message stringlengths 2 30.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 237 109k | cluster float64 10 10 | __index_level_0__ int64 474 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Sometimes it is not easy to come to an agreement in a bargain. Right now Sasha and Vova can't come to an agreement: Sasha names a price as high as possible, then Vova wants to remove as many digits from the price as possible. In more details, Sasha names some integer price n, Vova removes a non-empty substring of (consecutive) digits from the price, the remaining digits close the gap, and the resulting integer is the price.
For example, is Sasha names 1213121, Vova can remove the substring 1312, and the result is 121.
It is allowed for result to contain leading zeros. If Vova removes all digits, the price is considered to be 0.
Sasha wants to come up with some constraints so that Vova can't just remove all digits, but he needs some arguments supporting the constraints. To start with, he wants to compute the sum of all possible resulting prices after Vova's move.
Help Sasha to compute this sum. Since the answer can be very large, print it modulo 10^9 + 7.
Input
The first and only line contains a single integer n (1 ≤ n < 10^{10^5}).
Output
In the only line print the required sum modulo 10^9 + 7.
Examples
Input
107
Output
42
Input
100500100500
Output
428101984
Note
Consider the first example.
Vova can choose to remove 1, 0, 7, 10, 07, or 107. The results are 07, 17, 10, 7, 1, 0. Their sum is 42. | instruction | 0 | 54,137 | 10 | 108,274 |
Tags: combinatorics, dp, math
Correct Solution:
```
n = input()
m = len(n)
M = int(1e9)+7
s = 0
ans = 0
for i, c in enumerate(n):
x = int(c)
a = (s * (m-i) % M)
b = (x * ((i*(i+1)//2) % M)) % M
ans = (ans + pow(10, m-i-1, M) * (a + b)) % M
s = (s + x) % M
print(ans)
``` | output | 1 | 54,137 | 10 | 108,275 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sometimes it is not easy to come to an agreement in a bargain. Right now Sasha and Vova can't come to an agreement: Sasha names a price as high as possible, then Vova wants to remove as many digits from the price as possible. In more details, Sasha names some integer price n, Vova removes a non-empty substring of (consecutive) digits from the price, the remaining digits close the gap, and the resulting integer is the price.
For example, is Sasha names 1213121, Vova can remove the substring 1312, and the result is 121.
It is allowed for result to contain leading zeros. If Vova removes all digits, the price is considered to be 0.
Sasha wants to come up with some constraints so that Vova can't just remove all digits, but he needs some arguments supporting the constraints. To start with, he wants to compute the sum of all possible resulting prices after Vova's move.
Help Sasha to compute this sum. Since the answer can be very large, print it modulo 10^9 + 7.
Input
The first and only line contains a single integer n (1 ≤ n < 10^{10^5}).
Output
In the only line print the required sum modulo 10^9 + 7.
Examples
Input
107
Output
42
Input
100500100500
Output
428101984
Note
Consider the first example.
Vova can choose to remove 1, 0, 7, 10, 07, or 107. The results are 07, 17, 10, 7, 1, 0. Their sum is 42. | instruction | 0 | 54,138 | 10 | 108,276 |
Tags: combinatorics, dp, math
Correct Solution:
```
MOD = 10 ** 9 + 7
def mul(x, y):
global MOD
return ((x % MOD) * (y % MOD)) % MOD
def add(x, y):
global MOD
return ((x % MOD) + (y % MOD)) % MOD
def testcase():
n = input()
dp = [0 for _ in n]
dp[-1] = 0
temp = 1
for i in range(len(n) - 2, -1, -1):
dp[i] = add(dp[i + 1], mul(temp, len(n) - i - 1))
temp = mul(temp, 10)
ans = 0
for i in range(len(n)):
d = int(n[i])
r = mul(dp[i], d)
ans = add(ans, r)
temp = 1
for i in range(len(n) - 1, 0, -1):
d = int(n[i])
l = d
l = mul(d, mul(mul(i, i + 1), 500000004))
l = mul(l, temp)
temp = mul(temp, 10)
ans = add(ans, l)
print(ans)
return
t = 1#int(input())
for _ in range(t):
testcase()
``` | output | 1 | 54,138 | 10 | 108,277 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sometimes it is not easy to come to an agreement in a bargain. Right now Sasha and Vova can't come to an agreement: Sasha names a price as high as possible, then Vova wants to remove as many digits from the price as possible. In more details, Sasha names some integer price n, Vova removes a non-empty substring of (consecutive) digits from the price, the remaining digits close the gap, and the resulting integer is the price.
For example, is Sasha names 1213121, Vova can remove the substring 1312, and the result is 121.
It is allowed for result to contain leading zeros. If Vova removes all digits, the price is considered to be 0.
Sasha wants to come up with some constraints so that Vova can't just remove all digits, but he needs some arguments supporting the constraints. To start with, he wants to compute the sum of all possible resulting prices after Vova's move.
Help Sasha to compute this sum. Since the answer can be very large, print it modulo 10^9 + 7.
Input
The first and only line contains a single integer n (1 ≤ n < 10^{10^5}).
Output
In the only line print the required sum modulo 10^9 + 7.
Examples
Input
107
Output
42
Input
100500100500
Output
428101984
Note
Consider the first example.
Vova can choose to remove 1, 0, 7, 10, 07, or 107. The results are 07, 17, 10, 7, 1, 0. Their sum is 42. | instruction | 0 | 54,139 | 10 | 108,278 |
Tags: combinatorics, dp, math
Correct Solution:
```
from collections import Counter, OrderedDict
from itertools import permutations as perm
from sys import setrecursionlimit
from collections import deque
from sys import stdin
from bisect import *
from heapq import *
import threading
import math
g = lambda : stdin.readline().strip()
gl = lambda : g().split()
gil = lambda : [int(var) for var in gl()]
gfl = lambda : [float(var) for var in gl()]
gcl = lambda : list(g())
gbs = lambda : [int(var) for var in g()]
mod = int(1e9)+7
inf = float("inf")
a = gbs()
n = len(a)
p10 = [1]
for i in range(1, n):
p10.append((10*p10[-1])%mod)
p = [1]
for i in range(1, n):
v = ((i+1)*p10[i])%mod + p[-1]
p.append(v%mod)
p.append(0)
ans = 0
for i in range(n):
l, r = i, n-1-i
c = (((((l*(l+1))//2)%mod)*p10[r])%mod + p[r-1])%mod
ans += (c*a[i])%mod
ans %= mod
print(ans)
``` | output | 1 | 54,139 | 10 | 108,279 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sometimes it is not easy to come to an agreement in a bargain. Right now Sasha and Vova can't come to an agreement: Sasha names a price as high as possible, then Vova wants to remove as many digits from the price as possible. In more details, Sasha names some integer price n, Vova removes a non-empty substring of (consecutive) digits from the price, the remaining digits close the gap, and the resulting integer is the price.
For example, is Sasha names 1213121, Vova can remove the substring 1312, and the result is 121.
It is allowed for result to contain leading zeros. If Vova removes all digits, the price is considered to be 0.
Sasha wants to come up with some constraints so that Vova can't just remove all digits, but he needs some arguments supporting the constraints. To start with, he wants to compute the sum of all possible resulting prices after Vova's move.
Help Sasha to compute this sum. Since the answer can be very large, print it modulo 10^9 + 7.
Input
The first and only line contains a single integer n (1 ≤ n < 10^{10^5}).
Output
In the only line print the required sum modulo 10^9 + 7.
Examples
Input
107
Output
42
Input
100500100500
Output
428101984
Note
Consider the first example.
Vova can choose to remove 1, 0, 7, 10, 07, or 107. The results are 07, 17, 10, 7, 1, 0. Their sum is 42.
Submitted Solution:
```
# ========== //\\ //|| ||====//||
# || // \\ || || // ||
# || //====\\ || || // ||
# || // \\ || || // ||
# ========== // \\ ======== ||//====||
# code
def solve():
n = input()
n = n[::-1]
MOD = 10 ** 9 + 7
l = len(n)
cnt = 0
p = 1
ans = 0
for i in range(0, l):
m = l - i - 1
ans += cnt * (ord(n[i]) - 48)
ans %= MOD
ans += p * (m * (m + 1)) // 2 * (ord(n[i]) - 48)
cnt = (i + 1) * p + cnt
p *= 10
cnt %= MOD
p %= MOD
# print(ans, cnt)
print(ans)
return
def main():
t = 1
# t = int(input())
for _ in range(t):
solve()
if __name__ == "__main__":
main()
``` | instruction | 0 | 54,140 | 10 | 108,280 |
Yes | output | 1 | 54,140 | 10 | 108,281 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sometimes it is not easy to come to an agreement in a bargain. Right now Sasha and Vova can't come to an agreement: Sasha names a price as high as possible, then Vova wants to remove as many digits from the price as possible. In more details, Sasha names some integer price n, Vova removes a non-empty substring of (consecutive) digits from the price, the remaining digits close the gap, and the resulting integer is the price.
For example, is Sasha names 1213121, Vova can remove the substring 1312, and the result is 121.
It is allowed for result to contain leading zeros. If Vova removes all digits, the price is considered to be 0.
Sasha wants to come up with some constraints so that Vova can't just remove all digits, but he needs some arguments supporting the constraints. To start with, he wants to compute the sum of all possible resulting prices after Vova's move.
Help Sasha to compute this sum. Since the answer can be very large, print it modulo 10^9 + 7.
Input
The first and only line contains a single integer n (1 ≤ n < 10^{10^5}).
Output
In the only line print the required sum modulo 10^9 + 7.
Examples
Input
107
Output
42
Input
100500100500
Output
428101984
Note
Consider the first example.
Vova can choose to remove 1, 0, 7, 10, 07, or 107. The results are 07, 17, 10, 7, 1, 0. Their sum is 42.
Submitted Solution:
```
mod = int(1e9 + 7)
s = input()
ans = 0
cur = 0
p = 1
m = 0
for n in s[::-1]:
m += 1
n = int(n)
ans += n * cur
ans %= mod
# if m > 1:
# ans += n * p
left = len(s) - m
ans += (n * p) % mod * (1 + left) * left // 2
ans %= mod
# print(m, p, cur)
# print(n, ans)
cur += m * p
cur %= mod
p *= 10
p %= mod
print(ans % mod)
``` | instruction | 0 | 54,141 | 10 | 108,282 |
Yes | output | 1 | 54,141 | 10 | 108,283 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sometimes it is not easy to come to an agreement in a bargain. Right now Sasha and Vova can't come to an agreement: Sasha names a price as high as possible, then Vova wants to remove as many digits from the price as possible. In more details, Sasha names some integer price n, Vova removes a non-empty substring of (consecutive) digits from the price, the remaining digits close the gap, and the resulting integer is the price.
For example, is Sasha names 1213121, Vova can remove the substring 1312, and the result is 121.
It is allowed for result to contain leading zeros. If Vova removes all digits, the price is considered to be 0.
Sasha wants to come up with some constraints so that Vova can't just remove all digits, but he needs some arguments supporting the constraints. To start with, he wants to compute the sum of all possible resulting prices after Vova's move.
Help Sasha to compute this sum. Since the answer can be very large, print it modulo 10^9 + 7.
Input
The first and only line contains a single integer n (1 ≤ n < 10^{10^5}).
Output
In the only line print the required sum modulo 10^9 + 7.
Examples
Input
107
Output
42
Input
100500100500
Output
428101984
Note
Consider the first example.
Vova can choose to remove 1, 0, 7, 10, 07, or 107. The results are 07, 17, 10, 7, 1, 0. Their sum is 42.
Submitted Solution:
```
from sys import stdin, stdout
input = stdin.readline
print = lambda x:stdout.write(str(x)+'\n')
N = input().strip()
l = len(N)
mod = 10**9+7
pow10 = [1]
for i in range(1, l + 1):
pow10.append(pow10[-1] * 10 % mod)
cache = [0]
for i in range(l-1):
cache.append((cache[-1]+(i+1)*pow10[i])%mod)
cache = cache[::-1]
for i, ele in enumerate(N):
if 0<i<l-1:
ans += (int(ele) * cache[i] + int(ele) * pow10[l - i - 1] * (1 + i) * i // 2) % mod
ans %= mod
elif i==0:
ans = int(N[0]) * cache[0] % mod
else:
ans += (int(ele) * pow10[l - i - 1] * (1 + i) * i // 2) % mod
ans %= mod
print(ans)
``` | instruction | 0 | 54,142 | 10 | 108,284 |
Yes | output | 1 | 54,142 | 10 | 108,285 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sometimes it is not easy to come to an agreement in a bargain. Right now Sasha and Vova can't come to an agreement: Sasha names a price as high as possible, then Vova wants to remove as many digits from the price as possible. In more details, Sasha names some integer price n, Vova removes a non-empty substring of (consecutive) digits from the price, the remaining digits close the gap, and the resulting integer is the price.
For example, is Sasha names 1213121, Vova can remove the substring 1312, and the result is 121.
It is allowed for result to contain leading zeros. If Vova removes all digits, the price is considered to be 0.
Sasha wants to come up with some constraints so that Vova can't just remove all digits, but he needs some arguments supporting the constraints. To start with, he wants to compute the sum of all possible resulting prices after Vova's move.
Help Sasha to compute this sum. Since the answer can be very large, print it modulo 10^9 + 7.
Input
The first and only line contains a single integer n (1 ≤ n < 10^{10^5}).
Output
In the only line print the required sum modulo 10^9 + 7.
Examples
Input
107
Output
42
Input
100500100500
Output
428101984
Note
Consider the first example.
Vova can choose to remove 1, 0, 7, 10, 07, or 107. The results are 07, 17, 10, 7, 1, 0. Their sum is 42.
Submitted Solution:
```
from __future__ import division, print_function
import bisect
import math
import heapq
import itertools
import sys
from collections import deque
from atexit import register
from collections import Counter
from functools import reduce
if sys.version_info[0] < 3:
from io import BytesIO as stream
else:
from io import StringIO as stream
if sys.version_info[0] < 3:
class dict(dict):
"""dict() -> new empty dictionary"""
def items(self):
"""D.items() -> a set-like object providing a view on D's items"""
return dict.iteritems(self)
def keys(self):
"""D.keys() -> a set-like object providing a view on D's keys"""
return dict.iterkeys(self)
def values(self):
"""D.values() -> an object providing a view on D's values"""
return dict.itervalues(self)
input = raw_input
range = xrange
filter = itertools.ifilter
map = itertools.imap
zip = itertools.izip
def sync_with_stdio(sync=True):
"""Set whether the standard Python streams are allowed to buffer their I/O.
Args:
sync (bool, optional): The new synchronization setting.
"""
global input, flush
if sync:
flush = sys.stdout.flush
else:
sys.stdin = stream(sys.stdin.read())
input = lambda: sys.stdin.readline().rstrip('\r\n')
sys.stdout = stream()
register(lambda: sys.__stdout__.write(sys.stdout.getvalue()))
def main():
# sys.stdin = open("input.txt")
mod=10**9
mod+=7
s=input()
n=len(s)
peeche=[0]*n
aage=[0]*n
for i in range(1,n):
peeche[i]=(i*(i+1))//2
peeche[i]*=pow(10,n-i-1,mod)
peeche[i]%=mod
for i in range(n-1):
aage[0]+=(i+1)*(pow(10,i,mod))
aage[i]%=mod
for i in range(1,n):
aage[i]=aage[i-1]
aage[i]-=(n-i)*(pow(10,n-i-1,mod))
aage[i]%=mod
ans=0
for i in range(n):
ans+=(int(s[i])*(peeche[i]+aage[i]))
ans%=mod
print(ans%mod)
if __name__ == '__main__':
sync_with_stdio(False)
main()
``` | instruction | 0 | 54,143 | 10 | 108,286 |
Yes | output | 1 | 54,143 | 10 | 108,287 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sometimes it is not easy to come to an agreement in a bargain. Right now Sasha and Vova can't come to an agreement: Sasha names a price as high as possible, then Vova wants to remove as many digits from the price as possible. In more details, Sasha names some integer price n, Vova removes a non-empty substring of (consecutive) digits from the price, the remaining digits close the gap, and the resulting integer is the price.
For example, is Sasha names 1213121, Vova can remove the substring 1312, and the result is 121.
It is allowed for result to contain leading zeros. If Vova removes all digits, the price is considered to be 0.
Sasha wants to come up with some constraints so that Vova can't just remove all digits, but he needs some arguments supporting the constraints. To start with, he wants to compute the sum of all possible resulting prices after Vova's move.
Help Sasha to compute this sum. Since the answer can be very large, print it modulo 10^9 + 7.
Input
The first and only line contains a single integer n (1 ≤ n < 10^{10^5}).
Output
In the only line print the required sum modulo 10^9 + 7.
Examples
Input
107
Output
42
Input
100500100500
Output
428101984
Note
Consider the first example.
Vova can choose to remove 1, 0, 7, 10, 07, or 107. The results are 07, 17, 10, 7, 1, 0. Their sum is 42.
Submitted Solution:
```
import sys
import math as m
MOD = 1e9 + 7
n = input()
l = len(n)
f = [0]*(1000001)
f[0] = 0
f[1] = 1
pos = 10
for i in range(2,l):
f[i] = (f[i-1]+(i*pos)%MOD)%MOD
pos = (pos*10)%MOD
ans_ = int(0)
pos_ = 1
for i in range(l-1,-1,-1):
if i == 0:
tmp1 = 0
else:
tmp1 = (((i*(i+1)//2)%MOD*int(n[i])%MOD)%MOD*pos_)%MOD
tmp2 = (int(n[i])*f[l-1-i])%MOD
ans_ = ((ans_+tmp1)%MOD+tmp2)%MOD
pos_ = (pos_*10)%MOD
print(int(ans_))
``` | instruction | 0 | 54,144 | 10 | 108,288 |
No | output | 1 | 54,144 | 10 | 108,289 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sometimes it is not easy to come to an agreement in a bargain. Right now Sasha and Vova can't come to an agreement: Sasha names a price as high as possible, then Vova wants to remove as many digits from the price as possible. In more details, Sasha names some integer price n, Vova removes a non-empty substring of (consecutive) digits from the price, the remaining digits close the gap, and the resulting integer is the price.
For example, is Sasha names 1213121, Vova can remove the substring 1312, and the result is 121.
It is allowed for result to contain leading zeros. If Vova removes all digits, the price is considered to be 0.
Sasha wants to come up with some constraints so that Vova can't just remove all digits, but he needs some arguments supporting the constraints. To start with, he wants to compute the sum of all possible resulting prices after Vova's move.
Help Sasha to compute this sum. Since the answer can be very large, print it modulo 10^9 + 7.
Input
The first and only line contains a single integer n (1 ≤ n < 10^{10^5}).
Output
In the only line print the required sum modulo 10^9 + 7.
Examples
Input
107
Output
42
Input
100500100500
Output
428101984
Note
Consider the first example.
Vova can choose to remove 1, 0, 7, 10, 07, or 107. The results are 07, 17, 10, 7, 1, 0. Their sum is 42.
Submitted Solution:
```
s = input()
n = len(s)
a = [0]*n
mod = 10**9 + 7
for i in range (n-2,-1,-1):
a[i] = a[i+1] + (n-i-1)*pow(10,n-i-2,mod)
a[i]%=mod
ans = 0
for i in range (n):
val = int(s[i])*pow(10,n-i-1,mod)
times = i*(i+1)//2
ans += val*times
ans += int(s[i])*a[i]
print(ans)
``` | instruction | 0 | 54,145 | 10 | 108,290 |
No | output | 1 | 54,145 | 10 | 108,291 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sometimes it is not easy to come to an agreement in a bargain. Right now Sasha and Vova can't come to an agreement: Sasha names a price as high as possible, then Vova wants to remove as many digits from the price as possible. In more details, Sasha names some integer price n, Vova removes a non-empty substring of (consecutive) digits from the price, the remaining digits close the gap, and the resulting integer is the price.
For example, is Sasha names 1213121, Vova can remove the substring 1312, and the result is 121.
It is allowed for result to contain leading zeros. If Vova removes all digits, the price is considered to be 0.
Sasha wants to come up with some constraints so that Vova can't just remove all digits, but he needs some arguments supporting the constraints. To start with, he wants to compute the sum of all possible resulting prices after Vova's move.
Help Sasha to compute this sum. Since the answer can be very large, print it modulo 10^9 + 7.
Input
The first and only line contains a single integer n (1 ≤ n < 10^{10^5}).
Output
In the only line print the required sum modulo 10^9 + 7.
Examples
Input
107
Output
42
Input
100500100500
Output
428101984
Note
Consider the first example.
Vova can choose to remove 1, 0, 7, 10, 07, or 107. The results are 07, 17, 10, 7, 1, 0. Their sum is 42.
Submitted Solution:
```
import sys
import math as m
MOD = 1000000007
n = input()
l = len(n)
f = [0]*(1000000)
f[0] = 0
f[1] = 1
pos = 10
for i in range(2,l):
f[i] = (f[i-1]+i*pos%MOD)%MOD
pos = (pos*10)%MOD
ans_ = int(0)
pos_ = 1
for i in range(l-1,-1,-1):
if i == 0:
ans_ += 0
else:
ans_ = (ans_+((int(n[i])*(((i+1)%MOD)*(i%MOD)/2)%MOD)%MOD*(pos_%MOD)))%MOD
ans_ = (ans_%MOD+int(n[i])*f[l-1-i]%MOD)%MOD
pos_ = (pos_*10)%MOD
print(int(ans_))
``` | instruction | 0 | 54,146 | 10 | 108,292 |
No | output | 1 | 54,146 | 10 | 108,293 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sometimes it is not easy to come to an agreement in a bargain. Right now Sasha and Vova can't come to an agreement: Sasha names a price as high as possible, then Vova wants to remove as many digits from the price as possible. In more details, Sasha names some integer price n, Vova removes a non-empty substring of (consecutive) digits from the price, the remaining digits close the gap, and the resulting integer is the price.
For example, is Sasha names 1213121, Vova can remove the substring 1312, and the result is 121.
It is allowed for result to contain leading zeros. If Vova removes all digits, the price is considered to be 0.
Sasha wants to come up with some constraints so that Vova can't just remove all digits, but he needs some arguments supporting the constraints. To start with, he wants to compute the sum of all possible resulting prices after Vova's move.
Help Sasha to compute this sum. Since the answer can be very large, print it modulo 10^9 + 7.
Input
The first and only line contains a single integer n (1 ≤ n < 10^{10^5}).
Output
In the only line print the required sum modulo 10^9 + 7.
Examples
Input
107
Output
42
Input
100500100500
Output
428101984
Note
Consider the first example.
Vova can choose to remove 1, 0, 7, 10, 07, or 107. The results are 07, 17, 10, 7, 1, 0. Their sum is 42.
Submitted Solution:
```
x=input()
list={}
total=0
def add(x):
n=len(x)
ans=0
for i in range(n):
ans+=int(x[i])*(10**(n-1-i))
return ans
for i in range(len(x)):
for j in range(i,len(x)):
s1=x[:i]
if j+1<len(x):s2=x[j+1:]
else:s2=""
if len(x)==1:
s=x
else:s=s1+s2
total+=add(s)
print(total%(10**9+7))
``` | instruction | 0 | 54,147 | 10 | 108,294 |
No | output | 1 | 54,147 | 10 | 108,295 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gerald has been selling state secrets at leisure. All the secrets cost the same: n marks. The state which secrets Gerald is selling, has no paper money, only coins. But there are coins of all positive integer denominations that are powers of three: 1 mark, 3 marks, 9 marks, 27 marks and so on. There are no coins of other denominations. Of course, Gerald likes it when he gets money without the change. And all buyers respect him and try to give the desired sum without change, if possible. But this does not always happen.
One day an unlucky buyer came. He did not have the desired sum without change. Then he took out all his coins and tried to give Gerald a larger than necessary sum with as few coins as possible. What is the maximum number of coins he could get?
The formal explanation of the previous paragraph: we consider all the possible combinations of coins for which the buyer can not give Gerald the sum of n marks without change. For each such combination calculate the minimum number of coins that can bring the buyer at least n marks. Among all combinations choose the maximum of the minimum number of coins. This is the number we want.
Input
The single line contains a single integer n (1 ≤ n ≤ 1017).
Please, do not use the %lld specifier to read or write 64 bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
In a single line print an integer: the maximum number of coins the unlucky buyer could have paid with.
Examples
Input
1
Output
1
Input
4
Output
2
Note
In the first test case, if a buyer has exactly one coin of at least 3 marks, then, to give Gerald one mark, he will have to give this coin. In this sample, the customer can not have a coin of one mark, as in this case, he will be able to give the money to Gerald without any change.
In the second test case, if the buyer had exactly three coins of 3 marks, then, to give Gerald 4 marks, he will have to give two of these coins. The buyer cannot give three coins as he wants to minimize the number of coins that he gives. | instruction | 0 | 54,280 | 10 | 108,560 |
Tags: greedy
Correct Solution:
```
n,a=int(input()),1
while n%a==0:a*=3
print(n//a+1)
# Made By Mostafa_Khaled
``` | output | 1 | 54,280 | 10 | 108,561 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gerald has been selling state secrets at leisure. All the secrets cost the same: n marks. The state which secrets Gerald is selling, has no paper money, only coins. But there are coins of all positive integer denominations that are powers of three: 1 mark, 3 marks, 9 marks, 27 marks and so on. There are no coins of other denominations. Of course, Gerald likes it when he gets money without the change. And all buyers respect him and try to give the desired sum without change, if possible. But this does not always happen.
One day an unlucky buyer came. He did not have the desired sum without change. Then he took out all his coins and tried to give Gerald a larger than necessary sum with as few coins as possible. What is the maximum number of coins he could get?
The formal explanation of the previous paragraph: we consider all the possible combinations of coins for which the buyer can not give Gerald the sum of n marks without change. For each such combination calculate the minimum number of coins that can bring the buyer at least n marks. Among all combinations choose the maximum of the minimum number of coins. This is the number we want.
Input
The single line contains a single integer n (1 ≤ n ≤ 1017).
Please, do not use the %lld specifier to read or write 64 bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
In a single line print an integer: the maximum number of coins the unlucky buyer could have paid with.
Examples
Input
1
Output
1
Input
4
Output
2
Note
In the first test case, if a buyer has exactly one coin of at least 3 marks, then, to give Gerald one mark, he will have to give this coin. In this sample, the customer can not have a coin of one mark, as in this case, he will be able to give the money to Gerald without any change.
In the second test case, if the buyer had exactly three coins of 3 marks, then, to give Gerald 4 marks, he will have to give two of these coins. The buyer cannot give three coins as he wants to minimize the number of coins that he gives. | instruction | 0 | 54,281 | 10 | 108,562 |
Tags: greedy
Correct Solution:
```
n = int(input());
while n % 3 == 0:
n = n // 3;
print(n // 3 + 1);
``` | output | 1 | 54,281 | 10 | 108,563 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gerald has been selling state secrets at leisure. All the secrets cost the same: n marks. The state which secrets Gerald is selling, has no paper money, only coins. But there are coins of all positive integer denominations that are powers of three: 1 mark, 3 marks, 9 marks, 27 marks and so on. There are no coins of other denominations. Of course, Gerald likes it when he gets money without the change. And all buyers respect him and try to give the desired sum without change, if possible. But this does not always happen.
One day an unlucky buyer came. He did not have the desired sum without change. Then he took out all his coins and tried to give Gerald a larger than necessary sum with as few coins as possible. What is the maximum number of coins he could get?
The formal explanation of the previous paragraph: we consider all the possible combinations of coins for which the buyer can not give Gerald the sum of n marks without change. For each such combination calculate the minimum number of coins that can bring the buyer at least n marks. Among all combinations choose the maximum of the minimum number of coins. This is the number we want.
Input
The single line contains a single integer n (1 ≤ n ≤ 1017).
Please, do not use the %lld specifier to read or write 64 bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
In a single line print an integer: the maximum number of coins the unlucky buyer could have paid with.
Examples
Input
1
Output
1
Input
4
Output
2
Note
In the first test case, if a buyer has exactly one coin of at least 3 marks, then, to give Gerald one mark, he will have to give this coin. In this sample, the customer can not have a coin of one mark, as in this case, he will be able to give the money to Gerald without any change.
In the second test case, if the buyer had exactly three coins of 3 marks, then, to give Gerald 4 marks, he will have to give two of these coins. The buyer cannot give three coins as he wants to minimize the number of coins that he gives. | instruction | 0 | 54,282 | 10 | 108,564 |
Tags: greedy
Correct Solution:
```
def f(n):
def toTri(x):
if x <= 2:
return [x]
l = toTri(x // 3)
l.append(x % 3)
return l
digits = toTri(n)
for i in range(len(digits) - 1, -1, -1):
if digits[i] != 0:
break
coins = 1
mul = 1
for j in range(i - 1, -1, -1):
coins += mul * digits[j]
mul *= 3
return coins
n = int(input())
print(f(n))
``` | output | 1 | 54,282 | 10 | 108,565 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gerald has been selling state secrets at leisure. All the secrets cost the same: n marks. The state which secrets Gerald is selling, has no paper money, only coins. But there are coins of all positive integer denominations that are powers of three: 1 mark, 3 marks, 9 marks, 27 marks and so on. There are no coins of other denominations. Of course, Gerald likes it when he gets money without the change. And all buyers respect him and try to give the desired sum without change, if possible. But this does not always happen.
One day an unlucky buyer came. He did not have the desired sum without change. Then he took out all his coins and tried to give Gerald a larger than necessary sum with as few coins as possible. What is the maximum number of coins he could get?
The formal explanation of the previous paragraph: we consider all the possible combinations of coins for which the buyer can not give Gerald the sum of n marks without change. For each such combination calculate the minimum number of coins that can bring the buyer at least n marks. Among all combinations choose the maximum of the minimum number of coins. This is the number we want.
Input
The single line contains a single integer n (1 ≤ n ≤ 1017).
Please, do not use the %lld specifier to read or write 64 bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
In a single line print an integer: the maximum number of coins the unlucky buyer could have paid with.
Examples
Input
1
Output
1
Input
4
Output
2
Note
In the first test case, if a buyer has exactly one coin of at least 3 marks, then, to give Gerald one mark, he will have to give this coin. In this sample, the customer can not have a coin of one mark, as in this case, he will be able to give the money to Gerald without any change.
In the second test case, if the buyer had exactly three coins of 3 marks, then, to give Gerald 4 marks, he will have to give two of these coins. The buyer cannot give three coins as he wants to minimize the number of coins that he gives. | instruction | 0 | 54,283 | 10 | 108,566 |
Tags: greedy
Correct Solution:
```
n = int(input())
while n % 3 == 0:
n //= 3
print(n // 3 + 1)
``` | output | 1 | 54,283 | 10 | 108,567 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gerald has been selling state secrets at leisure. All the secrets cost the same: n marks. The state which secrets Gerald is selling, has no paper money, only coins. But there are coins of all positive integer denominations that are powers of three: 1 mark, 3 marks, 9 marks, 27 marks and so on. There are no coins of other denominations. Of course, Gerald likes it when he gets money without the change. And all buyers respect him and try to give the desired sum without change, if possible. But this does not always happen.
One day an unlucky buyer came. He did not have the desired sum without change. Then he took out all his coins and tried to give Gerald a larger than necessary sum with as few coins as possible. What is the maximum number of coins he could get?
The formal explanation of the previous paragraph: we consider all the possible combinations of coins for which the buyer can not give Gerald the sum of n marks without change. For each such combination calculate the minimum number of coins that can bring the buyer at least n marks. Among all combinations choose the maximum of the minimum number of coins. This is the number we want.
Input
The single line contains a single integer n (1 ≤ n ≤ 1017).
Please, do not use the %lld specifier to read or write 64 bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
In a single line print an integer: the maximum number of coins the unlucky buyer could have paid with.
Examples
Input
1
Output
1
Input
4
Output
2
Note
In the first test case, if a buyer has exactly one coin of at least 3 marks, then, to give Gerald one mark, he will have to give this coin. In this sample, the customer can not have a coin of one mark, as in this case, he will be able to give the money to Gerald without any change.
In the second test case, if the buyer had exactly three coins of 3 marks, then, to give Gerald 4 marks, he will have to give two of these coins. The buyer cannot give three coins as he wants to minimize the number of coins that he gives. | instruction | 0 | 54,284 | 10 | 108,568 |
Tags: greedy
Correct Solution:
```
n = int(input())
k = 1
while n % k == 0:
k *= 3
print(n // k + 1)
``` | output | 1 | 54,284 | 10 | 108,569 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gerald has been selling state secrets at leisure. All the secrets cost the same: n marks. The state which secrets Gerald is selling, has no paper money, only coins. But there are coins of all positive integer denominations that are powers of three: 1 mark, 3 marks, 9 marks, 27 marks and so on. There are no coins of other denominations. Of course, Gerald likes it when he gets money without the change. And all buyers respect him and try to give the desired sum without change, if possible. But this does not always happen.
One day an unlucky buyer came. He did not have the desired sum without change. Then he took out all his coins and tried to give Gerald a larger than necessary sum with as few coins as possible. What is the maximum number of coins he could get?
The formal explanation of the previous paragraph: we consider all the possible combinations of coins for which the buyer can not give Gerald the sum of n marks without change. For each such combination calculate the minimum number of coins that can bring the buyer at least n marks. Among all combinations choose the maximum of the minimum number of coins. This is the number we want.
Input
The single line contains a single integer n (1 ≤ n ≤ 1017).
Please, do not use the %lld specifier to read or write 64 bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
In a single line print an integer: the maximum number of coins the unlucky buyer could have paid with.
Examples
Input
1
Output
1
Input
4
Output
2
Note
In the first test case, if a buyer has exactly one coin of at least 3 marks, then, to give Gerald one mark, he will have to give this coin. In this sample, the customer can not have a coin of one mark, as in this case, he will be able to give the money to Gerald without any change.
In the second test case, if the buyer had exactly three coins of 3 marks, then, to give Gerald 4 marks, he will have to give two of these coins. The buyer cannot give three coins as he wants to minimize the number of coins that he gives. | instruction | 0 | 54,285 | 10 | 108,570 |
Tags: greedy
Correct Solution:
```
from fractions import Decimal
import math
n=Decimal(input())
d=Decimal('1')
while(n%d==0):
d*=3
if(d>n):
print(1)
else:
print(math.ceil(n/d))
``` | output | 1 | 54,285 | 10 | 108,571 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gerald has been selling state secrets at leisure. All the secrets cost the same: n marks. The state which secrets Gerald is selling, has no paper money, only coins. But there are coins of all positive integer denominations that are powers of three: 1 mark, 3 marks, 9 marks, 27 marks and so on. There are no coins of other denominations. Of course, Gerald likes it when he gets money without the change. And all buyers respect him and try to give the desired sum without change, if possible. But this does not always happen.
One day an unlucky buyer came. He did not have the desired sum without change. Then he took out all his coins and tried to give Gerald a larger than necessary sum with as few coins as possible. What is the maximum number of coins he could get?
The formal explanation of the previous paragraph: we consider all the possible combinations of coins for which the buyer can not give Gerald the sum of n marks without change. For each such combination calculate the minimum number of coins that can bring the buyer at least n marks. Among all combinations choose the maximum of the minimum number of coins. This is the number we want.
Input
The single line contains a single integer n (1 ≤ n ≤ 1017).
Please, do not use the %lld specifier to read or write 64 bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
In a single line print an integer: the maximum number of coins the unlucky buyer could have paid with.
Examples
Input
1
Output
1
Input
4
Output
2
Note
In the first test case, if a buyer has exactly one coin of at least 3 marks, then, to give Gerald one mark, he will have to give this coin. In this sample, the customer can not have a coin of one mark, as in this case, he will be able to give the money to Gerald without any change.
In the second test case, if the buyer had exactly three coins of 3 marks, then, to give Gerald 4 marks, he will have to give two of these coins. The buyer cannot give three coins as he wants to minimize the number of coins that he gives. | instruction | 0 | 54,286 | 10 | 108,572 |
Tags: greedy
Correct Solution:
```
import math, decimal
n = int(input())
k = 0
while n % 3**k == 0:
k += 1
D = decimal.Decimal
print(math.ceil(D(n) / D(3**k)))
``` | output | 1 | 54,286 | 10 | 108,573 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gerald has been selling state secrets at leisure. All the secrets cost the same: n marks. The state which secrets Gerald is selling, has no paper money, only coins. But there are coins of all positive integer denominations that are powers of three: 1 mark, 3 marks, 9 marks, 27 marks and so on. There are no coins of other denominations. Of course, Gerald likes it when he gets money without the change. And all buyers respect him and try to give the desired sum without change, if possible. But this does not always happen.
One day an unlucky buyer came. He did not have the desired sum without change. Then he took out all his coins and tried to give Gerald a larger than necessary sum with as few coins as possible. What is the maximum number of coins he could get?
The formal explanation of the previous paragraph: we consider all the possible combinations of coins for which the buyer can not give Gerald the sum of n marks without change. For each such combination calculate the minimum number of coins that can bring the buyer at least n marks. Among all combinations choose the maximum of the minimum number of coins. This is the number we want.
Input
The single line contains a single integer n (1 ≤ n ≤ 1017).
Please, do not use the %lld specifier to read or write 64 bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
In a single line print an integer: the maximum number of coins the unlucky buyer could have paid with.
Examples
Input
1
Output
1
Input
4
Output
2
Note
In the first test case, if a buyer has exactly one coin of at least 3 marks, then, to give Gerald one mark, he will have to give this coin. In this sample, the customer can not have a coin of one mark, as in this case, he will be able to give the money to Gerald without any change.
In the second test case, if the buyer had exactly three coins of 3 marks, then, to give Gerald 4 marks, he will have to give two of these coins. The buyer cannot give three coins as he wants to minimize the number of coins that he gives. | instruction | 0 | 54,287 | 10 | 108,574 |
Tags: greedy
Correct Solution:
```
n = int(input())
t = 1
while n % t == 0:
t *= 3
print(n // t + 1)
``` | output | 1 | 54,287 | 10 | 108,575 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gerald has been selling state secrets at leisure. All the secrets cost the same: n marks. The state which secrets Gerald is selling, has no paper money, only coins. But there are coins of all positive integer denominations that are powers of three: 1 mark, 3 marks, 9 marks, 27 marks and so on. There are no coins of other denominations. Of course, Gerald likes it when he gets money without the change. And all buyers respect him and try to give the desired sum without change, if possible. But this does not always happen.
One day an unlucky buyer came. He did not have the desired sum without change. Then he took out all his coins and tried to give Gerald a larger than necessary sum with as few coins as possible. What is the maximum number of coins he could get?
The formal explanation of the previous paragraph: we consider all the possible combinations of coins for which the buyer can not give Gerald the sum of n marks without change. For each such combination calculate the minimum number of coins that can bring the buyer at least n marks. Among all combinations choose the maximum of the minimum number of coins. This is the number we want.
Input
The single line contains a single integer n (1 ≤ n ≤ 1017).
Please, do not use the %lld specifier to read or write 64 bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
In a single line print an integer: the maximum number of coins the unlucky buyer could have paid with.
Examples
Input
1
Output
1
Input
4
Output
2
Note
In the first test case, if a buyer has exactly one coin of at least 3 marks, then, to give Gerald one mark, he will have to give this coin. In this sample, the customer can not have a coin of one mark, as in this case, he will be able to give the money to Gerald without any change.
In the second test case, if the buyer had exactly three coins of 3 marks, then, to give Gerald 4 marks, he will have to give two of these coins. The buyer cannot give three coins as he wants to minimize the number of coins that he gives.
Submitted Solution:
```
n = int(input())
t = 3
while n % t == 0:
t *= 3
print(n // t + 1)
``` | instruction | 0 | 54,288 | 10 | 108,576 |
Yes | output | 1 | 54,288 | 10 | 108,577 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gerald has been selling state secrets at leisure. All the secrets cost the same: n marks. The state which secrets Gerald is selling, has no paper money, only coins. But there are coins of all positive integer denominations that are powers of three: 1 mark, 3 marks, 9 marks, 27 marks and so on. There are no coins of other denominations. Of course, Gerald likes it when he gets money without the change. And all buyers respect him and try to give the desired sum without change, if possible. But this does not always happen.
One day an unlucky buyer came. He did not have the desired sum without change. Then he took out all his coins and tried to give Gerald a larger than necessary sum with as few coins as possible. What is the maximum number of coins he could get?
The formal explanation of the previous paragraph: we consider all the possible combinations of coins for which the buyer can not give Gerald the sum of n marks without change. For each such combination calculate the minimum number of coins that can bring the buyer at least n marks. Among all combinations choose the maximum of the minimum number of coins. This is the number we want.
Input
The single line contains a single integer n (1 ≤ n ≤ 1017).
Please, do not use the %lld specifier to read or write 64 bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
In a single line print an integer: the maximum number of coins the unlucky buyer could have paid with.
Examples
Input
1
Output
1
Input
4
Output
2
Note
In the first test case, if a buyer has exactly one coin of at least 3 marks, then, to give Gerald one mark, he will have to give this coin. In this sample, the customer can not have a coin of one mark, as in this case, he will be able to give the money to Gerald without any change.
In the second test case, if the buyer had exactly three coins of 3 marks, then, to give Gerald 4 marks, he will have to give two of these coins. The buyer cannot give three coins as he wants to minimize the number of coins that he gives.
Submitted Solution:
```
from math import ceil
def seal(a, b):
return (a + b - 1) // b
n = int(input())
i = 0
while(n % pow(3,i) == 0):
i+=1
print(seal(n, pow(3,i)))
``` | instruction | 0 | 54,289 | 10 | 108,578 |
Yes | output | 1 | 54,289 | 10 | 108,579 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gerald has been selling state secrets at leisure. All the secrets cost the same: n marks. The state which secrets Gerald is selling, has no paper money, only coins. But there are coins of all positive integer denominations that are powers of three: 1 mark, 3 marks, 9 marks, 27 marks and so on. There are no coins of other denominations. Of course, Gerald likes it when he gets money without the change. And all buyers respect him and try to give the desired sum without change, if possible. But this does not always happen.
One day an unlucky buyer came. He did not have the desired sum without change. Then he took out all his coins and tried to give Gerald a larger than necessary sum with as few coins as possible. What is the maximum number of coins he could get?
The formal explanation of the previous paragraph: we consider all the possible combinations of coins for which the buyer can not give Gerald the sum of n marks without change. For each such combination calculate the minimum number of coins that can bring the buyer at least n marks. Among all combinations choose the maximum of the minimum number of coins. This is the number we want.
Input
The single line contains a single integer n (1 ≤ n ≤ 1017).
Please, do not use the %lld specifier to read or write 64 bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
In a single line print an integer: the maximum number of coins the unlucky buyer could have paid with.
Examples
Input
1
Output
1
Input
4
Output
2
Note
In the first test case, if a buyer has exactly one coin of at least 3 marks, then, to give Gerald one mark, he will have to give this coin. In this sample, the customer can not have a coin of one mark, as in this case, he will be able to give the money to Gerald without any change.
In the second test case, if the buyer had exactly three coins of 3 marks, then, to give Gerald 4 marks, he will have to give two of these coins. The buyer cannot give three coins as he wants to minimize the number of coins that he gives.
Submitted Solution:
```
n=int(input())
if(n%3==0):
x=9
while(n%x==0):
x*=3
print(n//x+1)
else:
print(n//3+1)
``` | instruction | 0 | 54,290 | 10 | 108,580 |
Yes | output | 1 | 54,290 | 10 | 108,581 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gerald has been selling state secrets at leisure. All the secrets cost the same: n marks. The state which secrets Gerald is selling, has no paper money, only coins. But there are coins of all positive integer denominations that are powers of three: 1 mark, 3 marks, 9 marks, 27 marks and so on. There are no coins of other denominations. Of course, Gerald likes it when he gets money without the change. And all buyers respect him and try to give the desired sum without change, if possible. But this does not always happen.
One day an unlucky buyer came. He did not have the desired sum without change. Then he took out all his coins and tried to give Gerald a larger than necessary sum with as few coins as possible. What is the maximum number of coins he could get?
The formal explanation of the previous paragraph: we consider all the possible combinations of coins for which the buyer can not give Gerald the sum of n marks without change. For each such combination calculate the minimum number of coins that can bring the buyer at least n marks. Among all combinations choose the maximum of the minimum number of coins. This is the number we want.
Input
The single line contains a single integer n (1 ≤ n ≤ 1017).
Please, do not use the %lld specifier to read or write 64 bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
In a single line print an integer: the maximum number of coins the unlucky buyer could have paid with.
Examples
Input
1
Output
1
Input
4
Output
2
Note
In the first test case, if a buyer has exactly one coin of at least 3 marks, then, to give Gerald one mark, he will have to give this coin. In this sample, the customer can not have a coin of one mark, as in this case, he will be able to give the money to Gerald without any change.
In the second test case, if the buyer had exactly three coins of 3 marks, then, to give Gerald 4 marks, he will have to give two of these coins. The buyer cannot give three coins as he wants to minimize the number of coins that he gives.
Submitted Solution:
```
n = int(input())
p = 1
while n % p == 0: p *= 3
print(n//p + 1)
``` | instruction | 0 | 54,291 | 10 | 108,582 |
Yes | output | 1 | 54,291 | 10 | 108,583 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gerald has been selling state secrets at leisure. All the secrets cost the same: n marks. The state which secrets Gerald is selling, has no paper money, only coins. But there are coins of all positive integer denominations that are powers of three: 1 mark, 3 marks, 9 marks, 27 marks and so on. There are no coins of other denominations. Of course, Gerald likes it when he gets money without the change. And all buyers respect him and try to give the desired sum without change, if possible. But this does not always happen.
One day an unlucky buyer came. He did not have the desired sum without change. Then he took out all his coins and tried to give Gerald a larger than necessary sum with as few coins as possible. What is the maximum number of coins he could get?
The formal explanation of the previous paragraph: we consider all the possible combinations of coins for which the buyer can not give Gerald the sum of n marks without change. For each such combination calculate the minimum number of coins that can bring the buyer at least n marks. Among all combinations choose the maximum of the minimum number of coins. This is the number we want.
Input
The single line contains a single integer n (1 ≤ n ≤ 1017).
Please, do not use the %lld specifier to read or write 64 bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
In a single line print an integer: the maximum number of coins the unlucky buyer could have paid with.
Examples
Input
1
Output
1
Input
4
Output
2
Note
In the first test case, if a buyer has exactly one coin of at least 3 marks, then, to give Gerald one mark, he will have to give this coin. In this sample, the customer can not have a coin of one mark, as in this case, he will be able to give the money to Gerald without any change.
In the second test case, if the buyer had exactly three coins of 3 marks, then, to give Gerald 4 marks, he will have to give two of these coins. The buyer cannot give three coins as he wants to minimize the number of coins that he gives.
Submitted Solution:
```
import math
n = int(input())
t = 1
res = 1
while t < n:
if n % t != 0:
res = max(res, math.ceil(n / t))
t *= 3
print(res)
``` | instruction | 0 | 54,292 | 10 | 108,584 |
No | output | 1 | 54,292 | 10 | 108,585 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gerald has been selling state secrets at leisure. All the secrets cost the same: n marks. The state which secrets Gerald is selling, has no paper money, only coins. But there are coins of all positive integer denominations that are powers of three: 1 mark, 3 marks, 9 marks, 27 marks and so on. There are no coins of other denominations. Of course, Gerald likes it when he gets money without the change. And all buyers respect him and try to give the desired sum without change, if possible. But this does not always happen.
One day an unlucky buyer came. He did not have the desired sum without change. Then he took out all his coins and tried to give Gerald a larger than necessary sum with as few coins as possible. What is the maximum number of coins he could get?
The formal explanation of the previous paragraph: we consider all the possible combinations of coins for which the buyer can not give Gerald the sum of n marks without change. For each such combination calculate the minimum number of coins that can bring the buyer at least n marks. Among all combinations choose the maximum of the minimum number of coins. This is the number we want.
Input
The single line contains a single integer n (1 ≤ n ≤ 1017).
Please, do not use the %lld specifier to read or write 64 bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
In a single line print an integer: the maximum number of coins the unlucky buyer could have paid with.
Examples
Input
1
Output
1
Input
4
Output
2
Note
In the first test case, if a buyer has exactly one coin of at least 3 marks, then, to give Gerald one mark, he will have to give this coin. In this sample, the customer can not have a coin of one mark, as in this case, he will be able to give the money to Gerald without any change.
In the second test case, if the buyer had exactly three coins of 3 marks, then, to give Gerald 4 marks, he will have to give two of these coins. The buyer cannot give three coins as he wants to minimize the number of coins that he gives.
Submitted Solution:
```
from math import ceil, floor
n = int(input())
a = 1
while (n % a == 0):
a *= 3
print(floor(n / a))
``` | instruction | 0 | 54,293 | 10 | 108,586 |
No | output | 1 | 54,293 | 10 | 108,587 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gerald has been selling state secrets at leisure. All the secrets cost the same: n marks. The state which secrets Gerald is selling, has no paper money, only coins. But there are coins of all positive integer denominations that are powers of three: 1 mark, 3 marks, 9 marks, 27 marks and so on. There are no coins of other denominations. Of course, Gerald likes it when he gets money without the change. And all buyers respect him and try to give the desired sum without change, if possible. But this does not always happen.
One day an unlucky buyer came. He did not have the desired sum without change. Then he took out all his coins and tried to give Gerald a larger than necessary sum with as few coins as possible. What is the maximum number of coins he could get?
The formal explanation of the previous paragraph: we consider all the possible combinations of coins for which the buyer can not give Gerald the sum of n marks without change. For each such combination calculate the minimum number of coins that can bring the buyer at least n marks. Among all combinations choose the maximum of the minimum number of coins. This is the number we want.
Input
The single line contains a single integer n (1 ≤ n ≤ 1017).
Please, do not use the %lld specifier to read or write 64 bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
In a single line print an integer: the maximum number of coins the unlucky buyer could have paid with.
Examples
Input
1
Output
1
Input
4
Output
2
Note
In the first test case, if a buyer has exactly one coin of at least 3 marks, then, to give Gerald one mark, he will have to give this coin. In this sample, the customer can not have a coin of one mark, as in this case, he will be able to give the money to Gerald without any change.
In the second test case, if the buyer had exactly three coins of 3 marks, then, to give Gerald 4 marks, he will have to give two of these coins. The buyer cannot give three coins as he wants to minimize the number of coins that he gives.
Submitted Solution:
```
from math import ceil, floor
n = int(input())
a = 1
degree = 0
while (n % a == 0):
a *= 3
degree += 1
print(floor(n / degree))
``` | instruction | 0 | 54,294 | 10 | 108,588 |
No | output | 1 | 54,294 | 10 | 108,589 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gerald has been selling state secrets at leisure. All the secrets cost the same: n marks. The state which secrets Gerald is selling, has no paper money, only coins. But there are coins of all positive integer denominations that are powers of three: 1 mark, 3 marks, 9 marks, 27 marks and so on. There are no coins of other denominations. Of course, Gerald likes it when he gets money without the change. And all buyers respect him and try to give the desired sum without change, if possible. But this does not always happen.
One day an unlucky buyer came. He did not have the desired sum without change. Then he took out all his coins and tried to give Gerald a larger than necessary sum with as few coins as possible. What is the maximum number of coins he could get?
The formal explanation of the previous paragraph: we consider all the possible combinations of coins for which the buyer can not give Gerald the sum of n marks without change. For each such combination calculate the minimum number of coins that can bring the buyer at least n marks. Among all combinations choose the maximum of the minimum number of coins. This is the number we want.
Input
The single line contains a single integer n (1 ≤ n ≤ 1017).
Please, do not use the %lld specifier to read or write 64 bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
In a single line print an integer: the maximum number of coins the unlucky buyer could have paid with.
Examples
Input
1
Output
1
Input
4
Output
2
Note
In the first test case, if a buyer has exactly one coin of at least 3 marks, then, to give Gerald one mark, he will have to give this coin. In this sample, the customer can not have a coin of one mark, as in this case, he will be able to give the money to Gerald without any change.
In the second test case, if the buyer had exactly three coins of 3 marks, then, to give Gerald 4 marks, he will have to give two of these coins. The buyer cannot give three coins as he wants to minimize the number of coins that he gives.
Submitted Solution:
```
n = int(input())
c=[]
while n:
c+=[n%3]
n//=3
for i in range(len(c)):
if c[i]>0:
break
if i==len(c)-1:
print(1)
else:
c[i]=0
c[i+1]+=1
print(sum(c))
``` | instruction | 0 | 54,295 | 10 | 108,590 |
No | output | 1 | 54,295 | 10 | 108,591 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Pari wants to buy an expensive chocolate from Arya. She has n coins, the value of the i-th coin is ci. The price of the chocolate is k, so Pari will take a subset of her coins with sum equal to k and give it to Arya.
Looking at her coins, a question came to her mind: after giving the coins to Arya, what values does Arya can make with them? She is jealous and she doesn't want Arya to make a lot of values. So she wants to know all the values x, such that Arya will be able to make x using some subset of coins with the sum k.
Formally, Pari wants to know the values x such that there exists a subset of coins with the sum k such that some subset of this subset has the sum x, i.e. there is exists some way to pay for the chocolate, such that Arya will be able to make the sum x using these coins.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 500) — the number of coins and the price of the chocolate, respectively.
Next line will contain n integers c1, c2, ..., cn (1 ≤ ci ≤ 500) — the values of Pari's coins.
It's guaranteed that one can make value k using these coins.
Output
First line of the output must contain a single integer q— the number of suitable values x. Then print q integers in ascending order — the values that Arya can make for some subset of coins of Pari that pays for the chocolate.
Examples
Input
6 18
5 6 1 10 12 2
Output
16
0 1 2 3 5 6 7 8 10 11 12 13 15 16 17 18
Input
3 50
25 25 50
Output
3
0 25 50 | instruction | 0 | 54,440 | 10 | 108,880 |
Tags: dp
Correct Solution:
```
from sys import stdin,stdout
n,m=map(int,input().split())
s=[0]+list(map(int,stdin.readline().strip().split()))
dp=[[[False for i in range(m+1)] for j in range(m+1)] for k in range(2)]
dp[0][0][0]=True
cur=0
for i in range(1,n+1):
cur+=1
cur%=2
last=(cur+1)%2
for j in range(m+1):
for k in range(j+1):
if j-s[i]>-1 and k-s[i]>-1:
dp[cur][j][k]=dp[last][j][k] or dp[last][j-s[i]][k] or dp[last][j-s[i]][k-s[i]]
elif j-s[i]>-1 :
dp[cur][j][k]=dp[last][j][k] or dp[last][j-s[i]][k]
elif k-s[i]>-1:
dp[cur][j][k]=dp[last][j][k] or dp[last][j-s[i]][k-s[i]]
else:
dp[cur][j][k]=dp[last][j][k]
n1=0
ans=""
for i in range(k+1):
if dp[cur][k][i]:
ans+=str(i)+" "
n1+=1
stdout.write(str(n1)+chr(10))
stdout.write(ans)
``` | output | 1 | 54,440 | 10 | 108,881 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Pari wants to buy an expensive chocolate from Arya. She has n coins, the value of the i-th coin is ci. The price of the chocolate is k, so Pari will take a subset of her coins with sum equal to k and give it to Arya.
Looking at her coins, a question came to her mind: after giving the coins to Arya, what values does Arya can make with them? She is jealous and she doesn't want Arya to make a lot of values. So she wants to know all the values x, such that Arya will be able to make x using some subset of coins with the sum k.
Formally, Pari wants to know the values x such that there exists a subset of coins with the sum k such that some subset of this subset has the sum x, i.e. there is exists some way to pay for the chocolate, such that Arya will be able to make the sum x using these coins.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 500) — the number of coins and the price of the chocolate, respectively.
Next line will contain n integers c1, c2, ..., cn (1 ≤ ci ≤ 500) — the values of Pari's coins.
It's guaranteed that one can make value k using these coins.
Output
First line of the output must contain a single integer q— the number of suitable values x. Then print q integers in ascending order — the values that Arya can make for some subset of coins of Pari that pays for the chocolate.
Examples
Input
6 18
5 6 1 10 12 2
Output
16
0 1 2 3 5 6 7 8 10 11 12 13 15 16 17 18
Input
3 50
25 25 50
Output
3
0 25 50 | instruction | 0 | 54,441 | 10 | 108,882 |
Tags: dp
Correct Solution:
```
read = lambda: map(int, input().split())
n, k = read()
c = list(read())
dp = [0] * (k + 1)
dp[0] = 1
for x in c:
ndp = dp[:]
for i in range(k, x - 1, -1):
ndp[i] |= dp[i - x] | (dp[i - x] << x)
dp = ndp
b = bin(dp[-1])
ans = [i for i in range(k + 1) if b[-i - 1] == '1']
print(len(ans))
print(*ans)
``` | output | 1 | 54,441 | 10 | 108,883 |
Provide a correct Python 3 solution for this coding contest problem.
There are N stores called Store 1, Store 2, \cdots, Store N. Takahashi, who is at his house at time 0, is planning to visit some of these stores.
It takes Takahashi one unit of time to travel from his house to one of the stores, or between any two stores.
If Takahashi reaches Store i at time t, he can do shopping there after standing in a queue for a_i \times t + b_i units of time. (We assume that it takes no time other than waiting.)
All the stores close at time T + 0.5. If Takahashi is standing in a queue for some store then, he cannot do shopping there.
Takahashi does not do shopping more than once in the same store.
Find the maximum number of times he can do shopping before time T + 0.5.
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* 0 \leq a_i \leq 10^9
* 0 \leq b_i \leq 10^9
* 0 \leq T \leq 10^9
Input
Input is given from Standard Input in the following format:
N T
a_1 b_1
a_2 b_2
\vdots
a_N b_N
Output
Print the answer.
Examples
Input
3 7
2 0
3 2
0 3
Output
2
Input
1 3
0 3
Output
0
Input
5 21600
2 14
3 22
1 3
1 10
1 9
Output
5
Input
7 57
0 25
3 10
2 4
5 15
3 22
2 14
1 15
Output
3 | instruction | 0 | 54,561 | 10 | 109,122 |
"Correct Solution:
```
from bisect import bisect_right
import sys
sys.setrecursionlimit(10 ** 7)
input = sys.stdin.readline
n, t = map(int, input().split())
shop = []
shop_a0 = []
for i in range(n):
a, b = map(int, input().split())
if a != 0:
shop.append([a, b, a/(b+1)])
else:
shop_a0.append(b)
shop = sorted(shop, key=lambda x: -x[2])
shop_a0 = sorted(shop_a0)
n = len(shop)
INF = 10**10
dp = [INF for _ in range(30)]
dp[0] = 0
m = min(n+1, 30)
for i in range(n):
for j in range(m)[::-1]:
dp[j] = min(dp[j], dp[j-1] + 1 +
shop[i][0]*(dp[j-1]+1) + shop[i][1])
ss = []
if shop_a0:
ss.append(1+shop_a0[0])
for i in range(1, len(shop_a0)):
ss.append(ss[-1]+1+shop_a0[i])
ans = 0
for i in range(m):
if t - dp[i] >= 0:
ans = max(ans, bisect_right(ss, t-dp[i]) + i)
print(ans)
``` | output | 1 | 54,561 | 10 | 109,123 |
Provide a correct Python 3 solution for this coding contest problem.
There are N stores called Store 1, Store 2, \cdots, Store N. Takahashi, who is at his house at time 0, is planning to visit some of these stores.
It takes Takahashi one unit of time to travel from his house to one of the stores, or between any two stores.
If Takahashi reaches Store i at time t, he can do shopping there after standing in a queue for a_i \times t + b_i units of time. (We assume that it takes no time other than waiting.)
All the stores close at time T + 0.5. If Takahashi is standing in a queue for some store then, he cannot do shopping there.
Takahashi does not do shopping more than once in the same store.
Find the maximum number of times he can do shopping before time T + 0.5.
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* 0 \leq a_i \leq 10^9
* 0 \leq b_i \leq 10^9
* 0 \leq T \leq 10^9
Input
Input is given from Standard Input in the following format:
N T
a_1 b_1
a_2 b_2
\vdots
a_N b_N
Output
Print the answer.
Examples
Input
3 7
2 0
3 2
0 3
Output
2
Input
1 3
0 3
Output
0
Input
5 21600
2 14
3 22
1 3
1 10
1 9
Output
5
Input
7 57
0 25
3 10
2 4
5 15
3 22
2 14
1 15
Output
3 | instruction | 0 | 54,562 | 10 | 109,124 |
"Correct Solution:
```
# https://atcoder.jp/contests/hitachi2020/submissions/10704509
# PyPyなら通る
def main():
import sys
input = sys.stdin.readline
inf = 1 << 30
L = 30
N, T = map(int, input().split())
ps = []
coef_0 = []
for _ in range(N):
a, b = map(int, input().split())
if a == 0:
coef_0.append(b + 1)
else:
ps.append((a, b)) # namedtupleやめた,coef,const
ps.sort(key=lambda p: p[0] / (p[1] + 1), reverse=True)
coef_0.sort()
dp = [inf] * (L + 1)
dp[0] = 0
for coef, const in ps:
for visited in reversed(range(L)):
cand = (dp[visited] + 1) * (coef + 1) + const
if dp[visited + 1] > cand:
dp[visited + 1] = cand # 不等式で評価する
def accumulate(a):
s = 0
yield s # start=x->0
for x in a:
s += x
yield s
*acc, = accumulate(coef_0)
ans = 0
cnt = len(acc) - 1
for visited, consumption_time in enumerate(dp):
rest = T - consumption_time
if rest < 0: break # le->lt,continue->break,restは狭義単調減少
while cnt > 0 and acc[cnt] > rest: # bisect_right->しゃくとり
cnt -= 1
visited += cnt
if visited > ans:
ans = visited # 関数呼び出しとどちらが速いか(set->maxよりは逐次が速かった）
print(ans)
if __name__ == '__main__':
main()
``` | output | 1 | 54,562 | 10 | 109,125 |
Provide a correct Python 3 solution for this coding contest problem.
There are N stores called Store 1, Store 2, \cdots, Store N. Takahashi, who is at his house at time 0, is planning to visit some of these stores.
It takes Takahashi one unit of time to travel from his house to one of the stores, or between any two stores.
If Takahashi reaches Store i at time t, he can do shopping there after standing in a queue for a_i \times t + b_i units of time. (We assume that it takes no time other than waiting.)
All the stores close at time T + 0.5. If Takahashi is standing in a queue for some store then, he cannot do shopping there.
Takahashi does not do shopping more than once in the same store.
Find the maximum number of times he can do shopping before time T + 0.5.
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* 0 \leq a_i \leq 10^9
* 0 \leq b_i \leq 10^9
* 0 \leq T \leq 10^9
Input
Input is given from Standard Input in the following format:
N T
a_1 b_1
a_2 b_2
\vdots
a_N b_N
Output
Print the answer.
Examples
Input
3 7
2 0
3 2
0 3
Output
2
Input
1 3
0 3
Output
0
Input
5 21600
2 14
3 22
1 3
1 10
1 9
Output
5
Input
7 57
0 25
3 10
2 4
5 15
3 22
2 14
1 15
Output
3 | instruction | 0 | 54,563 | 10 | 109,126 |
"Correct Solution:
```
# https://atcoder.jp/contests/hitachi2020/submissions/10704509
# PyPyなら通る
def main():
import sys
input = sys.stdin.readline
inf = 1 << 30
L = 30
N, T = map(int, input().split())
ps = []
coef_0 = []
for _ in range(N):
a, b = map(int, input().split())
if a == 0:
coef_0.append(b + 1)
else:
ps.append((a, b)) # namedtupleやめた,coef,const
ps.sort(key=lambda p: p[0] / (p[1] + 1), reverse=True)
coef_0.sort()
dp = [inf] * (L + 1)
dp[0] = 0
for coef, const in ps:
for visited in reversed(range(L)):
dp[visited + 1] = min(
dp[visited + 1],
(dp[visited] + 1) * (coef + 1) + const
)
def accumulate(a):
s = 0
yield s # start=x->0
for x in a:
s += x
yield s
*acc, = accumulate(coef_0)
ans = 0
cnt = len(acc) - 1
for visited, consumption_time in enumerate(dp):
rest = T - consumption_time
if rest < 0: break # le->lt,continue->break,restは狭義単調減少
while cnt > 0 and acc[cnt] > rest: # bisect_right->しゃくとり
cnt -= 1
visited += cnt
if visited > ans:
ans = visited # 関数呼び出しとどちらが速いか(set->maxよりは逐次が速かった）
print(ans)
if __name__ == '__main__':
main()
``` | output | 1 | 54,563 | 10 | 109,127 |
Provide a correct Python 3 solution for this coding contest problem.
There are N stores called Store 1, Store 2, \cdots, Store N. Takahashi, who is at his house at time 0, is planning to visit some of these stores.
It takes Takahashi one unit of time to travel from his house to one of the stores, or between any two stores.
If Takahashi reaches Store i at time t, he can do shopping there after standing in a queue for a_i \times t + b_i units of time. (We assume that it takes no time other than waiting.)
All the stores close at time T + 0.5. If Takahashi is standing in a queue for some store then, he cannot do shopping there.
Takahashi does not do shopping more than once in the same store.
Find the maximum number of times he can do shopping before time T + 0.5.
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* 0 \leq a_i \leq 10^9
* 0 \leq b_i \leq 10^9
* 0 \leq T \leq 10^9
Input
Input is given from Standard Input in the following format:
N T
a_1 b_1
a_2 b_2
\vdots
a_N b_N
Output
Print the answer.
Examples
Input
3 7
2 0
3 2
0 3
Output
2
Input
1 3
0 3
Output
0
Input
5 21600
2 14
3 22
1 3
1 10
1 9
Output
5
Input
7 57
0 25
3 10
2 4
5 15
3 22
2 14
1 15
Output
3 | instruction | 0 | 54,564 | 10 | 109,128 |
"Correct Solution:
```
import sys
sys.setrecursionlimit(2147483647)
INF = 1 << 60
MOD = 10**9 + 7 # 998244353
input = lambda:sys.stdin.readline().rstrip()
from bisect import bisect_right
from itertools import accumulate
from functools import cmp_to_key
def cmp(store1, store2):
if store1 == store2:
return 0
a1, b1 = store1
a2, b2 = store2
return -1 if a2 * (b1 + 1) <= a1 * (b2 + 1) else 1
def resolve():
n, T = map(int, input().split())
AB = []
zero = []
for _ in range(n):
a, b = map(int, input().split())
if a != 0:
AB.append((a, b))
else:
zero.append(b + 1)
AB.sort(key = cmp_to_key(cmp))
zero.sort()
zero = list(accumulate(zero))
max_store = 30
dp = [T + 1] * max_store
dp[0] = 0
for a, b in AB:
for i in range(max_store - 2, -1, -1):
t = dp[i]
dp[i + 1] = min(dp[i + 1], (a + 1) * (t + 1) + b)
res = 0
for i in range(max_store):
t = dp[i]
if t > T:
break
rest = T - t
res = max(res, i + bisect_right(zero, rest))
print(res)
resolve()
``` | output | 1 | 54,564 | 10 | 109,129 |
Provide a correct Python 3 solution for this coding contest problem.
There are N stores called Store 1, Store 2, \cdots, Store N. Takahashi, who is at his house at time 0, is planning to visit some of these stores.
It takes Takahashi one unit of time to travel from his house to one of the stores, or between any two stores.
If Takahashi reaches Store i at time t, he can do shopping there after standing in a queue for a_i \times t + b_i units of time. (We assume that it takes no time other than waiting.)
All the stores close at time T + 0.5. If Takahashi is standing in a queue for some store then, he cannot do shopping there.
Takahashi does not do shopping more than once in the same store.
Find the maximum number of times he can do shopping before time T + 0.5.
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* 0 \leq a_i \leq 10^9
* 0 \leq b_i \leq 10^9
* 0 \leq T \leq 10^9
Input
Input is given from Standard Input in the following format:
N T
a_1 b_1
a_2 b_2
\vdots
a_N b_N
Output
Print the answer.
Examples
Input
3 7
2 0
3 2
0 3
Output
2
Input
1 3
0 3
Output
0
Input
5 21600
2 14
3 22
1 3
1 10
1 9
Output
5
Input
7 57
0 25
3 10
2 4
5 15
3 22
2 14
1 15
Output
3 | instruction | 0 | 54,565 | 10 | 109,130 |
"Correct Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
from functools import cmp_to_key
import itertools
N, T = map(int, readline().split())
m = map(int, read().split())
AB0 = []
AB1 = []
for a,b in zip(m,m):
if a == 0:
AB0.append(b)
else:
AB1.append((a,b))
def sort_func(ab0, ab1):
a0, b0 = ab0
a1, b1 = ab1
return a1*(1+b0) - a0*(1+b1)
AB0.sort()
AB1.sort(key=cmp_to_key(sort_func))
AB0, AB1
def compute_0():
INF = T + 1
dp = [INF] * 40
dp[0] = 0
for a,b in AB1:
for n in range(39, -1, -1):
if dp[n] > T:
continue
x = (dp[n] + 1) * (a + 1) + b
if x < dp[n+1]:
dp[n+1] = x
return dp
def compute_1():
return [0] + list(itertools.accumulate(x + 1 for x in AB0))
dp0 = compute_0()
dp1 = compute_1()
R = len(dp1) - 1
answer = 0
for i,x in enumerate(dp0):
if x > T:
break
while x + dp1[R] > T:
R -= 1
value = i + R
if answer < value:
answer = value
print(answer)
``` | output | 1 | 54,565 | 10 | 109,131 |
Provide a correct Python 3 solution for this coding contest problem.
There are N stores called Store 1, Store 2, \cdots, Store N. Takahashi, who is at his house at time 0, is planning to visit some of these stores.
It takes Takahashi one unit of time to travel from his house to one of the stores, or between any two stores.
If Takahashi reaches Store i at time t, he can do shopping there after standing in a queue for a_i \times t + b_i units of time. (We assume that it takes no time other than waiting.)
All the stores close at time T + 0.5. If Takahashi is standing in a queue for some store then, he cannot do shopping there.
Takahashi does not do shopping more than once in the same store.
Find the maximum number of times he can do shopping before time T + 0.5.
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* 0 \leq a_i \leq 10^9
* 0 \leq b_i \leq 10^9
* 0 \leq T \leq 10^9
Input
Input is given from Standard Input in the following format:
N T
a_1 b_1
a_2 b_2
\vdots
a_N b_N
Output
Print the answer.
Examples
Input
3 7
2 0
3 2
0 3
Output
2
Input
1 3
0 3
Output
0
Input
5 21600
2 14
3 22
1 3
1 10
1 9
Output
5
Input
7 57
0 25
3 10
2 4
5 15
3 22
2 14
1 15
Output
3 | instruction | 0 | 54,566 | 10 | 109,132 |
"Correct Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
from functools import cmp_to_key
import itertools
N, T = map(int, readline().split())
m = map(int, read().split())
AB0 = []
AB1 = []
for a,b in zip(m,m):
if a == 0:
AB0.append(b)
else:
AB1.append((a,b))
def sort_func(ab0, ab1):
a0, b0 = ab0
a1, b1 = ab1
return a1*(1+b0) - a0*(1+b1)
AB0.sort()
AB1.sort(key=lambda x: x[0] / (x[1] + 1), reverse=True)
AB0, AB1
def compute_0():
INF = T + 1
dp = [INF] * 40
dp[0] = 0
for a,b in AB1:
for n in range(39, -1, -1):
if dp[n] > T:
continue
x = (dp[n] + 1) * (a + 1) + b
if x < dp[n+1]:
dp[n+1] = x
return dp
def compute_1():
return [0] + list(itertools.accumulate(x + 1 for x in AB0))
dp0 = compute_0()
dp1 = compute_1()
R = len(dp1) - 1
answer = 0
for i,x in enumerate(dp0):
if x > T:
break
while x + dp1[R] > T:
R -= 1
value = i + R
if answer < value:
answer = value
print(answer)
``` | output | 1 | 54,566 | 10 | 109,133 |
Provide a correct Python 3 solution for this coding contest problem.
There are N stores called Store 1, Store 2, \cdots, Store N. Takahashi, who is at his house at time 0, is planning to visit some of these stores.
It takes Takahashi one unit of time to travel from his house to one of the stores, or between any two stores.
If Takahashi reaches Store i at time t, he can do shopping there after standing in a queue for a_i \times t + b_i units of time. (We assume that it takes no time other than waiting.)
All the stores close at time T + 0.5. If Takahashi is standing in a queue for some store then, he cannot do shopping there.
Takahashi does not do shopping more than once in the same store.
Find the maximum number of times he can do shopping before time T + 0.5.
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* 0 \leq a_i \leq 10^9
* 0 \leq b_i \leq 10^9
* 0 \leq T \leq 10^9
Input
Input is given from Standard Input in the following format:
N T
a_1 b_1
a_2 b_2
\vdots
a_N b_N
Output
Print the answer.
Examples
Input
3 7
2 0
3 2
0 3
Output
2
Input
1 3
0 3
Output
0
Input
5 21600
2 14
3 22
1 3
1 10
1 9
Output
5
Input
7 57
0 25
3 10
2 4
5 15
3 22
2 14
1 15
Output
3 | instruction | 0 | 54,567 | 10 | 109,134 |
"Correct Solution:
```
from operator import itemgetter
import bisect
n, t = map(int, input().split())
info = [list(map(int, input().split())) for i in range(n)]
INF = t + 1
LIMIT = 60
# mrkt_a0: a == 0 の店
# mrkt_others: a != 0 の店
mrkt_a0 = []
mrkt_others = []
for i in range(n):
a, b = info[i]
if a == 0:
mrkt_a0.append((a, b))
else:
comp = a / (1 + b)
mrkt_others.append((a, b, comp))
mrkt_a0.sort(key=itemgetter(1))
mrkt_others.sort(key=itemgetter(2), reverse=True)
# mrkt_a0について
# もしmarkt_a0を尋ねるなら尋ねるのを後回しにしたほうが良い
times_a0 = [0]
for a, b in mrkt_a0:
times_a0.append(times_a0[-1] + 1 + b)
# mrkt_othersについて
len_others = len(mrkt_others)
dp = [[INF] * LIMIT for i in range(len_others + 1)]
dp[0][0] = 0
for i in range(len_others):
a, b, _ = mrkt_others[i]
for cnt in range(LIMIT):
# i番目の店を尋ねないとき
dp[i + 1][cnt] = min(dp[i][cnt], dp[i + 1][cnt])
# i番目の店を尋ねるとき
if cnt + 1 < LIMIT:
dp[i + 1][cnt + 1] = min(a * (dp[i][cnt] + 1) + b + dp[i][cnt] + 1, dp[i + 1][cnt + 1])
ans = 0
for cnt in range(LIMIT):
if dp[-1][cnt] == t + 1:
continue
nokori = t - dp[-1][cnt]
cnt_a0 = bisect.bisect_right(times_a0, nokori) - 1
ans = max(ans, cnt + cnt_a0)
print(ans)
``` | output | 1 | 54,567 | 10 | 109,135 |
Provide a correct Python 3 solution for this coding contest problem.
There are N stores called Store 1, Store 2, \cdots, Store N. Takahashi, who is at his house at time 0, is planning to visit some of these stores.
It takes Takahashi one unit of time to travel from his house to one of the stores, or between any two stores.
If Takahashi reaches Store i at time t, he can do shopping there after standing in a queue for a_i \times t + b_i units of time. (We assume that it takes no time other than waiting.)
All the stores close at time T + 0.5. If Takahashi is standing in a queue for some store then, he cannot do shopping there.
Takahashi does not do shopping more than once in the same store.
Find the maximum number of times he can do shopping before time T + 0.5.
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* 0 \leq a_i \leq 10^9
* 0 \leq b_i \leq 10^9
* 0 \leq T \leq 10^9
Input
Input is given from Standard Input in the following format:
N T
a_1 b_1
a_2 b_2
\vdots
a_N b_N
Output
Print the answer.
Examples
Input
3 7
2 0
3 2
0 3
Output
2
Input
1 3
0 3
Output
0
Input
5 21600
2 14
3 22
1 3
1 10
1 9
Output
5
Input
7 57
0 25
3 10
2 4
5 15
3 22
2 14
1 15
Output
3 | instruction | 0 | 54,568 | 10 | 109,136 |
"Correct Solution:
```
# https://atcoder.jp/contests/hitachi2020/submissions/10704509
# PyPyなら通る
def main():
import sys
input = sys.stdin.readline
inf = 1 << 30
L = 30
N, T = map(int, input().split())
ps = []
coef_0 = []
for _ in range(N):
a, b = map(int, input().split())
if a == 0:
coef_0.append(b + 1)
else:
ps.append((a, b)) # namedtupleやめた,coef,const
ps.sort(key=lambda p: p[0] / (p[1] + 1), reverse=True)
coef_0.sort()
dp = [inf] * (L + 1)
dp[0] = 0
for coef, const in ps:
for visited in reversed(range(L)):
dp[visited + 1] = min(
dp[visited + 1],
(dp[visited] + 1) * (coef + 1) + const
)
def accumulate(a):
s = 0
yield s # start=x->0
for x in a:
s += x
yield s
*acc, = accumulate(coef_0)
ans = 0
cnt = len(acc) - 1
for visited, consumption_time in enumerate(dp):
rest = T - consumption_time
if rest < 0: break # le->lt,continue->break,restは狭義単調減少
while cnt > 0 and acc[cnt] > rest: # bisect_right->しゃくとり
cnt -= 1
ans = max(ans, visited + cnt)
print(ans) # setのminと逐次minどちらが速い
if __name__ == '__main__':
main()
``` | output | 1 | 54,568 | 10 | 109,137 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N stores called Store 1, Store 2, \cdots, Store N. Takahashi, who is at his house at time 0, is planning to visit some of these stores.
It takes Takahashi one unit of time to travel from his house to one of the stores, or between any two stores.
If Takahashi reaches Store i at time t, he can do shopping there after standing in a queue for a_i \times t + b_i units of time. (We assume that it takes no time other than waiting.)
All the stores close at time T + 0.5. If Takahashi is standing in a queue for some store then, he cannot do shopping there.
Takahashi does not do shopping more than once in the same store.
Find the maximum number of times he can do shopping before time T + 0.5.
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* 0 \leq a_i \leq 10^9
* 0 \leq b_i \leq 10^9
* 0 \leq T \leq 10^9
Input
Input is given from Standard Input in the following format:
N T
a_1 b_1
a_2 b_2
\vdots
a_N b_N
Output
Print the answer.
Examples
Input
3 7
2 0
3 2
0 3
Output
2
Input
1 3
0 3
Output
0
Input
5 21600
2 14
3 22
1 3
1 10
1 9
Output
5
Input
7 57
0 25
3 10
2 4
5 15
3 22
2 14
1 15
Output
3
Submitted Solution:
```
"""
Writer: SPD_9X2
dpか?(なるほど)
左からi番目までの店でk個行くと決めた時の最小の時間？
at+b と xt+y があった時、at+bを先に取ったほうが最終的に小さくなる条件は
頑張って手計算すると
x/(1+y) < a/(1+b) なので、a/(1+b)が大きい方から貪欲していくのが正解っぽい？
→ただの貪欲は×
ただ、いくつか取るやつを選べば、その中では↑の式によって最適な取り方が決まる
→a/(1+b)の大きい方からしかとらないので、その順で使う、使わないでdpする
→O(N**2)
どうやってdpするか…
=========答えを見た========
a != 0なら指数関数的に増えるので、(最低で2倍)
回る店の数はせいぜい30にしかならない。
よって、a != 0に関して回る店の数とその最小値をdpした後、a == 0に関して調べればいい…
むずいけど分かるなぁ…(悔しい)
"""
N,T = map(int,input().split())
kab = []
azero = []
for i in range(N):
a,b = map(int,input().split())
if a == 0:
azero.append(b)
else:
k = a/(1+b)
kab.append([-1*k,a,b])
nt = 0
kab.sort()
dp = [float("inf")] * 29
dp[0] = 0
for k,a,b in kab:
for i in range(27,-1,-1):
dp[i+1] = min(dp[i+1] , 1 + dp[i] + a*(1+dp[i]) + b )
azero.sort()
dsum = [0]
for i in azero:
dsum.append(dsum[-1] + 1 + i)
ans = 0
import bisect
for i in range(29):
if dp[i] > T:
break
else:
ans = max(i + bisect.bisect_right(dsum,T-dp[i]) - 1 , ans )
print (ans)
``` | instruction | 0 | 54,569 | 10 | 109,138 |
Yes | output | 1 | 54,569 | 10 | 109,139 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N stores called Store 1, Store 2, \cdots, Store N. Takahashi, who is at his house at time 0, is planning to visit some of these stores.
It takes Takahashi one unit of time to travel from his house to one of the stores, or between any two stores.
If Takahashi reaches Store i at time t, he can do shopping there after standing in a queue for a_i \times t + b_i units of time. (We assume that it takes no time other than waiting.)
All the stores close at time T + 0.5. If Takahashi is standing in a queue for some store then, he cannot do shopping there.
Takahashi does not do shopping more than once in the same store.
Find the maximum number of times he can do shopping before time T + 0.5.
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* 0 \leq a_i \leq 10^9
* 0 \leq b_i \leq 10^9
* 0 \leq T \leq 10^9
Input
Input is given from Standard Input in the following format:
N T
a_1 b_1
a_2 b_2
\vdots
a_N b_N
Output
Print the answer.
Examples
Input
3 7
2 0
3 2
0 3
Output
2
Input
1 3
0 3
Output
0
Input
5 21600
2 14
3 22
1 3
1 10
1 9
Output
5
Input
7 57
0 25
3 10
2 4
5 15
3 22
2 14
1 15
Output
3
Submitted Solution:
```
import sys
input = sys.stdin.readline
n, t = map(int, input().split())
shop = []
shop_0 = []
for i in range(n):
a, b = map(int, input().split())
if a == 0:
shop_0.append((a, b))
n -= 1
else:
shop.append((a, b))
shop.sort(key = lambda x: (x[0])/(1+x[1]), reverse = True)
shop_0.sort()
inf = 10**18
dp = [[inf for i in range(30)] for j in range(n+1)]
dp[0][0] = 0
for i in range(n):
for j in range(29):
dp[i+1][j] = min(dp[i][j], dp[i+1][j])
dp[i+1][j+1] = min(dp[i+1][j+1], dp[i][j] + (dp[i][j]+1)*(shop[i][0])+shop[i][1]+1)
ans = 0
for i in range(30):
tmp = i
T = dp[n][i]
if T > t:
continue
else:
for j in range(len(shop_0)):
if T+shop_0[j][1]+1 <= t:
T += shop_0[j][1]+1
tmp += 1
else:
break
ans = max(tmp, ans)
print(ans)
``` | instruction | 0 | 54,570 | 10 | 109,140 |
Yes | output | 1 | 54,570 | 10 | 109,141 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N stores called Store 1, Store 2, \cdots, Store N. Takahashi, who is at his house at time 0, is planning to visit some of these stores.
It takes Takahashi one unit of time to travel from his house to one of the stores, or between any two stores.
If Takahashi reaches Store i at time t, he can do shopping there after standing in a queue for a_i \times t + b_i units of time. (We assume that it takes no time other than waiting.)
All the stores close at time T + 0.5. If Takahashi is standing in a queue for some store then, he cannot do shopping there.
Takahashi does not do shopping more than once in the same store.
Find the maximum number of times he can do shopping before time T + 0.5.
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* 0 \leq a_i \leq 10^9
* 0 \leq b_i \leq 10^9
* 0 \leq T \leq 10^9
Input
Input is given from Standard Input in the following format:
N T
a_1 b_1
a_2 b_2
\vdots
a_N b_N
Output
Print the answer.
Examples
Input
3 7
2 0
3 2
0 3
Output
2
Input
1 3
0 3
Output
0
Input
5 21600
2 14
3 22
1 3
1 10
1 9
Output
5
Input
7 57
0 25
3 10
2 4
5 15
3 22
2 14
1 15
Output
3
Submitted Solution:
```
def solve(n, t, ab_list):
a0b_list = [ab[1] for ab in ab_list if ab[0] == 0]
apb_list = [ab for ab in ab_list if ab[0] > 0]
a0b_list = sorted(a0b_list)
apb_list = sorted(apb_list, key=lambda x: - x[0] / (x[1] + 1))
# DP
m = len(apb_list)
dp = [[t + 1] * 30 for _ in range(m + 1)]
for i in range(m + 1):
dp[i][0] = 0
for i in range(m):
for k in range(1, 30):
ti = dp[i][k - 1] + 1
a, b = apb_list[i]
dp[i + 1][k] = min(dp[i][k], ti + a * ti + b)
a0b_list_cum_sum = [0]
s = 0
for b in a0b_list:
s += b + 1
a0b_list_cum_sum.append(s)
res = 0
for k in range(len(a0b_list_cum_sum)):
for p in range(30):
if dp[m][p] + a0b_list_cum_sum[k] <= t:
res = max(res, k + p)
return res
def main():
n, t = map(int, input().split())
ab_list = [list(map(int, input().split())) for _ in range(n)]
res = solve(n, t, ab_list)
print(res)
def test():
assert solve(3, 7, [[2, 0], [3, 2], [0, 3]]) == 2
assert solve(1, 3, [[0, 3]]) == 0
assert solve(5, 21600, [[2, 14], [3, 22], [1, 3], [1, 10], [1, 9]]) == 5
if __name__ == "__main__":
test()
main()
``` | instruction | 0 | 54,571 | 10 | 109,142 |
Yes | output | 1 | 54,571 | 10 | 109,143 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N stores called Store 1, Store 2, \cdots, Store N. Takahashi, who is at his house at time 0, is planning to visit some of these stores.
It takes Takahashi one unit of time to travel from his house to one of the stores, or between any two stores.
If Takahashi reaches Store i at time t, he can do shopping there after standing in a queue for a_i \times t + b_i units of time. (We assume that it takes no time other than waiting.)
All the stores close at time T + 0.5. If Takahashi is standing in a queue for some store then, he cannot do shopping there.
Takahashi does not do shopping more than once in the same store.
Find the maximum number of times he can do shopping before time T + 0.5.
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* 0 \leq a_i \leq 10^9
* 0 \leq b_i \leq 10^9
* 0 \leq T \leq 10^9
Input
Input is given from Standard Input in the following format:
N T
a_1 b_1
a_2 b_2
\vdots
a_N b_N
Output
Print the answer.
Examples
Input
3 7
2 0
3 2
0 3
Output
2
Input
1 3
0 3
Output
0
Input
5 21600
2 14
3 22
1 3
1 10
1 9
Output
5
Input
7 57
0 25
3 10
2 4
5 15
3 22
2 14
1 15
Output
3
Submitted Solution:
```
import sys
from bisect import bisect_right
N,T = map(int,input().split())
ab = [list(map(int,input().split())) for i in range(N)]
b0 = []
ab1 = []
#aが0とそれ以外で分ける
for a,b in ab:
if a == 0:
#+1は店間の移動を先に考慮
b0.append(b+1)
else:
ab1.append([(b+1)/a, a, b])
N0 = len(b0)
N1 = len(ab1)
b0.sort()
#bは累積和で時間計算可能（最後にどこまで足せるかを調べる）
if N0>0:
b_ruiseki = [0]*N0
b_ruiseki[0] = b0[0]
for i in range(1,N0):
b_ruiseki[i] = b_ruiseki[i-1] + b0[i]
else:
b_ruiseki = []
#探索範囲が28店まででいいからソートする
ab1.sort()
INF = T + 100
dp = [[INF]*31 for i in range(N1+1)]
dp[0][0] = 0
ans = 0
for i in range(N1):
# dp[i+1][j] : 0~i番目までの店を見て，j個選ぶときの最小時間
for j in range(31):
if j > i+1:
break
if j == 0:
dp[i+1][j] = 0
else:
dp[i+1][j] = min(dp[i][j], dp[i][j-1]+1 + ab1[i][1]*(dp[i][j-1]+1) + ab1[i][2])
for j in range(31):
if dp[N1][j] > T:
break
# a>0をj店舗めぐった後にa=0を何店舗めぐれるか
num0 = bisect_right(b_ruiseki, T - dp[N1][j])
ans = max(ans, j + num0)
print(ans)
``` | instruction | 0 | 54,572 | 10 | 109,144 |
Yes | output | 1 | 54,572 | 10 | 109,145 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N stores called Store 1, Store 2, \cdots, Store N. Takahashi, who is at his house at time 0, is planning to visit some of these stores.
It takes Takahashi one unit of time to travel from his house to one of the stores, or between any two stores.
If Takahashi reaches Store i at time t, he can do shopping there after standing in a queue for a_i \times t + b_i units of time. (We assume that it takes no time other than waiting.)
All the stores close at time T + 0.5. If Takahashi is standing in a queue for some store then, he cannot do shopping there.
Takahashi does not do shopping more than once in the same store.
Find the maximum number of times he can do shopping before time T + 0.5.
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* 0 \leq a_i \leq 10^9
* 0 \leq b_i \leq 10^9
* 0 \leq T \leq 10^9
Input
Input is given from Standard Input in the following format:
N T
a_1 b_1
a_2 b_2
\vdots
a_N b_N
Output
Print the answer.
Examples
Input
3 7
2 0
3 2
0 3
Output
2
Input
1 3
0 3
Output
0
Input
5 21600
2 14
3 22
1 3
1 10
1 9
Output
5
Input
7 57
0 25
3 10
2 4
5 15
3 22
2 14
1 15
Output
3
Submitted Solution:
```
N, T = map(int, input().split())
P = [list(map(int,input().split())) for i in range(N)]
t = 0
for a, b in P:
if a*t + b < T + 0.5:
t += 1
print(t)
``` | instruction | 0 | 54,573 | 10 | 109,146 |
No | output | 1 | 54,573 | 10 | 109,147 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N stores called Store 1, Store 2, \cdots, Store N. Takahashi, who is at his house at time 0, is planning to visit some of these stores.
It takes Takahashi one unit of time to travel from his house to one of the stores, or between any two stores.
If Takahashi reaches Store i at time t, he can do shopping there after standing in a queue for a_i \times t + b_i units of time. (We assume that it takes no time other than waiting.)
All the stores close at time T + 0.5. If Takahashi is standing in a queue for some store then, he cannot do shopping there.
Takahashi does not do shopping more than once in the same store.
Find the maximum number of times he can do shopping before time T + 0.5.
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* 0 \leq a_i \leq 10^9
* 0 \leq b_i \leq 10^9
* 0 \leq T \leq 10^9
Input
Input is given from Standard Input in the following format:
N T
a_1 b_1
a_2 b_2
\vdots
a_N b_N
Output
Print the answer.
Examples
Input
3 7
2 0
3 2
0 3
Output
2
Input
1 3
0 3
Output
0
Input
5 21600
2 14
3 22
1 3
1 10
1 9
Output
5
Input
7 57
0 25
3 10
2 4
5 15
3 22
2 14
1 15
Output
3
Submitted Solution:
```
# -*- coding: utf-8 -*-
N, T = map(int, input().split())
a = [0] * N
b = [0] * N
for i in range(N):
a[i], b[i] = map(int, input().split())
t=0
arrived_store=[]
while t <= T:
t+=1
c = [[0,0]] * N
for i in range(N):
c[i] = [a[i] * t + b[i], i]
c.sort()
j=0
while c[j][1] in arrived_store:
j += 1
t+=c[j][0]
if t > T:
break
arrived_store.append(c[j][1])
if (len(arrived_store) == N):
break
print(len(arrived_store))
``` | instruction | 0 | 54,574 | 10 | 109,148 |
No | output | 1 | 54,574 | 10 | 109,149 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N stores called Store 1, Store 2, \cdots, Store N. Takahashi, who is at his house at time 0, is planning to visit some of these stores.
It takes Takahashi one unit of time to travel from his house to one of the stores, or between any two stores.
If Takahashi reaches Store i at time t, he can do shopping there after standing in a queue for a_i \times t + b_i units of time. (We assume that it takes no time other than waiting.)
All the stores close at time T + 0.5. If Takahashi is standing in a queue for some store then, he cannot do shopping there.
Takahashi does not do shopping more than once in the same store.
Find the maximum number of times he can do shopping before time T + 0.5.
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* 0 \leq a_i \leq 10^9
* 0 \leq b_i \leq 10^9
* 0 \leq T \leq 10^9
Input
Input is given from Standard Input in the following format:
N T
a_1 b_1
a_2 b_2
\vdots
a_N b_N
Output
Print the answer.
Examples
Input
3 7
2 0
3 2
0 3
Output
2
Input
1 3
0 3
Output
0
Input
5 21600
2 14
3 22
1 3
1 10
1 9
Output
5
Input
7 57
0 25
3 10
2 4
5 15
3 22
2 14
1 15
Output
3
Submitted Solution:
```
from bisect import bisect_right
import sys
sys.setrecursionlimit(10 ** 7)
input = sys.stdin.readline
n, t = map(int, input().split())
shop = []
shop_a0 = []
for i in range(n):
a, b = map(int, input().split())
if a != 0:
shop.append([a, b, a/(b+1)])
else:
shop_a0.append(b)
shop = sorted(shop, key=lambda x: -x[2])
shop_a0 = sorted(shop_a0)
n = len(shop)
INF = 10**10
dp = [INF for _ in range(30)]
dp[0] = 0
m = min(n+1, 30)
for i in range(n):
for j in range(m):
dp[j] = min(dp[j], dp[j-1] + 1 +
shop[i][0]*(dp[j-1]+1) + shop[i][1])
ss = []
if shop_a0:
ss.append(1+shop_a0[0])
for i in range(1, len(shop_a0)):
ss.append(ss[-1]+1+shop_a0[i])
ans = 0
for i in range(m):
if t - dp[i] >= 0:
ans = max(ans, bisect_right(ss, t-dp[i]) + i)
print(ans)
``` | instruction | 0 | 54,575 | 10 | 109,150 |
No | output | 1 | 54,575 | 10 | 109,151 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N stores called Store 1, Store 2, \cdots, Store N. Takahashi, who is at his house at time 0, is planning to visit some of these stores.
It takes Takahashi one unit of time to travel from his house to one of the stores, or between any two stores.
If Takahashi reaches Store i at time t, he can do shopping there after standing in a queue for a_i \times t + b_i units of time. (We assume that it takes no time other than waiting.)
All the stores close at time T + 0.5. If Takahashi is standing in a queue for some store then, he cannot do shopping there.
Takahashi does not do shopping more than once in the same store.
Find the maximum number of times he can do shopping before time T + 0.5.
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* 0 \leq a_i \leq 10^9
* 0 \leq b_i \leq 10^9
* 0 \leq T \leq 10^9
Input
Input is given from Standard Input in the following format:
N T
a_1 b_1
a_2 b_2
\vdots
a_N b_N
Output
Print the answer.
Examples
Input
3 7
2 0
3 2
0 3
Output
2
Input
1 3
0 3
Output
0
Input
5 21600
2 14
3 22
1 3
1 10
1 9
Output
5
Input
7 57
0 25
3 10
2 4
5 15
3 22
2 14
1 15
Output
3
Submitted Solution:
```
# coding: utf-8
import numpy as np
import sys
sys.setrecursionlimit(10 ** 9)
N, T = list(map(int, input().split()))
arr = [[0,0]]+ [list(map(int, input().split())) for _ in range(N)]
arr= np.array(arr)
ab_arr = arr[1:,0] / ((arr)[1:,1]+1)
perm = np.array([0] + list(np.array(ab_arr).argsort()[::-1]+1))
dp = (np.zeros((N+1, min(N+1, int(np.log2(T))+2)))).astype(np.float32)
dp[1, 1] = arr[perm[1], 0] * 1 + arr[perm[1], 1] + 1
pos = int(np.log2(T))
for j in range(1, pos+2):
tmp = np.minimum((dp[j-1:-1, j-1] + 1) * (arr[perm[j:], 0] + 1) + arr[perm[j:], 1], 10 ** 9 + 1)
dp[j:,j] = np.minimum.accumulate(tmp)
dp = np.array(dp)
arr = np.array(arr)
b = np.sort(arr[arr[:,0] == 0, 1])
t = np.max(dp[N, dp[N,:] < T + 0.5])
n_stores = (dp[N, 1:] < T + 0.5).sum()
i = 1
for b_ in b[1:]:
_t = t
t += b_ + 1
n_stores += 1
if t > T + 0.5:
t = _t
n_stores -= 1
break
print(n_stores)
``` | instruction | 0 | 54,576 | 10 | 109,152 |
No | output | 1 | 54,576 | 10 | 109,153 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gerald has been selling state secrets at leisure. All the secrets cost the same: n marks. The state which secrets Gerald is selling, has no paper money, only coins. But there are coins of all positive integer denominations that are powers of three: 1 mark, 3 marks, 9 marks, 27 marks and so on. There are no coins of other denominations. Of course, Gerald likes it when he gets money without the change. And all buyers respect him and try to give the desired sum without change, if possible. But this does not always happen.
One day an unlucky buyer came. He did not have the desired sum without change. Then he took out all his coins and tried to give Gerald a larger than necessary sum with as few coins as possible. What is the maximum number of coins he could get?
The formal explanation of the previous paragraph: we consider all the possible combinations of coins for which the buyer can not give Gerald the sum of n marks without change. For each such combination calculate the minimum number of coins that can bring the buyer at least n marks. Among all combinations choose the maximum of the minimum number of coins. This is the number we want.
Input
The single line contains a single integer n (1 ≤ n ≤ 1017).
Please, do not use the %lld specifier to read or write 64 bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
In a single line print an integer: the maximum number of coins the unlucky buyer could have paid with.
Examples
Input
1
Output
1
Input
4
Output
2
Note
In the first test case, if a buyer has exactly one coin of at least 3 marks, then, to give Gerald one mark, he will have to give this coin. In this sample, the customer can not have a coin of one mark, as in this case, he will be able to give the money to Gerald without any change.
In the second test case, if the buyer had exactly three coins of 3 marks, then, to give Gerald 4 marks, he will have to give two of these coins. The buyer cannot give three coins as he wants to minimize the number of coins that he gives. | instruction | 0 | 55,103 | 10 | 110,206 |
Tags: greedy
Correct Solution:
```
base,n=1,int(input())
while(n%base==0):base*=3
print(n//base+1)
``` | output | 1 | 55,103 | 10 | 110,207 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gerald has been selling state secrets at leisure. All the secrets cost the same: n marks. The state which secrets Gerald is selling, has no paper money, only coins. But there are coins of all positive integer denominations that are powers of three: 1 mark, 3 marks, 9 marks, 27 marks and so on. There are no coins of other denominations. Of course, Gerald likes it when he gets money without the change. And all buyers respect him and try to give the desired sum without change, if possible. But this does not always happen.
One day an unlucky buyer came. He did not have the desired sum without change. Then he took out all his coins and tried to give Gerald a larger than necessary sum with as few coins as possible. What is the maximum number of coins he could get?
The formal explanation of the previous paragraph: we consider all the possible combinations of coins for which the buyer can not give Gerald the sum of n marks without change. For each such combination calculate the minimum number of coins that can bring the buyer at least n marks. Among all combinations choose the maximum of the minimum number of coins. This is the number we want.
Input
The single line contains a single integer n (1 ≤ n ≤ 1017).
Please, do not use the %lld specifier to read or write 64 bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
In a single line print an integer: the maximum number of coins the unlucky buyer could have paid with.
Examples
Input
1
Output
1
Input
4
Output
2
Note
In the first test case, if a buyer has exactly one coin of at least 3 marks, then, to give Gerald one mark, he will have to give this coin. In this sample, the customer can not have a coin of one mark, as in this case, he will be able to give the money to Gerald without any change.
In the second test case, if the buyer had exactly three coins of 3 marks, then, to give Gerald 4 marks, he will have to give two of these coins. The buyer cannot give three coins as he wants to minimize the number of coins that he gives. | instruction | 0 | 55,104 | 10 | 110,208 |
Tags: greedy
Correct Solution:
```
n = int(input())
k = 1
while n % k == 0:
k *= 3
print(n // k + 1)
``` | output | 1 | 55,104 | 10 | 110,209 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gerald has been selling state secrets at leisure. All the secrets cost the same: n marks. The state which secrets Gerald is selling, has no paper money, only coins. But there are coins of all positive integer denominations that are powers of three: 1 mark, 3 marks, 9 marks, 27 marks and so on. There are no coins of other denominations. Of course, Gerald likes it when he gets money without the change. And all buyers respect him and try to give the desired sum without change, if possible. But this does not always happen.
One day an unlucky buyer came. He did not have the desired sum without change. Then he took out all his coins and tried to give Gerald a larger than necessary sum with as few coins as possible. What is the maximum number of coins he could get?
The formal explanation of the previous paragraph: we consider all the possible combinations of coins for which the buyer can not give Gerald the sum of n marks without change. For each such combination calculate the minimum number of coins that can bring the buyer at least n marks. Among all combinations choose the maximum of the minimum number of coins. This is the number we want.
Input
The single line contains a single integer n (1 ≤ n ≤ 1017).
Please, do not use the %lld specifier to read or write 64 bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
In a single line print an integer: the maximum number of coins the unlucky buyer could have paid with.
Examples
Input
1
Output
1
Input
4
Output
2
Note
In the first test case, if a buyer has exactly one coin of at least 3 marks, then, to give Gerald one mark, he will have to give this coin. In this sample, the customer can not have a coin of one mark, as in this case, he will be able to give the money to Gerald without any change.
In the second test case, if the buyer had exactly three coins of 3 marks, then, to give Gerald 4 marks, he will have to give two of these coins. The buyer cannot give three coins as he wants to minimize the number of coins that he gives. | instruction | 0 | 55,105 | 10 | 110,210 |
Tags: greedy
Correct Solution:
```
n = int(input())
while n % 3 == 0:
n //= 3
print(n // 3 + 1)
``` | output | 1 | 55,105 | 10 | 110,211 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gerald has been selling state secrets at leisure. All the secrets cost the same: n marks. The state which secrets Gerald is selling, has no paper money, only coins. But there are coins of all positive integer denominations that are powers of three: 1 mark, 3 marks, 9 marks, 27 marks and so on. There are no coins of other denominations. Of course, Gerald likes it when he gets money without the change. And all buyers respect him and try to give the desired sum without change, if possible. But this does not always happen.
One day an unlucky buyer came. He did not have the desired sum without change. Then he took out all his coins and tried to give Gerald a larger than necessary sum with as few coins as possible. What is the maximum number of coins he could get?
The formal explanation of the previous paragraph: we consider all the possible combinations of coins for which the buyer can not give Gerald the sum of n marks without change. For each such combination calculate the minimum number of coins that can bring the buyer at least n marks. Among all combinations choose the maximum of the minimum number of coins. This is the number we want.
Input
The single line contains a single integer n (1 ≤ n ≤ 1017).
Please, do not use the %lld specifier to read or write 64 bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
In a single line print an integer: the maximum number of coins the unlucky buyer could have paid with.
Examples
Input
1
Output
1
Input
4
Output
2
Note
In the first test case, if a buyer has exactly one coin of at least 3 marks, then, to give Gerald one mark, he will have to give this coin. In this sample, the customer can not have a coin of one mark, as in this case, he will be able to give the money to Gerald without any change.
In the second test case, if the buyer had exactly three coins of 3 marks, then, to give Gerald 4 marks, he will have to give two of these coins. The buyer cannot give three coins as he wants to minimize the number of coins that he gives. | instruction | 0 | 55,106 | 10 | 110,212 |
Tags: greedy
Correct Solution:
```
#-------------------------------------------------------------------------------
# Name: Codeforces
# Author: Gogol
#-------------------------------------------------------------------------------
import sys
from math import *
def solve():
n = int(input())
p = 1
while (n % p == 0):
p*=3
print(n//p+1)
solve()
``` | output | 1 | 55,106 | 10 | 110,213 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gerald has been selling state secrets at leisure. All the secrets cost the same: n marks. The state which secrets Gerald is selling, has no paper money, only coins. But there are coins of all positive integer denominations that are powers of three: 1 mark, 3 marks, 9 marks, 27 marks and so on. There are no coins of other denominations. Of course, Gerald likes it when he gets money without the change. And all buyers respect him and try to give the desired sum without change, if possible. But this does not always happen.
One day an unlucky buyer came. He did not have the desired sum without change. Then he took out all his coins and tried to give Gerald a larger than necessary sum with as few coins as possible. What is the maximum number of coins he could get?
The formal explanation of the previous paragraph: we consider all the possible combinations of coins for which the buyer can not give Gerald the sum of n marks without change. For each such combination calculate the minimum number of coins that can bring the buyer at least n marks. Among all combinations choose the maximum of the minimum number of coins. This is the number we want.
Input
The single line contains a single integer n (1 ≤ n ≤ 1017).
Please, do not use the %lld specifier to read or write 64 bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
In a single line print an integer: the maximum number of coins the unlucky buyer could have paid with.
Examples
Input
1
Output
1
Input
4
Output
2
Note
In the first test case, if a buyer has exactly one coin of at least 3 marks, then, to give Gerald one mark, he will have to give this coin. In this sample, the customer can not have a coin of one mark, as in this case, he will be able to give the money to Gerald without any change.
In the second test case, if the buyer had exactly three coins of 3 marks, then, to give Gerald 4 marks, he will have to give two of these coins. The buyer cannot give three coins as he wants to minimize the number of coins that he gives. | instruction | 0 | 55,107 | 10 | 110,214 |
Tags: greedy
Correct Solution:
```
from fractions import Decimal
import math
n=Decimal(input())
d=Decimal('1')
while(n%d==0):
d*=3
if(d>n):
print(1)
else:
print(math.ceil(n/d))
``` | output | 1 | 55,107 | 10 | 110,215 |
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