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 |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Berland it is the holiday of equality. In honor of the holiday the king decided to equalize the welfare of all citizens in Berland by the expense of the state treasury.
Totally in Berland there are n citizens, the welfare of each of them is estimated as the integer in ai burles (burle is the currency in Berland).
You are the royal treasurer, which needs to count the minimum charges of the kingdom on the king's present. The king can only give money, he hasn't a power to take away them.
Input
The first line contains the integer n (1 ≤ n ≤ 100) — the number of citizens in the kingdom.
The second line contains n integers a1, a2, ..., an, where ai (0 ≤ ai ≤ 106) — the welfare of the i-th citizen.
Output
In the only line print the integer S — the minimum number of burles which are had to spend.
Examples
Input
5
0 1 2 3 4
Output
10
Input
5
1 1 0 1 1
Output
1
Input
3
1 3 1
Output
4
Input
1
12
Output
0
Note
In the first example if we add to the first citizen 4 burles, to the second 3, to the third 2 and to the fourth 1, then the welfare of all citizens will equal 4.
In the second example it is enough to give one burle to the third citizen.
In the third example it is necessary to give two burles to the first and the third citizens to make the welfare of citizens equal 3.
In the fourth example it is possible to give nothing to everyone because all citizens have 12 burles.
Submitted Solution:
```
num = int(input())
l = input().split(" ")
m = max(l)
s = 0
for i in range(num) :
if (int(l[i]) != m) :
s += int(m) - int(l[i])
print(s)
``` | instruction | 0 | 73,821 | 10 | 147,642 |
No | output | 1 | 73,821 | 10 | 147,643 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Berland it is the holiday of equality. In honor of the holiday the king decided to equalize the welfare of all citizens in Berland by the expense of the state treasury.
Totally in Berland there are n citizens, the welfare of each of them is estimated as the integer in ai burles (burle is the currency in Berland).
You are the royal treasurer, which needs to count the minimum charges of the kingdom on the king's present. The king can only give money, he hasn't a power to take away them.
Input
The first line contains the integer n (1 ≤ n ≤ 100) — the number of citizens in the kingdom.
The second line contains n integers a1, a2, ..., an, where ai (0 ≤ ai ≤ 106) — the welfare of the i-th citizen.
Output
In the only line print the integer S — the minimum number of burles which are had to spend.
Examples
Input
5
0 1 2 3 4
Output
10
Input
5
1 1 0 1 1
Output
1
Input
3
1 3 1
Output
4
Input
1
12
Output
0
Note
In the first example if we add to the first citizen 4 burles, to the second 3, to the third 2 and to the fourth 1, then the welfare of all citizens will equal 4.
In the second example it is enough to give one burle to the third citizen.
In the third example it is necessary to give two burles to the first and the third citizens to make the welfare of citizens equal 3.
In the fourth example it is possible to give nothing to everyone because all citizens have 12 burles.
Submitted Solution:
```
num_citizens = int(input())
welfares = [int(a) for a in input().strip().split()]
welfares.sort()
maximum = welfares[-1]
total = maximum * num_citizens
print(sum(welfares) - total)
``` | instruction | 0 | 73,822 | 10 | 147,644 |
No | output | 1 | 73,822 | 10 | 147,645 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Berland it is the holiday of equality. In honor of the holiday the king decided to equalize the welfare of all citizens in Berland by the expense of the state treasury.
Totally in Berland there are n citizens, the welfare of each of them is estimated as the integer in ai burles (burle is the currency in Berland).
You are the royal treasurer, which needs to count the minimum charges of the kingdom on the king's present. The king can only give money, he hasn't a power to take away them.
Input
The first line contains the integer n (1 ≤ n ≤ 100) — the number of citizens in the kingdom.
The second line contains n integers a1, a2, ..., an, where ai (0 ≤ ai ≤ 106) — the welfare of the i-th citizen.
Output
In the only line print the integer S — the minimum number of burles which are had to spend.
Examples
Input
5
0 1 2 3 4
Output
10
Input
5
1 1 0 1 1
Output
1
Input
3
1 3 1
Output
4
Input
1
12
Output
0
Note
In the first example if we add to the first citizen 4 burles, to the second 3, to the third 2 and to the fourth 1, then the welfare of all citizens will equal 4.
In the second example it is enough to give one burle to the third citizen.
In the third example it is necessary to give two burles to the first and the third citizens to make the welfare of citizens equal 3.
In the fourth example it is possible to give nothing to everyone because all citizens have 12 burles.
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
S = 0
if len(a) == 1:
S = a[0]
else:
max_a = max(a)
for i in range(n):
diff = max_a - a[i]
if diff > 0:
S += diff
print (S)
``` | instruction | 0 | 73,823 | 10 | 147,646 |
No | output | 1 | 73,823 | 10 | 147,647 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi, who is A years old, is riding a Ferris wheel.
It costs B yen (B is an even number) to ride the Ferris wheel if you are 13 years old or older, but children between 6 and 12 years old (inclusive) can ride it for half the cost, and children who are 5 years old or younger are free of charge. (Yen is the currency of Japan.)
Find the cost of the Ferris wheel for Takahashi.
Constraints
* 0 ≤ A ≤ 100
* 2 ≤ B ≤ 1000
* B is an even number.
Input
Input is given from Standard Input in the following format:
A B
Output
Print the cost of the Ferris wheel for Takahashi.
Examples
Input
30 100
Output
100
Input
12 100
Output
50
Input
0 100
Output
0 | instruction | 0 | 73,992 | 10 | 147,984 |
"Correct Solution:
```
a,b=map(int,input().split())
if 5<a<13:print(b//2)
elif 12<a:print(b)
else:print(0)
``` | output | 1 | 73,992 | 10 | 147,985 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi, who is A years old, is riding a Ferris wheel.
It costs B yen (B is an even number) to ride the Ferris wheel if you are 13 years old or older, but children between 6 and 12 years old (inclusive) can ride it for half the cost, and children who are 5 years old or younger are free of charge. (Yen is the currency of Japan.)
Find the cost of the Ferris wheel for Takahashi.
Constraints
* 0 ≤ A ≤ 100
* 2 ≤ B ≤ 1000
* B is an even number.
Input
Input is given from Standard Input in the following format:
A B
Output
Print the cost of the Ferris wheel for Takahashi.
Examples
Input
30 100
Output
100
Input
12 100
Output
50
Input
0 100
Output
0 | instruction | 0 | 73,993 | 10 | 147,986 |
"Correct Solution:
```
a, b = [int(x) for x in input().split()]
if a < 6:
b = 0
elif a < 13:
b //= 2
print(b)
``` | output | 1 | 73,993 | 10 | 147,987 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi, who is A years old, is riding a Ferris wheel.
It costs B yen (B is an even number) to ride the Ferris wheel if you are 13 years old or older, but children between 6 and 12 years old (inclusive) can ride it for half the cost, and children who are 5 years old or younger are free of charge. (Yen is the currency of Japan.)
Find the cost of the Ferris wheel for Takahashi.
Constraints
* 0 ≤ A ≤ 100
* 2 ≤ B ≤ 1000
* B is an even number.
Input
Input is given from Standard Input in the following format:
A B
Output
Print the cost of the Ferris wheel for Takahashi.
Examples
Input
30 100
Output
100
Input
12 100
Output
50
Input
0 100
Output
0 | instruction | 0 | 73,994 | 10 | 147,988 |
"Correct Solution:
```
a, b=map(int, input().split())
if(a >=6):
print(b if a>=13 else b//2)
else:
print(0)
``` | output | 1 | 73,994 | 10 | 147,989 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi, who is A years old, is riding a Ferris wheel.
It costs B yen (B is an even number) to ride the Ferris wheel if you are 13 years old or older, but children between 6 and 12 years old (inclusive) can ride it for half the cost, and children who are 5 years old or younger are free of charge. (Yen is the currency of Japan.)
Find the cost of the Ferris wheel for Takahashi.
Constraints
* 0 ≤ A ≤ 100
* 2 ≤ B ≤ 1000
* B is an even number.
Input
Input is given from Standard Input in the following format:
A B
Output
Print the cost of the Ferris wheel for Takahashi.
Examples
Input
30 100
Output
100
Input
12 100
Output
50
Input
0 100
Output
0 | instruction | 0 | 73,995 | 10 | 147,990 |
"Correct Solution:
```
print((lambda a, b: b if a > 12 else b // 2 if a > 5 else 0)(*map(int, input().split())))
``` | output | 1 | 73,995 | 10 | 147,991 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi, who is A years old, is riding a Ferris wheel.
It costs B yen (B is an even number) to ride the Ferris wheel if you are 13 years old or older, but children between 6 and 12 years old (inclusive) can ride it for half the cost, and children who are 5 years old or younger are free of charge. (Yen is the currency of Japan.)
Find the cost of the Ferris wheel for Takahashi.
Constraints
* 0 ≤ A ≤ 100
* 2 ≤ B ≤ 1000
* B is an even number.
Input
Input is given from Standard Input in the following format:
A B
Output
Print the cost of the Ferris wheel for Takahashi.
Examples
Input
30 100
Output
100
Input
12 100
Output
50
Input
0 100
Output
0 | instruction | 0 | 73,996 | 10 | 147,992 |
"Correct Solution:
```
a,b=list((map(int,input().split())));print(b) if a>12 else print(int(b/2*bool(int(a/6))))
``` | output | 1 | 73,996 | 10 | 147,993 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi, who is A years old, is riding a Ferris wheel.
It costs B yen (B is an even number) to ride the Ferris wheel if you are 13 years old or older, but children between 6 and 12 years old (inclusive) can ride it for half the cost, and children who are 5 years old or younger are free of charge. (Yen is the currency of Japan.)
Find the cost of the Ferris wheel for Takahashi.
Constraints
* 0 ≤ A ≤ 100
* 2 ≤ B ≤ 1000
* B is an even number.
Input
Input is given from Standard Input in the following format:
A B
Output
Print the cost of the Ferris wheel for Takahashi.
Examples
Input
30 100
Output
100
Input
12 100
Output
50
Input
0 100
Output
0 | instruction | 0 | 73,997 | 10 | 147,994 |
"Correct Solution:
```
A,B=map(int,input().split());print(B if A>=13 else B//2 if 12 >= A >= 6 else 0)
``` | output | 1 | 73,997 | 10 | 147,995 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi, who is A years old, is riding a Ferris wheel.
It costs B yen (B is an even number) to ride the Ferris wheel if you are 13 years old or older, but children between 6 and 12 years old (inclusive) can ride it for half the cost, and children who are 5 years old or younger are free of charge. (Yen is the currency of Japan.)
Find the cost of the Ferris wheel for Takahashi.
Constraints
* 0 ≤ A ≤ 100
* 2 ≤ B ≤ 1000
* B is an even number.
Input
Input is given from Standard Input in the following format:
A B
Output
Print the cost of the Ferris wheel for Takahashi.
Examples
Input
30 100
Output
100
Input
12 100
Output
50
Input
0 100
Output
0 | instruction | 0 | 73,998 | 10 | 147,996 |
"Correct Solution:
```
x , y = map(int, input().split())
ans = y if x >= 13 else y//2 if 6<=x<=12 else 0
print(ans)
``` | output | 1 | 73,998 | 10 | 147,997 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi, who is A years old, is riding a Ferris wheel.
It costs B yen (B is an even number) to ride the Ferris wheel if you are 13 years old or older, but children between 6 and 12 years old (inclusive) can ride it for half the cost, and children who are 5 years old or younger are free of charge. (Yen is the currency of Japan.)
Find the cost of the Ferris wheel for Takahashi.
Constraints
* 0 ≤ A ≤ 100
* 2 ≤ B ≤ 1000
* B is an even number.
Input
Input is given from Standard Input in the following format:
A B
Output
Print the cost of the Ferris wheel for Takahashi.
Examples
Input
30 100
Output
100
Input
12 100
Output
50
Input
0 100
Output
0 | instruction | 0 | 73,999 | 10 | 147,998 |
"Correct Solution:
```
a,b=map(int,input().split())
if(6<=a<=12):
b//=2
elif(a<6):
b=0
print(b)
``` | output | 1 | 73,999 | 10 | 147,999 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi, who is A years old, is riding a Ferris wheel.
It costs B yen (B is an even number) to ride the Ferris wheel if you are 13 years old or older, but children between 6 and 12 years old (inclusive) can ride it for half the cost, and children who are 5 years old or younger are free of charge. (Yen is the currency of Japan.)
Find the cost of the Ferris wheel for Takahashi.
Constraints
* 0 ≤ A ≤ 100
* 2 ≤ B ≤ 1000
* B is an even number.
Input
Input is given from Standard Input in the following format:
A B
Output
Print the cost of the Ferris wheel for Takahashi.
Examples
Input
30 100
Output
100
Input
12 100
Output
50
Input
0 100
Output
0
Submitted Solution:
```
a,b=map(int,input().split())
print(b if a>=13 else (a>=6)*b//2)
``` | instruction | 0 | 74,000 | 10 | 148,000 |
Yes | output | 1 | 74,000 | 10 | 148,001 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi, who is A years old, is riding a Ferris wheel.
It costs B yen (B is an even number) to ride the Ferris wheel if you are 13 years old or older, but children between 6 and 12 years old (inclusive) can ride it for half the cost, and children who are 5 years old or younger are free of charge. (Yen is the currency of Japan.)
Find the cost of the Ferris wheel for Takahashi.
Constraints
* 0 ≤ A ≤ 100
* 2 ≤ B ≤ 1000
* B is an even number.
Input
Input is given from Standard Input in the following format:
A B
Output
Print the cost of the Ferris wheel for Takahashi.
Examples
Input
30 100
Output
100
Input
12 100
Output
50
Input
0 100
Output
0
Submitted Solution:
```
a,b=map(int, input().split())
print(b if a > 12 else int(b/2) if 6 <= a <= 12 else 0)
``` | instruction | 0 | 74,001 | 10 | 148,002 |
Yes | output | 1 | 74,001 | 10 | 148,003 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi, who is A years old, is riding a Ferris wheel.
It costs B yen (B is an even number) to ride the Ferris wheel if you are 13 years old or older, but children between 6 and 12 years old (inclusive) can ride it for half the cost, and children who are 5 years old or younger are free of charge. (Yen is the currency of Japan.)
Find the cost of the Ferris wheel for Takahashi.
Constraints
* 0 ≤ A ≤ 100
* 2 ≤ B ≤ 1000
* B is an even number.
Input
Input is given from Standard Input in the following format:
A B
Output
Print the cost of the Ferris wheel for Takahashi.
Examples
Input
30 100
Output
100
Input
12 100
Output
50
Input
0 100
Output
0
Submitted Solution:
```
A,B=map(int, input().split())
print(B if A>=13 else (B//2 if A>=6 else 0))
``` | instruction | 0 | 74,002 | 10 | 148,004 |
Yes | output | 1 | 74,002 | 10 | 148,005 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi, who is A years old, is riding a Ferris wheel.
It costs B yen (B is an even number) to ride the Ferris wheel if you are 13 years old or older, but children between 6 and 12 years old (inclusive) can ride it for half the cost, and children who are 5 years old or younger are free of charge. (Yen is the currency of Japan.)
Find the cost of the Ferris wheel for Takahashi.
Constraints
* 0 ≤ A ≤ 100
* 2 ≤ B ≤ 1000
* B is an even number.
Input
Input is given from Standard Input in the following format:
A B
Output
Print the cost of the Ferris wheel for Takahashi.
Examples
Input
30 100
Output
100
Input
12 100
Output
50
Input
0 100
Output
0
Submitted Solution:
```
a,b=map(int,input().split())
print(b if a>12 else 0 if a<6 else b//2)
``` | instruction | 0 | 74,003 | 10 | 148,006 |
Yes | output | 1 | 74,003 | 10 | 148,007 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi, who is A years old, is riding a Ferris wheel.
It costs B yen (B is an even number) to ride the Ferris wheel if you are 13 years old or older, but children between 6 and 12 years old (inclusive) can ride it for half the cost, and children who are 5 years old or younger are free of charge. (Yen is the currency of Japan.)
Find the cost of the Ferris wheel for Takahashi.
Constraints
* 0 ≤ A ≤ 100
* 2 ≤ B ≤ 1000
* B is an even number.
Input
Input is given from Standard Input in the following format:
A B
Output
Print the cost of the Ferris wheel for Takahashi.
Examples
Input
30 100
Output
100
Input
12 100
Output
50
Input
0 100
Output
0
Submitted Solution:
```
A=int(input())
B=int(input())
if A>=13: print(B)
elif A>=6: print(B/2)
else : print (0)
``` | instruction | 0 | 74,004 | 10 | 148,008 |
No | output | 1 | 74,004 | 10 | 148,009 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi, who is A years old, is riding a Ferris wheel.
It costs B yen (B is an even number) to ride the Ferris wheel if you are 13 years old or older, but children between 6 and 12 years old (inclusive) can ride it for half the cost, and children who are 5 years old or younger are free of charge. (Yen is the currency of Japan.)
Find the cost of the Ferris wheel for Takahashi.
Constraints
* 0 ≤ A ≤ 100
* 2 ≤ B ≤ 1000
* B is an even number.
Input
Input is given from Standard Input in the following format:
A B
Output
Print the cost of the Ferris wheel for Takahashi.
Examples
Input
30 100
Output
100
Input
12 100
Output
50
Input
0 100
Output
0
Submitted Solution:
```
N = int(input())
I = int(input())
if I <= 13:
print(N)
elif: 6 <= I <=12:
print(N/2)
elif: I <= 5
print("0")
``` | instruction | 0 | 74,005 | 10 | 148,010 |
No | output | 1 | 74,005 | 10 | 148,011 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi, who is A years old, is riding a Ferris wheel.
It costs B yen (B is an even number) to ride the Ferris wheel if you are 13 years old or older, but children between 6 and 12 years old (inclusive) can ride it for half the cost, and children who are 5 years old or younger are free of charge. (Yen is the currency of Japan.)
Find the cost of the Ferris wheel for Takahashi.
Constraints
* 0 ≤ A ≤ 100
* 2 ≤ B ≤ 1000
* B is an even number.
Input
Input is given from Standard Input in the following format:
A B
Output
Print the cost of the Ferris wheel for Takahashi.
Examples
Input
30 100
Output
100
Input
12 100
Output
50
Input
0 100
Output
0
Submitted Solution:
```
A,B=map(int,input().split())
if A>=13:
print(B)
elif A>5:
print(B/2)
else:
print(0)
``` | instruction | 0 | 74,006 | 10 | 148,012 |
No | output | 1 | 74,006 | 10 | 148,013 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi, who is A years old, is riding a Ferris wheel.
It costs B yen (B is an even number) to ride the Ferris wheel if you are 13 years old or older, but children between 6 and 12 years old (inclusive) can ride it for half the cost, and children who are 5 years old or younger are free of charge. (Yen is the currency of Japan.)
Find the cost of the Ferris wheel for Takahashi.
Constraints
* 0 ≤ A ≤ 100
* 2 ≤ B ≤ 1000
* B is an even number.
Input
Input is given from Standard Input in the following format:
A B
Output
Print the cost of the Ferris wheel for Takahashi.
Examples
Input
30 100
Output
100
Input
12 100
Output
50
Input
0 100
Output
0
Submitted Solution:
```
a,b = map(int,input().split())
if a >= 13:
print(b)
if 6<=a & a<=12:
print(b/2)
if a<=5:
print(0)
``` | instruction | 0 | 74,007 | 10 | 148,014 |
No | output | 1 | 74,007 | 10 | 148,015 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given four integers n, c_0, c_1 and h and a binary string s of length n.
A binary string is a string consisting of characters 0 and 1.
You can change any character of the string s (the string should be still binary after the change). You should pay h coins for each change.
After some changes (possibly zero) you want to buy the string. To buy the string you should buy all its characters. To buy the character 0 you should pay c_0 coins, to buy the character 1 you should pay c_1 coins.
Find the minimum number of coins needed to buy the string.
Input
The first line contains a single integer t (1 ≤ t ≤ 10) — the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of the description of each test case contains four integers n, c_{0}, c_{1}, h (1 ≤ n, c_{0}, c_{1}, h ≤ 1000).
The second line of the description of each test case contains the binary string s of length n.
Output
For each test case print a single integer — the minimum number of coins needed to buy the string.
Example
Input
6
3 1 1 1
100
5 10 100 1
01010
5 10 1 1
11111
5 1 10 1
11111
12 2 1 10
101110110101
2 100 1 10
00
Output
3
52
5
10
16
22
Note
In the first test case, you can buy all characters and pay 3 coins, because both characters 0 and 1 costs 1 coin.
In the second test case, you can firstly change 2-nd and 4-th symbols of the string from 1 to 0 and pay 2 coins for that. Your string will be 00000. After that, you can buy the string and pay 5 ⋅ 10 = 50 coins for that. The total number of coins paid will be 2 + 50 = 52. | instruction | 0 | 74,369 | 10 | 148,738 |
Tags: implementation, math
Correct Solution:
```
for _ in range(int(input())):
n,c0,c1,h=map(int,input().split())
s=input()
s0=s.count("0")
s1=s.count("1")
a1=(c0*s0)+(c1*s1)
a2=(c0*s0)+(s1*(c0+h))
a3=(c1*s1)+(s0*(c1+h))
ans=min(a1,a2,a3)
print(ans)
``` | output | 1 | 74,369 | 10 | 148,739 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given four integers n, c_0, c_1 and h and a binary string s of length n.
A binary string is a string consisting of characters 0 and 1.
You can change any character of the string s (the string should be still binary after the change). You should pay h coins for each change.
After some changes (possibly zero) you want to buy the string. To buy the string you should buy all its characters. To buy the character 0 you should pay c_0 coins, to buy the character 1 you should pay c_1 coins.
Find the minimum number of coins needed to buy the string.
Input
The first line contains a single integer t (1 ≤ t ≤ 10) — the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of the description of each test case contains four integers n, c_{0}, c_{1}, h (1 ≤ n, c_{0}, c_{1}, h ≤ 1000).
The second line of the description of each test case contains the binary string s of length n.
Output
For each test case print a single integer — the minimum number of coins needed to buy the string.
Example
Input
6
3 1 1 1
100
5 10 100 1
01010
5 10 1 1
11111
5 1 10 1
11111
12 2 1 10
101110110101
2 100 1 10
00
Output
3
52
5
10
16
22
Note
In the first test case, you can buy all characters and pay 3 coins, because both characters 0 and 1 costs 1 coin.
In the second test case, you can firstly change 2-nd and 4-th symbols of the string from 1 to 0 and pay 2 coins for that. Your string will be 00000. After that, you can buy the string and pay 5 ⋅ 10 = 50 coins for that. The total number of coins paid will be 2 + 50 = 52. | instruction | 0 | 74,370 | 10 | 148,740 |
Tags: implementation, math
Correct Solution:
```
for _ in range(int(input())):
n,c0,c1,h=map(int,input().split())
s=input()
if c0 < c1:
if (h+c0) >= c1:
flag=False
else:
c1=h+c0
elif c0 > c1:
if (h+c1) >= c0:
flag=False
else:
c0=h+c1
else:
pass
ans=0
for i in s:
if(i=='0'):
ans=ans+c0
else:
ans=ans+c1
print(ans)
``` | output | 1 | 74,370 | 10 | 148,741 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given four integers n, c_0, c_1 and h and a binary string s of length n.
A binary string is a string consisting of characters 0 and 1.
You can change any character of the string s (the string should be still binary after the change). You should pay h coins for each change.
After some changes (possibly zero) you want to buy the string. To buy the string you should buy all its characters. To buy the character 0 you should pay c_0 coins, to buy the character 1 you should pay c_1 coins.
Find the minimum number of coins needed to buy the string.
Input
The first line contains a single integer t (1 ≤ t ≤ 10) — the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of the description of each test case contains four integers n, c_{0}, c_{1}, h (1 ≤ n, c_{0}, c_{1}, h ≤ 1000).
The second line of the description of each test case contains the binary string s of length n.
Output
For each test case print a single integer — the minimum number of coins needed to buy the string.
Example
Input
6
3 1 1 1
100
5 10 100 1
01010
5 10 1 1
11111
5 1 10 1
11111
12 2 1 10
101110110101
2 100 1 10
00
Output
3
52
5
10
16
22
Note
In the first test case, you can buy all characters and pay 3 coins, because both characters 0 and 1 costs 1 coin.
In the second test case, you can firstly change 2-nd and 4-th symbols of the string from 1 to 0 and pay 2 coins for that. Your string will be 00000. After that, you can buy the string and pay 5 ⋅ 10 = 50 coins for that. The total number of coins paid will be 2 + 50 = 52. | instruction | 0 | 74,371 | 10 | 148,742 |
Tags: implementation, math
Correct Solution:
```
t = int(input())
for j in range(t):
n, c0, c1, h = map(int, input().split())
b = input()
summ = 0
if c0 > c1 + h:
for i in range(n):
if b[i] == '0':
summ += c1
summ += h
else:
summ += c1
elif c1 > c0 + h:
for i in range(n):
if b[i] == '1':
summ += c0
summ += h
else:
summ += c0
else:
for i in range(n):
if b[i] == '0':
summ += c0
else:
summ += c1
print(summ)
``` | output | 1 | 74,371 | 10 | 148,743 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given four integers n, c_0, c_1 and h and a binary string s of length n.
A binary string is a string consisting of characters 0 and 1.
You can change any character of the string s (the string should be still binary after the change). You should pay h coins for each change.
After some changes (possibly zero) you want to buy the string. To buy the string you should buy all its characters. To buy the character 0 you should pay c_0 coins, to buy the character 1 you should pay c_1 coins.
Find the minimum number of coins needed to buy the string.
Input
The first line contains a single integer t (1 ≤ t ≤ 10) — the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of the description of each test case contains four integers n, c_{0}, c_{1}, h (1 ≤ n, c_{0}, c_{1}, h ≤ 1000).
The second line of the description of each test case contains the binary string s of length n.
Output
For each test case print a single integer — the minimum number of coins needed to buy the string.
Example
Input
6
3 1 1 1
100
5 10 100 1
01010
5 10 1 1
11111
5 1 10 1
11111
12 2 1 10
101110110101
2 100 1 10
00
Output
3
52
5
10
16
22
Note
In the first test case, you can buy all characters and pay 3 coins, because both characters 0 and 1 costs 1 coin.
In the second test case, you can firstly change 2-nd and 4-th symbols of the string from 1 to 0 and pay 2 coins for that. Your string will be 00000. After that, you can buy the string and pay 5 ⋅ 10 = 50 coins for that. The total number of coins paid will be 2 + 50 = 52. | instruction | 0 | 74,372 | 10 | 148,744 |
Tags: implementation, math
Correct Solution:
```
for _ in range(int(input())):
a,b,c,d = map(int,input().split())
n = input()
ans = 0
for i in n:
if i=="0":
ans+=min(d+c,b)
elif i=="1":
ans+=min(d+b,c)
print(ans)
``` | output | 1 | 74,372 | 10 | 148,745 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given four integers n, c_0, c_1 and h and a binary string s of length n.
A binary string is a string consisting of characters 0 and 1.
You can change any character of the string s (the string should be still binary after the change). You should pay h coins for each change.
After some changes (possibly zero) you want to buy the string. To buy the string you should buy all its characters. To buy the character 0 you should pay c_0 coins, to buy the character 1 you should pay c_1 coins.
Find the minimum number of coins needed to buy the string.
Input
The first line contains a single integer t (1 ≤ t ≤ 10) — the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of the description of each test case contains four integers n, c_{0}, c_{1}, h (1 ≤ n, c_{0}, c_{1}, h ≤ 1000).
The second line of the description of each test case contains the binary string s of length n.
Output
For each test case print a single integer — the minimum number of coins needed to buy the string.
Example
Input
6
3 1 1 1
100
5 10 100 1
01010
5 10 1 1
11111
5 1 10 1
11111
12 2 1 10
101110110101
2 100 1 10
00
Output
3
52
5
10
16
22
Note
In the first test case, you can buy all characters and pay 3 coins, because both characters 0 and 1 costs 1 coin.
In the second test case, you can firstly change 2-nd and 4-th symbols of the string from 1 to 0 and pay 2 coins for that. Your string will be 00000. After that, you can buy the string and pay 5 ⋅ 10 = 50 coins for that. The total number of coins paid will be 2 + 50 = 52. | instruction | 0 | 74,373 | 10 | 148,746 |
Tags: implementation, math
Correct Solution:
```
for t in range(int(input())):
n,c0,c1,h=map(int,input().split())
s=input()
ze=s.count("0")
on=s.count("1")
noch=ze*c0+on*c1
ch1=ze*(h)+n*c1
ch2=on*h+n*c0
print(min(noch,ch1,ch2))
``` | output | 1 | 74,373 | 10 | 148,747 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given four integers n, c_0, c_1 and h and a binary string s of length n.
A binary string is a string consisting of characters 0 and 1.
You can change any character of the string s (the string should be still binary after the change). You should pay h coins for each change.
After some changes (possibly zero) you want to buy the string. To buy the string you should buy all its characters. To buy the character 0 you should pay c_0 coins, to buy the character 1 you should pay c_1 coins.
Find the minimum number of coins needed to buy the string.
Input
The first line contains a single integer t (1 ≤ t ≤ 10) — the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of the description of each test case contains four integers n, c_{0}, c_{1}, h (1 ≤ n, c_{0}, c_{1}, h ≤ 1000).
The second line of the description of each test case contains the binary string s of length n.
Output
For each test case print a single integer — the minimum number of coins needed to buy the string.
Example
Input
6
3 1 1 1
100
5 10 100 1
01010
5 10 1 1
11111
5 1 10 1
11111
12 2 1 10
101110110101
2 100 1 10
00
Output
3
52
5
10
16
22
Note
In the first test case, you can buy all characters and pay 3 coins, because both characters 0 and 1 costs 1 coin.
In the second test case, you can firstly change 2-nd and 4-th symbols of the string from 1 to 0 and pay 2 coins for that. Your string will be 00000. After that, you can buy the string and pay 5 ⋅ 10 = 50 coins for that. The total number of coins paid will be 2 + 50 = 52. | instruction | 0 | 74,374 | 10 | 148,748 |
Tags: implementation, math
Correct Solution:
```
t = int(input())
for _ in range(t):
n, c0, c1, h = list(map(int, input().split()))
s = input()
cost = 0
if c1 == c0:
cost += c0 * n
elif c0 > c1:
cost += c1 * s.count('1')
if c0 > h+c1:
cost += (h+c1)*s.count('0')
else:
cost+= (c0)*s.count('0')
else:
cost+=c0*s.count('0')
if c1 > h+c0:
cost += (h+c0)*s.count('1')
else:
cost+= (c1)*s.count('1')
print(cost)
``` | output | 1 | 74,374 | 10 | 148,749 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given four integers n, c_0, c_1 and h and a binary string s of length n.
A binary string is a string consisting of characters 0 and 1.
You can change any character of the string s (the string should be still binary after the change). You should pay h coins for each change.
After some changes (possibly zero) you want to buy the string. To buy the string you should buy all its characters. To buy the character 0 you should pay c_0 coins, to buy the character 1 you should pay c_1 coins.
Find the minimum number of coins needed to buy the string.
Input
The first line contains a single integer t (1 ≤ t ≤ 10) — the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of the description of each test case contains four integers n, c_{0}, c_{1}, h (1 ≤ n, c_{0}, c_{1}, h ≤ 1000).
The second line of the description of each test case contains the binary string s of length n.
Output
For each test case print a single integer — the minimum number of coins needed to buy the string.
Example
Input
6
3 1 1 1
100
5 10 100 1
01010
5 10 1 1
11111
5 1 10 1
11111
12 2 1 10
101110110101
2 100 1 10
00
Output
3
52
5
10
16
22
Note
In the first test case, you can buy all characters and pay 3 coins, because both characters 0 and 1 costs 1 coin.
In the second test case, you can firstly change 2-nd and 4-th symbols of the string from 1 to 0 and pay 2 coins for that. Your string will be 00000. After that, you can buy the string and pay 5 ⋅ 10 = 50 coins for that. The total number of coins paid will be 2 + 50 = 52. | instruction | 0 | 74,375 | 10 | 148,750 |
Tags: implementation, math
Correct Solution:
```
for t in range(int(input())):
n, c0, c1, h = map(int, input().split())
s = input()
_0 = s.count('0')
_1 = s.count('1')
print(min(c0 * _0 + c1 * _1, c0 * len(s) + h * _1, c1 * len(s) + h * _0))
``` | output | 1 | 74,375 | 10 | 148,751 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given four integers n, c_0, c_1 and h and a binary string s of length n.
A binary string is a string consisting of characters 0 and 1.
You can change any character of the string s (the string should be still binary after the change). You should pay h coins for each change.
After some changes (possibly zero) you want to buy the string. To buy the string you should buy all its characters. To buy the character 0 you should pay c_0 coins, to buy the character 1 you should pay c_1 coins.
Find the minimum number of coins needed to buy the string.
Input
The first line contains a single integer t (1 ≤ t ≤ 10) — the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of the description of each test case contains four integers n, c_{0}, c_{1}, h (1 ≤ n, c_{0}, c_{1}, h ≤ 1000).
The second line of the description of each test case contains the binary string s of length n.
Output
For each test case print a single integer — the minimum number of coins needed to buy the string.
Example
Input
6
3 1 1 1
100
5 10 100 1
01010
5 10 1 1
11111
5 1 10 1
11111
12 2 1 10
101110110101
2 100 1 10
00
Output
3
52
5
10
16
22
Note
In the first test case, you can buy all characters and pay 3 coins, because both characters 0 and 1 costs 1 coin.
In the second test case, you can firstly change 2-nd and 4-th symbols of the string from 1 to 0 and pay 2 coins for that. Your string will be 00000. After that, you can buy the string and pay 5 ⋅ 10 = 50 coins for that. The total number of coins paid will be 2 + 50 = 52. | instruction | 0 | 74,376 | 10 | 148,752 |
Tags: implementation, math
Correct Solution:
```
n=int(input())
for i in range(n):
n1,c0,c1,h=map(int,input().split())
s=input()
cnt0=s.count('0')
cnt1=s.count('1')
if c0==c1:
print(c0*n1)
elif c0<c1:
s0=(c0*n1)+(h*cnt1)
s00=(c0*cnt0)+(c1*cnt1)
print(min(s0,s00))
elif c0>c1:
s1=(c1*n1)+(h*cnt0)
s11=(c1*cnt1)+(c0*cnt0)
print(min(s1,s11))
``` | output | 1 | 74,376 | 10 | 148,753 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given four integers n, c_0, c_1 and h and a binary string s of length n.
A binary string is a string consisting of characters 0 and 1.
You can change any character of the string s (the string should be still binary after the change). You should pay h coins for each change.
After some changes (possibly zero) you want to buy the string. To buy the string you should buy all its characters. To buy the character 0 you should pay c_0 coins, to buy the character 1 you should pay c_1 coins.
Find the minimum number of coins needed to buy the string.
Input
The first line contains a single integer t (1 ≤ t ≤ 10) — the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of the description of each test case contains four integers n, c_{0}, c_{1}, h (1 ≤ n, c_{0}, c_{1}, h ≤ 1000).
The second line of the description of each test case contains the binary string s of length n.
Output
For each test case print a single integer — the minimum number of coins needed to buy the string.
Example
Input
6
3 1 1 1
100
5 10 100 1
01010
5 10 1 1
11111
5 1 10 1
11111
12 2 1 10
101110110101
2 100 1 10
00
Output
3
52
5
10
16
22
Note
In the first test case, you can buy all characters and pay 3 coins, because both characters 0 and 1 costs 1 coin.
In the second test case, you can firstly change 2-nd and 4-th symbols of the string from 1 to 0 and pay 2 coins for that. Your string will be 00000. After that, you can buy the string and pay 5 ⋅ 10 = 50 coins for that. The total number of coins paid will be 2 + 50 = 52.
Submitted Solution:
```
from collections import Counter
def fun(n,c0,c1,h,st):
dct=Counter(st)
# h for each charecter change
# c0 for buying 0
# c1 for buying 1
count1= dct.get('1') if dct.get('1')!=None else 0
count0= dct.get('0') if dct.get('0')!=None else 0
if(c0>c1):
print(min(((count1+count0)*c1 + count0*h),(count1*c1+count0*c0)))
else:
print(min(((count1+count0)*c0 + count1*h),(count1*c1+count0*c0)))
T= int(input())
for i in range(T):
# t= int(input())
n,c0,c1,h= list(map(int,input().split()))
st=input()
fun(n,c0,c1,h,st)
``` | instruction | 0 | 74,377 | 10 | 148,754 |
Yes | output | 1 | 74,377 | 10 | 148,755 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given four integers n, c_0, c_1 and h and a binary string s of length n.
A binary string is a string consisting of characters 0 and 1.
You can change any character of the string s (the string should be still binary after the change). You should pay h coins for each change.
After some changes (possibly zero) you want to buy the string. To buy the string you should buy all its characters. To buy the character 0 you should pay c_0 coins, to buy the character 1 you should pay c_1 coins.
Find the minimum number of coins needed to buy the string.
Input
The first line contains a single integer t (1 ≤ t ≤ 10) — the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of the description of each test case contains four integers n, c_{0}, c_{1}, h (1 ≤ n, c_{0}, c_{1}, h ≤ 1000).
The second line of the description of each test case contains the binary string s of length n.
Output
For each test case print a single integer — the minimum number of coins needed to buy the string.
Example
Input
6
3 1 1 1
100
5 10 100 1
01010
5 10 1 1
11111
5 1 10 1
11111
12 2 1 10
101110110101
2 100 1 10
00
Output
3
52
5
10
16
22
Note
In the first test case, you can buy all characters and pay 3 coins, because both characters 0 and 1 costs 1 coin.
In the second test case, you can firstly change 2-nd and 4-th symbols of the string from 1 to 0 and pay 2 coins for that. Your string will be 00000. After that, you can buy the string and pay 5 ⋅ 10 = 50 coins for that. The total number of coins paid will be 2 + 50 = 52.
Submitted Solution:
```
t=int(input())
for _ in range(t):
n,c0,c1,h=list(map(int,input().split()))
s=input()
n0=s.count('0')
n1=n-n0
if abs(c0-c1)<=h:
print(n0*c0+n1*c1)
else:
if c0>c1:
print(n*c1+n0*h)
else:
print(n*c0+n1*h)
``` | instruction | 0 | 74,378 | 10 | 148,756 |
Yes | output | 1 | 74,378 | 10 | 148,757 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given four integers n, c_0, c_1 and h and a binary string s of length n.
A binary string is a string consisting of characters 0 and 1.
You can change any character of the string s (the string should be still binary after the change). You should pay h coins for each change.
After some changes (possibly zero) you want to buy the string. To buy the string you should buy all its characters. To buy the character 0 you should pay c_0 coins, to buy the character 1 you should pay c_1 coins.
Find the minimum number of coins needed to buy the string.
Input
The first line contains a single integer t (1 ≤ t ≤ 10) — the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of the description of each test case contains four integers n, c_{0}, c_{1}, h (1 ≤ n, c_{0}, c_{1}, h ≤ 1000).
The second line of the description of each test case contains the binary string s of length n.
Output
For each test case print a single integer — the minimum number of coins needed to buy the string.
Example
Input
6
3 1 1 1
100
5 10 100 1
01010
5 10 1 1
11111
5 1 10 1
11111
12 2 1 10
101110110101
2 100 1 10
00
Output
3
52
5
10
16
22
Note
In the first test case, you can buy all characters and pay 3 coins, because both characters 0 and 1 costs 1 coin.
In the second test case, you can firstly change 2-nd and 4-th symbols of the string from 1 to 0 and pay 2 coins for that. Your string will be 00000. After that, you can buy the string and pay 5 ⋅ 10 = 50 coins for that. The total number of coins paid will be 2 + 50 = 52.
Submitted Solution:
```
t = int(input())
for i in range(t):
n, c0, c1, h = [int(elem) for elem in input().split(' ')]
s = input()
if c0 > c1:
if c0 - c1 > h:
print(h * s.count('0') + n*c1)
else:
print(c0*s.count('0') + c1*s.count('1'))
elif c1 > c0:
if c1 - c0 > h:
print(h * s.count('1') + n*c0)
else:
print(c0*s.count('0') + c1*s.count('1'))
else:
print(c0*s.count('0') + c1*s.count('1'))
``` | instruction | 0 | 74,379 | 10 | 148,758 |
Yes | output | 1 | 74,379 | 10 | 148,759 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given four integers n, c_0, c_1 and h and a binary string s of length n.
A binary string is a string consisting of characters 0 and 1.
You can change any character of the string s (the string should be still binary after the change). You should pay h coins for each change.
After some changes (possibly zero) you want to buy the string. To buy the string you should buy all its characters. To buy the character 0 you should pay c_0 coins, to buy the character 1 you should pay c_1 coins.
Find the minimum number of coins needed to buy the string.
Input
The first line contains a single integer t (1 ≤ t ≤ 10) — the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of the description of each test case contains four integers n, c_{0}, c_{1}, h (1 ≤ n, c_{0}, c_{1}, h ≤ 1000).
The second line of the description of each test case contains the binary string s of length n.
Output
For each test case print a single integer — the minimum number of coins needed to buy the string.
Example
Input
6
3 1 1 1
100
5 10 100 1
01010
5 10 1 1
11111
5 1 10 1
11111
12 2 1 10
101110110101
2 100 1 10
00
Output
3
52
5
10
16
22
Note
In the first test case, you can buy all characters and pay 3 coins, because both characters 0 and 1 costs 1 coin.
In the second test case, you can firstly change 2-nd and 4-th symbols of the string from 1 to 0 and pay 2 coins for that. Your string will be 00000. After that, you can buy the string and pay 5 ⋅ 10 = 50 coins for that. The total number of coins paid will be 2 + 50 = 52.
Submitted Solution:
```
for x in range(int(input())):
n,a,b,h=map(int,input().split())
s=input()
money=0
for y in s:
if y=="0":
money+=min(a,h+b)
if y=="1":
money+=min(b,a+h)
print(money)
``` | instruction | 0 | 74,380 | 10 | 148,760 |
Yes | output | 1 | 74,380 | 10 | 148,761 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given four integers n, c_0, c_1 and h and a binary string s of length n.
A binary string is a string consisting of characters 0 and 1.
You can change any character of the string s (the string should be still binary after the change). You should pay h coins for each change.
After some changes (possibly zero) you want to buy the string. To buy the string you should buy all its characters. To buy the character 0 you should pay c_0 coins, to buy the character 1 you should pay c_1 coins.
Find the minimum number of coins needed to buy the string.
Input
The first line contains a single integer t (1 ≤ t ≤ 10) — the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of the description of each test case contains four integers n, c_{0}, c_{1}, h (1 ≤ n, c_{0}, c_{1}, h ≤ 1000).
The second line of the description of each test case contains the binary string s of length n.
Output
For each test case print a single integer — the minimum number of coins needed to buy the string.
Example
Input
6
3 1 1 1
100
5 10 100 1
01010
5 10 1 1
11111
5 1 10 1
11111
12 2 1 10
101110110101
2 100 1 10
00
Output
3
52
5
10
16
22
Note
In the first test case, you can buy all characters and pay 3 coins, because both characters 0 and 1 costs 1 coin.
In the second test case, you can firstly change 2-nd and 4-th symbols of the string from 1 to 0 and pay 2 coins for that. Your string will be 00000. After that, you can buy the string and pay 5 ⋅ 10 = 50 coins for that. The total number of coins paid will be 2 + 50 = 52.
Submitted Solution:
```
t=int(input())
for z in range(t):
n,c0,c1,h=map(int,input().split())
s=input()
total=0
count0=s.count("0")
count1=s.count("1")
if c0<c1+h:
s=s.replace("1","0")
total=total+(count1*h)
elif c1<c0+h:
s=s.replace("0","1")
total=total+(count0*h)
x0=s.count("0")
x1=s.count("1")
total=total+(x0*c0)+(x1*c1)
print(total)
``` | instruction | 0 | 74,381 | 10 | 148,762 |
No | output | 1 | 74,381 | 10 | 148,763 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given four integers n, c_0, c_1 and h and a binary string s of length n.
A binary string is a string consisting of characters 0 and 1.
You can change any character of the string s (the string should be still binary after the change). You should pay h coins for each change.
After some changes (possibly zero) you want to buy the string. To buy the string you should buy all its characters. To buy the character 0 you should pay c_0 coins, to buy the character 1 you should pay c_1 coins.
Find the minimum number of coins needed to buy the string.
Input
The first line contains a single integer t (1 ≤ t ≤ 10) — the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of the description of each test case contains four integers n, c_{0}, c_{1}, h (1 ≤ n, c_{0}, c_{1}, h ≤ 1000).
The second line of the description of each test case contains the binary string s of length n.
Output
For each test case print a single integer — the minimum number of coins needed to buy the string.
Example
Input
6
3 1 1 1
100
5 10 100 1
01010
5 10 1 1
11111
5 1 10 1
11111
12 2 1 10
101110110101
2 100 1 10
00
Output
3
52
5
10
16
22
Note
In the first test case, you can buy all characters and pay 3 coins, because both characters 0 and 1 costs 1 coin.
In the second test case, you can firstly change 2-nd and 4-th symbols of the string from 1 to 0 and pay 2 coins for that. Your string will be 00000. After that, you can buy the string and pay 5 ⋅ 10 = 50 coins for that. The total number of coins paid will be 2 + 50 = 52.
Submitted Solution:
```
t=int(input())
for _ in range(t):
n,c0,c1,h = map(int, input().split())
s=input()
l0=s.count('0')
l1=s.count('1')
if h*n>(c0*l0+c1*l1):
print(c0*l0+c1*l1)
elif c0>c1:
if l0*h+n*c1<=l0*c0+l1*c1:
print(l0*h+n*c1)
else:
print(l0*c0+l1*c1)
elif c1>c0:
if l1*h+n*c0<l0*c0+l1*c1:
print(l1*h+n*c0)
else:
print(l0*c0+l1*c1)
else:
print(l1*c1+l0*c0)
``` | instruction | 0 | 74,382 | 10 | 148,764 |
No | output | 1 | 74,382 | 10 | 148,765 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given four integers n, c_0, c_1 and h and a binary string s of length n.
A binary string is a string consisting of characters 0 and 1.
You can change any character of the string s (the string should be still binary after the change). You should pay h coins for each change.
After some changes (possibly zero) you want to buy the string. To buy the string you should buy all its characters. To buy the character 0 you should pay c_0 coins, to buy the character 1 you should pay c_1 coins.
Find the minimum number of coins needed to buy the string.
Input
The first line contains a single integer t (1 ≤ t ≤ 10) — the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of the description of each test case contains four integers n, c_{0}, c_{1}, h (1 ≤ n, c_{0}, c_{1}, h ≤ 1000).
The second line of the description of each test case contains the binary string s of length n.
Output
For each test case print a single integer — the minimum number of coins needed to buy the string.
Example
Input
6
3 1 1 1
100
5 10 100 1
01010
5 10 1 1
11111
5 1 10 1
11111
12 2 1 10
101110110101
2 100 1 10
00
Output
3
52
5
10
16
22
Note
In the first test case, you can buy all characters and pay 3 coins, because both characters 0 and 1 costs 1 coin.
In the second test case, you can firstly change 2-nd and 4-th symbols of the string from 1 to 0 and pay 2 coins for that. Your string will be 00000. After that, you can buy the string and pay 5 ⋅ 10 = 50 coins for that. The total number of coins paid will be 2 + 50 = 52.
Submitted Solution:
```
t = int(input())
for _ in range(t):
n,c0,c1,h = [int(x) for x in input().split()]
s = list(input())
x1 = s.count('1')
x0 = s.count('0')
a = c0*x0+c1*x1
a = min(a,c0*x0+h*x1,c1*x1+h*x0)
print(a)
``` | instruction | 0 | 74,383 | 10 | 148,766 |
No | output | 1 | 74,383 | 10 | 148,767 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given four integers n, c_0, c_1 and h and a binary string s of length n.
A binary string is a string consisting of characters 0 and 1.
You can change any character of the string s (the string should be still binary after the change). You should pay h coins for each change.
After some changes (possibly zero) you want to buy the string. To buy the string you should buy all its characters. To buy the character 0 you should pay c_0 coins, to buy the character 1 you should pay c_1 coins.
Find the minimum number of coins needed to buy the string.
Input
The first line contains a single integer t (1 ≤ t ≤ 10) — the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of the description of each test case contains four integers n, c_{0}, c_{1}, h (1 ≤ n, c_{0}, c_{1}, h ≤ 1000).
The second line of the description of each test case contains the binary string s of length n.
Output
For each test case print a single integer — the minimum number of coins needed to buy the string.
Example
Input
6
3 1 1 1
100
5 10 100 1
01010
5 10 1 1
11111
5 1 10 1
11111
12 2 1 10
101110110101
2 100 1 10
00
Output
3
52
5
10
16
22
Note
In the first test case, you can buy all characters and pay 3 coins, because both characters 0 and 1 costs 1 coin.
In the second test case, you can firstly change 2-nd and 4-th symbols of the string from 1 to 0 and pay 2 coins for that. Your string will be 00000. After that, you can buy the string and pay 5 ⋅ 10 = 50 coins for that. The total number of coins paid will be 2 + 50 = 52.
Submitted Solution:
```
l=[]
def func(n,c0,c1,h,s):
a=0
if(c0==c1):
for i in range(len(s)):
a1=int(s[i])
a=a+(c0)
l.append(a)
return l
if(c0<c1):
for k in range(len(s)):
if(s[k]=='1'):
a=a+h
for j in range(len(s)):
a1=int(s[j])
a=a+(c0)
l.append(a)
return l
if(c1<c0):
for k in range(len(s)):
if(s[k]=='0'):
a=a+h
for j in range(len(s)):
a1=int(s[j])
a=a+(c1)
l.append(a)
return l
T=int(input())
for i in range(T):
n,c0,c1,h=map(int,input().split())
s=input()
d=func(n,c0,c1,h,s)
for v in l:
print(v)
``` | instruction | 0 | 74,384 | 10 | 148,768 |
No | output | 1 | 74,384 | 10 | 148,769 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n incoming messages for Vasya. The i-th message is going to be received after ti minutes. Each message has a cost, which equals to A initially. After being received, the cost of a message decreases by B each minute (it can become negative). Vasya can read any message after receiving it at any moment of time. After reading the message, Vasya's bank account receives the current cost of this message. Initially, Vasya's bank account is at 0.
Also, each minute Vasya's bank account receives C·k, where k is the amount of received but unread messages.
Vasya's messages are very important to him, and because of that he wants to have all messages read after T minutes.
Determine the maximum amount of money Vasya's bank account can hold after T minutes.
Input
The first line contains five integers n, A, B, C and T (1 ≤ n, A, B, C, T ≤ 1000).
The second string contains n integers ti (1 ≤ ti ≤ T).
Output
Output one integer — the answer to the problem.
Examples
Input
4 5 5 3 5
1 5 5 4
Output
20
Input
5 3 1 1 3
2 2 2 1 1
Output
15
Input
5 5 3 4 5
1 2 3 4 5
Output
35
Note
In the first sample the messages must be read immediately after receiving, Vasya receives A points for each message, n·A = 20 in total.
In the second sample the messages can be read at any integer moment.
In the third sample messages must be read at the moment T. This way Vasya has 1, 2, 3, 4 and 0 unread messages at the corresponding minutes, he gets 40 points for them. When reading messages, he receives (5 - 4·3) + (5 - 3·3) + (5 - 2·3) + (5 - 1·3) + 5 = - 5 points. This is 35 in total. | instruction | 0 | 74,722 | 10 | 149,444 |
Tags: math
Correct Solution:
```
#########################################################################################################\
#########################################################################################################
###################################The_Apurv_Rathore#####################################################
#########################################################################################################
#########################################################################################################
import sys,os,io
from sys import stdin
from math import log, gcd, ceil
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop, heapify
from bisect import bisect_left , bisect_right
import math
alphabets = list('abcdefghijklmnopqrstuvwxyz')
#for deep recursion__________________________________________-
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den,p - 2, p)) % p
def primeFactors(n):
l = []
while n % 2 == 0:
l.append(2)
n = n / 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
l.append(int(i))
n = n / i
if n > 2:
l.append(n)
c = dict(Counter(l))
return list(set(l))
# return c
def power(x, y, p) :
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
#____________________GetPrimeFactors in log(n)________________________________________
def sieveForSmallestPrimeFactor():
MAXN = 100001
spf = [0 for i in range(MAXN)]
spf[1] = 1
for i in range(2, MAXN):
spf[i] = i
for i in range(4, MAXN, 2):
spf[i] = 2
for i in range(3, math.ceil(math.sqrt(MAXN))):
if (spf[i] == i):
for j in range(i * i, MAXN, i):
if (spf[j] == j):
spf[j] = i
return spf
def getPrimeFactorizationLOGN(x):
spf = sieveForSmallestPrimeFactor()
ret = list()
while (x != 1):
ret.append(spf[x])
x = x // spf[x]
return ret
#____________________________________________________________
def SieveOfEratosthenes(n):
#time complexity = nlog(log(n))
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
def si():
return input()
def divideCeil(n,x):
if (n%x==0):
return n//x
return n//x+1
def ii():
return int(input())
def li():
return list(map(int,input().split()))
#__________________________TEMPLATE__________________OVER_______________________________________________________
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r") ; sys.stdout = open("output.txt","w")
else:
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def solve():
n,a,b,c,t = li()
ti = li()
if b>c:
print(n*a)
else:
ans = 0
ti.sort()
for i in ti:
ans+=(t-i)*(c-b)+a
print(ans)
t = 1
# t = ii()
for _ in range(t):
solve()
``` | output | 1 | 74,722 | 10 | 149,445 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n incoming messages for Vasya. The i-th message is going to be received after ti minutes. Each message has a cost, which equals to A initially. After being received, the cost of a message decreases by B each minute (it can become negative). Vasya can read any message after receiving it at any moment of time. After reading the message, Vasya's bank account receives the current cost of this message. Initially, Vasya's bank account is at 0.
Also, each minute Vasya's bank account receives C·k, where k is the amount of received but unread messages.
Vasya's messages are very important to him, and because of that he wants to have all messages read after T minutes.
Determine the maximum amount of money Vasya's bank account can hold after T minutes.
Input
The first line contains five integers n, A, B, C and T (1 ≤ n, A, B, C, T ≤ 1000).
The second string contains n integers ti (1 ≤ ti ≤ T).
Output
Output one integer — the answer to the problem.
Examples
Input
4 5 5 3 5
1 5 5 4
Output
20
Input
5 3 1 1 3
2 2 2 1 1
Output
15
Input
5 5 3 4 5
1 2 3 4 5
Output
35
Note
In the first sample the messages must be read immediately after receiving, Vasya receives A points for each message, n·A = 20 in total.
In the second sample the messages can be read at any integer moment.
In the third sample messages must be read at the moment T. This way Vasya has 1, 2, 3, 4 and 0 unread messages at the corresponding minutes, he gets 40 points for them. When reading messages, he receives (5 - 4·3) + (5 - 3·3) + (5 - 2·3) + (5 - 1·3) + 5 = - 5 points. This is 35 in total. | instruction | 0 | 74,723 | 10 | 149,446 |
Tags: math
Correct Solution:
```
n,a1,b1,c1,t=map(int,input().split())
r=list(map(int,input().split()))
if (c1<=b1):
print(n*a1)
else:
s=0
for i in r:
s+=(c1-b1)*(t-i)
print(s+n*a1)
#pilotezebi
``` | output | 1 | 74,723 | 10 | 149,447 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n incoming messages for Vasya. The i-th message is going to be received after ti minutes. Each message has a cost, which equals to A initially. After being received, the cost of a message decreases by B each minute (it can become negative). Vasya can read any message after receiving it at any moment of time. After reading the message, Vasya's bank account receives the current cost of this message. Initially, Vasya's bank account is at 0.
Also, each minute Vasya's bank account receives C·k, where k is the amount of received but unread messages.
Vasya's messages are very important to him, and because of that he wants to have all messages read after T minutes.
Determine the maximum amount of money Vasya's bank account can hold after T minutes.
Input
The first line contains five integers n, A, B, C and T (1 ≤ n, A, B, C, T ≤ 1000).
The second string contains n integers ti (1 ≤ ti ≤ T).
Output
Output one integer — the answer to the problem.
Examples
Input
4 5 5 3 5
1 5 5 4
Output
20
Input
5 3 1 1 3
2 2 2 1 1
Output
15
Input
5 5 3 4 5
1 2 3 4 5
Output
35
Note
In the first sample the messages must be read immediately after receiving, Vasya receives A points for each message, n·A = 20 in total.
In the second sample the messages can be read at any integer moment.
In the third sample messages must be read at the moment T. This way Vasya has 1, 2, 3, 4 and 0 unread messages at the corresponding minutes, he gets 40 points for them. When reading messages, he receives (5 - 4·3) + (5 - 3·3) + (5 - 2·3) + (5 - 1·3) + 5 = - 5 points. This is 35 in total. | instruction | 0 | 74,724 | 10 | 149,448 |
Tags: math
Correct Solution:
```
n, A, B, C, T = map(int, input().split())
a = list(map(int, input().split()))
money = 0
def shouldReadSoon(t):
later = A - B * (T - t)
laterCost = later + C * (T - t)
if laterCost > A:
return laterCost
else:
return A
for j in range(len(a)):
money += shouldReadSoon(a[j])
print(money)
``` | output | 1 | 74,724 | 10 | 149,449 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n incoming messages for Vasya. The i-th message is going to be received after ti minutes. Each message has a cost, which equals to A initially. After being received, the cost of a message decreases by B each minute (it can become negative). Vasya can read any message after receiving it at any moment of time. After reading the message, Vasya's bank account receives the current cost of this message. Initially, Vasya's bank account is at 0.
Also, each minute Vasya's bank account receives C·k, where k is the amount of received but unread messages.
Vasya's messages are very important to him, and because of that he wants to have all messages read after T minutes.
Determine the maximum amount of money Vasya's bank account can hold after T minutes.
Input
The first line contains five integers n, A, B, C and T (1 ≤ n, A, B, C, T ≤ 1000).
The second string contains n integers ti (1 ≤ ti ≤ T).
Output
Output one integer — the answer to the problem.
Examples
Input
4 5 5 3 5
1 5 5 4
Output
20
Input
5 3 1 1 3
2 2 2 1 1
Output
15
Input
5 5 3 4 5
1 2 3 4 5
Output
35
Note
In the first sample the messages must be read immediately after receiving, Vasya receives A points for each message, n·A = 20 in total.
In the second sample the messages can be read at any integer moment.
In the third sample messages must be read at the moment T. This way Vasya has 1, 2, 3, 4 and 0 unread messages at the corresponding minutes, he gets 40 points for them. When reading messages, he receives (5 - 4·3) + (5 - 3·3) + (5 - 2·3) + (5 - 1·3) + 5 = - 5 points. This is 35 in total. | instruction | 0 | 74,725 | 10 | 149,450 |
Tags: math
Correct Solution:
```
n,A,B,C,T=list(map(int, input().split(' ')))
a=list(map(int, input().split(' ')))
res=n*A
if C-B<=0:
print(res)
else:
res+=sum([(C-B)*(T-x) for x in a])
print(res)
``` | output | 1 | 74,725 | 10 | 149,451 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n incoming messages for Vasya. The i-th message is going to be received after ti minutes. Each message has a cost, which equals to A initially. After being received, the cost of a message decreases by B each minute (it can become negative). Vasya can read any message after receiving it at any moment of time. After reading the message, Vasya's bank account receives the current cost of this message. Initially, Vasya's bank account is at 0.
Also, each minute Vasya's bank account receives C·k, where k is the amount of received but unread messages.
Vasya's messages are very important to him, and because of that he wants to have all messages read after T minutes.
Determine the maximum amount of money Vasya's bank account can hold after T minutes.
Input
The first line contains five integers n, A, B, C and T (1 ≤ n, A, B, C, T ≤ 1000).
The second string contains n integers ti (1 ≤ ti ≤ T).
Output
Output one integer — the answer to the problem.
Examples
Input
4 5 5 3 5
1 5 5 4
Output
20
Input
5 3 1 1 3
2 2 2 1 1
Output
15
Input
5 5 3 4 5
1 2 3 4 5
Output
35
Note
In the first sample the messages must be read immediately after receiving, Vasya receives A points for each message, n·A = 20 in total.
In the second sample the messages can be read at any integer moment.
In the third sample messages must be read at the moment T. This way Vasya has 1, 2, 3, 4 and 0 unread messages at the corresponding minutes, he gets 40 points for them. When reading messages, he receives (5 - 4·3) + (5 - 3·3) + (5 - 2·3) + (5 - 1·3) + 5 = - 5 points. This is 35 in total. | instruction | 0 | 74,726 | 10 | 149,452 |
Tags: math
Correct Solution:
```
n, A, B, C, T = map(int, input().split())
t = sum(list(map(int, input().split())))
if C - B < 0:
print(n * A)
else:
print(n * A + (C - B) * (n * T - t))
``` | output | 1 | 74,726 | 10 | 149,453 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n incoming messages for Vasya. The i-th message is going to be received after ti minutes. Each message has a cost, which equals to A initially. After being received, the cost of a message decreases by B each minute (it can become negative). Vasya can read any message after receiving it at any moment of time. After reading the message, Vasya's bank account receives the current cost of this message. Initially, Vasya's bank account is at 0.
Also, each minute Vasya's bank account receives C·k, where k is the amount of received but unread messages.
Vasya's messages are very important to him, and because of that he wants to have all messages read after T minutes.
Determine the maximum amount of money Vasya's bank account can hold after T minutes.
Input
The first line contains five integers n, A, B, C and T (1 ≤ n, A, B, C, T ≤ 1000).
The second string contains n integers ti (1 ≤ ti ≤ T).
Output
Output one integer — the answer to the problem.
Examples
Input
4 5 5 3 5
1 5 5 4
Output
20
Input
5 3 1 1 3
2 2 2 1 1
Output
15
Input
5 5 3 4 5
1 2 3 4 5
Output
35
Note
In the first sample the messages must be read immediately after receiving, Vasya receives A points for each message, n·A = 20 in total.
In the second sample the messages can be read at any integer moment.
In the third sample messages must be read at the moment T. This way Vasya has 1, 2, 3, 4 and 0 unread messages at the corresponding minutes, he gets 40 points for them. When reading messages, he receives (5 - 4·3) + (5 - 3·3) + (5 - 2·3) + (5 - 1·3) + 5 = - 5 points. This is 35 in total. | instruction | 0 | 74,727 | 10 | 149,454 |
Tags: math
Correct Solution:
```
n, A, B, C, T = [int(x) for x in input().split()]
t = sorted([int(x) for x in input().split()], reverse=True)
k = 0
c = 0
standby = B < C
for i in range(1, T+1):
m = 0
while t != [] and t[-1] == i:
m += 1
t.pop()
if standby and i != T:
k += m
c += A*m-(B-C)*k
else:
c += A*m
print(c)
``` | output | 1 | 74,727 | 10 | 149,455 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n incoming messages for Vasya. The i-th message is going to be received after ti minutes. Each message has a cost, which equals to A initially. After being received, the cost of a message decreases by B each minute (it can become negative). Vasya can read any message after receiving it at any moment of time. After reading the message, Vasya's bank account receives the current cost of this message. Initially, Vasya's bank account is at 0.
Also, each minute Vasya's bank account receives C·k, where k is the amount of received but unread messages.
Vasya's messages are very important to him, and because of that he wants to have all messages read after T minutes.
Determine the maximum amount of money Vasya's bank account can hold after T minutes.
Input
The first line contains five integers n, A, B, C and T (1 ≤ n, A, B, C, T ≤ 1000).
The second string contains n integers ti (1 ≤ ti ≤ T).
Output
Output one integer — the answer to the problem.
Examples
Input
4 5 5 3 5
1 5 5 4
Output
20
Input
5 3 1 1 3
2 2 2 1 1
Output
15
Input
5 5 3 4 5
1 2 3 4 5
Output
35
Note
In the first sample the messages must be read immediately after receiving, Vasya receives A points for each message, n·A = 20 in total.
In the second sample the messages can be read at any integer moment.
In the third sample messages must be read at the moment T. This way Vasya has 1, 2, 3, 4 and 0 unread messages at the corresponding minutes, he gets 40 points for them. When reading messages, he receives (5 - 4·3) + (5 - 3·3) + (5 - 2·3) + (5 - 1·3) + 5 = - 5 points. This is 35 in total. | instruction | 0 | 74,728 | 10 | 149,456 |
Tags: math
Correct Solution:
```
n, A, B, C, T = map(int, input().split())
t = list(map(int, input().split()))
score = 0
if (C > B):
for tmp in t:
score += A + (T - tmp) * (C - B)
else:
score += n * A
print(score)
``` | output | 1 | 74,728 | 10 | 149,457 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n incoming messages for Vasya. The i-th message is going to be received after ti minutes. Each message has a cost, which equals to A initially. After being received, the cost of a message decreases by B each minute (it can become negative). Vasya can read any message after receiving it at any moment of time. After reading the message, Vasya's bank account receives the current cost of this message. Initially, Vasya's bank account is at 0.
Also, each minute Vasya's bank account receives C·k, where k is the amount of received but unread messages.
Vasya's messages are very important to him, and because of that he wants to have all messages read after T minutes.
Determine the maximum amount of money Vasya's bank account can hold after T minutes.
Input
The first line contains five integers n, A, B, C and T (1 ≤ n, A, B, C, T ≤ 1000).
The second string contains n integers ti (1 ≤ ti ≤ T).
Output
Output one integer — the answer to the problem.
Examples
Input
4 5 5 3 5
1 5 5 4
Output
20
Input
5 3 1 1 3
2 2 2 1 1
Output
15
Input
5 5 3 4 5
1 2 3 4 5
Output
35
Note
In the first sample the messages must be read immediately after receiving, Vasya receives A points for each message, n·A = 20 in total.
In the second sample the messages can be read at any integer moment.
In the third sample messages must be read at the moment T. This way Vasya has 1, 2, 3, 4 and 0 unread messages at the corresponding minutes, he gets 40 points for them. When reading messages, he receives (5 - 4·3) + (5 - 3·3) + (5 - 2·3) + (5 - 1·3) + 5 = - 5 points. This is 35 in total. | instruction | 0 | 74,729 | 10 | 149,458 |
Tags: math
Correct Solution:
```
n, A, B, C, T = map(int, input().split())
t = list(map(int, input().split()))
if B >= C:
print(A * n)
else:
ans = A * n
for time in t:
ans += (C - B) * (T - time)
print(ans)
``` | output | 1 | 74,729 | 10 | 149,459 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n incoming messages for Vasya. The i-th message is going to be received after ti minutes. Each message has a cost, which equals to A initially. After being received, the cost of a message decreases by B each minute (it can become negative). Vasya can read any message after receiving it at any moment of time. After reading the message, Vasya's bank account receives the current cost of this message. Initially, Vasya's bank account is at 0.
Also, each minute Vasya's bank account receives C·k, where k is the amount of received but unread messages.
Vasya's messages are very important to him, and because of that he wants to have all messages read after T minutes.
Determine the maximum amount of money Vasya's bank account can hold after T minutes.
Input
The first line contains five integers n, A, B, C and T (1 ≤ n, A, B, C, T ≤ 1000).
The second string contains n integers ti (1 ≤ ti ≤ T).
Output
Output one integer — the answer to the problem.
Examples
Input
4 5 5 3 5
1 5 5 4
Output
20
Input
5 3 1 1 3
2 2 2 1 1
Output
15
Input
5 5 3 4 5
1 2 3 4 5
Output
35
Note
In the first sample the messages must be read immediately after receiving, Vasya receives A points for each message, n·A = 20 in total.
In the second sample the messages can be read at any integer moment.
In the third sample messages must be read at the moment T. This way Vasya has 1, 2, 3, 4 and 0 unread messages at the corresponding minutes, he gets 40 points for them. When reading messages, he receives (5 - 4·3) + (5 - 3·3) + (5 - 2·3) + (5 - 1·3) + 5 = - 5 points. This is 35 in total.
Submitted Solution:
```
def solution(n, a, b, c, t, messages):
account = 0
for m in messages:
if b >= c:
account += a
else:
account += (c - b) * (t - m) + a
return account
n, a, b, c, t = map(int, input().split())
messages = list(map(int, input().split()))
print(solution(n, a, b, c, t, messages))
``` | instruction | 0 | 74,730 | 10 | 149,460 |
Yes | output | 1 | 74,730 | 10 | 149,461 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n incoming messages for Vasya. The i-th message is going to be received after ti minutes. Each message has a cost, which equals to A initially. After being received, the cost of a message decreases by B each minute (it can become negative). Vasya can read any message after receiving it at any moment of time. After reading the message, Vasya's bank account receives the current cost of this message. Initially, Vasya's bank account is at 0.
Also, each minute Vasya's bank account receives C·k, where k is the amount of received but unread messages.
Vasya's messages are very important to him, and because of that he wants to have all messages read after T minutes.
Determine the maximum amount of money Vasya's bank account can hold after T minutes.
Input
The first line contains five integers n, A, B, C and T (1 ≤ n, A, B, C, T ≤ 1000).
The second string contains n integers ti (1 ≤ ti ≤ T).
Output
Output one integer — the answer to the problem.
Examples
Input
4 5 5 3 5
1 5 5 4
Output
20
Input
5 3 1 1 3
2 2 2 1 1
Output
15
Input
5 5 3 4 5
1 2 3 4 5
Output
35
Note
In the first sample the messages must be read immediately after receiving, Vasya receives A points for each message, n·A = 20 in total.
In the second sample the messages can be read at any integer moment.
In the third sample messages must be read at the moment T. This way Vasya has 1, 2, 3, 4 and 0 unread messages at the corresponding minutes, he gets 40 points for them. When reading messages, he receives (5 - 4·3) + (5 - 3·3) + (5 - 2·3) + (5 - 1·3) + 5 = - 5 points. This is 35 in total.
Submitted Solution:
```
n, a, b, c, t = [int(x) for x in input().split()]
array = [int(x) for x in input().split()]
ret = 0
if b < c:
for i in array:
ret += c*(t-i) + a-(t-i)*b
print(ret)
else:
print(n*a)
``` | instruction | 0 | 74,731 | 10 | 149,462 |
Yes | output | 1 | 74,731 | 10 | 149,463 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n incoming messages for Vasya. The i-th message is going to be received after ti minutes. Each message has a cost, which equals to A initially. After being received, the cost of a message decreases by B each minute (it can become negative). Vasya can read any message after receiving it at any moment of time. After reading the message, Vasya's bank account receives the current cost of this message. Initially, Vasya's bank account is at 0.
Also, each minute Vasya's bank account receives C·k, where k is the amount of received but unread messages.
Vasya's messages are very important to him, and because of that he wants to have all messages read after T minutes.
Determine the maximum amount of money Vasya's bank account can hold after T minutes.
Input
The first line contains five integers n, A, B, C and T (1 ≤ n, A, B, C, T ≤ 1000).
The second string contains n integers ti (1 ≤ ti ≤ T).
Output
Output one integer — the answer to the problem.
Examples
Input
4 5 5 3 5
1 5 5 4
Output
20
Input
5 3 1 1 3
2 2 2 1 1
Output
15
Input
5 5 3 4 5
1 2 3 4 5
Output
35
Note
In the first sample the messages must be read immediately after receiving, Vasya receives A points for each message, n·A = 20 in total.
In the second sample the messages can be read at any integer moment.
In the third sample messages must be read at the moment T. This way Vasya has 1, 2, 3, 4 and 0 unread messages at the corresponding minutes, he gets 40 points for them. When reading messages, he receives (5 - 4·3) + (5 - 3·3) + (5 - 2·3) + (5 - 1·3) + 5 = - 5 points. This is 35 in total.
Submitted Solution:
```
n, a, b, c, t = map(int,input().split())
sum = n * a
s = list(map(int,input().split()))
for i in range(n):
if b < c:
sum += (c - b) * (t - s[i])
print(sum)
``` | instruction | 0 | 74,732 | 10 | 149,464 |
Yes | output | 1 | 74,732 | 10 | 149,465 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n incoming messages for Vasya. The i-th message is going to be received after ti minutes. Each message has a cost, which equals to A initially. After being received, the cost of a message decreases by B each minute (it can become negative). Vasya can read any message after receiving it at any moment of time. After reading the message, Vasya's bank account receives the current cost of this message. Initially, Vasya's bank account is at 0.
Also, each minute Vasya's bank account receives C·k, where k is the amount of received but unread messages.
Vasya's messages are very important to him, and because of that he wants to have all messages read after T minutes.
Determine the maximum amount of money Vasya's bank account can hold after T minutes.
Input
The first line contains five integers n, A, B, C and T (1 ≤ n, A, B, C, T ≤ 1000).
The second string contains n integers ti (1 ≤ ti ≤ T).
Output
Output one integer — the answer to the problem.
Examples
Input
4 5 5 3 5
1 5 5 4
Output
20
Input
5 3 1 1 3
2 2 2 1 1
Output
15
Input
5 5 3 4 5
1 2 3 4 5
Output
35
Note
In the first sample the messages must be read immediately after receiving, Vasya receives A points for each message, n·A = 20 in total.
In the second sample the messages can be read at any integer moment.
In the third sample messages must be read at the moment T. This way Vasya has 1, 2, 3, 4 and 0 unread messages at the corresponding minutes, he gets 40 points for them. When reading messages, he receives (5 - 4·3) + (5 - 3·3) + (5 - 2·3) + (5 - 1·3) + 5 = - 5 points. This is 35 in total.
Submitted Solution:
```
def be(k, delta, T, A):
s=len(k)
print(s*A)
def ce(k, delta, T, A):
s=0
for i in k:
s=s-delta*(T-i)+A
print(s)
s1=input()
s2=input()
s1=s1.split(" ")
k1=[int(a) for a in s1]
s2=s2.split(" ")
k2=[int(a) for a in s2]
delta=k1[2]-k1[3]
if delta>=0:
be(k2, delta, k1[4], k1[1])
else:
ce(k2, delta, k1[4], k1[1])
``` | instruction | 0 | 74,733 | 10 | 149,466 |
Yes | output | 1 | 74,733 | 10 | 149,467 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n incoming messages for Vasya. The i-th message is going to be received after ti minutes. Each message has a cost, which equals to A initially. After being received, the cost of a message decreases by B each minute (it can become negative). Vasya can read any message after receiving it at any moment of time. After reading the message, Vasya's bank account receives the current cost of this message. Initially, Vasya's bank account is at 0.
Also, each minute Vasya's bank account receives C·k, where k is the amount of received but unread messages.
Vasya's messages are very important to him, and because of that he wants to have all messages read after T minutes.
Determine the maximum amount of money Vasya's bank account can hold after T minutes.
Input
The first line contains five integers n, A, B, C and T (1 ≤ n, A, B, C, T ≤ 1000).
The second string contains n integers ti (1 ≤ ti ≤ T).
Output
Output one integer — the answer to the problem.
Examples
Input
4 5 5 3 5
1 5 5 4
Output
20
Input
5 3 1 1 3
2 2 2 1 1
Output
15
Input
5 5 3 4 5
1 2 3 4 5
Output
35
Note
In the first sample the messages must be read immediately after receiving, Vasya receives A points for each message, n·A = 20 in total.
In the second sample the messages can be read at any integer moment.
In the third sample messages must be read at the moment T. This way Vasya has 1, 2, 3, 4 and 0 unread messages at the corresponding minutes, he gets 40 points for them. When reading messages, he receives (5 - 4·3) + (5 - 3·3) + (5 - 2·3) + (5 - 1·3) + 5 = - 5 points. This is 35 in total.
Submitted Solution:
```
n=list(map(int,input().split()))
t=list(map(int,input().split()))
if (n[3]>n[2]):
print(n[1]*n[0])
else:
print(n[0]*n[1] +(n[0]*n[4]-sum(t)))
``` | instruction | 0 | 74,734 | 10 | 149,468 |
No | output | 1 | 74,734 | 10 | 149,469 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n incoming messages for Vasya. The i-th message is going to be received after ti minutes. Each message has a cost, which equals to A initially. After being received, the cost of a message decreases by B each minute (it can become negative). Vasya can read any message after receiving it at any moment of time. After reading the message, Vasya's bank account receives the current cost of this message. Initially, Vasya's bank account is at 0.
Also, each minute Vasya's bank account receives C·k, where k is the amount of received but unread messages.
Vasya's messages are very important to him, and because of that he wants to have all messages read after T minutes.
Determine the maximum amount of money Vasya's bank account can hold after T minutes.
Input
The first line contains five integers n, A, B, C and T (1 ≤ n, A, B, C, T ≤ 1000).
The second string contains n integers ti (1 ≤ ti ≤ T).
Output
Output one integer — the answer to the problem.
Examples
Input
4 5 5 3 5
1 5 5 4
Output
20
Input
5 3 1 1 3
2 2 2 1 1
Output
15
Input
5 5 3 4 5
1 2 3 4 5
Output
35
Note
In the first sample the messages must be read immediately after receiving, Vasya receives A points for each message, n·A = 20 in total.
In the second sample the messages can be read at any integer moment.
In the third sample messages must be read at the moment T. This way Vasya has 1, 2, 3, 4 and 0 unread messages at the corresponding minutes, he gets 40 points for them. When reading messages, he receives (5 - 4·3) + (5 - 3·3) + (5 - 2·3) + (5 - 1·3) + 5 = - 5 points. This is 35 in total.
Submitted Solution:
```
n=input()
``` | instruction | 0 | 74,735 | 10 | 149,470 |
No | output | 1 | 74,735 | 10 | 149,471 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n incoming messages for Vasya. The i-th message is going to be received after ti minutes. Each message has a cost, which equals to A initially. After being received, the cost of a message decreases by B each minute (it can become negative). Vasya can read any message after receiving it at any moment of time. After reading the message, Vasya's bank account receives the current cost of this message. Initially, Vasya's bank account is at 0.
Also, each minute Vasya's bank account receives C·k, where k is the amount of received but unread messages.
Vasya's messages are very important to him, and because of that he wants to have all messages read after T minutes.
Determine the maximum amount of money Vasya's bank account can hold after T minutes.
Input
The first line contains five integers n, A, B, C and T (1 ≤ n, A, B, C, T ≤ 1000).
The second string contains n integers ti (1 ≤ ti ≤ T).
Output
Output one integer — the answer to the problem.
Examples
Input
4 5 5 3 5
1 5 5 4
Output
20
Input
5 3 1 1 3
2 2 2 1 1
Output
15
Input
5 5 3 4 5
1 2 3 4 5
Output
35
Note
In the first sample the messages must be read immediately after receiving, Vasya receives A points for each message, n·A = 20 in total.
In the second sample the messages can be read at any integer moment.
In the third sample messages must be read at the moment T. This way Vasya has 1, 2, 3, 4 and 0 unread messages at the corresponding minutes, he gets 40 points for them. When reading messages, he receives (5 - 4·3) + (5 - 3·3) + (5 - 2·3) + (5 - 1·3) + 5 = - 5 points. This is 35 in total.
Submitted Solution:
```
n,A,B,C,T=map(int,input().split())
arr=map(int,input().split())
print(sum(arr)+A*n+((C-B)*(n*T-sum(arr)) if B<C else 0))
``` | instruction | 0 | 74,736 | 10 | 149,472 |
No | output | 1 | 74,736 | 10 | 149,473 |
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