = max( m , count +2)
print(m)
else :
if a[0] <= p :
print(1)
else :
print(0)
```
No
| 93,641 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of this problem. The only difference is the constraint on k β the number of gifts in the offer. In this version: k=2.
Vasya came to the store to buy goods for his friends for the New Year. It turned out that he was very lucky β today the offer "k of goods for the price of one" is held in store. Remember, that in this problem k=2.
Using this offer, Vasya can buy exactly k of any goods, paying only for the most expensive of them. Vasya decided to take this opportunity and buy as many goods as possible for his friends with the money he has.
More formally, for each good, its price is determined by a_i β the number of coins it costs. Initially, Vasya has p coins. He wants to buy the maximum number of goods. Vasya can perform one of the following operations as many times as necessary:
* Vasya can buy one good with the index i if he currently has enough coins (i.e p β₯ a_i). After buying this good, the number of Vasya's coins will decrease by a_i, (i.e it becomes p := p - a_i).
* Vasya can buy a good with the index i, and also choose exactly k-1 goods, the price of which does not exceed a_i, if he currently has enough coins (i.e p β₯ a_i). Thus, he buys all these k goods, and his number of coins decreases by a_i (i.e it becomes p := p - a_i).
Please note that each good can be bought no more than once.
For example, if the store now has n=5 goods worth a_1=2, a_2=4, a_3=3, a_4=5, a_5=7, respectively, k=2, and Vasya has 6 coins, then he can buy 3 goods. A good with the index 1 will be bought by Vasya without using the offer and he will pay 2 coins. Goods with the indices 2 and 3 Vasya will buy using the offer and he will pay 4 coins. It can be proved that Vasya can not buy more goods with six coins.
Help Vasya to find out the maximum number of goods he can buy.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases in the test.
The next lines contain a description of t test cases.
The first line of each test case contains three integers n, p, k (2 β€ n β€ 2 β
10^5, 1 β€ p β€ 2β
10^9, k=2) β the number of goods in the store, the number of coins Vasya has and the number of goods that can be bought by the price of the most expensive of them.
The second line of each test case contains n integers a_i (1 β€ a_i β€ 10^4) β the prices of goods.
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5. It is guaranteed that in this version of the problem k=2 for all test cases.
Output
For each test case in a separate line print one integer m β the maximum number of goods that Vasya can buy.
Example
Input
6
5 6 2
2 4 3 5 7
5 11 2
2 4 3 5 7
2 10000 2
10000 10000
2 9999 2
10000 10000
5 13 2
8 2 8 2 5
3 18 2
1 2 3
Output
3
4
2
0
4
3
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Fri Jan 3 10:18:30 2020
@author: 20122
"""
case = int(input())
number = []
for i in range(case):
a = input()
a = a.split()
n = int(a[0])
p = int(a[1])
k = int(a[2])
cost = input()
cost = list(map(int, cost.split()))
cost.sort()
total = 0
for j in range(0, k, k-1):
sum_p = 0
mount = 0
mark = 0
l = 0
if sum_p + cost[j] <= p:
sum_p += cost[j]
mount += j+1
for l in range(j+k, n, k):
if sum_p + cost[l] <= p:
sum_p += cost[l]
mount += k
else:
mark = 1
break
if mark == 0 and l < n-1 and j < n-1:
for m in range(l+1, n):
if sum_p + cost[m] <= p:
sum_p += cost[m]
mount += 1
else:
break
total = max(total, mount)
number.append(total)
for i in range(case):
print(number[i])
```
No
| 93,642 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of this problem. The only difference is the constraint on k β the number of gifts in the offer. In this version: k=2.
Vasya came to the store to buy goods for his friends for the New Year. It turned out that he was very lucky β today the offer "k of goods for the price of one" is held in store. Remember, that in this problem k=2.
Using this offer, Vasya can buy exactly k of any goods, paying only for the most expensive of them. Vasya decided to take this opportunity and buy as many goods as possible for his friends with the money he has.
More formally, for each good, its price is determined by a_i β the number of coins it costs. Initially, Vasya has p coins. He wants to buy the maximum number of goods. Vasya can perform one of the following operations as many times as necessary:
* Vasya can buy one good with the index i if he currently has enough coins (i.e p β₯ a_i). After buying this good, the number of Vasya's coins will decrease by a_i, (i.e it becomes p := p - a_i).
* Vasya can buy a good with the index i, and also choose exactly k-1 goods, the price of which does not exceed a_i, if he currently has enough coins (i.e p β₯ a_i). Thus, he buys all these k goods, and his number of coins decreases by a_i (i.e it becomes p := p - a_i).
Please note that each good can be bought no more than once.
For example, if the store now has n=5 goods worth a_1=2, a_2=4, a_3=3, a_4=5, a_5=7, respectively, k=2, and Vasya has 6 coins, then he can buy 3 goods. A good with the index 1 will be bought by Vasya without using the offer and he will pay 2 coins. Goods with the indices 2 and 3 Vasya will buy using the offer and he will pay 4 coins. It can be proved that Vasya can not buy more goods with six coins.
Help Vasya to find out the maximum number of goods he can buy.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases in the test.
The next lines contain a description of t test cases.
The first line of each test case contains three integers n, p, k (2 β€ n β€ 2 β
10^5, 1 β€ p β€ 2β
10^9, k=2) β the number of goods in the store, the number of coins Vasya has and the number of goods that can be bought by the price of the most expensive of them.
The second line of each test case contains n integers a_i (1 β€ a_i β€ 10^4) β the prices of goods.
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5. It is guaranteed that in this version of the problem k=2 for all test cases.
Output
For each test case in a separate line print one integer m β the maximum number of goods that Vasya can buy.
Example
Input
6
5 6 2
2 4 3 5 7
5 11 2
2 4 3 5 7
2 10000 2
10000 10000
2 9999 2
10000 10000
5 13 2
8 2 8 2 5
3 18 2
1 2 3
Output
3
4
2
0
4
3
Submitted Solution:
```
### a,b,c = map(int,input().split())
def calc(lst,p):
cnt = 0
i = 1
while i < len(lst):
if lst[i] > p:
if lst[i - 1] <= p:
cnt += 1
p -= lst[i - 1]
break
cnt += 2
p -= lst[i]
i += 2
if len(lst) % 2 == 1:
if p >= lst[len(lst) - 1]:
cnt+=1
p-=lst[len(lst)-1]
if cnt == 0:
if lst[0] <= p:
cnt = 1
p-=lst[0]
return cnt,p
def solve():
n,p,k = map(int,input().split())
lst = map(int,input().split())
lst = sorted(lst)
if p>=lst[0]:
cn=1
tp = p-lst[0]
res1,tmp = calc(lst,p)
res2,tmp = calc(lst[1:],p)
if res2%2==1:
res2+=1
else:
if tmp>=lst[0]:
res2+=1
print(max(res1,res2))
test = 1
test = int(input())
for i in range(test):
solve()
```
No
| 93,643 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alicia has an array, a_1, a_2, β¦, a_n, of non-negative integers. For each 1 β€ i β€ n, she has found a non-negative integer x_i = max(0, a_1, β¦, a_{i-1}). Note that for i=1, x_i = 0.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, then x = \{0, 0, 1, 2, 2\}.
Then, she calculated an array, b_1, b_2, β¦, b_n: b_i = a_i - x_i.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, b = \{0-0, 1-0, 2-1, 0-2, 3-2\} = \{0, 1, 1, -2, 1\}.
Alicia gives you the values b_1, b_2, β¦, b_n and asks you to restore the values a_1, a_2, β¦, a_n. Can you help her solve the problem?
Input
The first line contains one integer n (3 β€ n β€ 200 000) β the number of elements in Alicia's array.
The next line contains n integers, b_1, b_2, β¦, b_n (-10^9 β€ b_i β€ 10^9).
It is guaranteed that for the given array b there is a solution a_1, a_2, β¦, a_n, for all elements of which the following is true: 0 β€ a_i β€ 10^9.
Output
Print n integers, a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9), such that if you calculate x according to the statement, b_1 will be equal to a_1 - x_1, b_2 will be equal to a_2 - x_2, ..., and b_n will be equal to a_n - x_n.
It is guaranteed that there exists at least one solution for the given tests. It can be shown that the solution is unique.
Examples
Input
5
0 1 1 -2 1
Output
0 1 2 0 3
Input
3
1000 999999000 -1000000000
Output
1000 1000000000 0
Input
5
2 1 2 2 3
Output
2 3 5 7 10
Note
The first test was described in the problem statement.
In the second test, if Alicia had an array a = \{1000, 1000000000, 0\}, then x = \{0, 1000, 1000000000\} and b = \{1000-0, 1000000000-1000, 0-1000000000\} = \{1000, 999999000, -1000000000\}.
Tags: implementation, math
Correct Solution:
```
a = int(input())
a = list(map(int, input().split()))
x = max(a[0], 0 )
print(a[0], end=" ")
for i in range(1, len(a)):
print(x+a[i], end=" ")
x = max(x,x+a[i])
```
| 93,644 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alicia has an array, a_1, a_2, β¦, a_n, of non-negative integers. For each 1 β€ i β€ n, she has found a non-negative integer x_i = max(0, a_1, β¦, a_{i-1}). Note that for i=1, x_i = 0.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, then x = \{0, 0, 1, 2, 2\}.
Then, she calculated an array, b_1, b_2, β¦, b_n: b_i = a_i - x_i.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, b = \{0-0, 1-0, 2-1, 0-2, 3-2\} = \{0, 1, 1, -2, 1\}.
Alicia gives you the values b_1, b_2, β¦, b_n and asks you to restore the values a_1, a_2, β¦, a_n. Can you help her solve the problem?
Input
The first line contains one integer n (3 β€ n β€ 200 000) β the number of elements in Alicia's array.
The next line contains n integers, b_1, b_2, β¦, b_n (-10^9 β€ b_i β€ 10^9).
It is guaranteed that for the given array b there is a solution a_1, a_2, β¦, a_n, for all elements of which the following is true: 0 β€ a_i β€ 10^9.
Output
Print n integers, a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9), such that if you calculate x according to the statement, b_1 will be equal to a_1 - x_1, b_2 will be equal to a_2 - x_2, ..., and b_n will be equal to a_n - x_n.
It is guaranteed that there exists at least one solution for the given tests. It can be shown that the solution is unique.
Examples
Input
5
0 1 1 -2 1
Output
0 1 2 0 3
Input
3
1000 999999000 -1000000000
Output
1000 1000000000 0
Input
5
2 1 2 2 3
Output
2 3 5 7 10
Note
The first test was described in the problem statement.
In the second test, if Alicia had an array a = \{1000, 1000000000, 0\}, then x = \{0, 1000, 1000000000\} and b = \{1000-0, 1000000000-1000, 0-1000000000\} = \{1000, 999999000, -1000000000\}.
Tags: implementation, math
Correct Solution:
```
import sys
from math import sqrt
inp = sys.stdin.readline
# arr = list(map(int, inp().split()))
def is_prime(n):
for i in range(2, int(sqrt(n))+1):
if n % i == 0:
return(False)
return(True)
def a():
ans = ""
for _ in range(int(inp())):
n = int(inp())
s = ""
if n == 1:
s = "-1"
else:
s = ("2"*(n-1) + "3")
ans += (s+"\n")
print(ans)
def b():
n = int(inp())
arr = list(map(int, inp().split()))
arr2 = [arr[0]]
maxi = arr[0]
for i in range(1, n):
arr2.append(maxi + arr[i])
maxi = max(maxi, arr2[i])
print(*arr2)
if __name__ == "__main__":
# a()
b()
# c()
```
| 93,645 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alicia has an array, a_1, a_2, β¦, a_n, of non-negative integers. For each 1 β€ i β€ n, she has found a non-negative integer x_i = max(0, a_1, β¦, a_{i-1}). Note that for i=1, x_i = 0.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, then x = \{0, 0, 1, 2, 2\}.
Then, she calculated an array, b_1, b_2, β¦, b_n: b_i = a_i - x_i.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, b = \{0-0, 1-0, 2-1, 0-2, 3-2\} = \{0, 1, 1, -2, 1\}.
Alicia gives you the values b_1, b_2, β¦, b_n and asks you to restore the values a_1, a_2, β¦, a_n. Can you help her solve the problem?
Input
The first line contains one integer n (3 β€ n β€ 200 000) β the number of elements in Alicia's array.
The next line contains n integers, b_1, b_2, β¦, b_n (-10^9 β€ b_i β€ 10^9).
It is guaranteed that for the given array b there is a solution a_1, a_2, β¦, a_n, for all elements of which the following is true: 0 β€ a_i β€ 10^9.
Output
Print n integers, a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9), such that if you calculate x according to the statement, b_1 will be equal to a_1 - x_1, b_2 will be equal to a_2 - x_2, ..., and b_n will be equal to a_n - x_n.
It is guaranteed that there exists at least one solution for the given tests. It can be shown that the solution is unique.
Examples
Input
5
0 1 1 -2 1
Output
0 1 2 0 3
Input
3
1000 999999000 -1000000000
Output
1000 1000000000 0
Input
5
2 1 2 2 3
Output
2 3 5 7 10
Note
The first test was described in the problem statement.
In the second test, if Alicia had an array a = \{1000, 1000000000, 0\}, then x = \{0, 1000, 1000000000\} and b = \{1000-0, 1000000000-1000, 0-1000000000\} = \{1000, 999999000, -1000000000\}.
Tags: implementation, math
Correct Solution:
```
n=int(input())
l=[int(i) for i in input().split()]
m=0
for i in range(0,n):
k=l[i]+m
print(k,end=' ')
if m<k:
m=k
```
| 93,646 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alicia has an array, a_1, a_2, β¦, a_n, of non-negative integers. For each 1 β€ i β€ n, she has found a non-negative integer x_i = max(0, a_1, β¦, a_{i-1}). Note that for i=1, x_i = 0.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, then x = \{0, 0, 1, 2, 2\}.
Then, she calculated an array, b_1, b_2, β¦, b_n: b_i = a_i - x_i.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, b = \{0-0, 1-0, 2-1, 0-2, 3-2\} = \{0, 1, 1, -2, 1\}.
Alicia gives you the values b_1, b_2, β¦, b_n and asks you to restore the values a_1, a_2, β¦, a_n. Can you help her solve the problem?
Input
The first line contains one integer n (3 β€ n β€ 200 000) β the number of elements in Alicia's array.
The next line contains n integers, b_1, b_2, β¦, b_n (-10^9 β€ b_i β€ 10^9).
It is guaranteed that for the given array b there is a solution a_1, a_2, β¦, a_n, for all elements of which the following is true: 0 β€ a_i β€ 10^9.
Output
Print n integers, a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9), such that if you calculate x according to the statement, b_1 will be equal to a_1 - x_1, b_2 will be equal to a_2 - x_2, ..., and b_n will be equal to a_n - x_n.
It is guaranteed that there exists at least one solution for the given tests. It can be shown that the solution is unique.
Examples
Input
5
0 1 1 -2 1
Output
0 1 2 0 3
Input
3
1000 999999000 -1000000000
Output
1000 1000000000 0
Input
5
2 1 2 2 3
Output
2 3 5 7 10
Note
The first test was described in the problem statement.
In the second test, if Alicia had an array a = \{1000, 1000000000, 0\}, then x = \{0, 1000, 1000000000\} and b = \{1000-0, 1000000000-1000, 0-1000000000\} = \{1000, 999999000, -1000000000\}.
Tags: implementation, math
Correct Solution:
```
n=int(input())
l1=[int(x) for x in input().split()]
l=[]
l.append(l1[0])
m=l[0]
for j in range(1,n):
if l1[j]>0:
m=m+l1[j]
l.append(m)
else:
l.append(m+l1[j])
print(*l)
```
| 93,647 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alicia has an array, a_1, a_2, β¦, a_n, of non-negative integers. For each 1 β€ i β€ n, she has found a non-negative integer x_i = max(0, a_1, β¦, a_{i-1}). Note that for i=1, x_i = 0.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, then x = \{0, 0, 1, 2, 2\}.
Then, she calculated an array, b_1, b_2, β¦, b_n: b_i = a_i - x_i.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, b = \{0-0, 1-0, 2-1, 0-2, 3-2\} = \{0, 1, 1, -2, 1\}.
Alicia gives you the values b_1, b_2, β¦, b_n and asks you to restore the values a_1, a_2, β¦, a_n. Can you help her solve the problem?
Input
The first line contains one integer n (3 β€ n β€ 200 000) β the number of elements in Alicia's array.
The next line contains n integers, b_1, b_2, β¦, b_n (-10^9 β€ b_i β€ 10^9).
It is guaranteed that for the given array b there is a solution a_1, a_2, β¦, a_n, for all elements of which the following is true: 0 β€ a_i β€ 10^9.
Output
Print n integers, a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9), such that if you calculate x according to the statement, b_1 will be equal to a_1 - x_1, b_2 will be equal to a_2 - x_2, ..., and b_n will be equal to a_n - x_n.
It is guaranteed that there exists at least one solution for the given tests. It can be shown that the solution is unique.
Examples
Input
5
0 1 1 -2 1
Output
0 1 2 0 3
Input
3
1000 999999000 -1000000000
Output
1000 1000000000 0
Input
5
2 1 2 2 3
Output
2 3 5 7 10
Note
The first test was described in the problem statement.
In the second test, if Alicia had an array a = \{1000, 1000000000, 0\}, then x = \{0, 1000, 1000000000\} and b = \{1000-0, 1000000000-1000, 0-1000000000\} = \{1000, 999999000, -1000000000\}.
Tags: implementation, math
Correct Solution:
```
# your code goes here
n=int(input())
lst=list(map(int,input().split()))
sum1=0
a=[]
x=0
a.append(lst[0])
for i in range(1,n):
x=max(x,a[i-1])
a.append(lst[i]+x)
print(*a)
```
| 93,648 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alicia has an array, a_1, a_2, β¦, a_n, of non-negative integers. For each 1 β€ i β€ n, she has found a non-negative integer x_i = max(0, a_1, β¦, a_{i-1}). Note that for i=1, x_i = 0.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, then x = \{0, 0, 1, 2, 2\}.
Then, she calculated an array, b_1, b_2, β¦, b_n: b_i = a_i - x_i.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, b = \{0-0, 1-0, 2-1, 0-2, 3-2\} = \{0, 1, 1, -2, 1\}.
Alicia gives you the values b_1, b_2, β¦, b_n and asks you to restore the values a_1, a_2, β¦, a_n. Can you help her solve the problem?
Input
The first line contains one integer n (3 β€ n β€ 200 000) β the number of elements in Alicia's array.
The next line contains n integers, b_1, b_2, β¦, b_n (-10^9 β€ b_i β€ 10^9).
It is guaranteed that for the given array b there is a solution a_1, a_2, β¦, a_n, for all elements of which the following is true: 0 β€ a_i β€ 10^9.
Output
Print n integers, a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9), such that if you calculate x according to the statement, b_1 will be equal to a_1 - x_1, b_2 will be equal to a_2 - x_2, ..., and b_n will be equal to a_n - x_n.
It is guaranteed that there exists at least one solution for the given tests. It can be shown that the solution is unique.
Examples
Input
5
0 1 1 -2 1
Output
0 1 2 0 3
Input
3
1000 999999000 -1000000000
Output
1000 1000000000 0
Input
5
2 1 2 2 3
Output
2 3 5 7 10
Note
The first test was described in the problem statement.
In the second test, if Alicia had an array a = \{1000, 1000000000, 0\}, then x = \{0, 1000, 1000000000\} and b = \{1000-0, 1000000000-1000, 0-1000000000\} = \{1000, 999999000, -1000000000\}.
Tags: implementation, math
Correct Solution:
```
n=int(input())
b=list(map(int,input().split()))
ans=[b[0]]
m=b[0]
for i in range(1,n):
ans.append(m+b[i])
m=max(m,ans[-1])
print (*ans)
```
| 93,649 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alicia has an array, a_1, a_2, β¦, a_n, of non-negative integers. For each 1 β€ i β€ n, she has found a non-negative integer x_i = max(0, a_1, β¦, a_{i-1}). Note that for i=1, x_i = 0.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, then x = \{0, 0, 1, 2, 2\}.
Then, she calculated an array, b_1, b_2, β¦, b_n: b_i = a_i - x_i.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, b = \{0-0, 1-0, 2-1, 0-2, 3-2\} = \{0, 1, 1, -2, 1\}.
Alicia gives you the values b_1, b_2, β¦, b_n and asks you to restore the values a_1, a_2, β¦, a_n. Can you help her solve the problem?
Input
The first line contains one integer n (3 β€ n β€ 200 000) β the number of elements in Alicia's array.
The next line contains n integers, b_1, b_2, β¦, b_n (-10^9 β€ b_i β€ 10^9).
It is guaranteed that for the given array b there is a solution a_1, a_2, β¦, a_n, for all elements of which the following is true: 0 β€ a_i β€ 10^9.
Output
Print n integers, a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9), such that if you calculate x according to the statement, b_1 will be equal to a_1 - x_1, b_2 will be equal to a_2 - x_2, ..., and b_n will be equal to a_n - x_n.
It is guaranteed that there exists at least one solution for the given tests. It can be shown that the solution is unique.
Examples
Input
5
0 1 1 -2 1
Output
0 1 2 0 3
Input
3
1000 999999000 -1000000000
Output
1000 1000000000 0
Input
5
2 1 2 2 3
Output
2 3 5 7 10
Note
The first test was described in the problem statement.
In the second test, if Alicia had an array a = \{1000, 1000000000, 0\}, then x = \{0, 1000, 1000000000\} and b = \{1000-0, 1000000000-1000, 0-1000000000\} = \{1000, 999999000, -1000000000\}.
Tags: implementation, math
Correct Solution:
```
def solve():
n=input()
l=list(map(int,input().split()))
sum=0
sum2=l[0]
print(l[0],end=" ")
for i in range(1,len(l)):
sum=abs(sum2)+l[i]
print(sum,end=" ")
if(abs(sum)>sum2):
sum2=sum
print("")
solve()
```
| 93,650 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alicia has an array, a_1, a_2, β¦, a_n, of non-negative integers. For each 1 β€ i β€ n, she has found a non-negative integer x_i = max(0, a_1, β¦, a_{i-1}). Note that for i=1, x_i = 0.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, then x = \{0, 0, 1, 2, 2\}.
Then, she calculated an array, b_1, b_2, β¦, b_n: b_i = a_i - x_i.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, b = \{0-0, 1-0, 2-1, 0-2, 3-2\} = \{0, 1, 1, -2, 1\}.
Alicia gives you the values b_1, b_2, β¦, b_n and asks you to restore the values a_1, a_2, β¦, a_n. Can you help her solve the problem?
Input
The first line contains one integer n (3 β€ n β€ 200 000) β the number of elements in Alicia's array.
The next line contains n integers, b_1, b_2, β¦, b_n (-10^9 β€ b_i β€ 10^9).
It is guaranteed that for the given array b there is a solution a_1, a_2, β¦, a_n, for all elements of which the following is true: 0 β€ a_i β€ 10^9.
Output
Print n integers, a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9), such that if you calculate x according to the statement, b_1 will be equal to a_1 - x_1, b_2 will be equal to a_2 - x_2, ..., and b_n will be equal to a_n - x_n.
It is guaranteed that there exists at least one solution for the given tests. It can be shown that the solution is unique.
Examples
Input
5
0 1 1 -2 1
Output
0 1 2 0 3
Input
3
1000 999999000 -1000000000
Output
1000 1000000000 0
Input
5
2 1 2 2 3
Output
2 3 5 7 10
Note
The first test was described in the problem statement.
In the second test, if Alicia had an array a = \{1000, 1000000000, 0\}, then x = \{0, 1000, 1000000000\} and b = \{1000-0, 1000000000-1000, 0-1000000000\} = \{1000, 999999000, -1000000000\}.
Tags: implementation, math
Correct Solution:
```
n = int(input())
b = list(map(int, input().split()))
x = [0]
a = []
a.append(x[0] + b[0])
x.append(max(x[0], a[0]))
a.append(x[1] + b[1])
for i in range(2, n):
x.append(max(x[i-1], a[i-1]))
a.append(x[i] + b[i])
print(*a)
```
| 93,651 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alicia has an array, a_1, a_2, β¦, a_n, of non-negative integers. For each 1 β€ i β€ n, she has found a non-negative integer x_i = max(0, a_1, β¦, a_{i-1}). Note that for i=1, x_i = 0.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, then x = \{0, 0, 1, 2, 2\}.
Then, she calculated an array, b_1, b_2, β¦, b_n: b_i = a_i - x_i.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, b = \{0-0, 1-0, 2-1, 0-2, 3-2\} = \{0, 1, 1, -2, 1\}.
Alicia gives you the values b_1, b_2, β¦, b_n and asks you to restore the values a_1, a_2, β¦, a_n. Can you help her solve the problem?
Input
The first line contains one integer n (3 β€ n β€ 200 000) β the number of elements in Alicia's array.
The next line contains n integers, b_1, b_2, β¦, b_n (-10^9 β€ b_i β€ 10^9).
It is guaranteed that for the given array b there is a solution a_1, a_2, β¦, a_n, for all elements of which the following is true: 0 β€ a_i β€ 10^9.
Output
Print n integers, a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9), such that if you calculate x according to the statement, b_1 will be equal to a_1 - x_1, b_2 will be equal to a_2 - x_2, ..., and b_n will be equal to a_n - x_n.
It is guaranteed that there exists at least one solution for the given tests. It can be shown that the solution is unique.
Examples
Input
5
0 1 1 -2 1
Output
0 1 2 0 3
Input
3
1000 999999000 -1000000000
Output
1000 1000000000 0
Input
5
2 1 2 2 3
Output
2 3 5 7 10
Note
The first test was described in the problem statement.
In the second test, if Alicia had an array a = \{1000, 1000000000, 0\}, then x = \{0, 1000, 1000000000\} and b = \{1000-0, 1000000000-1000, 0-1000000000\} = \{1000, 999999000, -1000000000\}.
Submitted Solution:
```
n = int(input())
arr = list(map(int, input().split()))
temp = 0
for i in arr:
print(temp + i, end=" ")
if(i > 0):
temp += i
```
Yes
| 93,652 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alicia has an array, a_1, a_2, β¦, a_n, of non-negative integers. For each 1 β€ i β€ n, she has found a non-negative integer x_i = max(0, a_1, β¦, a_{i-1}). Note that for i=1, x_i = 0.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, then x = \{0, 0, 1, 2, 2\}.
Then, she calculated an array, b_1, b_2, β¦, b_n: b_i = a_i - x_i.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, b = \{0-0, 1-0, 2-1, 0-2, 3-2\} = \{0, 1, 1, -2, 1\}.
Alicia gives you the values b_1, b_2, β¦, b_n and asks you to restore the values a_1, a_2, β¦, a_n. Can you help her solve the problem?
Input
The first line contains one integer n (3 β€ n β€ 200 000) β the number of elements in Alicia's array.
The next line contains n integers, b_1, b_2, β¦, b_n (-10^9 β€ b_i β€ 10^9).
It is guaranteed that for the given array b there is a solution a_1, a_2, β¦, a_n, for all elements of which the following is true: 0 β€ a_i β€ 10^9.
Output
Print n integers, a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9), such that if you calculate x according to the statement, b_1 will be equal to a_1 - x_1, b_2 will be equal to a_2 - x_2, ..., and b_n will be equal to a_n - x_n.
It is guaranteed that there exists at least one solution for the given tests. It can be shown that the solution is unique.
Examples
Input
5
0 1 1 -2 1
Output
0 1 2 0 3
Input
3
1000 999999000 -1000000000
Output
1000 1000000000 0
Input
5
2 1 2 2 3
Output
2 3 5 7 10
Note
The first test was described in the problem statement.
In the second test, if Alicia had an array a = \{1000, 1000000000, 0\}, then x = \{0, 1000, 1000000000\} and b = \{1000-0, 1000000000-1000, 0-1000000000\} = \{1000, 999999000, -1000000000\}.
Submitted Solution:
```
n = int(input())
b=[]
a=[0]*n
x=[0]*n
b = [int(i) for i in input().split()]
#print(x)
for i in range(1,n):
if i<=(n-1):
a[i-1]=b[i-1]+x[i-1]
x[i]=max(a[i-1],x[i-1])
a[n-1]=b[n-1]+x[n-1]
for i in range(n):
print(a[i], end=" ")
```
Yes
| 93,653 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alicia has an array, a_1, a_2, β¦, a_n, of non-negative integers. For each 1 β€ i β€ n, she has found a non-negative integer x_i = max(0, a_1, β¦, a_{i-1}). Note that for i=1, x_i = 0.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, then x = \{0, 0, 1, 2, 2\}.
Then, she calculated an array, b_1, b_2, β¦, b_n: b_i = a_i - x_i.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, b = \{0-0, 1-0, 2-1, 0-2, 3-2\} = \{0, 1, 1, -2, 1\}.
Alicia gives you the values b_1, b_2, β¦, b_n and asks you to restore the values a_1, a_2, β¦, a_n. Can you help her solve the problem?
Input
The first line contains one integer n (3 β€ n β€ 200 000) β the number of elements in Alicia's array.
The next line contains n integers, b_1, b_2, β¦, b_n (-10^9 β€ b_i β€ 10^9).
It is guaranteed that for the given array b there is a solution a_1, a_2, β¦, a_n, for all elements of which the following is true: 0 β€ a_i β€ 10^9.
Output
Print n integers, a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9), such that if you calculate x according to the statement, b_1 will be equal to a_1 - x_1, b_2 will be equal to a_2 - x_2, ..., and b_n will be equal to a_n - x_n.
It is guaranteed that there exists at least one solution for the given tests. It can be shown that the solution is unique.
Examples
Input
5
0 1 1 -2 1
Output
0 1 2 0 3
Input
3
1000 999999000 -1000000000
Output
1000 1000000000 0
Input
5
2 1 2 2 3
Output
2 3 5 7 10
Note
The first test was described in the problem statement.
In the second test, if Alicia had an array a = \{1000, 1000000000, 0\}, then x = \{0, 1000, 1000000000\} and b = \{1000-0, 1000000000-1000, 0-1000000000\} = \{1000, 999999000, -1000000000\}.
Submitted Solution:
```
length = int(input())
b = list(map(int, input().split()))
x = 0
for i in range(length):
a = b[i] + x
print(a, end=' ')
x = max(a, x)
```
Yes
| 93,654 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alicia has an array, a_1, a_2, β¦, a_n, of non-negative integers. For each 1 β€ i β€ n, she has found a non-negative integer x_i = max(0, a_1, β¦, a_{i-1}). Note that for i=1, x_i = 0.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, then x = \{0, 0, 1, 2, 2\}.
Then, she calculated an array, b_1, b_2, β¦, b_n: b_i = a_i - x_i.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, b = \{0-0, 1-0, 2-1, 0-2, 3-2\} = \{0, 1, 1, -2, 1\}.
Alicia gives you the values b_1, b_2, β¦, b_n and asks you to restore the values a_1, a_2, β¦, a_n. Can you help her solve the problem?
Input
The first line contains one integer n (3 β€ n β€ 200 000) β the number of elements in Alicia's array.
The next line contains n integers, b_1, b_2, β¦, b_n (-10^9 β€ b_i β€ 10^9).
It is guaranteed that for the given array b there is a solution a_1, a_2, β¦, a_n, for all elements of which the following is true: 0 β€ a_i β€ 10^9.
Output
Print n integers, a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9), such that if you calculate x according to the statement, b_1 will be equal to a_1 - x_1, b_2 will be equal to a_2 - x_2, ..., and b_n will be equal to a_n - x_n.
It is guaranteed that there exists at least one solution for the given tests. It can be shown that the solution is unique.
Examples
Input
5
0 1 1 -2 1
Output
0 1 2 0 3
Input
3
1000 999999000 -1000000000
Output
1000 1000000000 0
Input
5
2 1 2 2 3
Output
2 3 5 7 10
Note
The first test was described in the problem statement.
In the second test, if Alicia had an array a = \{1000, 1000000000, 0\}, then x = \{0, 1000, 1000000000\} and b = \{1000-0, 1000000000-1000, 0-1000000000\} = \{1000, 999999000, -1000000000\}.
Submitted Solution:
```
n=int(input())
b=list(map(int,input().split()))
a=[b[0],b[1]+b[0]]
ma=max(a)
for x in range(2,n):
num=b[x]+ma
a.append(num)
if num>ma:ma=num
print(*a)
```
Yes
| 93,655 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alicia has an array, a_1, a_2, β¦, a_n, of non-negative integers. For each 1 β€ i β€ n, she has found a non-negative integer x_i = max(0, a_1, β¦, a_{i-1}). Note that for i=1, x_i = 0.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, then x = \{0, 0, 1, 2, 2\}.
Then, she calculated an array, b_1, b_2, β¦, b_n: b_i = a_i - x_i.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, b = \{0-0, 1-0, 2-1, 0-2, 3-2\} = \{0, 1, 1, -2, 1\}.
Alicia gives you the values b_1, b_2, β¦, b_n and asks you to restore the values a_1, a_2, β¦, a_n. Can you help her solve the problem?
Input
The first line contains one integer n (3 β€ n β€ 200 000) β the number of elements in Alicia's array.
The next line contains n integers, b_1, b_2, β¦, b_n (-10^9 β€ b_i β€ 10^9).
It is guaranteed that for the given array b there is a solution a_1, a_2, β¦, a_n, for all elements of which the following is true: 0 β€ a_i β€ 10^9.
Output
Print n integers, a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9), such that if you calculate x according to the statement, b_1 will be equal to a_1 - x_1, b_2 will be equal to a_2 - x_2, ..., and b_n will be equal to a_n - x_n.
It is guaranteed that there exists at least one solution for the given tests. It can be shown that the solution is unique.
Examples
Input
5
0 1 1 -2 1
Output
0 1 2 0 3
Input
3
1000 999999000 -1000000000
Output
1000 1000000000 0
Input
5
2 1 2 2 3
Output
2 3 5 7 10
Note
The first test was described in the problem statement.
In the second test, if Alicia had an array a = \{1000, 1000000000, 0\}, then x = \{0, 1000, 1000000000\} and b = \{1000-0, 1000000000-1000, 0-1000000000\} = \{1000, 999999000, -1000000000\}.
Submitted Solution:
```
N = int(input())
b = list(map(int,input().split()))
largest = 0
for i in range(N):
temp = b[i]
b[i]+=largest
if b[i]>0:
largest+=temp
for i in b:
print(i,end=' ')
```
No
| 93,656 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alicia has an array, a_1, a_2, β¦, a_n, of non-negative integers. For each 1 β€ i β€ n, she has found a non-negative integer x_i = max(0, a_1, β¦, a_{i-1}). Note that for i=1, x_i = 0.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, then x = \{0, 0, 1, 2, 2\}.
Then, she calculated an array, b_1, b_2, β¦, b_n: b_i = a_i - x_i.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, b = \{0-0, 1-0, 2-1, 0-2, 3-2\} = \{0, 1, 1, -2, 1\}.
Alicia gives you the values b_1, b_2, β¦, b_n and asks you to restore the values a_1, a_2, β¦, a_n. Can you help her solve the problem?
Input
The first line contains one integer n (3 β€ n β€ 200 000) β the number of elements in Alicia's array.
The next line contains n integers, b_1, b_2, β¦, b_n (-10^9 β€ b_i β€ 10^9).
It is guaranteed that for the given array b there is a solution a_1, a_2, β¦, a_n, for all elements of which the following is true: 0 β€ a_i β€ 10^9.
Output
Print n integers, a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9), such that if you calculate x according to the statement, b_1 will be equal to a_1 - x_1, b_2 will be equal to a_2 - x_2, ..., and b_n will be equal to a_n - x_n.
It is guaranteed that there exists at least one solution for the given tests. It can be shown that the solution is unique.
Examples
Input
5
0 1 1 -2 1
Output
0 1 2 0 3
Input
3
1000 999999000 -1000000000
Output
1000 1000000000 0
Input
5
2 1 2 2 3
Output
2 3 5 7 10
Note
The first test was described in the problem statement.
In the second test, if Alicia had an array a = \{1000, 1000000000, 0\}, then x = \{0, 1000, 1000000000\} and b = \{1000-0, 1000000000-1000, 0-1000000000\} = \{1000, 999999000, -1000000000\}.
Submitted Solution:
```
n=int(input())
a=[int(i) for i in input().split()]
b=[]
c=[]
for i in range(n):
if(i==0):
b.append(a[i])
c.append(a[i])
else:
if(a[i]>0):
b.append(a[i]+c[i-1])
c.append(a[i]+c[i-1])
else:
b.append(0)
c.append(abs(a[i]))
print(" ".join(str(i) for i in b))
```
No
| 93,657 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alicia has an array, a_1, a_2, β¦, a_n, of non-negative integers. For each 1 β€ i β€ n, she has found a non-negative integer x_i = max(0, a_1, β¦, a_{i-1}). Note that for i=1, x_i = 0.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, then x = \{0, 0, 1, 2, 2\}.
Then, she calculated an array, b_1, b_2, β¦, b_n: b_i = a_i - x_i.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, b = \{0-0, 1-0, 2-1, 0-2, 3-2\} = \{0, 1, 1, -2, 1\}.
Alicia gives you the values b_1, b_2, β¦, b_n and asks you to restore the values a_1, a_2, β¦, a_n. Can you help her solve the problem?
Input
The first line contains one integer n (3 β€ n β€ 200 000) β the number of elements in Alicia's array.
The next line contains n integers, b_1, b_2, β¦, b_n (-10^9 β€ b_i β€ 10^9).
It is guaranteed that for the given array b there is a solution a_1, a_2, β¦, a_n, for all elements of which the following is true: 0 β€ a_i β€ 10^9.
Output
Print n integers, a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9), such that if you calculate x according to the statement, b_1 will be equal to a_1 - x_1, b_2 will be equal to a_2 - x_2, ..., and b_n will be equal to a_n - x_n.
It is guaranteed that there exists at least one solution for the given tests. It can be shown that the solution is unique.
Examples
Input
5
0 1 1 -2 1
Output
0 1 2 0 3
Input
3
1000 999999000 -1000000000
Output
1000 1000000000 0
Input
5
2 1 2 2 3
Output
2 3 5 7 10
Note
The first test was described in the problem statement.
In the second test, if Alicia had an array a = \{1000, 1000000000, 0\}, then x = \{0, 1000, 1000000000\} and b = \{1000-0, 1000000000-1000, 0-1000000000\} = \{1000, 999999000, -1000000000\}.
Submitted Solution:
```
n = int(input())
b = [int(i) for i in input().split()]
a = [b[0]]
print(b[0],end=' ')
ma = a[0]
for i in range(1, n):
a.append(b[i] + ma)
ma = max(ma, a[-1])
print(b[i]+ma)
```
No
| 93,658 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alicia has an array, a_1, a_2, β¦, a_n, of non-negative integers. For each 1 β€ i β€ n, she has found a non-negative integer x_i = max(0, a_1, β¦, a_{i-1}). Note that for i=1, x_i = 0.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, then x = \{0, 0, 1, 2, 2\}.
Then, she calculated an array, b_1, b_2, β¦, b_n: b_i = a_i - x_i.
For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, b = \{0-0, 1-0, 2-1, 0-2, 3-2\} = \{0, 1, 1, -2, 1\}.
Alicia gives you the values b_1, b_2, β¦, b_n and asks you to restore the values a_1, a_2, β¦, a_n. Can you help her solve the problem?
Input
The first line contains one integer n (3 β€ n β€ 200 000) β the number of elements in Alicia's array.
The next line contains n integers, b_1, b_2, β¦, b_n (-10^9 β€ b_i β€ 10^9).
It is guaranteed that for the given array b there is a solution a_1, a_2, β¦, a_n, for all elements of which the following is true: 0 β€ a_i β€ 10^9.
Output
Print n integers, a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9), such that if you calculate x according to the statement, b_1 will be equal to a_1 - x_1, b_2 will be equal to a_2 - x_2, ..., and b_n will be equal to a_n - x_n.
It is guaranteed that there exists at least one solution for the given tests. It can be shown that the solution is unique.
Examples
Input
5
0 1 1 -2 1
Output
0 1 2 0 3
Input
3
1000 999999000 -1000000000
Output
1000 1000000000 0
Input
5
2 1 2 2 3
Output
2 3 5 7 10
Note
The first test was described in the problem statement.
In the second test, if Alicia had an array a = \{1000, 1000000000, 0\}, then x = \{0, 1000, 1000000000\} and b = \{1000-0, 1000000000-1000, 0-1000000000\} = \{1000, 999999000, -1000000000\}.
Submitted Solution:
```
n=int(input())
arr=list(map(int,input().split()))
pref=[0]*n
pref[0]=arr[0]
for i in range(1,n):
if pref[i-1]==0:
if i!=1:
pref[i]=pref[i-2]+arr[i]
elif i==1:
pref[i]=pref[i-1]+arr[i]
else:
pref[i]=pref[i-1]+arr[i]
for i in range(n):
print(pref[i],end=' ')
print('')
```
No
| 93,659 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a special jigsaw puzzle consisting of nβ
m identical pieces. Every piece has three tabs and one blank, as pictured below.
<image>
The jigsaw puzzle is considered solved if the following conditions hold:
1. The pieces are arranged into a grid with n rows and m columns.
2. For any two pieces that share an edge in the grid, a tab of one piece fits perfectly into a blank of the other piece.
Through rotation and translation of the pieces, determine if it is possible to solve the jigsaw puzzle.
Input
The test consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next t lines contain descriptions of test cases.
Each test case contains two integers n and m (1 β€ n,m β€ 10^5).
Output
For each test case output a single line containing "YES" if it is possible to solve the jigsaw puzzle, or "NO" otherwise. You can print each letter in any case (upper or lower).
Example
Input
3
1 3
100000 100000
2 2
Output
YES
NO
YES
Note
For the first test case, this is an example solution:
<image>
For the second test case, we can show that no solution exists.
For the third test case, this is an example solution:
<image>
Tags: math
Correct Solution:
```
from math import *
from sys import stdin,stdout
# input = stdin.readline
for __ in range(int(input())):
n,m=map(int,input().split())
if(min(n,m)<=1 or (n==2 and m==2)):
print("YES")
else:
print("NO")
```
| 93,660 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a special jigsaw puzzle consisting of nβ
m identical pieces. Every piece has three tabs and one blank, as pictured below.
<image>
The jigsaw puzzle is considered solved if the following conditions hold:
1. The pieces are arranged into a grid with n rows and m columns.
2. For any two pieces that share an edge in the grid, a tab of one piece fits perfectly into a blank of the other piece.
Through rotation and translation of the pieces, determine if it is possible to solve the jigsaw puzzle.
Input
The test consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next t lines contain descriptions of test cases.
Each test case contains two integers n and m (1 β€ n,m β€ 10^5).
Output
For each test case output a single line containing "YES" if it is possible to solve the jigsaw puzzle, or "NO" otherwise. You can print each letter in any case (upper or lower).
Example
Input
3
1 3
100000 100000
2 2
Output
YES
NO
YES
Note
For the first test case, this is an example solution:
<image>
For the second test case, we can show that no solution exists.
For the third test case, this is an example solution:
<image>
Tags: math
Correct Solution:
```
for t in range(int(input())):
a,b = map(int,input().split())
if a==b==2:
print('Yes')
elif a == 1 or b == 1: print('Yes')
else: print('No')
```
| 93,661 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a special jigsaw puzzle consisting of nβ
m identical pieces. Every piece has three tabs and one blank, as pictured below.
<image>
The jigsaw puzzle is considered solved if the following conditions hold:
1. The pieces are arranged into a grid with n rows and m columns.
2. For any two pieces that share an edge in the grid, a tab of one piece fits perfectly into a blank of the other piece.
Through rotation and translation of the pieces, determine if it is possible to solve the jigsaw puzzle.
Input
The test consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next t lines contain descriptions of test cases.
Each test case contains two integers n and m (1 β€ n,m β€ 10^5).
Output
For each test case output a single line containing "YES" if it is possible to solve the jigsaw puzzle, or "NO" otherwise. You can print each letter in any case (upper or lower).
Example
Input
3
1 3
100000 100000
2 2
Output
YES
NO
YES
Note
For the first test case, this is an example solution:
<image>
For the second test case, we can show that no solution exists.
For the third test case, this is an example solution:
<image>
Tags: math
Correct Solution:
```
z=input
from math import *
for _ in range(int(z())):
a,b=map(int,z().split())
if a==b<=2 or a==1 or b==1:
print('YES')
else:
print('NO')
```
| 93,662 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a special jigsaw puzzle consisting of nβ
m identical pieces. Every piece has three tabs and one blank, as pictured below.
<image>
The jigsaw puzzle is considered solved if the following conditions hold:
1. The pieces are arranged into a grid with n rows and m columns.
2. For any two pieces that share an edge in the grid, a tab of one piece fits perfectly into a blank of the other piece.
Through rotation and translation of the pieces, determine if it is possible to solve the jigsaw puzzle.
Input
The test consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next t lines contain descriptions of test cases.
Each test case contains two integers n and m (1 β€ n,m β€ 10^5).
Output
For each test case output a single line containing "YES" if it is possible to solve the jigsaw puzzle, or "NO" otherwise. You can print each letter in any case (upper or lower).
Example
Input
3
1 3
100000 100000
2 2
Output
YES
NO
YES
Note
For the first test case, this is an example solution:
<image>
For the second test case, we can show that no solution exists.
For the third test case, this is an example solution:
<image>
Tags: math
Correct Solution:
```
t = int(input())
for x in range(t):
wejscie = str(input())
a, b = wejscie.split()
a = int(a)
b = int(b)
if a == 1 or b == 1:
print("YES")
else:
if a <= 2 and b <= 2:
print("YES")
else:
print("NO")
```
| 93,663 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a special jigsaw puzzle consisting of nβ
m identical pieces. Every piece has three tabs and one blank, as pictured below.
<image>
The jigsaw puzzle is considered solved if the following conditions hold:
1. The pieces are arranged into a grid with n rows and m columns.
2. For any two pieces that share an edge in the grid, a tab of one piece fits perfectly into a blank of the other piece.
Through rotation and translation of the pieces, determine if it is possible to solve the jigsaw puzzle.
Input
The test consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next t lines contain descriptions of test cases.
Each test case contains two integers n and m (1 β€ n,m β€ 10^5).
Output
For each test case output a single line containing "YES" if it is possible to solve the jigsaw puzzle, or "NO" otherwise. You can print each letter in any case (upper or lower).
Example
Input
3
1 3
100000 100000
2 2
Output
YES
NO
YES
Note
For the first test case, this is an example solution:
<image>
For the second test case, we can show that no solution exists.
For the third test case, this is an example solution:
<image>
Tags: math
Correct Solution:
```
for _ in range(int(input())):
# n = int(input())
n, m = list(map(int, input().split()))
# arr = list(map(str, input().split()))
# arr = list(input())
# temp1 = temp2 = stars = stars1 = flag = 0
if n == 1 or m == 1 :
print('YES')
elif n + m <= 4:
print('YES')
else:
print('NO')
```
| 93,664 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a special jigsaw puzzle consisting of nβ
m identical pieces. Every piece has three tabs and one blank, as pictured below.
<image>
The jigsaw puzzle is considered solved if the following conditions hold:
1. The pieces are arranged into a grid with n rows and m columns.
2. For any two pieces that share an edge in the grid, a tab of one piece fits perfectly into a blank of the other piece.
Through rotation and translation of the pieces, determine if it is possible to solve the jigsaw puzzle.
Input
The test consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next t lines contain descriptions of test cases.
Each test case contains two integers n and m (1 β€ n,m β€ 10^5).
Output
For each test case output a single line containing "YES" if it is possible to solve the jigsaw puzzle, or "NO" otherwise. You can print each letter in any case (upper or lower).
Example
Input
3
1 3
100000 100000
2 2
Output
YES
NO
YES
Note
For the first test case, this is an example solution:
<image>
For the second test case, we can show that no solution exists.
For the third test case, this is an example solution:
<image>
Tags: math
Correct Solution:
```
import sys
input = sys.stdin.readline
t = int(input())
nm = [list(map(int, input().split())) for _ in range(t)]
for n, m in nm:
if n == 1 or m == 1:
print('YES')
elif n == 2 and m == 2:
print('YES')
else:
print('NO')
```
| 93,665 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a special jigsaw puzzle consisting of nβ
m identical pieces. Every piece has three tabs and one blank, as pictured below.
<image>
The jigsaw puzzle is considered solved if the following conditions hold:
1. The pieces are arranged into a grid with n rows and m columns.
2. For any two pieces that share an edge in the grid, a tab of one piece fits perfectly into a blank of the other piece.
Through rotation and translation of the pieces, determine if it is possible to solve the jigsaw puzzle.
Input
The test consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next t lines contain descriptions of test cases.
Each test case contains two integers n and m (1 β€ n,m β€ 10^5).
Output
For each test case output a single line containing "YES" if it is possible to solve the jigsaw puzzle, or "NO" otherwise. You can print each letter in any case (upper or lower).
Example
Input
3
1 3
100000 100000
2 2
Output
YES
NO
YES
Note
For the first test case, this is an example solution:
<image>
For the second test case, we can show that no solution exists.
For the third test case, this is an example solution:
<image>
Tags: math
Correct Solution:
```
test=int(input())
for i in range(test):
n,m=input().split()
if n=="1" or m=="1" or (n=="2" and m=="2"):
print("YES")
else:
print("NO")
```
| 93,666 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a special jigsaw puzzle consisting of nβ
m identical pieces. Every piece has three tabs and one blank, as pictured below.
<image>
The jigsaw puzzle is considered solved if the following conditions hold:
1. The pieces are arranged into a grid with n rows and m columns.
2. For any two pieces that share an edge in the grid, a tab of one piece fits perfectly into a blank of the other piece.
Through rotation and translation of the pieces, determine if it is possible to solve the jigsaw puzzle.
Input
The test consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next t lines contain descriptions of test cases.
Each test case contains two integers n and m (1 β€ n,m β€ 10^5).
Output
For each test case output a single line containing "YES" if it is possible to solve the jigsaw puzzle, or "NO" otherwise. You can print each letter in any case (upper or lower).
Example
Input
3
1 3
100000 100000
2 2
Output
YES
NO
YES
Note
For the first test case, this is an example solution:
<image>
For the second test case, we can show that no solution exists.
For the third test case, this is an example solution:
<image>
Tags: math
Correct Solution:
```
Times=int(input())
j=0
while j<Times:
s=input().split()
m=int(s[0])
n=int(s[1])
if m>n:
m,n=n,m
if m*n<=m+n:
print("YES")
else:
print("NO")
j+=1
```
| 93,667 |
Provide tags and a correct Python 2 solution for this coding contest problem.
You are given a special jigsaw puzzle consisting of nβ
m identical pieces. Every piece has three tabs and one blank, as pictured below.
<image>
The jigsaw puzzle is considered solved if the following conditions hold:
1. The pieces are arranged into a grid with n rows and m columns.
2. For any two pieces that share an edge in the grid, a tab of one piece fits perfectly into a blank of the other piece.
Through rotation and translation of the pieces, determine if it is possible to solve the jigsaw puzzle.
Input
The test consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next t lines contain descriptions of test cases.
Each test case contains two integers n and m (1 β€ n,m β€ 10^5).
Output
For each test case output a single line containing "YES" if it is possible to solve the jigsaw puzzle, or "NO" otherwise. You can print each letter in any case (upper or lower).
Example
Input
3
1 3
100000 100000
2 2
Output
YES
NO
YES
Note
For the first test case, this is an example solution:
<image>
For the second test case, we can show that no solution exists.
For the third test case, this is an example solution:
<image>
Tags: math
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
def in_num():
return int(raw_input())
def in_arr():
return map(int,raw_input().split())
def pr_num(n):
stdout.write(str(n)+'\n')
def pr_arr(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return map(int,stdin.read().split())
range = xrange # not for python 3.0+
for t in range(input()):
n,m=in_arr()
if min(n,m)==1 or (n==2 and m==2):
pr('YES\n')
else:
pr('NO\n')
```
| 93,668 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a special jigsaw puzzle consisting of nβ
m identical pieces. Every piece has three tabs and one blank, as pictured below.
<image>
The jigsaw puzzle is considered solved if the following conditions hold:
1. The pieces are arranged into a grid with n rows and m columns.
2. For any two pieces that share an edge in the grid, a tab of one piece fits perfectly into a blank of the other piece.
Through rotation and translation of the pieces, determine if it is possible to solve the jigsaw puzzle.
Input
The test consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next t lines contain descriptions of test cases.
Each test case contains two integers n and m (1 β€ n,m β€ 10^5).
Output
For each test case output a single line containing "YES" if it is possible to solve the jigsaw puzzle, or "NO" otherwise. You can print each letter in any case (upper or lower).
Example
Input
3
1 3
100000 100000
2 2
Output
YES
NO
YES
Note
For the first test case, this is an example solution:
<image>
For the second test case, we can show that no solution exists.
For the third test case, this is an example solution:
<image>
Submitted Solution:
```
t = int(input())
for i in range(t):
n,m = map(int,input().split())
if n==1 or m==1 or n==m==2:
print("YES")
else:
print("NO")
```
Yes
| 93,669 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a special jigsaw puzzle consisting of nβ
m identical pieces. Every piece has three tabs and one blank, as pictured below.
<image>
The jigsaw puzzle is considered solved if the following conditions hold:
1. The pieces are arranged into a grid with n rows and m columns.
2. For any two pieces that share an edge in the grid, a tab of one piece fits perfectly into a blank of the other piece.
Through rotation and translation of the pieces, determine if it is possible to solve the jigsaw puzzle.
Input
The test consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next t lines contain descriptions of test cases.
Each test case contains two integers n and m (1 β€ n,m β€ 10^5).
Output
For each test case output a single line containing "YES" if it is possible to solve the jigsaw puzzle, or "NO" otherwise. You can print each letter in any case (upper or lower).
Example
Input
3
1 3
100000 100000
2 2
Output
YES
NO
YES
Note
For the first test case, this is an example solution:
<image>
For the second test case, we can show that no solution exists.
For the third test case, this is an example solution:
<image>
Submitted Solution:
```
def answer(n,m):
if n==1 or m==1:
return "YES"
elif n<=2 and m<=2:
return "YES"
else:
return "NO"
t=int(input())
for i in range(t):
n,m=map(int,input().split())
print(answer(n,m))
```
Yes
| 93,670 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a special jigsaw puzzle consisting of nβ
m identical pieces. Every piece has three tabs and one blank, as pictured below.
<image>
The jigsaw puzzle is considered solved if the following conditions hold:
1. The pieces are arranged into a grid with n rows and m columns.
2. For any two pieces that share an edge in the grid, a tab of one piece fits perfectly into a blank of the other piece.
Through rotation and translation of the pieces, determine if it is possible to solve the jigsaw puzzle.
Input
The test consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next t lines contain descriptions of test cases.
Each test case contains two integers n and m (1 β€ n,m β€ 10^5).
Output
For each test case output a single line containing "YES" if it is possible to solve the jigsaw puzzle, or "NO" otherwise. You can print each letter in any case (upper or lower).
Example
Input
3
1 3
100000 100000
2 2
Output
YES
NO
YES
Note
For the first test case, this is an example solution:
<image>
For the second test case, we can show that no solution exists.
For the third test case, this is an example solution:
<image>
Submitted Solution:
```
for _ in range(int(input())):
n,m=map(int,input().split())
if n==1 or m==1:
print("YES")
elif n*m==4:
print("YES")
else:
print("NO")
```
Yes
| 93,671 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a special jigsaw puzzle consisting of nβ
m identical pieces. Every piece has three tabs and one blank, as pictured below.
<image>
The jigsaw puzzle is considered solved if the following conditions hold:
1. The pieces are arranged into a grid with n rows and m columns.
2. For any two pieces that share an edge in the grid, a tab of one piece fits perfectly into a blank of the other piece.
Through rotation and translation of the pieces, determine if it is possible to solve the jigsaw puzzle.
Input
The test consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next t lines contain descriptions of test cases.
Each test case contains two integers n and m (1 β€ n,m β€ 10^5).
Output
For each test case output a single line containing "YES" if it is possible to solve the jigsaw puzzle, or "NO" otherwise. You can print each letter in any case (upper or lower).
Example
Input
3
1 3
100000 100000
2 2
Output
YES
NO
YES
Note
For the first test case, this is an example solution:
<image>
For the second test case, we can show that no solution exists.
For the third test case, this is an example solution:
<image>
Submitted Solution:
```
t = int(input())
for i in range(t):
n,m = map(int,input().split())
if(m*n>=2*m*n-m-n):
print("YES")
else:
print("NO")
```
Yes
| 93,672 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a special jigsaw puzzle consisting of nβ
m identical pieces. Every piece has three tabs and one blank, as pictured below.
<image>
The jigsaw puzzle is considered solved if the following conditions hold:
1. The pieces are arranged into a grid with n rows and m columns.
2. For any two pieces that share an edge in the grid, a tab of one piece fits perfectly into a blank of the other piece.
Through rotation and translation of the pieces, determine if it is possible to solve the jigsaw puzzle.
Input
The test consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next t lines contain descriptions of test cases.
Each test case contains two integers n and m (1 β€ n,m β€ 10^5).
Output
For each test case output a single line containing "YES" if it is possible to solve the jigsaw puzzle, or "NO" otherwise. You can print each letter in any case (upper or lower).
Example
Input
3
1 3
100000 100000
2 2
Output
YES
NO
YES
Note
For the first test case, this is an example solution:
<image>
For the second test case, we can show that no solution exists.
For the third test case, this is an example solution:
<image>
Submitted Solution:
```
for _ in range(int(input())):
n,m=map(int,input().split())
if n>2 and m==1 or m>2 and n==1:
print('YES')
elif n==2 and m==2:
print('YES')
else:
print('NO')
```
No
| 93,673 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a special jigsaw puzzle consisting of nβ
m identical pieces. Every piece has three tabs and one blank, as pictured below.
<image>
The jigsaw puzzle is considered solved if the following conditions hold:
1. The pieces are arranged into a grid with n rows and m columns.
2. For any two pieces that share an edge in the grid, a tab of one piece fits perfectly into a blank of the other piece.
Through rotation and translation of the pieces, determine if it is possible to solve the jigsaw puzzle.
Input
The test consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next t lines contain descriptions of test cases.
Each test case contains two integers n and m (1 β€ n,m β€ 10^5).
Output
For each test case output a single line containing "YES" if it is possible to solve the jigsaw puzzle, or "NO" otherwise. You can print each letter in any case (upper or lower).
Example
Input
3
1 3
100000 100000
2 2
Output
YES
NO
YES
Note
For the first test case, this is an example solution:
<image>
For the second test case, we can show that no solution exists.
For the third test case, this is an example solution:
<image>
Submitted Solution:
```
# import sys
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
for _ in range(int(input())):
n,m = map(int,input().split())
if (min(n,m)<=2):
print('YES')
else:
print('NO')
```
No
| 93,674 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a special jigsaw puzzle consisting of nβ
m identical pieces. Every piece has three tabs and one blank, as pictured below.
<image>
The jigsaw puzzle is considered solved if the following conditions hold:
1. The pieces are arranged into a grid with n rows and m columns.
2. For any two pieces that share an edge in the grid, a tab of one piece fits perfectly into a blank of the other piece.
Through rotation and translation of the pieces, determine if it is possible to solve the jigsaw puzzle.
Input
The test consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next t lines contain descriptions of test cases.
Each test case contains two integers n and m (1 β€ n,m β€ 10^5).
Output
For each test case output a single line containing "YES" if it is possible to solve the jigsaw puzzle, or "NO" otherwise. You can print each letter in any case (upper or lower).
Example
Input
3
1 3
100000 100000
2 2
Output
YES
NO
YES
Note
For the first test case, this is an example solution:
<image>
For the second test case, we can show that no solution exists.
For the third test case, this is an example solution:
<image>
Submitted Solution:
```
for _ in range(int(input())):
n,m = map(int,input().split())
if(n>2 and m>2):
print("NO")
else:
print("YES")
```
No
| 93,675 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a special jigsaw puzzle consisting of nβ
m identical pieces. Every piece has three tabs and one blank, as pictured below.
<image>
The jigsaw puzzle is considered solved if the following conditions hold:
1. The pieces are arranged into a grid with n rows and m columns.
2. For any two pieces that share an edge in the grid, a tab of one piece fits perfectly into a blank of the other piece.
Through rotation and translation of the pieces, determine if it is possible to solve the jigsaw puzzle.
Input
The test consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next t lines contain descriptions of test cases.
Each test case contains two integers n and m (1 β€ n,m β€ 10^5).
Output
For each test case output a single line containing "YES" if it is possible to solve the jigsaw puzzle, or "NO" otherwise. You can print each letter in any case (upper or lower).
Example
Input
3
1 3
100000 100000
2 2
Output
YES
NO
YES
Note
For the first test case, this is an example solution:
<image>
For the second test case, we can show that no solution exists.
For the third test case, this is an example solution:
<image>
Submitted Solution:
```
t = int(input())
for _ in range(t):
n, m = (int(i) for i in input().split())
if(n == 1 or n == 2 or m == 1 or m == 2):
print("YES")
else:
print("NO")
```
No
| 93,676 |
Evaluate the correctness of the submitted Python 2 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a special jigsaw puzzle consisting of nβ
m identical pieces. Every piece has three tabs and one blank, as pictured below.
<image>
The jigsaw puzzle is considered solved if the following conditions hold:
1. The pieces are arranged into a grid with n rows and m columns.
2. For any two pieces that share an edge in the grid, a tab of one piece fits perfectly into a blank of the other piece.
Through rotation and translation of the pieces, determine if it is possible to solve the jigsaw puzzle.
Input
The test consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next t lines contain descriptions of test cases.
Each test case contains two integers n and m (1 β€ n,m β€ 10^5).
Output
For each test case output a single line containing "YES" if it is possible to solve the jigsaw puzzle, or "NO" otherwise. You can print each letter in any case (upper or lower).
Example
Input
3
1 3
100000 100000
2 2
Output
YES
NO
YES
Note
For the first test case, this is an example solution:
<image>
For the second test case, we can show that no solution exists.
For the third test case, this is an example solution:
<image>
Submitted Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
def in_num():
return int(raw_input())
def in_arr():
return map(int,raw_input().split())
def pr_num(n):
stdout.write(str(n)+'\n')
def pr_arr(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return map(int,stdin.read().split())
range = xrange # not for python 3.0+
for t in range(input()):
if min(in_arr())>2:
pr('NO\n')
else:
pr('YES\n')
```
No
| 93,677 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's denote the function f(s) that takes a string s consisting of lowercase Latin letters and dots, and returns a string consisting of lowercase Latin letters as follows:
1. let r be an empty string;
2. process the characters of s from left to right. For each character c, do the following: if c is a lowercase Latin letter, append c at the end of the string r; otherwise, delete the last character from r (if r is empty before deleting the last character β the function crashes);
3. return r as the result of the function.
You are given two strings s and t. You have to delete the minimum possible number of characters from s so that f(s) = t (and the function does not crash). Note that you aren't allowed to insert new characters into s or reorder the existing ones.
Input
The input consists of two lines: the first one contains s β a string consisting of lowercase Latin letters and dots, the second one contains t β a string consisting of lowercase Latin letters (1 β€ |t| β€ |s| β€ 10000).
Additional constraint on the input: it is possible to remove some number of characters from s so that f(s) = t.
Output
Print one integer β the minimum possible number of characters you have to delete from s so f(s) does not crash and returns t as the result of the function.
Examples
Input
a.ba.b.
abb
Output
2
Input
.bbac..a.c.cd
bacd
Output
3
Input
c..code..c...o.d.de
code
Output
3
Submitted Solution:
```
s = input()
t = input()
count = 0
def f(s):
r = ""
for i in range(len(s)):
if s[i] != ".":
r+=s[i]
else:
if r!="":
r=r[:-1]
else:
r = r[1:]
globals()['count'] += 1
return r
r = f(s)
if len(r) == len(t):
for i in range(len(r)):
if r[i] != t[i]:
count +=2
else:
count += abs(len(r)-len(t))
print(count)
```
No
| 93,678 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's denote the function f(s) that takes a string s consisting of lowercase Latin letters and dots, and returns a string consisting of lowercase Latin letters as follows:
1. let r be an empty string;
2. process the characters of s from left to right. For each character c, do the following: if c is a lowercase Latin letter, append c at the end of the string r; otherwise, delete the last character from r (if r is empty before deleting the last character β the function crashes);
3. return r as the result of the function.
You are given two strings s and t. You have to delete the minimum possible number of characters from s so that f(s) = t (and the function does not crash). Note that you aren't allowed to insert new characters into s or reorder the existing ones.
Input
The input consists of two lines: the first one contains s β a string consisting of lowercase Latin letters and dots, the second one contains t β a string consisting of lowercase Latin letters (1 β€ |t| β€ |s| β€ 10000).
Additional constraint on the input: it is possible to remove some number of characters from s so that f(s) = t.
Output
Print one integer β the minimum possible number of characters you have to delete from s so f(s) does not crash and returns t as the result of the function.
Examples
Input
a.ba.b.
abb
Output
2
Input
.bbac..a.c.cd
bacd
Output
3
Input
c..code..c...o.d.de
code
Output
3
Submitted Solution:
```
s=input()
new_set=set(s)
aplha=len(new_set)
print(aplha-1)
```
No
| 93,679 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's denote the function f(s) that takes a string s consisting of lowercase Latin letters and dots, and returns a string consisting of lowercase Latin letters as follows:
1. let r be an empty string;
2. process the characters of s from left to right. For each character c, do the following: if c is a lowercase Latin letter, append c at the end of the string r; otherwise, delete the last character from r (if r is empty before deleting the last character β the function crashes);
3. return r as the result of the function.
You are given two strings s and t. You have to delete the minimum possible number of characters from s so that f(s) = t (and the function does not crash). Note that you aren't allowed to insert new characters into s or reorder the existing ones.
Input
The input consists of two lines: the first one contains s β a string consisting of lowercase Latin letters and dots, the second one contains t β a string consisting of lowercase Latin letters (1 β€ |t| β€ |s| β€ 10000).
Additional constraint on the input: it is possible to remove some number of characters from s so that f(s) = t.
Output
Print one integer β the minimum possible number of characters you have to delete from s so f(s) does not crash and returns t as the result of the function.
Examples
Input
a.ba.b.
abb
Output
2
Input
.bbac..a.c.cd
bacd
Output
3
Input
c..code..c...o.d.de
code
Output
3
Submitted Solution:
```
l=input()
q=input()
k=''
m=l.replace('.','')
for i in m.lower():
if i not in k:
k=k+i
c=''
for i in k:
if i not in q:
c=c+i
count=0
for i in q:
count+=1
print(count-3)
```
No
| 93,680 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's denote the function f(s) that takes a string s consisting of lowercase Latin letters and dots, and returns a string consisting of lowercase Latin letters as follows:
1. let r be an empty string;
2. process the characters of s from left to right. For each character c, do the following: if c is a lowercase Latin letter, append c at the end of the string r; otherwise, delete the last character from r (if r is empty before deleting the last character β the function crashes);
3. return r as the result of the function.
You are given two strings s and t. You have to delete the minimum possible number of characters from s so that f(s) = t (and the function does not crash). Note that you aren't allowed to insert new characters into s or reorder the existing ones.
Input
The input consists of two lines: the first one contains s β a string consisting of lowercase Latin letters and dots, the second one contains t β a string consisting of lowercase Latin letters (1 β€ |t| β€ |s| β€ 10000).
Additional constraint on the input: it is possible to remove some number of characters from s so that f(s) = t.
Output
Print one integer β the minimum possible number of characters you have to delete from s so f(s) does not crash and returns t as the result of the function.
Examples
Input
a.ba.b.
abb
Output
2
Input
.bbac..a.c.cd
bacd
Output
3
Input
c..code..c...o.d.de
code
Output
3
Submitted Solution:
```
def string(s):
r=''
for i in s:
if(i in r or i=='.'):
pass
else:
r+=i
return(r)
s=input()
print(string(s))
```
No
| 93,681 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n computers in the company network. They are numbered from 1 to n.
For each pair of two computers 1 β€ i < j β€ n you know the value a_{i,j}: the difficulty of sending data between computers i and j. All values a_{i,j} for i<j are different.
You want to separate all computers into k sets A_1, A_2, β¦, A_k, such that the following conditions are satisfied:
* for each computer 1 β€ i β€ n there is exactly one set A_j, such that i β A_j;
* for each two pairs of computers (s, f) and (x, y) (s β f, x β y), such that s, f, x are from the same set but x and y are from different sets, a_{s,f} < a_{x,y}.
For each 1 β€ k β€ n find the number of ways to divide computers into k groups, such that all required conditions are satisfied. These values can be large, so you need to find them by modulo 998 244 353.
Input
The first line contains a single integer n (1 β€ n β€ 1500): the number of computers.
The i-th of the next n lines contains n integers a_{i,1}, a_{i,2}, β¦, a_{i,n}(0 β€ a_{i,j} β€ (n (n-1))/(2)).
It is guaranteed that:
* for all 1 β€ i β€ n a_{i,i} = 0;
* for all 1 β€ i < j β€ n a_{i,j} > 0;
* for all 1 β€ i < j β€ n a_{i,j} = a_{j,i};
* all a_{i,j} for i <j are different.
Output
Print n integers: the k-th of them should be equal to the number of possible ways to divide computers into k groups, such that all required conditions are satisfied, modulo 998 244 353.
Examples
Input
4
0 3 4 6
3 0 2 1
4 2 0 5
6 1 5 0
Output
1 0 1 1
Input
7
0 1 18 15 19 12 21
1 0 16 13 17 20 14
18 16 0 2 7 10 9
15 13 2 0 6 8 11
19 17 7 6 0 4 5
12 20 10 8 4 0 3
21 14 9 11 5 3 0
Output
1 1 2 3 4 3 1
Note
Here are all possible ways to separate all computers into 4 groups in the second example:
* \{1, 2\}, \{3, 4\}, \{5\}, \{6, 7\};
* \{1\}, \{2\}, \{3, 4\}, \{5, 6, 7\};
* \{1, 2\}, \{3\}, \{4\}, \{5, 6, 7\}.
Tags: combinatorics, dp, dsu, fft, graphs, trees
Correct Solution:
```
# import itertools as it
# import functools as ft
import math
teststring = """4
0 3 4 6
3 0 2 1
4 2 0 5
6 1 5 0
"""
online = __file__ != "/home/jhli/py/Grakn/Problem_G2.py"
true, false = True, False
if True:
def spitout():
for c in teststring.splitlines():
yield c
_ito = spitout()
if not online:
def input():
return next(_ito)
def build_enum(*a):
built = dict()
for i, c in enumerate(a):
built[c] = i
return lambda x: built[x]
# T = 1
# T = int(input())
##-----------------start coding-----------------
n = int(input())
E = [(0, 0)] * (int(n*(n-1)/2))
P = 998244353
for i in range(n):
L = list(map(int, input().split(" ")))
for j in range(i+1, n):
E[L[j]-1] = (i, j)
R = list(range(n))
C = [[0, 1] for _ in range(n)]
Nv = [1] * n
Ne = [0] * n
def root(x):
if x == R[x]:
return x
else:
R[x] = y = root(R[x])
return y
def prod(A, B, da, db):
C = [0] * (min(da+db, n) + 1)
for i in range(da+1):
for j in range(db+1):
if i + j <= n:
C[i+j] += A[i] * B[j]
C[i+j] %= P
return C
# print(E)
# print("")
for (x, y) in E:
r = rx = root(x)
ry = root(y)
# print((x, y, w), (rx, ry))
if rx != ry:
if r > ry: r = ry
R[rx] = R[ry] = r
C[r] = prod(C[rx], C[ry], Nv[rx], Nv[ry])
Nv[r] = Nv[rx] + Nv[ry]
Ne[r] = Ne[rx] + Ne[ry] + 1
else:
Ne[r] += 1
if Ne[r]*2 == Nv[r] * (Nv[r] - 1):
C[r][1] = 1
# print("R", R)
# print("Nv", Nv)
# print("Ne", Ne)
# print("C", C)
# print("")
print(" ".join(map(str, C[0][1:n+1])))
# print('Case #{}: {}'.format(ti, '...'))
```
| 93,682 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored AΒ·BΒ·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks.
At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) Γ (B - 2) Γ (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 Γ 1 Γ 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B ΠΈ C.
Given number n, find the minimally possible and maximally possible number of stolen hay blocks.
Input
The only line contains integer n from the problem's statement (1 β€ n β€ 109).
Output
Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves.
Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in Π‘++. It is preferred to use cin, cout streams or the %I64d specificator.
Examples
Input
4
Output
28 41
Input
7
Output
47 65
Input
12
Output
48 105
Note
Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 Γ 4 Γ 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) Γ (4 - 2) Γ (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 Γ 3 Γ 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) Γ (3 - 2) Γ (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks.
Tags: brute force, math
Correct Solution:
```
n = int(input())
u = v = 9 + 8 * n
for a in range(1, int(n ** (1 / 3)) + 1):
if n % a: continue
m = n // a
for b in range(a, int(m ** 0.5) + 1):
if m % b: continue
p = (a + 1) * (b + 2) * (m // b + 2) - n
if u > p: u = p
print(u, v)
```
| 93,683 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored AΒ·BΒ·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks.
At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) Γ (B - 2) Γ (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 Γ 1 Γ 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B ΠΈ C.
Given number n, find the minimally possible and maximally possible number of stolen hay blocks.
Input
The only line contains integer n from the problem's statement (1 β€ n β€ 109).
Output
Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves.
Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in Π‘++. It is preferred to use cin, cout streams or the %I64d specificator.
Examples
Input
4
Output
28 41
Input
7
Output
47 65
Input
12
Output
48 105
Note
Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 Γ 4 Γ 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) Γ (4 - 2) Γ (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 Γ 3 Γ 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) Γ (3 - 2) Γ (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks.
Tags: brute force, math
Correct Solution:
```
from math import sqrt
p, n = [], int(input())
def f(x, y): return (x + 2) * (y + 2) + (2 * (x + y + 2) * n) // (x * y)
for x in range(2, int(sqrt(n)) + 1):
if n % x == 0: p.append(x)
p += [n // x for x in reversed(p)]
p.append(n)
u = v = f(1, 1)
for m in p:
for x in range(1, int(sqrt(m)) + 1):
if m % x == 0: u = min(u, f(x, m // x))
print(u, v)
# Made By Mostafa_Khaled
```
| 93,684 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored AΒ·BΒ·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks.
At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) Γ (B - 2) Γ (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 Γ 1 Γ 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B ΠΈ C.
Given number n, find the minimally possible and maximally possible number of stolen hay blocks.
Input
The only line contains integer n from the problem's statement (1 β€ n β€ 109).
Output
Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves.
Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in Π‘++. It is preferred to use cin, cout streams or the %I64d specificator.
Examples
Input
4
Output
28 41
Input
7
Output
47 65
Input
12
Output
48 105
Note
Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 Γ 4 Γ 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) Γ (4 - 2) Γ (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 Γ 3 Γ 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) Γ (3 - 2) Γ (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks.
Tags: brute force, math
Correct Solution:
```
from math import sqrt
p, n = [], int(input())
def f(x, y): return (x + 2) * (y + 2) + (2 * (x + y + 2) * n) // (x * y)
for x in range(2, int(sqrt(n)) + 1):
if n % x == 0: p.append(x)
p += [n // x for x in reversed(p)]
p.append(n)
u = v = f(1, 1)
for m in p:
for x in range(1, int(sqrt(m)) + 1):
if m % x == 0: u = min(u, f(x, m // x))
print(u, v)
```
| 93,685 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored AΒ·BΒ·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks.
At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) Γ (B - 2) Γ (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 Γ 1 Γ 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B ΠΈ C.
Given number n, find the minimally possible and maximally possible number of stolen hay blocks.
Input
The only line contains integer n from the problem's statement (1 β€ n β€ 109).
Output
Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves.
Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in Π‘++. It is preferred to use cin, cout streams or the %I64d specificator.
Examples
Input
4
Output
28 41
Input
7
Output
47 65
Input
12
Output
48 105
Note
Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 Γ 4 Γ 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) Γ (4 - 2) Γ (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 Γ 3 Γ 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) Γ (3 - 2) Γ (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks.
Tags: brute force, math
Correct Solution:
```
from math import sqrt
n = int(input())
def f(x, y):
return (x+2) * (y+2) + (2*n * (x+y+2)) // (x * y)
factors = []
for i in range(2, int(sqrt(n))+1):
if n % i == 0:
factors.append(i)
factors += [n // i for i in reversed(factors)] + [n]
res = f(1, 1)
for i in factors:
for j in range(1, int(sqrt(i))+1):
if i % j == 0:
res = min(res, f(j, i // j))
print(res, f(1, 1))
```
| 93,686 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored AΒ·BΒ·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks.
At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) Γ (B - 2) Γ (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 Γ 1 Γ 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B ΠΈ C.
Given number n, find the minimally possible and maximally possible number of stolen hay blocks.
Input
The only line contains integer n from the problem's statement (1 β€ n β€ 109).
Output
Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves.
Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in Π‘++. It is preferred to use cin, cout streams or the %I64d specificator.
Examples
Input
4
Output
28 41
Input
7
Output
47 65
Input
12
Output
48 105
Note
Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 Γ 4 Γ 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) Γ (4 - 2) Γ (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 Γ 3 Γ 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) Γ (3 - 2) Γ (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks.
Tags: brute force, math
Correct Solution:
```
n=int(input())
L=[]
i=1
while(i*i*i<=n):
if(n%i!=0):
i+=1
continue
x=n//i
j=i
while(j*j<=x):
if(x%j!=0):
j+=1
continue
L.append((i,j,x//j))
j+=1
i+=1
maxx=0
minn=10**20
E=[[1,2,2],[2,1,2],[2,2,1]]
for item in L:
x=item[0]
y=item[1]
z=item[2]
A=[x,y,z]
for item in E:
m=(A[0]+item[0])*(A[1]+item[1])*(A[2]+item[2])
if(m<minn):
minn=m
if(m>maxx):
maxx=m
print(minn-n,maxx-n)
```
| 93,687 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored AΒ·BΒ·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks.
At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) Γ (B - 2) Γ (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 Γ 1 Γ 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B ΠΈ C.
Given number n, find the minimally possible and maximally possible number of stolen hay blocks.
Input
The only line contains integer n from the problem's statement (1 β€ n β€ 109).
Output
Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves.
Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in Π‘++. It is preferred to use cin, cout streams or the %I64d specificator.
Examples
Input
4
Output
28 41
Input
7
Output
47 65
Input
12
Output
48 105
Note
Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 Γ 4 Γ 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) Γ (4 - 2) Γ (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 Γ 3 Γ 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) Γ (3 - 2) Γ (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks.
Tags: brute force, math
Correct Solution:
```
n=int(input())
k=n
a=set()
a.add(n)
a.add(1)
w=999999999999999999999999999999999999999
ans=[1,1,1]
for i in range(int(pow(n,1/3)),0,-1):
n=k
if n%i==0:
ans[0]=i
else:
continue
n=n//i
#print(n)
for i in range(int(pow(n,1/2)),0,-1):
if n%i==0:
ans[2]=i
ans[1]=n//i
else:
continue
#print(ans)
ans.sort()
ans[0]+=1
ans[1]+=2
ans[2]+=2
n=k
w=min(w,ans[0]*ans[-1]*ans[1]-n)
n=k
e=8*n+9
print(w,e)
```
| 93,688 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored AΒ·BΒ·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks.
At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) Γ (B - 2) Γ (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 Γ 1 Γ 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B ΠΈ C.
Given number n, find the minimally possible and maximally possible number of stolen hay blocks.
Input
The only line contains integer n from the problem's statement (1 β€ n β€ 109).
Output
Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves.
Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in Π‘++. It is preferred to use cin, cout streams or the %I64d specificator.
Examples
Input
4
Output
28 41
Input
7
Output
47 65
Input
12
Output
48 105
Note
Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 Γ 4 Γ 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) Γ (4 - 2) Γ (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 Γ 3 Γ 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) Γ (3 - 2) Γ (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks.
Tags: brute force, math
Correct Solution:
```
import itertools as it
from functools import reduce
def factor(n):
"""
>>> factor(2)
[(2, 1)]
>>> factor(6)
[(3, 1), (2, 1)]
>>> factor(98)
[(7, 2), (2, 1)]
>>> factor(1)
[]
"""
result = []
i = 2
while i * i <= n:
j = 0
while n % i == 0:
j += 1
n //= i
if j > 0:
result += [(i, j)]
i += 1
if n > 1:
result += [(n, 1)]
return result[::-1]
def all_divisors(prime_factorization):
"""
>>> all_divisors(factor(6))
[1, 2, 3, 6]
>>> all_divisors(factor(12))
[1, 2, 4, 3, 6, 12]
>>> all_divisors(factor(7))
[1, 7]
>>> all_divisors(factor(1))
[1]
"""
if len(prime_factorization) == 0:
return [1]
result = []
factor_with_mult = []
for pfactor in prime_factorization:
factor_with_mult += [[1]]
for _ in range(pfactor[1]):
factor_with_mult[-1] += [factor_with_mult[-1][-1] * pfactor[0]]
for divisor in it.product(*factor_with_mult):
result += [reduce(lambda x, y: x * y, divisor)]
return result
if __name__ == '__main__':
n = int(input())
min_result = 10**20
max_result = 0
for divisor in all_divisors(factor(n)):
for div2 in all_divisors(factor(divisor)):
a = n // divisor + 1
b = divisor // div2 + 2
c = div2 + 2
min_result = min([min_result,
c * b + (a - 1) * 2 *
(b + c - 2)])
max_result = max([max_result,
c * b + (a - 1) * 2 *
(b + c - 2)])
print(str(min_result) + " " + str(max_result))
```
| 93,689 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored AΒ·BΒ·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks.
At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) Γ (B - 2) Γ (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 Γ 1 Γ 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B ΠΈ C.
Given number n, find the minimally possible and maximally possible number of stolen hay blocks.
Input
The only line contains integer n from the problem's statement (1 β€ n β€ 109).
Output
Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves.
Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in Π‘++. It is preferred to use cin, cout streams or the %I64d specificator.
Examples
Input
4
Output
28 41
Input
7
Output
47 65
Input
12
Output
48 105
Note
Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 Γ 4 Γ 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) Γ (4 - 2) Γ (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 Γ 3 Γ 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) Γ (3 - 2) Γ (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks.
Tags: brute force, math
Correct Solution:
```
n = int(input())
a = 1
ans = (99999999999999999999999999999999999999999999, 0)
while a ** 3 <= n:
if n % a != 0:
a += 1
continue
n1 = n // a
b = 1
while b * b <= n1:
if n1 % b != 0:
b += 1
continue
c = n1 // b
x1 = (a + 1) * (b + 2) * (c + 2)
x2 = (a + 2) * (b + 1) * (c + 2)
x3 = (a + 2) * (b + 2) * (c + 1)
ans = (min(ans[0], x1, x2, x3), max(ans[1], x1, x2, x3))
b += 1
a += 1
print(ans[0] - n, ans[1] - n)
```
| 93,690 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored AΒ·BΒ·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks.
At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) Γ (B - 2) Γ (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 Γ 1 Γ 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B ΠΈ C.
Given number n, find the minimally possible and maximally possible number of stolen hay blocks.
Input
The only line contains integer n from the problem's statement (1 β€ n β€ 109).
Output
Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves.
Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in Π‘++. It is preferred to use cin, cout streams or the %I64d specificator.
Examples
Input
4
Output
28 41
Input
7
Output
47 65
Input
12
Output
48 105
Note
Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 Γ 4 Γ 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) Γ (4 - 2) Γ (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 Γ 3 Γ 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) Γ (3 - 2) Γ (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks.
Submitted Solution:
```
n = int(input())
a = 1
mn = 1000000000000000
mx = -1000000000000000
while a * a * a <= n:
if n % a != 0:
a += 1
continue
b = a
while a * b * b <= n:
if (n // a) % b != 0:
b += 1
continue
c = n // a // b
#print(a, b, c)
cur = (a + 1) * (b + 2) * (c + 2)
mn = min(mn, cur)
mx = max(mx, cur)
cur = (a + 2) * (b + 1) * (c + 2)
mn = min(mn, cur)
mx = max(mx, cur)
cur = (a + 2) * (b + 2) * (c + 1)
mn = min(mn, cur)
mx = max(mx, cur)
b += 1
a += 1
print(mn - n, mx - n)
```
Yes
| 93,691 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored AΒ·BΒ·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks.
At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) Γ (B - 2) Γ (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 Γ 1 Γ 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B ΠΈ C.
Given number n, find the minimally possible and maximally possible number of stolen hay blocks.
Input
The only line contains integer n from the problem's statement (1 β€ n β€ 109).
Output
Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves.
Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in Π‘++. It is preferred to use cin, cout streams or the %I64d specificator.
Examples
Input
4
Output
28 41
Input
7
Output
47 65
Input
12
Output
48 105
Note
Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 Γ 4 Γ 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) Γ (4 - 2) Γ (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 Γ 3 Γ 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) Γ (3 - 2) Γ (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks.
Submitted Solution:
```
n = int(input())
mn = 999999999999
for i in range(1, int(n**(1/3))+1):
if n%i==0:
for j in range(1, int((n//i)**(1/2))+1):
if (n/i) % j == 0:
k = (n//i)//j
mn = min(mn, (i+1)*(k+2)*(j+2))
print(mn-n, (9*n+9-n))
```
Yes
| 93,692 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored AΒ·BΒ·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks.
At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) Γ (B - 2) Γ (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 Γ 1 Γ 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B ΠΈ C.
Given number n, find the minimally possible and maximally possible number of stolen hay blocks.
Input
The only line contains integer n from the problem's statement (1 β€ n β€ 109).
Output
Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves.
Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in Π‘++. It is preferred to use cin, cout streams or the %I64d specificator.
Examples
Input
4
Output
28 41
Input
7
Output
47 65
Input
12
Output
48 105
Note
Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 Γ 4 Γ 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) Γ (4 - 2) Γ (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 Γ 3 Γ 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) Γ (3 - 2) Γ (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks.
Submitted Solution:
```
# ========= /\ /| |====/|
# | / \ | | / |
# | /____\ | | / |
# | / \ | | / |
# ========= / \ ===== |/====|
# code
def main():
n = int(input())
m1 = float('inf')
m2 = float('-inf')
i = 1
while i*i*i <= n:
if n%i != 0:
i += 1
continue
j = 1
while j*j <= n//i:
if (n//i)%j != 0:
j += 1
continue
k = (n // i)//j
# print(i,j,k)
m1 = min(m1 , (i + 1)*(j + 2)*(k + 2) - n)
m1 = min(m1 , (j + 1)*(i + 2)*(k + 2) - n)
m1 = min(m1 , (k + 1)*(j + 2)*(i + 2) - n)
m2 = max(m2 , (i + 1)*(j + 2)*(k + 2) - n)
m2 = max(m2 , (j + 1)*(i + 2)*(k + 2) - n)
m2 = max(m2 , (k + 1)*(j + 2)*(i + 2) - n)
j += 1
i += 1
print(m1 , m2)
return
if __name__ == "__main__":
main()
```
Yes
| 93,693 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored AΒ·BΒ·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks.
At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) Γ (B - 2) Γ (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 Γ 1 Γ 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B ΠΈ C.
Given number n, find the minimally possible and maximally possible number of stolen hay blocks.
Input
The only line contains integer n from the problem's statement (1 β€ n β€ 109).
Output
Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves.
Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in Π‘++. It is preferred to use cin, cout streams or the %I64d specificator.
Examples
Input
4
Output
28 41
Input
7
Output
47 65
Input
12
Output
48 105
Note
Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 Γ 4 Γ 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) Γ (4 - 2) Γ (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 Γ 3 Γ 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) Γ (3 - 2) Γ (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks.
Submitted Solution:
```
n=int(input())
i=1
l=[]
while(i*i<=n):
if(n%i==0):
l.append(i)
i+=1
lmax=[]
lmin=[]
for i in range(len(l)):
for j in range(i,len(l)):
if(l[i]*l[j]<=n):
t=l[i]*l[j]
if(n%t==0):
k=n//t
lmax.append((l[i]+1)*(l[j]+2)*(k+2))
lmin.append((l[i]+1)*(l[j]+2)*(k+2))
lmax.append((l[i]+2)*(l[j]+2)*(k+1))
lmin.append((l[i]+2)*(l[j]+2)*(k+1))
lmax.append((l[i]+2)*(l[j]+1)*(k+2))
lmin.append((l[i]+2)*(l[j]+1)*(k+2))
print(min(lmin)-n,max(lmax)-n)
```
Yes
| 93,694 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored AΒ·BΒ·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks.
At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) Γ (B - 2) Γ (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 Γ 1 Γ 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B ΠΈ C.
Given number n, find the minimally possible and maximally possible number of stolen hay blocks.
Input
The only line contains integer n from the problem's statement (1 β€ n β€ 109).
Output
Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves.
Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in Π‘++. It is preferred to use cin, cout streams or the %I64d specificator.
Examples
Input
4
Output
28 41
Input
7
Output
47 65
Input
12
Output
48 105
Note
Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 Γ 4 Γ 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) Γ (4 - 2) Γ (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 Γ 3 Γ 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) Γ (3 - 2) Γ (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks.
Submitted Solution:
```
import math
n = int(input())
mx, mn = 0, 0
mx += (3*n)
mx *= 3
mx += 9
mx-=n
s = int(n**(1/3))
while n%s !=0:
s-=1
n //= s
k = int((n//s)**(1/2))
while n%k !=0:
k-=1
n //= k
m = [n,k,s]
g1 = min(m)
m.remove(g1)
g2 = min(m)
m.remove(g2)
m = m[0]
#mn += (m*g1*g2)
mn += (m*g1*2)
g2 += 2
mn += (g2*g1*2)
m += 2
mn += (m*g2)
print(mn, mx)
```
No
| 93,695 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored AΒ·BΒ·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks.
At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) Γ (B - 2) Γ (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 Γ 1 Γ 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B ΠΈ C.
Given number n, find the minimally possible and maximally possible number of stolen hay blocks.
Input
The only line contains integer n from the problem's statement (1 β€ n β€ 109).
Output
Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves.
Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in Π‘++. It is preferred to use cin, cout streams or the %I64d specificator.
Examples
Input
4
Output
28 41
Input
7
Output
47 65
Input
12
Output
48 105
Note
Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 Γ 4 Γ 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) Γ (4 - 2) Γ (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 Γ 3 Γ 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) Γ (3 - 2) Γ (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks.
Submitted Solution:
```
import itertools as it
from functools import reduce
def factor(n):
"""
>>> factor(2)
[(2, 1)]
>>> factor(6)
[(3, 1), (2, 1)]
>>> factor(98)
[(7, 2), (2, 1)]
>>> factor(1)
[]
"""
result = []
i = 2
while i * i <= n:
j = 0
while n % i == 0:
j += 1
n //= i
if j > 0:
result += [(i, j)]
i += 1
if n > 1:
result += [(n, 1)]
return result[::-1]
def all_divisors(prime_factorization):
"""
>>> all_divisors(factor(6))
[1, 2, 3, 6]
>>> all_divisors(factor(12))
[1, 2, 4, 3, 6, 12]
>>> all_divisors(factor(7))
[1, 7]
>>> all_divisors(factor(1))
[1]
"""
if len(prime_factorization) == 0:
return [1]
result = []
factor_with_mult = []
for pfactor in prime_factorization:
factor_with_mult += [[1]]
for _ in range(pfactor[1]):
factor_with_mult[-1] += [factor_with_mult[-1][-1] * pfactor[0]]
for divisor in it.product(*factor_with_mult):
result += [reduce(lambda x, y: x * y, divisor)]
return result
if __name__ == '__main__':
n = int(input())
min_result = 10**9
max_result = 0
for divisor in all_divisors(factor(n)):
for div2 in all_divisors(factor(divisor)):
a = n // divisor + 1
b = divisor // div2 + 2
c = div2 + 2
min_result = min([min_result,
c * b + (a - 1) * 2 *
(b + c - 2)])
max_result = max([max_result,
c * b + (a - 1) * 2 *
(b + c - 2)])
print(str(min_result) + " " + str(max_result))
```
No
| 93,696 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored AΒ·BΒ·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks.
At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) Γ (B - 2) Γ (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 Γ 1 Γ 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B ΠΈ C.
Given number n, find the minimally possible and maximally possible number of stolen hay blocks.
Input
The only line contains integer n from the problem's statement (1 β€ n β€ 109).
Output
Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves.
Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in Π‘++. It is preferred to use cin, cout streams or the %I64d specificator.
Examples
Input
4
Output
28 41
Input
7
Output
47 65
Input
12
Output
48 105
Note
Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 Γ 4 Γ 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) Γ (4 - 2) Γ (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 Γ 3 Γ 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) Γ (3 - 2) Γ (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks.
Submitted Solution:
```
n=int(input())
a=[1]
for i in range(int(n**0.5)+1,0,-1):
if n%i==0:
a.append(i)
a.append(n//i)
break
a.sort()
a[0]+=1
a[1]+=2
a[2]+=2
w=a[0]*a[1]*a[-1]-n
e=8*n+9
print(w,e)
```
No
| 93,697 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored AΒ·BΒ·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks.
At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) Γ (B - 2) Γ (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 Γ 1 Γ 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B ΠΈ C.
Given number n, find the minimally possible and maximally possible number of stolen hay blocks.
Input
The only line contains integer n from the problem's statement (1 β€ n β€ 109).
Output
Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves.
Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in Π‘++. It is preferred to use cin, cout streams or the %I64d specificator.
Examples
Input
4
Output
28 41
Input
7
Output
47 65
Input
12
Output
48 105
Note
Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 Γ 4 Γ 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) Γ (4 - 2) Γ (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 Γ 3 Γ 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) Γ (3 - 2) Γ (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks.
Submitted Solution:
```
import math
n = int(input())
mx = 3 * 3 * (n+1) - n
mxDiv = 1
for i in range(1, n+1):
if n % i == 0:
mxDiv = i
if i * i >= n:
break
mn = 2 * (2 + mxDiv) * (2 + n//mxDiv) - n
print(str(mn) + " " + str(mx))
```
No
| 93,698 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Berland regional ICPC contest has just ended. There were m participants numbered from 1 to m, who competed on a problemset of n problems numbered from 1 to n.
Now the editorial is about to take place. There are two problem authors, each of them is going to tell the tutorial to exactly k consecutive tasks of the problemset. The authors choose the segment of k consecutive tasks for themselves independently of each other. The segments can coincide, intersect or not intersect at all.
The i-th participant is interested in listening to the tutorial of all consecutive tasks from l_i to r_i. Each participant always chooses to listen to only the problem author that tells the tutorials to the maximum number of tasks he is interested in. Let this maximum number be a_i. No participant can listen to both of the authors, even if their segments don't intersect.
The authors want to choose the segments of k consecutive tasks for themselves in such a way that the sum of a_i over all participants is maximized.
Input
The first line contains three integers n, m and k (1 β€ n, m β€ 2000, 1 β€ k β€ n) β the number of problems, the number of participants and the length of the segment of tasks each of the problem authors plans to tell the tutorial to.
The i-th of the next m lines contains two integers l_i and r_i (1 β€ l_i β€ r_i β€ n) β the segment of tasks the i-th participant is interested in listening to the tutorial to.
Output
Print a single integer β the maximum sum of a_i over all participants.
Examples
Input
10 5 3
1 3
2 4
6 9
6 9
1 8
Output
14
Input
10 3 3
2 4
4 6
3 5
Output
8
Input
4 4 1
3 3
1 1
2 2
4 4
Output
2
Input
5 4 5
1 2
2 3
3 4
4 5
Output
8
Note
In the first example the first author can tell the tutorial to problems from 1 to 3 and the second one β from 6 to 8. That way the sequence of a_i will be [3, 2, 3, 3, 3]. Notice that the last participant can't listen to both author, he only chooses the one that tells the maximum number of problems he's interested in.
In the second example the first one can tell problems 2 to 4, the second one β 4 to 6.
In the third example the first one can tell problems 1 to 1, the second one β 2 to 2. Or 4 to 4 and 3 to 3. Every pair of different problems will get the same sum of 2.
In the fourth example the first one can tell problems 1 to 5, the second one β 1 to 5 as well.
Tags: brute force, dp, greedy, sortings, two pointers
Correct Solution:
```
# Enter your code here. Read input from STDIN. Print output to STDOUT# ===============================================================================================
# importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from itertools import *
import bisect
from heapq import *
from math import ceil, floor
from copy import *
from collections import deque, defaultdict
from collections import Counter as counter # Counter(list) return a dict with {key: count}
from itertools import combinations # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
from operator import *
# If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect
# If the element is already present in the list,
# the right most position where element has to be inserted is returned
# ==============================================================================================
# fast I/O region
BUFSIZE = 8192
from sys import stderr
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")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
# ===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
#### END ITERATE RECURSION ####
###########################
# Sorted list
class SortedList:
def __init__(self, iterable=[], _load=200):
"""Initialize sorted list instance."""
values = sorted(iterable)
self._len = _len = len(values)
self._load = _load
self._lists = _lists = [values[i:i + _load] for i in range(0, _len, _load)]
self._list_lens = [len(_list) for _list in _lists]
self._mins = [_list[0] for _list in _lists]
self._fen_tree = []
self._rebuild = True
def _fen_build(self):
"""Build a fenwick tree instance."""
self._fen_tree[:] = self._list_lens
_fen_tree = self._fen_tree
for i in range(len(_fen_tree)):
if i | i + 1 < len(_fen_tree):
_fen_tree[i | i + 1] += _fen_tree[i]
self._rebuild = False
def _fen_update(self, index, value):
"""Update `fen_tree[index] += value`."""
if not self._rebuild:
_fen_tree = self._fen_tree
while index < len(_fen_tree):
_fen_tree[index] += value
index |= index + 1
def _fen_query(self, end):
"""Return `sum(_fen_tree[:end])`."""
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
x = 0
while end:
x += _fen_tree[end - 1]
end &= end - 1
return x
def _fen_findkth(self, k):
"""Return a pair of (the largest `idx` such that `sum(_fen_tree[:idx]) <= k`, `k - sum(_fen_tree[:idx])`)."""
_list_lens = self._list_lens
if k < _list_lens[0]:
return 0, k
if k >= self._len - _list_lens[-1]:
return len(_list_lens) - 1, k + _list_lens[-1] - self._len
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
idx = -1
for d in reversed(range(len(_fen_tree).bit_length())):
right_idx = idx + (1 << d)
if right_idx < len(_fen_tree) and k >= _fen_tree[right_idx]:
idx = right_idx
k -= _fen_tree[idx]
return idx + 1, k
def _delete(self, pos, idx):
"""Delete value at the given `(pos, idx)`."""
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len -= 1
self._fen_update(pos, -1)
del _lists[pos][idx]
_list_lens[pos] -= 1
if _list_lens[pos]:
_mins[pos] = _lists[pos][0]
else:
del _lists[pos]
del _list_lens[pos]
del _mins[pos]
self._rebuild = True
def _loc_left(self, value):
"""Return an index pair that corresponds to the first position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
lo, pos = -1, len(_lists) - 1
while lo + 1 < pos:
mi = (lo + pos) >> 1
if value <= _mins[mi]:
pos = mi
else:
lo = mi
if pos and value <= _lists[pos - 1][-1]:
pos -= 1
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value <= _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def _loc_right(self, value):
"""Return an index pair that corresponds to the last position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
pos, hi = 0, len(_lists)
while pos + 1 < hi:
mi = (pos + hi) >> 1
if value < _mins[mi]:
hi = mi
else:
pos = mi
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value < _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def add(self, value):
"""Add `value` to sorted list."""
_load = self._load
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len += 1
if _lists:
pos, idx = self._loc_right(value)
self._fen_update(pos, 1)
_list = _lists[pos]
_list.insert(idx, value)
_list_lens[pos] += 1
_mins[pos] = _list[0]
if _load + _load < len(_list):
_lists.insert(pos + 1, _list[_load:])
_list_lens.insert(pos + 1, len(_list) - _load)
_mins.insert(pos + 1, _list[_load])
_list_lens[pos] = _load
del _list[_load:]
self._rebuild = True
else:
_lists.append([value])
_mins.append(value)
_list_lens.append(1)
self._rebuild = True
def discard(self, value):
"""Remove `value` from sorted list if it is a member."""
_lists = self._lists
if _lists:
pos, idx = self._loc_right(value)
if idx and _lists[pos][idx - 1] == value:
self._delete(pos, idx - 1)
def remove(self, value):
"""Remove `value` from sorted list; `value` must be a member."""
_len = self._len
self.discard(value)
if _len == self._len:
raise ValueError('{0!r} not in list'.format(value))
def pop(self, index=-1):
"""Remove and return value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
value = self._lists[pos][idx]
self._delete(pos, idx)
return value
def bisect_left(self, value):
"""Return the first index to insert `value` in the sorted list."""
pos, idx = self._loc_left(value)
return self._fen_query(pos) + idx
def bisect_right(self, value):
"""Return the last index to insert `value` in the sorted list."""
pos, idx = self._loc_right(value)
return self._fen_query(pos) + idx
def count(self, value):
"""Return number of occurrences of `value` in the sorted list."""
return self.bisect_right(value) - self.bisect_left(value)
def __len__(self):
"""Return the size of the sorted list."""
return self._len
def __getitem__(self, index):
"""Lookup value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
return self._lists[pos][idx]
def __delitem__(self, index):
"""Remove value at `index` from sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
self._delete(pos, idx)
def __contains__(self, value):
"""Return true if `value` is an element of the sorted list."""
_lists = self._lists
if _lists:
pos, idx = self._loc_left(value)
return idx < len(_lists[pos]) and _lists[pos][idx] == value
return False
def __iter__(self):
"""Return an iterator over the sorted list."""
return (value for _list in self._lists for value in _list)
def __reversed__(self):
"""Return a reverse iterator over the sorted list."""
return (value for _list in reversed(self._lists) for value in reversed(_list))
def __repr__(self):
"""Return string representation of sorted list."""
return 'SortedList({0})'.format(list(self))
# ===============================================================================================
# some shortcuts
mod = 1000000007
def inp(): return sys.stdin.readline().rstrip("\r\n") # for fast input
def out(var): sys.stdout.write(str(var)) # for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def fsep(): return map(float, inp().split())
def nextline(): out("\n") # as stdout.write always print sring.
def testcase(t):
for p in range(t):
solve()
def pow(x, y, p):
res = 1 # Initialize result
x = x % p # Update x if it is more , than or equal to p
if (x == 0):
return 0
while (y > 0):
if ((y & 1) == 1): # If y is odd, multiply, x with result
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
from functools import reduce
def factors(n):
return set(reduce(list.__add__,
([i, n // i] for i in range(1, int(n ** 0.5) + 1) if n % i == 0)))
def gcd(a, b):
if a == b: return a
while b > 0: a, b = b, a % b
return a
# discrete binary search
# minimise:
# def search():
# l = 0
# r = 10 ** 15
#
# for i in range(200):
# if isvalid(l):
# return l
# if l == r:
# return l
# m = (l + r) // 2
# if isvalid(m) and not isvalid(m - 1):
# return m
# if isvalid(m):
# r = m + 1
# else:
# l = m
# return m
# maximise:
# def search():
# l = 0
# r = 10 ** 15
#
# for i in range(200):
# # print(l,r)
# if isvalid(r):
# return r
# if l == r:
# return l
# m = (l + r) // 2
# if isvalid(m) and not isvalid(m + 1):
# return m
# if isvalid(m):
# l = m
# else:
# r = m - 1
# return m
##############Find sum of product of subsets of size k in a array
# ar=[0,1,2,3]
# k=3
# n=len(ar)-1
# dp=[0]*(n+1)
# dp[0]=1
# for pos in range(1,n+1):
# dp[pos]=0
# l=max(1,k+pos-n-1)
# for j in range(min(pos,k),l-1,-1):
# dp[j]=dp[j]+ar[pos]*dp[j-1]
# print(dp[k])
def prefix_sum(ar): # [1,2,3,4]->[1,3,6,10]
return list(accumulate(ar))
def suffix_sum(ar): # [1,2,3,4]->[10,9,7,4]
return list(accumulate(ar[::-1]))[::-1]
def N():
return int(inp())
dx = [0, 0, 1, -1]
dy = [1, -1, 0, 0]
def YES():
print("YES")
def NO():
print("NO")
def Yes():
print("Yes")
def No():
print("No")
# =========================================================================================
from collections import defaultdict
def numberOfSetBits(i):
i = i - ((i >> 1) & 0x55555555)
i = (i & 0x33333333) + ((i >> 2) & 0x33333333)
return (((i + (i >> 4) & 0xF0F0F0F) * 0x1010101) & 0xffffffff) >> 24
def lcm(a, b):
return abs((a // gcd(a, b)) * b)
#
# # to find factorial and ncr
# tot = 400005
# mod = 10 ** 9 + 7
# fac = [1, 1]
# finv = [1, 1]
# inv = [0, 1]
#
# for i in range(2, tot + 1):
# fac.append((fac[-1] * i) % mod)
# inv.append(mod - (inv[mod % i] * (mod // i) % mod))
# finv.append(finv[-1] * inv[-1] % mod)
#
#
# def comb(n, r):
# if n < r:
# return 0
# else:
# return fac[n] * (finv[r] * finv[n - r] % mod) % mod
def solve():
n,m,k=sep()
a=[lis() for _ in range(m)]
def solv(lst):
res = sum(lst[:k]);
s = res
for i in range(n - k): s += lst[i + k] - lst[i];res = max(res, s)
return res
a.sort(key=lambda x: sum(x) / 2);
apr = [0] * n
for el in a:
for i in range(el[0] - 1, el[1]): apr[i] += 1
res = solv(apr);
bpr = [0] * n
for r, l in a:
for i in range(r - 1, l): apr[i] -= 1;bpr[i] += 1
nres = solv(apr) + solv(bpr);
res = max(res, nres)
print(res)
solve()
#testcase(int(inp()))
# 5
# 3 6
# 5 10
# 4 3
# 2 1
# 1 3
```
| 93,699 |
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