message stringlengths 2 43.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 853 107k | cluster float64 24 24 | __index_level_0__ int64 1.71k 214k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has a cat and his cat is a real gourmet! Dependent on a day of the week he eats certain type of food:
* on Mondays, Thursdays and Sundays he eats fish food;
* on Tuesdays and Saturdays he eats rabbit stew;
* on other days of week he eats chicken stake.
Polycarp plans to go on a trip and already packed his backpack. His backpack contains:
* a daily rations of fish food;
* b daily rations of rabbit stew;
* c daily rations of chicken stakes.
Polycarp has to choose such day of the week to start his trip that his cat can eat without additional food purchases as long as possible. Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Input
The first line of the input contains three positive integers a, b and c (1 ≤ a, b, c ≤ 7⋅10^8) — the number of daily rations of fish food, rabbit stew and chicken stakes in Polycarps backpack correspondingly.
Output
Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Examples
Input
2 1 1
Output
4
Input
3 2 2
Output
7
Input
1 100 1
Output
3
Input
30 20 10
Output
39
Note
In the first example the best day for start of the trip is Sunday. In this case, during Sunday and Monday the cat will eat fish food, during Tuesday — rabbit stew and during Wednesday — chicken stake. So, after four days of the trip all food will be eaten.
In the second example Polycarp can start his trip in any day of the week. In any case there are food supplies only for one week in Polycarps backpack.
In the third example Polycarp can start his trip in any day, excluding Wednesday, Saturday and Sunday. In this case, the cat will eat three different dishes in three days. Nevertheless that after three days of a trip there will be 99 portions of rabbit stew in a backpack, can cannot eat anything in fourth day of a trip. | instruction | 0 | 105,310 | 24 | 210,620 |
Tags: implementation, math
Correct Solution:
```
[a,b,c] = [int(x) for x in input().split()]
def test(a,b,c) :
k = 0
if min(a,b,c) == c:
if a == 1 and b == 1 :
k = 3
elif a == 1 and b == 2 :
k = 3
elif a == 1 and b > 2 :
k = 3
elif a == 2 and b == 1 :
k = 4
elif a == 2 and b > 1 :
k = 5
elif a > 2 and b == 1 :
k = 5
else :
k = 6
elif min(a,b,c) == b :
if a == 1 and c == 1 :
k = 3
elif a == 1 and c ==2 :
k = 4
elif a == 1 and c >=2 :
k = 4
elif a == 2 and c == 1:
k = 4
elif a == 2 and c > 1 :
k = 4
elif a > 2 and c == 1 :
k = 5
else :
k = 6
elif min(a,b,c) == a :
if a == 1:
if b == 1 and c == 1 :
k = 3
elif b == 1 and c ==2 :
k = 4
elif b == 1 and c >=2 :
k = 4
elif b == 2 and c == 1:
k = 3
elif b == 2 and c > 1 :
k = 5
elif b > 2 and c == 1 :
k = 3
else :
k = 5
elif a == 2 :
if b == 1 and c == 1 :
k = 4
elif b == 1 and c ==2 :
k = 5
elif b == 1 and c >=2 :
k = 5
elif b == 2 and c == 1:
k = 5
elif b == 2 and c > 1 :
k = 6
elif b > 2 and c == 1 :
k = 5
else :
k = 6
return k
if a < 3 or b < 2 or c < 2 :
print(test(a,b,c))
else :
m = (min((a//3) ,(b//2), (c//2)) )
a = a - (m*3)
b = b - (m*2)
c = c - (m*2)
m = m*7
if a == 0 :
if min(b,c) >= 1:
print(m+2)
elif min(b,c) == 0 and max(b,c)== 1 :
print(m+1)
else :
print(m)
elif b == 0:
if min(a,c) >= 2 :
print(m+3)
elif c == 2 and a == 1:
print(m+3)
elif c== 1 and a >= 1 :
print(m+2)
elif c== 1 and a == 0:
print(m+1)
elif c== 0 and a >= 2:
print(m+2)
elif c==0 and a == 1:
print(m+1)
elif c==0 and a == 0:
print(m)
elif c == 0 :
if min(a,b) >= 2 :
print(m+4)
elif b == 2 and a == 1:
print(m+2)
elif b== 1 and a >= 1 :
print(m+3)
elif b== 1 and a == 0:
print(m+1)
elif b== 0 and a >= 2:
print(m+1)
elif b==0 and a == 1:
print(m+1)
elif b==0 and a == 0:
print(m)
else :
print(test(a,b,c) + m)
``` | output | 1 | 105,310 | 24 | 210,621 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has a cat and his cat is a real gourmet! Dependent on a day of the week he eats certain type of food:
* on Mondays, Thursdays and Sundays he eats fish food;
* on Tuesdays and Saturdays he eats rabbit stew;
* on other days of week he eats chicken stake.
Polycarp plans to go on a trip and already packed his backpack. His backpack contains:
* a daily rations of fish food;
* b daily rations of rabbit stew;
* c daily rations of chicken stakes.
Polycarp has to choose such day of the week to start his trip that his cat can eat without additional food purchases as long as possible. Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Input
The first line of the input contains three positive integers a, b and c (1 ≤ a, b, c ≤ 7⋅10^8) — the number of daily rations of fish food, rabbit stew and chicken stakes in Polycarps backpack correspondingly.
Output
Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Examples
Input
2 1 1
Output
4
Input
3 2 2
Output
7
Input
1 100 1
Output
3
Input
30 20 10
Output
39
Note
In the first example the best day for start of the trip is Sunday. In this case, during Sunday and Monday the cat will eat fish food, during Tuesday — rabbit stew and during Wednesday — chicken stake. So, after four days of the trip all food will be eaten.
In the second example Polycarp can start his trip in any day of the week. In any case there are food supplies only for one week in Polycarps backpack.
In the third example Polycarp can start his trip in any day, excluding Wednesday, Saturday and Sunday. In this case, the cat will eat three different dishes in three days. Nevertheless that after three days of a trip there will be 99 portions of rabbit stew in a backpack, can cannot eat anything in fourth day of a trip. | instruction | 0 | 105,311 | 24 | 210,622 |
Tags: implementation, math
Correct Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
##################################
# University of Wisconsin-Madison
# Author: Yaqi Zhang
##################################
# This module contains
##################################
# standard library
import sys
import math
def main():
nums = list(map(int, input().split()))
days = [{0, 3, 6}, {1, 5}, {2, 4}]
ans = -math.inf
for i in range(7):
mn = math.inf
for num, day in zip(nums, days):
cnt = 0
cur = i
while True:
if cur == 0:
cnt += num // len(day) * 7
num = num % len(day)
if cur in day:
num -= 1
if num < 0:
break
cnt += 1
cur = (cur + 1) % 7
mn = min(cnt, mn)
ans = max(ans, mn)
print(ans)
if __name__ == "__main__":
main()
``` | output | 1 | 105,311 | 24 | 210,623 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has a cat and his cat is a real gourmet! Dependent on a day of the week he eats certain type of food:
* on Mondays, Thursdays and Sundays he eats fish food;
* on Tuesdays and Saturdays he eats rabbit stew;
* on other days of week he eats chicken stake.
Polycarp plans to go on a trip and already packed his backpack. His backpack contains:
* a daily rations of fish food;
* b daily rations of rabbit stew;
* c daily rations of chicken stakes.
Polycarp has to choose such day of the week to start his trip that his cat can eat without additional food purchases as long as possible. Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Input
The first line of the input contains three positive integers a, b and c (1 ≤ a, b, c ≤ 7⋅10^8) — the number of daily rations of fish food, rabbit stew and chicken stakes in Polycarps backpack correspondingly.
Output
Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Examples
Input
2 1 1
Output
4
Input
3 2 2
Output
7
Input
1 100 1
Output
3
Input
30 20 10
Output
39
Note
In the first example the best day for start of the trip is Sunday. In this case, during Sunday and Monday the cat will eat fish food, during Tuesday — rabbit stew and during Wednesday — chicken stake. So, after four days of the trip all food will be eaten.
In the second example Polycarp can start his trip in any day of the week. In any case there are food supplies only for one week in Polycarps backpack.
In the third example Polycarp can start his trip in any day, excluding Wednesday, Saturday and Sunday. In this case, the cat will eat three different dishes in three days. Nevertheless that after three days of a trip there will be 99 portions of rabbit stew in a backpack, can cannot eat anything in fourth day of a trip. | instruction | 0 | 105,312 | 24 | 210,624 |
Tags: implementation, math
Correct Solution:
```
'''
Author: Ken
Problems: 1154C
'''
eat = []
eat = list(map(int, input().split()))
temp = min(int(eat[0]/3), int(eat[1]/2), int(eat[2]/2))
ans = temp*7
eat[0] -= temp*3
eat[1] -= temp*2
eat[2] -= temp*2
day = [0, 1, 2, 0, 2, 1, 0]
cir = -1
for i in range(7):
count = 0
temp = list.copy(eat)
for j in range(7):
# print(eat)
if temp[day[(i+j) % 7]] == 0:
cir = max(cir, count)
break
count += 1
temp[day[(i+j) % 7]] -= 1
print(ans+cir)
# 1 4 7
# 2 6
# 3 5
# abcacba
``` | output | 1 | 105,312 | 24 | 210,625 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has a cat and his cat is a real gourmet! Dependent on a day of the week he eats certain type of food:
* on Mondays, Thursdays and Sundays he eats fish food;
* on Tuesdays and Saturdays he eats rabbit stew;
* on other days of week he eats chicken stake.
Polycarp plans to go on a trip and already packed his backpack. His backpack contains:
* a daily rations of fish food;
* b daily rations of rabbit stew;
* c daily rations of chicken stakes.
Polycarp has to choose such day of the week to start his trip that his cat can eat without additional food purchases as long as possible. Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Input
The first line of the input contains three positive integers a, b and c (1 ≤ a, b, c ≤ 7⋅10^8) — the number of daily rations of fish food, rabbit stew and chicken stakes in Polycarps backpack correspondingly.
Output
Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Examples
Input
2 1 1
Output
4
Input
3 2 2
Output
7
Input
1 100 1
Output
3
Input
30 20 10
Output
39
Note
In the first example the best day for start of the trip is Sunday. In this case, during Sunday and Monday the cat will eat fish food, during Tuesday — rabbit stew and during Wednesday — chicken stake. So, after four days of the trip all food will be eaten.
In the second example Polycarp can start his trip in any day of the week. In any case there are food supplies only for one week in Polycarps backpack.
In the third example Polycarp can start his trip in any day, excluding Wednesday, Saturday and Sunday. In this case, the cat will eat three different dishes in three days. Nevertheless that after three days of a trip there will be 99 portions of rabbit stew in a backpack, can cannot eat anything in fourth day of a trip.
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Wed Apr 17 06:57:09 2019
@author: sharath
"""
l = list(map(int, input().split()))
x = min(l[0]//3, l[1]//2, l[2]//2)
l[0] -= 3*x
l[1] -= 2*x
l[2] -= 2*x
x *= 7
lista=[0, 1, 2, 0, 2, 1, 0]*2
m=c=0
a=[i for i in l]
for i in lista:
if l[i]:
c+=1
l[i]-=1
else:
c=0
l=[j for j in a]
m=max(m,c)
print(x+m)
``` | instruction | 0 | 105,313 | 24 | 210,626 |
Yes | output | 1 | 105,313 | 24 | 210,627 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has a cat and his cat is a real gourmet! Dependent on a day of the week he eats certain type of food:
* on Mondays, Thursdays and Sundays he eats fish food;
* on Tuesdays and Saturdays he eats rabbit stew;
* on other days of week he eats chicken stake.
Polycarp plans to go on a trip and already packed his backpack. His backpack contains:
* a daily rations of fish food;
* b daily rations of rabbit stew;
* c daily rations of chicken stakes.
Polycarp has to choose such day of the week to start his trip that his cat can eat without additional food purchases as long as possible. Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Input
The first line of the input contains three positive integers a, b and c (1 ≤ a, b, c ≤ 7⋅10^8) — the number of daily rations of fish food, rabbit stew and chicken stakes in Polycarps backpack correspondingly.
Output
Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Examples
Input
2 1 1
Output
4
Input
3 2 2
Output
7
Input
1 100 1
Output
3
Input
30 20 10
Output
39
Note
In the first example the best day for start of the trip is Sunday. In this case, during Sunday and Monday the cat will eat fish food, during Tuesday — rabbit stew and during Wednesday — chicken stake. So, after four days of the trip all food will be eaten.
In the second example Polycarp can start his trip in any day of the week. In any case there are food supplies only for one week in Polycarps backpack.
In the third example Polycarp can start his trip in any day, excluding Wednesday, Saturday and Sunday. In this case, the cat will eat three different dishes in three days. Nevertheless that after three days of a trip there will be 99 portions of rabbit stew in a backpack, can cannot eat anything in fourth day of a trip.
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Tue Apr 16 16:37:40 2019
@author: Acer
"""
def solve(a,b,c):
"""
input is a list s
"""
small = min(int(a/3),int(b/2),int(c/2));
totalDay = small*7;
a = a - small*3;
b = b - small*2;
c = c - small*2;
if (a == 0):
totalDay = totalDay + int(b > 0) + int(c > 0);
elif(a == 1):
totalDay = totalDay + 1;
if (c > 1):
totalDay = totalDay + 2;
totalDay = totalDay + int(b > 0) + int(b > 1);
elif (c > 0):
totalDay = totalDay + 1;
totalDay = totalDay + int(b > 0);
else:
totalDay = totalDay + int(b > 0);
elif(a == 2):
totalDay = totalDay + 2;
if (b > 1):
totalDay = totalDay + 2 + int(c > 0) + int(c > 1);
elif (b > 0):
totalDay = totalDay + 1 + int(c > 1) + int(c > 0);
else:
if (c > 2):
totalDay = totalDay + 1;
else:
if (b != 0) and (c != 0):
totalDay = totalDay + 3 + 2 + int(b > 1) + int(c > 1);
elif (b == 0):
totalDay = max(totalDay + 1 + int(c > 0) + int(c > 1), totalDay + 2);
elif (c == 0):
totalDay = totalDay + 2 + int(b > 0) + int(b > 1);
else:
totalDay = totalDay + 2;
return totalDay;
if __name__ == "__main__":
"""the solve(*args) structure is needed for testing purporses"""
a,b,c = [int(s) for s in input().split()];
print(solve(a,b,c));
``` | instruction | 0 | 105,314 | 24 | 210,628 |
Yes | output | 1 | 105,314 | 24 | 210,629 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has a cat and his cat is a real gourmet! Dependent on a day of the week he eats certain type of food:
* on Mondays, Thursdays and Sundays he eats fish food;
* on Tuesdays and Saturdays he eats rabbit stew;
* on other days of week he eats chicken stake.
Polycarp plans to go on a trip and already packed his backpack. His backpack contains:
* a daily rations of fish food;
* b daily rations of rabbit stew;
* c daily rations of chicken stakes.
Polycarp has to choose such day of the week to start his trip that his cat can eat without additional food purchases as long as possible. Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Input
The first line of the input contains three positive integers a, b and c (1 ≤ a, b, c ≤ 7⋅10^8) — the number of daily rations of fish food, rabbit stew and chicken stakes in Polycarps backpack correspondingly.
Output
Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Examples
Input
2 1 1
Output
4
Input
3 2 2
Output
7
Input
1 100 1
Output
3
Input
30 20 10
Output
39
Note
In the first example the best day for start of the trip is Sunday. In this case, during Sunday and Monday the cat will eat fish food, during Tuesday — rabbit stew and during Wednesday — chicken stake. So, after four days of the trip all food will be eaten.
In the second example Polycarp can start his trip in any day of the week. In any case there are food supplies only for one week in Polycarps backpack.
In the third example Polycarp can start his trip in any day, excluding Wednesday, Saturday and Sunday. In this case, the cat will eat three different dishes in three days. Nevertheless that after three days of a trip there will be 99 portions of rabbit stew in a backpack, can cannot eat anything in fourth day of a trip.
Submitted Solution:
```
a,b,c = map(int,input().split())
A = ["a","a","b","c","a","c","b"]
x = a//3;y=b//2;z=c//2
ans = min(x,y,z)*7
#print(x,y,z,min(x,y,z))
s = min(x,y,z)
x -= s
y -= s
z -= s
#print(x,y,z)
a = x*3 + a%3;b = y*2 +b%2;c = z*2+c%2
#print(a,b,c)
i = 0
fin_ans = 0
while(i<7):
j=i;t = 0
p = a;q=b;r=c
while(True):
#print(p,q,r,t)
if(A[j%7]=="a"):
p-=1
elif(A[j%7]=="b"):
q-=1
elif(A[j%7]=="c"):
r-=1
if(p<0 or q<0 or r<0):
break
j+=1;t+=1
fin_ans = max(fin_ans,t)
i+=1
print(ans+fin_ans)
``` | instruction | 0 | 105,315 | 24 | 210,630 |
Yes | output | 1 | 105,315 | 24 | 210,631 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has a cat and his cat is a real gourmet! Dependent on a day of the week he eats certain type of food:
* on Mondays, Thursdays and Sundays he eats fish food;
* on Tuesdays and Saturdays he eats rabbit stew;
* on other days of week he eats chicken stake.
Polycarp plans to go on a trip and already packed his backpack. His backpack contains:
* a daily rations of fish food;
* b daily rations of rabbit stew;
* c daily rations of chicken stakes.
Polycarp has to choose such day of the week to start his trip that his cat can eat without additional food purchases as long as possible. Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Input
The first line of the input contains three positive integers a, b and c (1 ≤ a, b, c ≤ 7⋅10^8) — the number of daily rations of fish food, rabbit stew and chicken stakes in Polycarps backpack correspondingly.
Output
Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Examples
Input
2 1 1
Output
4
Input
3 2 2
Output
7
Input
1 100 1
Output
3
Input
30 20 10
Output
39
Note
In the first example the best day for start of the trip is Sunday. In this case, during Sunday and Monday the cat will eat fish food, during Tuesday — rabbit stew and during Wednesday — chicken stake. So, after four days of the trip all food will be eaten.
In the second example Polycarp can start his trip in any day of the week. In any case there are food supplies only for one week in Polycarps backpack.
In the third example Polycarp can start his trip in any day, excluding Wednesday, Saturday and Sunday. In this case, the cat will eat three different dishes in three days. Nevertheless that after three days of a trip there will be 99 portions of rabbit stew in a backpack, can cannot eat anything in fourth day of a trip.
Submitted Solution:
```
a,b,c = map(int , input().split())
fish = [0,1,4]
rabbit = [2,6]
chick = [3,5]
x,y,z = a,b,c
maxdayCount = 0
count = 0
for start in range(7) :
day = start
count = 0
while(x > -1 and y > -1 and z > -1):
weeks = int(min(x/3 , y/2, z/2))
# x = int(x/3) + x%3
# y = int(y/2) + y%2
# z = int(z/2) + z%2
count += 7 * weeks
x -= 3 * weeks
y -= 2 * weeks
z -= 2 * weeks
# print(count , x, y ,z)
if day in fish :
x -= 1
elif day in rabbit :
y -= 1
else:
z -= 1
if (x < 0 or y < 0 or z < 0):
break
day += 1
count += 1
day = day % 7
if count > maxdayCount :
maxdayCount = count
x,y,z = a,b,c
print(maxdayCount)
``` | instruction | 0 | 105,316 | 24 | 210,632 |
Yes | output | 1 | 105,316 | 24 | 210,633 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has a cat and his cat is a real gourmet! Dependent on a day of the week he eats certain type of food:
* on Mondays, Thursdays and Sundays he eats fish food;
* on Tuesdays and Saturdays he eats rabbit stew;
* on other days of week he eats chicken stake.
Polycarp plans to go on a trip and already packed his backpack. His backpack contains:
* a daily rations of fish food;
* b daily rations of rabbit stew;
* c daily rations of chicken stakes.
Polycarp has to choose such day of the week to start his trip that his cat can eat without additional food purchases as long as possible. Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Input
The first line of the input contains three positive integers a, b and c (1 ≤ a, b, c ≤ 7⋅10^8) — the number of daily rations of fish food, rabbit stew and chicken stakes in Polycarps backpack correspondingly.
Output
Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Examples
Input
2 1 1
Output
4
Input
3 2 2
Output
7
Input
1 100 1
Output
3
Input
30 20 10
Output
39
Note
In the first example the best day for start of the trip is Sunday. In this case, during Sunday and Monday the cat will eat fish food, during Tuesday — rabbit stew and during Wednesday — chicken stake. So, after four days of the trip all food will be eaten.
In the second example Polycarp can start his trip in any day of the week. In any case there are food supplies only for one week in Polycarps backpack.
In the third example Polycarp can start his trip in any day, excluding Wednesday, Saturday and Sunday. In this case, the cat will eat three different dishes in three days. Nevertheless that after three days of a trip there will be 99 portions of rabbit stew in a backpack, can cannot eat anything in fourth day of a trip.
Submitted Solution:
```
x = list(map(int, input().split()))
# if x == [0, 1, 1] or x == [1, 0, 1]:
# print(2)
# elif x == [0, 0, 1] or x == [0, 1, 0]:
# print(1)
# elif x == [1, 2, 2]:
# # print(5)
# pass
# elif x == [1, 1, 2]:
# # print(4)
# pass
# else:
d = {0: 3, 1: 2, 2: 2}
a = [x[i] // d[i] for i in range(3)]
weeks = min(a)
least = a.index(min(a))
b = [x[i] - d[i] * weeks for i in range(3)]
days = weeks * 7
p = [0, 1, 2, 0, 2, 1, 0]
if weeks > 0:
i = len(p) - 1
while i >= 0 and b[p[i]] > 0:
days += 1
b[p[i]] -= 1
i -= 1
i = 0
while i < len(p) and b[p[i]] > 0:
days += 1
b[p[i]] -= 1
i += 1
else:
for j in range(7):
b = [x[i] - d[i] * weeks for i in range(3)]
cd = 0
while j < len(p) and b[p[j]] > 0:
cd += 1
b[p[j]] -= 1
j += 1
if j == len(p) and b[0] > 0:
cd += 1
if b[1] > 0:
cd += 1
if b[2] > 0:
cd += 1
days = max(days, cd)
# if x == [3, 1, 1]:
# print(5)
# else:
# print(days)
# 3 2 1 -> 6
print(days)
``` | instruction | 0 | 105,317 | 24 | 210,634 |
No | output | 1 | 105,317 | 24 | 210,635 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has a cat and his cat is a real gourmet! Dependent on a day of the week he eats certain type of food:
* on Mondays, Thursdays and Sundays he eats fish food;
* on Tuesdays and Saturdays he eats rabbit stew;
* on other days of week he eats chicken stake.
Polycarp plans to go on a trip and already packed his backpack. His backpack contains:
* a daily rations of fish food;
* b daily rations of rabbit stew;
* c daily rations of chicken stakes.
Polycarp has to choose such day of the week to start his trip that his cat can eat without additional food purchases as long as possible. Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Input
The first line of the input contains three positive integers a, b and c (1 ≤ a, b, c ≤ 7⋅10^8) — the number of daily rations of fish food, rabbit stew and chicken stakes in Polycarps backpack correspondingly.
Output
Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Examples
Input
2 1 1
Output
4
Input
3 2 2
Output
7
Input
1 100 1
Output
3
Input
30 20 10
Output
39
Note
In the first example the best day for start of the trip is Sunday. In this case, during Sunday and Monday the cat will eat fish food, during Tuesday — rabbit stew and during Wednesday — chicken stake. So, after four days of the trip all food will be eaten.
In the second example Polycarp can start his trip in any day of the week. In any case there are food supplies only for one week in Polycarps backpack.
In the third example Polycarp can start his trip in any day, excluding Wednesday, Saturday and Sunday. In this case, the cat will eat three different dishes in three days. Nevertheless that after three days of a trip there will be 99 portions of rabbit stew in a backpack, can cannot eat anything in fourth day of a trip.
Submitted Solution:
```
import math
list1=list(map(int,input().split()))
list2=[list1[0]/3,list1[1]/2,list1[2]/2]
list3=[]
n=list2.index(min(list2))
m=math.floor(list2[n])
list3.append(list1[0]-3*m)
list3.append(list1[1]-2*m)
list3.append(list1[2]-2*m)
food=[0,1,2,0,2,1,0]
day = [0,0,0,0,0,0,0]
for i in range(7):
for j in range(7):
if list3[food[(j+i)%7]] != 0:
list3[food[(j+i)%7]] -= 1
day[i] += 1
else:
break
print(m*7+max(day))
``` | instruction | 0 | 105,318 | 24 | 210,636 |
No | output | 1 | 105,318 | 24 | 210,637 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has a cat and his cat is a real gourmet! Dependent on a day of the week he eats certain type of food:
* on Mondays, Thursdays and Sundays he eats fish food;
* on Tuesdays and Saturdays he eats rabbit stew;
* on other days of week he eats chicken stake.
Polycarp plans to go on a trip and already packed his backpack. His backpack contains:
* a daily rations of fish food;
* b daily rations of rabbit stew;
* c daily rations of chicken stakes.
Polycarp has to choose such day of the week to start his trip that his cat can eat without additional food purchases as long as possible. Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Input
The first line of the input contains three positive integers a, b and c (1 ≤ a, b, c ≤ 7⋅10^8) — the number of daily rations of fish food, rabbit stew and chicken stakes in Polycarps backpack correspondingly.
Output
Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Examples
Input
2 1 1
Output
4
Input
3 2 2
Output
7
Input
1 100 1
Output
3
Input
30 20 10
Output
39
Note
In the first example the best day for start of the trip is Sunday. In this case, during Sunday and Monday the cat will eat fish food, during Tuesday — rabbit stew and during Wednesday — chicken stake. So, after four days of the trip all food will be eaten.
In the second example Polycarp can start his trip in any day of the week. In any case there are food supplies only for one week in Polycarps backpack.
In the third example Polycarp can start his trip in any day, excluding Wednesday, Saturday and Sunday. In this case, the cat will eat three different dishes in three days. Nevertheless that after three days of a trip there will be 99 portions of rabbit stew in a backpack, can cannot eat anything in fourth day of a trip.
Submitted Solution:
```
ip = [int(i) for i in input().split()]
answers = []
store = ip.copy()
def fun(start):
ans = 0
if start == 0:
for i in range(sum(ip)):
day = i%7
if day == 0 or day == 3 or day == 6:
if ip[0] == 0:
break
ip[0]-=1
ans+=1
elif day == 1 or day == 5:
if ip[1] == 0:
break
ip[1]-=1
ans+=1
else:
if ip[2] == 0:
break
ip[2]-=1
ans+=1
return ans
if start == 1:
for i in range(sum(ip)):
day = i%7
if day == 2 or day == 5 or day == 6:
if ip[0] == 0:
break
ip[0]-=1
ans+=1
elif day == 0 or day == 4:
if ip[1] == 0:
break
ip[1]-=1
ans+=1
else:
if ip[2] == 0:
break
ip[2]-=1
ans+=1
return ans
if start == 2:
for i in range(sum(ip)):
day = i%7
if day == 1 or day == 4 or day == 5:
if ip[0] == 0:
break
ip[0]-=1
ans+=1
elif day == 3 or day == 6:
if ip[1] == 0:
break
ip[1]-=1
ans+=1
else:
if ip[2] == 0:
break
ip[2]-=1
ans+=1
return ans
if start == 3:
for i in range(sum(ip)):
day = i%7
if day == 0 or day == 3 or day == 4:
if ip[0] == 0:
break
ip[0]-=1
ans+=1
elif day == 2 or day == 5:
if ip[1] == 0:
break
ip[1]-=1
ans+=1
else:
if ip[2] == 0:
break
ip[2]-=1
ans+=1
return ans
if start == 4:
for i in range(sum(ip)):
day = i%7
if day == 6 or day == 2 or day == 3:
if ip[0] == 0:
break
ip[0]-=1
ans+=1
elif day == 1 or day == 4:
if ip[1] == 0:
break
ip[1]-=1
ans+=1
else:
if ip[2] == 0:
break
ip[2]-=1
ans+=1
return ans
if start == 5:
for i in range(sum(ip)):
day = i%7
if day == 5 or day == 1 or day == 2:
if ip[0] == 0:
break
ip[0]-=1
ans+=1
elif day == 0 or day == 3:
if ip[1] == 0:
break
ip[1]-=1
ans+=1
else:
if ip[2] == 0:
break
ip[2]-=1
ans+=1
return ans
if start == 6:
for i in range(sum(ip)):
day = i%7
if day == 4 or day == 0 or day == 1:
if ip[0] == 0:
break
ip[0]-=1
ans+=1
elif day == 6 or day == 2:
if ip[1] == 0:
break
ip[1]-=1
ans+=1
else:
if ip[2] == 0:
break
ip[2]-=1
ans+=1
return ans
ip = store.copy()
for i in range(7):
answers.append(fun(i))
print(max(answers))
``` | instruction | 0 | 105,319 | 24 | 210,638 |
No | output | 1 | 105,319 | 24 | 210,639 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has a cat and his cat is a real gourmet! Dependent on a day of the week he eats certain type of food:
* on Mondays, Thursdays and Sundays he eats fish food;
* on Tuesdays and Saturdays he eats rabbit stew;
* on other days of week he eats chicken stake.
Polycarp plans to go on a trip and already packed his backpack. His backpack contains:
* a daily rations of fish food;
* b daily rations of rabbit stew;
* c daily rations of chicken stakes.
Polycarp has to choose such day of the week to start his trip that his cat can eat without additional food purchases as long as possible. Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Input
The first line of the input contains three positive integers a, b and c (1 ≤ a, b, c ≤ 7⋅10^8) — the number of daily rations of fish food, rabbit stew and chicken stakes in Polycarps backpack correspondingly.
Output
Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Examples
Input
2 1 1
Output
4
Input
3 2 2
Output
7
Input
1 100 1
Output
3
Input
30 20 10
Output
39
Note
In the first example the best day for start of the trip is Sunday. In this case, during Sunday and Monday the cat will eat fish food, during Tuesday — rabbit stew and during Wednesday — chicken stake. So, after four days of the trip all food will be eaten.
In the second example Polycarp can start his trip in any day of the week. In any case there are food supplies only for one week in Polycarps backpack.
In the third example Polycarp can start his trip in any day, excluding Wednesday, Saturday and Sunday. In this case, the cat will eat three different dishes in three days. Nevertheless that after three days of a trip there will be 99 portions of rabbit stew in a backpack, can cannot eat anything in fourth day of a trip.
Submitted Solution:
```
import math
x = list(map(int,input().split()))
if sum(x) <= 7:
print(sum(x))
else:
if min(x) * 3 <= 7:
print(min(x) * 3)
else:
r = []
r.append(x[0]/3)
r.append(x[1]/2)
r.append(x[2]/2)
if r[0] == min(r):
hasil = math.floor(x[0]/3)
sisa1 = x[1] - (hasil * 2)
sisa2 = x[2] - (hasil * 2)
#print(sisa1,sisa2)
rem = x[0] % 3
if rem == 0:
if sisa1 > 1 and sisa2 > 1:
print(7 * hasil + 2)
else:
print(7 * hasil + 1)
elif rem == 1:
if sisa1 > 2 and sisa2 > 2:
print(7 * hasil + 5)
else:
print(7 * hasil + 3)
else:
print(7 * hasil + 6)
elif r[1] == min(r):
hasil = math.floor(x[1]/2)
sisa0 = x[0] - (hasil * 3)
sisa2 = x[2] - (hasil * 2)
#print(sisa0,sisa2)
rem = x[1] % 2
if rem == 0:
if sisa0 > 1 and sisa2 > 2:
print(7 * hasil + 3)
else:
print(7 * hasil + sisa0 + sisa2)
else:
if sisa0 > 3 and sisa2 > 2:
print(7 * hasil + 6)
else:
print(7 * hasil + rem + sisa0 + sisa2)
else:
hasil = math.floor(x[2]/2)
sisa0 = x[0] - (hasil * 3)
sisa1 = x[1] - (hasil * 2)
#print(sisa0,sisa1)
rem = x[2] % 2
if rem == 0:
if sisa0 > 2 and sisa1 > 2:
print(7 * hasil + 4)
else:
print(7 * hasil + sisa0 + sisa1)
else:
if sisa0 > 2 and sisa1 > 3:
print(7 * hasil + 6)
else:
print(7 * hasil + rem + sisa0 + sisa1)
``` | instruction | 0 | 105,320 | 24 | 210,640 |
No | output | 1 | 105,320 | 24 | 210,641 |
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. | instruction | 0 | 105,829 | 24 | 211,658 |
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)
``` | output | 1 | 105,829 | 24 | 211,659 |
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. | instruction | 0 | 105,830 | 24 | 211,660 |
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)
``` | output | 1 | 105,830 | 24 | 211,661 |
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. | instruction | 0 | 105,831 | 24 | 211,662 |
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))
``` | output | 1 | 105,831 | 24 | 211,663 |
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. | instruction | 0 | 105,832 | 24 | 211,664 |
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)))
``` | output | 1 | 105,832 | 24 | 211,665 |
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. | instruction | 0 | 105,833 | 24 | 211,666 |
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)
``` | output | 1 | 105,833 | 24 | 211,667 |
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. | instruction | 0 | 105,834 | 24 | 211,668 |
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))
``` | output | 1 | 105,834 | 24 | 211,669 |
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. | instruction | 0 | 105,835 | 24 | 211,670 |
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())
``` | output | 1 | 105,835 | 24 | 211,671 |
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. | instruction | 0 | 105,836 | 24 | 211,672 |
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)
``` | output | 1 | 105,836 | 24 | 211,673 |
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. | instruction | 0 | 105,837 | 24 | 211,674 |
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)
``` | output | 1 | 105,837 | 24 | 211,675 |
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))
``` | instruction | 0 | 105,838 | 24 | 211,676 |
Yes | output | 1 | 105,838 | 24 | 211,677 |
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)
``` | instruction | 0 | 105,839 | 24 | 211,678 |
Yes | output | 1 | 105,839 | 24 | 211,679 |
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)
``` | instruction | 0 | 105,840 | 24 | 211,680 |
Yes | output | 1 | 105,840 | 24 | 211,681 |
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)
``` | instruction | 0 | 105,841 | 24 | 211,682 |
Yes | output | 1 | 105,841 | 24 | 211,683 |
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)
``` | instruction | 0 | 105,842 | 24 | 211,684 |
No | output | 1 | 105,842 | 24 | 211,685 |
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)
``` | instruction | 0 | 105,843 | 24 | 211,686 |
No | output | 1 | 105,843 | 24 | 211,687 |
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))))
``` | instruction | 0 | 105,844 | 24 | 211,688 |
No | output | 1 | 105,844 | 24 | 211,689 |
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)
``` | instruction | 0 | 105,845 | 24 | 211,690 |
No | output | 1 | 105,845 | 24 | 211,691 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Monocarp has decided to buy a new TV set and hang it on the wall in his flat. The wall has enough free space so Monocarp can buy a TV set with screen width not greater than a and screen height not greater than b. Monocarp is also used to TV sets with a certain aspect ratio: formally, if the width of the screen is w, and the height of the screen is h, then the following condition should be met: w/h = x/y.
There are many different TV sets in the shop. Monocarp is sure that for any pair of positive integers w and h there is a TV set with screen width w and height h in the shop.
Monocarp isn't ready to choose the exact TV set he is going to buy. Firstly he wants to determine the optimal screen resolution. He has decided to try all possible variants of screen size. But he must count the number of pairs of positive integers w and h, beforehand, such that (w ≤ a), (h ≤ b) and (w/h = x/y).
In other words, Monocarp wants to determine the number of TV sets having aspect ratio x/y, screen width not exceeding a, and screen height not exceeding b. Two TV sets are considered different if they have different screen width or different screen height.
Input
The first line contains four integers a, b, x, y (1 ≤ a, b, x, y ≤ 10^{18}) — the constraints on the screen width and height, and on the aspect ratio.
Output
Print one integer — the number of different variants to choose TV screen width and screen height so that they meet the aforementioned constraints.
Examples
Input
17 15 5 3
Output
3
Input
14 16 7 22
Output
0
Input
4 2 6 4
Output
1
Input
1000000000000000000 1000000000000000000 999999866000004473 999999822000007597
Output
1000000063
Note
In the first example, there are 3 possible variants: (5, 3), (10, 6), (15, 9).
In the second example, there is no TV set meeting the constraints.
In the third example, there is only one variant: (3, 2). | instruction | 0 | 107,033 | 24 | 214,066 |
Tags: math
Correct Solution:
```
import math
I=lambda:map(int,input().split())
a,b,x,y=I()
g1=math.gcd(x,y)
x=x//g1
y=y//g1
print(min(a//x,b//y))
``` | output | 1 | 107,033 | 24 | 214,067 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Monocarp has decided to buy a new TV set and hang it on the wall in his flat. The wall has enough free space so Monocarp can buy a TV set with screen width not greater than a and screen height not greater than b. Monocarp is also used to TV sets with a certain aspect ratio: formally, if the width of the screen is w, and the height of the screen is h, then the following condition should be met: w/h = x/y.
There are many different TV sets in the shop. Monocarp is sure that for any pair of positive integers w and h there is a TV set with screen width w and height h in the shop.
Monocarp isn't ready to choose the exact TV set he is going to buy. Firstly he wants to determine the optimal screen resolution. He has decided to try all possible variants of screen size. But he must count the number of pairs of positive integers w and h, beforehand, such that (w ≤ a), (h ≤ b) and (w/h = x/y).
In other words, Monocarp wants to determine the number of TV sets having aspect ratio x/y, screen width not exceeding a, and screen height not exceeding b. Two TV sets are considered different if they have different screen width or different screen height.
Input
The first line contains four integers a, b, x, y (1 ≤ a, b, x, y ≤ 10^{18}) — the constraints on the screen width and height, and on the aspect ratio.
Output
Print one integer — the number of different variants to choose TV screen width and screen height so that they meet the aforementioned constraints.
Examples
Input
17 15 5 3
Output
3
Input
14 16 7 22
Output
0
Input
4 2 6 4
Output
1
Input
1000000000000000000 1000000000000000000 999999866000004473 999999822000007597
Output
1000000063
Note
In the first example, there are 3 possible variants: (5, 3), (10, 6), (15, 9).
In the second example, there is no TV set meeting the constraints.
In the third example, there is only one variant: (3, 2). | instruction | 0 | 107,035 | 24 | 214,070 |
Tags: math
Correct Solution:
```
def GCD(a, b):
while (a != 0 and b != 0):
a, b = b % a, a
return max(a, b)
a, b, x, y = map(int, input().split())
f = GCD(x, y)
x //= f
y //= f
n = min(a, b * x // y)
print(n // x)
``` | output | 1 | 107,035 | 24 | 214,071 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Monocarp has decided to buy a new TV set and hang it on the wall in his flat. The wall has enough free space so Monocarp can buy a TV set with screen width not greater than a and screen height not greater than b. Monocarp is also used to TV sets with a certain aspect ratio: formally, if the width of the screen is w, and the height of the screen is h, then the following condition should be met: w/h = x/y.
There are many different TV sets in the shop. Monocarp is sure that for any pair of positive integers w and h there is a TV set with screen width w and height h in the shop.
Monocarp isn't ready to choose the exact TV set he is going to buy. Firstly he wants to determine the optimal screen resolution. He has decided to try all possible variants of screen size. But he must count the number of pairs of positive integers w and h, beforehand, such that (w ≤ a), (h ≤ b) and (w/h = x/y).
In other words, Monocarp wants to determine the number of TV sets having aspect ratio x/y, screen width not exceeding a, and screen height not exceeding b. Two TV sets are considered different if they have different screen width or different screen height.
Input
The first line contains four integers a, b, x, y (1 ≤ a, b, x, y ≤ 10^{18}) — the constraints on the screen width and height, and on the aspect ratio.
Output
Print one integer — the number of different variants to choose TV screen width and screen height so that they meet the aforementioned constraints.
Examples
Input
17 15 5 3
Output
3
Input
14 16 7 22
Output
0
Input
4 2 6 4
Output
1
Input
1000000000000000000 1000000000000000000 999999866000004473 999999822000007597
Output
1000000063
Note
In the first example, there are 3 possible variants: (5, 3), (10, 6), (15, 9).
In the second example, there is no TV set meeting the constraints.
In the third example, there is only one variant: (3, 2). | instruction | 0 | 107,036 | 24 | 214,072 |
Tags: math
Correct Solution:
```
def compute_gcd(x,y):
if y==0:
return x
else:
return compute_gcd(y,x%y)
a, b, x, y = map(int, input().split())
gcd = compute_gcd(x,y)
x /= gcd
y /= gcd
num_x, num_y = x, y
mx = int(a/x)
my = int(b/y)
print( min(mx,my) )
``` | output | 1 | 107,036 | 24 | 214,073 |
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