message stringlengths 2 30.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 237 109k | cluster float64 10 10 | __index_level_0__ int64 474 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
There are two types of burgers in your restaurant β hamburgers and chicken burgers! To assemble a hamburger you need two buns and a beef patty. To assemble a chicken burger you need two buns and a chicken cutlet.
You have b buns, p beef patties and f chicken cutlets in your restaurant. You can sell one hamburger for h dollars and one chicken burger for c dollars. Calculate the maximum profit you can achieve.
You have to answer t independent queries.
Input
The first line contains one integer t (1 β€ t β€ 100) β the number of queries.
The first line of each query contains three integers b, p and f (1 β€ b, ~p, ~f β€ 100) β the number of buns, beef patties and chicken cutlets in your restaurant.
The second line of each query contains two integers h and c (1 β€ h, ~c β€ 100) β the hamburger and chicken burger prices in your restaurant.
Output
For each query print one integer β the maximum profit you can achieve.
Example
Input
3
15 2 3
5 10
7 5 2
10 12
1 100 100
100 100
Output
40
34
0
Note
In first query you have to sell two hamburgers and three chicken burgers. Your income is 2 β
5 + 3 β
10 = 40.
In second query you have to ell one hamburgers and two chicken burgers. Your income is 1 β
10 + 2 β
12 = 34.
In third query you can not create any type of burgers because because you have only one bun. So your income is zero. | instruction | 0 | 36,129 | 10 | 72,258 |
Tags: brute force, greedy, implementation, math
Correct Solution:
```
t=int(input())
for i in range(0,t):
b,p,f=map(int,input().split())
h,c=map(int,input().split())
if(b<2):
print(0)
else:
if(c>h):
k=b//2
if(k>=(p+f)):
print(f*c+p*h)
else:
if(f<=k):
print(c*f+(k-f)*h)
else:
print(k*c)
else:
k=b//2
if(k>=(p+f)):
print(f*c+p*h)
else:
if(p<=k):
print(h*p+(k-p)*c)
else:
print(k*h)
``` | output | 1 | 36,129 | 10 | 72,259 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are two types of burgers in your restaurant β hamburgers and chicken burgers! To assemble a hamburger you need two buns and a beef patty. To assemble a chicken burger you need two buns and a chicken cutlet.
You have b buns, p beef patties and f chicken cutlets in your restaurant. You can sell one hamburger for h dollars and one chicken burger for c dollars. Calculate the maximum profit you can achieve.
You have to answer t independent queries.
Input
The first line contains one integer t (1 β€ t β€ 100) β the number of queries.
The first line of each query contains three integers b, p and f (1 β€ b, ~p, ~f β€ 100) β the number of buns, beef patties and chicken cutlets in your restaurant.
The second line of each query contains two integers h and c (1 β€ h, ~c β€ 100) β the hamburger and chicken burger prices in your restaurant.
Output
For each query print one integer β the maximum profit you can achieve.
Example
Input
3
15 2 3
5 10
7 5 2
10 12
1 100 100
100 100
Output
40
34
0
Note
In first query you have to sell two hamburgers and three chicken burgers. Your income is 2 β
5 + 3 β
10 = 40.
In second query you have to ell one hamburgers and two chicken burgers. Your income is 1 β
10 + 2 β
12 = 34.
In third query you can not create any type of burgers because because you have only one bun. So your income is zero. | instruction | 0 | 36,130 | 10 | 72,260 |
Tags: brute force, greedy, implementation, math
Correct Solution:
```
''' Ψ¨ΩΨ³ΩΩ
Ω Ψ§ΩΩΩΩΩΩ Ψ§ΩΨ±ΩΩΨΩΩ
ΩΩ°ΩΩ Ψ§ΩΨ±ΩΩΨΩΩΩ
Ω '''
#codeforces1207A_live
gi = lambda : list(map(int,input().split()))
for k in range(gi()[0]):
b, p, f = gi()
h, c = gi()
ans = 0
if h > c:
tmep = min(b // 2, p)
ans += tmep * h
b -= tmep * 2
ans += min(b // 2, f) * c
else:
tmep = min(b // 2, f)
ans += tmep * c
b -= tmep * 2
ans += min(b // 2, p) * h
print(ans)
``` | output | 1 | 36,130 | 10 | 72,261 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are two types of burgers in your restaurant β hamburgers and chicken burgers! To assemble a hamburger you need two buns and a beef patty. To assemble a chicken burger you need two buns and a chicken cutlet.
You have b buns, p beef patties and f chicken cutlets in your restaurant. You can sell one hamburger for h dollars and one chicken burger for c dollars. Calculate the maximum profit you can achieve.
You have to answer t independent queries.
Input
The first line contains one integer t (1 β€ t β€ 100) β the number of queries.
The first line of each query contains three integers b, p and f (1 β€ b, ~p, ~f β€ 100) β the number of buns, beef patties and chicken cutlets in your restaurant.
The second line of each query contains two integers h and c (1 β€ h, ~c β€ 100) β the hamburger and chicken burger prices in your restaurant.
Output
For each query print one integer β the maximum profit you can achieve.
Example
Input
3
15 2 3
5 10
7 5 2
10 12
1 100 100
100 100
Output
40
34
0
Note
In first query you have to sell two hamburgers and three chicken burgers. Your income is 2 β
5 + 3 β
10 = 40.
In second query you have to ell one hamburgers and two chicken burgers. Your income is 1 β
10 + 2 β
12 = 34.
In third query you can not create any type of burgers because because you have only one bun. So your income is zero. | instruction | 0 | 36,131 | 10 | 72,262 |
Tags: brute force, greedy, implementation, math
Correct Solution:
```
p = int(input())
for i in range(p):
b, p, f = map(int, input().split())
h, c = map(int, input().split())
b //= 2
if c > h:
p, f = f, p
h, c = c, h
ans1 = min(b, p) * h
ans2 = min(max(b - p, 0), f) * c
print(ans1 + ans2)
``` | output | 1 | 36,131 | 10 | 72,263 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are two types of burgers in your restaurant β hamburgers and chicken burgers! To assemble a hamburger you need two buns and a beef patty. To assemble a chicken burger you need two buns and a chicken cutlet.
You have b buns, p beef patties and f chicken cutlets in your restaurant. You can sell one hamburger for h dollars and one chicken burger for c dollars. Calculate the maximum profit you can achieve.
You have to answer t independent queries.
Input
The first line contains one integer t (1 β€ t β€ 100) β the number of queries.
The first line of each query contains three integers b, p and f (1 β€ b, ~p, ~f β€ 100) β the number of buns, beef patties and chicken cutlets in your restaurant.
The second line of each query contains two integers h and c (1 β€ h, ~c β€ 100) β the hamburger and chicken burger prices in your restaurant.
Output
For each query print one integer β the maximum profit you can achieve.
Example
Input
3
15 2 3
5 10
7 5 2
10 12
1 100 100
100 100
Output
40
34
0
Note
In first query you have to sell two hamburgers and three chicken burgers. Your income is 2 β
5 + 3 β
10 = 40.
In second query you have to ell one hamburgers and two chicken burgers. Your income is 1 β
10 + 2 β
12 = 34.
In third query you can not create any type of burgers because because you have only one bun. So your income is zero. | instruction | 0 | 36,132 | 10 | 72,264 |
Tags: brute force, greedy, implementation, math
Correct Solution:
```
t=int(input())
v=[]
for i in range(t):
l1=list(map(int,input().split()))
l2=list(map(int,input().split()))
b,p,f=l1[0],l1[1],l1[2]
h,c=l2[0],l2[1]
if c>h:
c1=min(b//2,f)
h1=min((b-(c1*2))//2,p)
else:
h1=min(b//2,p)
c1=min((b-(h1*2))//2,f)
m=c1*c+h1*h
v.append(m)
for i in range(t):
print(v[i])
``` | output | 1 | 36,132 | 10 | 72,265 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are two types of burgers in your restaurant β hamburgers and chicken burgers! To assemble a hamburger you need two buns and a beef patty. To assemble a chicken burger you need two buns and a chicken cutlet.
You have b buns, p beef patties and f chicken cutlets in your restaurant. You can sell one hamburger for h dollars and one chicken burger for c dollars. Calculate the maximum profit you can achieve.
You have to answer t independent queries.
Input
The first line contains one integer t (1 β€ t β€ 100) β the number of queries.
The first line of each query contains three integers b, p and f (1 β€ b, ~p, ~f β€ 100) β the number of buns, beef patties and chicken cutlets in your restaurant.
The second line of each query contains two integers h and c (1 β€ h, ~c β€ 100) β the hamburger and chicken burger prices in your restaurant.
Output
For each query print one integer β the maximum profit you can achieve.
Example
Input
3
15 2 3
5 10
7 5 2
10 12
1 100 100
100 100
Output
40
34
0
Note
In first query you have to sell two hamburgers and three chicken burgers. Your income is 2 β
5 + 3 β
10 = 40.
In second query you have to ell one hamburgers and two chicken burgers. Your income is 1 β
10 + 2 β
12 = 34.
In third query you can not create any type of burgers because because you have only one bun. So your income is zero. | instruction | 0 | 36,133 | 10 | 72,266 |
Tags: brute force, greedy, implementation, math
Correct Solution:
```
n = int(input())
for _ in range(n):
b, p, f = map(int, input().split())
h, c = map(int, input().split())
res = 0
if h > c:
count = min(b // 2, p)
res += h * count
b -= 2 * count
count = min(b // 2, f)
res += c * count
else:
count = min(b // 2, f)
res += c * count
b -= 2 * count
count = min(b // 2, p)
res += h * count
print(res)
``` | output | 1 | 36,133 | 10 | 72,267 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two types of burgers in your restaurant β hamburgers and chicken burgers! To assemble a hamburger you need two buns and a beef patty. To assemble a chicken burger you need two buns and a chicken cutlet.
You have b buns, p beef patties and f chicken cutlets in your restaurant. You can sell one hamburger for h dollars and one chicken burger for c dollars. Calculate the maximum profit you can achieve.
You have to answer t independent queries.
Input
The first line contains one integer t (1 β€ t β€ 100) β the number of queries.
The first line of each query contains three integers b, p and f (1 β€ b, ~p, ~f β€ 100) β the number of buns, beef patties and chicken cutlets in your restaurant.
The second line of each query contains two integers h and c (1 β€ h, ~c β€ 100) β the hamburger and chicken burger prices in your restaurant.
Output
For each query print one integer β the maximum profit you can achieve.
Example
Input
3
15 2 3
5 10
7 5 2
10 12
1 100 100
100 100
Output
40
34
0
Note
In first query you have to sell two hamburgers and three chicken burgers. Your income is 2 β
5 + 3 β
10 = 40.
In second query you have to ell one hamburgers and two chicken burgers. Your income is 1 β
10 + 2 β
12 = 34.
In third query you can not create any type of burgers because because you have only one bun. So your income is zero.
Submitted Solution:
```
tests = int(input())
while tests!= 0:
b, p, f = [int(x) for x in input().split()]
h, c = [int(x) for x in input().split()]
income = 0
while b >= 2 and (p > 0 or f > 0):
if h > c:
if p > 0:
income += h
p -= 1
else:
income += c
f -= 1
else:
if f > 0:
income += c
f -= 1
else:
income += h
p -= 1
b -= 2
print(income)
tests -= 1
``` | instruction | 0 | 36,134 | 10 | 72,268 |
Yes | output | 1 | 36,134 | 10 | 72,269 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two types of burgers in your restaurant β hamburgers and chicken burgers! To assemble a hamburger you need two buns and a beef patty. To assemble a chicken burger you need two buns and a chicken cutlet.
You have b buns, p beef patties and f chicken cutlets in your restaurant. You can sell one hamburger for h dollars and one chicken burger for c dollars. Calculate the maximum profit you can achieve.
You have to answer t independent queries.
Input
The first line contains one integer t (1 β€ t β€ 100) β the number of queries.
The first line of each query contains three integers b, p and f (1 β€ b, ~p, ~f β€ 100) β the number of buns, beef patties and chicken cutlets in your restaurant.
The second line of each query contains two integers h and c (1 β€ h, ~c β€ 100) β the hamburger and chicken burger prices in your restaurant.
Output
For each query print one integer β the maximum profit you can achieve.
Example
Input
3
15 2 3
5 10
7 5 2
10 12
1 100 100
100 100
Output
40
34
0
Note
In first query you have to sell two hamburgers and three chicken burgers. Your income is 2 β
5 + 3 β
10 = 40.
In second query you have to ell one hamburgers and two chicken burgers. Your income is 1 β
10 + 2 β
12 = 34.
In third query you can not create any type of burgers because because you have only one bun. So your income is zero.
Submitted Solution:
```
t = int(input())
b = [0 for i in range (t)]
p = [0 for i in range (t)]
f = [0 for i in range (t)]
h = [0 for i in range (t)]
c = [0 for i in range (t)]
total = [0 for i in range (t)]
for i in range (t):
b[i], p[i], f[i] = input().split()
h[i], c[i] = input().split()
i = 0
for i in range (t):
b[i] = int(b[i])
p[i] = int(p[i])
f[i] = int(f[i])
h[i] = int(h[i])
c[i] = int(c[i])
for i in range (t):
total[i] = 0
if(h[i] > c[i]):
while((b[i] > 1) and (p[i] > 0)):
b[i] -= 2
p[i] -= 1
total[i] += h[i]
while((b[i] > 1) and (f[i] > 0)):
b[i] -= 2
f[i] -= 1
total[i] += c[i]
else:
while((b[i] > 1) and (f[i] > 0)):
b[i] -= 2
f[i] -= 1
total[i] += c[i]
while((b[i] > 1) and (p[i] > 0)):
b[i] -= 2
p[i] -= 1
total[i] += h[i]
print(total[i])
``` | instruction | 0 | 36,135 | 10 | 72,270 |
Yes | output | 1 | 36,135 | 10 | 72,271 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two types of burgers in your restaurant β hamburgers and chicken burgers! To assemble a hamburger you need two buns and a beef patty. To assemble a chicken burger you need two buns and a chicken cutlet.
You have b buns, p beef patties and f chicken cutlets in your restaurant. You can sell one hamburger for h dollars and one chicken burger for c dollars. Calculate the maximum profit you can achieve.
You have to answer t independent queries.
Input
The first line contains one integer t (1 β€ t β€ 100) β the number of queries.
The first line of each query contains three integers b, p and f (1 β€ b, ~p, ~f β€ 100) β the number of buns, beef patties and chicken cutlets in your restaurant.
The second line of each query contains two integers h and c (1 β€ h, ~c β€ 100) β the hamburger and chicken burger prices in your restaurant.
Output
For each query print one integer β the maximum profit you can achieve.
Example
Input
3
15 2 3
5 10
7 5 2
10 12
1 100 100
100 100
Output
40
34
0
Note
In first query you have to sell two hamburgers and three chicken burgers. Your income is 2 β
5 + 3 β
10 = 40.
In second query you have to ell one hamburgers and two chicken burgers. Your income is 1 β
10 + 2 β
12 = 34.
In third query you can not create any type of burgers because because you have only one bun. So your income is zero.
Submitted Solution:
```
for _ in range(int(input())):
b,p,f = map(int, input().split())
h,c = map(int, input().split())
if h<c:
k = 0
l = 0
if (b-(f*2))>=0:
k = f
b = b-(f*2)
if (b-(p*2))>=0:
l = p
else:
l = b//2
else:
k = b//2
print((l*h)+(k*c))
else:
k = 0
l = 0
if (b-(p*2))>=0:
k = p
b = b-(p*2)
if (b-(f*2))>=0:
l = f
else:
l = b//2
else:
k = b//2
print((k*h)+(l*c))
``` | instruction | 0 | 36,136 | 10 | 72,272 |
Yes | output | 1 | 36,136 | 10 | 72,273 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two types of burgers in your restaurant β hamburgers and chicken burgers! To assemble a hamburger you need two buns and a beef patty. To assemble a chicken burger you need two buns and a chicken cutlet.
You have b buns, p beef patties and f chicken cutlets in your restaurant. You can sell one hamburger for h dollars and one chicken burger for c dollars. Calculate the maximum profit you can achieve.
You have to answer t independent queries.
Input
The first line contains one integer t (1 β€ t β€ 100) β the number of queries.
The first line of each query contains three integers b, p and f (1 β€ b, ~p, ~f β€ 100) β the number of buns, beef patties and chicken cutlets in your restaurant.
The second line of each query contains two integers h and c (1 β€ h, ~c β€ 100) β the hamburger and chicken burger prices in your restaurant.
Output
For each query print one integer β the maximum profit you can achieve.
Example
Input
3
15 2 3
5 10
7 5 2
10 12
1 100 100
100 100
Output
40
34
0
Note
In first query you have to sell two hamburgers and three chicken burgers. Your income is 2 β
5 + 3 β
10 = 40.
In second query you have to ell one hamburgers and two chicken burgers. Your income is 1 β
10 + 2 β
12 = 34.
In third query you can not create any type of burgers because because you have only one bun. So your income is zero.
Submitted Solution:
```
# - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - #
# IN THE NAME OF GOD
# C++ : Done.!
# Python 3 : Loading...
import math
import random
import sys
#-----------------------------------> Programmer : Kadi <-----------------------------------#
t = int(input())
while t > 0 :
x, y, z=map(int,input().split())
h, c=map(int,input().split())
if h > c:
print(min(y, x // 2) * h + min(z, x // 2 - min(y, x // 2)) * c)
else:
print(min(z, x // 2) * c + min(y, x // 2 - min(z, x // 2)) * h)
t -= 1
# for what price?!
# - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - #
``` | instruction | 0 | 36,137 | 10 | 72,274 |
Yes | output | 1 | 36,137 | 10 | 72,275 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two types of burgers in your restaurant β hamburgers and chicken burgers! To assemble a hamburger you need two buns and a beef patty. To assemble a chicken burger you need two buns and a chicken cutlet.
You have b buns, p beef patties and f chicken cutlets in your restaurant. You can sell one hamburger for h dollars and one chicken burger for c dollars. Calculate the maximum profit you can achieve.
You have to answer t independent queries.
Input
The first line contains one integer t (1 β€ t β€ 100) β the number of queries.
The first line of each query contains three integers b, p and f (1 β€ b, ~p, ~f β€ 100) β the number of buns, beef patties and chicken cutlets in your restaurant.
The second line of each query contains two integers h and c (1 β€ h, ~c β€ 100) β the hamburger and chicken burger prices in your restaurant.
Output
For each query print one integer β the maximum profit you can achieve.
Example
Input
3
15 2 3
5 10
7 5 2
10 12
1 100 100
100 100
Output
40
34
0
Note
In first query you have to sell two hamburgers and three chicken burgers. Your income is 2 β
5 + 3 β
10 = 40.
In second query you have to ell one hamburgers and two chicken burgers. Your income is 1 β
10 + 2 β
12 = 34.
In third query you can not create any type of burgers because because you have only one bun. So your income is zero.
Submitted Solution:
```
num = int(input())
mas1 = []
mas2 = []
res = []
i = 0
while i < num:
for a in input().split():
mas1.append(int(a))
for b in input().split():
mas2.append(int(b))
if mas1[0] <= mas1[1] or mas1[0] <= mas1[2]:
res.append(0)
else:
if mas1[0] > 2*(mas1[1]+mas1[2]):
x = mas1[1] * mas2[0] + mas1[2] * mas2[1]
res.append(x)
else:
# while 2*(mas1[1]+mas1[2]) >= mas1[0]:
# if mas1[1] > mas1[2]:
# mas1[1] -= 1
# if mas1[2] > mas1[1]:
# mas1[2] -= 1
# if mas1[1] == mas1[2]:
# mas1[1] -= 1
# mas1[2] -= 1
# print("mas1[1] = " + str(mas1[1]) + "\nmas1[2] = " + str(mas1[2]))
# x = mas1[1] * mas2[0] + mas1[2] * mas2[1]
# res.append(x)
number = mas1[0] // 2
while (mas1[1] + mas1[2] > number):
if mas1[1] > mas1[2]:
mas1[1] -= 1
if mas1[2] > mas1[1]:
mas1[2] -= 1
if mas1[1] == mas1[2]:
mas1[1] -= 1
# mas1[2] -= 1
# mas1[1] -= 1
print("mas1[1] = " + str(mas1[1]) + "\nmas1[2] = " + str(mas1[2]))
x = mas1[1] * mas2[0] + mas1[2] * mas2[1]
res.append(x)
mas1.clear()
mas2.clear()
i += 1
for i in res:
print(i)
``` | instruction | 0 | 36,138 | 10 | 72,276 |
No | output | 1 | 36,138 | 10 | 72,277 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two types of burgers in your restaurant β hamburgers and chicken burgers! To assemble a hamburger you need two buns and a beef patty. To assemble a chicken burger you need two buns and a chicken cutlet.
You have b buns, p beef patties and f chicken cutlets in your restaurant. You can sell one hamburger for h dollars and one chicken burger for c dollars. Calculate the maximum profit you can achieve.
You have to answer t independent queries.
Input
The first line contains one integer t (1 β€ t β€ 100) β the number of queries.
The first line of each query contains three integers b, p and f (1 β€ b, ~p, ~f β€ 100) β the number of buns, beef patties and chicken cutlets in your restaurant.
The second line of each query contains two integers h and c (1 β€ h, ~c β€ 100) β the hamburger and chicken burger prices in your restaurant.
Output
For each query print one integer β the maximum profit you can achieve.
Example
Input
3
15 2 3
5 10
7 5 2
10 12
1 100 100
100 100
Output
40
34
0
Note
In first query you have to sell two hamburgers and three chicken burgers. Your income is 2 β
5 + 3 β
10 = 40.
In second query you have to ell one hamburgers and two chicken burgers. Your income is 1 β
10 + 2 β
12 = 34.
In third query you can not create any type of burgers because because you have only one bun. So your income is zero.
Submitted Solution:
```
n = int(input())
for _ in range(n):
b,p,f = map(int,input().split())
h,c = map(int,input().split())
if b<2:
print("0")
elif (p+f)*2 < b:
print(p*h+f*c)
else:
if max(h,c)==c:
if 2*f<b:
print(f*c+(b-2*f)//2*h)
else:
print((b//2)*c)
else:
if 2*c<b:
print(f*h+(b-2*f)//2*c)
else:
print((b//2)*h)
``` | instruction | 0 | 36,140 | 10 | 72,280 |
No | output | 1 | 36,140 | 10 | 72,281 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two types of burgers in your restaurant β hamburgers and chicken burgers! To assemble a hamburger you need two buns and a beef patty. To assemble a chicken burger you need two buns and a chicken cutlet.
You have b buns, p beef patties and f chicken cutlets in your restaurant. You can sell one hamburger for h dollars and one chicken burger for c dollars. Calculate the maximum profit you can achieve.
You have to answer t independent queries.
Input
The first line contains one integer t (1 β€ t β€ 100) β the number of queries.
The first line of each query contains three integers b, p and f (1 β€ b, ~p, ~f β€ 100) β the number of buns, beef patties and chicken cutlets in your restaurant.
The second line of each query contains two integers h and c (1 β€ h, ~c β€ 100) β the hamburger and chicken burger prices in your restaurant.
Output
For each query print one integer β the maximum profit you can achieve.
Example
Input
3
15 2 3
5 10
7 5 2
10 12
1 100 100
100 100
Output
40
34
0
Note
In first query you have to sell two hamburgers and three chicken burgers. Your income is 2 β
5 + 3 β
10 = 40.
In second query you have to ell one hamburgers and two chicken burgers. Your income is 1 β
10 + 2 β
12 = 34.
In third query you can not create any type of burgers because because you have only one bun. So your income is zero.
Submitted Solution:
```
q=int(input())
for i in range(q):
total=0
sobra=0
arr=list(map(int,input().split()))
arr2=list(map(int,input().split()))
if (arr[0]//2)>arr[1]+arr[2]:
if arr2[0]>=arr2[1]:
#if arr[1]<=(arr[0]//2):
total+=arr[1]*arr2[0]
sobra=(arr[0]//2)-arr[1]
if sobra>=arr[2]:
total+=arr[2]*arr2[1]
elif arr2[0]<arr2[1]:
#if arr[2]<=(arr[0]//2):
total+=arr[2]*arr2[1]
sobra=(arr[0]//2)-arr[2]
if sobra>=arr[1]:
total+=arr2[0]*arr[1]
print(total)
else:
if arr2[0]>arr2[1]:
total+=arr[1]*(arr2[0])
total+=((arr[0]//2)-arr[1])*arr2[1]
elif arr2[0]<arr2[1]:
total+=arr[2]*arr2[1]
total+=((arr[0]//2)-arr[2])*arr2[0]
print(total)
``` | instruction | 0 | 36,141 | 10 | 72,282 |
No | output | 1 | 36,141 | 10 | 72,283 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is constraints.
The BerTV channel every day broadcasts one episode of one of the k TV shows. You know the schedule for the next n days: a sequence of integers a_1, a_2, ..., a_n (1 β€ a_i β€ k), where a_i is the show, the episode of which will be shown in i-th day.
The subscription to the show is bought for the entire show (i.e. for all its episodes), for each show the subscription is bought separately.
How many minimum subscriptions do you need to buy in order to have the opportunity to watch episodes of purchased shows d (1 β€ d β€ n) days in a row? In other words, you want to buy the minimum number of TV shows so that there is some segment of d consecutive days in which all episodes belong to the purchased shows.
Input
The first line contains an integer t (1 β€ t β€ 10000) β the number of test cases in the input. Then t test case descriptions follow.
The first line of each test case contains three integers n, k and d (1 β€ n β€ 2β
10^5, 1 β€ k β€ 10^6, 1 β€ d β€ n). The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ k), where a_i is the show that is broadcasted on the i-th day.
It is guaranteed that the sum of the values ββof n for all test cases in the input does not exceed 2β
10^5.
Output
Print t integers β the answers to the test cases in the input in the order they follow. The answer to a test case is the minimum number of TV shows for which you need to purchase a subscription so that you can watch episodes of the purchased TV shows on BerTV for d consecutive days. Please note that it is permissible that you will be able to watch more than d days in a row.
Example
Input
4
5 2 2
1 2 1 2 1
9 3 3
3 3 3 2 2 2 1 1 1
4 10 4
10 8 6 4
16 9 8
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3
Output
2
1
4
5
Note
In the first test case to have an opportunity to watch shows for two consecutive days, you need to buy a subscription on show 1 and on show 2. So the answer is two.
In the second test case, you can buy a subscription to any show because for each show you can find a segment of three consecutive days, consisting only of episodes of this show.
In the third test case in the unique segment of four days, you have four different shows, so you need to buy a subscription to all these four shows.
In the fourth test case, you can buy subscriptions to shows 3,5,7,8,9, and you will be able to watch shows for the last eight days. | instruction | 0 | 36,142 | 10 | 72,284 |
Tags: implementation, two pointers
Correct Solution:
```
from collections import Counter as cnt
import sys
def count(n,k,d,a):
mat=cnt(a[:d])
i,prev=0,0
j,m=d,len(mat)
while j<n:
prev=a[i]
mat[a[j]]=mat.get(a[j],0)+1
mat[prev] = mat[prev]-1
if mat[prev] == 0:
del mat[prev]
m=min(m,len(mat))
i+=1
j+=1
return(m)
for i in range(int(input())):
n,k,d=map(int,input().split())
a=list(map(int,input().split()))
print(count(n,k,d,a))
``` | output | 1 | 36,142 | 10 | 72,285 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is constraints.
The BerTV channel every day broadcasts one episode of one of the k TV shows. You know the schedule for the next n days: a sequence of integers a_1, a_2, ..., a_n (1 β€ a_i β€ k), where a_i is the show, the episode of which will be shown in i-th day.
The subscription to the show is bought for the entire show (i.e. for all its episodes), for each show the subscription is bought separately.
How many minimum subscriptions do you need to buy in order to have the opportunity to watch episodes of purchased shows d (1 β€ d β€ n) days in a row? In other words, you want to buy the minimum number of TV shows so that there is some segment of d consecutive days in which all episodes belong to the purchased shows.
Input
The first line contains an integer t (1 β€ t β€ 10000) β the number of test cases in the input. Then t test case descriptions follow.
The first line of each test case contains three integers n, k and d (1 β€ n β€ 2β
10^5, 1 β€ k β€ 10^6, 1 β€ d β€ n). The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ k), where a_i is the show that is broadcasted on the i-th day.
It is guaranteed that the sum of the values ββof n for all test cases in the input does not exceed 2β
10^5.
Output
Print t integers β the answers to the test cases in the input in the order they follow. The answer to a test case is the minimum number of TV shows for which you need to purchase a subscription so that you can watch episodes of the purchased TV shows on BerTV for d consecutive days. Please note that it is permissible that you will be able to watch more than d days in a row.
Example
Input
4
5 2 2
1 2 1 2 1
9 3 3
3 3 3 2 2 2 1 1 1
4 10 4
10 8 6 4
16 9 8
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3
Output
2
1
4
5
Note
In the first test case to have an opportunity to watch shows for two consecutive days, you need to buy a subscription on show 1 and on show 2. So the answer is two.
In the second test case, you can buy a subscription to any show because for each show you can find a segment of three consecutive days, consisting only of episodes of this show.
In the third test case in the unique segment of four days, you have four different shows, so you need to buy a subscription to all these four shows.
In the fourth test case, you can buy subscriptions to shows 3,5,7,8,9, and you will be able to watch shows for the last eight days. | instruction | 0 | 36,143 | 10 | 72,286 |
Tags: implementation, two pointers
Correct Solution:
```
t = int(input())
for i in range(t):
n, k, d = map(int, input().split())
a = list(map(int, input().split()))
b = dict()
for i in range(n):
b[a[i]] = 0
count = 0
for i in range(d):
if b[a[i]] == 0:
count += 1
b[a[i]] += 1
ans = count
for i in range(n - d):
if b[a[i]] == 1:
count -=1
b[a[i]] -= 1
if b[a[i + d]] == 0:
count += 1
b[a[i + d]] += 1
ans = min(ans, count)
print(ans)
``` | output | 1 | 36,143 | 10 | 72,287 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is constraints.
The BerTV channel every day broadcasts one episode of one of the k TV shows. You know the schedule for the next n days: a sequence of integers a_1, a_2, ..., a_n (1 β€ a_i β€ k), where a_i is the show, the episode of which will be shown in i-th day.
The subscription to the show is bought for the entire show (i.e. for all its episodes), for each show the subscription is bought separately.
How many minimum subscriptions do you need to buy in order to have the opportunity to watch episodes of purchased shows d (1 β€ d β€ n) days in a row? In other words, you want to buy the minimum number of TV shows so that there is some segment of d consecutive days in which all episodes belong to the purchased shows.
Input
The first line contains an integer t (1 β€ t β€ 10000) β the number of test cases in the input. Then t test case descriptions follow.
The first line of each test case contains three integers n, k and d (1 β€ n β€ 2β
10^5, 1 β€ k β€ 10^6, 1 β€ d β€ n). The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ k), where a_i is the show that is broadcasted on the i-th day.
It is guaranteed that the sum of the values ββof n for all test cases in the input does not exceed 2β
10^5.
Output
Print t integers β the answers to the test cases in the input in the order they follow. The answer to a test case is the minimum number of TV shows for which you need to purchase a subscription so that you can watch episodes of the purchased TV shows on BerTV for d consecutive days. Please note that it is permissible that you will be able to watch more than d days in a row.
Example
Input
4
5 2 2
1 2 1 2 1
9 3 3
3 3 3 2 2 2 1 1 1
4 10 4
10 8 6 4
16 9 8
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3
Output
2
1
4
5
Note
In the first test case to have an opportunity to watch shows for two consecutive days, you need to buy a subscription on show 1 and on show 2. So the answer is two.
In the second test case, you can buy a subscription to any show because for each show you can find a segment of three consecutive days, consisting only of episodes of this show.
In the third test case in the unique segment of four days, you have four different shows, so you need to buy a subscription to all these four shows.
In the fourth test case, you can buy subscriptions to shows 3,5,7,8,9, and you will be able to watch shows for the last eight days. | instruction | 0 | 36,144 | 10 | 72,288 |
Tags: implementation, two pointers
Correct Solution:
```
for _ in range(int(input())):
n,k,d = list(map(int,input().split()))
lst = list(map(int,input().split()))
dct = {}
for i in lst[:d]:
if i not in dct:
dct[i] = 1
else:
dct[i] += 1
m = len(dct)
for i in range(d,n):
if lst[i] in dct:
dct[lst[i]] += 1
else:
dct[lst[i]] = 1
dct[lst[i-d]] -= 1
if dct[lst[i-d]] == 0:
del dct[lst[i-d]]
m = min(m,len(dct))
print(m)
``` | output | 1 | 36,144 | 10 | 72,289 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is constraints.
The BerTV channel every day broadcasts one episode of one of the k TV shows. You know the schedule for the next n days: a sequence of integers a_1, a_2, ..., a_n (1 β€ a_i β€ k), where a_i is the show, the episode of which will be shown in i-th day.
The subscription to the show is bought for the entire show (i.e. for all its episodes), for each show the subscription is bought separately.
How many minimum subscriptions do you need to buy in order to have the opportunity to watch episodes of purchased shows d (1 β€ d β€ n) days in a row? In other words, you want to buy the minimum number of TV shows so that there is some segment of d consecutive days in which all episodes belong to the purchased shows.
Input
The first line contains an integer t (1 β€ t β€ 10000) β the number of test cases in the input. Then t test case descriptions follow.
The first line of each test case contains three integers n, k and d (1 β€ n β€ 2β
10^5, 1 β€ k β€ 10^6, 1 β€ d β€ n). The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ k), where a_i is the show that is broadcasted on the i-th day.
It is guaranteed that the sum of the values ββof n for all test cases in the input does not exceed 2β
10^5.
Output
Print t integers β the answers to the test cases in the input in the order they follow. The answer to a test case is the minimum number of TV shows for which you need to purchase a subscription so that you can watch episodes of the purchased TV shows on BerTV for d consecutive days. Please note that it is permissible that you will be able to watch more than d days in a row.
Example
Input
4
5 2 2
1 2 1 2 1
9 3 3
3 3 3 2 2 2 1 1 1
4 10 4
10 8 6 4
16 9 8
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3
Output
2
1
4
5
Note
In the first test case to have an opportunity to watch shows for two consecutive days, you need to buy a subscription on show 1 and on show 2. So the answer is two.
In the second test case, you can buy a subscription to any show because for each show you can find a segment of three consecutive days, consisting only of episodes of this show.
In the third test case in the unique segment of four days, you have four different shows, so you need to buy a subscription to all these four shows.
In the fourth test case, you can buy subscriptions to shows 3,5,7,8,9, and you will be able to watch shows for the last eight days. | instruction | 0 | 36,145 | 10 | 72,290 |
Tags: implementation, two pointers
Correct Solution:
```
for _ in range(int(input())):
n,k,d=map(int,input().split())
l=list(map(int,input().split()))
dict={}
t=set()
for i in range(d):
eleman=l[i]
t.add(eleman)
try:
dict[eleman]+=1
except:
dict[eleman]=1
best=len(t)
p=d
kk=0
while p < n:
pre=l[kk]
if dict[pre] >1:
dict[pre]-=1
else:
t.remove(pre)
dict[pre]-=1
t.add(l[p])
try:
dict[l[p]]+=1
except:
dict[l[p]]=1
best=min(best,len(t))
p+=1
kk+=1
print(best)
``` | output | 1 | 36,145 | 10 | 72,291 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is constraints.
The BerTV channel every day broadcasts one episode of one of the k TV shows. You know the schedule for the next n days: a sequence of integers a_1, a_2, ..., a_n (1 β€ a_i β€ k), where a_i is the show, the episode of which will be shown in i-th day.
The subscription to the show is bought for the entire show (i.e. for all its episodes), for each show the subscription is bought separately.
How many minimum subscriptions do you need to buy in order to have the opportunity to watch episodes of purchased shows d (1 β€ d β€ n) days in a row? In other words, you want to buy the minimum number of TV shows so that there is some segment of d consecutive days in which all episodes belong to the purchased shows.
Input
The first line contains an integer t (1 β€ t β€ 10000) β the number of test cases in the input. Then t test case descriptions follow.
The first line of each test case contains three integers n, k and d (1 β€ n β€ 2β
10^5, 1 β€ k β€ 10^6, 1 β€ d β€ n). The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ k), where a_i is the show that is broadcasted on the i-th day.
It is guaranteed that the sum of the values ββof n for all test cases in the input does not exceed 2β
10^5.
Output
Print t integers β the answers to the test cases in the input in the order they follow. The answer to a test case is the minimum number of TV shows for which you need to purchase a subscription so that you can watch episodes of the purchased TV shows on BerTV for d consecutive days. Please note that it is permissible that you will be able to watch more than d days in a row.
Example
Input
4
5 2 2
1 2 1 2 1
9 3 3
3 3 3 2 2 2 1 1 1
4 10 4
10 8 6 4
16 9 8
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3
Output
2
1
4
5
Note
In the first test case to have an opportunity to watch shows for two consecutive days, you need to buy a subscription on show 1 and on show 2. So the answer is two.
In the second test case, you can buy a subscription to any show because for each show you can find a segment of three consecutive days, consisting only of episodes of this show.
In the third test case in the unique segment of four days, you have four different shows, so you need to buy a subscription to all these four shows.
In the fourth test case, you can buy subscriptions to shows 3,5,7,8,9, and you will be able to watch shows for the last eight days. | instruction | 0 | 36,146 | 10 | 72,292 |
Tags: implementation, two pointers
Correct Solution:
```
for __ in range(int(input())):
n, k, d = list(map(int, input().split()))
ar = list(map(int, input().split()))
A = dict()
num = 0
for i in range(d):
if ar[i] in A:
A[ar[i]] += 1
else:
A[ar[i]] = 1
num += 1
ans = num
for j in range(d, n):
A[ar[j - d]] -= 1
if A[ar[j - d]] == 0:
num -= 1
if ar[j] in A:
if A[ar[j]] == 0:
num += 1
A[ar[j]] += 1
else:
A[ar[j]] = 1
num += 1
ans = min(num, ans)
print(ans)
``` | output | 1 | 36,146 | 10 | 72,293 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is constraints.
The BerTV channel every day broadcasts one episode of one of the k TV shows. You know the schedule for the next n days: a sequence of integers a_1, a_2, ..., a_n (1 β€ a_i β€ k), where a_i is the show, the episode of which will be shown in i-th day.
The subscription to the show is bought for the entire show (i.e. for all its episodes), for each show the subscription is bought separately.
How many minimum subscriptions do you need to buy in order to have the opportunity to watch episodes of purchased shows d (1 β€ d β€ n) days in a row? In other words, you want to buy the minimum number of TV shows so that there is some segment of d consecutive days in which all episodes belong to the purchased shows.
Input
The first line contains an integer t (1 β€ t β€ 10000) β the number of test cases in the input. Then t test case descriptions follow.
The first line of each test case contains three integers n, k and d (1 β€ n β€ 2β
10^5, 1 β€ k β€ 10^6, 1 β€ d β€ n). The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ k), where a_i is the show that is broadcasted on the i-th day.
It is guaranteed that the sum of the values ββof n for all test cases in the input does not exceed 2β
10^5.
Output
Print t integers β the answers to the test cases in the input in the order they follow. The answer to a test case is the minimum number of TV shows for which you need to purchase a subscription so that you can watch episodes of the purchased TV shows on BerTV for d consecutive days. Please note that it is permissible that you will be able to watch more than d days in a row.
Example
Input
4
5 2 2
1 2 1 2 1
9 3 3
3 3 3 2 2 2 1 1 1
4 10 4
10 8 6 4
16 9 8
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3
Output
2
1
4
5
Note
In the first test case to have an opportunity to watch shows for two consecutive days, you need to buy a subscription on show 1 and on show 2. So the answer is two.
In the second test case, you can buy a subscription to any show because for each show you can find a segment of three consecutive days, consisting only of episodes of this show.
In the third test case in the unique segment of four days, you have four different shows, so you need to buy a subscription to all these four shows.
In the fourth test case, you can buy subscriptions to shows 3,5,7,8,9, and you will be able to watch shows for the last eight days. | instruction | 0 | 36,147 | 10 | 72,294 |
Tags: implementation, two pointers
Correct Solution:
```
for i in range(int(input())):
n, k, d = map(int, input().split())
a = list(map(int, input().split()))
s = d
b = {}
for j in range(d):
b[a[j]] = b.get(a[j], 0) + 1
s = len(b)
for j in range(d, n):
if b[a[j - d]] == 1:
del b[a[j - d]]
b[a[j]] = b.get(a[j], 0) + 1
else:
b[a[j -d]] -= 1
b[a[j]] = b.get(a[j], 0) + 1
s = min(s, len(b))
print(s)
``` | output | 1 | 36,147 | 10 | 72,295 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is constraints.
The BerTV channel every day broadcasts one episode of one of the k TV shows. You know the schedule for the next n days: a sequence of integers a_1, a_2, ..., a_n (1 β€ a_i β€ k), where a_i is the show, the episode of which will be shown in i-th day.
The subscription to the show is bought for the entire show (i.e. for all its episodes), for each show the subscription is bought separately.
How many minimum subscriptions do you need to buy in order to have the opportunity to watch episodes of purchased shows d (1 β€ d β€ n) days in a row? In other words, you want to buy the minimum number of TV shows so that there is some segment of d consecutive days in which all episodes belong to the purchased shows.
Input
The first line contains an integer t (1 β€ t β€ 10000) β the number of test cases in the input. Then t test case descriptions follow.
The first line of each test case contains three integers n, k and d (1 β€ n β€ 2β
10^5, 1 β€ k β€ 10^6, 1 β€ d β€ n). The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ k), where a_i is the show that is broadcasted on the i-th day.
It is guaranteed that the sum of the values ββof n for all test cases in the input does not exceed 2β
10^5.
Output
Print t integers β the answers to the test cases in the input in the order they follow. The answer to a test case is the minimum number of TV shows for which you need to purchase a subscription so that you can watch episodes of the purchased TV shows on BerTV for d consecutive days. Please note that it is permissible that you will be able to watch more than d days in a row.
Example
Input
4
5 2 2
1 2 1 2 1
9 3 3
3 3 3 2 2 2 1 1 1
4 10 4
10 8 6 4
16 9 8
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3
Output
2
1
4
5
Note
In the first test case to have an opportunity to watch shows for two consecutive days, you need to buy a subscription on show 1 and on show 2. So the answer is two.
In the second test case, you can buy a subscription to any show because for each show you can find a segment of three consecutive days, consisting only of episodes of this show.
In the third test case in the unique segment of four days, you have four different shows, so you need to buy a subscription to all these four shows.
In the fourth test case, you can buy subscriptions to shows 3,5,7,8,9, and you will be able to watch shows for the last eight days. | instruction | 0 | 36,148 | 10 | 72,296 |
Tags: implementation, two pointers
Correct Solution:
```
from collections import deque
t=int(input())
for i in range(t):
n,k,dp=[int(x) for x in input().split()]
d={}
i=0
p=deque()
cur=0
min=k
for el in input().split():
i+=1
if i<=dp:
p.append(el)
if el in d.keys():
d[el]+=1
else:
d[el]=1
cur+=1
else:
if cur<min:
min=cur
##deleting
exc=p.popleft()
if d[exc]==1:
d.pop(exc)
cur-=1
else:
d[exc]-=1
##adding
p.append(el)
if el in d.keys():
d[el]+=1
else:
d[el]=1
cur+=1
##print(d,p)
if min>cur:
min=cur
print(min)
``` | output | 1 | 36,148 | 10 | 72,297 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is constraints.
The BerTV channel every day broadcasts one episode of one of the k TV shows. You know the schedule for the next n days: a sequence of integers a_1, a_2, ..., a_n (1 β€ a_i β€ k), where a_i is the show, the episode of which will be shown in i-th day.
The subscription to the show is bought for the entire show (i.e. for all its episodes), for each show the subscription is bought separately.
How many minimum subscriptions do you need to buy in order to have the opportunity to watch episodes of purchased shows d (1 β€ d β€ n) days in a row? In other words, you want to buy the minimum number of TV shows so that there is some segment of d consecutive days in which all episodes belong to the purchased shows.
Input
The first line contains an integer t (1 β€ t β€ 10000) β the number of test cases in the input. Then t test case descriptions follow.
The first line of each test case contains three integers n, k and d (1 β€ n β€ 2β
10^5, 1 β€ k β€ 10^6, 1 β€ d β€ n). The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ k), where a_i is the show that is broadcasted on the i-th day.
It is guaranteed that the sum of the values ββof n for all test cases in the input does not exceed 2β
10^5.
Output
Print t integers β the answers to the test cases in the input in the order they follow. The answer to a test case is the minimum number of TV shows for which you need to purchase a subscription so that you can watch episodes of the purchased TV shows on BerTV for d consecutive days. Please note that it is permissible that you will be able to watch more than d days in a row.
Example
Input
4
5 2 2
1 2 1 2 1
9 3 3
3 3 3 2 2 2 1 1 1
4 10 4
10 8 6 4
16 9 8
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3
Output
2
1
4
5
Note
In the first test case to have an opportunity to watch shows for two consecutive days, you need to buy a subscription on show 1 and on show 2. So the answer is two.
In the second test case, you can buy a subscription to any show because for each show you can find a segment of three consecutive days, consisting only of episodes of this show.
In the third test case in the unique segment of four days, you have four different shows, so you need to buy a subscription to all these four shows.
In the fourth test case, you can buy subscriptions to shows 3,5,7,8,9, and you will be able to watch shows for the last eight days. | instruction | 0 | 36,149 | 10 | 72,298 |
Tags: implementation, two pointers
Correct Solution:
```
from os import path
import sys
# mod = int(1e9 + 7)
# import re
from math import ceil, floor,gcd,log,log2 ,factorial
from collections import defaultdict , Counter,deque
from itertools import permutations
# from bisect import bisect_left, bisect_right
#popping from the end is less taxing,since you don't have to shift any elements
maxx = float('inf')
I = lambda :int(sys.stdin.buffer.readline())
tup= lambda : map(int , sys.stdin.buffer.readline().split())
lint = lambda :[int(x) for x in sys.stdin.buffer.readline().split()]
S = lambda: sys.stdin.readline().replace('\n', '').strip()
def grid(r, c): return [lint() for i in range(r)]
# def debug(*args, c=6): print('\033[3{}m'.format(c), *args, '\033[0m', file=sys.stderr)
stpr = lambda x : sys.stdout.write(f'{x}' + '\n')
star = lambda x: print(' '.join(map(str, x)))
if (path.exists('input.txt')):
sys.stdin=open('input.txt','r');sys.stdout=open('output.txt','w');
#left shift --- num*(2**k) --(k - shift)
# input = sys.stdin.readline
for _ in range(I()):
n , k ,d =tup()
ls = lint()
x = ls[:d]
dd = defaultdict(int)
for i in x :dd[i]+=1
x = set(x)
ans = len(x)
c =0
for i in range(n-d):
if dd[ls[i+d]] ==0:
x.add(ls[i+d])
dd[ls[i+d]]+=1
dd[ls[c]]-=1
if dd[ls[c]] ==0:
x.remove(ls[c])
c+=1
ans = min(ans , len(x))
print(ans)
``` | output | 1 | 36,149 | 10 | 72,299 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is constraints.
The BerTV channel every day broadcasts one episode of one of the k TV shows. You know the schedule for the next n days: a sequence of integers a_1, a_2, ..., a_n (1 β€ a_i β€ k), where a_i is the show, the episode of which will be shown in i-th day.
The subscription to the show is bought for the entire show (i.e. for all its episodes), for each show the subscription is bought separately.
How many minimum subscriptions do you need to buy in order to have the opportunity to watch episodes of purchased shows d (1 β€ d β€ n) days in a row? In other words, you want to buy the minimum number of TV shows so that there is some segment of d consecutive days in which all episodes belong to the purchased shows.
Input
The first line contains an integer t (1 β€ t β€ 10000) β the number of test cases in the input. Then t test case descriptions follow.
The first line of each test case contains three integers n, k and d (1 β€ n β€ 2β
10^5, 1 β€ k β€ 10^6, 1 β€ d β€ n). The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ k), where a_i is the show that is broadcasted on the i-th day.
It is guaranteed that the sum of the values ββof n for all test cases in the input does not exceed 2β
10^5.
Output
Print t integers β the answers to the test cases in the input in the order they follow. The answer to a test case is the minimum number of TV shows for which you need to purchase a subscription so that you can watch episodes of the purchased TV shows on BerTV for d consecutive days. Please note that it is permissible that you will be able to watch more than d days in a row.
Example
Input
4
5 2 2
1 2 1 2 1
9 3 3
3 3 3 2 2 2 1 1 1
4 10 4
10 8 6 4
16 9 8
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3
Output
2
1
4
5
Note
In the first test case to have an opportunity to watch shows for two consecutive days, you need to buy a subscription on show 1 and on show 2. So the answer is two.
In the second test case, you can buy a subscription to any show because for each show you can find a segment of three consecutive days, consisting only of episodes of this show.
In the third test case in the unique segment of four days, you have four different shows, so you need to buy a subscription to all these four shows.
In the fourth test case, you can buy subscriptions to shows 3,5,7,8,9, and you will be able to watch shows for the last eight days.
Submitted Solution:
```
"""
d = 3
1 2 3 4 5 6
a 3 2 1 3 3 3
k 1 2 3 3 2 1
d 1 2 3 3 3 3
r 3 2
4
5 2 2
1 2 1 2 1
9 3 3
3 3 3 2 2 2 1 1 1
4 10 4
10 8 6 4
16 9 8
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3
"""
import collections
t = int(input())
for _ in range(t):
n, k, d = map(int, input().split())
a = list(map(int, input().split()))
s = collections.defaultdict(int) # [0] * (k+1)
r = 0
for x in a[:d]:
# if x in s:
# s[x] += 1
# else:
# s[x] = 1
if s[x] == 0:
r += 1
s[x] += 1
recorde = r #len(set(a[:d])) # s.values()
for i in range(d, len(a)):
if s[a[i-d]] == 1:
r -= 1
s[a[i - d]] -= 1
if s[a[i]] == 0:
r += 1
s[a[i]] += 1
if r < recorde:
recorde = r
print(recorde)
``` | instruction | 0 | 36,150 | 10 | 72,300 |
Yes | output | 1 | 36,150 | 10 | 72,301 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is constraints.
The BerTV channel every day broadcasts one episode of one of the k TV shows. You know the schedule for the next n days: a sequence of integers a_1, a_2, ..., a_n (1 β€ a_i β€ k), where a_i is the show, the episode of which will be shown in i-th day.
The subscription to the show is bought for the entire show (i.e. for all its episodes), for each show the subscription is bought separately.
How many minimum subscriptions do you need to buy in order to have the opportunity to watch episodes of purchased shows d (1 β€ d β€ n) days in a row? In other words, you want to buy the minimum number of TV shows so that there is some segment of d consecutive days in which all episodes belong to the purchased shows.
Input
The first line contains an integer t (1 β€ t β€ 10000) β the number of test cases in the input. Then t test case descriptions follow.
The first line of each test case contains three integers n, k and d (1 β€ n β€ 2β
10^5, 1 β€ k β€ 10^6, 1 β€ d β€ n). The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ k), where a_i is the show that is broadcasted on the i-th day.
It is guaranteed that the sum of the values ββof n for all test cases in the input does not exceed 2β
10^5.
Output
Print t integers β the answers to the test cases in the input in the order they follow. The answer to a test case is the minimum number of TV shows for which you need to purchase a subscription so that you can watch episodes of the purchased TV shows on BerTV for d consecutive days. Please note that it is permissible that you will be able to watch more than d days in a row.
Example
Input
4
5 2 2
1 2 1 2 1
9 3 3
3 3 3 2 2 2 1 1 1
4 10 4
10 8 6 4
16 9 8
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3
Output
2
1
4
5
Note
In the first test case to have an opportunity to watch shows for two consecutive days, you need to buy a subscription on show 1 and on show 2. So the answer is two.
In the second test case, you can buy a subscription to any show because for each show you can find a segment of three consecutive days, consisting only of episodes of this show.
In the third test case in the unique segment of four days, you have four different shows, so you need to buy a subscription to all these four shows.
In the fourth test case, you can buy subscriptions to shows 3,5,7,8,9, and you will be able to watch shows for the last eight days.
Submitted Solution:
```
import sys
def mapi(): return map(int,input().split())
def maps(): return map(str,input().split())
#--------------------------------------------------
for _ in range(int(input())):
n,k,d = mapi()
a = list(mapi())
mp = {}
for i in range(d):
try:
mp[a[i]]+=1
except KeyError:
mp[a[i]]=1
res = len(mp)
i = 1
j = d
while i<n and j<n:
try:
mp[a[j]]+=1
except:
mp[a[j]]=1
if mp[a[i-1]]==1:
del mp[a[i-1]]
else:
mp[a[i-1]]-=1
i+=1
j+=1
res = min(res, len(mp))
#print(i,j)
print(res)
``` | instruction | 0 | 36,151 | 10 | 72,302 |
Yes | output | 1 | 36,151 | 10 | 72,303 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is constraints.
The BerTV channel every day broadcasts one episode of one of the k TV shows. You know the schedule for the next n days: a sequence of integers a_1, a_2, ..., a_n (1 β€ a_i β€ k), where a_i is the show, the episode of which will be shown in i-th day.
The subscription to the show is bought for the entire show (i.e. for all its episodes), for each show the subscription is bought separately.
How many minimum subscriptions do you need to buy in order to have the opportunity to watch episodes of purchased shows d (1 β€ d β€ n) days in a row? In other words, you want to buy the minimum number of TV shows so that there is some segment of d consecutive days in which all episodes belong to the purchased shows.
Input
The first line contains an integer t (1 β€ t β€ 10000) β the number of test cases in the input. Then t test case descriptions follow.
The first line of each test case contains three integers n, k and d (1 β€ n β€ 2β
10^5, 1 β€ k β€ 10^6, 1 β€ d β€ n). The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ k), where a_i is the show that is broadcasted on the i-th day.
It is guaranteed that the sum of the values ββof n for all test cases in the input does not exceed 2β
10^5.
Output
Print t integers β the answers to the test cases in the input in the order they follow. The answer to a test case is the minimum number of TV shows for which you need to purchase a subscription so that you can watch episodes of the purchased TV shows on BerTV for d consecutive days. Please note that it is permissible that you will be able to watch more than d days in a row.
Example
Input
4
5 2 2
1 2 1 2 1
9 3 3
3 3 3 2 2 2 1 1 1
4 10 4
10 8 6 4
16 9 8
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3
Output
2
1
4
5
Note
In the first test case to have an opportunity to watch shows for two consecutive days, you need to buy a subscription on show 1 and on show 2. So the answer is two.
In the second test case, you can buy a subscription to any show because for each show you can find a segment of three consecutive days, consisting only of episodes of this show.
In the third test case in the unique segment of four days, you have four different shows, so you need to buy a subscription to all these four shows.
In the fourth test case, you can buy subscriptions to shows 3,5,7,8,9, and you will be able to watch shows for the last eight days.
Submitted Solution:
```
for _ in range(int(input())):
n, k, d = map(int, input().split())
a = list(map(int, input().split()))
s = {}
for q in range(d):
s[a[q]] = s.get(a[q], 0)+1
ans = len(s)
for q in range(d, n):
if s[a[q-d]] == 1:
del s[a[q-d]]
else:
s[a[q-d]] -= 1
s[a[q]] = s.get(a[q], 0)+1
ans = min(ans, len(s))
print(ans)
``` | instruction | 0 | 36,152 | 10 | 72,304 |
Yes | output | 1 | 36,152 | 10 | 72,305 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is constraints.
The BerTV channel every day broadcasts one episode of one of the k TV shows. You know the schedule for the next n days: a sequence of integers a_1, a_2, ..., a_n (1 β€ a_i β€ k), where a_i is the show, the episode of which will be shown in i-th day.
The subscription to the show is bought for the entire show (i.e. for all its episodes), for each show the subscription is bought separately.
How many minimum subscriptions do you need to buy in order to have the opportunity to watch episodes of purchased shows d (1 β€ d β€ n) days in a row? In other words, you want to buy the minimum number of TV shows so that there is some segment of d consecutive days in which all episodes belong to the purchased shows.
Input
The first line contains an integer t (1 β€ t β€ 10000) β the number of test cases in the input. Then t test case descriptions follow.
The first line of each test case contains three integers n, k and d (1 β€ n β€ 2β
10^5, 1 β€ k β€ 10^6, 1 β€ d β€ n). The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ k), where a_i is the show that is broadcasted on the i-th day.
It is guaranteed that the sum of the values ββof n for all test cases in the input does not exceed 2β
10^5.
Output
Print t integers β the answers to the test cases in the input in the order they follow. The answer to a test case is the minimum number of TV shows for which you need to purchase a subscription so that you can watch episodes of the purchased TV shows on BerTV for d consecutive days. Please note that it is permissible that you will be able to watch more than d days in a row.
Example
Input
4
5 2 2
1 2 1 2 1
9 3 3
3 3 3 2 2 2 1 1 1
4 10 4
10 8 6 4
16 9 8
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3
Output
2
1
4
5
Note
In the first test case to have an opportunity to watch shows for two consecutive days, you need to buy a subscription on show 1 and on show 2. So the answer is two.
In the second test case, you can buy a subscription to any show because for each show you can find a segment of three consecutive days, consisting only of episodes of this show.
In the third test case in the unique segment of four days, you have four different shows, so you need to buy a subscription to all these four shows.
In the fourth test case, you can buy subscriptions to shows 3,5,7,8,9, and you will be able to watch shows for the last eight days.
Submitted Solution:
```
t=int(input())
for p in range(t):
n,k,d=[int(x) for x in input().split()]
a=[int(x) for x in input().split()]
arr=10**100
test={}
counter=0
for i in range(n):
if a[i] not in test:
test[a[i]]=0
test[a[i]]+=1
if test[a[i]]==1:
counter+=1
if i==d-1:
arr=counter
if i>=d:
test[a[i-d]]-=1
if test[a[i-d]]==0:
counter-=1
arr=min(counter,arr)
print(arr)
``` | instruction | 0 | 36,153 | 10 | 72,306 |
Yes | output | 1 | 36,153 | 10 | 72,307 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is constraints.
The BerTV channel every day broadcasts one episode of one of the k TV shows. You know the schedule for the next n days: a sequence of integers a_1, a_2, ..., a_n (1 β€ a_i β€ k), where a_i is the show, the episode of which will be shown in i-th day.
The subscription to the show is bought for the entire show (i.e. for all its episodes), for each show the subscription is bought separately.
How many minimum subscriptions do you need to buy in order to have the opportunity to watch episodes of purchased shows d (1 β€ d β€ n) days in a row? In other words, you want to buy the minimum number of TV shows so that there is some segment of d consecutive days in which all episodes belong to the purchased shows.
Input
The first line contains an integer t (1 β€ t β€ 10000) β the number of test cases in the input. Then t test case descriptions follow.
The first line of each test case contains three integers n, k and d (1 β€ n β€ 2β
10^5, 1 β€ k β€ 10^6, 1 β€ d β€ n). The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ k), where a_i is the show that is broadcasted on the i-th day.
It is guaranteed that the sum of the values ββof n for all test cases in the input does not exceed 2β
10^5.
Output
Print t integers β the answers to the test cases in the input in the order they follow. The answer to a test case is the minimum number of TV shows for which you need to purchase a subscription so that you can watch episodes of the purchased TV shows on BerTV for d consecutive days. Please note that it is permissible that you will be able to watch more than d days in a row.
Example
Input
4
5 2 2
1 2 1 2 1
9 3 3
3 3 3 2 2 2 1 1 1
4 10 4
10 8 6 4
16 9 8
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3
Output
2
1
4
5
Note
In the first test case to have an opportunity to watch shows for two consecutive days, you need to buy a subscription on show 1 and on show 2. So the answer is two.
In the second test case, you can buy a subscription to any show because for each show you can find a segment of three consecutive days, consisting only of episodes of this show.
In the third test case in the unique segment of four days, you have four different shows, so you need to buy a subscription to all these four shows.
In the fourth test case, you can buy subscriptions to shows 3,5,7,8,9, and you will be able to watch shows for the last eight days.
Submitted Solution:
```
from collections import Counter as C,defaultdict as D,deque as Q
from operator import itemgetter as I
from itertools import product as P,permutations as PERMUT
from bisect import bisect_left as BL,bisect_right as BR,insort as INSORT
from heapq import heappush as HPUSH,heappop as HPOP
from math import floor as MF,ceil as MC, gcd as MG,factorial as F,sqrt as SQRT, inf as INFINITY,log as LOG
from sys import stdin, stdout
INPUT=stdin.readline
PRINT=stdout.write
L=list;M=map
def Player1():
print("")
def Player2():
print("")
def Yes():
PRINT("Yes\n")
def No():
PRINT("No\n")
def IsPrime(n):
for i in range(2,MC(SQRT(n))+1):
if n%i==0:
return False
return True
def Factors(x):
ans=[]
for i in range(1,MC(SQRT(x))+1):
if x%i==0:
ans.append(i)
if x%(x//i)==0:
ans.append(x//i)
return ans
def CheckPath(source,destination,g):
visited=[0]*101
q=Q()
q.append(source)
visited[source]=1
while q:
node=q.popleft()
if node==destination:
return 1
for v in g[node]:
if not visited[v]:
q.append(v)
visited[v]=1
return 0
def Sieve(n):
prime=[1]*(n+1)
p=2
while p*p<=n:
if prime[p]:
for i in range(p*p,n+1,p):
prime[i]=0
p+=1
primes=[]
for p in range(2,n+1):
if prime[p]:
primes.append(p)
return primes
def Prefix(a,n):
p=[]
for i in range(n):
if i==0:
p.append(a[0])
else:
p.append(p[-1]+a[i])
return p
def Suffix(a,n):
s=[0]*n
for i in range(n-1,-1,-1):
if i==n-1:
s[i]=a[i]
else:
s[i]=s[i+1]+a[i]
return s
def Spf(n):
spf=[0 for i in range(n)]
spf[1]=1
for i in range(2,n):
spf[i]=i
for i in range(4,n,2):
spf[i]=2
for i in range(3,MC(SQRT(n))+1):
if spf[i]==i:
for j in range(i*i,n,i):
if spf[j]==j:
spf[j]=i
return spf
def DFS(g,s,visited,ans):
visited[s]=1
for u,c in g[s]:
if visited[u]:
continue
if c==ans[s]:
if c==1:
ans[u]=2
else:
ans[u]=1
else:
ans[u]=c
DFS(g,u,visited,ans)
def lcm(a,b):
return (a*b)//(MG(a,b))
def Kadane(numbers):
max_so_far=-INFINITY
max_ending_here=0
max_element=-INFINITY
for i in range(len(numbers)):
max_ending_here=max(max_ending_here+numbers[i],0)
max_so_far=max(max_ending_here,max_so_far)
max_element=max(max_element,numbers[i])
if max_so_far==0:
max_so_far=max_element
return max_so_far
def subset(start,a,res,ans):
res.append(ans)
for i in range(start,9):
subset(i+1,a,res,ans+[a[i]])
def Main():
for _ in range(int(INPUT())):
n,k,d=M(int,INPUT().split( ))
a=L(M(int,INPUT().split( )))
s=set(a[:d])
ans=len(s);j=d;m=INFINITY
for i in range(n-d+1):
if a[i] in s:
s.remove(a[i])
ans-=1
if j<n and a[j] not in s:
ans+=1
s.add(a[j])
j+=1
m=min(m,ans)
PRINT("%d\n"%(m+1))
Main()
'mysql -u root -p'
``` | instruction | 0 | 36,154 | 10 | 72,308 |
No | output | 1 | 36,154 | 10 | 72,309 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is constraints.
The BerTV channel every day broadcasts one episode of one of the k TV shows. You know the schedule for the next n days: a sequence of integers a_1, a_2, ..., a_n (1 β€ a_i β€ k), where a_i is the show, the episode of which will be shown in i-th day.
The subscription to the show is bought for the entire show (i.e. for all its episodes), for each show the subscription is bought separately.
How many minimum subscriptions do you need to buy in order to have the opportunity to watch episodes of purchased shows d (1 β€ d β€ n) days in a row? In other words, you want to buy the minimum number of TV shows so that there is some segment of d consecutive days in which all episodes belong to the purchased shows.
Input
The first line contains an integer t (1 β€ t β€ 10000) β the number of test cases in the input. Then t test case descriptions follow.
The first line of each test case contains three integers n, k and d (1 β€ n β€ 2β
10^5, 1 β€ k β€ 10^6, 1 β€ d β€ n). The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ k), where a_i is the show that is broadcasted on the i-th day.
It is guaranteed that the sum of the values ββof n for all test cases in the input does not exceed 2β
10^5.
Output
Print t integers β the answers to the test cases in the input in the order they follow. The answer to a test case is the minimum number of TV shows for which you need to purchase a subscription so that you can watch episodes of the purchased TV shows on BerTV for d consecutive days. Please note that it is permissible that you will be able to watch more than d days in a row.
Example
Input
4
5 2 2
1 2 1 2 1
9 3 3
3 3 3 2 2 2 1 1 1
4 10 4
10 8 6 4
16 9 8
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3
Output
2
1
4
5
Note
In the first test case to have an opportunity to watch shows for two consecutive days, you need to buy a subscription on show 1 and on show 2. So the answer is two.
In the second test case, you can buy a subscription to any show because for each show you can find a segment of three consecutive days, consisting only of episodes of this show.
In the third test case in the unique segment of four days, you have four different shows, so you need to buy a subscription to all these four shows.
In the fourth test case, you can buy subscriptions to shows 3,5,7,8,9, and you will be able to watch shows for the last eight days.
Submitted Solution:
```
def STR(): return list(input())
def INT(): return int(input())
def MAP(): return map(int, input().split())
def MAP2():return map(float,input().split())
def LIST(): return list(map(int, input().split()))
def STRING(): return input()
import string
import sys
from heapq import heappop , heappush
from bisect import *
from collections import deque , Counter , defaultdict
from math import *
from itertools import permutations , accumulate
dx = [-1 , 1 , 0 , 0 ]
dy = [0 , 0 , 1 , - 1]
#visited = [[False for i in range(m)] for j in range(n)]
# primes = [2,11,101,1009,10007,100003,1000003,10000019,102345689]
#sys.stdin = open(r'input.txt' , 'r')
#sys.stdout = open(r'output.txt' , 'w')
#for tt in range(INT()):
#Code
for tt in range(INT()):
n,k,d = MAP()
arr = LIST()
l = []
v = []
for i in arr:
if len(l) == d :
v.append(len(set(l)))
l.clear()
l.append(i)
else:
l.append(i)
if len(l) == d :
v.append(len(set(l)))
print(min(v))
``` | instruction | 0 | 36,155 | 10 | 72,310 |
No | output | 1 | 36,155 | 10 | 72,311 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is constraints.
The BerTV channel every day broadcasts one episode of one of the k TV shows. You know the schedule for the next n days: a sequence of integers a_1, a_2, ..., a_n (1 β€ a_i β€ k), where a_i is the show, the episode of which will be shown in i-th day.
The subscription to the show is bought for the entire show (i.e. for all its episodes), for each show the subscription is bought separately.
How many minimum subscriptions do you need to buy in order to have the opportunity to watch episodes of purchased shows d (1 β€ d β€ n) days in a row? In other words, you want to buy the minimum number of TV shows so that there is some segment of d consecutive days in which all episodes belong to the purchased shows.
Input
The first line contains an integer t (1 β€ t β€ 10000) β the number of test cases in the input. Then t test case descriptions follow.
The first line of each test case contains three integers n, k and d (1 β€ n β€ 2β
10^5, 1 β€ k β€ 10^6, 1 β€ d β€ n). The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ k), where a_i is the show that is broadcasted on the i-th day.
It is guaranteed that the sum of the values ββof n for all test cases in the input does not exceed 2β
10^5.
Output
Print t integers β the answers to the test cases in the input in the order they follow. The answer to a test case is the minimum number of TV shows for which you need to purchase a subscription so that you can watch episodes of the purchased TV shows on BerTV for d consecutive days. Please note that it is permissible that you will be able to watch more than d days in a row.
Example
Input
4
5 2 2
1 2 1 2 1
9 3 3
3 3 3 2 2 2 1 1 1
4 10 4
10 8 6 4
16 9 8
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3
Output
2
1
4
5
Note
In the first test case to have an opportunity to watch shows for two consecutive days, you need to buy a subscription on show 1 and on show 2. So the answer is two.
In the second test case, you can buy a subscription to any show because for each show you can find a segment of three consecutive days, consisting only of episodes of this show.
In the third test case in the unique segment of four days, you have four different shows, so you need to buy a subscription to all these four shows.
In the fourth test case, you can buy subscriptions to shows 3,5,7,8,9, and you will be able to watch shows for the last eight days.
Submitted Solution:
```
#sys.stdout=open("output.txt", 'w')
#sys.stdout.write("Yes" + '\n')
#from sys import stdin
#input=stdin.readline
#a = sorted([(n, i) for i, n in enumerate(map(int, input().split()))])
# from collections import Counter
# import sys
#s="abcdefghijklmnopqrstuvwxyz"
#n=int(input())
#n,k=map(int,input().split())
#arr=list(map(int,input().split()))
#arr=list(map(int,input().split
'''n=int(input())
abc=[]
for i in range(n):
abc.append(list(map(int,input().split())))
dp=[[-1,-1,-1] for i in range(n)]
for i in range(n):
if i==0:
for j in range(3):
dp[i][j]=abc[i][j]
else:
dp[i][0]=max(dp[i-1][1]+abc[i][0],dp[i-1][2]+abc[i][0])
dp[i][1]=max(dp[i-1][2]+abc[i][1],dp[i-1][0]+abc[i][1])
dp[i][2]=max(dp[i-1][0]+abc[i][2],dp[i-1][1]+abc[i][2])
print(max(dp[n-1]))'''
from collections import Counter
for _ in range(int(input())):
n,k,d= map(int, input().split())
arr=list(map(int,input().split()))
ans=len(set(arr[:d]))
#print(ans)
q=Counter(arr[:d])
#print(q)
cnt=0
m=d
for i in range(d,n):
if arr[i] not in q:
q[arr[i]]=1
ans+=1
elif q[arr[i]]>0:
q[arr[i]]+=1
elif q[arr[i]]==0:
q[arr[i]]=1
ans+=1
q[arr[cnt]]-=1
if q[arr[cnt]]==0:
ans-=1
#print("ans",ans)
cnt+=1
m=min(m,ans)
print(m)
``` | instruction | 0 | 36,156 | 10 | 72,312 |
No | output | 1 | 36,156 | 10 | 72,313 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is constraints.
The BerTV channel every day broadcasts one episode of one of the k TV shows. You know the schedule for the next n days: a sequence of integers a_1, a_2, ..., a_n (1 β€ a_i β€ k), where a_i is the show, the episode of which will be shown in i-th day.
The subscription to the show is bought for the entire show (i.e. for all its episodes), for each show the subscription is bought separately.
How many minimum subscriptions do you need to buy in order to have the opportunity to watch episodes of purchased shows d (1 β€ d β€ n) days in a row? In other words, you want to buy the minimum number of TV shows so that there is some segment of d consecutive days in which all episodes belong to the purchased shows.
Input
The first line contains an integer t (1 β€ t β€ 10000) β the number of test cases in the input. Then t test case descriptions follow.
The first line of each test case contains three integers n, k and d (1 β€ n β€ 2β
10^5, 1 β€ k β€ 10^6, 1 β€ d β€ n). The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ k), where a_i is the show that is broadcasted on the i-th day.
It is guaranteed that the sum of the values ββof n for all test cases in the input does not exceed 2β
10^5.
Output
Print t integers β the answers to the test cases in the input in the order they follow. The answer to a test case is the minimum number of TV shows for which you need to purchase a subscription so that you can watch episodes of the purchased TV shows on BerTV for d consecutive days. Please note that it is permissible that you will be able to watch more than d days in a row.
Example
Input
4
5 2 2
1 2 1 2 1
9 3 3
3 3 3 2 2 2 1 1 1
4 10 4
10 8 6 4
16 9 8
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3
Output
2
1
4
5
Note
In the first test case to have an opportunity to watch shows for two consecutive days, you need to buy a subscription on show 1 and on show 2. So the answer is two.
In the second test case, you can buy a subscription to any show because for each show you can find a segment of three consecutive days, consisting only of episodes of this show.
In the third test case in the unique segment of four days, you have four different shows, so you need to buy a subscription to all these four shows.
In the fourth test case, you can buy subscriptions to shows 3,5,7,8,9, and you will be able to watch shows for the last eight days.
Submitted Solution:
```
from collections import Counter as cnt
def count(n,k,d,a):
mat=cnt(a[:d])
i,prev=1,0
j,m=d,n+1
if d==n:
return(len(mat))
while j<n:
prev=a[i-1]
mat[a[j]]=mat.get(a[j],0)+1
mat[prev] = mat.get(prev)-1
if mat[prev] == 0:
del mat[prev]
m=min(m,len(mat))
i+=1
j+=1
return(m)
for i in range(int(input())):
n,k,d=map(int,input().split())
a=list(map(int,input().split()))
print(count(n,k,d,a))
``` | instruction | 0 | 36,157 | 10 | 72,314 |
No | output | 1 | 36,157 | 10 | 72,315 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arkady and Masha want to choose decorations for thier aquarium in Fishdom game. They have n decorations to choose from, each of them has some cost. To complete a task Arkady and Masha need to choose exactly m decorations from given, and they want to spend as little money as possible.
There is one difficulty: Masha likes some a of the given decorations, Arkady likes some b of the given decorations. Some decorations may be liked by both Arkady and Masha, or not be liked by both. The friends want to choose such decorations so that each of them likes at least k decorations among the chosen. Help Masha and Arkady find the minimum sum of money they need to spend.
Input
The first line contains three integers n, m and k (1 β€ n β€ 200000, 1 β€ m β€ n, 1 β€ k β€ n) β the number of decorations, how many decorations the friends should choose, how many decorations each of them should like among the chosen.
The second line contains n integers c1, c2, ..., cn (1 β€ ci β€ 109) β decorations costs.
The third line contains single integer a (1 β€ a β€ n) β the number of decorations liked by Masha. The fourth line contains a distinct integers x1, x2, ..., xa (1 β€ xi β€ n) β the ids of decorations liked by Masha.
The next two lines describe decorations liked by Arkady in the same format.
Output
Print single integer: the minimum sum of money the friends should spend to fulfill all constraints. If it is not possible, print -1.
Examples
Input
4 3 2
3 2 2 1
2
1 2
2
1 3
Output
7
Input
4 3 2
3 2 2 1
2
1 2
3
4 1 3
Output
6
Input
4 2 2
3 2 2 1
2
1 2
3
4 1 3
Output
-1
Note
In the first example the only possible variant to choose 3 decorations having all conditions satisfied is to choose decorations 1, 2, 3.
In the second example friends can choose decoration 4 instead of decoration 3, because this one is one coin cheaper.
In the third example it's not possible to choose 2 decorations in a way that both are liked by both Masha and Arkady.
Submitted Solution:
```
n,m,k=list(map(int,input().strip().split(' ')))
cost=list(map(int,input().strip().split(' ')))
a=int(input())
la=list(map(int,input().strip().split(' ')))
b=int(input())
lb=list(map(int,input().strip().split(' ')))
La=[0 for i in range(n)]
Lb=[0 for i in range(n)]
for x in la:
La[x-1]=1
for x in lb:
Lb[x-1]=1
Lboth=[0 for i in range(n)]
for i in range(n):
if La[i]==1 and Lb[i]==1:
Lboth[i]=1
numboth=sum(Lboth)
check=0
if k>m or sum(La)<k or sum(Lb)<k:
check=1
else:
if numboth<k:
if (k-numboth)+(k-numboth)+numboth>m:
check=1
def findout(remain,alike,blike,bothlike,needa,needb):
check=0
if needa>alike or needb>blike or remain<needa or remain<needb:
return 0
else:
if remain>=needa-min(needa,bothlike)+needb-min(needb,bothlike)+max(min(needa,bothlike),min(needb,bothlike)):
return 1
else:
return 0
if check==1:
print(-1)
else:
COST=[]
for i in range(len(cost)):
COST+=[[cost[i],i]]
COST=sorted(COST)
likedbya=[]
likedbyb=[]
likedbyboth=[]
tempa=0
tempb=0
tempboth=0
for i in range(len(COST)):
if La[COST[i][1]]==1:
tempa+=1
if Lb[COST[i][1]]==1:
tempb+=1
if La[COST[i][1]]==1 and Lb[COST[i][1]]==1:
tempboth+=1
likedbya+=[tempa]
likedbyb+=[tempb]
likedbyboth+=[tempboth]
needa=k
needb=k
total=0
used=0
for i in range(len(cost)):
mon,position=COST[i]
if Lboth[position]==1:
used+=1
total+=COST[i][0]
needa-=1
needb-=1
needa=max(needa,0)
needb=max(needb,0)
elif La[position]==1:
if needb==0:
total+=COST[i][0]
needa-=1
used+=1
needa=max(0,needa)
else:
remain=m-used-1
tempneeda=needa-1
tempneedb=needb
remainalike=sum(La)-likedbya[i]
remainblike=sum(Lb)-likedbyb[i]
remainbothlike=sum(Lboth)-likedbyboth[i]
if findout(remain,remainalike,remainblike,remainbothlike,tempneeda,tempneedb)==1:
total+=COST[i][0]
needa-=1
used+=1
needa=max(needa,0)
elif Lb[position]==1:
if needa==0:
total+=COST[i][0]
needb-=1
used+=1
needb=max(0,needb)
else:
remain=m-used-1
tempneeda=needa
tempneedb=needb-1
remainalike=sum(La)-likedbya[i]
remainblike=sum(Lb)-likedbyb[i]
remainbothlike=sum(Lboth)-likedbyboth[i]
if findout(remain,remainalike,remainblike,remainbothlike,tempneeda,tempneedb)==1:
total+=COST[i][0]
needb-=1
used+=1
needb=max(needb,0)
else:
remain=m-used-1
if 1==1:
tempneeda=needa
tempneedb=needb
remainalike=sum(La)-likedbya[i]
remainblike=sum(Lb)-likedbyb[i]
remainbothlike=sum(Lboth)-likedbyboth[i]
if findout(remain,remainalike,remainblike,remainbothlike,tempneeda,tempneedb)==1:
total+=COST[i][0]
#needb-=1
used+=1
#needb=max(needb,0)
if used==m:
print(total)
break
``` | instruction | 0 | 36,578 | 10 | 73,156 |
No | output | 1 | 36,578 | 10 | 73,157 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arkady and Masha want to choose decorations for thier aquarium in Fishdom game. They have n decorations to choose from, each of them has some cost. To complete a task Arkady and Masha need to choose exactly m decorations from given, and they want to spend as little money as possible.
There is one difficulty: Masha likes some a of the given decorations, Arkady likes some b of the given decorations. Some decorations may be liked by both Arkady and Masha, or not be liked by both. The friends want to choose such decorations so that each of them likes at least k decorations among the chosen. Help Masha and Arkady find the minimum sum of money they need to spend.
Input
The first line contains three integers n, m and k (1 β€ n β€ 200000, 1 β€ m β€ n, 1 β€ k β€ n) β the number of decorations, how many decorations the friends should choose, how many decorations each of them should like among the chosen.
The second line contains n integers c1, c2, ..., cn (1 β€ ci β€ 109) β decorations costs.
The third line contains single integer a (1 β€ a β€ n) β the number of decorations liked by Masha. The fourth line contains a distinct integers x1, x2, ..., xa (1 β€ xi β€ n) β the ids of decorations liked by Masha.
The next two lines describe decorations liked by Arkady in the same format.
Output
Print single integer: the minimum sum of money the friends should spend to fulfill all constraints. If it is not possible, print -1.
Examples
Input
4 3 2
3 2 2 1
2
1 2
2
1 3
Output
7
Input
4 3 2
3 2 2 1
2
1 2
3
4 1 3
Output
6
Input
4 2 2
3 2 2 1
2
1 2
3
4 1 3
Output
-1
Note
In the first example the only possible variant to choose 3 decorations having all conditions satisfied is to choose decorations 1, 2, 3.
In the second example friends can choose decoration 4 instead of decoration 3, because this one is one coin cheaper.
In the third example it's not possible to choose 2 decorations in a way that both are liked by both Masha and Arkady.
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
"""
def less_than(x_like, y_like):
n_x = len(x_like)
n_y = len(y_like)
if n_x == 0:
return False
if n_y == 0:
return True
else:
return x_like[0] <= y_like[0]
def satisfy_k(ab_like, a_like, b_like, m, k, cost):
ak = k
bk = k
#all vectors should already be sorted
for i in range(len(ab_like)):
ab_like[i] /= 2
while ak > 0 and bk > 0:
if less_than(ab_like, a_like) and less_than(ab_like, b_like):
cost += (ab_like.pop(0) * 2)
ak -= 1
bk -= 1
m -= 1
elif less_than(a_like, b_like):
cost += a_like.pop(0)
ak -= 1
m -= 1
elif less_than(b_like, a_like): #can't use just else becaues b_like could be empty
cost += b_like.pop(0)
bk -= 1
m -= 1
else:
break
for i in range(len(ab_like)):
ab_like[i] *= 2
while ak > 0 or bk > 0:
if less_than(ab_like, a_like) and less_than(ab_like, b_like):
cost += ab_like.pop(0)
ak -= 1
bk -= 1
m -= 1
elif less_than(a_like, b_like):
cost += a_like.pop(0)
ak -= 1
m -= 1
elif less_than(b_like, a_like): #can't use just else becaues b_like could be empty
cost += b_like.pop(0)
bk -= 1
m -= 1
else: #all vectors are empty
break
return cost, m
def satisfy_m(nobody_like, ab_like, a_like, b_like, m, cost):
for i in range(m):
if less_than(nobody_like, ab_like) and less_than(nobody_like, a_like) and less_than(nobody_like, b_like):
cost += nobody_like.pop(0)
elif less_than(ab_like, a_like) and less_than(ab_like, b_like):
cost += ab_like.pop(0)
elif less_than(a_like, b_like):
cost += a_like.pop(0)
elif less_than(b_like, a_like):
cost += b_like.pop(0)
else:
break
return cost
def solve():
cost = 0
#n, m, k = [4, 3, 2]
#cost_vec = [3, 2, 2, 1]
#player_one = set([0, 1])
#player_two = set([3, 0, 2])
n, m, k = [int(x) for x in input().split()]
cost_vec = [int(x) for x in input().split()]
#--skip input
input()
#-- use set
player_one = set([int(x)-1 for x in input().split()])
#--skip input
input()
#-- use set
player_two = set([int(x)-1 for x in input().split()])
ab_like = []
a_like = []
b_like = []
nobody_like = []
for i, c in enumerate(cost_vec):
if (i in player_one) and (i in player_two):
ab_like.append(c)
elif (i in player_one):
a_like.append(c)
elif (i in player_two):
b_like.append(c)
else:
nobody_like.append(c)
ab_like = sorted(ab_like)
a_like = sorted(a_like)
b_like = sorted(b_like)
nobody_like = sorted(nobody_like)
n_both_must_like = max(0, (k*2)-m)
if n_both_must_like > len(ab_like):
return -1
for _ in range(n_both_must_like):
cost += ab_like.pop(0)
m -= 1
k -= 1
cost, m = satisfy_k(ab_like, a_like, b_like, m, k, cost)
cost = satisfy_m(ab_like, a_like, b_like, nobody_like, m, cost)
print(cost)
solve()
``` | instruction | 0 | 36,579 | 10 | 73,158 |
No | output | 1 | 36,579 | 10 | 73,159 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arkady and Masha want to choose decorations for thier aquarium in Fishdom game. They have n decorations to choose from, each of them has some cost. To complete a task Arkady and Masha need to choose exactly m decorations from given, and they want to spend as little money as possible.
There is one difficulty: Masha likes some a of the given decorations, Arkady likes some b of the given decorations. Some decorations may be liked by both Arkady and Masha, or not be liked by both. The friends want to choose such decorations so that each of them likes at least k decorations among the chosen. Help Masha and Arkady find the minimum sum of money they need to spend.
Input
The first line contains three integers n, m and k (1 β€ n β€ 200000, 1 β€ m β€ n, 1 β€ k β€ n) β the number of decorations, how many decorations the friends should choose, how many decorations each of them should like among the chosen.
The second line contains n integers c1, c2, ..., cn (1 β€ ci β€ 109) β decorations costs.
The third line contains single integer a (1 β€ a β€ n) β the number of decorations liked by Masha. The fourth line contains a distinct integers x1, x2, ..., xa (1 β€ xi β€ n) β the ids of decorations liked by Masha.
The next two lines describe decorations liked by Arkady in the same format.
Output
Print single integer: the minimum sum of money the friends should spend to fulfill all constraints. If it is not possible, print -1.
Examples
Input
4 3 2
3 2 2 1
2
1 2
2
1 3
Output
7
Input
4 3 2
3 2 2 1
2
1 2
3
4 1 3
Output
6
Input
4 2 2
3 2 2 1
2
1 2
3
4 1 3
Output
-1
Note
In the first example the only possible variant to choose 3 decorations having all conditions satisfied is to choose decorations 1, 2, 3.
In the second example friends can choose decoration 4 instead of decoration 3, because this one is one coin cheaper.
In the third example it's not possible to choose 2 decorations in a way that both are liked by both Masha and Arkady.
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
"""
def less_than(x_like, y_like):
n_x = len(x_like)
n_y = len(y_like)
if n_x == 0:
return False
if n_y == 0:
return True
else:
return x_like[0] <= y_like[0]
def satisfy_k(ab_like, a_like, b_like, m, k, cost):
ak = k
bk = k
#all vectors should already be sorted
for i in range(len(ab_like)):
ab_like[i] /= 2
while ak > 0 and bk > 0:
if less_than(ab_like, a_like) and less_than(ab_like, b_like):
cost += (ab_like.pop(0) * 2)
ak -= 1
bk -= 1
m -= 1
elif less_than(a_like, b_like):
cost += a_like.pop(0)
ak -= 1
m -= 1
elif less_than(b_like, a_like): #can't use just else becaues b_like could be empty
cost += b_like.pop(0)
bk -= 1
m -= 1
else:
break
for i in range(len(ab_like)):
ab_like[i] *= 2
while ak > 0 or bk > 0:
if less_than(ab_like, a_like) and less_than(ab_like, b_like):
cost += ab_like.pop(0)
ak -= 1
bk -= 1
m -= 1
elif less_than(a_like, b_like):
cost += a_like.pop(0)
ak -= 1
m -= 1
elif less_than(b_like, a_like): #can't use just else becaues b_like could be empty
cost += b_like.pop(0)
bk -= 1
m -= 1
else: #all vectors are empty
break
return cost, m
def satisfy_m(nobody_like, ab_like, a_like, b_like, m, cost):
for i in range(m):
if less_than(nobody_like, ab_like) and less_than(nobody_like, a_like) and less_than(nobody_like, b_like):
cost += nobody_like.pop(0)
elif less_than(ab_like, a_like) and less_than(ab_like, b_like):
cost += ab_like.pop(0)
elif less_than(a_like, b_like):
cost += a_like.pop(0)
elif less_than(b_like, a_like):
cost += b_like.pop(0)
else:
break
return cost
def solve():
cost = 0
#n, m, k = [4, 3, 2]
#cost_vec = [3, 2, 2, 1]
#player_one = set([0, 1])
#player_two = set([3, 0, 2])
n, m, k = [int(x) for x in input().split()]
cost_vec = [int(x) for x in input().split()]
#--skip input
input()
#-- use set
player_one = set([int(x)-1 for x in input().split()])
#--skip input
input()
#-- use set
player_two = set([int(x)-1 for x in input().split()])
ab_like = []
a_like = []
b_like = []
nobody_like = []
for i, c in enumerate(cost_vec):
if (i in player_one) and (i in player_two):
ab_like.append(c)
elif (i in player_one):
a_like.append(c)
elif (i in player_two):
b_like.append(c)
else:
nobody_like.append(c)
ab_like = sorted(ab_like)
a_like = sorted(a_like)
b_like = sorted(b_like)
nobody_like = sorted(nobody_like)
n_both_must_like = max(0, (k*2)-m)
if n_both_must_like > len(ab_like):
return -1
for _ in range(n_both_must_like):
cost += ab_like.pop(0)
m -= 1
k -= 1
cost, m = satisfy_k(ab_like, a_like, b_like, m, k, cost)
cost = satisfy_m(ab_like, a_like, b_like, nobody_like, m, cost)
return int(cost)
print(solve())
``` | instruction | 0 | 36,580 | 10 | 73,160 |
No | output | 1 | 36,580 | 10 | 73,161 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arkady and Masha want to choose decorations for thier aquarium in Fishdom game. They have n decorations to choose from, each of them has some cost. To complete a task Arkady and Masha need to choose exactly m decorations from given, and they want to spend as little money as possible.
There is one difficulty: Masha likes some a of the given decorations, Arkady likes some b of the given decorations. Some decorations may be liked by both Arkady and Masha, or not be liked by both. The friends want to choose such decorations so that each of them likes at least k decorations among the chosen. Help Masha and Arkady find the minimum sum of money they need to spend.
Input
The first line contains three integers n, m and k (1 β€ n β€ 200000, 1 β€ m β€ n, 1 β€ k β€ n) β the number of decorations, how many decorations the friends should choose, how many decorations each of them should like among the chosen.
The second line contains n integers c1, c2, ..., cn (1 β€ ci β€ 109) β decorations costs.
The third line contains single integer a (1 β€ a β€ n) β the number of decorations liked by Masha. The fourth line contains a distinct integers x1, x2, ..., xa (1 β€ xi β€ n) β the ids of decorations liked by Masha.
The next two lines describe decorations liked by Arkady in the same format.
Output
Print single integer: the minimum sum of money the friends should spend to fulfill all constraints. If it is not possible, print -1.
Examples
Input
4 3 2
3 2 2 1
2
1 2
2
1 3
Output
7
Input
4 3 2
3 2 2 1
2
1 2
3
4 1 3
Output
6
Input
4 2 2
3 2 2 1
2
1 2
3
4 1 3
Output
-1
Note
In the first example the only possible variant to choose 3 decorations having all conditions satisfied is to choose decorations 1, 2, 3.
In the second example friends can choose decoration 4 instead of decoration 3, because this one is one coin cheaper.
In the third example it's not possible to choose 2 decorations in a way that both are liked by both Masha and Arkady.
Submitted Solution:
```
import math
def less_than(x_like, y_like, x_i, y_i):
if x_i >= len(x_like):
return False
if y_i >= len(y_like):
return True
else:
return x_like[x_i] <= y_like[y_i]
def satisfy_k_helper(ab_like, a_like, b_like, ab_i, a_i, b_i, m, ak, bk, cost, solution_dict):
if cost >= solution_dict["min_cost"]:
return
if ak <= 0 and bk <= 0:
solution_dict["min_cost"] = cost
solution_dict["m"] = m
solution_dict["ab_i"] = ab_i
solution_dict["a_i"] = a_i
solution_dict["b_i"] = b_i
else:
if ab_i < len(ab_like):
satisfy_k_helper(ab_like, a_like, b_like, ab_i+1, a_i, b_i, m-1, ak-1, bk-1, cost + ab_like[ab_i], solution_dict)
if a_i < len(a_like):
satisfy_k_helper(ab_like, a_like, b_like, ab_i, a_i+1, b_i, m-1, ak-1, bk, cost + a_like[a_i], solution_dict)
if b_i < len(b_like):
satisfy_k_helper(ab_like, a_like, b_like, ab_i, a_i, b_i+1, m-1, ak, bk-1, cost + b_like[b_i], solution_dict)
def satisfy_k(ab_like, a_like, b_like, m, k, cost):
ak = k
bk = k
solution_dict = {"min_cost":math.inf, "m":m, "ab_i":0, "a_i":0, "b_i":0}
satisfy_k_helper(ab_like, a_like, b_like, 0, 0, 0, m, ak, bk, cost, solution_dict)
return solution_dict
def satisfy_m(nobody_like, ab_like, a_like, b_like, ab_i, a_i, b_i, m, cost):
nb_i = 0
for i in range(m):
if less_than(nobody_like, ab_like, nb_i, ab_i) and less_than(nobody_like, a_like, nb_i, a_i) and less_than(nobody_like, b_like, nb_i, b_i):
cost += nobody_like[nb_i]
nb_i += 1
elif less_than(ab_like, a_like, ab_i, a_i) and less_than(ab_like, b_like, ab_i, b_i):
cost += ab_like[ab_i]
ab_i += 1
elif less_than(a_like, b_like, a_i, b_i):
cost += a_like[a_i]
a_i += 1
elif less_than(b_like, a_like):
cost += b_like[b_i]
b_i += 1
else:
break
return cost
def solve():
# cost = 0
# n, m, k = [4, 3, 2]
# cost_vec = [3, 2, 2, 1]
# player_one = set([0, 1])
# player_two = set([3, 0, 2])
n, m, k = [int(x) for x in input().split()]
cost_vec = [int(x) for x in input().split()]
#--skip input
input()
#-- use set
player_one = set([int(x)-1 for x in input().split()])
#--skip input
input()
#-- use set
player_two = set([int(x)-1 for x in input().split()])
ab_like = []
a_like = []
b_like = []
nobody_like = []
for i, c in enumerate(cost_vec):
if (i in player_one) and (i in player_two):
ab_like.append(c)
elif (i in player_one):
a_like.append(c)
elif (i in player_two):
b_like.append(c)
else:
nobody_like.append(c)
ab_like = sorted(ab_like)
a_like = sorted(a_like)
b_like = sorted(b_like)
nobody_like = sorted(nobody_like)
n_both_must_like = max(0, (k*2)-m)
if n_both_must_like > len(ab_like):
return -1
cost = 0
for _ in range(n_both_must_like):
cost += ab_like.pop(0)
m -= 1
k -= 1
solution_dict = satisfy_k(ab_like, a_like, b_like, m, k, cost)
cost = satisfy_m(nobody_like, ab_like, a_like, b_like, solution_dict["ab_i"],
solution_dict["a_i"], solution_dict["b_i"], solution_dict["m"], solution_dict["min_cost"])
return cost
print(solve())
``` | instruction | 0 | 36,581 | 10 | 73,162 |
No | output | 1 | 36,581 | 10 | 73,163 |
Provide a correct Python 3 solution for this coding contest problem.
There is a long-established secondhand bookstore called JOI secondhand bookstore in your town, and you often use the JOI secondhand bookstore. Each book has a standard price, and you can buy it at that price if you go to the JOI secondhand bookstore.
At the JOI secondhand bookstore, books are classified into 10 genres such as novels, manga, and magazines. Genres are numbered from 1 to 10. The JOI secondhand bookstore has a service that if you buy books of the same genre in bulk, they will buy them at a high price. Specifically, when books of the same genre are purchased together in T books, the purchase price per book of that genre is T -1 yen higher than the standard price. For example, if books of the same genre with standard prices of 100 yen, 120 yen, and 150 yen are sold together at the JOI secondhand bookstore, the purchase prices will be 102 yen, 122 yen, and 152 yen, respectively.
By the way, you have to move in a hurry due to personal reasons. You have N books, but it is difficult to bring all the books to your new home, so I decided to sell K of the N books to the JOI secondhand bookstore.
input
Read the following input from standard input.
* The integers N and K are written on the first line, separated by blanks, indicating that the number of books you have is N, of which K books will be sold to JOI secondhand bookstores.
* The following N lines contain information about your book. On the first line of i + (1 β€ i β€ N), the integers Ci and Gi are written separated by blanks, indicating that the base price of the i-th book is Ci and the genre number is Gi. ..
output
Output an integer representing the maximum value of the total purchase price to the standard output on one line.
Example
Input
7 4
14 1
13 2
12 3
14 2
8 2
16 3
11 2
Output
60 | instruction | 0 | 36,784 | 10 | 73,568 |
"Correct Solution:
```
def solve():
n, k = map(int,input().split())
group_num = 10
book_map = [[] for i in range(group_num)]
acc_map = [[0] for i in range(group_num)]
for i in range(n):
c, g = map(int,input().split())
book_map[g - 1].append(c)
for i in range(group_num):
bmi = book_map[i]
bmi.sort(reverse=True)
ami = acc_map[i]
acc = 0
for j in range(len(bmi)):
acc += (bmi[j] + j * 2)
ami.append(acc)
dp = [[0] * (k + 1) for i in range(group_num + 1)]
for y in range(1, k + 1):
for x in range(1, group_num + 1):
accs = acc_map[x - 1]
dp_pre = dp[x - 1]
dp[x][y] = max([dp_pre[y - z] + accs[z] for z in range(min(y + 1, len(accs)))])
print(dp[group_num][k])
solve()
``` | output | 1 | 36,784 | 10 | 73,569 |
Provide a correct Python 3 solution for this coding contest problem.
There is a long-established secondhand bookstore called JOI secondhand bookstore in your town, and you often use the JOI secondhand bookstore. Each book has a standard price, and you can buy it at that price if you go to the JOI secondhand bookstore.
At the JOI secondhand bookstore, books are classified into 10 genres such as novels, manga, and magazines. Genres are numbered from 1 to 10. The JOI secondhand bookstore has a service that if you buy books of the same genre in bulk, they will buy them at a high price. Specifically, when books of the same genre are purchased together in T books, the purchase price per book of that genre is T -1 yen higher than the standard price. For example, if books of the same genre with standard prices of 100 yen, 120 yen, and 150 yen are sold together at the JOI secondhand bookstore, the purchase prices will be 102 yen, 122 yen, and 152 yen, respectively.
By the way, you have to move in a hurry due to personal reasons. You have N books, but it is difficult to bring all the books to your new home, so I decided to sell K of the N books to the JOI secondhand bookstore.
input
Read the following input from standard input.
* The integers N and K are written on the first line, separated by blanks, indicating that the number of books you have is N, of which K books will be sold to JOI secondhand bookstores.
* The following N lines contain information about your book. On the first line of i + (1 β€ i β€ N), the integers Ci and Gi are written separated by blanks, indicating that the base price of the i-th book is Ci and the genre number is Gi. ..
output
Output an integer representing the maximum value of the total purchase price to the standard output on one line.
Example
Input
7 4
14 1
13 2
12 3
14 2
8 2
16 3
11 2
Output
60 | instruction | 0 | 36,785 | 10 | 73,570 |
"Correct Solution:
```
def solve():
n, k = map(int,input().split())
group_num = 10
book_map = [[] for i in range(group_num)]
acc_map = []
for i in range(n):
c, g = map(int,input().split())
book_map[g - 1].append(c)
for i in range(group_num):
bmi = book_map[i]
bmi.sort(reverse=True)
ami = [0 for i in range(len(bmi) + 1)]
for j in range(len(bmi)):
ami[j + 1] = ami[j] + bmi[j] + j * 2
acc_map.append(ami)
dp = [[0] * (k + 1) for i in range(group_num + 1)]
for y in range(1, k + 1):
for x in range(1, group_num + 1):
accs = acc_map[x - 1]
dp_pre = dp[x - 1]
dp[x][y] = max([dp_pre[y - z] + accs[z] for z in range(min(y + 1, len(accs)))])
print(dp[group_num][k])
solve()
``` | output | 1 | 36,785 | 10 | 73,571 |
Provide a correct Python 3 solution for this coding contest problem.
There is a long-established secondhand bookstore called JOI secondhand bookstore in your town, and you often use the JOI secondhand bookstore. Each book has a standard price, and you can buy it at that price if you go to the JOI secondhand bookstore.
At the JOI secondhand bookstore, books are classified into 10 genres such as novels, manga, and magazines. Genres are numbered from 1 to 10. The JOI secondhand bookstore has a service that if you buy books of the same genre in bulk, they will buy them at a high price. Specifically, when books of the same genre are purchased together in T books, the purchase price per book of that genre is T -1 yen higher than the standard price. For example, if books of the same genre with standard prices of 100 yen, 120 yen, and 150 yen are sold together at the JOI secondhand bookstore, the purchase prices will be 102 yen, 122 yen, and 152 yen, respectively.
By the way, you have to move in a hurry due to personal reasons. You have N books, but it is difficult to bring all the books to your new home, so I decided to sell K of the N books to the JOI secondhand bookstore.
input
Read the following input from standard input.
* The integers N and K are written on the first line, separated by blanks, indicating that the number of books you have is N, of which K books will be sold to JOI secondhand bookstores.
* The following N lines contain information about your book. On the first line of i + (1 β€ i β€ N), the integers Ci and Gi are written separated by blanks, indicating that the base price of the i-th book is Ci and the genre number is Gi. ..
output
Output an integer representing the maximum value of the total purchase price to the standard output on one line.
Example
Input
7 4
14 1
13 2
12 3
14 2
8 2
16 3
11 2
Output
60 | instruction | 0 | 36,786 | 10 | 73,572 |
"Correct Solution:
```
def main():
import sys
input=sys.stdin.readline
n,k=map(int,input().split())
ab=[list(map(int,input().split())) for _ in [0]*n]
g=[[] for _ in [0]*10]
[g[b-1].append(a) for a,b in ab]
[g[c].sort(reverse=True) for c in range(10)]
for c in range(10):g[c]=[0]+g[c]
for c in range(10):
for i in range(2,len(g[c])):
g[c][i]+=g[c][i-1]+2*(i-1)
dp=[0]*(k+1)
for c in range(10):
dp2=[0]*(k+1)
for i in range(len(g[c])):
for j in range(k+1-i):
dp2[i+j]=max(dp2[i+j],dp[j]+g[c][i])
dp=dp2
print(max(dp))
if __name__=='__main__':
main()
``` | output | 1 | 36,786 | 10 | 73,573 |
Provide a correct Python 3 solution for this coding contest problem.
There is a long-established secondhand bookstore called JOI secondhand bookstore in your town, and you often use the JOI secondhand bookstore. Each book has a standard price, and you can buy it at that price if you go to the JOI secondhand bookstore.
At the JOI secondhand bookstore, books are classified into 10 genres such as novels, manga, and magazines. Genres are numbered from 1 to 10. The JOI secondhand bookstore has a service that if you buy books of the same genre in bulk, they will buy them at a high price. Specifically, when books of the same genre are purchased together in T books, the purchase price per book of that genre is T -1 yen higher than the standard price. For example, if books of the same genre with standard prices of 100 yen, 120 yen, and 150 yen are sold together at the JOI secondhand bookstore, the purchase prices will be 102 yen, 122 yen, and 152 yen, respectively.
By the way, you have to move in a hurry due to personal reasons. You have N books, but it is difficult to bring all the books to your new home, so I decided to sell K of the N books to the JOI secondhand bookstore.
input
Read the following input from standard input.
* The integers N and K are written on the first line, separated by blanks, indicating that the number of books you have is N, of which K books will be sold to JOI secondhand bookstores.
* The following N lines contain information about your book. On the first line of i + (1 β€ i β€ N), the integers Ci and Gi are written separated by blanks, indicating that the base price of the i-th book is Ci and the genre number is Gi. ..
output
Output an integer representing the maximum value of the total purchase price to the standard output on one line.
Example
Input
7 4
14 1
13 2
12 3
14 2
8 2
16 3
11 2
Output
60 | instruction | 0 | 36,787 | 10 | 73,574 |
"Correct Solution:
```
def solve():
n, k = map(int,input().split())
group_num = 10
book_map = [[] for i in range(group_num)]
acc_map = [[0] for i in range(group_num)]
for i in range(n):
c,g = map(int,input().split())
book_map[g - 1].append(c)
for i in range(group_num):
book_map[i].sort(reverse=True)
acc = 0
for j in range(len(book_map[i])):
book_map[i][j] += j * 2
acc += book_map[i][j]
acc_map[i].append(acc)
# print(acc_map)
dp = [[0] * (k + 1) for i in range(group_num + 1)]
for y in range(1,k + 1):
for x in range(1,group_num + 1):
for z in range(min(y + 1, len(acc_map[x - 1]))):
dp[x][y] = max(dp[x][y], dp[x - 1][y - z] + acc_map[x - 1][z])
print(dp[group_num][k])
solve()
``` | output | 1 | 36,787 | 10 | 73,575 |
Provide a correct Python 3 solution for this coding contest problem.
There is a long-established secondhand bookstore called JOI secondhand bookstore in your town, and you often use the JOI secondhand bookstore. Each book has a standard price, and you can buy it at that price if you go to the JOI secondhand bookstore.
At the JOI secondhand bookstore, books are classified into 10 genres such as novels, manga, and magazines. Genres are numbered from 1 to 10. The JOI secondhand bookstore has a service that if you buy books of the same genre in bulk, they will buy them at a high price. Specifically, when books of the same genre are purchased together in T books, the purchase price per book of that genre is T -1 yen higher than the standard price. For example, if books of the same genre with standard prices of 100 yen, 120 yen, and 150 yen are sold together at the JOI secondhand bookstore, the purchase prices will be 102 yen, 122 yen, and 152 yen, respectively.
By the way, you have to move in a hurry due to personal reasons. You have N books, but it is difficult to bring all the books to your new home, so I decided to sell K of the N books to the JOI secondhand bookstore.
input
Read the following input from standard input.
* The integers N and K are written on the first line, separated by blanks, indicating that the number of books you have is N, of which K books will be sold to JOI secondhand bookstores.
* The following N lines contain information about your book. On the first line of i + (1 β€ i β€ N), the integers Ci and Gi are written separated by blanks, indicating that the base price of the i-th book is Ci and the genre number is Gi. ..
output
Output an integer representing the maximum value of the total purchase price to the standard output on one line.
Example
Input
7 4
14 1
13 2
12 3
14 2
8 2
16 3
11 2
Output
60 | instruction | 0 | 36,788 | 10 | 73,576 |
"Correct Solution:
```
from heapq import heappush as push
from heapq import heappop as pop
def solve():
n, k = map(int,input().split())
group_num = 10
book_map = [[] for i in range(group_num)]
acc_map = [[0] for i in range(group_num)]
for i in range(n):
c,g = map(int,input().split())
push(book_map[g - 1], -c)
for i in range(group_num):
bmi = book_map[i]
append = acc_map[i].append
acc = 0
for j in range(len(bmi)):
acc += (pop(bmi) - j * 2)
append(acc)
dp = [[0] * (k + 1) for i in range(group_num + 1)]
for y in range(1,k + 1):
for x in range(1,group_num + 1):
mp = acc_map[x - 1]
be = dp[x - 1]
for z in range(min(y + 1, len(mp))):
v1 = be[y - z]
v2 = mp[z]
if dp[x][y] > v1 + v2:
dp[x][y] = v1 + v2
print(-dp[group_num][k])
solve()
``` | output | 1 | 36,788 | 10 | 73,577 |
Provide a correct Python 3 solution for this coding contest problem.
There is a long-established secondhand bookstore called JOI secondhand bookstore in your town, and you often use the JOI secondhand bookstore. Each book has a standard price, and you can buy it at that price if you go to the JOI secondhand bookstore.
At the JOI secondhand bookstore, books are classified into 10 genres such as novels, manga, and magazines. Genres are numbered from 1 to 10. The JOI secondhand bookstore has a service that if you buy books of the same genre in bulk, they will buy them at a high price. Specifically, when books of the same genre are purchased together in T books, the purchase price per book of that genre is T -1 yen higher than the standard price. For example, if books of the same genre with standard prices of 100 yen, 120 yen, and 150 yen are sold together at the JOI secondhand bookstore, the purchase prices will be 102 yen, 122 yen, and 152 yen, respectively.
By the way, you have to move in a hurry due to personal reasons. You have N books, but it is difficult to bring all the books to your new home, so I decided to sell K of the N books to the JOI secondhand bookstore.
input
Read the following input from standard input.
* The integers N and K are written on the first line, separated by blanks, indicating that the number of books you have is N, of which K books will be sold to JOI secondhand bookstores.
* The following N lines contain information about your book. On the first line of i + (1 β€ i β€ N), the integers Ci and Gi are written separated by blanks, indicating that the base price of the i-th book is Ci and the genre number is Gi. ..
output
Output an integer representing the maximum value of the total purchase price to the standard output on one line.
Example
Input
7 4
14 1
13 2
12 3
14 2
8 2
16 3
11 2
Output
60 | instruction | 0 | 36,789 | 10 | 73,578 |
"Correct Solution:
```
def solve():
n, k = map(int,input().split())
group_num = 10
book_map = [[] for i in range(group_num)]
acc_map = [[0] for i in range(group_num)]
for i in range(n):
c,g = map(int,input().split())
book_map[g - 1].append(c)
for i in range(group_num):
bmi = book_map[i]
bmi.sort(reverse=True)
append = acc_map[i].append
acc = 0
for j in range(len(bmi)):
acc += (bmi[j] + j * 2)
append(acc)
dp = [[0] * (k + 1) for i in range(group_num + 1)]
for y in range(1,k + 1):
for x in range(1,group_num + 1):
mp = acc_map[x - 1]
be = dp[x - 1]
dp[x][y] = max([be[y - z] + mp[z] for z in range(min(y + 1, len(mp)))])
print(dp[group_num][k])
solve()
``` | output | 1 | 36,789 | 10 | 73,579 |
Provide a correct Python 3 solution for this coding contest problem.
There is a long-established secondhand bookstore called JOI secondhand bookstore in your town, and you often use the JOI secondhand bookstore. Each book has a standard price, and you can buy it at that price if you go to the JOI secondhand bookstore.
At the JOI secondhand bookstore, books are classified into 10 genres such as novels, manga, and magazines. Genres are numbered from 1 to 10. The JOI secondhand bookstore has a service that if you buy books of the same genre in bulk, they will buy them at a high price. Specifically, when books of the same genre are purchased together in T books, the purchase price per book of that genre is T -1 yen higher than the standard price. For example, if books of the same genre with standard prices of 100 yen, 120 yen, and 150 yen are sold together at the JOI secondhand bookstore, the purchase prices will be 102 yen, 122 yen, and 152 yen, respectively.
By the way, you have to move in a hurry due to personal reasons. You have N books, but it is difficult to bring all the books to your new home, so I decided to sell K of the N books to the JOI secondhand bookstore.
input
Read the following input from standard input.
* The integers N and K are written on the first line, separated by blanks, indicating that the number of books you have is N, of which K books will be sold to JOI secondhand bookstores.
* The following N lines contain information about your book. On the first line of i + (1 β€ i β€ N), the integers Ci and Gi are written separated by blanks, indicating that the base price of the i-th book is Ci and the genre number is Gi. ..
output
Output an integer representing the maximum value of the total purchase price to the standard output on one line.
Example
Input
7 4
14 1
13 2
12 3
14 2
8 2
16 3
11 2
Output
60 | instruction | 0 | 36,790 | 10 | 73,580 |
"Correct Solution:
```
import sys
def solve():
file_input = sys.stdin
N, K = map(int, file_input.readline().split())
G = [[] for i in range(10)]
for line in file_input:
c, g = map(int, line.split())
G[g - 1].append(c)
for genre in G:
genre.sort(reverse=True)
for i, p in zip(range(1, len(genre)), genre):
genre[i] += p + 2 * i # Recording cumulative price
C = [0] * (K + 1)
for genre in G:
pre_C = C.copy()
for n, p in zip(range(len(genre), 0, -1), genre[::-1]):
for i, vals in enumerate(zip(pre_C[n:], C)):
v1, v2 = vals
v1 += p
if v1 > v2:
C[i] = v1
print(C[0])
solve()
``` | output | 1 | 36,790 | 10 | 73,581 |
Provide a correct Python 3 solution for this coding contest problem.
There is a long-established secondhand bookstore called JOI secondhand bookstore in your town, and you often use the JOI secondhand bookstore. Each book has a standard price, and you can buy it at that price if you go to the JOI secondhand bookstore.
At the JOI secondhand bookstore, books are classified into 10 genres such as novels, manga, and magazines. Genres are numbered from 1 to 10. The JOI secondhand bookstore has a service that if you buy books of the same genre in bulk, they will buy them at a high price. Specifically, when books of the same genre are purchased together in T books, the purchase price per book of that genre is T -1 yen higher than the standard price. For example, if books of the same genre with standard prices of 100 yen, 120 yen, and 150 yen are sold together at the JOI secondhand bookstore, the purchase prices will be 102 yen, 122 yen, and 152 yen, respectively.
By the way, you have to move in a hurry due to personal reasons. You have N books, but it is difficult to bring all the books to your new home, so I decided to sell K of the N books to the JOI secondhand bookstore.
input
Read the following input from standard input.
* The integers N and K are written on the first line, separated by blanks, indicating that the number of books you have is N, of which K books will be sold to JOI secondhand bookstores.
* The following N lines contain information about your book. On the first line of i + (1 β€ i β€ N), the integers Ci and Gi are written separated by blanks, indicating that the base price of the i-th book is Ci and the genre number is Gi. ..
output
Output an integer representing the maximum value of the total purchase price to the standard output on one line.
Example
Input
7 4
14 1
13 2
12 3
14 2
8 2
16 3
11 2
Output
60 | instruction | 0 | 36,791 | 10 | 73,582 |
"Correct Solution:
```
def solve():
N, K = map(int, input().split())
a = [[] for _ in [0]*10]
l = []
for c, g in (tuple(map(int, input().split())) for _ in [0]*N):
a[g-1].append(c)
for i, prices in enumerate(a):
ln = len(prices)
if ln == 0:
continue
dp = [0]*(K+1)
for j, price in enumerate(prices):
_k = j if j < K-1 else K-1
for k, prev, cur in zip(range(_k, -1, -1), dp[_k::-1], dp[_k+1::-1]):
if prev+price+k*2 > cur:
dp[k+1] = prev+price+k*2
l.append(dp[1:ln+1])
dp = [float("-inf")]*(K+1)
dp[0] = 0
for prices in l:
cdp = dp[:]
for i, price in enumerate(prices, start=1):
if i > K:
break
for j, cur, prev in zip(range(K, -1, -1), cdp[::-1], dp[K-i::-1]):
if 0 < prev+price > cur:
cdp[j] = prev+price
dp = cdp
print(max(dp))
if __name__ == "__main__":
solve()
``` | output | 1 | 36,791 | 10 | 73,583 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a long-established secondhand bookstore called JOI secondhand bookstore in your town, and you often use the JOI secondhand bookstore. Each book has a standard price, and you can buy it at that price if you go to the JOI secondhand bookstore.
At the JOI secondhand bookstore, books are classified into 10 genres such as novels, manga, and magazines. Genres are numbered from 1 to 10. The JOI secondhand bookstore has a service that if you buy books of the same genre in bulk, they will buy them at a high price. Specifically, when books of the same genre are purchased together in T books, the purchase price per book of that genre is T -1 yen higher than the standard price. For example, if books of the same genre with standard prices of 100 yen, 120 yen, and 150 yen are sold together at the JOI secondhand bookstore, the purchase prices will be 102 yen, 122 yen, and 152 yen, respectively.
By the way, you have to move in a hurry due to personal reasons. You have N books, but it is difficult to bring all the books to your new home, so I decided to sell K of the N books to the JOI secondhand bookstore.
input
Read the following input from standard input.
* The integers N and K are written on the first line, separated by blanks, indicating that the number of books you have is N, of which K books will be sold to JOI secondhand bookstores.
* The following N lines contain information about your book. On the first line of i + (1 β€ i β€ N), the integers Ci and Gi are written separated by blanks, indicating that the base price of the i-th book is Ci and the genre number is Gi. ..
output
Output an integer representing the maximum value of the total purchase price to the standard output on one line.
Example
Input
7 4
14 1
13 2
12 3
14 2
8 2
16 3
11 2
Output
60
Submitted Solution:
```
def solve():
n, k = map(int,input().split())
group_num = 10
book_map = [[] for i in range(group_num)]
acc_map = [[0] for i in range(group_num)]
for i in range(n):
c,g = map(int,input().split())
book_map[g - 1].append(c)
for i in range(group_num):
book_map[i].sort(reverse=True)
acc = 0
bmi = book_map[i]
append = acc_map[i].append
for j in range(len(bmi)):
acc += (bmi[j] + j * 2)
append(acc)
dp = [[0] * (k + 1) for i in range(group_num + 1)]
for y in range(1,k + 1):
for x in range(1,group_num + 1):
mp = acc_map[x - 1]
be = dp[x - 1]
for z in range(min(y + 1, len(mp))):
v1 = be[y - z]
v2 = mp[z]
if dp[x][y] < v1 + v2:
dp[x][y] = v1 + v2
print(dp[group_num][k])
solve()
``` | instruction | 0 | 36,792 | 10 | 73,584 |
Yes | output | 1 | 36,792 | 10 | 73,585 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a long-established secondhand bookstore called JOI secondhand bookstore in your town, and you often use the JOI secondhand bookstore. Each book has a standard price, and you can buy it at that price if you go to the JOI secondhand bookstore.
At the JOI secondhand bookstore, books are classified into 10 genres such as novels, manga, and magazines. Genres are numbered from 1 to 10. The JOI secondhand bookstore has a service that if you buy books of the same genre in bulk, they will buy them at a high price. Specifically, when books of the same genre are purchased together in T books, the purchase price per book of that genre is T -1 yen higher than the standard price. For example, if books of the same genre with standard prices of 100 yen, 120 yen, and 150 yen are sold together at the JOI secondhand bookstore, the purchase prices will be 102 yen, 122 yen, and 152 yen, respectively.
By the way, you have to move in a hurry due to personal reasons. You have N books, but it is difficult to bring all the books to your new home, so I decided to sell K of the N books to the JOI secondhand bookstore.
input
Read the following input from standard input.
* The integers N and K are written on the first line, separated by blanks, indicating that the number of books you have is N, of which K books will be sold to JOI secondhand bookstores.
* The following N lines contain information about your book. On the first line of i + (1 β€ i β€ N), the integers Ci and Gi are written separated by blanks, indicating that the base price of the i-th book is Ci and the genre number is Gi. ..
output
Output an integer representing the maximum value of the total purchase price to the standard output on one line.
Example
Input
7 4
14 1
13 2
12 3
14 2
8 2
16 3
11 2
Output
60
Submitted Solution:
```
import sys
def solve():
file_input = sys.stdin
N, K = map(int, file_input.readline().split())
G = [[] for i in range(10)]
for line in file_input:
c, g = map(int, line.split())
G[g - 1].append(c)
for genre in G:
genre.sort(reverse=True)
for p, i in zip(genre, range(1, len(genre))):
genre[i] += p + 2 * i # Recording cumulative price
C = [0] * (K + 1)
for genre in G:
pre_C = C.copy()
for n, p in enumerate(genre, start=1):
for i, vals in enumerate(zip(pre_C[n:], C)):
v1, v2 = vals
v1 += p
if v1 > v2:
C[i] = v1
print(C[0])
solve()
``` | instruction | 0 | 36,793 | 10 | 73,586 |
Yes | output | 1 | 36,793 | 10 | 73,587 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a long-established secondhand bookstore called JOI secondhand bookstore in your town, and you often use the JOI secondhand bookstore. Each book has a standard price, and you can buy it at that price if you go to the JOI secondhand bookstore.
At the JOI secondhand bookstore, books are classified into 10 genres such as novels, manga, and magazines. Genres are numbered from 1 to 10. The JOI secondhand bookstore has a service that if you buy books of the same genre in bulk, they will buy them at a high price. Specifically, when books of the same genre are purchased together in T books, the purchase price per book of that genre is T -1 yen higher than the standard price. For example, if books of the same genre with standard prices of 100 yen, 120 yen, and 150 yen are sold together at the JOI secondhand bookstore, the purchase prices will be 102 yen, 122 yen, and 152 yen, respectively.
By the way, you have to move in a hurry due to personal reasons. You have N books, but it is difficult to bring all the books to your new home, so I decided to sell K of the N books to the JOI secondhand bookstore.
input
Read the following input from standard input.
* The integers N and K are written on the first line, separated by blanks, indicating that the number of books you have is N, of which K books will be sold to JOI secondhand bookstores.
* The following N lines contain information about your book. On the first line of i + (1 β€ i β€ N), the integers Ci and Gi are written separated by blanks, indicating that the base price of the i-th book is Ci and the genre number is Gi. ..
output
Output an integer representing the maximum value of the total purchase price to the standard output on one line.
Example
Input
7 4
14 1
13 2
12 3
14 2
8 2
16 3
11 2
Output
60
Submitted Solution:
```
from itertools import accumulate
n, k = map(int, input().split())
books = [[] for _ in range(10)]
while n:
c, g = map(int, input().split())
books[g - 1].append(c)
n -= 1
books_acc = [[0] + list(accumulate(c + i * 2 for i, c in enumerate(sorted(q, reverse=True)))) for q in books]
def memoize(f):
memo = [[-1] * (k + 1) for _ in range(10)]
def main(x, y):
if x > 9:
return 0
result = memo[x][y]
if result < 0:
result = memo[x][y] = f(x, y)
return result
return main
@memoize
def combi(g, remain):
book_acc = list(books_acc[g])
salable = min(remain + 1, len(book_acc))
return max([book_acc[i] + combi(g + 1, remain - i) for i in range(salable)], default=0)
print(combi(0, k))
``` | instruction | 0 | 36,794 | 10 | 73,588 |
Yes | output | 1 | 36,794 | 10 | 73,589 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a long-established secondhand bookstore called JOI secondhand bookstore in your town, and you often use the JOI secondhand bookstore. Each book has a standard price, and you can buy it at that price if you go to the JOI secondhand bookstore.
At the JOI secondhand bookstore, books are classified into 10 genres such as novels, manga, and magazines. Genres are numbered from 1 to 10. The JOI secondhand bookstore has a service that if you buy books of the same genre in bulk, they will buy them at a high price. Specifically, when books of the same genre are purchased together in T books, the purchase price per book of that genre is T -1 yen higher than the standard price. For example, if books of the same genre with standard prices of 100 yen, 120 yen, and 150 yen are sold together at the JOI secondhand bookstore, the purchase prices will be 102 yen, 122 yen, and 152 yen, respectively.
By the way, you have to move in a hurry due to personal reasons. You have N books, but it is difficult to bring all the books to your new home, so I decided to sell K of the N books to the JOI secondhand bookstore.
input
Read the following input from standard input.
* The integers N and K are written on the first line, separated by blanks, indicating that the number of books you have is N, of which K books will be sold to JOI secondhand bookstores.
* The following N lines contain information about your book. On the first line of i + (1 β€ i β€ N), the integers Ci and Gi are written separated by blanks, indicating that the base price of the i-th book is Ci and the genre number is Gi. ..
output
Output an integer representing the maximum value of the total purchase price to the standard output on one line.
Example
Input
7 4
14 1
13 2
12 3
14 2
8 2
16 3
11 2
Output
60
Submitted Solution:
```
def solve():
N, K = map(int, input().split())
a = [[] for _ in [0]*10]
l = []
for c, g in (tuple(map(int, input().split())) for _ in [0]*N):
a[g-1].append(c)
for i, prices in enumerate(a):
ln = len(prices)
if ln == 0:
continue
dp = [0]*(K+1)
for j, price in enumerate(prices):
_k = j if j < K-1 else K-1
for k, prev, cur in zip(range(_k, -1, -1), dp[_k::-1], dp[_k+1::-1]):
if prev+price+k*2 > cur:
dp[k+1] = prev+price+k*2
l.append(dp[1:ln+1])
dp = [0]*(K+1)
for prices in l:
for i, price in enumerate(prices, start=1):
for j, cur, prev in zip(range(K, -1, -1), dp[::-1], dp[K-i::-1]):
if prev+price > cur:
dp[j] = prev+price
print(max(dp))
if __name__ == "__main__":
solve()
``` | instruction | 0 | 36,795 | 10 | 73,590 |
No | output | 1 | 36,795 | 10 | 73,591 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a long-established secondhand bookstore called JOI secondhand bookstore in your town, and you often use the JOI secondhand bookstore. Each book has a standard price, and you can buy it at that price if you go to the JOI secondhand bookstore.
At the JOI secondhand bookstore, books are classified into 10 genres such as novels, manga, and magazines. Genres are numbered from 1 to 10. The JOI secondhand bookstore has a service that if you buy books of the same genre in bulk, they will buy them at a high price. Specifically, when books of the same genre are purchased together in T books, the purchase price per book of that genre is T -1 yen higher than the standard price. For example, if books of the same genre with standard prices of 100 yen, 120 yen, and 150 yen are sold together at the JOI secondhand bookstore, the purchase prices will be 102 yen, 122 yen, and 152 yen, respectively.
By the way, you have to move in a hurry due to personal reasons. You have N books, but it is difficult to bring all the books to your new home, so I decided to sell K of the N books to the JOI secondhand bookstore.
input
Read the following input from standard input.
* The integers N and K are written on the first line, separated by blanks, indicating that the number of books you have is N, of which K books will be sold to JOI secondhand bookstores.
* The following N lines contain information about your book. On the first line of i + (1 β€ i β€ N), the integers Ci and Gi are written separated by blanks, indicating that the base price of the i-th book is Ci and the genre number is Gi. ..
output
Output an integer representing the maximum value of the total purchase price to the standard output on one line.
Example
Input
7 4
14 1
13 2
12 3
14 2
8 2
16 3
11 2
Output
60
Submitted Solution:
```
from collections import deque
from itertools import chain, combinations
n, k = map(int, input().split())
books = [[] for _ in range(10)]
while n:
c, g = map(int, input().split())
books[g - 1].append(c)
n -= 1
for q in books:
q.sort(reverse=True)
done = set()
max_cost = 0
for book_combi in combinations(chain.from_iterable([[i] * min(k, len(q)) for i, q in enumerate(books) if q]), k):
if book_combi in done:
continue
done.add(book_combi)
cost = 0
counts = [0] * 10
for i in book_combi:
cost += books[i][counts[i]] + counts[i] * 2
counts[i] += 1
if cost > max_cost:
max_cost = cost
print(max_cost)
``` | instruction | 0 | 36,796 | 10 | 73,592 |
No | output | 1 | 36,796 | 10 | 73,593 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a long-established secondhand bookstore called JOI secondhand bookstore in your town, and you often use the JOI secondhand bookstore. Each book has a standard price, and you can buy it at that price if you go to the JOI secondhand bookstore.
At the JOI secondhand bookstore, books are classified into 10 genres such as novels, manga, and magazines. Genres are numbered from 1 to 10. The JOI secondhand bookstore has a service that if you buy books of the same genre in bulk, they will buy them at a high price. Specifically, when books of the same genre are purchased together in T books, the purchase price per book of that genre is T -1 yen higher than the standard price. For example, if books of the same genre with standard prices of 100 yen, 120 yen, and 150 yen are sold together at the JOI secondhand bookstore, the purchase prices will be 102 yen, 122 yen, and 152 yen, respectively.
By the way, you have to move in a hurry due to personal reasons. You have N books, but it is difficult to bring all the books to your new home, so I decided to sell K of the N books to the JOI secondhand bookstore.
input
Read the following input from standard input.
* The integers N and K are written on the first line, separated by blanks, indicating that the number of books you have is N, of which K books will be sold to JOI secondhand bookstores.
* The following N lines contain information about your book. On the first line of i + (1 β€ i β€ N), the integers Ci and Gi are written separated by blanks, indicating that the base price of the i-th book is Ci and the genre number is Gi. ..
output
Output an integer representing the maximum value of the total purchase price to the standard output on one line.
Example
Input
7 4
14 1
13 2
12 3
14 2
8 2
16 3
11 2
Output
60
Submitted Solution:
```
from itertools import accumulate
n, k = map(int, input().split())
books = [[] for _ in range(10)]
while n:
c, g = map(int, input().split())
books[g - 1].append(c)
n -= 1
for q in books:
q.sort(reverse=True)
books_acc = [[0] + list(accumulate(c + i * 2 for i, c in enumerate(q))) for q in books]
def memoize(f):
memo = [[-1] * (k + 1) for _ in range(10)]
def main(x, y):
result = memo[x][y]
if result < 0:
result = memo[x][y] = f(x, y)
return result
return main
@memoize
def combi(g, remain):
if g > 9:
return 0
salable = min(remain + 1, len(books[g]))
book_acc = list(books_acc[g])
return max([book_acc[i] + combi(g + 1, k - i) for i in range(salable)], default=0)
print(combi(0, k))
``` | instruction | 0 | 36,797 | 10 | 73,594 |
No | output | 1 | 36,797 | 10 | 73,595 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a long-established secondhand bookstore called JOI secondhand bookstore in your town, and you often use the JOI secondhand bookstore. Each book has a standard price, and you can buy it at that price if you go to the JOI secondhand bookstore.
At the JOI secondhand bookstore, books are classified into 10 genres such as novels, manga, and magazines. Genres are numbered from 1 to 10. The JOI secondhand bookstore has a service that if you buy books of the same genre in bulk, they will buy them at a high price. Specifically, when books of the same genre are purchased together in T books, the purchase price per book of that genre is T -1 yen higher than the standard price. For example, if books of the same genre with standard prices of 100 yen, 120 yen, and 150 yen are sold together at the JOI secondhand bookstore, the purchase prices will be 102 yen, 122 yen, and 152 yen, respectively.
By the way, you have to move in a hurry due to personal reasons. You have N books, but it is difficult to bring all the books to your new home, so I decided to sell K of the N books to the JOI secondhand bookstore.
input
Read the following input from standard input.
* The integers N and K are written on the first line, separated by blanks, indicating that the number of books you have is N, of which K books will be sold to JOI secondhand bookstores.
* The following N lines contain information about your book. On the first line of i + (1 β€ i β€ N), the integers Ci and Gi are written separated by blanks, indicating that the base price of the i-th book is Ci and the genre number is Gi. ..
output
Output an integer representing the maximum value of the total purchase price to the standard output on one line.
Example
Input
7 4
14 1
13 2
12 3
14 2
8 2
16 3
11 2
Output
60
Submitted Solution:
```
import sys
def solve():
file_input = sys.stdin
N, K = map(int, file_input.readline().split())
G = [[] for i in range(10)]
for line in file_input:
c, g = map(int, line.split())
G[g - 1].append(c)
for genre in G:
genre.sort(reverse=True)
for p, i in zip(genre, range(1, len(genre))):
genre[i] += p + 2 * i # Recording cumulative price
C = [0] * (K + 1)
for genre in G:
for n, p in enumerate(genre, start=1):
for i, vals in enumerate(zip(C[n:], C)):
v1, v2 = vals
v1 += p
if v1 > v2:
C[i] = v1
print(C[0])
solve()
``` | instruction | 0 | 36,798 | 10 | 73,596 |
No | output | 1 | 36,798 | 10 | 73,597 |
Provide a correct Python 3 solution for this coding contest problem.
You want to make change for $ n $ cents. Assuming that you have infinite supply of coins of 1, 5, 10 and / or 25 cents coins respectively, find the minimum number of coins you need.
Constraints
* $ 1 \ le n \ le 10 ^ 9 $
Input
$ n $
The integer $ n $ is given in a line.
output
Print the minimum number of coins you need in a line.
Examples
Input
100
Output
4
Input
54321
Output
2175 | instruction | 0 | 36,818 | 10 | 73,636 |
"Correct Solution:
```
N = int(input())
ans = N // 25; N %= 25
ans += N // 10; N %= 10
ans += N // 5; N %= 5
ans += N
print(ans)
``` | output | 1 | 36,818 | 10 | 73,637 |
Provide a correct Python 3 solution for this coding contest problem.
You want to make change for $ n $ cents. Assuming that you have infinite supply of coins of 1, 5, 10 and / or 25 cents coins respectively, find the minimum number of coins you need.
Constraints
* $ 1 \ le n \ le 10 ^ 9 $
Input
$ n $
The integer $ n $ is given in a line.
output
Print the minimum number of coins you need in a line.
Examples
Input
100
Output
4
Input
54321
Output
2175 | instruction | 0 | 36,819 | 10 | 73,638 |
"Correct Solution:
```
def num_coins(cents):
coins = [25, 10, 5, 1]
count = 0
for coin in coins:
while cents >= coin:
cents = cents - coin
count = count + 1
return count
print(num_coins(int(input())))
``` | output | 1 | 36,819 | 10 | 73,639 |
Provide a correct Python 3 solution for this coding contest problem.
You want to make change for $ n $ cents. Assuming that you have infinite supply of coins of 1, 5, 10 and / or 25 cents coins respectively, find the minimum number of coins you need.
Constraints
* $ 1 \ le n \ le 10 ^ 9 $
Input
$ n $
The integer $ n $ is given in a line.
output
Print the minimum number of coins you need in a line.
Examples
Input
100
Output
4
Input
54321
Output
2175 | instruction | 0 | 36,820 | 10 | 73,640 |
"Correct Solution:
```
n = int(input())
a1 = n//25
a2 = (n - a1*25)//10
a3 = (n - a1*25 - a2*10)//5
a4 = n - a1*25 - a2*10 - a3*5
print(a1+a2+a3+a4)
``` | output | 1 | 36,820 | 10 | 73,641 |
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