text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109 + 7.
The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.
Input
The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.
Output
Print the minimum number of steps modulo 109 + 7.
Examples
Input
ab
Output
1
Input
aab
Output
3
Note
The first example: "ab" → "bba".
The second example: "aab" → "abba" → "bbaba" → "bbbbaa".
Submitted Solution:
```
n=input()
i=0
while "ab" in n:
n=n.replace("ab","bba")
i+=1
print(i%(10**6+7))
```
No
| 105,800 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109 + 7.
The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.
Input
The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.
Output
Print the minimum number of steps modulo 109 + 7.
Examples
Input
ab
Output
1
Input
aab
Output
3
Note
The first example: "ab" → "bba".
The second example: "aab" → "abba" → "bbaba" → "bbbbaa".
Submitted Solution:
```
n = input()
r = 0
c = 0
t = []
for i in range(len(n)):
if n[i]=="a":
c+=1
else:
t.append(c)
c = 0
for i in t:
if i==1:
r+=1
elif i>1:
r+=i+1
print(r%(10**9+7))
```
No
| 105,801 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109 + 7.
The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.
Input
The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.
Output
Print the minimum number of steps modulo 109 + 7.
Examples
Input
ab
Output
1
Input
aab
Output
3
Note
The first example: "ab" → "bba".
The second example: "aab" → "abba" → "bbaba" → "bbbbaa".
Submitted Solution:
```
s=input()
ans=0
j=0
for i in range(0,len(s)):
if i>=j:
cnt1=0
while i<len(s) and s[i]=='a':
cnt1=cnt1+1
i=i+1
cnt2=0
while i<len(s) and s[i]=='b':
cnt2=cnt2+1
i=i+1
j=i;
if cnt1>0 and cnt2>0:
ans=ans + cnt2
bb= cnt2*2
cnt1=cnt1-1
ans = ans + bb*(pow(2,cnt1,1000000007)-1)
#print(ans)
print(ans%1000000007)
```
No
| 105,802 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109 + 7.
The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.
Input
The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.
Output
Print the minimum number of steps modulo 109 + 7.
Examples
Input
ab
Output
1
Input
aab
Output
3
Note
The first example: "ab" → "bba".
The second example: "aab" → "abba" → "bbaba" → "bbbbaa".
Submitted Solution:
```
s = input()
r = 0
na = 0
for i in range(len(s)):
if s[i] == 'a':
na += 1
else:
r += (2**na - 1)%1000000007
print(r)
```
No
| 105,803 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the n-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
Input
The input data consist of a single integer n (1 ≤ n ≤ 109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
Output
Print the single line — the name of the person who drinks the n-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
Examples
Input
1
Output
Sheldon
Input
6
Output
Sheldon
Input
1802
Output
Penny
Tags: implementation, math
Correct Solution:
```
import math
name=['Sheldon','Leonard','Penny','Rajesh','Howard']
n=int(input())
for i in range(100000000):
if math.floor(n/5)-(2**i-1)<2**i:
h=i
#print(h)
print(name[math.ceil((n-5*(2**h-1))/2**h-1)])
exit()
```
| 105,804 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the n-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
Input
The input data consist of a single integer n (1 ≤ n ≤ 109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
Output
Print the single line — the name of the person who drinks the n-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
Examples
Input
1
Output
Sheldon
Input
6
Output
Sheldon
Input
1802
Output
Penny
Tags: implementation, math
Correct Solution:
```
n = int(input())
while(n>5):
n = n-4
n = n/2
n = int(n)
l = ["Sheldon", "Leonard", "Penny", "Rajesh", "Howard"]
print(l[n-1])
```
| 105,805 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the n-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
Input
The input data consist of a single integer n (1 ≤ n ≤ 109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
Output
Print the single line — the name of the person who drinks the n-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
Examples
Input
1
Output
Sheldon
Input
6
Output
Sheldon
Input
1802
Output
Penny
Tags: implementation, math
Correct Solution:
```
n = int(input())
s = 1+float(n)/5
import math
k = math.ceil((math.log(s,2)))
deta = n-5*(2**(k-1)-1)
test = math.ceil(deta/(2**(k-1)))
if test == 1:
print("Sheldon")
if test == 2:
print("Leonard")
if test == 3:
print("Penny")
if test == 4:
print("Rajesh")
if test == 5:
print("Howard")
```
| 105,806 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the n-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
Input
The input data consist of a single integer n (1 ≤ n ≤ 109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
Output
Print the single line — the name of the person who drinks the n-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
Examples
Input
1
Output
Sheldon
Input
6
Output
Sheldon
Input
1802
Output
Penny
Tags: implementation, math
Correct Solution:
```
import sys
class Scanner:
def __init__(self):
self.current_tokens = []
def remaining_tokens(self):
return len(self.current_tokens)
def nextline(self):
assert self.remaining_tokens() == 0, "Reading next line with remaining tokens"
return input()
def nexttokens(self):
return self.nextline().split()
def nexttoken(self):
if len(self.current_tokens) == 0:
self.current_tokens = self.nexttokens()
assert self.remaining_tokens() > 0, "Not enough tokens to parse."
return self.current_tokens.pop(0)
def nextints(self, n=-1):
if n == -1:
return list(map(int, self.nexttokens()))
else:
return (self.nextint() for i in range(n))
def nextint(self):
return int(self.nexttoken())
def quit():
sys.exit(0)
stdin = Scanner()
nextint = stdin.nextint
nextints = stdin.nextints
nextline = stdin.nextline
n = nextint() - 1
itr = 0
i = 0
r = ['Sheldon', 'Leonard', 'Penny', 'Rajesh', 'Howard']
while i <= n:
for j in range(5):
if n in range(i, i + pow(2, itr)):
print(r[j])
quit()
i += pow(2, itr)
itr += 1
```
| 105,807 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the n-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
Input
The input data consist of a single integer n (1 ≤ n ≤ 109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
Output
Print the single line — the name of the person who drinks the n-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
Examples
Input
1
Output
Sheldon
Input
6
Output
Sheldon
Input
1802
Output
Penny
Tags: implementation, math
Correct Solution:
```
import math
def doubleCola(n: int) -> str:
dictionary = {
1:'Sheldon',
2:'Leonard',
3:'Penny',
4:'Rajesh',
5:'Howard'
}
lowerBound = 1
upperBound = 5
currentNumber = 1
while n > upperBound:
currentNumber *= 2
lowerBound = 1 + upperBound
upperBound += 5 * currentNumber
difference = lowerBound - 1
return dictionary.get(math.ceil((n-difference)/currentNumber))
def main():
n = int(input().strip())
print(doubleCola(n))
main()
```
| 105,808 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the n-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
Input
The input data consist of a single integer n (1 ≤ n ≤ 109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
Output
Print the single line — the name of the person who drinks the n-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
Examples
Input
1
Output
Sheldon
Input
6
Output
Sheldon
Input
1802
Output
Penny
Tags: implementation, math
Correct Solution:
```
# A. Double Cola
# 82A
names = ['Sheldon', 'Leonard', 'Penny', 'Rajesh', 'Howard']
n = int(input()) - 1
# print(f'n is initially {n}')
r = 0
while n >= 5 * 2**r:
# print()
n -= 5 * 2**r
# print(f'subtracting n by {5 * 2**r}')
r += 1
# print(f'Now n is {n} and r is {r}')
n //= 2**r
# print(f'Since 2**r is {2**r} and r is {r}, changing n to {n}')
assert(n < 5)
print(names[n])
# print(names)
```
| 105,809 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the n-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
Input
The input data consist of a single integer n (1 ≤ n ≤ 109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
Output
Print the single line — the name of the person who drinks the n-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
Examples
Input
1
Output
Sheldon
Input
6
Output
Sheldon
Input
1802
Output
Penny
Tags: implementation, math
Correct Solution:
```
import math
n = int(input())
for i in range(1, 30):
if 5 * (2 ** (i - 1) - 1) < n <= 5 * (2 ** i - 1):
k = i
break
man = ["null","Sheldon","Leonard","Penny","Rajesh","Howard"]
c = math.ceil((n - 5 * (2 ** (k - 1) - 1)) / (2 ** (k - 1)))
print(man[c])
```
| 105,810 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the n-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
Input
The input data consist of a single integer n (1 ≤ n ≤ 109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
Output
Print the single line — the name of the person who drinks the n-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
Examples
Input
1
Output
Sheldon
Input
6
Output
Sheldon
Input
1802
Output
Penny
Tags: implementation, math
Correct Solution:
```
from math import ceil
d = {0: "Sheldon", 1: "Leonard", 2: "Penny", 3: "Rajesh", 4: "Howard"}
n = int(input())
i = 1
while n > i*5:
n -= i*5
i *= 2
print(d[ceil(n/i)-1])
```
| 105,811 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the n-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
Input
The input data consist of a single integer n (1 ≤ n ≤ 109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
Output
Print the single line — the name of the person who drinks the n-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
Examples
Input
1
Output
Sheldon
Input
6
Output
Sheldon
Input
1802
Output
Penny
Submitted Solution:
```
a=["Sheldon", "Leonard", "Penny", "Rajesh", "Howard"]
n=int(input())
n-=1
while(n>=5):
n=(n-5)//2
print(a[n])
```
Yes
| 105,812 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the n-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
Input
The input data consist of a single integer n (1 ≤ n ≤ 109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
Output
Print the single line — the name of the person who drinks the n-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
Examples
Input
1
Output
Sheldon
Input
6
Output
Sheldon
Input
1802
Output
Penny
Submitted Solution:
```
import math
n =int(input())
a=["Sheldon", "Leonard", "Penny", "Rajesh", "Howard"]
if n<=5:
print(a[n-1])
else:
sum=5
z=0
while(sum<n):
z+=1
sum+=2**z*5
# # print(z)
# if(n>=2**(z)*5):
# else:
# k=n-2**(z-1)*5
# print(sum,z)
k=n-sum+(2**(z)*5)
# print(k)
l=math.ceil(k/2**(z))
# print(l)
print(a[l-1])
```
Yes
| 105,813 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the n-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
Input
The input data consist of a single integer n (1 ≤ n ≤ 109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
Output
Print the single line — the name of the person who drinks the n-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
Examples
Input
1
Output
Sheldon
Input
6
Output
Sheldon
Input
1802
Output
Penny
Submitted Solution:
```
from collections import deque
def f(n):
q = ["Sheldon", "Leonard", "Penny", "Rajesh", "Howard"]
while n >5:
if n%2 == 1:
n -= 5
n //= 2
else:
n -=4
n //=2
return q[n-1]
n = int(input())
print(f(n))
```
Yes
| 105,814 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the n-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
Input
The input data consist of a single integer n (1 ≤ n ≤ 109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
Output
Print the single line — the name of the person who drinks the n-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
Examples
Input
1
Output
Sheldon
Input
6
Output
Sheldon
Input
1802
Output
Penny
Submitted Solution:
```
# WHERE: https://codeforces.com/problemset/page/8?order=BY_RATING_ASC
# Taxi
# https://codeforces.com/problemset/problem/158/B
# Input
# The first line contains integer n (1 ≤ n ≤ 105) — the number of groups of schoolchildren. The second line contains a sequence of integers s1, s2, ..., sn (1 ≤ si ≤ 4). The integers are separated by a space, si is the number of children in the i-th group.
# Output
# Print the single number — the minimum number of taxis necessary to drive all children to Polycarpus.
# ignoreInput = input()
# groups = list(map(int, input().split()))
# i = 0
# j = len(groups) - 1 # last position of the array
# counter = 0
# counters = [0, 0, 0, 0]
# for group in groups:
# counters[(group-1)] += 1
# fours = counters[3]
# threes = counters[2]
# ones = 0
# if counters[0] > threes:
# ones = counters[0] - threes
# twos = (counters[1]//2) + (((counters[1] % 2)*2+ones)//4)
# # left = 0
# if (((counters[1] % 2)*2+ones) % 4) != 0:
# twos += 1
# print(fours+threes+twos)
# ---------------------------------------------------------------------------------------------------------------------
# Fancy Fence
# https://codeforces.com/problemset/problem/270/A
# Input
# The first line of input contains an integer t (0 < t < 180) — the number of tests. Each of the following t lines contains a single integer a (0 < a < 180) — the angle the robot can make corners at measured in degrees.
# Output
# For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.
# times = int(input())
# answers = []
# for time in range(times):
# a = int(input())
# if 360 % (180 - a) == 0:
# answers.append("YES")
# else:
# answers.append("NO")
# for answer in answers:
# print(answer)
# ---------------------------------------------------------------------------------------------------------------------
# Interesting drink
# https://codeforces.com/problemset/problem/706/B
# Input
# The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of shops in the city that sell Vasiliy's favourite drink.
# The second line contains n integers xi (1 ≤ xi ≤ 100 000) — prices of the bottles of the drink in the i-th shop.
# The third line contains a single integer q (1 ≤ q ≤ 100 000) — the number of days Vasiliy plans to buy the drink.
# Then follow q lines each containing one integer mi (1 ≤ mi ≤ 109) — the number of coins Vasiliy can spent on the i-th day.
# Output
# Print q integers. The i-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the i-th day.
# nShops = int(input())
# shopPrices = list(map(int, input().split()))
# shopPrices.sort()
# times = int(input())
# answers = []
# def binarySearch(array, target, carry):
# index = len(array)
# if len(array) == 0:
# return carry
# if index == 1:
# if target < array[0]:
# return carry
# else:
# return carry + 1
# # position in the middle
# index = index//2
# if target < array[index]:
# # return the left
# newPrices = array[0:index]
# return binarySearch(newPrices, target, carry)
# else:
# # return the right
# carry += (index)
# newPrices = array[index:]
# return binarySearch(newPrices, target, carry)
# def iterativeBinary(array, target):
# low = 0
# high = len(array) - 1
# while (low <= high):
# mid = low + ((high-low)//2)
# if array[mid] > target:
# high = mid - 1
# else:
# low = mid + 1
# return low
# for time in range(times):
# money = int(input())
# # looks like the way i implemented the binary search isnt logN :(
# # buys = binarySearch(shopPrices, money, 0)
# buys = iterativeBinary(shopPrices, money)
# answers.append(buys)
# for answer in answers:
# print(answer)
# ---------------------------------------------------------------------------------------------------------------------
# A and B and Compilation Errors
# https://codeforces.com/problemset/problem/519/B
# Output
# Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
# compilationTimes = int(input())
# errors1 = list(map(int, input().split()))
# errors2 = list(map(int, input().split()))
# errors3 = list(map(int, input().split()))
# errors1.sort()
# errors2.sort()
# errors3.sort()
# n = -1
# isNFound = False
# m = 1
# isMFound = False
# for time in range(compilationTimes):
# if isNFound and isMFound:
# break
# if not isNFound:
# if time < len(errors2):
# if errors1[time] != errors2[time]:
# n = errors1[time]
# isNFound = True
# else:
# n = errors1[time]
# isNFound = True
# if not isMFound:
# if time < len(errors3):
# if errors2[time] != errors3[time]:
# m = errors2[time]
# isMFound = True
# else:
# m = errors2[time]
# isNFound = True
# print(n)
# print(m)
# ---------------------------------------------------------------------------------------------------------------------
target = int(input())
people = ["Sheldon", "Leonard", "Penny", "Rajesh", "Howard"]
target -= 1
k = 0
while True:
if 5 * (2**k - 1) > target:
break
k += 1
k -= 1
# print(k)
# print((5 * (2**k - 1)))
corrimiento = target - (5 * (2**k - 1))
realK = 2**(k)
positionArray = (corrimiento)//realK
# print(realK)
# print(corrimiento)
print(people[positionArray])
# ---------------------------------------------------------------------------------------------------------------------
```
Yes
| 105,815 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the n-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
Input
The input data consist of a single integer n (1 ≤ n ≤ 109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
Output
Print the single line — the name of the person who drinks the n-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
Examples
Input
1
Output
Sheldon
Input
6
Output
Sheldon
Input
1802
Output
Penny
Submitted Solution:
```
n = int(input())
k = 1
max = 0
while True:
max = max + 2**k * 5
min = max - (2**k * 5)
if min <= n <= max:
break
k += 1
if (n - min) % 2**k == 0:
who = (n - min) // 2**k
elif (n - min) % 2**k != 0:
who = ((n - min) // 2**k) + 1
if who == 1:
print("Sheldon")
elif who == 2:
print("Leonard")
elif who == 3:
print("Penny")
elif who == 4:
print("Rajesh")
elif who == 5:
print("Howard" )
```
No
| 105,816 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the n-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
Input
The input data consist of a single integer n (1 ≤ n ≤ 109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
Output
Print the single line — the name of the person who drinks the n-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
Examples
Input
1
Output
Sheldon
Input
6
Output
Sheldon
Input
1802
Output
Penny
Submitted Solution:
```
n = int(input())
text = ["Sheldon", "Leonard", "Penny", "Rajesh", "Howard"]
while n > len(text):
n -= len(text)
print(text[n - 1])
```
No
| 105,817 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the n-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
Input
The input data consist of a single integer n (1 ≤ n ≤ 109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
Output
Print the single line — the name of the person who drinks the n-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
Examples
Input
1
Output
Sheldon
Input
6
Output
Sheldon
Input
1802
Output
Penny
Submitted Solution:
```
import sys
import math
import os
names=["Sheldon", "Leonard", "Penny", "Rajesh", "Howard"]
n=int(input())
k=1
cur=5*k
while cur < n:
k+=1
cur+=5*k
#print(cur)
n-=(cur-k*5)
#print(n,k)
#print(int(n/k))
delta=0
if n%k>0:
delta=1
print(names[int(n/k)+delta-1])
```
No
| 105,818 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the n-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
Input
The input data consist of a single integer n (1 ≤ n ≤ 109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
Output
Print the single line — the name of the person who drinks the n-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
Examples
Input
1
Output
Sheldon
Input
6
Output
Sheldon
Input
1802
Output
Penny
Submitted Solution:
```
ans=["Sheldon", "Leonard", "Penny", "Rajesh" ,"Howard"]
x=int(input())
cur=int(1)
while cur*5<=x:
x-=(cur*5)
cur*=2
print(ans[x%5-1])
```
No
| 105,819 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mojtaba and Arpa are playing a game. They have a list of n numbers in the game.
In a player's turn, he chooses a number pk (where p is a prime number and k is a positive integer) such that pk divides at least one number in the list. For each number in the list divisible by pk, call it x, the player will delete x and add <image> to the list. The player who can not make a valid choice of p and k loses.
Mojtaba starts the game and the players alternatively make moves. Determine which one of players will be the winner if both players play optimally.
Input
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of elements in the list.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the elements of the list.
Output
If Mojtaba wins, print "Mojtaba", otherwise print "Arpa" (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
4
1 1 1 1
Output
Arpa
Input
4
1 1 17 17
Output
Mojtaba
Input
4
1 1 17 289
Output
Arpa
Input
5
1 2 3 4 5
Output
Arpa
Note
In the first sample test, Mojtaba can't move.
In the second sample test, Mojtaba chooses p = 17 and k = 1, then the list changes to [1, 1, 1, 1].
In the third sample test, if Mojtaba chooses p = 17 and k = 1, then Arpa chooses p = 17 and k = 1 and wins, if Mojtaba chooses p = 17 and k = 2, then Arpa chooses p = 17 and k = 1 and wins.
Tags: bitmasks, dp, games
Correct Solution:
```
import sys
input = sys.stdin.buffer.readline
from collections import Counter
games = Counter() # prime : bitmask of if that power of the prime exists
def add_primes(a):
i = 2
while i*i <= a:
cnt = 0
while a % i == 0:
a //= i
cnt += 1
if cnt:
games[i] |= 1 << cnt
i += 1
if a != 1:
games[a] |= 1 << 1
def mex(a):
i = 0
while True:
if i not in a:
return i
i += 1
grundy_val = {}
def grundy_value(mask):
if mask in grundy_val:
return grundy_val[mask]
else:
transition_masks = set()
k = 1
while 1<<k <= mask:
new_mask = (mask&((1<<k)-1))|(mask>>k)
transition_masks.add(grundy_value(new_mask))
k += 1
grundy_val[mask] = mex(transition_masks)
return grundy_val[mask]
n = int(input())
a = list(map(int,input().split()))
for i in a:
add_primes(i)
xor_sum = 0
for game in games.values():
grundy_val = {}
xor_sum ^= grundy_value(game)
if xor_sum:
print("Mojtaba")
else:
print("Arpa")
```
| 105,820 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mojtaba and Arpa are playing a game. They have a list of n numbers in the game.
In a player's turn, he chooses a number pk (where p is a prime number and k is a positive integer) such that pk divides at least one number in the list. For each number in the list divisible by pk, call it x, the player will delete x and add <image> to the list. The player who can not make a valid choice of p and k loses.
Mojtaba starts the game and the players alternatively make moves. Determine which one of players will be the winner if both players play optimally.
Input
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of elements in the list.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the elements of the list.
Output
If Mojtaba wins, print "Mojtaba", otherwise print "Arpa" (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
4
1 1 1 1
Output
Arpa
Input
4
1 1 17 17
Output
Mojtaba
Input
4
1 1 17 289
Output
Arpa
Input
5
1 2 3 4 5
Output
Arpa
Note
In the first sample test, Mojtaba can't move.
In the second sample test, Mojtaba chooses p = 17 and k = 1, then the list changes to [1, 1, 1, 1].
In the third sample test, if Mojtaba chooses p = 17 and k = 1, then Arpa chooses p = 17 and k = 1 and wins, if Mojtaba chooses p = 17 and k = 2, then Arpa chooses p = 17 and k = 1 and wins.
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
b=[]
c=[]
e=[]
ans = 0
for i in range(30):
e.append(0)
j=2
if (j == 2):
for i in range(n):
l=0
while a[i] % j == 0:
a[i]=a[i]/j
l+=1
if (l!=0):
e[l]=1
k=0
for i in range(30):
if (e[i]==1):
ans=ans^(i-k)
k=i
j+=1
for i in range(30):
e[i]=0
if (j==3):
for i in range(n):
l=0
while a[i] % j == 0:
a[i]=a[i]/j
l+=1
if (l!=0):
e[l]=1
k=0
for i in range(30):
if (e[i]==1):
ans=ans^(i-k)
k=i
j+=2
for i in range(20):
e[i]=0
if (j==5):
for i in range(n):
l=0
while a[i] % j == 0:
a[i]=a[i]/j
l+=1
if (l!=0):
e[l]=1
k=0
for i in range(30):
if (e[i]==1):
ans=ans^(i-k)
k=i
j+=2
for i in range(20):
e[i]=0
if (j==7):
for i in range(n):
l=0
while a[i] % j == 0:
a[i]=a[i]/j
l+=1
if (l!=0):
e[l]=1
k=0
for i in range(30):
if (e[i]==1):
ans=ans^(i-k)
k=i
j+=2
for i in range(20):
e[i]=0
while (j< 31700):
for i in range(11):
e[i]=0
for i in range(n):
l=0
while a[i] % j == 0:
a[i]=a[i]/j
l+=1
if (l!=0):
e[l]=1
k=0
for i in range(11):
if (e[i]==1):
ans=ans^(i-k)
k=i
j+=2
for i in range(n):
if a[i]!=1:
ans=ans^1
for j in range(i,n):
if a[i]==a[j]:
a[j]=1
a[i]=1
if (ans !=0):
print("Mojtaba")
else:
print("Arpa")
```
No
| 105,821 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mojtaba and Arpa are playing a game. They have a list of n numbers in the game.
In a player's turn, he chooses a number pk (where p is a prime number and k is a positive integer) such that pk divides at least one number in the list. For each number in the list divisible by pk, call it x, the player will delete x and add <image> to the list. The player who can not make a valid choice of p and k loses.
Mojtaba starts the game and the players alternatively make moves. Determine which one of players will be the winner if both players play optimally.
Input
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of elements in the list.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the elements of the list.
Output
If Mojtaba wins, print "Mojtaba", otherwise print "Arpa" (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
4
1 1 1 1
Output
Arpa
Input
4
1 1 17 17
Output
Mojtaba
Input
4
1 1 17 289
Output
Arpa
Input
5
1 2 3 4 5
Output
Arpa
Note
In the first sample test, Mojtaba can't move.
In the second sample test, Mojtaba chooses p = 17 and k = 1, then the list changes to [1, 1, 1, 1].
In the third sample test, if Mojtaba chooses p = 17 and k = 1, then Arpa chooses p = 17 and k = 1 and wins, if Mojtaba chooses p = 17 and k = 2, then Arpa chooses p = 17 and k = 1 and wins.
Submitted Solution:
```
n=int(input())
a= [int(x) for x in input().split(' ')]
if a[-1] in [17,10]:
print("Mojtaba")
else:
print("Arpa")
```
No
| 105,822 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mojtaba and Arpa are playing a game. They have a list of n numbers in the game.
In a player's turn, he chooses a number pk (where p is a prime number and k is a positive integer) such that pk divides at least one number in the list. For each number in the list divisible by pk, call it x, the player will delete x and add <image> to the list. The player who can not make a valid choice of p and k loses.
Mojtaba starts the game and the players alternatively make moves. Determine which one of players will be the winner if both players play optimally.
Input
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of elements in the list.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the elements of the list.
Output
If Mojtaba wins, print "Mojtaba", otherwise print "Arpa" (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
4
1 1 1 1
Output
Arpa
Input
4
1 1 17 17
Output
Mojtaba
Input
4
1 1 17 289
Output
Arpa
Input
5
1 2 3 4 5
Output
Arpa
Note
In the first sample test, Mojtaba can't move.
In the second sample test, Mojtaba chooses p = 17 and k = 1, then the list changes to [1, 1, 1, 1].
In the third sample test, if Mojtaba chooses p = 17 and k = 1, then Arpa chooses p = 17 and k = 1 and wins, if Mojtaba chooses p = 17 and k = 2, then Arpa chooses p = 17 and k = 1 and wins.
Submitted Solution:
```
def prime_factors(n):
i = 2
while i * i <= n:
if n % i:
i += 1
else:
n //= i
yield i
if n > 1:
yield n
from bisect import bisect
def move_generator(p):
for i in range(p.bit_length()):
yield (p & ((1 << i) - 1)) | (p >> (i+1))
def helper(L):
memo = {0:False}
def rec(P):
try:
return memo[P]
except KeyError:
return not all(rec(Q) for Q in move_generator(P))
return rec(L)
from collections import defaultdict
from itertools import chain
memo = defaultdict(set)
n = int(input())
for a in map(int,input().split()):
f = list(prime_factors(a))
p = 1
c = 0
for x in f:
if x == p:
c += 1
else:
memo[p].add(c)
p = x
c = 1
memo[p].add(c)
del memo[1]
def convert(s):
b = 0
for i in s:
b |= 1 << i-1
return b
c = 0
for s in memo.values():
c += helper(convert(s))
print('Mojtaba' if c%2 else 'Arpa')
```
No
| 105,823 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mojtaba and Arpa are playing a game. They have a list of n numbers in the game.
In a player's turn, he chooses a number pk (where p is a prime number and k is a positive integer) such that pk divides at least one number in the list. For each number in the list divisible by pk, call it x, the player will delete x and add <image> to the list. The player who can not make a valid choice of p and k loses.
Mojtaba starts the game and the players alternatively make moves. Determine which one of players will be the winner if both players play optimally.
Input
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of elements in the list.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the elements of the list.
Output
If Mojtaba wins, print "Mojtaba", otherwise print "Arpa" (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
4
1 1 1 1
Output
Arpa
Input
4
1 1 17 17
Output
Mojtaba
Input
4
1 1 17 289
Output
Arpa
Input
5
1 2 3 4 5
Output
Arpa
Note
In the first sample test, Mojtaba can't move.
In the second sample test, Mojtaba chooses p = 17 and k = 1, then the list changes to [1, 1, 1, 1].
In the third sample test, if Mojtaba chooses p = 17 and k = 1, then Arpa chooses p = 17 and k = 1 and wins, if Mojtaba chooses p = 17 and k = 2, then Arpa chooses p = 17 and k = 1 and wins.
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
b=[]
c=[]
e=[]
ans = 0
for i in range(30):
e.append(0)
j=2
if (j == 2):
for i in range(n):
l=0
while a[i] % j == 0:
a[i]=a[i]/j
l+=1
if (l!=0):
e[l]=1
k=0
for i in range(30):
if (e[i]==1):
ans=ans^(i-k)
k=i
j+=1
for i in range(30):
e[i]=0
if (j==3):
for i in range(n):
l=0
while a[i] % j == 0:
a[i]=a[i]/j
l+=1
if (l!=0):
e[l]=1
k=0
for i in range(30):
if (e[i]==1):
ans=ans^(i-k)
k=i
j+=2
for i in range(20):
e[i]=0
if (j==5):
for i in range(n):
l=0
while a[i] % j == 0:
a[i]=a[i]/j
l+=1
if (l!=0):
e[l]=1
k=0
for i in range(30):
if (e[i]==1):
ans=ans^(i-k)
k=i
j+=2
for i in range(20):
e[i]=0
if (j==7):
for i in range(n):
l=0
while a[i] % j == 0:
a[i]=a[i]/j
l+=1
if (l!=0):
e[l]=1
k=0
for i in range(30):
if (e[i]==1):
ans=ans^(i-k)
k=i
j+=2
for i in range(20):
e[i]=0
while (j< 31623):
for i in range(10):
e[i]=0
for i in range(n):
l=0
while a[i] % j == 0:
a[i]=a[i]/j
l+=1
if (l!=0):
e[l]=1
k=0
for i in range(10):
if (e[i]==1):
ans=ans^(i-k)
k=i
j+=2
for i in range(n):
if a[i]!=1:
ans=ans^1
for j in range(i,n):
if a[i]==a[j]:
a[j]=1
a[i]=1
if (ans !=0):
print("Mojtaba")
else:
print("Arpa")
```
No
| 105,824 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alexey recently held a programming contest for students from Berland. n students participated in a contest, i-th of them solved ai problems. Now he wants to award some contestants. Alexey can award the students with diplomas of three different degrees. Each student either will receive one diploma of some degree, or won't receive any diplomas at all. Let cntx be the number of students that are awarded with diplomas of degree x (1 ≤ x ≤ 3). The following conditions must hold:
* For each x (1 ≤ x ≤ 3) cntx > 0;
* For any two degrees x and y cntx ≤ 2·cnty.
Of course, there are a lot of ways to distribute the diplomas. Let bi be the degree of diploma i-th student will receive (or - 1 if i-th student won't receive any diplomas). Also for any x such that 1 ≤ x ≤ 3 let cx be the maximum number of problems solved by a student that receives a diploma of degree x, and dx be the minimum number of problems solved by a student that receives a diploma of degree x. Alexey wants to distribute the diplomas in such a way that:
1. If student i solved more problems than student j, then he has to be awarded not worse than student j (it's impossible that student j receives a diploma and i doesn't receive any, and also it's impossible that both of them receive a diploma, but bj < bi);
2. d1 - c2 is maximum possible;
3. Among all ways that maximize the previous expression, d2 - c3 is maximum possible;
4. Among all ways that correspond to the two previous conditions, d3 - c - 1 is maximum possible, where c - 1 is the maximum number of problems solved by a student that doesn't receive any diploma (or 0 if each student is awarded with some diploma).
Help Alexey to find a way to award the contestants!
Input
The first line contains one integer number n (3 ≤ n ≤ 3000).
The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 5000).
Output
Output n numbers. i-th number must be equal to the degree of diploma i-th contestant will receive (or - 1 if he doesn't receive any diploma).
If there are multiple optimal solutions, print any of them. It is guaranteed that the answer always exists.
Examples
Input
4
1 2 3 4
Output
3 3 2 1
Input
6
1 4 3 1 1 2
Output
-1 1 2 -1 -1 3
Submitted Solution:
```
print("-1 3 2 1")
```
No
| 105,825 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alexey recently held a programming contest for students from Berland. n students participated in a contest, i-th of them solved ai problems. Now he wants to award some contestants. Alexey can award the students with diplomas of three different degrees. Each student either will receive one diploma of some degree, or won't receive any diplomas at all. Let cntx be the number of students that are awarded with diplomas of degree x (1 ≤ x ≤ 3). The following conditions must hold:
* For each x (1 ≤ x ≤ 3) cntx > 0;
* For any two degrees x and y cntx ≤ 2·cnty.
Of course, there are a lot of ways to distribute the diplomas. Let bi be the degree of diploma i-th student will receive (or - 1 if i-th student won't receive any diplomas). Also for any x such that 1 ≤ x ≤ 3 let cx be the maximum number of problems solved by a student that receives a diploma of degree x, and dx be the minimum number of problems solved by a student that receives a diploma of degree x. Alexey wants to distribute the diplomas in such a way that:
1. If student i solved more problems than student j, then he has to be awarded not worse than student j (it's impossible that student j receives a diploma and i doesn't receive any, and also it's impossible that both of them receive a diploma, but bj < bi);
2. d1 - c2 is maximum possible;
3. Among all ways that maximize the previous expression, d2 - c3 is maximum possible;
4. Among all ways that correspond to the two previous conditions, d3 - c - 1 is maximum possible, where c - 1 is the maximum number of problems solved by a student that doesn't receive any diploma (or 0 if each student is awarded with some diploma).
Help Alexey to find a way to award the contestants!
Input
The first line contains one integer number n (3 ≤ n ≤ 3000).
The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 5000).
Output
Output n numbers. i-th number must be equal to the degree of diploma i-th contestant will receive (or - 1 if he doesn't receive any diploma).
If there are multiple optimal solutions, print any of them. It is guaranteed that the answer always exists.
Examples
Input
4
1 2 3 4
Output
3 3 2 1
Input
6
1 4 3 1 1 2
Output
-1 1 2 -1 -1 3
Submitted Solution:
```
n = int(input())
a = [int(x) for x in input().split()]
b = sorted(a)
dip1 = {}
dip2 = {}
dip3 = {}
ndip = {}
first_value = 0
first_index = 2
for i in range(first_index, n):
if b[i] - b[i-1] >= first_value:
first_value = b[i] - b[i-1]
first_index = i
second_value = 0
second_index = 1
for i in range(second_index, first_index):
if b[i] - b[i-1] >= second_value:
second_value = b[i] - b[i-1]
second_index = i
third_value = 0
third_index = 0
for i in range(third_index, second_index):
if i == 0:
continue
if b[i] - b[i-1] >= third_value:
third_value = b[i] - b[i-1]
third_index = i
for i in range(0, third_index):
if ndip.__contains__(b[i]):
ndip[b[i]] += 1
else:
ndip[b[i]] = 1
for i in range(third_index, second_index):
if dip3.__contains__(b[i]):
dip3[b[i]] += 1
else:
dip3[b[i]] = 1
for i in range(second_index, first_index):
if dip2.__contains__(b[i]):
dip2[b[i]] += 1
else:
dip2[b[i]] = 1
for i in range(first_index, n):
if dip1.__contains__(b[i]):
dip1[b[i]] += 1
else:
dip1[b[i]] = 1
# print(first_index, second_index, third_index, )
# print(dip)
#
# print(a)
for x in a:
if ndip.__contains__(x) and ndip[x]:
print('-1', end=' ')
ndip[x] -= 1
elif dip1.__contains__(x) and dip1[x]:
print('1', end=' ')
dip1[x] -= 1
elif dip2.__contains__(x) and dip2[x]:
print('2', end=' ')
dip2[x] -= 1
elif dip3.__contains__(x) and dip3[x]:
print('3', end=' ')
dip3[x] -= 1
```
No
| 105,826 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alexey recently held a programming contest for students from Berland. n students participated in a contest, i-th of them solved ai problems. Now he wants to award some contestants. Alexey can award the students with diplomas of three different degrees. Each student either will receive one diploma of some degree, or won't receive any diplomas at all. Let cntx be the number of students that are awarded with diplomas of degree x (1 ≤ x ≤ 3). The following conditions must hold:
* For each x (1 ≤ x ≤ 3) cntx > 0;
* For any two degrees x and y cntx ≤ 2·cnty.
Of course, there are a lot of ways to distribute the diplomas. Let bi be the degree of diploma i-th student will receive (or - 1 if i-th student won't receive any diplomas). Also for any x such that 1 ≤ x ≤ 3 let cx be the maximum number of problems solved by a student that receives a diploma of degree x, and dx be the minimum number of problems solved by a student that receives a diploma of degree x. Alexey wants to distribute the diplomas in such a way that:
1. If student i solved more problems than student j, then he has to be awarded not worse than student j (it's impossible that student j receives a diploma and i doesn't receive any, and also it's impossible that both of them receive a diploma, but bj < bi);
2. d1 - c2 is maximum possible;
3. Among all ways that maximize the previous expression, d2 - c3 is maximum possible;
4. Among all ways that correspond to the two previous conditions, d3 - c - 1 is maximum possible, where c - 1 is the maximum number of problems solved by a student that doesn't receive any diploma (or 0 if each student is awarded with some diploma).
Help Alexey to find a way to award the contestants!
Input
The first line contains one integer number n (3 ≤ n ≤ 3000).
The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 5000).
Output
Output n numbers. i-th number must be equal to the degree of diploma i-th contestant will receive (or - 1 if he doesn't receive any diploma).
If there are multiple optimal solutions, print any of them. It is guaranteed that the answer always exists.
Examples
Input
4
1 2 3 4
Output
3 3 2 1
Input
6
1 4 3 1 1 2
Output
-1 1 2 -1 -1 3
Submitted Solution:
```
n = int(input())
a = [int(x) for x in input().split()]
b = sorted(a)
dip = {}
first_value = 0
first_index = 2
for i in range(first_index, n):
if b[i] - b[i-1] >= first_value:
first_value = b[i] - b[i-1]
first_index = i
second_value = 0
second_index = 1
for i in range(second_index, first_index):
if b[i] - b[i-1] >= second_value:
second_value = b[i] - b[i-1]
second_index = i
third_value = 0
third_index = 0
for i in range(third_index, second_index):
if i == 0:
continue
if b[i] - b[i-1] >= third_value:
third_value = b[i] - b[i-1]
third_index = i
for i in range(0, third_index):
dip[b[i]] = -1
for i in range(third_index, second_index):
dip[b[i]] = 3
for i in range(second_index, first_index):
dip[b[i]] = 2
for i in range(first_index, n):
dip[b[i]] = 1
# print(first_index, second_index, third_index, )
# print(dip)
#
# print(a)
for x in a:
print(dip[x], end=' ')
```
No
| 105,827 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alexey recently held a programming contest for students from Berland. n students participated in a contest, i-th of them solved ai problems. Now he wants to award some contestants. Alexey can award the students with diplomas of three different degrees. Each student either will receive one diploma of some degree, or won't receive any diplomas at all. Let cntx be the number of students that are awarded with diplomas of degree x (1 ≤ x ≤ 3). The following conditions must hold:
* For each x (1 ≤ x ≤ 3) cntx > 0;
* For any two degrees x and y cntx ≤ 2·cnty.
Of course, there are a lot of ways to distribute the diplomas. Let bi be the degree of diploma i-th student will receive (or - 1 if i-th student won't receive any diplomas). Also for any x such that 1 ≤ x ≤ 3 let cx be the maximum number of problems solved by a student that receives a diploma of degree x, and dx be the minimum number of problems solved by a student that receives a diploma of degree x. Alexey wants to distribute the diplomas in such a way that:
1. If student i solved more problems than student j, then he has to be awarded not worse than student j (it's impossible that student j receives a diploma and i doesn't receive any, and also it's impossible that both of them receive a diploma, but bj < bi);
2. d1 - c2 is maximum possible;
3. Among all ways that maximize the previous expression, d2 - c3 is maximum possible;
4. Among all ways that correspond to the two previous conditions, d3 - c - 1 is maximum possible, where c - 1 is the maximum number of problems solved by a student that doesn't receive any diploma (or 0 if each student is awarded with some diploma).
Help Alexey to find a way to award the contestants!
Input
The first line contains one integer number n (3 ≤ n ≤ 3000).
The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 5000).
Output
Output n numbers. i-th number must be equal to the degree of diploma i-th contestant will receive (or - 1 if he doesn't receive any diploma).
If there are multiple optimal solutions, print any of them. It is guaranteed that the answer always exists.
Examples
Input
4
1 2 3 4
Output
3 3 2 1
Input
6
1 4 3 1 1 2
Output
-1 1 2 -1 -1 3
Submitted Solution:
```
n = int(input())
a = [int(x) for x in input().split()]
b = sorted(a)
dip1 = {}
dip2 = {}
dip3 = {}
ndip = {}
first_value = 0
first_index = 2
for i in range(first_index, n):
if b[i] - b[i-1] >= first_value:
first_value = b[i] - b[i-1]
first_index = i
second_value = 0
second_index = 1
for i in range(second_index, first_index):
if b[i] - b[i-1] >= second_value:
second_value = b[i] - b[i-1]
second_index = i
third_value = 0
third_index = 0
for i in range(third_index, second_index):
if i == 0:
continue
if b[i] - b[i-1] >= third_value:
third_value = b[i] - b[i-1]
third_index = i
for i in range(0, third_index):
if ndip.__contains__(b[i]):
ndip[b[i]] += 1
else:
ndip[b[i]] = 1
for i in range(third_index, second_index):
if dip3.__contains__(b[i]):
dip3[b[i]] += 1
else:
dip3[b[i]] = 1
for i in range(second_index, first_index):
if dip2.__contains__(b[i]):
dip2[b[i]] += 1
else:
dip2[b[i]] = 1
for i in range(first_index, n):
if dip1.__contains__(b[i]):
dip1[b[i]] += 1
else:
dip1[b[i]] = 1
# print(first_index, second_index, third_index, )
# print(dip)
#
# print(a)
for i in range(n):
if ndip.__contains__(a[i]) and ndip[a[i]]:
print('-1', end=' ')
ndip[a[i]] -= 1
elif dip1.__contains__(a[i]) and dip1[a[i]]:
print('1', end=' ')
dip1[a[i]] -= 1
elif dip2.__contains__(a[i]) and dip2[a[i]]:
print('2', end=' ')
dip2[a[i]] -= 1
elif dip3.__contains__(a[i]) and dip3[a[i]]:
print('3', end=' ')
dip3[a[i]] -= 1
```
No
| 105,828 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n shovels in Polycarp's shop. The i-th shovel costs i burles, that is, the first shovel costs 1 burle, the second shovel costs 2 burles, the third shovel costs 3 burles, and so on. Polycarps wants to sell shovels in pairs.
Visitors are more likely to buy a pair of shovels if their total cost ends with several 9s. Because of this, Polycarp wants to choose a pair of shovels to sell in such a way that the sum of their costs ends with maximum possible number of nines. For example, if he chooses shovels with costs 12345 and 37454, their total cost is 49799, it ends with two nines.
You are to compute the number of pairs of shovels such that their total cost ends with maximum possible number of nines. Two pairs are considered different if there is a shovel presented in one pair, but not in the other.
Input
The first line contains a single integer n (2 ≤ n ≤ 109) — the number of shovels in Polycarp's shop.
Output
Print the number of pairs of shovels such that their total cost ends with maximum possible number of nines.
Note that it is possible that the largest number of 9s at the end is 0, then you should count all such ways.
It is guaranteed that for every n ≤ 109 the answer doesn't exceed 2·109.
Examples
Input
7
Output
3
Input
14
Output
9
Input
50
Output
1
Note
In the first example the maximum possible number of nines at the end is one. Polycarp cah choose the following pairs of shovels for that purpose:
* 2 and 7;
* 3 and 6;
* 4 and 5.
In the second example the maximum number of nines at the end of total cost of two shovels is one. The following pairs of shovels suit Polycarp:
* 1 and 8;
* 2 and 7;
* 3 and 6;
* 4 and 5;
* 5 and 14;
* 6 and 13;
* 7 and 12;
* 8 and 11;
* 9 and 10.
In the third example it is necessary to choose shovels 49 and 50, because the sum of their cost is 99, that means that the total number of nines is equal to two, which is maximum possible for n = 50.
Tags: constructive algorithms, math
Correct Solution:
```
n = int(input())
largest = n + n - 1
possible = [0, 9, 99, 999, 9999, 99999, 999999, 9999999, 99999999, 999999999, 999999999]
maximum9 = 0
indx1 = 0
i = 0
for p in possible:
if p <= largest and p > maximum9:
maximum9 = p
indx1 = i
i += 1
indx2 = 0
for i in range(9):
if largest >= i*10**indx1+maximum9:
indx2 = i
else:
break
count = 0
for i in range(indx2+1):
count += max((2*min(n, i*10**indx1+maximum9-1)- max(1,i*10**indx1+maximum9)+1)//2, 0)
print(count)
```
| 105,829 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n shovels in Polycarp's shop. The i-th shovel costs i burles, that is, the first shovel costs 1 burle, the second shovel costs 2 burles, the third shovel costs 3 burles, and so on. Polycarps wants to sell shovels in pairs.
Visitors are more likely to buy a pair of shovels if their total cost ends with several 9s. Because of this, Polycarp wants to choose a pair of shovels to sell in such a way that the sum of their costs ends with maximum possible number of nines. For example, if he chooses shovels with costs 12345 and 37454, their total cost is 49799, it ends with two nines.
You are to compute the number of pairs of shovels such that their total cost ends with maximum possible number of nines. Two pairs are considered different if there is a shovel presented in one pair, but not in the other.
Input
The first line contains a single integer n (2 ≤ n ≤ 109) — the number of shovels in Polycarp's shop.
Output
Print the number of pairs of shovels such that their total cost ends with maximum possible number of nines.
Note that it is possible that the largest number of 9s at the end is 0, then you should count all such ways.
It is guaranteed that for every n ≤ 109 the answer doesn't exceed 2·109.
Examples
Input
7
Output
3
Input
14
Output
9
Input
50
Output
1
Note
In the first example the maximum possible number of nines at the end is one. Polycarp cah choose the following pairs of shovels for that purpose:
* 2 and 7;
* 3 and 6;
* 4 and 5.
In the second example the maximum number of nines at the end of total cost of two shovels is one. The following pairs of shovels suit Polycarp:
* 1 and 8;
* 2 and 7;
* 3 and 6;
* 4 and 5;
* 5 and 14;
* 6 and 13;
* 7 and 12;
* 8 and 11;
* 9 and 10.
In the third example it is necessary to choose shovels 49 and 50, because the sum of their cost is 99, that means that the total number of nines is equal to two, which is maximum possible for n = 50.
Tags: constructive algorithms, math
Correct Solution:
```
n=int(input())
import sys
num9=len(str(2*n))-1
s=[str(i) for i in range(10)]
ans=0
if(num9==0):
print(n*(n-1)//2)
sys.exit()
for i in s:
x=i+'9'*num9
x=int(x)
ans+=max(0,min((n-x//2),x//2))
print(ans)
```
| 105,830 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n shovels in Polycarp's shop. The i-th shovel costs i burles, that is, the first shovel costs 1 burle, the second shovel costs 2 burles, the third shovel costs 3 burles, and so on. Polycarps wants to sell shovels in pairs.
Visitors are more likely to buy a pair of shovels if their total cost ends with several 9s. Because of this, Polycarp wants to choose a pair of shovels to sell in such a way that the sum of their costs ends with maximum possible number of nines. For example, if he chooses shovels with costs 12345 and 37454, their total cost is 49799, it ends with two nines.
You are to compute the number of pairs of shovels such that their total cost ends with maximum possible number of nines. Two pairs are considered different if there is a shovel presented in one pair, but not in the other.
Input
The first line contains a single integer n (2 ≤ n ≤ 109) — the number of shovels in Polycarp's shop.
Output
Print the number of pairs of shovels such that their total cost ends with maximum possible number of nines.
Note that it is possible that the largest number of 9s at the end is 0, then you should count all such ways.
It is guaranteed that for every n ≤ 109 the answer doesn't exceed 2·109.
Examples
Input
7
Output
3
Input
14
Output
9
Input
50
Output
1
Note
In the first example the maximum possible number of nines at the end is one. Polycarp cah choose the following pairs of shovels for that purpose:
* 2 and 7;
* 3 and 6;
* 4 and 5.
In the second example the maximum number of nines at the end of total cost of two shovels is one. The following pairs of shovels suit Polycarp:
* 1 and 8;
* 2 and 7;
* 3 and 6;
* 4 and 5;
* 5 and 14;
* 6 and 13;
* 7 and 12;
* 8 and 11;
* 9 and 10.
In the third example it is necessary to choose shovels 49 and 50, because the sum of their cost is 99, that means that the total number of nines is equal to two, which is maximum possible for n = 50.
Tags: constructive algorithms, math
Correct Solution:
```
from math import *
import sys
#sys.stdin = open('in.txt')
n = int(input())
def closest9(n):
s = '9'*(len(str(n+1))-1)
return 0 if len(s) == 0 else int(s)
def solve(n):
if n == 2: return 1
if n == 3: return 3
if n == 4: return 6
s = n+n-1
c = closest9(s)
if c*10 + 9 == s: return 1
p = c
res = 0
for i in range(10):
if p <= n+1:
res += p//2
elif p > s:
break
else:
res += 1+(s - p)//2
#print(p, v)
p += c+1
return res
print(solve(n))
```
| 105,831 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n shovels in Polycarp's shop. The i-th shovel costs i burles, that is, the first shovel costs 1 burle, the second shovel costs 2 burles, the third shovel costs 3 burles, and so on. Polycarps wants to sell shovels in pairs.
Visitors are more likely to buy a pair of shovels if their total cost ends with several 9s. Because of this, Polycarp wants to choose a pair of shovels to sell in such a way that the sum of their costs ends with maximum possible number of nines. For example, if he chooses shovels with costs 12345 and 37454, their total cost is 49799, it ends with two nines.
You are to compute the number of pairs of shovels such that their total cost ends with maximum possible number of nines. Two pairs are considered different if there is a shovel presented in one pair, but not in the other.
Input
The first line contains a single integer n (2 ≤ n ≤ 109) — the number of shovels in Polycarp's shop.
Output
Print the number of pairs of shovels such that their total cost ends with maximum possible number of nines.
Note that it is possible that the largest number of 9s at the end is 0, then you should count all such ways.
It is guaranteed that for every n ≤ 109 the answer doesn't exceed 2·109.
Examples
Input
7
Output
3
Input
14
Output
9
Input
50
Output
1
Note
In the first example the maximum possible number of nines at the end is one. Polycarp cah choose the following pairs of shovels for that purpose:
* 2 and 7;
* 3 and 6;
* 4 and 5.
In the second example the maximum number of nines at the end of total cost of two shovels is one. The following pairs of shovels suit Polycarp:
* 1 and 8;
* 2 and 7;
* 3 and 6;
* 4 and 5;
* 5 and 14;
* 6 and 13;
* 7 and 12;
* 8 and 11;
* 9 and 10.
In the third example it is necessary to choose shovels 49 and 50, because the sum of their cost is 99, that means that the total number of nines is equal to two, which is maximum possible for n = 50.
Tags: constructive algorithms, math
Correct Solution:
```
n = int(input())
v = min(n, 5)
if v < 5:
print(n*(n - 1) // 2)
exit()
while v * 10 <= n:
v *= 10
print(sum(min(n - i * v + 1, v * i - 1) for i in range(1, n // v + 1)))
```
| 105,832 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n shovels in Polycarp's shop. The i-th shovel costs i burles, that is, the first shovel costs 1 burle, the second shovel costs 2 burles, the third shovel costs 3 burles, and so on. Polycarps wants to sell shovels in pairs.
Visitors are more likely to buy a pair of shovels if their total cost ends with several 9s. Because of this, Polycarp wants to choose a pair of shovels to sell in such a way that the sum of their costs ends with maximum possible number of nines. For example, if he chooses shovels with costs 12345 and 37454, their total cost is 49799, it ends with two nines.
You are to compute the number of pairs of shovels such that their total cost ends with maximum possible number of nines. Two pairs are considered different if there is a shovel presented in one pair, but not in the other.
Input
The first line contains a single integer n (2 ≤ n ≤ 109) — the number of shovels in Polycarp's shop.
Output
Print the number of pairs of shovels such that their total cost ends with maximum possible number of nines.
Note that it is possible that the largest number of 9s at the end is 0, then you should count all such ways.
It is guaranteed that for every n ≤ 109 the answer doesn't exceed 2·109.
Examples
Input
7
Output
3
Input
14
Output
9
Input
50
Output
1
Note
In the first example the maximum possible number of nines at the end is one. Polycarp cah choose the following pairs of shovels for that purpose:
* 2 and 7;
* 3 and 6;
* 4 and 5.
In the second example the maximum number of nines at the end of total cost of two shovels is one. The following pairs of shovels suit Polycarp:
* 1 and 8;
* 2 and 7;
* 3 and 6;
* 4 and 5;
* 5 and 14;
* 6 and 13;
* 7 and 12;
* 8 and 11;
* 9 and 10.
In the third example it is necessary to choose shovels 49 and 50, because the sum of their cost is 99, that means that the total number of nines is equal to two, which is maximum possible for n = 50.
Tags: constructive algorithms, math
Correct Solution:
```
n = int(input())
a= 5
while a * 10 <= n:
a *= 10
print(sum(min(n - i * a + 1, a * i - 1) for i in range(1, n // a + 1)) if n>=5 else n*(n-1)//2)
```
| 105,833 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n shovels in Polycarp's shop. The i-th shovel costs i burles, that is, the first shovel costs 1 burle, the second shovel costs 2 burles, the third shovel costs 3 burles, and so on. Polycarps wants to sell shovels in pairs.
Visitors are more likely to buy a pair of shovels if their total cost ends with several 9s. Because of this, Polycarp wants to choose a pair of shovels to sell in such a way that the sum of their costs ends with maximum possible number of nines. For example, if he chooses shovels with costs 12345 and 37454, their total cost is 49799, it ends with two nines.
You are to compute the number of pairs of shovels such that their total cost ends with maximum possible number of nines. Two pairs are considered different if there is a shovel presented in one pair, but not in the other.
Input
The first line contains a single integer n (2 ≤ n ≤ 109) — the number of shovels in Polycarp's shop.
Output
Print the number of pairs of shovels such that their total cost ends with maximum possible number of nines.
Note that it is possible that the largest number of 9s at the end is 0, then you should count all such ways.
It is guaranteed that for every n ≤ 109 the answer doesn't exceed 2·109.
Examples
Input
7
Output
3
Input
14
Output
9
Input
50
Output
1
Note
In the first example the maximum possible number of nines at the end is one. Polycarp cah choose the following pairs of shovels for that purpose:
* 2 and 7;
* 3 and 6;
* 4 and 5.
In the second example the maximum number of nines at the end of total cost of two shovels is one. The following pairs of shovels suit Polycarp:
* 1 and 8;
* 2 and 7;
* 3 and 6;
* 4 and 5;
* 5 and 14;
* 6 and 13;
* 7 and 12;
* 8 and 11;
* 9 and 10.
In the third example it is necessary to choose shovels 49 and 50, because the sum of their cost is 99, that means that the total number of nines is equal to two, which is maximum possible for n = 50.
Tags: constructive algorithms, math
Correct Solution:
```
def check9(x):
i = len(x) - 1
while (i >= 0):
if (x[i] != '9'):
return len(x) - i - 1
i -= 1
return len(x) - i - 1
def solve(n):
if (n < 5):
return (n*(n-1)//2)
res = 0
x = str(n+n-1)
length = len(x)
if (check9(x) == length):
return 1
cur = '9'*(length-1)
for i in range(9):
c = str(i)
p = int(c+cur)
if (p <= n+1):
res += p//2
elif (p > n+n-1):
res += 0
else:
res += 1 + (n + n - 1 - p)//2
return res
n = int(input())
print(solve(n))
```
| 105,834 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n shovels in Polycarp's shop. The i-th shovel costs i burles, that is, the first shovel costs 1 burle, the second shovel costs 2 burles, the third shovel costs 3 burles, and so on. Polycarps wants to sell shovels in pairs.
Visitors are more likely to buy a pair of shovels if their total cost ends with several 9s. Because of this, Polycarp wants to choose a pair of shovels to sell in such a way that the sum of their costs ends with maximum possible number of nines. For example, if he chooses shovels with costs 12345 and 37454, their total cost is 49799, it ends with two nines.
You are to compute the number of pairs of shovels such that their total cost ends with maximum possible number of nines. Two pairs are considered different if there is a shovel presented in one pair, but not in the other.
Input
The first line contains a single integer n (2 ≤ n ≤ 109) — the number of shovels in Polycarp's shop.
Output
Print the number of pairs of shovels such that their total cost ends with maximum possible number of nines.
Note that it is possible that the largest number of 9s at the end is 0, then you should count all such ways.
It is guaranteed that for every n ≤ 109 the answer doesn't exceed 2·109.
Examples
Input
7
Output
3
Input
14
Output
9
Input
50
Output
1
Note
In the first example the maximum possible number of nines at the end is one. Polycarp cah choose the following pairs of shovels for that purpose:
* 2 and 7;
* 3 and 6;
* 4 and 5.
In the second example the maximum number of nines at the end of total cost of two shovels is one. The following pairs of shovels suit Polycarp:
* 1 and 8;
* 2 and 7;
* 3 and 6;
* 4 and 5;
* 5 and 14;
* 6 and 13;
* 7 and 12;
* 8 and 11;
* 9 and 10.
In the third example it is necessary to choose shovels 49 and 50, because the sum of their cost is 99, that means that the total number of nines is equal to two, which is maximum possible for n = 50.
Tags: constructive algorithms, math
Correct Solution:
```
n = int(input())
def length(obj):
return len(str(obj))
def num_nine(num):
i = 0
j = 9
while num >= j:
i += 1
j += 9 * (10 ** i)
return i
def split():
blocks = (n + 1) // count
extra = (n + 1) % count
b = blocks // 2
f = blocks - b
return [f, b, extra]
def ans():
if nines == 0:
return int(n * (n - 1) / 2)
if n == 10 ** nines - 1:
return n - (10 ** nines) // 2
if lent == nines:
return n - (10 ** nines) // 2 + 1
ls = split()
return ls[0] * ls[1] * count - ls[1] + ls[0] * ls[2]
num = 2 * n - 1
lent = length(n)
nines = num_nine(num)
count = (10 ** nines) // 2
#print('length', lent)
#print('nines', nines)
print(ans())
```
| 105,835 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n shovels in Polycarp's shop. The i-th shovel costs i burles, that is, the first shovel costs 1 burle, the second shovel costs 2 burles, the third shovel costs 3 burles, and so on. Polycarps wants to sell shovels in pairs.
Visitors are more likely to buy a pair of shovels if their total cost ends with several 9s. Because of this, Polycarp wants to choose a pair of shovels to sell in such a way that the sum of their costs ends with maximum possible number of nines. For example, if he chooses shovels with costs 12345 and 37454, their total cost is 49799, it ends with two nines.
You are to compute the number of pairs of shovels such that their total cost ends with maximum possible number of nines. Two pairs are considered different if there is a shovel presented in one pair, but not in the other.
Input
The first line contains a single integer n (2 ≤ n ≤ 109) — the number of shovels in Polycarp's shop.
Output
Print the number of pairs of shovels such that their total cost ends with maximum possible number of nines.
Note that it is possible that the largest number of 9s at the end is 0, then you should count all such ways.
It is guaranteed that for every n ≤ 109 the answer doesn't exceed 2·109.
Examples
Input
7
Output
3
Input
14
Output
9
Input
50
Output
1
Note
In the first example the maximum possible number of nines at the end is one. Polycarp cah choose the following pairs of shovels for that purpose:
* 2 and 7;
* 3 and 6;
* 4 and 5.
In the second example the maximum number of nines at the end of total cost of two shovels is one. The following pairs of shovels suit Polycarp:
* 1 and 8;
* 2 and 7;
* 3 and 6;
* 4 and 5;
* 5 and 14;
* 6 and 13;
* 7 and 12;
* 8 and 11;
* 9 and 10.
In the third example it is necessary to choose shovels 49 and 50, because the sum of their cost is 99, that means that the total number of nines is equal to two, which is maximum possible for n = 50.
Tags: constructive algorithms, math
Correct Solution:
```
from collections import defaultdict, deque, Counter
from sys import stdin, stdout
from heapq import heappush, heappop
import math
import io
import os
import math
import bisect
#?############################################################
def isPrime(x):
for i in range(2, x):
if i*i > x:
break
if (x % i == 0):
return False
return True
#?############################################################
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den, p - 2, p)) % p
#?############################################################
def primeFactors(n):
l = []
while n % 2 == 0:
l.append(2)
n = n / 2
for i in range(3, int(math.sqrt(n))+1, 2):
while n % i == 0:
l.append(int(i))
n = n / i
if n > 2:
l.append(n)
return list(set(l))
#?############################################################
def power(x, y, p):
res = 1
x = x % p
if (x == 0):
return 0
while (y > 0):
if ((y & 1) == 1):
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
#?############################################################
def sieve(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
#?############################################################
def digits(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
#?############################################################
def ceil(n, x):
if (n % x == 0):
return n//x
return n//x+1
#?############################################################
def mapin():
return map(int, input().split())
#?############################################################
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
# python3 15.py<in>op
n = int(input())
s = len(str(n))
if(n <5):
print(n*(n-1)//2)
elif(2*n == (10**s)-1):
print(1)
else:
nn = 5
while n >= nn * 10:
nn *= 10
# print(nn)
ans = 0
temp = nn
while temp <= n:
ans += min(n - (temp-1), temp - 1)
temp += nn
# print(temp)
print(ans)
```
| 105,836 |
Provide tags and a correct Python 2 solution for this coding contest problem.
There are n shovels in Polycarp's shop. The i-th shovel costs i burles, that is, the first shovel costs 1 burle, the second shovel costs 2 burles, the third shovel costs 3 burles, and so on. Polycarps wants to sell shovels in pairs.
Visitors are more likely to buy a pair of shovels if their total cost ends with several 9s. Because of this, Polycarp wants to choose a pair of shovels to sell in such a way that the sum of their costs ends with maximum possible number of nines. For example, if he chooses shovels with costs 12345 and 37454, their total cost is 49799, it ends with two nines.
You are to compute the number of pairs of shovels such that their total cost ends with maximum possible number of nines. Two pairs are considered different if there is a shovel presented in one pair, but not in the other.
Input
The first line contains a single integer n (2 ≤ n ≤ 109) — the number of shovels in Polycarp's shop.
Output
Print the number of pairs of shovels such that their total cost ends with maximum possible number of nines.
Note that it is possible that the largest number of 9s at the end is 0, then you should count all such ways.
It is guaranteed that for every n ≤ 109 the answer doesn't exceed 2·109.
Examples
Input
7
Output
3
Input
14
Output
9
Input
50
Output
1
Note
In the first example the maximum possible number of nines at the end is one. Polycarp cah choose the following pairs of shovels for that purpose:
* 2 and 7;
* 3 and 6;
* 4 and 5.
In the second example the maximum number of nines at the end of total cost of two shovels is one. The following pairs of shovels suit Polycarp:
* 1 and 8;
* 2 and 7;
* 3 and 6;
* 4 and 5;
* 5 and 14;
* 6 and 13;
* 7 and 12;
* 8 and 11;
* 9 and 10.
In the third example it is necessary to choose shovels 49 and 50, because the sum of their cost is 99, that means that the total number of nines is equal to two, which is maximum possible for n = 50.
Tags: constructive algorithms, math
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
def in_num():
return int(raw_input())
def in_arr():
return map(int,raw_input().split())
def pr_num(n):
stdout.write(str(n)+'\n')
def pr_arr(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return map(int,stdin.read().split())
range = xrange # not for python 3.0+
n=input()
x=5
while x<=n:
x*=10
x/=10
if n>=5:
ans=0
for i in range(x-1,n,x):
ans+=min(i,n-i)
pr_num(ans)
else:
ans=(n*(n-1))/2
pr_num(ans)
```
| 105,837 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n shovels in Polycarp's shop. The i-th shovel costs i burles, that is, the first shovel costs 1 burle, the second shovel costs 2 burles, the third shovel costs 3 burles, and so on. Polycarps wants to sell shovels in pairs.
Visitors are more likely to buy a pair of shovels if their total cost ends with several 9s. Because of this, Polycarp wants to choose a pair of shovels to sell in such a way that the sum of their costs ends with maximum possible number of nines. For example, if he chooses shovels with costs 12345 and 37454, their total cost is 49799, it ends with two nines.
You are to compute the number of pairs of shovels such that their total cost ends with maximum possible number of nines. Two pairs are considered different if there is a shovel presented in one pair, but not in the other.
Input
The first line contains a single integer n (2 ≤ n ≤ 109) — the number of shovels in Polycarp's shop.
Output
Print the number of pairs of shovels such that their total cost ends with maximum possible number of nines.
Note that it is possible that the largest number of 9s at the end is 0, then you should count all such ways.
It is guaranteed that for every n ≤ 109 the answer doesn't exceed 2·109.
Examples
Input
7
Output
3
Input
14
Output
9
Input
50
Output
1
Note
In the first example the maximum possible number of nines at the end is one. Polycarp cah choose the following pairs of shovels for that purpose:
* 2 and 7;
* 3 and 6;
* 4 and 5.
In the second example the maximum number of nines at the end of total cost of two shovels is one. The following pairs of shovels suit Polycarp:
* 1 and 8;
* 2 and 7;
* 3 and 6;
* 4 and 5;
* 5 and 14;
* 6 and 13;
* 7 and 12;
* 8 and 11;
* 9 and 10.
In the third example it is necessary to choose shovels 49 and 50, because the sum of their cost is 99, that means that the total number of nines is equal to two, which is maximum possible for n = 50.
Submitted Solution:
```
from math import factorial as fac
def solve(n):
if n <= 4:
return fac(n) // (2 * fac(n - 2))
m = n + (n - 1)
x = '9'
while(int(x + '9') <= m):
x += '9'
l = []
for i in range(10):
if int(str(i) + x) <= m:
l.append(int(str(i) + x))
res = 0
for p in l:
y = min(p - 1, n)
res += (y - (p - y) + 1) // 2
return res
n = int(input())
print(solve(n))
```
Yes
| 105,838 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n shovels in Polycarp's shop. The i-th shovel costs i burles, that is, the first shovel costs 1 burle, the second shovel costs 2 burles, the third shovel costs 3 burles, and so on. Polycarps wants to sell shovels in pairs.
Visitors are more likely to buy a pair of shovels if their total cost ends with several 9s. Because of this, Polycarp wants to choose a pair of shovels to sell in such a way that the sum of their costs ends with maximum possible number of nines. For example, if he chooses shovels with costs 12345 and 37454, their total cost is 49799, it ends with two nines.
You are to compute the number of pairs of shovels such that their total cost ends with maximum possible number of nines. Two pairs are considered different if there is a shovel presented in one pair, but not in the other.
Input
The first line contains a single integer n (2 ≤ n ≤ 109) — the number of shovels in Polycarp's shop.
Output
Print the number of pairs of shovels such that their total cost ends with maximum possible number of nines.
Note that it is possible that the largest number of 9s at the end is 0, then you should count all such ways.
It is guaranteed that for every n ≤ 109 the answer doesn't exceed 2·109.
Examples
Input
7
Output
3
Input
14
Output
9
Input
50
Output
1
Note
In the first example the maximum possible number of nines at the end is one. Polycarp cah choose the following pairs of shovels for that purpose:
* 2 and 7;
* 3 and 6;
* 4 and 5.
In the second example the maximum number of nines at the end of total cost of two shovels is one. The following pairs of shovels suit Polycarp:
* 1 and 8;
* 2 and 7;
* 3 and 6;
* 4 and 5;
* 5 and 14;
* 6 and 13;
* 7 and 12;
* 8 and 11;
* 9 and 10.
In the third example it is necessary to choose shovels 49 and 50, because the sum of their cost is 99, that means that the total number of nines is equal to two, which is maximum possible for n = 50.
Submitted Solution:
```
n = int(input())
if (n <= 4):
print(int(n * (n - 1) / 2))
else:
v = 9
dig = 10
x = n * 2 - 1
while (v + dig * 9 <= x):
v += dig * 9
dig *= 10
ans = 0
f = True
while (f):
f = False
y = v // 2
pairs = y - max(0, v - n - 1)
if (pairs > 0):
f = True
ans += pairs
v += dig
if (n * 2 - 1 < v):
break
print(ans)
```
Yes
| 105,839 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n shovels in Polycarp's shop. The i-th shovel costs i burles, that is, the first shovel costs 1 burle, the second shovel costs 2 burles, the third shovel costs 3 burles, and so on. Polycarps wants to sell shovels in pairs.
Visitors are more likely to buy a pair of shovels if their total cost ends with several 9s. Because of this, Polycarp wants to choose a pair of shovels to sell in such a way that the sum of their costs ends with maximum possible number of nines. For example, if he chooses shovels with costs 12345 and 37454, their total cost is 49799, it ends with two nines.
You are to compute the number of pairs of shovels such that their total cost ends with maximum possible number of nines. Two pairs are considered different if there is a shovel presented in one pair, but not in the other.
Input
The first line contains a single integer n (2 ≤ n ≤ 109) — the number of shovels in Polycarp's shop.
Output
Print the number of pairs of shovels such that their total cost ends with maximum possible number of nines.
Note that it is possible that the largest number of 9s at the end is 0, then you should count all such ways.
It is guaranteed that for every n ≤ 109 the answer doesn't exceed 2·109.
Examples
Input
7
Output
3
Input
14
Output
9
Input
50
Output
1
Note
In the first example the maximum possible number of nines at the end is one. Polycarp cah choose the following pairs of shovels for that purpose:
* 2 and 7;
* 3 and 6;
* 4 and 5.
In the second example the maximum number of nines at the end of total cost of two shovels is one. The following pairs of shovels suit Polycarp:
* 1 and 8;
* 2 and 7;
* 3 and 6;
* 4 and 5;
* 5 and 14;
* 6 and 13;
* 7 and 12;
* 8 and 11;
* 9 and 10.
In the third example it is necessary to choose shovels 49 and 50, because the sum of their cost is 99, that means that the total number of nines is equal to two, which is maximum possible for n = 50.
Submitted Solution:
```
n = int(input())
max9 = 1
while (int('9' * max9) + 1) // 2 <= n:
max9 += 1
max9 -= 1
k = 0
ans = 0
f = True
while f:
number = int(str(k) + '9' * max9)
b = min(number - 1, n)
a = number // 2 + 1
if a <= b:
m = b - a + 1
ans += m
k += 1
else:
f = False
if n == 2:
print(1)
elif n == 3:
print(3)
elif n == 4:
print(6)
else:
print(ans)
```
Yes
| 105,840 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n shovels in Polycarp's shop. The i-th shovel costs i burles, that is, the first shovel costs 1 burle, the second shovel costs 2 burles, the third shovel costs 3 burles, and so on. Polycarps wants to sell shovels in pairs.
Visitors are more likely to buy a pair of shovels if their total cost ends with several 9s. Because of this, Polycarp wants to choose a pair of shovels to sell in such a way that the sum of their costs ends with maximum possible number of nines. For example, if he chooses shovels with costs 12345 and 37454, their total cost is 49799, it ends with two nines.
You are to compute the number of pairs of shovels such that their total cost ends with maximum possible number of nines. Two pairs are considered different if there is a shovel presented in one pair, but not in the other.
Input
The first line contains a single integer n (2 ≤ n ≤ 109) — the number of shovels in Polycarp's shop.
Output
Print the number of pairs of shovels such that their total cost ends with maximum possible number of nines.
Note that it is possible that the largest number of 9s at the end is 0, then you should count all such ways.
It is guaranteed that for every n ≤ 109 the answer doesn't exceed 2·109.
Examples
Input
7
Output
3
Input
14
Output
9
Input
50
Output
1
Note
In the first example the maximum possible number of nines at the end is one. Polycarp cah choose the following pairs of shovels for that purpose:
* 2 and 7;
* 3 and 6;
* 4 and 5.
In the second example the maximum number of nines at the end of total cost of two shovels is one. The following pairs of shovels suit Polycarp:
* 1 and 8;
* 2 and 7;
* 3 and 6;
* 4 and 5;
* 5 and 14;
* 6 and 13;
* 7 and 12;
* 8 and 11;
* 9 and 10.
In the third example it is necessary to choose shovels 49 and 50, because the sum of their cost is 99, that means that the total number of nines is equal to two, which is maximum possible for n = 50.
Submitted Solution:
```
from collections import Counter
n=int(input())
if n<5:
d={2:1,3:3,4:6}
print(d[n])
exit()
z=len(str(n//5))
nn=5*10**(z-1)
n0=n-nn+1
if n0<nn:
print(n0)
elif n0==nn:
print(n0-1)
elif n0<=2*nn:
print(n0-1)
elif n0<3*nn:
print(n0*2-2*nn-1)
elif n0==3*nn:
print(n0*2-2*nn-2)
elif n0<=4*nn:
print(n0*2-2*nn-2)
elif n0<5*nn:
print(n0*3-6*nn-2)
elif n0==5*nn:
print(n0*3-6*nn-3)
elif n0<=6*nn:
print(n0*3-6*nn-3)
elif n0<7*nn:
print(n0*4-12*nn-3)
elif n0==8*nn:
print(n0*4-12*nn-4)
elif n0<=8*nn:
print(n0*4-12*nn-4)
elif n0<9*nn:
print(n0*5-20*nn-4)
elif n0==9*nn:
print(n0*5-20*nn-5)
```
Yes
| 105,841 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n shovels in Polycarp's shop. The i-th shovel costs i burles, that is, the first shovel costs 1 burle, the second shovel costs 2 burles, the third shovel costs 3 burles, and so on. Polycarps wants to sell shovels in pairs.
Visitors are more likely to buy a pair of shovels if their total cost ends with several 9s. Because of this, Polycarp wants to choose a pair of shovels to sell in such a way that the sum of their costs ends with maximum possible number of nines. For example, if he chooses shovels with costs 12345 and 37454, their total cost is 49799, it ends with two nines.
You are to compute the number of pairs of shovels such that their total cost ends with maximum possible number of nines. Two pairs are considered different if there is a shovel presented in one pair, but not in the other.
Input
The first line contains a single integer n (2 ≤ n ≤ 109) — the number of shovels in Polycarp's shop.
Output
Print the number of pairs of shovels such that their total cost ends with maximum possible number of nines.
Note that it is possible that the largest number of 9s at the end is 0, then you should count all such ways.
It is guaranteed that for every n ≤ 109 the answer doesn't exceed 2·109.
Examples
Input
7
Output
3
Input
14
Output
9
Input
50
Output
1
Note
In the first example the maximum possible number of nines at the end is one. Polycarp cah choose the following pairs of shovels for that purpose:
* 2 and 7;
* 3 and 6;
* 4 and 5.
In the second example the maximum number of nines at the end of total cost of two shovels is one. The following pairs of shovels suit Polycarp:
* 1 and 8;
* 2 and 7;
* 3 and 6;
* 4 and 5;
* 5 and 14;
* 6 and 13;
* 7 and 12;
* 8 and 11;
* 9 and 10.
In the third example it is necessary to choose shovels 49 and 50, because the sum of their cost is 99, that means that the total number of nines is equal to two, which is maximum possible for n = 50.
Submitted Solution:
```
n=int(input())
list=[]
2<=n<=109
for i in range(1,n+1):
for j in range(1,n+1):
x = i + j
if i!=j and x%10==9 and i<j:
list.append(i)
list.append(j)
a=len(list)
print(a//2)
```
No
| 105,842 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n shovels in Polycarp's shop. The i-th shovel costs i burles, that is, the first shovel costs 1 burle, the second shovel costs 2 burles, the third shovel costs 3 burles, and so on. Polycarps wants to sell shovels in pairs.
Visitors are more likely to buy a pair of shovels if their total cost ends with several 9s. Because of this, Polycarp wants to choose a pair of shovels to sell in such a way that the sum of their costs ends with maximum possible number of nines. For example, if he chooses shovels with costs 12345 and 37454, their total cost is 49799, it ends with two nines.
You are to compute the number of pairs of shovels such that their total cost ends with maximum possible number of nines. Two pairs are considered different if there is a shovel presented in one pair, but not in the other.
Input
The first line contains a single integer n (2 ≤ n ≤ 109) — the number of shovels in Polycarp's shop.
Output
Print the number of pairs of shovels such that their total cost ends with maximum possible number of nines.
Note that it is possible that the largest number of 9s at the end is 0, then you should count all such ways.
It is guaranteed that for every n ≤ 109 the answer doesn't exceed 2·109.
Examples
Input
7
Output
3
Input
14
Output
9
Input
50
Output
1
Note
In the first example the maximum possible number of nines at the end is one. Polycarp cah choose the following pairs of shovels for that purpose:
* 2 and 7;
* 3 and 6;
* 4 and 5.
In the second example the maximum number of nines at the end of total cost of two shovels is one. The following pairs of shovels suit Polycarp:
* 1 and 8;
* 2 and 7;
* 3 and 6;
* 4 and 5;
* 5 and 14;
* 6 and 13;
* 7 and 12;
* 8 and 11;
* 9 and 10.
In the third example it is necessary to choose shovels 49 and 50, because the sum of their cost is 99, that means that the total number of nines is equal to two, which is maximum possible for n = 50.
Submitted Solution:
```
t=1
#t=int(input())
for _ in range (t):
n=int(input())
#n,k=map(int,input().split())
#a=list(map(int,input().split()))
#b=list(map(int,input().split()))
#s=input()
#n=len(s)
c=len(str(n))
limit=10**c-1
if limit==n:
print((n+1)//2)
continue
#print(limit)
if limit-n<n:
ans=abs(limit-2*n)+1
print(ans//2)
else:
limit//=10
ans=(limit//2)+n-limit
print(ans)
```
No
| 105,843 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n shovels in Polycarp's shop. The i-th shovel costs i burles, that is, the first shovel costs 1 burle, the second shovel costs 2 burles, the third shovel costs 3 burles, and so on. Polycarps wants to sell shovels in pairs.
Visitors are more likely to buy a pair of shovels if their total cost ends with several 9s. Because of this, Polycarp wants to choose a pair of shovels to sell in such a way that the sum of their costs ends with maximum possible number of nines. For example, if he chooses shovels with costs 12345 and 37454, their total cost is 49799, it ends with two nines.
You are to compute the number of pairs of shovels such that their total cost ends with maximum possible number of nines. Two pairs are considered different if there is a shovel presented in one pair, but not in the other.
Input
The first line contains a single integer n (2 ≤ n ≤ 109) — the number of shovels in Polycarp's shop.
Output
Print the number of pairs of shovels such that their total cost ends with maximum possible number of nines.
Note that it is possible that the largest number of 9s at the end is 0, then you should count all such ways.
It is guaranteed that for every n ≤ 109 the answer doesn't exceed 2·109.
Examples
Input
7
Output
3
Input
14
Output
9
Input
50
Output
1
Note
In the first example the maximum possible number of nines at the end is one. Polycarp cah choose the following pairs of shovels for that purpose:
* 2 and 7;
* 3 and 6;
* 4 and 5.
In the second example the maximum number of nines at the end of total cost of two shovels is one. The following pairs of shovels suit Polycarp:
* 1 and 8;
* 2 and 7;
* 3 and 6;
* 4 and 5;
* 5 and 14;
* 6 and 13;
* 7 and 12;
* 8 and 11;
* 9 and 10.
In the third example it is necessary to choose shovels 49 and 50, because the sum of their cost is 99, that means that the total number of nines is equal to two, which is maximum possible for n = 50.
Submitted Solution:
```
from math import log
n = int(input())
if n<5:
print((n*(n-1))//2)
else:
s = int(log(n//5)/log(10))
c = 5*10**s
f = n//c+1
print((f//2)*((f+1)//2)*c-((f*c-n-1)*(f//2))-((n+1)//(10**(s+1))))
```
No
| 105,844 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n shovels in Polycarp's shop. The i-th shovel costs i burles, that is, the first shovel costs 1 burle, the second shovel costs 2 burles, the third shovel costs 3 burles, and so on. Polycarps wants to sell shovels in pairs.
Visitors are more likely to buy a pair of shovels if their total cost ends with several 9s. Because of this, Polycarp wants to choose a pair of shovels to sell in such a way that the sum of their costs ends with maximum possible number of nines. For example, if he chooses shovels with costs 12345 and 37454, their total cost is 49799, it ends with two nines.
You are to compute the number of pairs of shovels such that their total cost ends with maximum possible number of nines. Two pairs are considered different if there is a shovel presented in one pair, but not in the other.
Input
The first line contains a single integer n (2 ≤ n ≤ 109) — the number of shovels in Polycarp's shop.
Output
Print the number of pairs of shovels such that their total cost ends with maximum possible number of nines.
Note that it is possible that the largest number of 9s at the end is 0, then you should count all such ways.
It is guaranteed that for every n ≤ 109 the answer doesn't exceed 2·109.
Examples
Input
7
Output
3
Input
14
Output
9
Input
50
Output
1
Note
In the first example the maximum possible number of nines at the end is one. Polycarp cah choose the following pairs of shovels for that purpose:
* 2 and 7;
* 3 and 6;
* 4 and 5.
In the second example the maximum number of nines at the end of total cost of two shovels is one. The following pairs of shovels suit Polycarp:
* 1 and 8;
* 2 and 7;
* 3 and 6;
* 4 and 5;
* 5 and 14;
* 6 and 13;
* 7 and 12;
* 8 and 11;
* 9 and 10.
In the third example it is necessary to choose shovels 49 and 50, because the sum of their cost is 99, that means that the total number of nines is equal to two, which is maximum possible for n = 50.
Submitted Solution:
```
n=int(input())
if n<5:
print(0)
exit()
s0=str(n+n-1)
k=len(s0)
if s0!=k*'9':
k-=1
s=k*'9'
def cnt(s):
v=int(s)
#print(v)
if v>n*2-1:
return 0
if v==2*n-1:
return 1
if v>n:
return n-v//2
if v<=n:
return v//2
ans=cnt(s)
for i in range(1,9):
ans+=cnt(str(i)+s)
print(ans)
```
No
| 105,845 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today the North Pole hosts an Olympiad in a sport called... toy igloo skyscrapers' building!
There are n walruses taking part in the contest. Each walrus is given a unique number from 1 to n. After start each walrus begins to build his own igloo skyscraper. Initially, at the moment of time equal to 0, the height of the skyscraper i-th walrus is equal to ai. Each minute the i-th walrus finishes building bi floors.
The journalists that are reporting from the spot where the Olympiad is taking place, make q queries to the organizers. Each query is characterized by a group of three numbers li, ri, ti. The organizers respond to each query with a number x, such that:
1. Number x lies on the interval from li to ri inclusive (li ≤ x ≤ ri).
2. The skyscraper of the walrus number x possesses the maximum height among the skyscrapers of all walruses from the interval [li, ri] at the moment of time ti.
For each journalists' query print the number of the walrus x that meets the above-given criteria. If there are several possible answers, print any of them.
Input
The first line contains numbers n and q (1 ≤ n, q ≤ 105). Next n lines contain pairs of numbers ai, bi (1 ≤ ai, bi ≤ 109). Then follow q queries i the following format li, ri, ti, one per each line (1 ≤ li ≤ ri ≤ n, 0 ≤ ti ≤ 106). All input numbers are integers.
Output
For each journalists' query print the number of the walrus x that meets the criteria, given in the statement. Print one number per line.
Examples
Input
5 4
4 1
3 5
6 2
3 5
6 5
1 5 2
1 3 5
1 1 0
1 5 0
Output
5
2
1
5
Input
5 4
6 1
5 1
2 5
4 3
6 1
2 4 1
3 4 5
1 4 5
1 2 0
Output
3
3
3
1
Submitted Solution:
```
from sys import stdin, stdout
EPS = 10 ** (-20)
def merge(ind):
if tree[ind * 2][0][1] > tree[ind * 2 + 1][0][1]:
tree[ind].append(tree[ind * 2][0])
i = 1
j = 0
elif tree[ind * 2][0][1] < tree[ind * 2 + 1][0][1]:
tree[ind].append(tree[ind * 2 + 1][0])
i = 0
j = 1
elif tree[ind * 2][0][2] > tree[ind * 2 + 1][0][2]:
tree[ind].append(tree[ind * 2][0])
i = 1
j = 1
else:
tree[ind].append(tree[ind * 2 + 1][0])
i = 1
j = 1
time = 10 ** 6 + 6
while i < len(tree[ind * 2]) and tree[ind * 2][i][2] <= tree[ind][-1][2] and tree[ind][-1][1] + tree[ind][-1][2] * time >= tree[ind * 2][i][1] + tree[ind * 2][i][2] * time:
i += 1
while j < len(tree[ind * 2 + 1]) and tree[ind * 2 + 1][j][2] <= tree[ind][-1][2] and tree[ind][-1][1] + tree[ind][-1][2] * time >= tree[ind * 2 + 1][j][1] + tree[ind * 2 + 1][j][2] * time:
j += 1
while i < len(tree[ind * 2]) and j < len(tree[ind * 2 + 1]):
t1, t2 = (tree[ind][-1][1] - tree[ind * 2][i][1]) / (tree[ind * 2][i][2] - tree[ind][-1][2]), (tree[ind][-1][1] - tree[ind * 2 + 1][j][1]) / (tree[ind * 2 + 1][j][2] - tree[ind][-1][2])
if t1 < t2:
tree[ind].append((t1, tree[ind * 2][i][1], tree[ind * 2][i][2], tree[ind * 2][i][3]))
i += 1
else:
tree[ind].append((t2, tree[ind * 2 + 1][j][1], tree[ind * 2 + 1][j][2], tree[ind * 2 + 1][j][3]))
j += 1
while i < len(tree[ind * 2]) and tree[ind * 2][i][2] <= tree[ind][-1][2] and tree[ind][-1][1] + tree[ind][-1][2] * time >= tree[ind * 2][i][1] + tree[ind * 2][i][2] * time:
i += 1
while j < len(tree[ind * 2 + 1]) and tree[ind * 2 + 1][j][2] <= tree[ind][-1][2] and tree[ind][-1][1] + tree[ind][-1][2] * time >= tree[ind * 2 + 1][j][1] + tree[ind * 2 + 1][j][2] * time:
j += 1
while i < len(tree[ind * 2]):
if (tree[ind * 2][i][2] > tree[ind][-1][2] or tree[ind * 2][i][1] + tree[ind * 2][i][2] * time > tree[ind][-1][1] + tree[ind][-1][2] * time):
tree[ind].append(((tree[ind][-1][1] - tree[ind * 2][i][1]) / (tree[ind * 2][i][2] - tree[ind][-1][2]), tree[ind * 2][i][1], tree[ind * 2][i][2], tree[ind * 2][i][3]))
i += 1
while j < len(tree[ind * 2 + 1]):
if (tree[ind * 2 + 1][j][2] > tree[ind][-1][2] or tree[ind * 2 + 1][j][1] + tree[ind * 2 + 1][j][2] * time > tree[ind][-1][1] + tree[ind][-1][2] * time):
tree[ind].append(((tree[ind][-1][1] - tree[ind * 2 + 1][j][1]) / (tree[ind * 2 + 1][j][2] - tree[ind][-1][2]), tree[ind * 2 + 1][j][1], tree[ind * 2 + 1][j][2], tree[ind * 2 + 1][j][3]))
j += 1
def get(l, r, lb, rb, ind, time):
if l == lb and r == rb:
l, r = -1, len(tree[ind])
while r - l > 1:
m = (r + l) // 2
if tree[ind][m][0] <= time + EPS:
l = m
else:
r = m
if l == -1:
return 0
else:
return tree[ind][l]
m = (lb + rb) // 2
first, second = 0, 0
if l <= m:
first = get(l, min(m, r), lb, m, ind * 2, time)
if r > m:
second = get(max(l, m + 1), r, m + 1, rb, ind * 2 + 1, time)
if not first:
return second
elif not second:
return first
else:
return max(second, first, key = lambda x: x[1] + x[2] * time)
n, q = map(int, stdin.readline().split())
start = []
rise = []
for i in range(n):
a, b = map(int, stdin.readline().split())
start.append(a)
rise.append(b)
power = 1
while power <= n:
power *= 2
tree = [[] for i in range(power * 2)]
for i in range(n):
tree[power + i].append((0, start[i], rise[i], i + 1))
for i in range(power - 1, 0, -1):
if not tree[i * 2]:
tree[i] = tree[i * 2 + 1]
elif not tree[i * 2 + 1]:
tree[i] = tree[i * 2]
else:
merge(i)
for i in range(q):
l, r, t = map(int, stdin.readline().split())
stdout.write(str(get(l, r, 1, power, 1, t)[3]) + '\n')
```
No
| 105,846 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today the North Pole hosts an Olympiad in a sport called... toy igloo skyscrapers' building!
There are n walruses taking part in the contest. Each walrus is given a unique number from 1 to n. After start each walrus begins to build his own igloo skyscraper. Initially, at the moment of time equal to 0, the height of the skyscraper i-th walrus is equal to ai. Each minute the i-th walrus finishes building bi floors.
The journalists that are reporting from the spot where the Olympiad is taking place, make q queries to the organizers. Each query is characterized by a group of three numbers li, ri, ti. The organizers respond to each query with a number x, such that:
1. Number x lies on the interval from li to ri inclusive (li ≤ x ≤ ri).
2. The skyscraper of the walrus number x possesses the maximum height among the skyscrapers of all walruses from the interval [li, ri] at the moment of time ti.
For each journalists' query print the number of the walrus x that meets the above-given criteria. If there are several possible answers, print any of them.
Input
The first line contains numbers n and q (1 ≤ n, q ≤ 105). Next n lines contain pairs of numbers ai, bi (1 ≤ ai, bi ≤ 109). Then follow q queries i the following format li, ri, ti, one per each line (1 ≤ li ≤ ri ≤ n, 0 ≤ ti ≤ 106). All input numbers are integers.
Output
For each journalists' query print the number of the walrus x that meets the criteria, given in the statement. Print one number per line.
Examples
Input
5 4
4 1
3 5
6 2
3 5
6 5
1 5 2
1 3 5
1 1 0
1 5 0
Output
5
2
1
5
Input
5 4
6 1
5 1
2 5
4 3
6 1
2 4 1
3 4 5
1 4 5
1 2 0
Output
3
3
3
1
Submitted Solution:
```
from sys import stdin, stdout
from decimal import Decimal
EPS = Decimal(10) ** Decimal(-30)
def merge(ind):
if tree[ind * 2][0][1] > tree[ind * 2 + 1][0][1]:
tree[ind].append(tree[ind * 2][0])
i = 1
j = 0
elif tree[ind * 2][0][1] < tree[ind * 2 + 1][0][1]:
tree[ind].append(tree[ind * 2 + 1][0])
i = 0
j = 1
elif tree[ind * 2][0][2] > tree[ind * 2 + 1][0][2]:
tree[ind].append(tree[ind * 2][0])
i = 1
j = 1
else:
tree[ind].append(tree[ind * 2 + 1][0])
i = 1
j = 1
time = Decimal(10 ** 6 + 6)
while i < len(tree[ind * 2]) and tree[ind * 2][i][2] <= tree[ind][-1][2] and tree[ind][-1][1] + tree[ind][-1][2] * time >= tree[ind * 2][i][1] + tree[ind * 2][i][2] * time:
i += 1
while j < len(tree[ind * 2 + 1]) and tree[ind * 2 + 1][j][2] <= tree[ind][-1][2] and tree[ind][-1][1] + tree[ind][-1][2] * time >= tree[ind * 2 + 1][j][1] + tree[ind * 2 + 1][j][2] * time:
j += 1
while i < len(tree[ind * 2]) and j < len(tree[ind * 2 + 1]):
t1, t2 = (tree[ind][-1][1] - tree[ind * 2][i][1]) / (tree[ind * 2][i][2] - tree[ind][-1][2]), (tree[ind][-1][1] - tree[ind * 2 + 1][j][1]) / (tree[ind * 2 + 1][j][2] - tree[ind][-1][2])
if t1 < t2:
tree[ind].append((t1, tree[ind * 2][i][1], tree[ind * 2][i][2], tree[ind * 2][i][3]))
i += 1
else:
tree[ind].append((t2, tree[ind * 2 + 1][j][1], tree[ind * 2 + 1][j][2], tree[ind * 2 + 1][j][3]))
j += 1
while i < len(tree[ind * 2]) and tree[ind * 2][i][2] <= tree[ind][-1][2] and tree[ind][-1][1] + tree[ind][-1][2] * time >= tree[ind * 2][i][1] + tree[ind * 2][i][2] * time:
i += 1
while j < len(tree[ind * 2 + 1]) and tree[ind * 2 + 1][j][2] <= tree[ind][-1][2] and tree[ind][-1][1] + tree[ind][-1][2] * time >= tree[ind * 2 + 1][j][1] + tree[ind * 2 + 1][j][2] * time:
j += 1
while i < len(tree[ind * 2]):
if (tree[ind * 2][i][2] > tree[ind][-1][2] or tree[ind * 2][i][1] + tree[ind * 2][i][2] * time > tree[ind][-1][1] + tree[ind][-1][2] * time):
tree[ind].append(((tree[ind][-1][1] - tree[ind * 2][i][1]) / (tree[ind * 2][i][2] - tree[ind][-1][2]), tree[ind * 2][i][1], tree[ind * 2][i][2], tree[ind * 2][i][3]))
i += 1
while j < len(tree[ind * 2 + 1]):
if (tree[ind * 2 + 1][j][2] > tree[ind][-1][2] or tree[ind * 2 + 1][j][1] + tree[ind * 2 + 1][j][2] * time > tree[ind][-1][1] + tree[ind][-1][2] * time):
tree[ind].append(((tree[ind][-1][1] - tree[ind * 2 + 1][j][1]) / (tree[ind * 2 + 1][j][2] - tree[ind][-1][2]), tree[ind * 2 + 1][j][1], tree[ind * 2 + 1][j][2], tree[ind * 2 + 1][j][3]))
j += 1
def get(l, r, lb, rb, ind, time):
if l == lb and r == rb:
l, r = -1, len(tree[ind])
while r - l > 1:
m = (r + l) // 2
if tree[ind][m][0] <= time + EPS:
l = m
else:
r = m
if l == -1:
return 0
else:
return tree[ind][l]
m = (lb + rb) // 2
first, second = 0, 0
if l <= m:
first = get(l, min(m, r), lb, m, ind * 2, time)
if r > m:
second = get(max(l, m + 1), r, m + 1, rb, ind * 2 + 1, time)
if not first:
return second
elif not second:
return first
else:
return max(second, first, key = lambda x: (x[1] + x[2] * time, x[2]))
n, q = map(int, stdin.readline().split())
start = []
rise = []
for i in range(n):
a, b = map(int, stdin.readline().split())
start.append(a)
rise.append(b)
power = 1
while power <= n:
power *= 2
tree = [[] for i in range(power * 2)]
for i in range(n):
tree[power + i].append((Decimal(0), Decimal(start[i]), Decimal(rise[i]), i + 1))
for i in range(power - 1, 0, -1):
if not tree[i * 2]:
tree[i] = tree[i * 2 + 1]
elif not tree[i * 2 + 1]:
tree[i] = tree[i * 2]
else:
merge(i)
for i in range(q):
l, r, t = map(int, stdin.readline().split())
stdout.write(str(get(l, r, 1, power, 1, t)[3]) + '\n')
```
No
| 105,847 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today the North Pole hosts an Olympiad in a sport called... toy igloo skyscrapers' building!
There are n walruses taking part in the contest. Each walrus is given a unique number from 1 to n. After start each walrus begins to build his own igloo skyscraper. Initially, at the moment of time equal to 0, the height of the skyscraper i-th walrus is equal to ai. Each minute the i-th walrus finishes building bi floors.
The journalists that are reporting from the spot where the Olympiad is taking place, make q queries to the organizers. Each query is characterized by a group of three numbers li, ri, ti. The organizers respond to each query with a number x, such that:
1. Number x lies on the interval from li to ri inclusive (li ≤ x ≤ ri).
2. The skyscraper of the walrus number x possesses the maximum height among the skyscrapers of all walruses from the interval [li, ri] at the moment of time ti.
For each journalists' query print the number of the walrus x that meets the above-given criteria. If there are several possible answers, print any of them.
Input
The first line contains numbers n and q (1 ≤ n, q ≤ 105). Next n lines contain pairs of numbers ai, bi (1 ≤ ai, bi ≤ 109). Then follow q queries i the following format li, ri, ti, one per each line (1 ≤ li ≤ ri ≤ n, 0 ≤ ti ≤ 106). All input numbers are integers.
Output
For each journalists' query print the number of the walrus x that meets the criteria, given in the statement. Print one number per line.
Examples
Input
5 4
4 1
3 5
6 2
3 5
6 5
1 5 2
1 3 5
1 1 0
1 5 0
Output
5
2
1
5
Input
5 4
6 1
5 1
2 5
4 3
6 1
2 4 1
3 4 5
1 4 5
1 2 0
Output
3
3
3
1
Submitted Solution:
```
from sys import stdin, stdout
EPS = 10 ** (-20)
def merge(ind):
if tree[ind * 2][0][1] > tree[ind * 2 + 1][0][1]:
tree[ind].append(tree[ind * 2][0])
i = 1
j = 0
elif tree[ind * 2][0][1] < tree[ind * 2 + 1][0][1]:
tree[ind].append(tree[ind * 2 + 1][0])
i = 0
j = 1
elif tree[ind * 2][0][2] > tree[ind * 2 + 1][0][2]:
tree[ind].append(tree[ind * 2][0])
i = 1
j = 1
else:
tree[ind].append(tree[ind * 2 + 1][0])
i = 1
j = 1
time = 10 ** 6 + 6
while i < len(tree[ind * 2]) and tree[ind * 2][i][2] <= tree[ind][-1][2] and tree[ind][-1][1] + tree[ind][-1][2] * time >= tree[ind * 2][i][1] + tree[ind * 2][i][2] * time:
i += 1
while j < len(tree[ind * 2 + 1]) and tree[ind * 2 + 1][j][2] <= tree[ind][-1][2] and tree[ind][-1][1] + tree[ind][-1][2] * time >= tree[ind * 2 + 1][j][1] + tree[ind * 2 + 1][j][2] * time:
j += 1
while i < len(tree[ind * 2]) and j < len(tree[ind * 2 + 1]):
t1, t2 = (tree[ind][-1][1] - tree[ind * 2][i][1]) / (tree[ind * 2][i][2] - tree[ind][-1][2]), (tree[ind][-1][1] - tree[ind * 2 + 1][j][1]) / (tree[ind * 2 + 1][j][2] - tree[ind][-1][2])
if t1 < t2:
tree[ind].append((t1, tree[ind * 2][i][1], tree[ind * 2][i][2], tree[ind * 2][i][3]))
i += 1
else:
tree[ind].append((t2, tree[ind * 2 + 1][j][1], tree[ind * 2 + 1][j][2], tree[ind * 2 + 1][j][3]))
j += 1
while i < len(tree[ind * 2]) and tree[ind * 2][i][2] <= tree[ind][-1][2] and tree[ind][-1][1] + tree[ind][-1][2] * time >= tree[ind * 2][i][1] + tree[ind * 2][i][2] * time:
i += 1
while j < len(tree[ind * 2 + 1]) and tree[ind * 2 + 1][j][2] <= tree[ind][-1][2] and tree[ind][-1][1] + tree[ind][-1][2] * time >= tree[ind * 2 + 1][j][1] + tree[ind * 2 + 1][j][2] * time:
j += 1
while i < len(tree[ind * 2]):
if (tree[ind * 2][i][2] > tree[ind][-1][2] or tree[ind * 2][i][1] + tree[ind * 2][i][2] * time > tree[ind][-1][1] + tree[ind][-1][2] * time):
tree[ind].append(((tree[ind][-1][1] - tree[ind * 2][i][1]) / (tree[ind * 2][i][2] - tree[ind][-1][2]), tree[ind * 2][i][1], tree[ind * 2][i][2], tree[ind * 2][i][3]))
i += 1
while j < len(tree[ind * 2 + 1]):
if (tree[ind * 2 + 1][j][2] > tree[ind][-1][2] or tree[ind * 2 + 1][j][1] + tree[ind * 2 + 1][j][2] * time > tree[ind][-1][1] + tree[ind][-1][2] * time):
tree[ind].append(((tree[ind][-1][1] - tree[ind * 2 + 1][j][1]) / (tree[ind * 2 + 1][j][2] - tree[ind][-1][2]), tree[ind * 2 + 1][j][1], tree[ind * 2 + 1][j][2], tree[ind * 2 + 1][j][3]))
j += 1
def get(l, r, lb, rb, ind, time):
if l == lb and r == rb:
l, r = -1, len(tree[ind])
while r - l > 1:
m = (r + l) // 2
if tree[ind][m][0] <= time + EPS:
l = m
else:
r = m
if l == -1:
return 0
else:
return tree[ind][l]
m = (lb + rb) // 2
first, second = 0, 0
if l <= m:
first = get(l, min(m, r), lb, m, ind * 2, time)
if r > m:
second = get(max(l, m + 1), r, m + 1, rb, ind * 2 + 1, time)
if not first:
return second
elif not second:
return first
else:
return max(second, first, key = lambda x: (x[1] + x[2] * time, x[2]))
n, q = map(int, stdin.readline().split())
start = []
rise = []
for i in range(n):
a, b = map(int, stdin.readline().split())
start.append(a)
rise.append(b)
power = 1
while power <= n:
power *= 2
tree = [[] for i in range(power * 2)]
for i in range(n):
tree[power + i].append((0, start[i], rise[i], i + 1))
for i in range(power - 1, 0, -1):
if not tree[i * 2]:
tree[i] = tree[i * 2 + 1]
elif not tree[i * 2 + 1]:
tree[i] = tree[i * 2]
else:
merge(i)
for i in range(q):
l, r, t = map(int, stdin.readline().split())
stdout.write(str(get(l, r, 1, power, 1, t)[3]) + '\n')
```
No
| 105,848 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today the North Pole hosts an Olympiad in a sport called... toy igloo skyscrapers' building!
There are n walruses taking part in the contest. Each walrus is given a unique number from 1 to n. After start each walrus begins to build his own igloo skyscraper. Initially, at the moment of time equal to 0, the height of the skyscraper i-th walrus is equal to ai. Each minute the i-th walrus finishes building bi floors.
The journalists that are reporting from the spot where the Olympiad is taking place, make q queries to the organizers. Each query is characterized by a group of three numbers li, ri, ti. The organizers respond to each query with a number x, such that:
1. Number x lies on the interval from li to ri inclusive (li ≤ x ≤ ri).
2. The skyscraper of the walrus number x possesses the maximum height among the skyscrapers of all walruses from the interval [li, ri] at the moment of time ti.
For each journalists' query print the number of the walrus x that meets the above-given criteria. If there are several possible answers, print any of them.
Input
The first line contains numbers n and q (1 ≤ n, q ≤ 105). Next n lines contain pairs of numbers ai, bi (1 ≤ ai, bi ≤ 109). Then follow q queries i the following format li, ri, ti, one per each line (1 ≤ li ≤ ri ≤ n, 0 ≤ ti ≤ 106). All input numbers are integers.
Output
For each journalists' query print the number of the walrus x that meets the criteria, given in the statement. Print one number per line.
Examples
Input
5 4
4 1
3 5
6 2
3 5
6 5
1 5 2
1 3 5
1 1 0
1 5 0
Output
5
2
1
5
Input
5 4
6 1
5 1
2 5
4 3
6 1
2 4 1
3 4 5
1 4 5
1 2 0
Output
3
3
3
1
Submitted Solution:
```
from sys import stdin, stdout
EPS = 10 ** (-20)
def merge(ind):
if tree[ind * 2][0][1] > tree[ind * 2 + 1][0][1]:
tree[ind].append(tree[ind * 2][0])
i = 1
j = 0
elif tree[ind * 2][0][1] < tree[ind * 2 + 1][0][1]:
tree[ind].append(tree[ind * 2 + 1][0])
i = 0
j = 1
elif tree[ind * 2][0][2] > tree[ind * 2 + 1][0][2]:
tree[ind].append(tree[ind * 2][0])
i = 1
j = 1
else:
tree[ind].append(tree[ind * 2 + 1][0])
i = 1
j = 1
while i < len(tree[ind * 2]) and (tree[ind * 2][i][2] - tree[ind][-1][2]):
time = (tree[ind][-1][1] - tree[ind * 2][i][1]) / (tree[ind * 2][i][2] - tree[ind][-1][2]) + 5
if tree[ind][-1][1] + tree[ind][-1][2] * time >= tree[ind * 2][i][1] + tree[ind * 2][i][2] * time:
i += 1
else:
break
while j < len(tree[ind * 2 + 1]) and (tree[ind * 2 + 1][j][2] - tree[ind][-1][2]):
time = (tree[ind][-1][1] - tree[ind * 2 + 1][j][1]) / (tree[ind * 2 + 1][j][2] - tree[ind][-1][2]) + 5
if tree[ind][-1][1] + tree[ind][-1][2] * time >= tree[ind * 2 + 1][j][1] + tree[ind * 2 + 1][j][2] * time:
j += 1
else:
break
while i < len(tree[ind * 2]) and j < len(tree[ind * 2 + 1]):
t1, t2 = (tree[ind][-1][1] - tree[ind * 2][i][1]) / (tree[ind * 2][i][2] - tree[ind][-1][2] + EPS), (tree[ind][-1][1] - tree[ind * 2 + 1][j][1]) / (tree[ind * 2 + 1][j][2] - tree[ind][-1][2] + EPS)
if t1 < t2:
tree[ind].append((t1, tree[ind * 2][i][1], tree[ind * 2][i][2], tree[ind * 2][i][3]))
i += 1
else:
tree[ind].append((t2, tree[ind * 2 + 1][j][1], tree[ind * 2 + 1][j][2], tree[ind * 2 + 1][j][3]))
j += 1
while i < len(tree[ind * 2]) and (tree[ind * 2][i][2] - tree[ind][-1][2]):
time = (tree[ind][-1][1] - tree[ind * 2][i][1]) / (tree[ind * 2][i][2] - tree[ind][-1][2]) + 5
if tree[ind][-1][1] + tree[ind][-1][2] * time >= tree[ind * 2][i][1] + tree[ind * 2][i][2] * time:
i += 1
else:
break
while j < len(tree[ind * 2 + 1]) and (tree[ind * 2 + 1][j][2] - tree[ind][-1][2]):
time = (tree[ind][-1][1] - tree[ind * 2 + 1][j][1]) / (tree[ind * 2 + 1][j][2] - tree[ind][-1][2]) + 5
if tree[ind][-1][1] + tree[ind][-1][2] * time >= tree[ind * 2 + 1][j][1] + tree[ind * 2 + 1][j][2] * time:
j += 1
else:
break
while i < len(tree[ind * 2]) and tree[ind * 2][i][2] > tree[ind][-1][2]:
tree[ind].append(((tree[ind][-1][1] - tree[ind * 2][i][1]) / (tree[ind * 2][i][2] - tree[ind][-1][2] + EPS), tree[ind * 2][i][1], tree[ind * 2][i][2], tree[ind * 2][i][3]))
i += 1
while j < len(tree[ind * 2 + 1]) and tree[ind * 2 + 1][j][2] > tree[ind][-1][2]:
tree[ind].append(((tree[ind][-1][1] - tree[ind * 2 + 1][j][1]) / (tree[ind * 2 + 1][j][2] - tree[ind][-1][2]) + EPS, tree[ind * 2 + 1][j][1], tree[ind * 2 + 1][j][2], tree[ind * 2 + 1][j][3]))
j += 1
def get(l, r, lb, rb, ind, time):
if l == lb and r == rb:
l, r = -1, len(tree[ind])
while r - l > 1:
m = (r + l) // 2
if tree[ind][m][0] <= time + EPS:
l = m
else:
r = m
if l == -1:
return 0
else:
return tree[ind][l]
m = (lb + rb) // 2
first, second = 0, 0
if l <= m:
first = get(l, min(m, r), lb, m, ind * 2, time)
if r > m:
second = get(max(l, m + 1), r, m + 1, rb, ind * 2 + 1, time)
if not first:
return second
elif not second:
return first
else:
return max(second, first, key = lambda x: x[1] + x[2] * time)
n, q = map(int, stdin.readline().split())
start = []
rise = []
for i in range(n):
a, b = map(int, stdin.readline().split())
start.append(a)
rise.append(b)
power = 1
while power <= n:
power *= 2
tree = [[] for i in range(power * 2)]
for i in range(n):
tree[power + i].append((0, start[i], rise[i], i + 1))
for i in range(power - 1, 0, -1):
if not tree[i * 2]:
tree[i] = tree[i * 2 + 1]
elif not tree[i * 2 + 1]:
tree[i] = tree[i * 2]
else:
merge(i)
for i in range(q):
l, r, t = map(int, stdin.readline().split())
stdout.write(str(get(l, r, 1, power, 1, t)[3]) + '\n')
```
No
| 105,849 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice likes snow a lot! Unfortunately, this year's winter is already over, and she can't expect to have any more of it. Bob has thus bought her a gift — a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.
Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other.
Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.
You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day.
Input
The first line contains a single integer N (1 ≤ N ≤ 105) — the number of days.
The second line contains N integers V1, V2, ..., VN (0 ≤ Vi ≤ 109), where Vi is the initial size of a snow pile made on the day i.
The third line contains N integers T1, T2, ..., TN (0 ≤ Ti ≤ 109), where Ti is the temperature on the day i.
Output
Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.
Examples
Input
3
10 10 5
5 7 2
Output
5 12 4
Input
5
30 25 20 15 10
9 10 12 4 13
Output
9 20 35 11 25
Note
In the first sample, Bob first makes a snow pile of volume 10, which melts to the size of 5 on the same day. On the second day, he makes another pile of size 10. Since it is a bit warmer than the day before, the first pile disappears completely while the second pile shrinks to 3. At the end of the second day, he has only a single pile of size 3. On the third day he makes a smaller pile than usual, but as the temperature dropped too, both piles survive till the end of the day.
Tags: binary search, data structures
Correct Solution:
```
import itertools
import operator
from bisect import bisect
def main():
N = int(input())
V = tuple(map(int, input().split()))
T = tuple(map(int, input().split()))
templ = [0] + list(itertools.accumulate(T, operator.add))
M = [0] * (N + 2)
lost = M[:]
for j, v in enumerate(V):
i = bisect(templ, v + templ[j])
M[i] += v + templ[j] - templ[i-1]
lost[i] += 1
F = [0] * (N + 1)
for i in range(N):
F[i+1] += F[i] + 1 - lost[i+1]
print(*[F[i] * T[i-1] + M[i] for i in range(1, N + 1)], sep=' ')
main()
```
| 105,850 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice likes snow a lot! Unfortunately, this year's winter is already over, and she can't expect to have any more of it. Bob has thus bought her a gift — a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.
Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other.
Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.
You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day.
Input
The first line contains a single integer N (1 ≤ N ≤ 105) — the number of days.
The second line contains N integers V1, V2, ..., VN (0 ≤ Vi ≤ 109), where Vi is the initial size of a snow pile made on the day i.
The third line contains N integers T1, T2, ..., TN (0 ≤ Ti ≤ 109), where Ti is the temperature on the day i.
Output
Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.
Examples
Input
3
10 10 5
5 7 2
Output
5 12 4
Input
5
30 25 20 15 10
9 10 12 4 13
Output
9 20 35 11 25
Note
In the first sample, Bob first makes a snow pile of volume 10, which melts to the size of 5 on the same day. On the second day, he makes another pile of size 10. Since it is a bit warmer than the day before, the first pile disappears completely while the second pile shrinks to 3. At the end of the second day, he has only a single pile of size 3. On the third day he makes a smaller pile than usual, but as the temperature dropped too, both piles survive till the end of the day.
Tags: binary search, data structures
Correct Solution:
```
import math,sys,bisect,heapq
from collections import defaultdict,Counter,deque
from itertools import groupby,accumulate
#sys.setrecursionlimit(200000000)
int1 = lambda x: int(x) - 1
input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
ilele = lambda: map(int,input().split())
alele = lambda: list(map(int, input().split()))
ilelec = lambda: map(int1,input().split())
alelec = lambda: list(map(int1, input().split()))
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
#MOD = 1000000000 + 7
def Y(c): print(["NO","YES"][c])
def y(c): print(["no","yes"][c])
def Yy(c): print(["No","Yes"][c])
N = int(input())
V = alele()
T= alele()
B = [0] + list(accumulate(T))
Ans1 = [0]*(N+2)
Ans2 = [0]*(N+2)
#print(B)
for i in range(N):
x = V[i]
if T[i] < V[i]:
a = bisect.bisect_left(B,x + B[i])
#print(x+B[i],a)
Ans2[i] += 1
#print(i,a)
Ans2[a-1] -= 1
if T[i] < V[i]:
z = B[a-1] - B[i]
zz = V[i] -z
Ans1[a-1] +=zz
#print(zz)
#print(Ans2)
#print(Ans1)
else:
Ans1[i] += V[i]
C = list(accumulate(Ans2))
#print(C)
#print(Ans1)
for i in range(N):
print((C[i]*T[i]) + Ans1[i],end = ' ')
```
| 105,851 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice likes snow a lot! Unfortunately, this year's winter is already over, and she can't expect to have any more of it. Bob has thus bought her a gift — a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.
Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other.
Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.
You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day.
Input
The first line contains a single integer N (1 ≤ N ≤ 105) — the number of days.
The second line contains N integers V1, V2, ..., VN (0 ≤ Vi ≤ 109), where Vi is the initial size of a snow pile made on the day i.
The third line contains N integers T1, T2, ..., TN (0 ≤ Ti ≤ 109), where Ti is the temperature on the day i.
Output
Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.
Examples
Input
3
10 10 5
5 7 2
Output
5 12 4
Input
5
30 25 20 15 10
9 10 12 4 13
Output
9 20 35 11 25
Note
In the first sample, Bob first makes a snow pile of volume 10, which melts to the size of 5 on the same day. On the second day, he makes another pile of size 10. Since it is a bit warmer than the day before, the first pile disappears completely while the second pile shrinks to 3. At the end of the second day, he has only a single pile of size 3. On the third day he makes a smaller pile than usual, but as the temperature dropped too, both piles survive till the end of the day.
Tags: binary search, data structures
Correct Solution:
```
import heapq
heap = []
n = int(input())
V = list(map(int, input().split()))
T = list(map(int, input().split()))
tmp = 0
for i in range(n):
ans = 0
heapq.heappush(heap, tmp+V[i])
while len(heap) and heap[0]<=tmp+T[i]:
ans += heapq.heappop(heap)-tmp
tmp += T[i]
ans += T[i]*len(heap)
print(ans, end=' ')
```
| 105,852 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice likes snow a lot! Unfortunately, this year's winter is already over, and she can't expect to have any more of it. Bob has thus bought her a gift — a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.
Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other.
Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.
You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day.
Input
The first line contains a single integer N (1 ≤ N ≤ 105) — the number of days.
The second line contains N integers V1, V2, ..., VN (0 ≤ Vi ≤ 109), where Vi is the initial size of a snow pile made on the day i.
The third line contains N integers T1, T2, ..., TN (0 ≤ Ti ≤ 109), where Ti is the temperature on the day i.
Output
Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.
Examples
Input
3
10 10 5
5 7 2
Output
5 12 4
Input
5
30 25 20 15 10
9 10 12 4 13
Output
9 20 35 11 25
Note
In the first sample, Bob first makes a snow pile of volume 10, which melts to the size of 5 on the same day. On the second day, he makes another pile of size 10. Since it is a bit warmer than the day before, the first pile disappears completely while the second pile shrinks to 3. At the end of the second day, he has only a single pile of size 3. On the third day he makes a smaller pile than usual, but as the temperature dropped too, both piles survive till the end of the day.
Tags: binary search, data structures
Correct Solution:
```
n = int(input())
import heapq as hq
heap = []
# total = 0
temp = 0
ans = [-1 for _ in range(n)]
volumes = [int(x) for x in input().split()]
temps = [int(x) for x in input().split()]
# prevtemp = 0
for i in range(n):
# hq.heappush(heap, volumes[i])
prevtemp = temp
temp += temps[i]
hq.heappush(heap, volumes[i] + prevtemp)
# total += 1
curr = 0
while len(heap) and heap[0] <= temp:
m = hq.heappop(heap)
curr += m - prevtemp
curr += (len(heap) * temps[i])
ans[i] = curr
print(' '.join([str(x) for x in ans]))
```
| 105,853 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice likes snow a lot! Unfortunately, this year's winter is already over, and she can't expect to have any more of it. Bob has thus bought her a gift — a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.
Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other.
Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.
You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day.
Input
The first line contains a single integer N (1 ≤ N ≤ 105) — the number of days.
The second line contains N integers V1, V2, ..., VN (0 ≤ Vi ≤ 109), where Vi is the initial size of a snow pile made on the day i.
The third line contains N integers T1, T2, ..., TN (0 ≤ Ti ≤ 109), where Ti is the temperature on the day i.
Output
Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.
Examples
Input
3
10 10 5
5 7 2
Output
5 12 4
Input
5
30 25 20 15 10
9 10 12 4 13
Output
9 20 35 11 25
Note
In the first sample, Bob first makes a snow pile of volume 10, which melts to the size of 5 on the same day. On the second day, he makes another pile of size 10. Since it is a bit warmer than the day before, the first pile disappears completely while the second pile shrinks to 3. At the end of the second day, he has only a single pile of size 3. On the third day he makes a smaller pile than usual, but as the temperature dropped too, both piles survive till the end of the day.
Tags: binary search, data structures
Correct Solution:
```
# import sys
# from io import StringIO
#
# sys.stdin = StringIO(open(__file__.replace('.py', '.in')).read())
def bin_search_base(arr, val, base_index):
lo = base_index
hi = len(arr) - 1
while lo < hi:
if hi - lo == 1 and arr[lo] - arr[base_index] < val < arr[hi] - arr[base_index]:
return lo
if val >= arr[hi] - arr[base_index]:
return hi
c = (hi + lo) // 2
if val < arr[c] - arr[base_index]:
hi = c
elif val > arr[c] - arr[base_index]:
lo = c
else:
return c
return lo
n = int(input())
volumes = list(map(int, input().split()))
temps = list(map(int, input().split()))
temp_prefix = [0]
for i in range(n):
temp_prefix.append(temps[i] + temp_prefix[i])
gone_count = [0] * (n + 1)
remaining_count = [0] * n
gone_volume = [0] * (n + 1)
for i in range(n):
j = bin_search_base(temp_prefix, volumes[i], i)
gone_count[j] += 1
gone_volume[j] += volumes[i] - (temp_prefix[j] - temp_prefix[i])
for i in range(n):
remaining_count[i] = 1 + remaining_count[i-1] - gone_count[i]
for i in range(n):
gone_volume[i] += remaining_count[i] * temps[i]
print(' '.join(map(str, gone_volume[:n])))
```
| 105,854 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice likes snow a lot! Unfortunately, this year's winter is already over, and she can't expect to have any more of it. Bob has thus bought her a gift — a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.
Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other.
Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.
You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day.
Input
The first line contains a single integer N (1 ≤ N ≤ 105) — the number of days.
The second line contains N integers V1, V2, ..., VN (0 ≤ Vi ≤ 109), where Vi is the initial size of a snow pile made on the day i.
The third line contains N integers T1, T2, ..., TN (0 ≤ Ti ≤ 109), where Ti is the temperature on the day i.
Output
Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.
Examples
Input
3
10 10 5
5 7 2
Output
5 12 4
Input
5
30 25 20 15 10
9 10 12 4 13
Output
9 20 35 11 25
Note
In the first sample, Bob first makes a snow pile of volume 10, which melts to the size of 5 on the same day. On the second day, he makes another pile of size 10. Since it is a bit warmer than the day before, the first pile disappears completely while the second pile shrinks to 3. At the end of the second day, he has only a single pile of size 3. On the third day he makes a smaller pile than usual, but as the temperature dropped too, both piles survive till the end of the day.
Tags: binary search, data structures
Correct Solution:
```
import heapq
n = int(input())
V = list(map(int, input().split()))
T = list(map(int, input().split()))
a = []
ans = []
tmp = 0
for i in range(0, n):
heapq.heappush(a, tmp + V[i])
tmpAns = 0
while(len(a) and a[0] <= tmp + T[i]):
tmpAns += heapq.heappop(a) - tmp
tmpAns += len(a) * T[i]
ans.append(tmpAns)
tmp += T[i]
print(" ".join(map(str,ans)))
```
| 105,855 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice likes snow a lot! Unfortunately, this year's winter is already over, and she can't expect to have any more of it. Bob has thus bought her a gift — a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.
Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other.
Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.
You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day.
Input
The first line contains a single integer N (1 ≤ N ≤ 105) — the number of days.
The second line contains N integers V1, V2, ..., VN (0 ≤ Vi ≤ 109), where Vi is the initial size of a snow pile made on the day i.
The third line contains N integers T1, T2, ..., TN (0 ≤ Ti ≤ 109), where Ti is the temperature on the day i.
Output
Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.
Examples
Input
3
10 10 5
5 7 2
Output
5 12 4
Input
5
30 25 20 15 10
9 10 12 4 13
Output
9 20 35 11 25
Note
In the first sample, Bob first makes a snow pile of volume 10, which melts to the size of 5 on the same day. On the second day, he makes another pile of size 10. Since it is a bit warmer than the day before, the first pile disappears completely while the second pile shrinks to 3. At the end of the second day, he has only a single pile of size 3. On the third day he makes a smaller pile than usual, but as the temperature dropped too, both piles survive till the end of the day.
Tags: binary search, data structures
Correct Solution:
```
n = int(input())
vs = [int(x) for x in input().split()]
ts = [int(x) for x in input().split()]
from heapq import *
q = []
heapify(q)
sts = ts[:]
for i in range(1, n):
sts[i] += sts[i-1]
res = []
for i in range(n):
v = vs[i]
t = ts[i]
v += sts[i] - ts[i]
heappush(q, v)
minv = heappop(q)
count = 0
while minv <= sts[i]:
count += minv - sts[i] + ts[i]
if not q: break
minv = heappop(q)
else:
heappush(q, minv)
res.append(count + len(q) * t)
print(" ".join(map(str, res)))
```
| 105,856 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice likes snow a lot! Unfortunately, this year's winter is already over, and she can't expect to have any more of it. Bob has thus bought her a gift — a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.
Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other.
Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.
You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day.
Input
The first line contains a single integer N (1 ≤ N ≤ 105) — the number of days.
The second line contains N integers V1, V2, ..., VN (0 ≤ Vi ≤ 109), where Vi is the initial size of a snow pile made on the day i.
The third line contains N integers T1, T2, ..., TN (0 ≤ Ti ≤ 109), where Ti is the temperature on the day i.
Output
Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.
Examples
Input
3
10 10 5
5 7 2
Output
5 12 4
Input
5
30 25 20 15 10
9 10 12 4 13
Output
9 20 35 11 25
Note
In the first sample, Bob first makes a snow pile of volume 10, which melts to the size of 5 on the same day. On the second day, he makes another pile of size 10. Since it is a bit warmer than the day before, the first pile disappears completely while the second pile shrinks to 3. At the end of the second day, he has only a single pile of size 3. On the third day he makes a smaller pile than usual, but as the temperature dropped too, both piles survive till the end of the day.
Tags: binary search, data structures
Correct Solution:
```
from itertools import accumulate
from bisect import bisect
n = int(input())
vs = list(map(int, input().split()))
ts = list(map(int, input().split()))
fully_melt = [0] * n
result = [0] * n
dprint = lambda *args: None
#dprint = print
ts_accum = list(accumulate(ts))
dprint('ts_accum', ts_accum)
for i, v in enumerate(vs):
offset = ts_accum[i - 1] if i > 0 else 0
p = bisect(ts_accum, v + offset, lo=i)
dprint('i = ', i, ', p =', p, ', offset = ', offset)
if p < n:
fully_melt[p] += 1
result[p] -= ts_accum[p] - v - offset
dprint('fully_melt', fully_melt)
dprint('result', result)
multiplier = 1
for i, t in enumerate(ts):
dprint('i =', i, ', mult =', multiplier)
result[i] += t * multiplier
multiplier += 1 - fully_melt[i]
print(' '.join(str(r) for r in result))
```
| 105,857 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice likes snow a lot! Unfortunately, this year's winter is already over, and she can't expect to have any more of it. Bob has thus bought her a gift — a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.
Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other.
Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.
You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day.
Input
The first line contains a single integer N (1 ≤ N ≤ 105) — the number of days.
The second line contains N integers V1, V2, ..., VN (0 ≤ Vi ≤ 109), where Vi is the initial size of a snow pile made on the day i.
The third line contains N integers T1, T2, ..., TN (0 ≤ Ti ≤ 109), where Ti is the temperature on the day i.
Output
Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.
Examples
Input
3
10 10 5
5 7 2
Output
5 12 4
Input
5
30 25 20 15 10
9 10 12 4 13
Output
9 20 35 11 25
Note
In the first sample, Bob first makes a snow pile of volume 10, which melts to the size of 5 on the same day. On the second day, he makes another pile of size 10. Since it is a bit warmer than the day before, the first pile disappears completely while the second pile shrinks to 3. At the end of the second day, he has only a single pile of size 3. On the third day he makes a smaller pile than usual, but as the temperature dropped too, both piles survive till the end of the day.
Submitted Solution:
```
import configparser
import math
import sys
input = sys.stdin.readline
def get(l, r, cumsum):
res = cumsum[r]
if l >= 1:
res -= cumsum[l-1]
return res
def can(l, r, cumsum, am):
return am >= get(l, r, cumsum)
def main():
n = int(input())
v = [int(x) for x in input().split(' ')]
t = [int(x) for x in input().split(' ')]
cumsum = [0 for i in range(n)]
cumsum[0] = t[0]
for i in range(1, n):
cumsum[i] = t[i] + cumsum[i-1]
ans = [0 for i in range(n)]
alive = [[] for i in range(n)]
for i in range(n):
pos = i-1
for j in range(25, -1, -1):
jump = 2**j
if pos+jump < n and can(i, pos+jump, cumsum, v[i]):
pos += jump
if pos == i - 1:
alive[i].append(v[i])
else:
ans[i] += 1
if pos + 1 < n:
ans[pos+1] -= 1
if pos != n-1 and v[i] > get(i, pos, cumsum):
alive[pos+1].append(v[i] - get(i, pos, cumsum))
for i in range(1, n):
ans[i] += ans[i-1]
res = [0 for i in range(n)]
for i in range(n):
res[i] = ans[i]*t[i]
for j in alive[i]:
res[i] += j
for i in res:
print(i, end=' ')
if __name__ == '__main__':
main()
```
Yes
| 105,858 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice likes snow a lot! Unfortunately, this year's winter is already over, and she can't expect to have any more of it. Bob has thus bought her a gift — a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.
Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other.
Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.
You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day.
Input
The first line contains a single integer N (1 ≤ N ≤ 105) — the number of days.
The second line contains N integers V1, V2, ..., VN (0 ≤ Vi ≤ 109), where Vi is the initial size of a snow pile made on the day i.
The third line contains N integers T1, T2, ..., TN (0 ≤ Ti ≤ 109), where Ti is the temperature on the day i.
Output
Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.
Examples
Input
3
10 10 5
5 7 2
Output
5 12 4
Input
5
30 25 20 15 10
9 10 12 4 13
Output
9 20 35 11 25
Note
In the first sample, Bob first makes a snow pile of volume 10, which melts to the size of 5 on the same day. On the second day, he makes another pile of size 10. Since it is a bit warmer than the day before, the first pile disappears completely while the second pile shrinks to 3. At the end of the second day, he has only a single pile of size 3. On the third day he makes a smaller pile than usual, but as the temperature dropped too, both piles survive till the end of the day.
Submitted Solution:
```
from sys import stdin
def binary_search(l, r, val):
global prefix
i, j = l, r
while i < j:
mid = (i + j) // 2
ptr = prefix[mid]
if ptr <= val:
i = mid + 1
else:
j = mid
return i
n = int(stdin.buffer.readline())
V = list(map(int, stdin.buffer.readline().split()))
T = list(map(int, stdin.buffer.readline().split()))
cnt = [0] * (n + 1)
res = [0] * (n + 1)
prefix = [0] * (n + 1)
for i in range(n):
if i == 0:
prefix[i] = T[i]
else:
prefix[i] = prefix[i - 1] + T[i]
for i in range(n):
id = None
if i == 0:
id = binary_search(i, n - 1, V[i])
else:
id = binary_search(i, n - 1, V[i] + prefix[i - 1])
if id == n - 1 and V[i] + prefix[i - 1] > prefix[id]:
id += 1
cnt[i] += 1
cnt[id] -= 1
if id == n:
continue
if i == 0:
res[id] += (T[id] - prefix[id] + V[i])
else:
res[id] += (T[id] - prefix[id] + V[i] + prefix[i - 1])
for i in range(n):
if i > 0:
cnt[i] += cnt[i - 1]
res[i] += cnt[i] * T[i]
for i in range(n):
if i == n - 1:
print(res[i])
else:
print(res[i], end = ' ')
```
Yes
| 105,859 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice likes snow a lot! Unfortunately, this year's winter is already over, and she can't expect to have any more of it. Bob has thus bought her a gift — a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.
Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other.
Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.
You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day.
Input
The first line contains a single integer N (1 ≤ N ≤ 105) — the number of days.
The second line contains N integers V1, V2, ..., VN (0 ≤ Vi ≤ 109), where Vi is the initial size of a snow pile made on the day i.
The third line contains N integers T1, T2, ..., TN (0 ≤ Ti ≤ 109), where Ti is the temperature on the day i.
Output
Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.
Examples
Input
3
10 10 5
5 7 2
Output
5 12 4
Input
5
30 25 20 15 10
9 10 12 4 13
Output
9 20 35 11 25
Note
In the first sample, Bob first makes a snow pile of volume 10, which melts to the size of 5 on the same day. On the second day, he makes another pile of size 10. Since it is a bit warmer than the day before, the first pile disappears completely while the second pile shrinks to 3. At the end of the second day, he has only a single pile of size 3. On the third day he makes a smaller pile than usual, but as the temperature dropped too, both piles survive till the end of the day.
Submitted Solution:
```
from math import *
from collections import *
def bs(x,i):
l = i
r = n-1
if(i != 0):
while(l <= r):
mid = (l+r)//2
if(mid == n-1):
return mid
if(pre[mid]-pre[i-1] > x and pre[mid-1]-pre[i-1] < x):
return mid
if(pre[mid]-pre[i-1] == x):
return mid
if(pre[mid]-pre[i-1] > x):
r = mid - 1
else:
l = mid + 1
else:
while(l <= r):
mid = (l+r)//2
if(mid == n-1):
return mid
if(pre[mid] > x and pre[mid-1] < x):
return mid
if(pre[mid] == x):
return mid
if(pre[mid] > x):
r = mid - 1
else:
l = mid + 1
return mid
n = int(input())
l = list(map(int,input().split()))
t = list(map(int,input().split()))
pre = [t[0]]
for i in range(1,n):
pre.append(pre[i-1] + t[i])
dis = [0 for i in range(n)]
val = [0 for i in range(n)]
for i in range(n):
dis[i] = bs(l[i],i)
#print(dis[i])
if dis[i] > 0:
if i > 0:
if(l[i] > pre[dis[i]] - pre[i-1]):
#print(i)
val[dis[i]] += t[dis[i]]
else:
val[dis[i]] += l[i] - pre[dis[i]-1] + pre[i-1]
else:
if(l[i] > pre[dis[i]]):
val[dis[i]] += t[dis[i]]
else:
val[dis[i]] += l[i] - pre[dis[i]-1]
else:
if(l[i] > pre[dis[i]]):
val[dis[i]] += t[dis[i]]
else:
val[dis[i]] += l[0]
ans = [0 for i in range(n)]
for i in range(n):
ans[i] += 1
ans[dis[i]] -= 1
for i in range(1,n):
ans[i] += ans[i-1]
for i in range(n):
ans[i] *= t[i]
ans[i] += val[i]
'''
print(pre)
print(dis)
print(val)
'''
for i in ans:
print(i,end = " ")
print()
```
Yes
| 105,860 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice likes snow a lot! Unfortunately, this year's winter is already over, and she can't expect to have any more of it. Bob has thus bought her a gift — a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.
Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other.
Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.
You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day.
Input
The first line contains a single integer N (1 ≤ N ≤ 105) — the number of days.
The second line contains N integers V1, V2, ..., VN (0 ≤ Vi ≤ 109), where Vi is the initial size of a snow pile made on the day i.
The third line contains N integers T1, T2, ..., TN (0 ≤ Ti ≤ 109), where Ti is the temperature on the day i.
Output
Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.
Examples
Input
3
10 10 5
5 7 2
Output
5 12 4
Input
5
30 25 20 15 10
9 10 12 4 13
Output
9 20 35 11 25
Note
In the first sample, Bob first makes a snow pile of volume 10, which melts to the size of 5 on the same day. On the second day, he makes another pile of size 10. Since it is a bit warmer than the day before, the first pile disappears completely while the second pile shrinks to 3. At the end of the second day, he has only a single pile of size 3. On the third day he makes a smaller pile than usual, but as the temperature dropped too, both piles survive till the end of the day.
Submitted Solution:
```
#!/usr/bin/env python3
from sys import stdin, stdout
from bisect import bisect
def rint():
return map(int, stdin.readline().split())
#lines = stdin.readlines()
N = int(input())
V = list(rint())
T = list(rint())
TT = [T[0]]
for i in range(1,N):
TT.append(TT[i-1] + T[i])
D = [ 0 for i in range(N)]
M = [ 0 for i in range(N)]
TT.append(0)
for i in range(0,N):
j = bisect(TT, V[i] + TT[i-1], i, N)
if j >= N:
continue
D[j] += 1
M[j] = M[j]+ V[i] - (TT[j-1] - TT[i-1])
for i in range(1,N):
D[i] += D[i-1]
ans = [ 0 for i in range(N)]
for i in range(N):
ans[i] = T[i]*(i+1 - D[i]) + M[i]
print(*ans)
```
Yes
| 105,861 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice likes snow a lot! Unfortunately, this year's winter is already over, and she can't expect to have any more of it. Bob has thus bought her a gift — a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.
Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other.
Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.
You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day.
Input
The first line contains a single integer N (1 ≤ N ≤ 105) — the number of days.
The second line contains N integers V1, V2, ..., VN (0 ≤ Vi ≤ 109), where Vi is the initial size of a snow pile made on the day i.
The third line contains N integers T1, T2, ..., TN (0 ≤ Ti ≤ 109), where Ti is the temperature on the day i.
Output
Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.
Examples
Input
3
10 10 5
5 7 2
Output
5 12 4
Input
5
30 25 20 15 10
9 10 12 4 13
Output
9 20 35 11 25
Note
In the first sample, Bob first makes a snow pile of volume 10, which melts to the size of 5 on the same day. On the second day, he makes another pile of size 10. Since it is a bit warmer than the day before, the first pile disappears completely while the second pile shrinks to 3. At the end of the second day, he has only a single pile of size 3. On the third day he makes a smaller pile than usual, but as the temperature dropped too, both piles survive till the end of the day.
Submitted Solution:
```
n = int(input())
vs = [int(x) for x in input().split()]
ts = [int(x) for x in input().split()]
from heapq import *
q = []
heapify(q)
sts = ts[:]
for i in range(1, n):
sts[i] += sts[i-1]
res = []
for i in range(n):
v = vs[i]
t = ts[i]
if i: v += sts[i-1]
heappush(q, v)
minv = heappop(q)
if minv > sts[i]:
heappush(q, minv)
res.append(len(q) * t)
else:
count = 0
while minv <= sts[i]:
count += minv - sts[i-1]
if not q: break
minv = heappop(q)
else:
heappush(q, minv)
res.append(count + len(q) * t)
print(" ".join(map(str, res)))
```
No
| 105,862 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice likes snow a lot! Unfortunately, this year's winter is already over, and she can't expect to have any more of it. Bob has thus bought her a gift — a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.
Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other.
Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.
You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day.
Input
The first line contains a single integer N (1 ≤ N ≤ 105) — the number of days.
The second line contains N integers V1, V2, ..., VN (0 ≤ Vi ≤ 109), where Vi is the initial size of a snow pile made on the day i.
The third line contains N integers T1, T2, ..., TN (0 ≤ Ti ≤ 109), where Ti is the temperature on the day i.
Output
Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.
Examples
Input
3
10 10 5
5 7 2
Output
5 12 4
Input
5
30 25 20 15 10
9 10 12 4 13
Output
9 20 35 11 25
Note
In the first sample, Bob first makes a snow pile of volume 10, which melts to the size of 5 on the same day. On the second day, he makes another pile of size 10. Since it is a bit warmer than the day before, the first pile disappears completely while the second pile shrinks to 3. At the end of the second day, he has only a single pile of size 3. On the third day he makes a smaller pile than usual, but as the temperature dropped too, both piles survive till the end of the day.
Submitted Solution:
```
import heapq
heap = []
n = int(input())
V = list(map(int, input().split()))
T = list(map(int, input().split()))
tmp = 0
for i in range(n):
ans = 0
heapq.heappush(heap, V[i])
while len(heap) and heap[0]<=T[i]:
ans += heapq.heappop(heap)
tmp += T[i]
ans += T[i]*len(heap)
print(ans, end=' ')
```
No
| 105,863 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice likes snow a lot! Unfortunately, this year's winter is already over, and she can't expect to have any more of it. Bob has thus bought her a gift — a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.
Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other.
Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.
You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day.
Input
The first line contains a single integer N (1 ≤ N ≤ 105) — the number of days.
The second line contains N integers V1, V2, ..., VN (0 ≤ Vi ≤ 109), where Vi is the initial size of a snow pile made on the day i.
The third line contains N integers T1, T2, ..., TN (0 ≤ Ti ≤ 109), where Ti is the temperature on the day i.
Output
Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.
Examples
Input
3
10 10 5
5 7 2
Output
5 12 4
Input
5
30 25 20 15 10
9 10 12 4 13
Output
9 20 35 11 25
Note
In the first sample, Bob first makes a snow pile of volume 10, which melts to the size of 5 on the same day. On the second day, he makes another pile of size 10. Since it is a bit warmer than the day before, the first pile disappears completely while the second pile shrinks to 3. At the end of the second day, he has only a single pile of size 3. On the third day he makes a smaller pile than usual, but as the temperature dropped too, both piles survive till the end of the day.
Submitted Solution:
```
def binary_search(a, l, r, num):
mid = (l+r)//2
# print("l==" + str(l) + ", r==" + str(r) + ", mid==" + str(mid))
if(mid == num):
return mid+1
if(l > r):
return l
if(a[mid] < num):
return binary_search(a, l+1, r, num)
else:
return binary_search(a, l, r-1, num)
n = int(input())
V = list(map(int, input().split()))
T = list(map(int, input().split()))
a = []
ans = []
for i in range(0, n):
index = binary_search(a, 0, len(a)-1, V[i])
# print("index==" + str(index))
a.insert(index, V[i])
# print(a)
index0 = binary_search(a, 0, len(a)-1, T[i])
# print("index0==" + str(index0))
# print("sum==" + str(sum(a[:index0])))
ans.append(sum(a[:index0]) + (len(a) - index0)*T[i])
del a[:index0]
for j in range(0, len(a)):
a[j] -= T[i]
print(" ".join(map(str,ans)))
```
No
| 105,864 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice likes snow a lot! Unfortunately, this year's winter is already over, and she can't expect to have any more of it. Bob has thus bought her a gift — a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.
Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other.
Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.
You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day.
Input
The first line contains a single integer N (1 ≤ N ≤ 105) — the number of days.
The second line contains N integers V1, V2, ..., VN (0 ≤ Vi ≤ 109), where Vi is the initial size of a snow pile made on the day i.
The third line contains N integers T1, T2, ..., TN (0 ≤ Ti ≤ 109), where Ti is the temperature on the day i.
Output
Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.
Examples
Input
3
10 10 5
5 7 2
Output
5 12 4
Input
5
30 25 20 15 10
9 10 12 4 13
Output
9 20 35 11 25
Note
In the first sample, Bob first makes a snow pile of volume 10, which melts to the size of 5 on the same day. On the second day, he makes another pile of size 10. Since it is a bit warmer than the day before, the first pile disappears completely while the second pile shrinks to 3. At the end of the second day, he has only a single pile of size 3. On the third day he makes a smaller pile than usual, but as the temperature dropped too, both piles survive till the end of the day.
Submitted Solution:
```
import configparser
import math
import sys
input = sys.stdin.readline
def get(l, r, cumsum):
res = cumsum[r]
if l >= 1:
res -= cumsum[l-1]
return res
def can(l, r, cumsum, am):
return am >= get(l, r, cumsum)
def main():
n = int(input())
v = [int(x) for x in input().split(' ')]
t = [int(x) for x in input().split(' ')]
cumsum = [0 for i in range(n)]
cumsum[0] = t[0]
for i in range(1, n):
cumsum[i] = t[i] + cumsum[i-1]
ans = [0 for i in range(n)]
alive = [[] for i in range(n)]
for i in range(n):
pos = i-1
for j in range(6, -1, -1):
jump = 2**j
if pos+jump < n and can(i, pos+jump, cumsum, v[i]):
pos += jump
if pos == i - 1:
alive[i].append(v[i])
else:
ans[i] += 1
if pos + 1 < n:
ans[pos+1] -= 1
if pos != n-1 and v[i] > get(i, pos, cumsum):
alive[pos+1].append(v[i] - get(i, pos, cumsum))
for i in range(1, n):
ans[i] += ans[i-1]
res = [0 for i in range(n)]
for i in range(n):
res[i] = ans[i]*t[i]
for j in alive[i]:
res[i] += j
for i in res:
print(i, end=' ')
if __name__ == '__main__':
main()
```
No
| 105,865 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence a1, a2, ..., an of one-dimensional segments numbered 1 through n. Your task is to find two distinct indices i and j such that segment ai lies within segment aj.
Segment [l1, r1] lies within segment [l2, r2] iff l1 ≥ l2 and r1 ≤ r2.
Print indices i and j. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Input
The first line contains one integer n (1 ≤ n ≤ 3·105) — the number of segments.
Each of the next n lines contains two integers li and ri (1 ≤ li ≤ ri ≤ 109) — the i-th segment.
Output
Print two distinct indices i and j such that segment ai lies within segment aj. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Examples
Input
5
1 10
2 9
3 9
2 3
2 9
Output
2 1
Input
3
1 5
2 6
6 20
Output
-1 -1
Note
In the first example the following pairs are considered correct:
* (2, 1), (3, 1), (4, 1), (5, 1) — not even touching borders;
* (3, 2), (4, 2), (3, 5), (4, 5) — touch one border;
* (5, 2), (2, 5) — match exactly.
Tags: greedy, implementation, sortings
Correct Solution:
```
n=int(input())
segs=[]
for i in range(n):
a,b=map(int,input().split())
segs.append([a,b,i])
segs.sort(key=lambda x:[x[0],-x[1]])
for i in range(1,n):
if segs[i][1]<=segs[i-1][1]:
print(segs[i][2]+1,segs[i-1][2]+1)
exit()
segs.sort(key=lambda x:[-x[1],x[0]])
for i in range(1,n):
if segs[i][0]>=segs[0][0]:
print(segs[i][2]+1,segs[0][2]+1)
exit()
print('-1 -1')
```
| 105,866 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence a1, a2, ..., an of one-dimensional segments numbered 1 through n. Your task is to find two distinct indices i and j such that segment ai lies within segment aj.
Segment [l1, r1] lies within segment [l2, r2] iff l1 ≥ l2 and r1 ≤ r2.
Print indices i and j. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Input
The first line contains one integer n (1 ≤ n ≤ 3·105) — the number of segments.
Each of the next n lines contains two integers li and ri (1 ≤ li ≤ ri ≤ 109) — the i-th segment.
Output
Print two distinct indices i and j such that segment ai lies within segment aj. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Examples
Input
5
1 10
2 9
3 9
2 3
2 9
Output
2 1
Input
3
1 5
2 6
6 20
Output
-1 -1
Note
In the first example the following pairs are considered correct:
* (2, 1), (3, 1), (4, 1), (5, 1) — not even touching borders;
* (3, 2), (4, 2), (3, 5), (4, 5) — touch one border;
* (5, 2), (2, 5) — match exactly.
Tags: greedy, implementation, sortings
Correct Solution:
```
import sys
input=sys.stdin.readline
n = int(input())
ls = []
for i in range(n):
l, r = list(map(int, input().strip().split()))
ls.append([l, r, i])
ls = sorted(ls, key=lambda x: (x[0], -x[1]))
flag = False
for i in range(n-1):
cur_l, cur_r, mother = ls[i]
if cur_r >= ls[i+1][1]:
print(ls[i+1][2] + 1, mother + 1)
flag = True
break
if not flag:
print(-1, -1)
```
| 105,867 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence a1, a2, ..., an of one-dimensional segments numbered 1 through n. Your task is to find two distinct indices i and j such that segment ai lies within segment aj.
Segment [l1, r1] lies within segment [l2, r2] iff l1 ≥ l2 and r1 ≤ r2.
Print indices i and j. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Input
The first line contains one integer n (1 ≤ n ≤ 3·105) — the number of segments.
Each of the next n lines contains two integers li and ri (1 ≤ li ≤ ri ≤ 109) — the i-th segment.
Output
Print two distinct indices i and j such that segment ai lies within segment aj. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Examples
Input
5
1 10
2 9
3 9
2 3
2 9
Output
2 1
Input
3
1 5
2 6
6 20
Output
-1 -1
Note
In the first example the following pairs are considered correct:
* (2, 1), (3, 1), (4, 1), (5, 1) — not even touching borders;
* (3, 2), (4, 2), (3, 5), (4, 5) — touch one border;
* (5, 2), (2, 5) — match exactly.
Tags: greedy, implementation, sortings
Correct Solution:
```
n = int(input())
arr = []
for i in range(n):
l, r = map(int, input().split())
arr.append((l, -r, i))
ans = '-1 -1'
a = sorted(arr)
for i in range(len(a)):
a[i] = (a[i][0], -a[i][1], a[i][2])
for i in range(len(a)-1):
if a[i][0] <= a[i+1][0] and a[i][1] >= a[i+1][1]:
ans = str(a[i+1][2] + 1) + ' ' + str(a[i][2] + 1)
print(ans)
```
| 105,868 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence a1, a2, ..., an of one-dimensional segments numbered 1 through n. Your task is to find two distinct indices i and j such that segment ai lies within segment aj.
Segment [l1, r1] lies within segment [l2, r2] iff l1 ≥ l2 and r1 ≤ r2.
Print indices i and j. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Input
The first line contains one integer n (1 ≤ n ≤ 3·105) — the number of segments.
Each of the next n lines contains two integers li and ri (1 ≤ li ≤ ri ≤ 109) — the i-th segment.
Output
Print two distinct indices i and j such that segment ai lies within segment aj. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Examples
Input
5
1 10
2 9
3 9
2 3
2 9
Output
2 1
Input
3
1 5
2 6
6 20
Output
-1 -1
Note
In the first example the following pairs are considered correct:
* (2, 1), (3, 1), (4, 1), (5, 1) — not even touching borders;
* (3, 2), (4, 2), (3, 5), (4, 5) — touch one border;
* (5, 2), (2, 5) — match exactly.
Tags: greedy, implementation, sortings
Correct Solution:
```
n = int(input())
a = []
for i in range(1, n+1):
l, r = map(int, input().split())
a.append((l, r, i))
a.sort()
for i in range(n-1):
if (a[i][0] == a[i+1][0]):
print(str(a[i][2]) + ' ' + str(a[i+1][2]))
break
if (a[i][1] >= a[i+1][1]):
print(str(a[i+1][2]) + ' ' + str(a[i][2]))
break
else:
print('-1 -1')
```
| 105,869 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence a1, a2, ..., an of one-dimensional segments numbered 1 through n. Your task is to find two distinct indices i and j such that segment ai lies within segment aj.
Segment [l1, r1] lies within segment [l2, r2] iff l1 ≥ l2 and r1 ≤ r2.
Print indices i and j. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Input
The first line contains one integer n (1 ≤ n ≤ 3·105) — the number of segments.
Each of the next n lines contains two integers li and ri (1 ≤ li ≤ ri ≤ 109) — the i-th segment.
Output
Print two distinct indices i and j such that segment ai lies within segment aj. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Examples
Input
5
1 10
2 9
3 9
2 3
2 9
Output
2 1
Input
3
1 5
2 6
6 20
Output
-1 -1
Note
In the first example the following pairs are considered correct:
* (2, 1), (3, 1), (4, 1), (5, 1) — not even touching borders;
* (3, 2), (4, 2), (3, 5), (4, 5) — touch one border;
* (5, 2), (2, 5) — match exactly.
Tags: greedy, implementation, sortings
Correct Solution:
```
from sys import stdin
class segment :
def __init__(self,x,y,n) :
self.a = x
self.b = y
self.no = n
def __gt__(self,other):
if(self.a == other.a):
return self.b < other.b
return self.a > other.a
def __lt__(self,other):
if(self.a == other.a):
return self.b > other.b
return self.a < other.a
def __eq__(self,other):
if(self.a==other.a and self.b == self.b):
return True
else:
return False
def __str__(self) :
return str(self.no)
n = int(stdin.readline())
arr = []
for i in range(n) :
r,z = map(int,stdin.readline().split())
s = segment(r,z,i+1)
arr.append(s)
arr.sort()
for i in range(1,n):
if(arr[i].b <= arr[i-1].b) :
print (arr[i],arr[i-1])
exit()
print (-1,-1)
```
| 105,870 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence a1, a2, ..., an of one-dimensional segments numbered 1 through n. Your task is to find two distinct indices i and j such that segment ai lies within segment aj.
Segment [l1, r1] lies within segment [l2, r2] iff l1 ≥ l2 and r1 ≤ r2.
Print indices i and j. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Input
The first line contains one integer n (1 ≤ n ≤ 3·105) — the number of segments.
Each of the next n lines contains two integers li and ri (1 ≤ li ≤ ri ≤ 109) — the i-th segment.
Output
Print two distinct indices i and j such that segment ai lies within segment aj. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Examples
Input
5
1 10
2 9
3 9
2 3
2 9
Output
2 1
Input
3
1 5
2 6
6 20
Output
-1 -1
Note
In the first example the following pairs are considered correct:
* (2, 1), (3, 1), (4, 1), (5, 1) — not even touching borders;
* (3, 2), (4, 2), (3, 5), (4, 5) — touch one border;
* (5, 2), (2, 5) — match exactly.
Tags: greedy, implementation, sortings
Correct Solution:
```
n = int(input())
l = [i for i in range(n)]
d = [tuple(map(int, input().split())) for _ in range(n)]
l.sort(key = lambda x: (d[x][0], -d[x][1]))
for i in range(n - 1):
if d[l[i + 1]][1] <= d[l[i]][1]:
print(l[i + 1] + 1, l[i] + 1)
exit()
print(-1, -1)
```
| 105,871 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence a1, a2, ..., an of one-dimensional segments numbered 1 through n. Your task is to find two distinct indices i and j such that segment ai lies within segment aj.
Segment [l1, r1] lies within segment [l2, r2] iff l1 ≥ l2 and r1 ≤ r2.
Print indices i and j. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Input
The first line contains one integer n (1 ≤ n ≤ 3·105) — the number of segments.
Each of the next n lines contains two integers li and ri (1 ≤ li ≤ ri ≤ 109) — the i-th segment.
Output
Print two distinct indices i and j such that segment ai lies within segment aj. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Examples
Input
5
1 10
2 9
3 9
2 3
2 9
Output
2 1
Input
3
1 5
2 6
6 20
Output
-1 -1
Note
In the first example the following pairs are considered correct:
* (2, 1), (3, 1), (4, 1), (5, 1) — not even touching borders;
* (3, 2), (4, 2), (3, 5), (4, 5) — touch one border;
* (5, 2), (2, 5) — match exactly.
Tags: greedy, implementation, sortings
Correct Solution:
```
n=int(input())
a=[list(map(int,input().split())) for i in range(n)]
for i in range(n):
a[i].append(i+1)
a.sort()
for i in range(n-1):
if a[i][1]>=a[i+1][1]:
exit(print(a[i+1][2],a[i][2]))
if a[i][0]==a[i+1][0] and a[i][1]<a[i+1][1]:
exit(print(a[i][2],a[i+1][2]))
print('-1','-1')
```
| 105,872 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence a1, a2, ..., an of one-dimensional segments numbered 1 through n. Your task is to find two distinct indices i and j such that segment ai lies within segment aj.
Segment [l1, r1] lies within segment [l2, r2] iff l1 ≥ l2 and r1 ≤ r2.
Print indices i and j. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Input
The first line contains one integer n (1 ≤ n ≤ 3·105) — the number of segments.
Each of the next n lines contains two integers li and ri (1 ≤ li ≤ ri ≤ 109) — the i-th segment.
Output
Print two distinct indices i and j such that segment ai lies within segment aj. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Examples
Input
5
1 10
2 9
3 9
2 3
2 9
Output
2 1
Input
3
1 5
2 6
6 20
Output
-1 -1
Note
In the first example the following pairs are considered correct:
* (2, 1), (3, 1), (4, 1), (5, 1) — not even touching borders;
* (3, 2), (4, 2), (3, 5), (4, 5) — touch one border;
* (5, 2), (2, 5) — match exactly.
Tags: greedy, implementation, sortings
Correct Solution:
```
N = int(input())
A = []
for i in range(1, N+1):
l, r = map(int, input().split())
A.append([l, r, i])
A.sort(key=lambda x:x[0])
if N == 1:
print(-1, -1)
quit()
a = A[0][1]
for i in range(1, N):
if A[i][0] == A[i-1][0] and A[i][1] > A[i-1][1]:
print(A[i-1][2], A[i][2])
quit()
elif A[i][1] <= a:
print(A[i][2], A[i-1][2])
quit()
else:
if i == N-1:
print(-1, -1)
quit()
else:
a = A[i][1]
```
| 105,873 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a1, a2, ..., an of one-dimensional segments numbered 1 through n. Your task is to find two distinct indices i and j such that segment ai lies within segment aj.
Segment [l1, r1] lies within segment [l2, r2] iff l1 ≥ l2 and r1 ≤ r2.
Print indices i and j. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Input
The first line contains one integer n (1 ≤ n ≤ 3·105) — the number of segments.
Each of the next n lines contains two integers li and ri (1 ≤ li ≤ ri ≤ 109) — the i-th segment.
Output
Print two distinct indices i and j such that segment ai lies within segment aj. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Examples
Input
5
1 10
2 9
3 9
2 3
2 9
Output
2 1
Input
3
1 5
2 6
6 20
Output
-1 -1
Note
In the first example the following pairs are considered correct:
* (2, 1), (3, 1), (4, 1), (5, 1) — not even touching borders;
* (3, 2), (4, 2), (3, 5), (4, 5) — touch one border;
* (5, 2), (2, 5) — match exactly.
Submitted Solution:
```
#!/usr/bin/env python3
n = int(input().strip())
ais = [tuple(map(int, input().strip().split())) for _ in range(n)]
def solve(ais):
bis = [(l, r, i + 1) for i, (l, r) in enumerate(ais)]
bis.sort(key=lambda t: (t[0], -t[1]))
rr = bis[0][1] - 1
ir = bis[0][2]
for l, r, i in bis:
if r <= rr:
return (i, ir)
else:
rr = r
ir = i
return (-1, -1)
i, j = solve(ais)
print (i, j)
```
Yes
| 105,874 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a1, a2, ..., an of one-dimensional segments numbered 1 through n. Your task is to find two distinct indices i and j such that segment ai lies within segment aj.
Segment [l1, r1] lies within segment [l2, r2] iff l1 ≥ l2 and r1 ≤ r2.
Print indices i and j. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Input
The first line contains one integer n (1 ≤ n ≤ 3·105) — the number of segments.
Each of the next n lines contains two integers li and ri (1 ≤ li ≤ ri ≤ 109) — the i-th segment.
Output
Print two distinct indices i and j such that segment ai lies within segment aj. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Examples
Input
5
1 10
2 9
3 9
2 3
2 9
Output
2 1
Input
3
1 5
2 6
6 20
Output
-1 -1
Note
In the first example the following pairs are considered correct:
* (2, 1), (3, 1), (4, 1), (5, 1) — not even touching borders;
* (3, 2), (4, 2), (3, 5), (4, 5) — touch one border;
* (5, 2), (2, 5) — match exactly.
Submitted Solution:
```
import sys
n = int(sys.stdin.readline())
intervals = list(map(lambda x : (int(x[0]), int(x[1])), map(str.split, sys.stdin.readlines())))
intervals = list(enumerate(intervals))
intervals.sort(key=lambda x : 1000000009 * x[1][0] - x[1][1])
r = 0
ans1 = -1
ans2 = -1
for interval in intervals:
if interval[1][1] <= r:
ans1 = interval[0]
break
else:
ans2 = interval[0]
r = interval[1][1]
if ans1 == -1:
print('-1 -1')
else:
print(ans1 +1, ans2 +1)
```
Yes
| 105,875 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a1, a2, ..., an of one-dimensional segments numbered 1 through n. Your task is to find two distinct indices i and j such that segment ai lies within segment aj.
Segment [l1, r1] lies within segment [l2, r2] iff l1 ≥ l2 and r1 ≤ r2.
Print indices i and j. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Input
The first line contains one integer n (1 ≤ n ≤ 3·105) — the number of segments.
Each of the next n lines contains two integers li and ri (1 ≤ li ≤ ri ≤ 109) — the i-th segment.
Output
Print two distinct indices i and j such that segment ai lies within segment aj. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Examples
Input
5
1 10
2 9
3 9
2 3
2 9
Output
2 1
Input
3
1 5
2 6
6 20
Output
-1 -1
Note
In the first example the following pairs are considered correct:
* (2, 1), (3, 1), (4, 1), (5, 1) — not even touching borders;
* (3, 2), (4, 2), (3, 5), (4, 5) — touch one border;
* (5, 2), (2, 5) — match exactly.
Submitted Solution:
```
def main():
n = int(input())
arr = []
for i in range(n):
arr.append(list(map(int, input().split())) + [i + 1])
arr.sort(key=lambda x:[x[0], -x[1]])
ind = 0
for i in range(1, len(arr)):
if arr[ind][1] >= arr[i][1]:
print(arr[i][2], arr[ind][2])
return
else:
ind = i
print(-1, -1)
main()
```
Yes
| 105,876 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a1, a2, ..., an of one-dimensional segments numbered 1 through n. Your task is to find two distinct indices i and j such that segment ai lies within segment aj.
Segment [l1, r1] lies within segment [l2, r2] iff l1 ≥ l2 and r1 ≤ r2.
Print indices i and j. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Input
The first line contains one integer n (1 ≤ n ≤ 3·105) — the number of segments.
Each of the next n lines contains two integers li and ri (1 ≤ li ≤ ri ≤ 109) — the i-th segment.
Output
Print two distinct indices i and j such that segment ai lies within segment aj. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Examples
Input
5
1 10
2 9
3 9
2 3
2 9
Output
2 1
Input
3
1 5
2 6
6 20
Output
-1 -1
Note
In the first example the following pairs are considered correct:
* (2, 1), (3, 1), (4, 1), (5, 1) — not even touching borders;
* (3, 2), (4, 2), (3, 5), (4, 5) — touch one border;
* (5, 2), (2, 5) — match exactly.
Submitted Solution:
```
n = int(input())
L = []
for i in range(n):
l,r = map(int,input().split())
L.append((l,-r,i+1))
L = sorted(L)
f = 0
for i in range(1,n):
if L[i-1][1] <= L[i][1]:
f = 1
print(L[i][2],L[i-1][2])
break
if f == 0:
print(-1,-1)
```
Yes
| 105,877 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a1, a2, ..., an of one-dimensional segments numbered 1 through n. Your task is to find two distinct indices i and j such that segment ai lies within segment aj.
Segment [l1, r1] lies within segment [l2, r2] iff l1 ≥ l2 and r1 ≤ r2.
Print indices i and j. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Input
The first line contains one integer n (1 ≤ n ≤ 3·105) — the number of segments.
Each of the next n lines contains two integers li and ri (1 ≤ li ≤ ri ≤ 109) — the i-th segment.
Output
Print two distinct indices i and j such that segment ai lies within segment aj. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Examples
Input
5
1 10
2 9
3 9
2 3
2 9
Output
2 1
Input
3
1 5
2 6
6 20
Output
-1 -1
Note
In the first example the following pairs are considered correct:
* (2, 1), (3, 1), (4, 1), (5, 1) — not even touching borders;
* (3, 2), (4, 2), (3, 5), (4, 5) — touch one border;
* (5, 2), (2, 5) — match exactly.
Submitted Solution:
```
n = int(input())
segments = []
for _ in range(n):
x, y = map(int, input().split())
segments.append((x,y))
segments = sorted(segments, key=lambda seg: seg[1], reverse=True)
segments = sorted(segments, key=lambda seg: seg[0], reverse=False)
max_seg = (-1,-1)
output = (-1, -1)
# print(segments)
for i in range(n):
if segments[i][1] > max_seg[0]:
max_seg = (segments[i][1], i)
else:
fst = segments.index(segments[i]) + 1
snd = segments.index(segments[max_seg[1]]) + 1
# print(str(max_seg[1]) + ' ' + str(i))
print(str(fst) + ' ' + str(snd))
break
if max_seg == (-1, -1):
print('-1 -1')
```
No
| 105,878 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a1, a2, ..., an of one-dimensional segments numbered 1 through n. Your task is to find two distinct indices i and j such that segment ai lies within segment aj.
Segment [l1, r1] lies within segment [l2, r2] iff l1 ≥ l2 and r1 ≤ r2.
Print indices i and j. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Input
The first line contains one integer n (1 ≤ n ≤ 3·105) — the number of segments.
Each of the next n lines contains two integers li and ri (1 ≤ li ≤ ri ≤ 109) — the i-th segment.
Output
Print two distinct indices i and j such that segment ai lies within segment aj. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Examples
Input
5
1 10
2 9
3 9
2 3
2 9
Output
2 1
Input
3
1 5
2 6
6 20
Output
-1 -1
Note
In the first example the following pairs are considered correct:
* (2, 1), (3, 1), (4, 1), (5, 1) — not even touching borders;
* (3, 2), (4, 2), (3, 5), (4, 5) — touch one border;
* (5, 2), (2, 5) — match exactly.
Submitted Solution:
```
#Failed
n = int(input())
def divconq(seq):
if(len(seq) <= 1):
return False
eraser = seq[len(seq)//2]
skip_index = len(seq)//2
left = []
left_out = False
right_out = False
left_border = False
right_border = False
# l_c = 0
right = []
#r_c = 0
for i in range(len(seq)):
if not i == skip_index:
if seq[i][0] < eraser[0]:
left_out = True
if seq[i][1] > eraser[1]:
right_out = True
if seq[i][0] == eraser[0]:
left_border = True
if seq[i][1] == eraser[1]:
right_border = True
if ((left_out or left_border) and (right_border or right_out)) or ((not left_out) and (not right_out)):
if eraser[1]-eraser[0] >= seq[i][1] - seq[i][0]:
print(eraser[2],seq[i][2])
else:
print(seq[i][2],eraser[2])
return True
elif left_out:
left.append(seq[i])
else:
right.append(seq[i])
left_out = False
right_out = False
left_border = False
right_border = False
return divconq(left) or divconq(right)
seq = []
for i in range(n):
seq.append(list(map(int,input().split())))
seq[i].append(i+1)
seq = sorted(seq,key = lambda x : (x[0],-x[1],x[2]))
if not divconq(seq):
print(-1,-1)
```
No
| 105,879 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a1, a2, ..., an of one-dimensional segments numbered 1 through n. Your task is to find two distinct indices i and j such that segment ai lies within segment aj.
Segment [l1, r1] lies within segment [l2, r2] iff l1 ≥ l2 and r1 ≤ r2.
Print indices i and j. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Input
The first line contains one integer n (1 ≤ n ≤ 3·105) — the number of segments.
Each of the next n lines contains two integers li and ri (1 ≤ li ≤ ri ≤ 109) — the i-th segment.
Output
Print two distinct indices i and j such that segment ai lies within segment aj. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Examples
Input
5
1 10
2 9
3 9
2 3
2 9
Output
2 1
Input
3
1 5
2 6
6 20
Output
-1 -1
Note
In the first example the following pairs are considered correct:
* (2, 1), (3, 1), (4, 1), (5, 1) — not even touching borders;
* (3, 2), (4, 2), (3, 5), (4, 5) — touch one border;
* (5, 2), (2, 5) — match exactly.
Submitted Solution:
```
n=int(input())
segs=[]
for i in range(n):
a,b=map(int,input().split())
segs.append([a,b,i])
segs.sort(key=lambda x:x[0])
for i in range(1,n):
if segs[i][1]<=segs[0][1]:
print(segs[i][2]+1,segs[0][2]+1)
exit()
segs.sort(key=lambda x:-x[1])
for i in range(1,n):
if segs[i][0]>=segs[0][0]:
print(segs[i][2]+1,segs[0][2]+1)
print('-1 -1')
```
No
| 105,880 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a1, a2, ..., an of one-dimensional segments numbered 1 through n. Your task is to find two distinct indices i and j such that segment ai lies within segment aj.
Segment [l1, r1] lies within segment [l2, r2] iff l1 ≥ l2 and r1 ≤ r2.
Print indices i and j. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Input
The first line contains one integer n (1 ≤ n ≤ 3·105) — the number of segments.
Each of the next n lines contains two integers li and ri (1 ≤ li ≤ ri ≤ 109) — the i-th segment.
Output
Print two distinct indices i and j such that segment ai lies within segment aj. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Examples
Input
5
1 10
2 9
3 9
2 3
2 9
Output
2 1
Input
3
1 5
2 6
6 20
Output
-1 -1
Note
In the first example the following pairs are considered correct:
* (2, 1), (3, 1), (4, 1), (5, 1) — not even touching borders;
* (3, 2), (4, 2), (3, 5), (4, 5) — touch one border;
* (5, 2), (2, 5) — match exactly.
Submitted Solution:
```
N = int(input())
A = []
for i in range(1, N+1):
l, r = map(int, input().split())
A.append([l, r, i])
A.sort(key=lambda x:x[0])
if N == 1:
print(-1, -1)
quit()
a = A[0][1]
for i in range(1, N):
if i == 1 and A[1][0] == A[0][0] and A[0][1] > A[1][1]:
print(A[0][2], A[1][2])
elif A[i][1] <= a:
print(A[i][2], A[i-1][2])
quit()
else:
if i == N-1:
print(-1, -1)
quit()
else:
a = A[i][1]
```
No
| 105,881 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Allen has a LOT of money. He has n dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are 1, 5, 10, 20, 100. What is the minimum number of bills Allen could receive after withdrawing his entire balance?
Input
The first and only line of input contains a single integer n (1 ≤ n ≤ 10^9).
Output
Output the minimum number of bills that Allen could receive.
Examples
Input
125
Output
3
Input
43
Output
5
Input
1000000000
Output
10000000
Note
In the first sample case, Allen can withdraw this with a 100 dollar bill, a 20 dollar bill, and a 5 dollar bill. There is no way for Allen to receive 125 dollars in one or two bills.
In the second sample case, Allen can withdraw two 20 dollar bills and three 1 dollar bills.
In the third sample case, Allen can withdraw 100000000 (ten million!) 100 dollar bills.
Tags: dp, greedy
Correct Solution:
```
l=[100,20,10,5,1]
n=int(input())
s=0
for i in l:
s+=n//i
n=n%i
print(s)
```
| 105,882 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Allen has a LOT of money. He has n dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are 1, 5, 10, 20, 100. What is the minimum number of bills Allen could receive after withdrawing his entire balance?
Input
The first and only line of input contains a single integer n (1 ≤ n ≤ 10^9).
Output
Output the minimum number of bills that Allen could receive.
Examples
Input
125
Output
3
Input
43
Output
5
Input
1000000000
Output
10000000
Note
In the first sample case, Allen can withdraw this with a 100 dollar bill, a 20 dollar bill, and a 5 dollar bill. There is no way for Allen to receive 125 dollars in one or two bills.
In the second sample case, Allen can withdraw two 20 dollar bills and three 1 dollar bills.
In the third sample case, Allen can withdraw 100000000 (ten million!) 100 dollar bills.
Tags: dp, greedy
Correct Solution:
```
N = int(input())
F, tmp = [0]*(N%100+1), 0
F[0] = 0
if N >= 100:
tmp = N // 100
#print(N)
for i in range(1, N%100+1):
if i >= 100: F[i] = 1 + min(F[i-1], F[i-5], F[i-10], F[i-20], F[i-100])
elif i >= 20: F[i] = 1 + min(F[i-1], F[i-5], F[i-10], F[i-20])
elif i >= 10: F[i] = 1 + min(F[i-1], F[i-5], F[i-10])
elif i >= 5: F[i] = 1 + min(F[i-1], F[i-5])
elif i >= 1: F[i] = 1 + F[i-1]
print(F[N%100] + tmp)
```
| 105,883 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Allen has a LOT of money. He has n dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are 1, 5, 10, 20, 100. What is the minimum number of bills Allen could receive after withdrawing his entire balance?
Input
The first and only line of input contains a single integer n (1 ≤ n ≤ 10^9).
Output
Output the minimum number of bills that Allen could receive.
Examples
Input
125
Output
3
Input
43
Output
5
Input
1000000000
Output
10000000
Note
In the first sample case, Allen can withdraw this with a 100 dollar bill, a 20 dollar bill, and a 5 dollar bill. There is no way for Allen to receive 125 dollars in one or two bills.
In the second sample case, Allen can withdraw two 20 dollar bills and three 1 dollar bills.
In the third sample case, Allen can withdraw 100000000 (ten million!) 100 dollar bills.
Tags: dp, greedy
Correct Solution:
```
import math
n = int(input())
ans = n//100
n=n%100
ans += n//20
n = n%20
ans += n//10
n = n%10
ans += n//5
n = n%5
ans += n
print(ans)
```
| 105,884 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Allen has a LOT of money. He has n dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are 1, 5, 10, 20, 100. What is the minimum number of bills Allen could receive after withdrawing his entire balance?
Input
The first and only line of input contains a single integer n (1 ≤ n ≤ 10^9).
Output
Output the minimum number of bills that Allen could receive.
Examples
Input
125
Output
3
Input
43
Output
5
Input
1000000000
Output
10000000
Note
In the first sample case, Allen can withdraw this with a 100 dollar bill, a 20 dollar bill, and a 5 dollar bill. There is no way for Allen to receive 125 dollars in one or two bills.
In the second sample case, Allen can withdraw two 20 dollar bills and three 1 dollar bills.
In the third sample case, Allen can withdraw 100000000 (ten million!) 100 dollar bills.
Tags: dp, greedy
Correct Solution:
```
n=int(input().replace(',',''))
c=0
if(n>=100):
c+=n//100
n=n%100
if(n>=20):
c+=n//20
n=n%20
if(n>=10):
c+=n//10
n=n%10
if(n>=5):
c+=n//5
n=n%5
print(c+n)
```
| 105,885 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Allen has a LOT of money. He has n dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are 1, 5, 10, 20, 100. What is the minimum number of bills Allen could receive after withdrawing his entire balance?
Input
The first and only line of input contains a single integer n (1 ≤ n ≤ 10^9).
Output
Output the minimum number of bills that Allen could receive.
Examples
Input
125
Output
3
Input
43
Output
5
Input
1000000000
Output
10000000
Note
In the first sample case, Allen can withdraw this with a 100 dollar bill, a 20 dollar bill, and a 5 dollar bill. There is no way for Allen to receive 125 dollars in one or two bills.
In the second sample case, Allen can withdraw two 20 dollar bills and three 1 dollar bills.
In the third sample case, Allen can withdraw 100000000 (ten million!) 100 dollar bills.
Tags: dp, greedy
Correct Solution:
```
def main():
n = int(input())
bill = 0
if n >= 100:
bill += n//100
n %= 100
if n >= 20:
bill+= n//20
n %= 20
if n >= 10:
bill+= n//10
n %= 10
if n >= 5:
bill+= n//5
n %= 5
bill += n
print(bill)
if __name__ == "__main__":
main()
```
| 105,886 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Allen has a LOT of money. He has n dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are 1, 5, 10, 20, 100. What is the minimum number of bills Allen could receive after withdrawing his entire balance?
Input
The first and only line of input contains a single integer n (1 ≤ n ≤ 10^9).
Output
Output the minimum number of bills that Allen could receive.
Examples
Input
125
Output
3
Input
43
Output
5
Input
1000000000
Output
10000000
Note
In the first sample case, Allen can withdraw this with a 100 dollar bill, a 20 dollar bill, and a 5 dollar bill. There is no way for Allen to receive 125 dollars in one or two bills.
In the second sample case, Allen can withdraw two 20 dollar bills and three 1 dollar bills.
In the third sample case, Allen can withdraw 100000000 (ten million!) 100 dollar bills.
Tags: dp, greedy
Correct Solution:
```
# https://codeforces.com/problemset/problem/996/A
# variation of unbounded knapsack and also has greedy appraoch to it
# O(Nd) time - TLE for 10^9 input
# def lottery(total, coins):
# dp = [float("inf")] * (total + 1)
# dp[0] = 0
# for coin in coins:
# for amount in range(1, total + 1):
# if coin <= amount:
# dp[amount] = min(dp[amount], 1 + dp[amount - coin])
# return dp[total]
# n = int(input())
# coins = [1, 5, 10, 20, 100]
# print(lottery(n, coins))
# O(k) - for fixed lenght of demoninations
n = int(input())
a = 0
for i in [100, 20, 10, 5, 1]:
a += n // i
n %= i
print(a)
# print(n // 100 % 100 // 10)
```
| 105,887 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Allen has a LOT of money. He has n dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are 1, 5, 10, 20, 100. What is the minimum number of bills Allen could receive after withdrawing his entire balance?
Input
The first and only line of input contains a single integer n (1 ≤ n ≤ 10^9).
Output
Output the minimum number of bills that Allen could receive.
Examples
Input
125
Output
3
Input
43
Output
5
Input
1000000000
Output
10000000
Note
In the first sample case, Allen can withdraw this with a 100 dollar bill, a 20 dollar bill, and a 5 dollar bill. There is no way for Allen to receive 125 dollars in one or two bills.
In the second sample case, Allen can withdraw two 20 dollar bills and three 1 dollar bills.
In the third sample case, Allen can withdraw 100000000 (ten million!) 100 dollar bills.
Tags: dp, greedy
Correct Solution:
```
n = int(input())
ans=0
ans+=int(n/100)
n%=100
ans+=int(n/20)
n%=20
ans+=int(n/10)
n%=10
ans+=int(n/5)
n%=5
ans+=n
print(ans)
```
| 105,888 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Allen has a LOT of money. He has n dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are 1, 5, 10, 20, 100. What is the minimum number of bills Allen could receive after withdrawing his entire balance?
Input
The first and only line of input contains a single integer n (1 ≤ n ≤ 10^9).
Output
Output the minimum number of bills that Allen could receive.
Examples
Input
125
Output
3
Input
43
Output
5
Input
1000000000
Output
10000000
Note
In the first sample case, Allen can withdraw this with a 100 dollar bill, a 20 dollar bill, and a 5 dollar bill. There is no way for Allen to receive 125 dollars in one or two bills.
In the second sample case, Allen can withdraw two 20 dollar bills and three 1 dollar bills.
In the third sample case, Allen can withdraw 100000000 (ten million!) 100 dollar bills.
Tags: dp, greedy
Correct Solution:
```
n=int(input())
c=0
c=c+n//100
n=n%100
c=c+n//20
n=n%20
c=c+n//10
n=n%10
c=c+n//5
n=n%5
c=c+n
print(c)
```
| 105,889 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Allen has a LOT of money. He has n dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are 1, 5, 10, 20, 100. What is the minimum number of bills Allen could receive after withdrawing his entire balance?
Input
The first and only line of input contains a single integer n (1 ≤ n ≤ 10^9).
Output
Output the minimum number of bills that Allen could receive.
Examples
Input
125
Output
3
Input
43
Output
5
Input
1000000000
Output
10000000
Note
In the first sample case, Allen can withdraw this with a 100 dollar bill, a 20 dollar bill, and a 5 dollar bill. There is no way for Allen to receive 125 dollars in one or two bills.
In the second sample case, Allen can withdraw two 20 dollar bills and three 1 dollar bills.
In the third sample case, Allen can withdraw 100000000 (ten million!) 100 dollar bills.
Submitted Solution:
```
#coding:utf-8
n = int(input())
a = n // 100
r = (n % 100) // 20
c = (n % 20) // 10
l = (n % 10) // 5
b = (n % 5) // 1
f = l + c + r + a + b
print(f)
```
Yes
| 105,890 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Allen has a LOT of money. He has n dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are 1, 5, 10, 20, 100. What is the minimum number of bills Allen could receive after withdrawing his entire balance?
Input
The first and only line of input contains a single integer n (1 ≤ n ≤ 10^9).
Output
Output the minimum number of bills that Allen could receive.
Examples
Input
125
Output
3
Input
43
Output
5
Input
1000000000
Output
10000000
Note
In the first sample case, Allen can withdraw this with a 100 dollar bill, a 20 dollar bill, and a 5 dollar bill. There is no way for Allen to receive 125 dollars in one or two bills.
In the second sample case, Allen can withdraw two 20 dollar bills and three 1 dollar bills.
In the third sample case, Allen can withdraw 100000000 (ten million!) 100 dollar bills.
Submitted Solution:
```
n = int(input())
count = 0
count = count + int(n/100)
n = n%100
count = count + int(n/20)
n = n%20
count = count + int(n/10)
n = n%10
count = count + int(n/5)
n = n%5
count = count + int(n/1)
print(int(count))
```
Yes
| 105,891 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Allen has a LOT of money. He has n dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are 1, 5, 10, 20, 100. What is the minimum number of bills Allen could receive after withdrawing his entire balance?
Input
The first and only line of input contains a single integer n (1 ≤ n ≤ 10^9).
Output
Output the minimum number of bills that Allen could receive.
Examples
Input
125
Output
3
Input
43
Output
5
Input
1000000000
Output
10000000
Note
In the first sample case, Allen can withdraw this with a 100 dollar bill, a 20 dollar bill, and a 5 dollar bill. There is no way for Allen to receive 125 dollars in one or two bills.
In the second sample case, Allen can withdraw two 20 dollar bills and three 1 dollar bills.
In the third sample case, Allen can withdraw 100000000 (ten million!) 100 dollar bills.
Submitted Solution:
```
n=int(input());print((n//100)+(n%100)//20+((n%100)%20)//10+(((n%100)%20)%10)//5+(((n%100)%20)%10)%5)
```
Yes
| 105,892 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Allen has a LOT of money. He has n dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are 1, 5, 10, 20, 100. What is the minimum number of bills Allen could receive after withdrawing his entire balance?
Input
The first and only line of input contains a single integer n (1 ≤ n ≤ 10^9).
Output
Output the minimum number of bills that Allen could receive.
Examples
Input
125
Output
3
Input
43
Output
5
Input
1000000000
Output
10000000
Note
In the first sample case, Allen can withdraw this with a 100 dollar bill, a 20 dollar bill, and a 5 dollar bill. There is no way for Allen to receive 125 dollars in one or two bills.
In the second sample case, Allen can withdraw two 20 dollar bills and three 1 dollar bills.
In the third sample case, Allen can withdraw 100000000 (ten million!) 100 dollar bills.
Submitted Solution:
```
t=int(input())
k=t
v=0
v=t//100
t=t%100
v=v+t//20
t=t%20
v=v+t//10
t=t%10
v=v+t//5
t=t%5
v=v+t
print(v)
```
Yes
| 105,893 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Allen has a LOT of money. He has n dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are 1, 5, 10, 20, 100. What is the minimum number of bills Allen could receive after withdrawing his entire balance?
Input
The first and only line of input contains a single integer n (1 ≤ n ≤ 10^9).
Output
Output the minimum number of bills that Allen could receive.
Examples
Input
125
Output
3
Input
43
Output
5
Input
1000000000
Output
10000000
Note
In the first sample case, Allen can withdraw this with a 100 dollar bill, a 20 dollar bill, and a 5 dollar bill. There is no way for Allen to receive 125 dollars in one or two bills.
In the second sample case, Allen can withdraw two 20 dollar bills and three 1 dollar bills.
In the third sample case, Allen can withdraw 100000000 (ten million!) 100 dollar bills.
Submitted Solution:
```
n = int(input())
cup =[100,20,10,5,1]
count = 0
for i in cup:
if n % i == 0:
count = n//i
break
while n >= i:
n-=i
count+=1
print(count)
```
No
| 105,894 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Allen has a LOT of money. He has n dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are 1, 5, 10, 20, 100. What is the minimum number of bills Allen could receive after withdrawing his entire balance?
Input
The first and only line of input contains a single integer n (1 ≤ n ≤ 10^9).
Output
Output the minimum number of bills that Allen could receive.
Examples
Input
125
Output
3
Input
43
Output
5
Input
1000000000
Output
10000000
Note
In the first sample case, Allen can withdraw this with a 100 dollar bill, a 20 dollar bill, and a 5 dollar bill. There is no way for Allen to receive 125 dollars in one or two bills.
In the second sample case, Allen can withdraw two 20 dollar bills and three 1 dollar bills.
In the third sample case, Allen can withdraw 100000000 (ten million!) 100 dollar bills.
Submitted Solution:
```
n = int(input())
deno = [1,5,10,20,100]
count = 0
while n != 0:
maxi = 0
for i in range(5):
if maxi < deno[i] and deno[i] < n:
maxi = deno[i]
count += 1
n = n % maxi
print(count)
```
No
| 105,895 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Allen has a LOT of money. He has n dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are 1, 5, 10, 20, 100. What is the minimum number of bills Allen could receive after withdrawing his entire balance?
Input
The first and only line of input contains a single integer n (1 ≤ n ≤ 10^9).
Output
Output the minimum number of bills that Allen could receive.
Examples
Input
125
Output
3
Input
43
Output
5
Input
1000000000
Output
10000000
Note
In the first sample case, Allen can withdraw this with a 100 dollar bill, a 20 dollar bill, and a 5 dollar bill. There is no way for Allen to receive 125 dollars in one or two bills.
In the second sample case, Allen can withdraw two 20 dollar bills and three 1 dollar bills.
In the third sample case, Allen can withdraw 100000000 (ten million!) 100 dollar bills.
Submitted Solution:
```
count =0
N =int(input())
if N>=100 and N%100==0:
count = N//100
print(count)
exit()
elif N>100:
while N>100:
count+=1
N=N-100
if N>=20 and N%20==0:
count+=N//20
print(count)
exit()
elif N>20:
while N>20:
count+=1
N-=20
if N>=10 and N%10==0:
count+=N//10
print(count)
exit()
elif N>10:
while N>10:
count+=1
N-=10
if N>=5 and N%5==0:
count+=N//5
print(count)
exit()
elif N>5:
while N>5:
count+=1
N-=5
print(count+N)
print("something")
```
No
| 105,896 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Allen has a LOT of money. He has n dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are 1, 5, 10, 20, 100. What is the minimum number of bills Allen could receive after withdrawing his entire balance?
Input
The first and only line of input contains a single integer n (1 ≤ n ≤ 10^9).
Output
Output the minimum number of bills that Allen could receive.
Examples
Input
125
Output
3
Input
43
Output
5
Input
1000000000
Output
10000000
Note
In the first sample case, Allen can withdraw this with a 100 dollar bill, a 20 dollar bill, and a 5 dollar bill. There is no way for Allen to receive 125 dollars in one or two bills.
In the second sample case, Allen can withdraw two 20 dollar bills and three 1 dollar bills.
In the third sample case, Allen can withdraw 100000000 (ten million!) 100 dollar bills.
Submitted Solution:
```
n=int(input())
c=0
l=[]
while True:
if n!=0:
if n>=100:
c+=1
n=n-100
elif n>=20 and n<=100:
c+=1
n=n-20
elif n>=10 and n<=20:
c+=1
n=n-10
elif n>=5 and n<=10:
c+=5
n=n-5
else:
c+=1
n=n-1
else:
break
print(c)
```
No
| 105,897 |
Provide a correct Python 3 solution for this coding contest problem.
We have N bulbs arranged on a number line, numbered 1 to N from left to right. Bulb i is at coordinate i.
Each bulb has a non-negative integer parameter called intensity. When there is a bulb of intensity d at coordinate x, the bulb illuminates the segment from coordinate x-d-0.5 to x+d+0.5. Initially, the intensity of Bulb i is A_i. We will now do the following operation K times in a row:
* For each integer i between 1 and N (inclusive), let B_i be the number of bulbs illuminating coordinate i. Then, change the intensity of each bulb i to B_i.
Find the intensity of each bulb after the K operations.
Constraints
* 1 \leq N \leq 2 \times 10^5
* 1 \leq K \leq 2 \times 10^5
* 0 \leq A_i \leq N
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 \ldots A_N
Output
Print the intensity A{'}_i of each bulb i after the K operations to Standard Output in the following format:
A{'}_1 A{'}_2 \ldots A{'}_N
Output
Print the intensity A{'}_i of each bulb i after the K operations to Standard Output in the following format:
A{'}_1 A{'}_2 \ldots A{'}_N
Examples
Input
5 1
1 0 0 1 0
Output
1 2 2 1 2
Input
5 2
1 0 0 1 0
Output
3 3 4 4 3
"Correct Solution:
```
n,k,*a,=map(int,open(0).read().split())
def f(li):
b=[0]*(n+1)
for i,x in enumerate(li):
b[max(0,i-x)]+=1
b[min(i+x+1,n)]-=1
for i in range(n):
b[i+1]+=b[i]
return b[:-1]
for _ in range(k):
a=f(a)
if a==[n]*n:
print(*a)
exit()
print(*a)
```
| 105,898 |
Provide a correct Python 3 solution for this coding contest problem.
We have N bulbs arranged on a number line, numbered 1 to N from left to right. Bulb i is at coordinate i.
Each bulb has a non-negative integer parameter called intensity. When there is a bulb of intensity d at coordinate x, the bulb illuminates the segment from coordinate x-d-0.5 to x+d+0.5. Initially, the intensity of Bulb i is A_i. We will now do the following operation K times in a row:
* For each integer i between 1 and N (inclusive), let B_i be the number of bulbs illuminating coordinate i. Then, change the intensity of each bulb i to B_i.
Find the intensity of each bulb after the K operations.
Constraints
* 1 \leq N \leq 2 \times 10^5
* 1 \leq K \leq 2 \times 10^5
* 0 \leq A_i \leq N
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 \ldots A_N
Output
Print the intensity A{'}_i of each bulb i after the K operations to Standard Output in the following format:
A{'}_1 A{'}_2 \ldots A{'}_N
Output
Print the intensity A{'}_i of each bulb i after the K operations to Standard Output in the following format:
A{'}_1 A{'}_2 \ldots A{'}_N
Examples
Input
5 1
1 0 0 1 0
Output
1 2 2 1 2
Input
5 2
1 0 0 1 0
Output
3 3 4 4 3
"Correct Solution:
```
import sys
input = sys.stdin.readline
from itertools import accumulate
n, k = map(int, input().split())
A = tuple(map(int, input().split()))
for _ in range(k):
C = [0]*n
for i, a in enumerate(A):
C[max(0, i-a)] += 1
if i+a < n-1:
C[i+a+1] -= 1
A = tuple(accumulate(C))
if min(A) == n:
break
print(*A)
```
| 105,899 |
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