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.
Ayush is a cashier at the shopping center. Recently his department has started a ''click and collect" service which allows users to shop online.
The store contains k items. n customers have already used the above service. Each user paid for m items. Let aij denote the j-th item in the i-th person's order.
Due to the space limitations all the items are arranged in one single row. When Ayush receives the i-th order he will find one by one all the items aij (1 β€ j β€ m) in the row. Let pos(x) denote the position of the item x in the row at the moment of its collection. Then Ayush takes time equal to pos(ai1) + pos(ai2) + ... + pos(aim) for the i-th customer.
When Ayush accesses the x-th element he keeps a new stock in the front of the row and takes away the x-th element. Thus the values are updating.
Your task is to calculate the total time it takes for Ayush to process all the orders.
You can assume that the market has endless stock.
Input
The first line contains three integers n, m and k (1 β€ n, k β€ 100, 1 β€ m β€ k) β the number of users, the number of items each user wants to buy and the total number of items at the market.
The next line contains k distinct integers pl (1 β€ pl β€ k) denoting the initial positions of the items in the store. The items are numbered with integers from 1 to k.
Each of the next n lines contains m distinct integers aij (1 β€ aij β€ k) β the order of the i-th person.
Output
Print the only integer t β the total time needed for Ayush to process all the orders.
Example
Input
2 2 5
3 4 1 2 5
1 5
3 1
Output
14
Note
Customer 1 wants the items 1 and 5.
pos(1) = 3, so the new positions are: [1, 3, 4, 2, 5].
pos(5) = 5, so the new positions are: [5, 1, 3, 4, 2].
Time taken for the first customer is 3 + 5 = 8.
Customer 2 wants the items 3 and 1.
pos(3) = 3, so the new positions are: [3, 5, 1, 4, 2].
pos(1) = 3, so the new positions are: [1, 3, 5, 4, 2].
Time taken for the second customer is 3 + 3 = 6.
Total time is 8 + 6 = 14.
Formally pos(x) is the index of x in the current row. | instruction | 0 | 49,290 | 10 | 98,580 |
Tags: brute force
Correct Solution:
```
n, m, k = map(int, input().split())
queue = list(map(int, input().split()))
mass = []
for i in range(n):
mass += list(map(int, input().split()))
count = 0
for elem in mass:
c = queue.index(elem)
count += c + 1
queue.pop(c)
queue.insert(0, elem)
print(count)
``` | output | 1 | 49,290 | 10 | 98,581 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ayush is a cashier at the shopping center. Recently his department has started a ''click and collect" service which allows users to shop online.
The store contains k items. n customers have already used the above service. Each user paid for m items. Let aij denote the j-th item in the i-th person's order.
Due to the space limitations all the items are arranged in one single row. When Ayush receives the i-th order he will find one by one all the items aij (1 β€ j β€ m) in the row. Let pos(x) denote the position of the item x in the row at the moment of its collection. Then Ayush takes time equal to pos(ai1) + pos(ai2) + ... + pos(aim) for the i-th customer.
When Ayush accesses the x-th element he keeps a new stock in the front of the row and takes away the x-th element. Thus the values are updating.
Your task is to calculate the total time it takes for Ayush to process all the orders.
You can assume that the market has endless stock.
Input
The first line contains three integers n, m and k (1 β€ n, k β€ 100, 1 β€ m β€ k) β the number of users, the number of items each user wants to buy and the total number of items at the market.
The next line contains k distinct integers pl (1 β€ pl β€ k) denoting the initial positions of the items in the store. The items are numbered with integers from 1 to k.
Each of the next n lines contains m distinct integers aij (1 β€ aij β€ k) β the order of the i-th person.
Output
Print the only integer t β the total time needed for Ayush to process all the orders.
Example
Input
2 2 5
3 4 1 2 5
1 5
3 1
Output
14
Note
Customer 1 wants the items 1 and 5.
pos(1) = 3, so the new positions are: [1, 3, 4, 2, 5].
pos(5) = 5, so the new positions are: [5, 1, 3, 4, 2].
Time taken for the first customer is 3 + 5 = 8.
Customer 2 wants the items 3 and 1.
pos(3) = 3, so the new positions are: [3, 5, 1, 4, 2].
pos(1) = 3, so the new positions are: [1, 3, 5, 4, 2].
Time taken for the second customer is 3 + 3 = 6.
Total time is 8 + 6 = 14.
Formally pos(x) is the index of x in the current row. | instruction | 0 | 49,291 | 10 | 98,582 |
Tags: brute force
Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Wed Apr 20 21:02:31 2016
@author: Kirill
"""
n, m, k = map(int, input().split())
order = list(map(int, input().split()))
items = []
for i in range(n):
items.append(list(map(int, input().split())))
ans = 0
for i in range(n):
for j in range(m):
item = order.index(items[i][j])
ans += item + 1
order.insert(0, order.pop(item))
print(ans)
``` | output | 1 | 49,291 | 10 | 98,583 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ayush is a cashier at the shopping center. Recently his department has started a ''click and collect" service which allows users to shop online.
The store contains k items. n customers have already used the above service. Each user paid for m items. Let aij denote the j-th item in the i-th person's order.
Due to the space limitations all the items are arranged in one single row. When Ayush receives the i-th order he will find one by one all the items aij (1 β€ j β€ m) in the row. Let pos(x) denote the position of the item x in the row at the moment of its collection. Then Ayush takes time equal to pos(ai1) + pos(ai2) + ... + pos(aim) for the i-th customer.
When Ayush accesses the x-th element he keeps a new stock in the front of the row and takes away the x-th element. Thus the values are updating.
Your task is to calculate the total time it takes for Ayush to process all the orders.
You can assume that the market has endless stock.
Input
The first line contains three integers n, m and k (1 β€ n, k β€ 100, 1 β€ m β€ k) β the number of users, the number of items each user wants to buy and the total number of items at the market.
The next line contains k distinct integers pl (1 β€ pl β€ k) denoting the initial positions of the items in the store. The items are numbered with integers from 1 to k.
Each of the next n lines contains m distinct integers aij (1 β€ aij β€ k) β the order of the i-th person.
Output
Print the only integer t β the total time needed for Ayush to process all the orders.
Example
Input
2 2 5
3 4 1 2 5
1 5
3 1
Output
14
Note
Customer 1 wants the items 1 and 5.
pos(1) = 3, so the new positions are: [1, 3, 4, 2, 5].
pos(5) = 5, so the new positions are: [5, 1, 3, 4, 2].
Time taken for the first customer is 3 + 5 = 8.
Customer 2 wants the items 3 and 1.
pos(3) = 3, so the new positions are: [3, 5, 1, 4, 2].
pos(1) = 3, so the new positions are: [1, 3, 5, 4, 2].
Time taken for the second customer is 3 + 3 = 6.
Total time is 8 + 6 = 14.
Formally pos(x) is the index of x in the current row.
Submitted Solution:
```
n, m, k = map(int, input().split())
a = list(map(int, input().split()))
ans = 0
for i in range(n):
orders = map(int, input().split())
for order in orders:
ans += a.index(order) + 1
a.remove(order)
a.insert(0, order)
print(ans)
``` | instruction | 0 | 49,292 | 10 | 98,584 |
Yes | output | 1 | 49,292 | 10 | 98,585 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ayush is a cashier at the shopping center. Recently his department has started a ''click and collect" service which allows users to shop online.
The store contains k items. n customers have already used the above service. Each user paid for m items. Let aij denote the j-th item in the i-th person's order.
Due to the space limitations all the items are arranged in one single row. When Ayush receives the i-th order he will find one by one all the items aij (1 β€ j β€ m) in the row. Let pos(x) denote the position of the item x in the row at the moment of its collection. Then Ayush takes time equal to pos(ai1) + pos(ai2) + ... + pos(aim) for the i-th customer.
When Ayush accesses the x-th element he keeps a new stock in the front of the row and takes away the x-th element. Thus the values are updating.
Your task is to calculate the total time it takes for Ayush to process all the orders.
You can assume that the market has endless stock.
Input
The first line contains three integers n, m and k (1 β€ n, k β€ 100, 1 β€ m β€ k) β the number of users, the number of items each user wants to buy and the total number of items at the market.
The next line contains k distinct integers pl (1 β€ pl β€ k) denoting the initial positions of the items in the store. The items are numbered with integers from 1 to k.
Each of the next n lines contains m distinct integers aij (1 β€ aij β€ k) β the order of the i-th person.
Output
Print the only integer t β the total time needed for Ayush to process all the orders.
Example
Input
2 2 5
3 4 1 2 5
1 5
3 1
Output
14
Note
Customer 1 wants the items 1 and 5.
pos(1) = 3, so the new positions are: [1, 3, 4, 2, 5].
pos(5) = 5, so the new positions are: [5, 1, 3, 4, 2].
Time taken for the first customer is 3 + 5 = 8.
Customer 2 wants the items 3 and 1.
pos(3) = 3, so the new positions are: [3, 5, 1, 4, 2].
pos(1) = 3, so the new positions are: [1, 3, 5, 4, 2].
Time taken for the second customer is 3 + 3 = 6.
Total time is 8 + 6 = 14.
Formally pos(x) is the index of x in the current row.
Submitted Solution:
```
n, m, k = map(int, input().split())
l = list(map(int, input().split()))
ans = 0
for i in range(n):
t = list(map(int, input().split()))
for j in range(m):
ans += l.index(t[j]) + 1
l.remove(t[j])
l.insert(0, t[j])
print (ans)
``` | instruction | 0 | 49,293 | 10 | 98,586 |
Yes | output | 1 | 49,293 | 10 | 98,587 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ayush is a cashier at the shopping center. Recently his department has started a ''click and collect" service which allows users to shop online.
The store contains k items. n customers have already used the above service. Each user paid for m items. Let aij denote the j-th item in the i-th person's order.
Due to the space limitations all the items are arranged in one single row. When Ayush receives the i-th order he will find one by one all the items aij (1 β€ j β€ m) in the row. Let pos(x) denote the position of the item x in the row at the moment of its collection. Then Ayush takes time equal to pos(ai1) + pos(ai2) + ... + pos(aim) for the i-th customer.
When Ayush accesses the x-th element he keeps a new stock in the front of the row and takes away the x-th element. Thus the values are updating.
Your task is to calculate the total time it takes for Ayush to process all the orders.
You can assume that the market has endless stock.
Input
The first line contains three integers n, m and k (1 β€ n, k β€ 100, 1 β€ m β€ k) β the number of users, the number of items each user wants to buy and the total number of items at the market.
The next line contains k distinct integers pl (1 β€ pl β€ k) denoting the initial positions of the items in the store. The items are numbered with integers from 1 to k.
Each of the next n lines contains m distinct integers aij (1 β€ aij β€ k) β the order of the i-th person.
Output
Print the only integer t β the total time needed for Ayush to process all the orders.
Example
Input
2 2 5
3 4 1 2 5
1 5
3 1
Output
14
Note
Customer 1 wants the items 1 and 5.
pos(1) = 3, so the new positions are: [1, 3, 4, 2, 5].
pos(5) = 5, so the new positions are: [5, 1, 3, 4, 2].
Time taken for the first customer is 3 + 5 = 8.
Customer 2 wants the items 3 and 1.
pos(3) = 3, so the new positions are: [3, 5, 1, 4, 2].
pos(1) = 3, so the new positions are: [1, 3, 5, 4, 2].
Time taken for the second customer is 3 + 3 = 6.
Total time is 8 + 6 = 14.
Formally pos(x) is the index of x in the current row.
Submitted Solution:
```
n, m, k = map(int, input().split())
A = list(map(int, input().split()))
time = 0
for a in range(n):
B = list(map(int, input().split()))
for i in B:
K = [i]
time += (A.index(i) + 1)
for a in A:
if a != i:
K.append(a)
A = K
print(time)
``` | instruction | 0 | 49,294 | 10 | 98,588 |
Yes | output | 1 | 49,294 | 10 | 98,589 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ayush is a cashier at the shopping center. Recently his department has started a ''click and collect" service which allows users to shop online.
The store contains k items. n customers have already used the above service. Each user paid for m items. Let aij denote the j-th item in the i-th person's order.
Due to the space limitations all the items are arranged in one single row. When Ayush receives the i-th order he will find one by one all the items aij (1 β€ j β€ m) in the row. Let pos(x) denote the position of the item x in the row at the moment of its collection. Then Ayush takes time equal to pos(ai1) + pos(ai2) + ... + pos(aim) for the i-th customer.
When Ayush accesses the x-th element he keeps a new stock in the front of the row and takes away the x-th element. Thus the values are updating.
Your task is to calculate the total time it takes for Ayush to process all the orders.
You can assume that the market has endless stock.
Input
The first line contains three integers n, m and k (1 β€ n, k β€ 100, 1 β€ m β€ k) β the number of users, the number of items each user wants to buy and the total number of items at the market.
The next line contains k distinct integers pl (1 β€ pl β€ k) denoting the initial positions of the items in the store. The items are numbered with integers from 1 to k.
Each of the next n lines contains m distinct integers aij (1 β€ aij β€ k) β the order of the i-th person.
Output
Print the only integer t β the total time needed for Ayush to process all the orders.
Example
Input
2 2 5
3 4 1 2 5
1 5
3 1
Output
14
Note
Customer 1 wants the items 1 and 5.
pos(1) = 3, so the new positions are: [1, 3, 4, 2, 5].
pos(5) = 5, so the new positions are: [5, 1, 3, 4, 2].
Time taken for the first customer is 3 + 5 = 8.
Customer 2 wants the items 3 and 1.
pos(3) = 3, so the new positions are: [3, 5, 1, 4, 2].
pos(1) = 3, so the new positions are: [1, 3, 5, 4, 2].
Time taken for the second customer is 3 + 3 = 6.
Total time is 8 + 6 = 14.
Formally pos(x) is the index of x in the current row.
Submitted Solution:
```
import io, sys, atexit, os
import math as ma
from sys import exit
from decimal import Decimal as dec
from itertools import permutations
from itertools import combinations
def li ():
return list (map (int, input ().split ()))
def num ():
return map (int, input ().split ())
def nu ():
return int (input ())
def find_gcd ( x, y ):
while (y):
x, y = y, x % y
return x
def lcm ( x, y ):
gg = find_gcd (x, y)
return (x * y // gg)
def printDivisors ( n ):
# Note that this loop runs till square root
i = 1
cc=0
while i <= ma.sqrt (n):
if (n % i == 0):
# If divisors are equal, print only one
if (n / i == i):
cc+=1
else:
# Otherwise print both
cc+=2
i = i + 1
return cc
mm = 1000000007
yp = 0
def solve ():
t = 1
for tt in range (t):
n,m,k=num()
a=li()
ff=a
ss=0
for i in range(n):
b=li()
for j in range(m):
for ll in range(k):
if(b[j]==ff[ll]):
ss+=ll+1
ff=[b[j]]+ff[0:ll]+ff[ll+1:]
break
print(ss)
if __name__ == "__main__":
solve ()
``` | instruction | 0 | 49,295 | 10 | 98,590 |
Yes | output | 1 | 49,295 | 10 | 98,591 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ayush is a cashier at the shopping center. Recently his department has started a ''click and collect" service which allows users to shop online.
The store contains k items. n customers have already used the above service. Each user paid for m items. Let aij denote the j-th item in the i-th person's order.
Due to the space limitations all the items are arranged in one single row. When Ayush receives the i-th order he will find one by one all the items aij (1 β€ j β€ m) in the row. Let pos(x) denote the position of the item x in the row at the moment of its collection. Then Ayush takes time equal to pos(ai1) + pos(ai2) + ... + pos(aim) for the i-th customer.
When Ayush accesses the x-th element he keeps a new stock in the front of the row and takes away the x-th element. Thus the values are updating.
Your task is to calculate the total time it takes for Ayush to process all the orders.
You can assume that the market has endless stock.
Input
The first line contains three integers n, m and k (1 β€ n, k β€ 100, 1 β€ m β€ k) β the number of users, the number of items each user wants to buy and the total number of items at the market.
The next line contains k distinct integers pl (1 β€ pl β€ k) denoting the initial positions of the items in the store. The items are numbered with integers from 1 to k.
Each of the next n lines contains m distinct integers aij (1 β€ aij β€ k) β the order of the i-th person.
Output
Print the only integer t β the total time needed for Ayush to process all the orders.
Example
Input
2 2 5
3 4 1 2 5
1 5
3 1
Output
14
Note
Customer 1 wants the items 1 and 5.
pos(1) = 3, so the new positions are: [1, 3, 4, 2, 5].
pos(5) = 5, so the new positions are: [5, 1, 3, 4, 2].
Time taken for the first customer is 3 + 5 = 8.
Customer 2 wants the items 3 and 1.
pos(3) = 3, so the new positions are: [3, 5, 1, 4, 2].
pos(1) = 3, so the new positions are: [1, 3, 5, 4, 2].
Time taken for the second customer is 3 + 3 = 6.
Total time is 8 + 6 = 14.
Formally pos(x) is the index of x in the current row.
Submitted Solution:
```
R = lambda : map(int, input().split())
n, m, k = R()
val = list(R())
res = 0
for i in range(n):
pos = list(R())
for p in pos:
res += val[p - 1]
val = [p] + [_ for _ in val if _ != p]
#print(*val)
print(res)
``` | instruction | 0 | 49,296 | 10 | 98,592 |
No | output | 1 | 49,296 | 10 | 98,593 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ayush is a cashier at the shopping center. Recently his department has started a ''click and collect" service which allows users to shop online.
The store contains k items. n customers have already used the above service. Each user paid for m items. Let aij denote the j-th item in the i-th person's order.
Due to the space limitations all the items are arranged in one single row. When Ayush receives the i-th order he will find one by one all the items aij (1 β€ j β€ m) in the row. Let pos(x) denote the position of the item x in the row at the moment of its collection. Then Ayush takes time equal to pos(ai1) + pos(ai2) + ... + pos(aim) for the i-th customer.
When Ayush accesses the x-th element he keeps a new stock in the front of the row and takes away the x-th element. Thus the values are updating.
Your task is to calculate the total time it takes for Ayush to process all the orders.
You can assume that the market has endless stock.
Input
The first line contains three integers n, m and k (1 β€ n, k β€ 100, 1 β€ m β€ k) β the number of users, the number of items each user wants to buy and the total number of items at the market.
The next line contains k distinct integers pl (1 β€ pl β€ k) denoting the initial positions of the items in the store. The items are numbered with integers from 1 to k.
Each of the next n lines contains m distinct integers aij (1 β€ aij β€ k) β the order of the i-th person.
Output
Print the only integer t β the total time needed for Ayush to process all the orders.
Example
Input
2 2 5
3 4 1 2 5
1 5
3 1
Output
14
Note
Customer 1 wants the items 1 and 5.
pos(1) = 3, so the new positions are: [1, 3, 4, 2, 5].
pos(5) = 5, so the new positions are: [5, 1, 3, 4, 2].
Time taken for the first customer is 3 + 5 = 8.
Customer 2 wants the items 3 and 1.
pos(3) = 3, so the new positions are: [3, 5, 1, 4, 2].
pos(1) = 3, so the new positions are: [1, 3, 5, 4, 2].
Time taken for the second customer is 3 + 3 = 6.
Total time is 8 + 6 = 14.
Formally pos(x) is the index of x in the current row.
Submitted Solution:
```
#!/usr/bin/python3
n, m, k = map(int, input().split())
positionOf = [x-1 for x in map(int, input().split())]
sol = 0
for i in range(n):
v = [x-1 for x in map(int, input().split())]
for x in v:
sol += positionOf[x] + 1
for y in range(k):
if positionOf[y] < positionOf[x]:
positionOf[y] += 1
positionOf[x] = 0
print(sol)
``` | instruction | 0 | 49,297 | 10 | 98,594 |
No | output | 1 | 49,297 | 10 | 98,595 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ayush is a cashier at the shopping center. Recently his department has started a ''click and collect" service which allows users to shop online.
The store contains k items. n customers have already used the above service. Each user paid for m items. Let aij denote the j-th item in the i-th person's order.
Due to the space limitations all the items are arranged in one single row. When Ayush receives the i-th order he will find one by one all the items aij (1 β€ j β€ m) in the row. Let pos(x) denote the position of the item x in the row at the moment of its collection. Then Ayush takes time equal to pos(ai1) + pos(ai2) + ... + pos(aim) for the i-th customer.
When Ayush accesses the x-th element he keeps a new stock in the front of the row and takes away the x-th element. Thus the values are updating.
Your task is to calculate the total time it takes for Ayush to process all the orders.
You can assume that the market has endless stock.
Input
The first line contains three integers n, m and k (1 β€ n, k β€ 100, 1 β€ m β€ k) β the number of users, the number of items each user wants to buy and the total number of items at the market.
The next line contains k distinct integers pl (1 β€ pl β€ k) denoting the initial positions of the items in the store. The items are numbered with integers from 1 to k.
Each of the next n lines contains m distinct integers aij (1 β€ aij β€ k) β the order of the i-th person.
Output
Print the only integer t β the total time needed for Ayush to process all the orders.
Example
Input
2 2 5
3 4 1 2 5
1 5
3 1
Output
14
Note
Customer 1 wants the items 1 and 5.
pos(1) = 3, so the new positions are: [1, 3, 4, 2, 5].
pos(5) = 5, so the new positions are: [5, 1, 3, 4, 2].
Time taken for the first customer is 3 + 5 = 8.
Customer 2 wants the items 3 and 1.
pos(3) = 3, so the new positions are: [3, 5, 1, 4, 2].
pos(1) = 3, so the new positions are: [1, 3, 5, 4, 2].
Time taken for the second customer is 3 + 3 = 6.
Total time is 8 + 6 = 14.
Formally pos(x) is the index of x in the current row.
Submitted Solution:
```
import io, sys, atexit, os
import math as ma
from sys import exit
from decimal import Decimal as dec
from itertools import permutations
from itertools import combinations
def li ():
return list (map (int, input ().split ()))
def num ():
return map (int, input ().split ())
def nu ():
return int (input ())
def find_gcd ( x, y ):
while (y):
x, y = y, x % y
return x
def lcm ( x, y ):
gg = find_gcd (x, y)
return (x * y // gg)
def printDivisors ( n ):
# Note that this loop runs till square root
i = 1
cc=0
while i <= ma.sqrt (n):
if (n % i == 0):
# If divisors are equal, print only one
if (n / i == i):
cc+=1
else:
# Otherwise print both
cc+=2
i = i + 1
return cc
mm = 1000000007
yp = 0
def solve ():
t = 1
for tt in range (t):
n,m,k=num()
a=li()
ff=[0]*k
for i in range(k):
ff[a[i]-1]=i
ss=0
for i in range(n):
b=li()
for j in range(m):
for ll in range(k):
if(b[j]-1==ff[ll]):
ss+=ll+1
ff=[b[j]-1]+ff[0:ll]+ff[ll+1:]
break
print(ss)
if __name__ == "__main__":
solve ()
``` | instruction | 0 | 49,298 | 10 | 98,596 |
No | output | 1 | 49,298 | 10 | 98,597 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ayush is a cashier at the shopping center. Recently his department has started a ''click and collect" service which allows users to shop online.
The store contains k items. n customers have already used the above service. Each user paid for m items. Let aij denote the j-th item in the i-th person's order.
Due to the space limitations all the items are arranged in one single row. When Ayush receives the i-th order he will find one by one all the items aij (1 β€ j β€ m) in the row. Let pos(x) denote the position of the item x in the row at the moment of its collection. Then Ayush takes time equal to pos(ai1) + pos(ai2) + ... + pos(aim) for the i-th customer.
When Ayush accesses the x-th element he keeps a new stock in the front of the row and takes away the x-th element. Thus the values are updating.
Your task is to calculate the total time it takes for Ayush to process all the orders.
You can assume that the market has endless stock.
Input
The first line contains three integers n, m and k (1 β€ n, k β€ 100, 1 β€ m β€ k) β the number of users, the number of items each user wants to buy and the total number of items at the market.
The next line contains k distinct integers pl (1 β€ pl β€ k) denoting the initial positions of the items in the store. The items are numbered with integers from 1 to k.
Each of the next n lines contains m distinct integers aij (1 β€ aij β€ k) β the order of the i-th person.
Output
Print the only integer t β the total time needed for Ayush to process all the orders.
Example
Input
2 2 5
3 4 1 2 5
1 5
3 1
Output
14
Note
Customer 1 wants the items 1 and 5.
pos(1) = 3, so the new positions are: [1, 3, 4, 2, 5].
pos(5) = 5, so the new positions are: [5, 1, 3, 4, 2].
Time taken for the first customer is 3 + 5 = 8.
Customer 2 wants the items 3 and 1.
pos(3) = 3, so the new positions are: [3, 5, 1, 4, 2].
pos(1) = 3, so the new positions are: [1, 3, 5, 4, 2].
Time taken for the second customer is 3 + 3 = 6.
Total time is 8 + 6 = 14.
Formally pos(x) is the index of x in the current row.
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Wed Apr 20 21:02:31 2016
@author: Kirill
"""
n, m, k = map(int, input('input n m k:\n').split())
order = list(map(int, input().split()))
items = []
for i in range(n):
items.append(list(map(int, input().split())))
ans = 0
for i in range(n):
for j in range(m):
ans += order.index(items[i][j]) + 1
order.insert(0, order.pop(items[i][j] - 1))
print(ans)
``` | instruction | 0 | 49,299 | 10 | 98,598 |
No | output | 1 | 49,299 | 10 | 98,599 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The good times at Heidi's library are over. Marmots finally got their internet connections and stopped coming to the library altogether. Not only that, but the bookstore has begun charging extortionate prices for some books. Namely, whereas in the previous versions each book could be bought for 1 CHF, now the price of book i is ci CHF.
Input
The first line of input will contain two integers n and k (<image>). The second line will contain n integers a1, a2, ..., an (1 β€ ai β€ n) β the sequence of book requests. The third line contains n integers c1, c2, ..., cn (0 β€ ci β€ 106) β the costs of the books.
Output
On a single line print the minimum cost of buying books at the store so as to satisfy all requests.
Examples
Input
4 80
1 2 2 1
1 1 1 1
Output
2
Input
4 1
1 2 2 1
1 1 1 1
Output
3
Input
4 2
1 2 3 1
1 1 1 1
Output
3
Input
7 2
1 2 3 1 1 1 2
1 10 1 0 0 0 0
Output
13
Note
The first three sample cases are repeated, but the fourth one is new.
In the fourth test case, when buying book 3, Heidi should discard either book 1 or 2. Even though book 2 will be requested later than book 1, she should keep it, because it is so expensive to buy again.
Submitted Solution:
```
print(2)
``` | instruction | 0 | 49,351 | 10 | 98,702 |
No | output | 1 | 49,351 | 10 | 98,703 |
Provide a correct Python 3 solution for this coding contest problem.
Find the price of a product before tax such that, when the consumption tax rate is 8 percent and 10 percent, the amount of consumption tax levied on it is A yen and B yen, respectively. (Yen is the currency of Japan.)
Here, the price before tax must be a positive integer, and the amount of consumption tax is rounded down to the nearest integer.
If multiple prices satisfy the condition, print the lowest such price; if no price satisfies the condition, print `-1`.
Constraints
* 1 \leq A \leq B \leq 100
* A and B are integers.
Input
Input is given from Standard Input in the following format:
A B
Output
If there is a price that satisfies the condition, print an integer representing the lowest such price; otherwise, print `-1`.
Examples
Input
2 2
Output
25
Input
8 10
Output
100
Input
19 99
Output
-1 | instruction | 0 | 49,413 | 10 | 98,826 |
"Correct Solution:
```
a, b = map(int, input().split())
for i in range(1, 1010):
if int(i * 0.08) == a and int(i * 0.1) == b: print(i); exit()
print(-1)
``` | output | 1 | 49,413 | 10 | 98,827 |
Provide a correct Python 3 solution for this coding contest problem.
Find the price of a product before tax such that, when the consumption tax rate is 8 percent and 10 percent, the amount of consumption tax levied on it is A yen and B yen, respectively. (Yen is the currency of Japan.)
Here, the price before tax must be a positive integer, and the amount of consumption tax is rounded down to the nearest integer.
If multiple prices satisfy the condition, print the lowest such price; if no price satisfies the condition, print `-1`.
Constraints
* 1 \leq A \leq B \leq 100
* A and B are integers.
Input
Input is given from Standard Input in the following format:
A B
Output
If there is a price that satisfies the condition, print an integer representing the lowest such price; otherwise, print `-1`.
Examples
Input
2 2
Output
25
Input
8 10
Output
100
Input
19 99
Output
-1 | instruction | 0 | 49,414 | 10 | 98,828 |
"Correct Solution:
```
A,B=map(int,input().split())
for i in range(10000):
if (i*0.08)//1==A and (i*0.1)//1==B:
print(i)
exit()
print(-1)
``` | output | 1 | 49,414 | 10 | 98,829 |
Provide a correct Python 3 solution for this coding contest problem.
Find the price of a product before tax such that, when the consumption tax rate is 8 percent and 10 percent, the amount of consumption tax levied on it is A yen and B yen, respectively. (Yen is the currency of Japan.)
Here, the price before tax must be a positive integer, and the amount of consumption tax is rounded down to the nearest integer.
If multiple prices satisfy the condition, print the lowest such price; if no price satisfies the condition, print `-1`.
Constraints
* 1 \leq A \leq B \leq 100
* A and B are integers.
Input
Input is given from Standard Input in the following format:
A B
Output
If there is a price that satisfies the condition, print an integer representing the lowest such price; otherwise, print `-1`.
Examples
Input
2 2
Output
25
Input
8 10
Output
100
Input
19 99
Output
-1 | instruction | 0 | 49,415 | 10 | 98,830 |
"Correct Solution:
```
a, b = map(int, input().split())
for i in range(1001):
if int(i * 0.08) == a and int(i * 0.1) == b:
print(i)
break
else:
print(-1)
``` | output | 1 | 49,415 | 10 | 98,831 |
Provide a correct Python 3 solution for this coding contest problem.
Find the price of a product before tax such that, when the consumption tax rate is 8 percent and 10 percent, the amount of consumption tax levied on it is A yen and B yen, respectively. (Yen is the currency of Japan.)
Here, the price before tax must be a positive integer, and the amount of consumption tax is rounded down to the nearest integer.
If multiple prices satisfy the condition, print the lowest such price; if no price satisfies the condition, print `-1`.
Constraints
* 1 \leq A \leq B \leq 100
* A and B are integers.
Input
Input is given from Standard Input in the following format:
A B
Output
If there is a price that satisfies the condition, print an integer representing the lowest such price; otherwise, print `-1`.
Examples
Input
2 2
Output
25
Input
8 10
Output
100
Input
19 99
Output
-1 | instruction | 0 | 49,416 | 10 | 98,832 |
"Correct Solution:
```
A, B =map(int, input().split())
r = -1
for i in range(1001):
if i *8//100==A and i *10//100==B:
r = i
break
print(r)
``` | output | 1 | 49,416 | 10 | 98,833 |
Provide a correct Python 3 solution for this coding contest problem.
Find the price of a product before tax such that, when the consumption tax rate is 8 percent and 10 percent, the amount of consumption tax levied on it is A yen and B yen, respectively. (Yen is the currency of Japan.)
Here, the price before tax must be a positive integer, and the amount of consumption tax is rounded down to the nearest integer.
If multiple prices satisfy the condition, print the lowest such price; if no price satisfies the condition, print `-1`.
Constraints
* 1 \leq A \leq B \leq 100
* A and B are integers.
Input
Input is given from Standard Input in the following format:
A B
Output
If there is a price that satisfies the condition, print an integer representing the lowest such price; otherwise, print `-1`.
Examples
Input
2 2
Output
25
Input
8 10
Output
100
Input
19 99
Output
-1 | instruction | 0 | 49,417 | 10 | 98,834 |
"Correct Solution:
```
A,B=map(int,input().split())
ans=-1
for i in range(1500):
if int(i*0.08)==A and int(i*0.1)==B:
ans=i
break
print(ans)
``` | output | 1 | 49,417 | 10 | 98,835 |
Provide a correct Python 3 solution for this coding contest problem.
Find the price of a product before tax such that, when the consumption tax rate is 8 percent and 10 percent, the amount of consumption tax levied on it is A yen and B yen, respectively. (Yen is the currency of Japan.)
Here, the price before tax must be a positive integer, and the amount of consumption tax is rounded down to the nearest integer.
If multiple prices satisfy the condition, print the lowest such price; if no price satisfies the condition, print `-1`.
Constraints
* 1 \leq A \leq B \leq 100
* A and B are integers.
Input
Input is given from Standard Input in the following format:
A B
Output
If there is a price that satisfies the condition, print an integer representing the lowest such price; otherwise, print `-1`.
Examples
Input
2 2
Output
25
Input
8 10
Output
100
Input
19 99
Output
-1 | instruction | 0 | 49,418 | 10 | 98,836 |
"Correct Solution:
```
a,b=map(int,input().split())
for x in range(10000):
if int(x*0.08)==a and int(x*0.1)==b:
print(x)
exit()
print(-1)
``` | output | 1 | 49,418 | 10 | 98,837 |
Provide a correct Python 3 solution for this coding contest problem.
Find the price of a product before tax such that, when the consumption tax rate is 8 percent and 10 percent, the amount of consumption tax levied on it is A yen and B yen, respectively. (Yen is the currency of Japan.)
Here, the price before tax must be a positive integer, and the amount of consumption tax is rounded down to the nearest integer.
If multiple prices satisfy the condition, print the lowest such price; if no price satisfies the condition, print `-1`.
Constraints
* 1 \leq A \leq B \leq 100
* A and B are integers.
Input
Input is given from Standard Input in the following format:
A B
Output
If there is a price that satisfies the condition, print an integer representing the lowest such price; otherwise, print `-1`.
Examples
Input
2 2
Output
25
Input
8 10
Output
100
Input
19 99
Output
-1 | instruction | 0 | 49,419 | 10 | 98,838 |
"Correct Solution:
```
a,b=map(int,input().split());print(([i for i in range(10**6)if a==i//12.5and b==i//10]+[-1])[0])
``` | output | 1 | 49,419 | 10 | 98,839 |
Provide a correct Python 3 solution for this coding contest problem.
Find the price of a product before tax such that, when the consumption tax rate is 8 percent and 10 percent, the amount of consumption tax levied on it is A yen and B yen, respectively. (Yen is the currency of Japan.)
Here, the price before tax must be a positive integer, and the amount of consumption tax is rounded down to the nearest integer.
If multiple prices satisfy the condition, print the lowest such price; if no price satisfies the condition, print `-1`.
Constraints
* 1 \leq A \leq B \leq 100
* A and B are integers.
Input
Input is given from Standard Input in the following format:
A B
Output
If there is a price that satisfies the condition, print an integer representing the lowest such price; otherwise, print `-1`.
Examples
Input
2 2
Output
25
Input
8 10
Output
100
Input
19 99
Output
-1 | instruction | 0 | 49,420 | 10 | 98,840 |
"Correct Solution:
```
A,B=map(int,input().split())
for i in range(10001):
if (i*8)//100==A and (i*10)//100==B:
print(i)
exit()
print(-1)
``` | output | 1 | 49,420 | 10 | 98,841 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the price of a product before tax such that, when the consumption tax rate is 8 percent and 10 percent, the amount of consumption tax levied on it is A yen and B yen, respectively. (Yen is the currency of Japan.)
Here, the price before tax must be a positive integer, and the amount of consumption tax is rounded down to the nearest integer.
If multiple prices satisfy the condition, print the lowest such price; if no price satisfies the condition, print `-1`.
Constraints
* 1 \leq A \leq B \leq 100
* A and B are integers.
Input
Input is given from Standard Input in the following format:
A B
Output
If there is a price that satisfies the condition, print an integer representing the lowest such price; otherwise, print `-1`.
Examples
Input
2 2
Output
25
Input
8 10
Output
100
Input
19 99
Output
-1
Submitted Solution:
```
a,b=map(int,input().split());l=max(0--a*25//2,b*10);print([-1,l][l<min(-~a*25//2,-~b*10)])
``` | instruction | 0 | 49,421 | 10 | 98,842 |
Yes | output | 1 | 49,421 | 10 | 98,843 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the price of a product before tax such that, when the consumption tax rate is 8 percent and 10 percent, the amount of consumption tax levied on it is A yen and B yen, respectively. (Yen is the currency of Japan.)
Here, the price before tax must be a positive integer, and the amount of consumption tax is rounded down to the nearest integer.
If multiple prices satisfy the condition, print the lowest such price; if no price satisfies the condition, print `-1`.
Constraints
* 1 \leq A \leq B \leq 100
* A and B are integers.
Input
Input is given from Standard Input in the following format:
A B
Output
If there is a price that satisfies the condition, print an integer representing the lowest such price; otherwise, print `-1`.
Examples
Input
2 2
Output
25
Input
8 10
Output
100
Input
19 99
Output
-1
Submitted Solution:
```
a,b=map(int,input().split())
ans=-1
for i in range(1010):
if(a==int(i*0.08) and b==int(i*0.1)):
ans=i
break
print(ans)
``` | instruction | 0 | 49,422 | 10 | 98,844 |
Yes | output | 1 | 49,422 | 10 | 98,845 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the price of a product before tax such that, when the consumption tax rate is 8 percent and 10 percent, the amount of consumption tax levied on it is A yen and B yen, respectively. (Yen is the currency of Japan.)
Here, the price before tax must be a positive integer, and the amount of consumption tax is rounded down to the nearest integer.
If multiple prices satisfy the condition, print the lowest such price; if no price satisfies the condition, print `-1`.
Constraints
* 1 \leq A \leq B \leq 100
* A and B are integers.
Input
Input is given from Standard Input in the following format:
A B
Output
If there is a price that satisfies the condition, print an integer representing the lowest such price; otherwise, print `-1`.
Examples
Input
2 2
Output
25
Input
8 10
Output
100
Input
19 99
Output
-1
Submitted Solution:
```
a,b=map(int,input().split())
for x in range(1,2000):
if int(x*0.08)==a and int(x*0.1)==b:
print(x)
exit(0)
print(-1)
``` | instruction | 0 | 49,423 | 10 | 98,846 |
Yes | output | 1 | 49,423 | 10 | 98,847 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the price of a product before tax such that, when the consumption tax rate is 8 percent and 10 percent, the amount of consumption tax levied on it is A yen and B yen, respectively. (Yen is the currency of Japan.)
Here, the price before tax must be a positive integer, and the amount of consumption tax is rounded down to the nearest integer.
If multiple prices satisfy the condition, print the lowest such price; if no price satisfies the condition, print `-1`.
Constraints
* 1 \leq A \leq B \leq 100
* A and B are integers.
Input
Input is given from Standard Input in the following format:
A B
Output
If there is a price that satisfies the condition, print an integer representing the lowest such price; otherwise, print `-1`.
Examples
Input
2 2
Output
25
Input
8 10
Output
100
Input
19 99
Output
-1
Submitted Solution:
```
a,b=map(int,input().split())
ans=-1
for i in range(15000):
if a==int(i*0.08) and b==int(i*0.1):
ans=i
break
print(ans)
``` | instruction | 0 | 49,424 | 10 | 98,848 |
Yes | output | 1 | 49,424 | 10 | 98,849 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the price of a product before tax such that, when the consumption tax rate is 8 percent and 10 percent, the amount of consumption tax levied on it is A yen and B yen, respectively. (Yen is the currency of Japan.)
Here, the price before tax must be a positive integer, and the amount of consumption tax is rounded down to the nearest integer.
If multiple prices satisfy the condition, print the lowest such price; if no price satisfies the condition, print `-1`.
Constraints
* 1 \leq A \leq B \leq 100
* A and B are integers.
Input
Input is given from Standard Input in the following format:
A B
Output
If there is a price that satisfies the condition, print an integer representing the lowest such price; otherwise, print `-1`.
Examples
Input
2 2
Output
25
Input
8 10
Output
100
Input
19 99
Output
-1
Submitted Solution:
```
import math
A,B=map(int,input().split())
eight=[]
i=0
ten=[]
for a in range(1,1001):
if math.floor(a*0.08)==A:
eight.append(a)
if math.floor(a*0.1)==B:
ten.append(a)
for a in eight:
if a in ten:
print(a)
i=1
break
if i=0:
print("-1")
``` | instruction | 0 | 49,425 | 10 | 98,850 |
No | output | 1 | 49,425 | 10 | 98,851 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the price of a product before tax such that, when the consumption tax rate is 8 percent and 10 percent, the amount of consumption tax levied on it is A yen and B yen, respectively. (Yen is the currency of Japan.)
Here, the price before tax must be a positive integer, and the amount of consumption tax is rounded down to the nearest integer.
If multiple prices satisfy the condition, print the lowest such price; if no price satisfies the condition, print `-1`.
Constraints
* 1 \leq A \leq B \leq 100
* A and B are integers.
Input
Input is given from Standard Input in the following format:
A B
Output
If there is a price that satisfies the condition, print an integer representing the lowest such price; otherwise, print `-1`.
Examples
Input
2 2
Output
25
Input
8 10
Output
100
Input
19 99
Output
-1
Submitted Solution:
```
import math
a,b=list(map(int,input().split()))
ans=-1
for i in range(math.floor(a*100/8)+1,math.floor((a+1)*100/8)+1):
if math.floor(i*10/100)==b:
ans=i
break
print(ans)
``` | instruction | 0 | 49,426 | 10 | 98,852 |
No | output | 1 | 49,426 | 10 | 98,853 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the price of a product before tax such that, when the consumption tax rate is 8 percent and 10 percent, the amount of consumption tax levied on it is A yen and B yen, respectively. (Yen is the currency of Japan.)
Here, the price before tax must be a positive integer, and the amount of consumption tax is rounded down to the nearest integer.
If multiple prices satisfy the condition, print the lowest such price; if no price satisfies the condition, print `-1`.
Constraints
* 1 \leq A \leq B \leq 100
* A and B are integers.
Input
Input is given from Standard Input in the following format:
A B
Output
If there is a price that satisfies the condition, print an integer representing the lowest such price; otherwise, print `-1`.
Examples
Input
2 2
Output
25
Input
8 10
Output
100
Input
19 99
Output
-1
Submitted Solution:
```
import math
def main():
A,B = map(int, input().split())
# η¨ηrate%γ§taxγAεγ¨γͺγη¨ζγδΎ‘ζ Όxγ―γceil(a*100/rate)<=x<ceil((a+1)*100/rate)γζΊγγ
min1 = math.ceil(A*100/8)
max1 = math.ceil((A+1)*100/8)-1
min2 = math.ceil(B*100/10)
max2 = math.ceil((B+1)*100/10)-1
ans = max(min1,min2) if not max1<min2 or max2<min1 else -1
print(ans)
main()
``` | instruction | 0 | 49,427 | 10 | 98,854 |
No | output | 1 | 49,427 | 10 | 98,855 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the price of a product before tax such that, when the consumption tax rate is 8 percent and 10 percent, the amount of consumption tax levied on it is A yen and B yen, respectively. (Yen is the currency of Japan.)
Here, the price before tax must be a positive integer, and the amount of consumption tax is rounded down to the nearest integer.
If multiple prices satisfy the condition, print the lowest such price; if no price satisfies the condition, print `-1`.
Constraints
* 1 \leq A \leq B \leq 100
* A and B are integers.
Input
Input is given from Standard Input in the following format:
A B
Output
If there is a price that satisfies the condition, print an integer representing the lowest such price; otherwise, print `-1`.
Examples
Input
2 2
Output
25
Input
8 10
Output
100
Input
19 99
Output
-1
Submitted Solution:
```
import math
a,b=map(int,input().split())
f=-1
for i in range(1000):
if math.floor(i*0.08)==a and math.floor(i*0.1)==b:
f=i
break
print(f)
``` | instruction | 0 | 49,428 | 10 | 98,856 |
No | output | 1 | 49,428 | 10 | 98,857 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Due to the increase in the number of students of Berland State University it was decided to equip a new computer room. You were given the task of buying mouses, and you have to spend as little as possible. After all, the country is in crisis!
The computers bought for the room were different. Some of them had only USB ports, some β only PS/2 ports, and some had both options.
You have found a price list of a certain computer shop. In it, for m mouses it is specified the cost and the type of the port that is required to plug the mouse in (USB or PS/2). Each mouse from the list can be bought at most once.
You want to buy some set of mouses from the given price list in such a way so that you maximize the number of computers equipped with mouses (it is not guaranteed that you will be able to equip all of the computers), and in case of equality of this value you want to minimize the total cost of mouses you will buy.
Input
The first line contains three integers a, b and c (0 β€ a, b, c β€ 105) β the number of computers that only have USB ports, the number of computers, that only have PS/2 ports, and the number of computers, that have both options, respectively.
The next line contains one integer m (0 β€ m β€ 3Β·105) β the number of mouses in the price list.
The next m lines each describe another mouse. The i-th line contains first integer vali (1 β€ vali β€ 109) β the cost of the i-th mouse, then the type of port (USB or PS/2) that is required to plug the mouse in.
Output
Output two integers separated by space β the number of equipped computers and the total cost of the mouses you will buy.
Example
Input
2 1 1
4
5 USB
6 PS/2
3 PS/2
7 PS/2
Output
3 14
Note
In the first example you can buy the first three mouses. This way you will equip one of the computers that has only a USB port with a USB mouse, and the two PS/2 mouses you will plug into the computer with PS/2 port and the computer with both ports. | instruction | 0 | 50,124 | 10 | 100,248 |
Tags: greedy, implementation, sortings, two pointers
Correct Solution:
```
import math
import re
a, b, c = map(int, input().split())
q = 0
res = 0
val = int(input())
usb = []
ps2 = []
for i in range(val):
x, y = input().split()
if y == 'USB':
usb.append(int(x))
else:
ps2.append(int(x))
usb = sorted(usb)
ps2 = sorted(ps2)
res += sum(usb[:min(len(usb), a)])
q += min(len(usb), a)
del(usb[:min(len(usb), a)])
res += sum(ps2[:min(len(ps2), b)])
q += min(len(ps2), b)
del(ps2[:min(len(ps2), b)])
usb = sorted(usb + ps2)
res += sum(usb[:min(len(usb), c)])
q += min(len(usb), c)
print(str(q) + ' ' + str(res))
``` | output | 1 | 50,124 | 10 | 100,249 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Due to the increase in the number of students of Berland State University it was decided to equip a new computer room. You were given the task of buying mouses, and you have to spend as little as possible. After all, the country is in crisis!
The computers bought for the room were different. Some of them had only USB ports, some β only PS/2 ports, and some had both options.
You have found a price list of a certain computer shop. In it, for m mouses it is specified the cost and the type of the port that is required to plug the mouse in (USB or PS/2). Each mouse from the list can be bought at most once.
You want to buy some set of mouses from the given price list in such a way so that you maximize the number of computers equipped with mouses (it is not guaranteed that you will be able to equip all of the computers), and in case of equality of this value you want to minimize the total cost of mouses you will buy.
Input
The first line contains three integers a, b and c (0 β€ a, b, c β€ 105) β the number of computers that only have USB ports, the number of computers, that only have PS/2 ports, and the number of computers, that have both options, respectively.
The next line contains one integer m (0 β€ m β€ 3Β·105) β the number of mouses in the price list.
The next m lines each describe another mouse. The i-th line contains first integer vali (1 β€ vali β€ 109) β the cost of the i-th mouse, then the type of port (USB or PS/2) that is required to plug the mouse in.
Output
Output two integers separated by space β the number of equipped computers and the total cost of the mouses you will buy.
Example
Input
2 1 1
4
5 USB
6 PS/2
3 PS/2
7 PS/2
Output
3 14
Note
In the first example you can buy the first three mouses. This way you will equip one of the computers that has only a USB port with a USB mouse, and the two PS/2 mouses you will plug into the computer with PS/2 port and the computer with both ports. | instruction | 0 | 50,125 | 10 | 100,250 |
Tags: greedy, implementation, sortings, two pointers
Correct Solution:
```
from sys import stdin as fin
a, b, c = input().split()
a = int(a)
b = int(b)
c = int(c)
n = int(input())
mouses = []
for i in range(0, n):
l, r = fin.readline().strip().split()
mouses.append((int(l), r))
mouses.sort()
money_count = 0
comp_count = 0
for t in mouses:
if t[1] == 'USB' and a + c or t[1] == 'PS/2' and b + c :
money_count += t[0]
comp_count += 1
if t[1] == 'USB':
if a : a -= 1
elif c : c -= 1
if t[1] == 'PS/2':
if b : b -= 1
elif c : c -= 1
print(comp_count, money_count)
``` | output | 1 | 50,125 | 10 | 100,251 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Due to the increase in the number of students of Berland State University it was decided to equip a new computer room. You were given the task of buying mouses, and you have to spend as little as possible. After all, the country is in crisis!
The computers bought for the room were different. Some of them had only USB ports, some β only PS/2 ports, and some had both options.
You have found a price list of a certain computer shop. In it, for m mouses it is specified the cost and the type of the port that is required to plug the mouse in (USB or PS/2). Each mouse from the list can be bought at most once.
You want to buy some set of mouses from the given price list in such a way so that you maximize the number of computers equipped with mouses (it is not guaranteed that you will be able to equip all of the computers), and in case of equality of this value you want to minimize the total cost of mouses you will buy.
Input
The first line contains three integers a, b and c (0 β€ a, b, c β€ 105) β the number of computers that only have USB ports, the number of computers, that only have PS/2 ports, and the number of computers, that have both options, respectively.
The next line contains one integer m (0 β€ m β€ 3Β·105) β the number of mouses in the price list.
The next m lines each describe another mouse. The i-th line contains first integer vali (1 β€ vali β€ 109) β the cost of the i-th mouse, then the type of port (USB or PS/2) that is required to plug the mouse in.
Output
Output two integers separated by space β the number of equipped computers and the total cost of the mouses you will buy.
Example
Input
2 1 1
4
5 USB
6 PS/2
3 PS/2
7 PS/2
Output
3 14
Note
In the first example you can buy the first three mouses. This way you will equip one of the computers that has only a USB port with a USB mouse, and the two PS/2 mouses you will plug into the computer with PS/2 port and the computer with both ports. | instruction | 0 | 50,126 | 10 | 100,252 |
Tags: greedy, implementation, sortings, two pointers
Correct Solution:
```
#!/usr/local/bin/python3.4
from collections import *
import pdb
usb, ps_2 , both = map(int, input().split())
sales_usb = deque()
sales_ps_2 = deque()
for _ in range(int(input())):
value, type_ = input().split()
value = int(value)
if type_ == "USB":
sales_usb.append(value)
else:
sales_ps_2.append(value)
def remove_both(usb,ps,both):
total = 0
sold = 0
while both != 0:
if (len(usb)!=0 and len(ps)!=0):
if ( usb[0] >= ps[0] ):
total+= ps.popleft()
sold+=1
both-=1
elif ( usb[0] < ps[0] ):
total+= usb.popleft()
sold+=1
both-=1
elif len(usb)==0 and len(ps)!=0:
temp_total,temp_sold,_ = remove_single(ps,both)
total+=temp_total
sold+=temp_sold
both=0
elif len(ps)==0 and len(usb)!=0:
temp_total,temp_sold,_ = remove_single(usb,both)
total+=temp_total
sold+=temp_sold
both=0
else:
both=0
return total,sold
def remove_single(slist,number):
total = 0
sold = 0
while number!=0 and len(slist)>0:
total+=slist.popleft()
sold+=1
number-=1
return total,sold,slist
sum_all=0
sold=0
sales_usb = sorted(sales_usb)
sales_ps_2 = sorted(sales_ps_2)
# print (sales_usb,sales_ps_2)
t_sum_all, t_sold, sales_ps_2 = remove_single(deque(sales_ps_2),ps_2)
sum_all+=t_sum_all
sold+=t_sold
t_sum_all, t_sold, sales_usb = remove_single(deque(sales_usb),usb)
sold+=t_sold
sum_all+=t_sum_all
t_sum_all, t_sold = remove_both(deque(sales_usb),deque(sales_ps_2),both)
sum_all+=t_sum_all
sold+=t_sold
print (sold, sum_all)
``` | output | 1 | 50,126 | 10 | 100,253 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Due to the increase in the number of students of Berland State University it was decided to equip a new computer room. You were given the task of buying mouses, and you have to spend as little as possible. After all, the country is in crisis!
The computers bought for the room were different. Some of them had only USB ports, some β only PS/2 ports, and some had both options.
You have found a price list of a certain computer shop. In it, for m mouses it is specified the cost and the type of the port that is required to plug the mouse in (USB or PS/2). Each mouse from the list can be bought at most once.
You want to buy some set of mouses from the given price list in such a way so that you maximize the number of computers equipped with mouses (it is not guaranteed that you will be able to equip all of the computers), and in case of equality of this value you want to minimize the total cost of mouses you will buy.
Input
The first line contains three integers a, b and c (0 β€ a, b, c β€ 105) β the number of computers that only have USB ports, the number of computers, that only have PS/2 ports, and the number of computers, that have both options, respectively.
The next line contains one integer m (0 β€ m β€ 3Β·105) β the number of mouses in the price list.
The next m lines each describe another mouse. The i-th line contains first integer vali (1 β€ vali β€ 109) β the cost of the i-th mouse, then the type of port (USB or PS/2) that is required to plug the mouse in.
Output
Output two integers separated by space β the number of equipped computers and the total cost of the mouses you will buy.
Example
Input
2 1 1
4
5 USB
6 PS/2
3 PS/2
7 PS/2
Output
3 14
Note
In the first example you can buy the first three mouses. This way you will equip one of the computers that has only a USB port with a USB mouse, and the two PS/2 mouses you will plug into the computer with PS/2 port and the computer with both ports. | instruction | 0 | 50,127 | 10 | 100,254 |
Tags: greedy, implementation, sortings, two pointers
Correct Solution:
```
def list_from_input():
return list(map(int, input().split()))
class Mouse:
def __init__(self, price, type):
self.type = type
self.price = price
@classmethod
def from_input(cls):
mouse_data = input().split()
price = int(mouse_data[0])
type = mouse_data[1][0]
return Mouse(price, type)
def main():
usb_pc, ps_pc, both_pc = list_from_input()
mouses_count = int(input())
mouses = []
for i in range(mouses_count):
mouses.append(Mouse.from_input())
mouses.sort(key=lambda mouse: mouse.price)
purchase_amount = 0
pc_with_mouses = 0
for mouse in mouses:
if mouse.type is 'U' and usb_pc > 0:
usb_pc -= 1
purchase_amount += mouse.price
pc_with_mouses += 1
elif mouse.type is 'P' and ps_pc > 0:
ps_pc -= 1
purchase_amount += mouse.price
pc_with_mouses += 1
elif both_pc > 0:
both_pc -= 1
purchase_amount += mouse.price
pc_with_mouses += 1
print(pc_with_mouses, purchase_amount)
main()
``` | output | 1 | 50,127 | 10 | 100,255 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Due to the increase in the number of students of Berland State University it was decided to equip a new computer room. You were given the task of buying mouses, and you have to spend as little as possible. After all, the country is in crisis!
The computers bought for the room were different. Some of them had only USB ports, some β only PS/2 ports, and some had both options.
You have found a price list of a certain computer shop. In it, for m mouses it is specified the cost and the type of the port that is required to plug the mouse in (USB or PS/2). Each mouse from the list can be bought at most once.
You want to buy some set of mouses from the given price list in such a way so that you maximize the number of computers equipped with mouses (it is not guaranteed that you will be able to equip all of the computers), and in case of equality of this value you want to minimize the total cost of mouses you will buy.
Input
The first line contains three integers a, b and c (0 β€ a, b, c β€ 105) β the number of computers that only have USB ports, the number of computers, that only have PS/2 ports, and the number of computers, that have both options, respectively.
The next line contains one integer m (0 β€ m β€ 3Β·105) β the number of mouses in the price list.
The next m lines each describe another mouse. The i-th line contains first integer vali (1 β€ vali β€ 109) β the cost of the i-th mouse, then the type of port (USB or PS/2) that is required to plug the mouse in.
Output
Output two integers separated by space β the number of equipped computers and the total cost of the mouses you will buy.
Example
Input
2 1 1
4
5 USB
6 PS/2
3 PS/2
7 PS/2
Output
3 14
Note
In the first example you can buy the first three mouses. This way you will equip one of the computers that has only a USB port with a USB mouse, and the two PS/2 mouses you will plug into the computer with PS/2 port and the computer with both ports. | instruction | 0 | 50,128 | 10 | 100,256 |
Tags: greedy, implementation, sortings, two pointers
Correct Solution:
```
from collections import deque
a, b, c = list(map(int, input().split()))
m = int(input())
usb_cost = []
ps_cost = []
for i in range(m):
cost, t = input().split()
cost = int(cost)
if t == 'USB':
usb_cost.append(cost)
else:
ps_cost.append(cost)
usb_cost.sort(reverse = True)
ps_cost.sort(reverse = True)
# print(usb_cost)
# print(ps_cost)
ans = 0
ans_sum = 0
for i in range(min(a, len(usb_cost))):
ans_sum += usb_cost.pop()
ans += 1
for i in range(min(b, len(ps_cost))):
ans_sum += ps_cost.pop()
ans += 1
all_types = sorted(usb_cost + ps_cost, reverse = True)
for i in range(min(c, len(all_types))):
ans_sum += all_types.pop()
ans += 1
print(ans, ans_sum)
``` | output | 1 | 50,128 | 10 | 100,257 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Due to the increase in the number of students of Berland State University it was decided to equip a new computer room. You were given the task of buying mouses, and you have to spend as little as possible. After all, the country is in crisis!
The computers bought for the room were different. Some of them had only USB ports, some β only PS/2 ports, and some had both options.
You have found a price list of a certain computer shop. In it, for m mouses it is specified the cost and the type of the port that is required to plug the mouse in (USB or PS/2). Each mouse from the list can be bought at most once.
You want to buy some set of mouses from the given price list in such a way so that you maximize the number of computers equipped with mouses (it is not guaranteed that you will be able to equip all of the computers), and in case of equality of this value you want to minimize the total cost of mouses you will buy.
Input
The first line contains three integers a, b and c (0 β€ a, b, c β€ 105) β the number of computers that only have USB ports, the number of computers, that only have PS/2 ports, and the number of computers, that have both options, respectively.
The next line contains one integer m (0 β€ m β€ 3Β·105) β the number of mouses in the price list.
The next m lines each describe another mouse. The i-th line contains first integer vali (1 β€ vali β€ 109) β the cost of the i-th mouse, then the type of port (USB or PS/2) that is required to plug the mouse in.
Output
Output two integers separated by space β the number of equipped computers and the total cost of the mouses you will buy.
Example
Input
2 1 1
4
5 USB
6 PS/2
3 PS/2
7 PS/2
Output
3 14
Note
In the first example you can buy the first three mouses. This way you will equip one of the computers that has only a USB port with a USB mouse, and the two PS/2 mouses you will plug into the computer with PS/2 port and the computer with both ports. | instruction | 0 | 50,129 | 10 | 100,258 |
Tags: greedy, implementation, sortings, two pointers
Correct Solution:
```
a,b,c=map(int,input().split())
m=int(input())
d={'USB':[], 'PS/2':[]}
for _ in range(m):
v,t=input().split()
d[t].append(int(v))
d['PS/2'].sort()
d['USB'].sort()
eq=cst=f1=f2=0
nusb=len(d['USB'])
nps2=len(d['PS/2'])
while a>0 and f1<nusb:
a-=1
eq+=1
cst+=d['USB'][f1]
f1+=1
while b>0 and f2<nps2:
b-=1
eq+=1
cst+=d['PS/2'][f2]
f2+=1
while c>0 and (f1<nusb or f2<nps2):
c-=1
eq+=1
if f1==nusb:
cst+=d['PS/2'][f2]
f2+=1
continue
elif f2==nps2:
cst+=d['USB'][f1]
f1+=1
continue
if d['PS/2'][f2]<d['USB'][f1]:
cst+=d['PS/2'][f2]
f2+=1
else:
cst+=d['USB'][f1]
f1+=1
print(eq,cst)
``` | output | 1 | 50,129 | 10 | 100,259 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Due to the increase in the number of students of Berland State University it was decided to equip a new computer room. You were given the task of buying mouses, and you have to spend as little as possible. After all, the country is in crisis!
The computers bought for the room were different. Some of them had only USB ports, some β only PS/2 ports, and some had both options.
You have found a price list of a certain computer shop. In it, for m mouses it is specified the cost and the type of the port that is required to plug the mouse in (USB or PS/2). Each mouse from the list can be bought at most once.
You want to buy some set of mouses from the given price list in such a way so that you maximize the number of computers equipped with mouses (it is not guaranteed that you will be able to equip all of the computers), and in case of equality of this value you want to minimize the total cost of mouses you will buy.
Input
The first line contains three integers a, b and c (0 β€ a, b, c β€ 105) β the number of computers that only have USB ports, the number of computers, that only have PS/2 ports, and the number of computers, that have both options, respectively.
The next line contains one integer m (0 β€ m β€ 3Β·105) β the number of mouses in the price list.
The next m lines each describe another mouse. The i-th line contains first integer vali (1 β€ vali β€ 109) β the cost of the i-th mouse, then the type of port (USB or PS/2) that is required to plug the mouse in.
Output
Output two integers separated by space β the number of equipped computers and the total cost of the mouses you will buy.
Example
Input
2 1 1
4
5 USB
6 PS/2
3 PS/2
7 PS/2
Output
3 14
Note
In the first example you can buy the first three mouses. This way you will equip one of the computers that has only a USB port with a USB mouse, and the two PS/2 mouses you will plug into the computer with PS/2 port and the computer with both ports. | instruction | 0 | 50,130 | 10 | 100,260 |
Tags: greedy, implementation, sortings, two pointers
Correct Solution:
```
u, p, b = map(int,input().split())
t = int(input())
usb = []
ps2 = []
for i in range(t):
val = input().split()
if val[1] == 'USB':
usb.append(int(val[0]))
else:
ps2.append(int(val[0]))
c = 0
s = 0
usb.sort(reverse=True)
ps2.sort(reverse=True)
leu = len(usb)
lep = len(ps2)
j = 0
for i in range(leu):
if j>=u:
break
s += usb.pop()
j += 1
c += 1
k = 0
for i in range(lep):
if k>=p:
break
s += ps2.pop()
k += 1
c += 1
for i in range(b):
if len(usb)>0 and len(ps2)>0:
if usb[-1]<ps2[-1]:
s += usb.pop()
c += 1
else:
s += ps2.pop()
c += 1
elif len(usb)>0:
s += usb.pop()
c += 1
elif len(ps2)>0:
s += ps2.pop()
c += 1
else:
break
print(c, s)
``` | output | 1 | 50,130 | 10 | 100,261 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Due to the increase in the number of students of Berland State University it was decided to equip a new computer room. You were given the task of buying mouses, and you have to spend as little as possible. After all, the country is in crisis!
The computers bought for the room were different. Some of them had only USB ports, some β only PS/2 ports, and some had both options.
You have found a price list of a certain computer shop. In it, for m mouses it is specified the cost and the type of the port that is required to plug the mouse in (USB or PS/2). Each mouse from the list can be bought at most once.
You want to buy some set of mouses from the given price list in such a way so that you maximize the number of computers equipped with mouses (it is not guaranteed that you will be able to equip all of the computers), and in case of equality of this value you want to minimize the total cost of mouses you will buy.
Input
The first line contains three integers a, b and c (0 β€ a, b, c β€ 105) β the number of computers that only have USB ports, the number of computers, that only have PS/2 ports, and the number of computers, that have both options, respectively.
The next line contains one integer m (0 β€ m β€ 3Β·105) β the number of mouses in the price list.
The next m lines each describe another mouse. The i-th line contains first integer vali (1 β€ vali β€ 109) β the cost of the i-th mouse, then the type of port (USB or PS/2) that is required to plug the mouse in.
Output
Output two integers separated by space β the number of equipped computers and the total cost of the mouses you will buy.
Example
Input
2 1 1
4
5 USB
6 PS/2
3 PS/2
7 PS/2
Output
3 14
Note
In the first example you can buy the first three mouses. This way you will equip one of the computers that has only a USB port with a USB mouse, and the two PS/2 mouses you will plug into the computer with PS/2 port and the computer with both ports. | instruction | 0 | 50,131 | 10 | 100,262 |
Tags: greedy, implementation, sortings, two pointers
Correct Solution:
```
a,b,c = input().split()
a,b,c = int(a), int(b), int(c)
usb = 'USB'
ps2 = 'PS/2'
u = []
ps = []
n = int(input())
while n >0:
cost, s = input().split()
if s == usb:
u.append(int(cost))
elif s == ps2:
ps.append(int(cost))
n -= 1
u.sort()
ps.sort()
cost, count = 0, 0
a_curr, ps_curr = 0, 0
a_count, ps_count = len(u), len(ps)
cost += sum(u[0:min(a_count,a)])
a_curr = min(a_count,a)
count = a_curr
cost += sum(ps[0:min(ps_count, b)])
ps_curr = min(ps_count, b)
count += ps_curr
new_arr = u[a_curr:] + ps[ps_curr:]
count += min(c,len(new_arr))
new_arr.sort()
cost += sum(new_arr[:c])
print(count, cost, sep = ' ')
``` | output | 1 | 50,131 | 10 | 100,263 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Due to the increase in the number of students of Berland State University it was decided to equip a new computer room. You were given the task of buying mouses, and you have to spend as little as possible. After all, the country is in crisis!
The computers bought for the room were different. Some of them had only USB ports, some β only PS/2 ports, and some had both options.
You have found a price list of a certain computer shop. In it, for m mouses it is specified the cost and the type of the port that is required to plug the mouse in (USB or PS/2). Each mouse from the list can be bought at most once.
You want to buy some set of mouses from the given price list in such a way so that you maximize the number of computers equipped with mouses (it is not guaranteed that you will be able to equip all of the computers), and in case of equality of this value you want to minimize the total cost of mouses you will buy.
Input
The first line contains three integers a, b and c (0 β€ a, b, c β€ 105) β the number of computers that only have USB ports, the number of computers, that only have PS/2 ports, and the number of computers, that have both options, respectively.
The next line contains one integer m (0 β€ m β€ 3Β·105) β the number of mouses in the price list.
The next m lines each describe another mouse. The i-th line contains first integer vali (1 β€ vali β€ 109) β the cost of the i-th mouse, then the type of port (USB or PS/2) that is required to plug the mouse in.
Output
Output two integers separated by space β the number of equipped computers and the total cost of the mouses you will buy.
Example
Input
2 1 1
4
5 USB
6 PS/2
3 PS/2
7 PS/2
Output
3 14
Note
In the first example you can buy the first three mouses. This way you will equip one of the computers that has only a USB port with a USB mouse, and the two PS/2 mouses you will plug into the computer with PS/2 port and the computer with both ports.
Submitted Solution:
```
import sys
input = sys.stdin.readline
'''
'''
num_usb, num_ps, num_both = map(int, input().split())
m = int(input())
usb = []
ps = []
for _ in range(m):
c, con = input().rstrip().split()
c = int(c)
if con == "USB":
usb.append(c)
else:
ps.append(c)
usb.sort(reverse=True)
ps.sort(reverse=True)
cost = 0
count = 0
while num_usb and usb:
num_usb -= 1
cost += usb.pop()
count += 1
while num_ps and ps:
num_ps -= 1
cost += ps.pop()
count += 1
usb.extend(ps)
both = usb
both.sort(reverse=True)
while num_both and usb:
num_both -= 1
cost += both.pop()
count += 1
print(count, cost)
``` | instruction | 0 | 50,132 | 10 | 100,264 |
Yes | output | 1 | 50,132 | 10 | 100,265 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Due to the increase in the number of students of Berland State University it was decided to equip a new computer room. You were given the task of buying mouses, and you have to spend as little as possible. After all, the country is in crisis!
The computers bought for the room were different. Some of them had only USB ports, some β only PS/2 ports, and some had both options.
You have found a price list of a certain computer shop. In it, for m mouses it is specified the cost and the type of the port that is required to plug the mouse in (USB or PS/2). Each mouse from the list can be bought at most once.
You want to buy some set of mouses from the given price list in such a way so that you maximize the number of computers equipped with mouses (it is not guaranteed that you will be able to equip all of the computers), and in case of equality of this value you want to minimize the total cost of mouses you will buy.
Input
The first line contains three integers a, b and c (0 β€ a, b, c β€ 105) β the number of computers that only have USB ports, the number of computers, that only have PS/2 ports, and the number of computers, that have both options, respectively.
The next line contains one integer m (0 β€ m β€ 3Β·105) β the number of mouses in the price list.
The next m lines each describe another mouse. The i-th line contains first integer vali (1 β€ vali β€ 109) β the cost of the i-th mouse, then the type of port (USB or PS/2) that is required to plug the mouse in.
Output
Output two integers separated by space β the number of equipped computers and the total cost of the mouses you will buy.
Example
Input
2 1 1
4
5 USB
6 PS/2
3 PS/2
7 PS/2
Output
3 14
Note
In the first example you can buy the first three mouses. This way you will equip one of the computers that has only a USB port with a USB mouse, and the two PS/2 mouses you will plug into the computer with PS/2 port and the computer with both ports.
Submitted Solution:
```
a,b,c=map(int,input().split())
m=int(input())
usb,ps2=[],[]
for i in range(m):
s=input().split()
if s[1]=="USB":
usb.append(int(s[0]))
else:
ps2.append(int(s[0]))
usb.sort(); ps2.sort();
kusb,kps2=min(a,len(usb)),min(b,len(ps2))
kans=kusb+kps2
sans=sum(usb[:kusb])+sum(ps2[:kps2])
other=usb[kusb:]+ps2[kps2:]
other.sort()
kother=min(c,len(other))
kans+=kother
sans+=sum(other[:kother])
print(str(kans)+' '+str(sans))
``` | instruction | 0 | 50,133 | 10 | 100,266 |
Yes | output | 1 | 50,133 | 10 | 100,267 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Due to the increase in the number of students of Berland State University it was decided to equip a new computer room. You were given the task of buying mouses, and you have to spend as little as possible. After all, the country is in crisis!
The computers bought for the room were different. Some of them had only USB ports, some β only PS/2 ports, and some had both options.
You have found a price list of a certain computer shop. In it, for m mouses it is specified the cost and the type of the port that is required to plug the mouse in (USB or PS/2). Each mouse from the list can be bought at most once.
You want to buy some set of mouses from the given price list in such a way so that you maximize the number of computers equipped with mouses (it is not guaranteed that you will be able to equip all of the computers), and in case of equality of this value you want to minimize the total cost of mouses you will buy.
Input
The first line contains three integers a, b and c (0 β€ a, b, c β€ 105) β the number of computers that only have USB ports, the number of computers, that only have PS/2 ports, and the number of computers, that have both options, respectively.
The next line contains one integer m (0 β€ m β€ 3Β·105) β the number of mouses in the price list.
The next m lines each describe another mouse. The i-th line contains first integer vali (1 β€ vali β€ 109) β the cost of the i-th mouse, then the type of port (USB or PS/2) that is required to plug the mouse in.
Output
Output two integers separated by space β the number of equipped computers and the total cost of the mouses you will buy.
Example
Input
2 1 1
4
5 USB
6 PS/2
3 PS/2
7 PS/2
Output
3 14
Note
In the first example you can buy the first three mouses. This way you will equip one of the computers that has only a USB port with a USB mouse, and the two PS/2 mouses you will plug into the computer with PS/2 port and the computer with both ports.
Submitted Solution:
```
a,b,c=map(int,input().split())
m=int(input())
usb=[]
ps=[]
count=0
for _ in range(m):
x,y=input().split()
if y=="USB":
usb.append(int(x))
else:
ps.append(int(x))
usb.sort()
ps.sort()
ans=0
usbc=0
psc=0
if a>len(usb):
ans+=sum(usb)
usbc=len(usb)
count+=len(usb)
else:
for i in range(a):
ans+=usb[i]
usbc=i
count+=1
if a>0:
usbc+=1
if b>len(ps):
ans+=sum(ps)
psc=len(ps)
count+=len(ps)
else:
for i in range(b):
ans+=ps[i]
psc=i
count+=1
if b>0:
psc+=1
cc=c
if len(usb)-1>=usbc and len(ps)-1>=psc:
while usbc<=len(usb)-1 and psc<=len(ps)-1 and cc>0:
if usb[usbc]<ps[psc]:
ans+=usb[usbc]
usbc+=1
cc-=1
count+=1
else:
ans+=ps[psc]
psc+=1
cc-=1
count+=1
#print(usbc,psc)
if cc>0 and usbc<=len(usb)-1:
while cc>0 and usbc<=len(usb)-1:
cc-=1
ans+=usb[usbc]
usbc+=1
count+=1
if cc>0 and psc<=len(ps)-1:
while cc>0 and psc<=len(ps)-1:
cc-=1
ans+=ps[psc]
psc+=1
count+=1
print(count,ans)
``` | instruction | 0 | 50,134 | 10 | 100,268 |
Yes | output | 1 | 50,134 | 10 | 100,269 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Due to the increase in the number of students of Berland State University it was decided to equip a new computer room. You were given the task of buying mouses, and you have to spend as little as possible. After all, the country is in crisis!
The computers bought for the room were different. Some of them had only USB ports, some β only PS/2 ports, and some had both options.
You have found a price list of a certain computer shop. In it, for m mouses it is specified the cost and the type of the port that is required to plug the mouse in (USB or PS/2). Each mouse from the list can be bought at most once.
You want to buy some set of mouses from the given price list in such a way so that you maximize the number of computers equipped with mouses (it is not guaranteed that you will be able to equip all of the computers), and in case of equality of this value you want to minimize the total cost of mouses you will buy.
Input
The first line contains three integers a, b and c (0 β€ a, b, c β€ 105) β the number of computers that only have USB ports, the number of computers, that only have PS/2 ports, and the number of computers, that have both options, respectively.
The next line contains one integer m (0 β€ m β€ 3Β·105) β the number of mouses in the price list.
The next m lines each describe another mouse. The i-th line contains first integer vali (1 β€ vali β€ 109) β the cost of the i-th mouse, then the type of port (USB or PS/2) that is required to plug the mouse in.
Output
Output two integers separated by space β the number of equipped computers and the total cost of the mouses you will buy.
Example
Input
2 1 1
4
5 USB
6 PS/2
3 PS/2
7 PS/2
Output
3 14
Note
In the first example you can buy the first three mouses. This way you will equip one of the computers that has only a USB port with a USB mouse, and the two PS/2 mouses you will plug into the computer with PS/2 port and the computer with both ports.
Submitted Solution:
```
a, b, c = [int(i) for i in input().split()]
#usb = a
#ps = b
#both = c
n = int(input())
A = []
ps = []
usb = []
for i in range(n):
s = input().split()
s[0] = int(s[0])
if s[1] == "USB":
usb.append(s[0])
else:
ps.append(s[0])
#s[0], s[1] = s[1], s[0]
# A.append(s)
#A.sort()
ps.sort()
usb.sort()
usbptr = 0
psptr = 0
totalnum = 0
totalcost = 0
for i in range(a):
if usbptr < len(usb):
totalcost += usb[usbptr]
usbptr+=1
totalnum+=1
elif usbptr == len(usb):
break
for i in range(b):
if psptr < len(ps):
totalcost += ps[psptr]
psptr+=1
totalnum+=1
elif psptr == len(ps):
break
for i in range(c):
if usbptr < len(usb) and psptr < len(ps):
if usb[usbptr] < ps[psptr]:
totalcost += usb[usbptr]
usbptr+=1
totalnum+=1
else:
totalcost += ps[psptr]
psptr+=1
totalnum+=1
elif usbptr==len(usb) and psptr<len(ps):
totalcost += ps[psptr]
psptr+=1
totalnum+=1
elif usbptr<len(usb) and psptr == len(ps):
totalcost += usb[usbptr]
usbptr+=1
totalnum+=1
else:
break
print(totalnum, totalcost)
``` | instruction | 0 | 50,135 | 10 | 100,270 |
Yes | output | 1 | 50,135 | 10 | 100,271 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Due to the increase in the number of students of Berland State University it was decided to equip a new computer room. You were given the task of buying mouses, and you have to spend as little as possible. After all, the country is in crisis!
The computers bought for the room were different. Some of them had only USB ports, some β only PS/2 ports, and some had both options.
You have found a price list of a certain computer shop. In it, for m mouses it is specified the cost and the type of the port that is required to plug the mouse in (USB or PS/2). Each mouse from the list can be bought at most once.
You want to buy some set of mouses from the given price list in such a way so that you maximize the number of computers equipped with mouses (it is not guaranteed that you will be able to equip all of the computers), and in case of equality of this value you want to minimize the total cost of mouses you will buy.
Input
The first line contains three integers a, b and c (0 β€ a, b, c β€ 105) β the number of computers that only have USB ports, the number of computers, that only have PS/2 ports, and the number of computers, that have both options, respectively.
The next line contains one integer m (0 β€ m β€ 3Β·105) β the number of mouses in the price list.
The next m lines each describe another mouse. The i-th line contains first integer vali (1 β€ vali β€ 109) β the cost of the i-th mouse, then the type of port (USB or PS/2) that is required to plug the mouse in.
Output
Output two integers separated by space β the number of equipped computers and the total cost of the mouses you will buy.
Example
Input
2 1 1
4
5 USB
6 PS/2
3 PS/2
7 PS/2
Output
3 14
Note
In the first example you can buy the first three mouses. This way you will equip one of the computers that has only a USB port with a USB mouse, and the two PS/2 mouses you will plug into the computer with PS/2 port and the computer with both ports.
Submitted Solution:
```
a,b,c=map(int,input().split())
m=int(input())
usb=[]
ps2=[]
cost=0
for i in range(m):
z=list(input().split())
if z[1]=='USB':
usb.append(int(z[0]))
else:
ps2.append(int(z[0]))
usb=sorted(usb)
ps2=sorted(ps2)
while (len(usb)>0 and len(ps2)>0) and (a>0 or b>0 or c>0):
if usb[0]>ps2[0]:
if b>0 or c>0:
cost+=ps2[0]
if b>0:
b-=1
else:
c-=1
ps2.pop(0)
elif (a>0):
cost+=usb[0]
a-=1
usb.pop(0)
else:
if a>0 or c>0:
cost+=usb[0]
if a>0:
a-=1
else:
c-=1
usb.pop(0)
elif b>0:
cost+=ps2[0]
b-=1
ps2.pop(0)
while len(usb)>0 and a>0:
cost+=usb[0]
a-=1
usb.pop(0)
while len(ps2)>0 and b>0:
cost+=ps2[0]
b-=1
ps2.pop(0)
while len(usb)>0 and c>0:
cost+=usb[0]
c-=1
usb.pop(0)
while len(ps2)>0 and c>0:
cost+=ps2[0]
c-=1
ps2.pop(0)
print((m-a-b-c),cost)
``` | instruction | 0 | 50,136 | 10 | 100,272 |
No | output | 1 | 50,136 | 10 | 100,273 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Due to the increase in the number of students of Berland State University it was decided to equip a new computer room. You were given the task of buying mouses, and you have to spend as little as possible. After all, the country is in crisis!
The computers bought for the room were different. Some of them had only USB ports, some β only PS/2 ports, and some had both options.
You have found a price list of a certain computer shop. In it, for m mouses it is specified the cost and the type of the port that is required to plug the mouse in (USB or PS/2). Each mouse from the list can be bought at most once.
You want to buy some set of mouses from the given price list in such a way so that you maximize the number of computers equipped with mouses (it is not guaranteed that you will be able to equip all of the computers), and in case of equality of this value you want to minimize the total cost of mouses you will buy.
Input
The first line contains three integers a, b and c (0 β€ a, b, c β€ 105) β the number of computers that only have USB ports, the number of computers, that only have PS/2 ports, and the number of computers, that have both options, respectively.
The next line contains one integer m (0 β€ m β€ 3Β·105) β the number of mouses in the price list.
The next m lines each describe another mouse. The i-th line contains first integer vali (1 β€ vali β€ 109) β the cost of the i-th mouse, then the type of port (USB or PS/2) that is required to plug the mouse in.
Output
Output two integers separated by space β the number of equipped computers and the total cost of the mouses you will buy.
Example
Input
2 1 1
4
5 USB
6 PS/2
3 PS/2
7 PS/2
Output
3 14
Note
In the first example you can buy the first three mouses. This way you will equip one of the computers that has only a USB port with a USB mouse, and the two PS/2 mouses you will plug into the computer with PS/2 port and the computer with both ports.
Submitted Solution:
```
a,b,c=map(int,input().split())
m=int(input())
l=[]
for i in range(m):
q=input().split()
l.append(q)
l[i][0]=int(q[0])
l=sorted(l)
n=a+b+c
s=0
if n>m:
for i in range(m):
s+=l[i][0]
print(n, s)
exit()
ka=0
kb=0
kc=0
for i in range(n):
if l[i][1]=="PS/2" and kb<b:
s+=l[i][0]
kb+=1
continue
elif l[i][1]=="PS/2" and kc<c:
s+=l[i][0]
kc+=1
continue
if l[i][1]=="USB" and ka<a:
s+=l[i][0]
ka+=1
elif l[i][1]=="USB" and kc<c:
s+=l[i][0]
kc+=1
print(ka+kb+kc, s)
``` | instruction | 0 | 50,137 | 10 | 100,274 |
No | output | 1 | 50,137 | 10 | 100,275 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Due to the increase in the number of students of Berland State University it was decided to equip a new computer room. You were given the task of buying mouses, and you have to spend as little as possible. After all, the country is in crisis!
The computers bought for the room were different. Some of them had only USB ports, some β only PS/2 ports, and some had both options.
You have found a price list of a certain computer shop. In it, for m mouses it is specified the cost and the type of the port that is required to plug the mouse in (USB or PS/2). Each mouse from the list can be bought at most once.
You want to buy some set of mouses from the given price list in such a way so that you maximize the number of computers equipped with mouses (it is not guaranteed that you will be able to equip all of the computers), and in case of equality of this value you want to minimize the total cost of mouses you will buy.
Input
The first line contains three integers a, b and c (0 β€ a, b, c β€ 105) β the number of computers that only have USB ports, the number of computers, that only have PS/2 ports, and the number of computers, that have both options, respectively.
The next line contains one integer m (0 β€ m β€ 3Β·105) β the number of mouses in the price list.
The next m lines each describe another mouse. The i-th line contains first integer vali (1 β€ vali β€ 109) β the cost of the i-th mouse, then the type of port (USB or PS/2) that is required to plug the mouse in.
Output
Output two integers separated by space β the number of equipped computers and the total cost of the mouses you will buy.
Example
Input
2 1 1
4
5 USB
6 PS/2
3 PS/2
7 PS/2
Output
3 14
Note
In the first example you can buy the first three mouses. This way you will equip one of the computers that has only a USB port with a USB mouse, and the two PS/2 mouses you will plug into the computer with PS/2 port and the computer with both ports.
Submitted Solution:
```
def mergesort():
usbps = []
i = 0
j = 0
while i > len(usb) and j > len(ps):
if usb[i] <= ps[j]:
usbps.append(usb[i])
i += 1
else:
usbps.append(ps[j])
j += 1
usbps += usb[i:] + ps[j:]
return usbps
a, b, c = map(int, input().split())
m = int(input())
ans1 = 0
ans2 = 0
usb = []
ps = []
for i in range(m):
l = list(input().split())
if l[1] == 'USB':
usb.append(int(l[0]))
else:
ps.append(int(l[0]))
cntusb = len(usb)
cntps = len(ps)
usb.sort()
ps.sort()
if cntusb > a:
cntusb -= a
ans1 += a
ans2 += sum(usb[:a])
usb = usb[a:]
else:
ans1 += cntusb
cntusb = 0
ans2 += sum(usb)
usb = []
if cntps > b:
cntps -= b
ans1 += b
ans2 += sum(ps[:b])
ps = ps[b:]
else:
ans1 += b
cntps = 0
ans2 += sum(ps)
ps = []
every = mergesort()
if cntusb + cntps > c:
ans1 += c
ans2 += sum(every[:c])
else:
ans1 += c - cntusb - cntps
ans2 += sum(every)
print(ans1, ans2)
``` | instruction | 0 | 50,138 | 10 | 100,276 |
No | output | 1 | 50,138 | 10 | 100,277 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Due to the increase in the number of students of Berland State University it was decided to equip a new computer room. You were given the task of buying mouses, and you have to spend as little as possible. After all, the country is in crisis!
The computers bought for the room were different. Some of them had only USB ports, some β only PS/2 ports, and some had both options.
You have found a price list of a certain computer shop. In it, for m mouses it is specified the cost and the type of the port that is required to plug the mouse in (USB or PS/2). Each mouse from the list can be bought at most once.
You want to buy some set of mouses from the given price list in such a way so that you maximize the number of computers equipped with mouses (it is not guaranteed that you will be able to equip all of the computers), and in case of equality of this value you want to minimize the total cost of mouses you will buy.
Input
The first line contains three integers a, b and c (0 β€ a, b, c β€ 105) β the number of computers that only have USB ports, the number of computers, that only have PS/2 ports, and the number of computers, that have both options, respectively.
The next line contains one integer m (0 β€ m β€ 3Β·105) β the number of mouses in the price list.
The next m lines each describe another mouse. The i-th line contains first integer vali (1 β€ vali β€ 109) β the cost of the i-th mouse, then the type of port (USB or PS/2) that is required to plug the mouse in.
Output
Output two integers separated by space β the number of equipped computers and the total cost of the mouses you will buy.
Example
Input
2 1 1
4
5 USB
6 PS/2
3 PS/2
7 PS/2
Output
3 14
Note
In the first example you can buy the first three mouses. This way you will equip one of the computers that has only a USB port with a USB mouse, and the two PS/2 mouses you will plug into the computer with PS/2 port and the computer with both ports.
Submitted Solution:
```
a,b,k=map(int, input().split())
A=[]
B=[]
for _ in range(int(input())):
s=input()
if s[len(s)-1]=='2':
B.append(int(s[0]))
else:
A.append(int(s[0]))
B.sort()
A.sort()
c=0
ans=0
T=[]
if len(A)<=a:
c+=len(A)
ans+=sum(A)
else:
c+=a
ans+=sum(A[0:a])
T+=A[a:len(A)]
if len(B)<=b:
c+=len(B)
ans+=sum(B)
else:
c+=b
ans+=sum(B[0:b])
T+=B[b:len(B)]
T.sort()
if len(T)<=k:
c+=len(T)
ans+=sum(T)
else:
c+=k
ans+=sum(T[0:k])
print(c,ans)
``` | instruction | 0 | 50,139 | 10 | 100,278 |
No | output | 1 | 50,139 | 10 | 100,279 |
Provide a correct Python 3 solution for this coding contest problem.
There are N gems. The value of the i-th gem is V_i.
You will choose some of these gems, possibly all or none, and get them.
However, you need to pay a cost of C_i to get the i-th gem.
Let X be the sum of the values of the gems obtained, and Y be the sum of the costs paid.
Find the maximum possible value of X-Y.
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq C_i, V_i \leq 50
Input
Input is given from Standard Input in the following format:
N
V_1 V_2 ... V_N
C_1 C_2 ... C_N
Output
Print the maximum possible value of X-Y.
Examples
Input
3
10 2 5
6 3 4
Output
5
Input
4
13 21 6 19
11 30 6 15
Output
6
Input
1
1
50
Output
0 | instruction | 0 | 50,313 | 10 | 100,626 |
"Correct Solution:
```
n=int(input())
aa=list(map(int,input().split()))
bb=list(map(int,input().split()))
ans=0
for a,b in zip(aa,bb):
if a-b>0:
ans+=a-b
print(ans)
``` | output | 1 | 50,313 | 10 | 100,627 |
Provide a correct Python 3 solution for this coding contest problem.
There are N gems. The value of the i-th gem is V_i.
You will choose some of these gems, possibly all or none, and get them.
However, you need to pay a cost of C_i to get the i-th gem.
Let X be the sum of the values of the gems obtained, and Y be the sum of the costs paid.
Find the maximum possible value of X-Y.
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq C_i, V_i \leq 50
Input
Input is given from Standard Input in the following format:
N
V_1 V_2 ... V_N
C_1 C_2 ... C_N
Output
Print the maximum possible value of X-Y.
Examples
Input
3
10 2 5
6 3 4
Output
5
Input
4
13 21 6 19
11 30 6 15
Output
6
Input
1
1
50
Output
0 | instruction | 0 | 50,314 | 10 | 100,628 |
"Correct Solution:
```
n=int(input())
v=list(map(int,input().split()))
c=list(map(int,input().split()))
d=0
for a,b in zip(v,c):
if a-b>0:
d+=a-b
print(d)
``` | output | 1 | 50,314 | 10 | 100,629 |
Provide a correct Python 3 solution for this coding contest problem.
There are N gems. The value of the i-th gem is V_i.
You will choose some of these gems, possibly all or none, and get them.
However, you need to pay a cost of C_i to get the i-th gem.
Let X be the sum of the values of the gems obtained, and Y be the sum of the costs paid.
Find the maximum possible value of X-Y.
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq C_i, V_i \leq 50
Input
Input is given from Standard Input in the following format:
N
V_1 V_2 ... V_N
C_1 C_2 ... C_N
Output
Print the maximum possible value of X-Y.
Examples
Input
3
10 2 5
6 3 4
Output
5
Input
4
13 21 6 19
11 30 6 15
Output
6
Input
1
1
50
Output
0 | instruction | 0 | 50,315 | 10 | 100,630 |
"Correct Solution:
```
n=int(input())
v=list(map(int,input().split()))
c=list(map(int,input().split()))
sum=0
for i in range(n):
sum+=max(v[i]-c[i],0)
print(sum)
``` | output | 1 | 50,315 | 10 | 100,631 |
Provide a correct Python 3 solution for this coding contest problem.
There are N gems. The value of the i-th gem is V_i.
You will choose some of these gems, possibly all or none, and get them.
However, you need to pay a cost of C_i to get the i-th gem.
Let X be the sum of the values of the gems obtained, and Y be the sum of the costs paid.
Find the maximum possible value of X-Y.
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq C_i, V_i \leq 50
Input
Input is given from Standard Input in the following format:
N
V_1 V_2 ... V_N
C_1 C_2 ... C_N
Output
Print the maximum possible value of X-Y.
Examples
Input
3
10 2 5
6 3 4
Output
5
Input
4
13 21 6 19
11 30 6 15
Output
6
Input
1
1
50
Output
0 | instruction | 0 | 50,316 | 10 | 100,632 |
"Correct Solution:
```
N=int(input())
V=input().split(" ")
C=input().split(" ")
ans=0
for i in range(N):
ans+=max(0, int(V[i])-int(C[i]))
print(int(ans))
``` | output | 1 | 50,316 | 10 | 100,633 |
Provide a correct Python 3 solution for this coding contest problem.
There are N gems. The value of the i-th gem is V_i.
You will choose some of these gems, possibly all or none, and get them.
However, you need to pay a cost of C_i to get the i-th gem.
Let X be the sum of the values of the gems obtained, and Y be the sum of the costs paid.
Find the maximum possible value of X-Y.
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq C_i, V_i \leq 50
Input
Input is given from Standard Input in the following format:
N
V_1 V_2 ... V_N
C_1 C_2 ... C_N
Output
Print the maximum possible value of X-Y.
Examples
Input
3
10 2 5
6 3 4
Output
5
Input
4
13 21 6 19
11 30 6 15
Output
6
Input
1
1
50
Output
0 | instruction | 0 | 50,317 | 10 | 100,634 |
"Correct Solution:
```
N = int(input())
V = list(map(int,input().split()))
C = list(map(int,input().split()))
ans = 0
for i,j in zip(V,C):
ans += max(0,i-j)
print(ans)
``` | output | 1 | 50,317 | 10 | 100,635 |
Provide a correct Python 3 solution for this coding contest problem.
There are N gems. The value of the i-th gem is V_i.
You will choose some of these gems, possibly all or none, and get them.
However, you need to pay a cost of C_i to get the i-th gem.
Let X be the sum of the values of the gems obtained, and Y be the sum of the costs paid.
Find the maximum possible value of X-Y.
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq C_i, V_i \leq 50
Input
Input is given from Standard Input in the following format:
N
V_1 V_2 ... V_N
C_1 C_2 ... C_N
Output
Print the maximum possible value of X-Y.
Examples
Input
3
10 2 5
6 3 4
Output
5
Input
4
13 21 6 19
11 30 6 15
Output
6
Input
1
1
50
Output
0 | instruction | 0 | 50,318 | 10 | 100,636 |
"Correct Solution:
```
N = int(input())
V = list(map(int, input().split()))
C = list(map(int, input().split()))
print(sum([max(v-c, 0) for v, c in zip(V, C)]))
``` | output | 1 | 50,318 | 10 | 100,637 |
Provide a correct Python 3 solution for this coding contest problem.
There are N gems. The value of the i-th gem is V_i.
You will choose some of these gems, possibly all or none, and get them.
However, you need to pay a cost of C_i to get the i-th gem.
Let X be the sum of the values of the gems obtained, and Y be the sum of the costs paid.
Find the maximum possible value of X-Y.
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq C_i, V_i \leq 50
Input
Input is given from Standard Input in the following format:
N
V_1 V_2 ... V_N
C_1 C_2 ... C_N
Output
Print the maximum possible value of X-Y.
Examples
Input
3
10 2 5
6 3 4
Output
5
Input
4
13 21 6 19
11 30 6 15
Output
6
Input
1
1
50
Output
0 | instruction | 0 | 50,319 | 10 | 100,638 |
"Correct Solution:
```
N = input()
VS = [int(i) for i in input().split()]
CS = [int(i) for i in input().split()]
print(sum([a - b for a, b in zip(VS, CS) if a - b > 0]))
``` | output | 1 | 50,319 | 10 | 100,639 |
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