message stringlengths 2 59.7k | message_type stringclasses 2
values | message_id int64 0 1 | conversation_id int64 37 108k | cluster float64 20 20 | __index_level_0__ int64 74 217k |
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Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a number ebne (even but not even) if and only if its sum of digits is divisible by 2 but the number itself is not divisible by 2. For example, 13, 1227, 185217 are ebne numbers, while 12, 2, 177013, 265918 are not. If you're still unsure what ebne numbers are, you can look at the sample notes for more clarification.
You are given a non-negative integer s, consisting of n digits. You can delete some digits (they are not necessary consecutive/successive) to make the given number ebne. You cannot change the order of the digits, that is, after deleting the digits the remaining digits collapse. The resulting number shouldn't contain leading zeros. You can delete any number of digits between 0 (do not delete any digits at all) and n-1.
For example, if you are given s=222373204424185217171912 then one of possible ways to make it ebne is: 222373204424185217171912 → 2237344218521717191. The sum of digits of 2237344218521717191 is equal to 70 and is divisible by 2, but number itself is not divisible by 2: it means that the resulting number is ebne.
Find any resulting number that is ebne. If it's impossible to create an ebne number from the given number report about it.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 3000) — the number of digits in the original number.
The second line of each test case contains a non-negative integer number s, consisting of n digits.
It is guaranteed that s does not contain leading zeros and the sum of n over all test cases does not exceed 3000.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create an ebne number, print "-1" (without quotes);
* Otherwise, print the resulting number after deleting some, possibly zero, but not all digits. This number should be ebne. If there are multiple answers, you can print any of them. Note that answers with leading zeros or empty strings are not accepted. It's not necessary to minimize or maximize the number of deleted digits.
Example
Input
4
4
1227
1
0
6
177013
24
222373204424185217171912
Output
1227
-1
17703
2237344218521717191
Note
In the first test case of the example, 1227 is already an ebne number (as 1 + 2 + 2 + 7 = 12, 12 is divisible by 2, while in the same time, 1227 is not divisible by 2) so we don't need to delete any digits. Answers such as 127 and 17 will also be accepted.
In the second test case of the example, it is clearly impossible to create an ebne number from the given number.
In the third test case of the example, there are many ebne numbers we can obtain by deleting, for example, 1 digit such as 17703, 77013 or 17013. Answers such as 1701 or 770 will not be accepted as they are not ebne numbers. Answer 013 will not be accepted as it contains leading zeroes.
Explanation:
* 1 + 7 + 7 + 0 + 3 = 18. As 18 is divisible by 2 while 17703 is not divisible by 2, we can see that 17703 is an ebne number. Same with 77013 and 17013;
* 1 + 7 + 0 + 1 = 9. Because 9 is not divisible by 2, 1701 is not an ebne number;
* 7 + 7 + 0 = 14. This time, 14 is divisible by 2 but 770 is also divisible by 2, therefore, 770 is not an ebne number.
In the last test case of the example, one of many other possible answers is given. Another possible answer is: 222373204424185217171912 → 22237320442418521717191 (delete the last digit). | instruction | 0 | 89,394 | 20 | 178,788 |
Tags: greedy, math, strings
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
l = list(input())
ans = ''
for i in range(n):
if int(l[i]) % 2 != 0:
ans += l[i]
if len(ans) == 2:
print(ans)
break
else:
print(-1)
``` | output | 1 | 89,394 | 20 | 178,789 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a number ebne (even but not even) if and only if its sum of digits is divisible by 2 but the number itself is not divisible by 2. For example, 13, 1227, 185217 are ebne numbers, while 12, 2, 177013, 265918 are not. If you're still unsure what ebne numbers are, you can look at the sample notes for more clarification.
You are given a non-negative integer s, consisting of n digits. You can delete some digits (they are not necessary consecutive/successive) to make the given number ebne. You cannot change the order of the digits, that is, after deleting the digits the remaining digits collapse. The resulting number shouldn't contain leading zeros. You can delete any number of digits between 0 (do not delete any digits at all) and n-1.
For example, if you are given s=222373204424185217171912 then one of possible ways to make it ebne is: 222373204424185217171912 → 2237344218521717191. The sum of digits of 2237344218521717191 is equal to 70 and is divisible by 2, but number itself is not divisible by 2: it means that the resulting number is ebne.
Find any resulting number that is ebne. If it's impossible to create an ebne number from the given number report about it.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 3000) — the number of digits in the original number.
The second line of each test case contains a non-negative integer number s, consisting of n digits.
It is guaranteed that s does not contain leading zeros and the sum of n over all test cases does not exceed 3000.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create an ebne number, print "-1" (without quotes);
* Otherwise, print the resulting number after deleting some, possibly zero, but not all digits. This number should be ebne. If there are multiple answers, you can print any of them. Note that answers with leading zeros or empty strings are not accepted. It's not necessary to minimize or maximize the number of deleted digits.
Example
Input
4
4
1227
1
0
6
177013
24
222373204424185217171912
Output
1227
-1
17703
2237344218521717191
Note
In the first test case of the example, 1227 is already an ebne number (as 1 + 2 + 2 + 7 = 12, 12 is divisible by 2, while in the same time, 1227 is not divisible by 2) so we don't need to delete any digits. Answers such as 127 and 17 will also be accepted.
In the second test case of the example, it is clearly impossible to create an ebne number from the given number.
In the third test case of the example, there are many ebne numbers we can obtain by deleting, for example, 1 digit such as 17703, 77013 or 17013. Answers such as 1701 or 770 will not be accepted as they are not ebne numbers. Answer 013 will not be accepted as it contains leading zeroes.
Explanation:
* 1 + 7 + 7 + 0 + 3 = 18. As 18 is divisible by 2 while 17703 is not divisible by 2, we can see that 17703 is an ebne number. Same with 77013 and 17013;
* 1 + 7 + 0 + 1 = 9. Because 9 is not divisible by 2, 1701 is not an ebne number;
* 7 + 7 + 0 = 14. This time, 14 is divisible by 2 but 770 is also divisible by 2, therefore, 770 is not an ebne number.
In the last test case of the example, one of many other possible answers is given. Another possible answer is: 222373204424185217171912 → 22237320442418521717191 (delete the last digit). | instruction | 0 | 89,395 | 20 | 178,790 |
Tags: greedy, math, strings
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
a = list(map(int, input()))
o = sum(a) % 2
while o or (a and a[-1] % 2 == 0):
o ^= a[-1] % 2
a.pop()
if a:
print(''.join(map(str, a)))
else:
print(-1)
``` | output | 1 | 89,395 | 20 | 178,791 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a number ebne (even but not even) if and only if its sum of digits is divisible by 2 but the number itself is not divisible by 2. For example, 13, 1227, 185217 are ebne numbers, while 12, 2, 177013, 265918 are not. If you're still unsure what ebne numbers are, you can look at the sample notes for more clarification.
You are given a non-negative integer s, consisting of n digits. You can delete some digits (they are not necessary consecutive/successive) to make the given number ebne. You cannot change the order of the digits, that is, after deleting the digits the remaining digits collapse. The resulting number shouldn't contain leading zeros. You can delete any number of digits between 0 (do not delete any digits at all) and n-1.
For example, if you are given s=222373204424185217171912 then one of possible ways to make it ebne is: 222373204424185217171912 → 2237344218521717191. The sum of digits of 2237344218521717191 is equal to 70 and is divisible by 2, but number itself is not divisible by 2: it means that the resulting number is ebne.
Find any resulting number that is ebne. If it's impossible to create an ebne number from the given number report about it.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 3000) — the number of digits in the original number.
The second line of each test case contains a non-negative integer number s, consisting of n digits.
It is guaranteed that s does not contain leading zeros and the sum of n over all test cases does not exceed 3000.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create an ebne number, print "-1" (without quotes);
* Otherwise, print the resulting number after deleting some, possibly zero, but not all digits. This number should be ebne. If there are multiple answers, you can print any of them. Note that answers with leading zeros or empty strings are not accepted. It's not necessary to minimize or maximize the number of deleted digits.
Example
Input
4
4
1227
1
0
6
177013
24
222373204424185217171912
Output
1227
-1
17703
2237344218521717191
Note
In the first test case of the example, 1227 is already an ebne number (as 1 + 2 + 2 + 7 = 12, 12 is divisible by 2, while in the same time, 1227 is not divisible by 2) so we don't need to delete any digits. Answers such as 127 and 17 will also be accepted.
In the second test case of the example, it is clearly impossible to create an ebne number from the given number.
In the third test case of the example, there are many ebne numbers we can obtain by deleting, for example, 1 digit such as 17703, 77013 or 17013. Answers such as 1701 or 770 will not be accepted as they are not ebne numbers. Answer 013 will not be accepted as it contains leading zeroes.
Explanation:
* 1 + 7 + 7 + 0 + 3 = 18. As 18 is divisible by 2 while 17703 is not divisible by 2, we can see that 17703 is an ebne number. Same with 77013 and 17013;
* 1 + 7 + 0 + 1 = 9. Because 9 is not divisible by 2, 1701 is not an ebne number;
* 7 + 7 + 0 = 14. This time, 14 is divisible by 2 but 770 is also divisible by 2, therefore, 770 is not an ebne number.
In the last test case of the example, one of many other possible answers is given. Another possible answer is: 222373204424185217171912 → 22237320442418521717191 (delete the last digit). | instruction | 0 | 89,396 | 20 | 178,792 |
Tags: greedy, math, strings
Correct Solution:
```
t = int(input())
for i in range(t):
summ = 0
bul = False
a = []
l_a = 0
n = int(input())
s = input()
for i in range(len(s) - 1, -1, -1):
if l_a < 2 and int(s[i]) % 2 == 1:
a.append(i)
bul = True
l_a += 1
if bul:
summ += int(s[i])
if l_a == 0 or a[0] == 0:
print(-1)
elif summ % 2 == 1 and l_a == 1:
print(-1)
elif summ % 2 == 0:
print(s[:a[0] + 1])
else:
print(s[:a[1]] + s[a[1] + 1:a[0] + 1])
``` | output | 1 | 89,396 | 20 | 178,793 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a number ebne (even but not even) if and only if its sum of digits is divisible by 2 but the number itself is not divisible by 2. For example, 13, 1227, 185217 are ebne numbers, while 12, 2, 177013, 265918 are not. If you're still unsure what ebne numbers are, you can look at the sample notes for more clarification.
You are given a non-negative integer s, consisting of n digits. You can delete some digits (they are not necessary consecutive/successive) to make the given number ebne. You cannot change the order of the digits, that is, after deleting the digits the remaining digits collapse. The resulting number shouldn't contain leading zeros. You can delete any number of digits between 0 (do not delete any digits at all) and n-1.
For example, if you are given s=222373204424185217171912 then one of possible ways to make it ebne is: 222373204424185217171912 → 2237344218521717191. The sum of digits of 2237344218521717191 is equal to 70 and is divisible by 2, but number itself is not divisible by 2: it means that the resulting number is ebne.
Find any resulting number that is ebne. If it's impossible to create an ebne number from the given number report about it.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 3000) — the number of digits in the original number.
The second line of each test case contains a non-negative integer number s, consisting of n digits.
It is guaranteed that s does not contain leading zeros and the sum of n over all test cases does not exceed 3000.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create an ebne number, print "-1" (without quotes);
* Otherwise, print the resulting number after deleting some, possibly zero, but not all digits. This number should be ebne. If there are multiple answers, you can print any of them. Note that answers with leading zeros or empty strings are not accepted. It's not necessary to minimize or maximize the number of deleted digits.
Example
Input
4
4
1227
1
0
6
177013
24
222373204424185217171912
Output
1227
-1
17703
2237344218521717191
Note
In the first test case of the example, 1227 is already an ebne number (as 1 + 2 + 2 + 7 = 12, 12 is divisible by 2, while in the same time, 1227 is not divisible by 2) so we don't need to delete any digits. Answers such as 127 and 17 will also be accepted.
In the second test case of the example, it is clearly impossible to create an ebne number from the given number.
In the third test case of the example, there are many ebne numbers we can obtain by deleting, for example, 1 digit such as 17703, 77013 or 17013. Answers such as 1701 or 770 will not be accepted as they are not ebne numbers. Answer 013 will not be accepted as it contains leading zeroes.
Explanation:
* 1 + 7 + 7 + 0 + 3 = 18. As 18 is divisible by 2 while 17703 is not divisible by 2, we can see that 17703 is an ebne number. Same with 77013 and 17013;
* 1 + 7 + 0 + 1 = 9. Because 9 is not divisible by 2, 1701 is not an ebne number;
* 7 + 7 + 0 = 14. This time, 14 is divisible by 2 but 770 is also divisible by 2, therefore, 770 is not an ebne number.
In the last test case of the example, one of many other possible answers is given. Another possible answer is: 222373204424185217171912 → 22237320442418521717191 (delete the last digit). | instruction | 0 | 89,397 | 20 | 178,794 |
Tags: greedy, math, strings
Correct Solution:
```
q=int(input())
for i in range(q):
s=''
t=''
n=0
ans=0
n=int(input())
s=input()
for j in range(n):
if (int(s[j])%2==1):
t+=s[j]
ans+=1
if (ans>=2):
print(t[0]+t[1])
else:
print(-1)
``` | output | 1 | 89,397 | 20 | 178,795 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a number ebne (even but not even) if and only if its sum of digits is divisible by 2 but the number itself is not divisible by 2. For example, 13, 1227, 185217 are ebne numbers, while 12, 2, 177013, 265918 are not. If you're still unsure what ebne numbers are, you can look at the sample notes for more clarification.
You are given a non-negative integer s, consisting of n digits. You can delete some digits (they are not necessary consecutive/successive) to make the given number ebne. You cannot change the order of the digits, that is, after deleting the digits the remaining digits collapse. The resulting number shouldn't contain leading zeros. You can delete any number of digits between 0 (do not delete any digits at all) and n-1.
For example, if you are given s=222373204424185217171912 then one of possible ways to make it ebne is: 222373204424185217171912 → 2237344218521717191. The sum of digits of 2237344218521717191 is equal to 70 and is divisible by 2, but number itself is not divisible by 2: it means that the resulting number is ebne.
Find any resulting number that is ebne. If it's impossible to create an ebne number from the given number report about it.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 3000) — the number of digits in the original number.
The second line of each test case contains a non-negative integer number s, consisting of n digits.
It is guaranteed that s does not contain leading zeros and the sum of n over all test cases does not exceed 3000.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create an ebne number, print "-1" (without quotes);
* Otherwise, print the resulting number after deleting some, possibly zero, but not all digits. This number should be ebne. If there are multiple answers, you can print any of them. Note that answers with leading zeros or empty strings are not accepted. It's not necessary to minimize or maximize the number of deleted digits.
Example
Input
4
4
1227
1
0
6
177013
24
222373204424185217171912
Output
1227
-1
17703
2237344218521717191
Note
In the first test case of the example, 1227 is already an ebne number (as 1 + 2 + 2 + 7 = 12, 12 is divisible by 2, while in the same time, 1227 is not divisible by 2) so we don't need to delete any digits. Answers such as 127 and 17 will also be accepted.
In the second test case of the example, it is clearly impossible to create an ebne number from the given number.
In the third test case of the example, there are many ebne numbers we can obtain by deleting, for example, 1 digit such as 17703, 77013 or 17013. Answers such as 1701 or 770 will not be accepted as they are not ebne numbers. Answer 013 will not be accepted as it contains leading zeroes.
Explanation:
* 1 + 7 + 7 + 0 + 3 = 18. As 18 is divisible by 2 while 17703 is not divisible by 2, we can see that 17703 is an ebne number. Same with 77013 and 17013;
* 1 + 7 + 0 + 1 = 9. Because 9 is not divisible by 2, 1701 is not an ebne number;
* 7 + 7 + 0 = 14. This time, 14 is divisible by 2 but 770 is also divisible by 2, therefore, 770 is not an ebne number.
In the last test case of the example, one of many other possible answers is given. Another possible answer is: 222373204424185217171912 → 22237320442418521717191 (delete the last digit).
Submitted Solution:
```
t = int(input())
for _ in range(t):
_ = int(input())
n = input()
ans = ""
cnt = 0
for i in n:
if int(i) % 2:
ans += i
cnt += 1
if cnt >= 2:
break
if cnt < 2:
print(-1)
else:
print(ans)
``` | instruction | 0 | 89,398 | 20 | 178,796 |
Yes | output | 1 | 89,398 | 20 | 178,797 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a number ebne (even but not even) if and only if its sum of digits is divisible by 2 but the number itself is not divisible by 2. For example, 13, 1227, 185217 are ebne numbers, while 12, 2, 177013, 265918 are not. If you're still unsure what ebne numbers are, you can look at the sample notes for more clarification.
You are given a non-negative integer s, consisting of n digits. You can delete some digits (they are not necessary consecutive/successive) to make the given number ebne. You cannot change the order of the digits, that is, after deleting the digits the remaining digits collapse. The resulting number shouldn't contain leading zeros. You can delete any number of digits between 0 (do not delete any digits at all) and n-1.
For example, if you are given s=222373204424185217171912 then one of possible ways to make it ebne is: 222373204424185217171912 → 2237344218521717191. The sum of digits of 2237344218521717191 is equal to 70 and is divisible by 2, but number itself is not divisible by 2: it means that the resulting number is ebne.
Find any resulting number that is ebne. If it's impossible to create an ebne number from the given number report about it.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 3000) — the number of digits in the original number.
The second line of each test case contains a non-negative integer number s, consisting of n digits.
It is guaranteed that s does not contain leading zeros and the sum of n over all test cases does not exceed 3000.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create an ebne number, print "-1" (without quotes);
* Otherwise, print the resulting number after deleting some, possibly zero, but not all digits. This number should be ebne. If there are multiple answers, you can print any of them. Note that answers with leading zeros or empty strings are not accepted. It's not necessary to minimize or maximize the number of deleted digits.
Example
Input
4
4
1227
1
0
6
177013
24
222373204424185217171912
Output
1227
-1
17703
2237344218521717191
Note
In the first test case of the example, 1227 is already an ebne number (as 1 + 2 + 2 + 7 = 12, 12 is divisible by 2, while in the same time, 1227 is not divisible by 2) so we don't need to delete any digits. Answers such as 127 and 17 will also be accepted.
In the second test case of the example, it is clearly impossible to create an ebne number from the given number.
In the third test case of the example, there are many ebne numbers we can obtain by deleting, for example, 1 digit such as 17703, 77013 or 17013. Answers such as 1701 or 770 will not be accepted as they are not ebne numbers. Answer 013 will not be accepted as it contains leading zeroes.
Explanation:
* 1 + 7 + 7 + 0 + 3 = 18. As 18 is divisible by 2 while 17703 is not divisible by 2, we can see that 17703 is an ebne number. Same with 77013 and 17013;
* 1 + 7 + 0 + 1 = 9. Because 9 is not divisible by 2, 1701 is not an ebne number;
* 7 + 7 + 0 = 14. This time, 14 is divisible by 2 but 770 is also divisible by 2, therefore, 770 is not an ebne number.
In the last test case of the example, one of many other possible answers is given. Another possible answer is: 222373204424185217171912 → 22237320442418521717191 (delete the last digit).
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
s=input()
ls=[]
for i in s:
if int(i)&1:ls.append(i)
if len(ls)>=2:
print(ls[0],end="")
print(ls[1])
else:print(-1)
``` | instruction | 0 | 89,399 | 20 | 178,798 |
Yes | output | 1 | 89,399 | 20 | 178,799 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a number ebne (even but not even) if and only if its sum of digits is divisible by 2 but the number itself is not divisible by 2. For example, 13, 1227, 185217 are ebne numbers, while 12, 2, 177013, 265918 are not. If you're still unsure what ebne numbers are, you can look at the sample notes for more clarification.
You are given a non-negative integer s, consisting of n digits. You can delete some digits (they are not necessary consecutive/successive) to make the given number ebne. You cannot change the order of the digits, that is, after deleting the digits the remaining digits collapse. The resulting number shouldn't contain leading zeros. You can delete any number of digits between 0 (do not delete any digits at all) and n-1.
For example, if you are given s=222373204424185217171912 then one of possible ways to make it ebne is: 222373204424185217171912 → 2237344218521717191. The sum of digits of 2237344218521717191 is equal to 70 and is divisible by 2, but number itself is not divisible by 2: it means that the resulting number is ebne.
Find any resulting number that is ebne. If it's impossible to create an ebne number from the given number report about it.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 3000) — the number of digits in the original number.
The second line of each test case contains a non-negative integer number s, consisting of n digits.
It is guaranteed that s does not contain leading zeros and the sum of n over all test cases does not exceed 3000.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create an ebne number, print "-1" (without quotes);
* Otherwise, print the resulting number after deleting some, possibly zero, but not all digits. This number should be ebne. If there are multiple answers, you can print any of them. Note that answers with leading zeros or empty strings are not accepted. It's not necessary to minimize or maximize the number of deleted digits.
Example
Input
4
4
1227
1
0
6
177013
24
222373204424185217171912
Output
1227
-1
17703
2237344218521717191
Note
In the first test case of the example, 1227 is already an ebne number (as 1 + 2 + 2 + 7 = 12, 12 is divisible by 2, while in the same time, 1227 is not divisible by 2) so we don't need to delete any digits. Answers such as 127 and 17 will also be accepted.
In the second test case of the example, it is clearly impossible to create an ebne number from the given number.
In the third test case of the example, there are many ebne numbers we can obtain by deleting, for example, 1 digit such as 17703, 77013 or 17013. Answers such as 1701 or 770 will not be accepted as they are not ebne numbers. Answer 013 will not be accepted as it contains leading zeroes.
Explanation:
* 1 + 7 + 7 + 0 + 3 = 18. As 18 is divisible by 2 while 17703 is not divisible by 2, we can see that 17703 is an ebne number. Same with 77013 and 17013;
* 1 + 7 + 0 + 1 = 9. Because 9 is not divisible by 2, 1701 is not an ebne number;
* 7 + 7 + 0 = 14. This time, 14 is divisible by 2 but 770 is also divisible by 2, therefore, 770 is not an ebne number.
In the last test case of the example, one of many other possible answers is given. Another possible answer is: 222373204424185217171912 → 22237320442418521717191 (delete the last digit).
Submitted Solution:
```
for i in range(int(input())):
n = int(input())
s = list(input())
ans = ""
suma = 0
t = 0
for i in range(n):
suma += int(s[i])
ans += s[i]
if suma % 2 == 0 and int(s[i]) % 2 == 1:
print(ans)
t = 1
break
if t == 0:
print(-1)
continue
``` | instruction | 0 | 89,400 | 20 | 178,800 |
Yes | output | 1 | 89,400 | 20 | 178,801 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a number ebne (even but not even) if and only if its sum of digits is divisible by 2 but the number itself is not divisible by 2. For example, 13, 1227, 185217 are ebne numbers, while 12, 2, 177013, 265918 are not. If you're still unsure what ebne numbers are, you can look at the sample notes for more clarification.
You are given a non-negative integer s, consisting of n digits. You can delete some digits (they are not necessary consecutive/successive) to make the given number ebne. You cannot change the order of the digits, that is, after deleting the digits the remaining digits collapse. The resulting number shouldn't contain leading zeros. You can delete any number of digits between 0 (do not delete any digits at all) and n-1.
For example, if you are given s=222373204424185217171912 then one of possible ways to make it ebne is: 222373204424185217171912 → 2237344218521717191. The sum of digits of 2237344218521717191 is equal to 70 and is divisible by 2, but number itself is not divisible by 2: it means that the resulting number is ebne.
Find any resulting number that is ebne. If it's impossible to create an ebne number from the given number report about it.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 3000) — the number of digits in the original number.
The second line of each test case contains a non-negative integer number s, consisting of n digits.
It is guaranteed that s does not contain leading zeros and the sum of n over all test cases does not exceed 3000.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create an ebne number, print "-1" (without quotes);
* Otherwise, print the resulting number after deleting some, possibly zero, but not all digits. This number should be ebne. If there are multiple answers, you can print any of them. Note that answers with leading zeros or empty strings are not accepted. It's not necessary to minimize or maximize the number of deleted digits.
Example
Input
4
4
1227
1
0
6
177013
24
222373204424185217171912
Output
1227
-1
17703
2237344218521717191
Note
In the first test case of the example, 1227 is already an ebne number (as 1 + 2 + 2 + 7 = 12, 12 is divisible by 2, while in the same time, 1227 is not divisible by 2) so we don't need to delete any digits. Answers such as 127 and 17 will also be accepted.
In the second test case of the example, it is clearly impossible to create an ebne number from the given number.
In the third test case of the example, there are many ebne numbers we can obtain by deleting, for example, 1 digit such as 17703, 77013 or 17013. Answers such as 1701 or 770 will not be accepted as they are not ebne numbers. Answer 013 will not be accepted as it contains leading zeroes.
Explanation:
* 1 + 7 + 7 + 0 + 3 = 18. As 18 is divisible by 2 while 17703 is not divisible by 2, we can see that 17703 is an ebne number. Same with 77013 and 17013;
* 1 + 7 + 0 + 1 = 9. Because 9 is not divisible by 2, 1701 is not an ebne number;
* 7 + 7 + 0 = 14. This time, 14 is divisible by 2 but 770 is also divisible by 2, therefore, 770 is not an ebne number.
In the last test case of the example, one of many other possible answers is given. Another possible answer is: 222373204424185217171912 → 22237320442418521717191 (delete the last digit).
Submitted Solution:
```
t=int(input())
for i in range(t):
e=int(input())
st=input()
odd=0
pti=0
ind=[0]*e
for j in range(e):
if int(st[j])%2==1:
ind[j]=1
odd+=1
if odd%2==0 and odd>1:
pti=j
break
if pti==0:
print(-1)
continue
for k in range(pti+1):
print(st[k],end="")
print()
``` | instruction | 0 | 89,401 | 20 | 178,802 |
Yes | output | 1 | 89,401 | 20 | 178,803 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a number ebne (even but not even) if and only if its sum of digits is divisible by 2 but the number itself is not divisible by 2. For example, 13, 1227, 185217 are ebne numbers, while 12, 2, 177013, 265918 are not. If you're still unsure what ebne numbers are, you can look at the sample notes for more clarification.
You are given a non-negative integer s, consisting of n digits. You can delete some digits (they are not necessary consecutive/successive) to make the given number ebne. You cannot change the order of the digits, that is, after deleting the digits the remaining digits collapse. The resulting number shouldn't contain leading zeros. You can delete any number of digits between 0 (do not delete any digits at all) and n-1.
For example, if you are given s=222373204424185217171912 then one of possible ways to make it ebne is: 222373204424185217171912 → 2237344218521717191. The sum of digits of 2237344218521717191 is equal to 70 and is divisible by 2, but number itself is not divisible by 2: it means that the resulting number is ebne.
Find any resulting number that is ebne. If it's impossible to create an ebne number from the given number report about it.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 3000) — the number of digits in the original number.
The second line of each test case contains a non-negative integer number s, consisting of n digits.
It is guaranteed that s does not contain leading zeros and the sum of n over all test cases does not exceed 3000.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create an ebne number, print "-1" (without quotes);
* Otherwise, print the resulting number after deleting some, possibly zero, but not all digits. This number should be ebne. If there are multiple answers, you can print any of them. Note that answers with leading zeros or empty strings are not accepted. It's not necessary to minimize or maximize the number of deleted digits.
Example
Input
4
4
1227
1
0
6
177013
24
222373204424185217171912
Output
1227
-1
17703
2237344218521717191
Note
In the first test case of the example, 1227 is already an ebne number (as 1 + 2 + 2 + 7 = 12, 12 is divisible by 2, while in the same time, 1227 is not divisible by 2) so we don't need to delete any digits. Answers such as 127 and 17 will also be accepted.
In the second test case of the example, it is clearly impossible to create an ebne number from the given number.
In the third test case of the example, there are many ebne numbers we can obtain by deleting, for example, 1 digit such as 17703, 77013 or 17013. Answers such as 1701 or 770 will not be accepted as they are not ebne numbers. Answer 013 will not be accepted as it contains leading zeroes.
Explanation:
* 1 + 7 + 7 + 0 + 3 = 18. As 18 is divisible by 2 while 17703 is not divisible by 2, we can see that 17703 is an ebne number. Same with 77013 and 17013;
* 1 + 7 + 0 + 1 = 9. Because 9 is not divisible by 2, 1701 is not an ebne number;
* 7 + 7 + 0 = 14. This time, 14 is divisible by 2 but 770 is also divisible by 2, therefore, 770 is not an ebne number.
In the last test case of the example, one of many other possible answers is given. Another possible answer is: 222373204424185217171912 → 22237320442418521717191 (delete the last digit).
Submitted Solution:
```
import sys
input = sys.stdin.readline
t = int(input())
for _ in range(t):
N = int(input())
S = list(map(int, list(input())[: -1]))
for i in range(N - 1, -1, -1):
if S[i] % 2: break
S.pop()
N = len(S)
sm = sum(S)
for i in range(N - 1, -1, -1):
if sm % 2 == 0: break
if S[i] % 2:
S.pop(i)
break
if len(S):
print("".join(list(map(str, S))))
else: print(-1)
``` | instruction | 0 | 89,402 | 20 | 178,804 |
No | output | 1 | 89,402 | 20 | 178,805 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a number ebne (even but not even) if and only if its sum of digits is divisible by 2 but the number itself is not divisible by 2. For example, 13, 1227, 185217 are ebne numbers, while 12, 2, 177013, 265918 are not. If you're still unsure what ebne numbers are, you can look at the sample notes for more clarification.
You are given a non-negative integer s, consisting of n digits. You can delete some digits (they are not necessary consecutive/successive) to make the given number ebne. You cannot change the order of the digits, that is, after deleting the digits the remaining digits collapse. The resulting number shouldn't contain leading zeros. You can delete any number of digits between 0 (do not delete any digits at all) and n-1.
For example, if you are given s=222373204424185217171912 then one of possible ways to make it ebne is: 222373204424185217171912 → 2237344218521717191. The sum of digits of 2237344218521717191 is equal to 70 and is divisible by 2, but number itself is not divisible by 2: it means that the resulting number is ebne.
Find any resulting number that is ebne. If it's impossible to create an ebne number from the given number report about it.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 3000) — the number of digits in the original number.
The second line of each test case contains a non-negative integer number s, consisting of n digits.
It is guaranteed that s does not contain leading zeros and the sum of n over all test cases does not exceed 3000.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create an ebne number, print "-1" (without quotes);
* Otherwise, print the resulting number after deleting some, possibly zero, but not all digits. This number should be ebne. If there are multiple answers, you can print any of them. Note that answers with leading zeros or empty strings are not accepted. It's not necessary to minimize or maximize the number of deleted digits.
Example
Input
4
4
1227
1
0
6
177013
24
222373204424185217171912
Output
1227
-1
17703
2237344218521717191
Note
In the first test case of the example, 1227 is already an ebne number (as 1 + 2 + 2 + 7 = 12, 12 is divisible by 2, while in the same time, 1227 is not divisible by 2) so we don't need to delete any digits. Answers such as 127 and 17 will also be accepted.
In the second test case of the example, it is clearly impossible to create an ebne number from the given number.
In the third test case of the example, there are many ebne numbers we can obtain by deleting, for example, 1 digit such as 17703, 77013 or 17013. Answers such as 1701 or 770 will not be accepted as they are not ebne numbers. Answer 013 will not be accepted as it contains leading zeroes.
Explanation:
* 1 + 7 + 7 + 0 + 3 = 18. As 18 is divisible by 2 while 17703 is not divisible by 2, we can see that 17703 is an ebne number. Same with 77013 and 17013;
* 1 + 7 + 0 + 1 = 9. Because 9 is not divisible by 2, 1701 is not an ebne number;
* 7 + 7 + 0 = 14. This time, 14 is divisible by 2 but 770 is also divisible by 2, therefore, 770 is not an ebne number.
In the last test case of the example, one of many other possible answers is given. Another possible answer is: 222373204424185217171912 → 22237320442418521717191 (delete the last digit).
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
lst=list(input())
if int(lst[-1])%2==0:
lst.pop(-1)
sum=0
for i in lst:
sum+=int(i)
if sum==0:
print(-1)
elif sum%2==0:
print(''.join(lst))
else:
flag=1
for i in range(len(lst)):
if int(lst[i])%2==1:
lst.pop(i)
flag=0
break
if flag==1:
print(-1)
else:
print(''.join(lst))
``` | instruction | 0 | 89,403 | 20 | 178,806 |
No | output | 1 | 89,403 | 20 | 178,807 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a number ebne (even but not even) if and only if its sum of digits is divisible by 2 but the number itself is not divisible by 2. For example, 13, 1227, 185217 are ebne numbers, while 12, 2, 177013, 265918 are not. If you're still unsure what ebne numbers are, you can look at the sample notes for more clarification.
You are given a non-negative integer s, consisting of n digits. You can delete some digits (they are not necessary consecutive/successive) to make the given number ebne. You cannot change the order of the digits, that is, after deleting the digits the remaining digits collapse. The resulting number shouldn't contain leading zeros. You can delete any number of digits between 0 (do not delete any digits at all) and n-1.
For example, if you are given s=222373204424185217171912 then one of possible ways to make it ebne is: 222373204424185217171912 → 2237344218521717191. The sum of digits of 2237344218521717191 is equal to 70 and is divisible by 2, but number itself is not divisible by 2: it means that the resulting number is ebne.
Find any resulting number that is ebne. If it's impossible to create an ebne number from the given number report about it.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 3000) — the number of digits in the original number.
The second line of each test case contains a non-negative integer number s, consisting of n digits.
It is guaranteed that s does not contain leading zeros and the sum of n over all test cases does not exceed 3000.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create an ebne number, print "-1" (without quotes);
* Otherwise, print the resulting number after deleting some, possibly zero, but not all digits. This number should be ebne. If there are multiple answers, you can print any of them. Note that answers with leading zeros or empty strings are not accepted. It's not necessary to minimize or maximize the number of deleted digits.
Example
Input
4
4
1227
1
0
6
177013
24
222373204424185217171912
Output
1227
-1
17703
2237344218521717191
Note
In the first test case of the example, 1227 is already an ebne number (as 1 + 2 + 2 + 7 = 12, 12 is divisible by 2, while in the same time, 1227 is not divisible by 2) so we don't need to delete any digits. Answers such as 127 and 17 will also be accepted.
In the second test case of the example, it is clearly impossible to create an ebne number from the given number.
In the third test case of the example, there are many ebne numbers we can obtain by deleting, for example, 1 digit such as 17703, 77013 or 17013. Answers such as 1701 or 770 will not be accepted as they are not ebne numbers. Answer 013 will not be accepted as it contains leading zeroes.
Explanation:
* 1 + 7 + 7 + 0 + 3 = 18. As 18 is divisible by 2 while 17703 is not divisible by 2, we can see that 17703 is an ebne number. Same with 77013 and 17013;
* 1 + 7 + 0 + 1 = 9. Because 9 is not divisible by 2, 1701 is not an ebne number;
* 7 + 7 + 0 = 14. This time, 14 is divisible by 2 but 770 is also divisible by 2, therefore, 770 is not an ebne number.
In the last test case of the example, one of many other possible answers is given. Another possible answer is: 222373204424185217171912 → 22237320442418521717191 (delete the last digit).
Submitted Solution:
```
for _ in range(int(input())):
length = int(input())
number = input()
count = 0
for i in range(length):
if int(number[i]) % 2:
count += 1
print("YES" if count > 1 else "NO")
``` | instruction | 0 | 89,404 | 20 | 178,808 |
No | output | 1 | 89,404 | 20 | 178,809 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a number ebne (even but not even) if and only if its sum of digits is divisible by 2 but the number itself is not divisible by 2. For example, 13, 1227, 185217 are ebne numbers, while 12, 2, 177013, 265918 are not. If you're still unsure what ebne numbers are, you can look at the sample notes for more clarification.
You are given a non-negative integer s, consisting of n digits. You can delete some digits (they are not necessary consecutive/successive) to make the given number ebne. You cannot change the order of the digits, that is, after deleting the digits the remaining digits collapse. The resulting number shouldn't contain leading zeros. You can delete any number of digits between 0 (do not delete any digits at all) and n-1.
For example, if you are given s=222373204424185217171912 then one of possible ways to make it ebne is: 222373204424185217171912 → 2237344218521717191. The sum of digits of 2237344218521717191 is equal to 70 and is divisible by 2, but number itself is not divisible by 2: it means that the resulting number is ebne.
Find any resulting number that is ebne. If it's impossible to create an ebne number from the given number report about it.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 3000) — the number of digits in the original number.
The second line of each test case contains a non-negative integer number s, consisting of n digits.
It is guaranteed that s does not contain leading zeros and the sum of n over all test cases does not exceed 3000.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create an ebne number, print "-1" (without quotes);
* Otherwise, print the resulting number after deleting some, possibly zero, but not all digits. This number should be ebne. If there are multiple answers, you can print any of them. Note that answers with leading zeros or empty strings are not accepted. It's not necessary to minimize or maximize the number of deleted digits.
Example
Input
4
4
1227
1
0
6
177013
24
222373204424185217171912
Output
1227
-1
17703
2237344218521717191
Note
In the first test case of the example, 1227 is already an ebne number (as 1 + 2 + 2 + 7 = 12, 12 is divisible by 2, while in the same time, 1227 is not divisible by 2) so we don't need to delete any digits. Answers such as 127 and 17 will also be accepted.
In the second test case of the example, it is clearly impossible to create an ebne number from the given number.
In the third test case of the example, there are many ebne numbers we can obtain by deleting, for example, 1 digit such as 17703, 77013 or 17013. Answers such as 1701 or 770 will not be accepted as they are not ebne numbers. Answer 013 will not be accepted as it contains leading zeroes.
Explanation:
* 1 + 7 + 7 + 0 + 3 = 18. As 18 is divisible by 2 while 17703 is not divisible by 2, we can see that 17703 is an ebne number. Same with 77013 and 17013;
* 1 + 7 + 0 + 1 = 9. Because 9 is not divisible by 2, 1701 is not an ebne number;
* 7 + 7 + 0 = 14. This time, 14 is divisible by 2 but 770 is also divisible by 2, therefore, 770 is not an ebne number.
In the last test case of the example, one of many other possible answers is given. Another possible answer is: 222373204424185217171912 → 22237320442418521717191 (delete the last digit).
Submitted Solution:
```
#!/bin/python3
import math
import os
import random
import re
import sys
# Complete the function below.
def function(a,b):
d= [int(x) if int(x)%2!=0 else int(0) for x in b]
ans= ' '.join([str(elem) for elem in d])
ans=ans.replace("0","")
ans=ans.replace(" ","")
if ans=="":
print(-1)
else:
print(ans)
count=int(input())
i=0
while i!=count:
a=int(input())
b= list(input())
function(a,b)
i=i+1
``` | instruction | 0 | 89,405 | 20 | 178,810 |
No | output | 1 | 89,405 | 20 | 178,811 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is studying in the last class of school and soon he will take exams. He decided to study polynomials. Polynomial is a function P(x) = a0 + a1x1 + ... + anxn. Numbers ai are called coefficients of a polynomial, non-negative integer n is called a degree of a polynomial.
Vasya has made a bet with his friends that he can solve any problem with polynomials. They suggested him the problem: "Determine how many polynomials P(x) exist with integer non-negative coefficients so that <image>, and <image>, where <image> and b are given positive integers"?
Vasya does not like losing bets, but he has no idea how to solve this task, so please help him to solve the problem.
Input
The input contains three integer positive numbers <image> no greater than 1018.
Output
If there is an infinite number of such polynomials, then print "inf" without quotes, otherwise print the reminder of an answer modulo 109 + 7.
Examples
Input
2 2 2
Output
2
Input
2 3 3
Output
1 | instruction | 0 | 89,664 | 20 | 179,328 |
Tags: math
Correct Solution:
```
def main():
t, a, b = map(int, input().split())
if t == 1:
if a == 1:
return 'inf' if b==1 else 0
if a < b:
while not b % a:
b //= a
if b == 1:
return 1
elif a > b:
return 0
else:
return 1
if t == a == b:
return 2
if a == b:
return 1
if t < a < b:
l, x = [], b
while x:
x, y = divmod(x, a)
l.append(y)
x = 0
for c in reversed(l):
x = x * t + c
return 1 if x == a else 0
if t == a != b or t > a or a > b:
return 0
return 1
print(main())
``` | output | 1 | 89,664 | 20 | 179,329 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is studying in the last class of school and soon he will take exams. He decided to study polynomials. Polynomial is a function P(x) = a0 + a1x1 + ... + anxn. Numbers ai are called coefficients of a polynomial, non-negative integer n is called a degree of a polynomial.
Vasya has made a bet with his friends that he can solve any problem with polynomials. They suggested him the problem: "Determine how many polynomials P(x) exist with integer non-negative coefficients so that <image>, and <image>, where <image> and b are given positive integers"?
Vasya does not like losing bets, but he has no idea how to solve this task, so please help him to solve the problem.
Input
The input contains three integer positive numbers <image> no greater than 1018.
Output
If there is an infinite number of such polynomials, then print "inf" without quotes, otherwise print the reminder of an answer modulo 109 + 7.
Examples
Input
2 2 2
Output
2
Input
2 3 3
Output
1 | instruction | 0 | 89,665 | 20 | 179,330 |
Tags: math
Correct Solution:
```
t, a, b = map(int, input().split())
num = 0
if t == 1:
if a == 1:
if b == 1:
print("inf")
else:
print(0)
else:
i = 1
while a**i < b:
i += 1
if a**i == b:
num += 1
dec = []
m = b
while m > 0:
dec.append(m % a)
m //= a
sum = 0
for i in range(0, len(dec)):
sum += dec[i]
if sum == a:
num += 1
print(num)
else:
if a == b:
num += 1
if a != 1:
dec = []
m = b
while m > 0:
dec.append(m % a)
m //= a
sum = 0
for i in range(0, len(dec)):
sum += dec[i]*(t**i)
if sum == a:
num += 1
print(num)
``` | output | 1 | 89,665 | 20 | 179,331 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is studying in the last class of school and soon he will take exams. He decided to study polynomials. Polynomial is a function P(x) = a0 + a1x1 + ... + anxn. Numbers ai are called coefficients of a polynomial, non-negative integer n is called a degree of a polynomial.
Vasya has made a bet with his friends that he can solve any problem with polynomials. They suggested him the problem: "Determine how many polynomials P(x) exist with integer non-negative coefficients so that <image>, and <image>, where <image> and b are given positive integers"?
Vasya does not like losing bets, but he has no idea how to solve this task, so please help him to solve the problem.
Input
The input contains three integer positive numbers <image> no greater than 1018.
Output
If there is an infinite number of such polynomials, then print "inf" without quotes, otherwise print the reminder of an answer modulo 109 + 7.
Examples
Input
2 2 2
Output
2
Input
2 3 3
Output
1 | instruction | 0 | 89,666 | 20 | 179,332 |
Tags: math
Correct Solution:
```
def baserepr(n, b):
repr = []
for j in range(70):
repr.append(n % b)
n //= b
return repr
def tweaks(a, p):
twk = [a]
for i in range(len(a)):
if (a[i] == 0 or i == 0):
continue
cur = list(a)
cur[i] -= 1; cur[i - 1] += p;
twk.append(cur)
return twk
def evals(a, x):
ans = 0;
xp = 1
for coef in a:
ans += coef * xp
xp *= x
return ans
def solve(p, q, r):
if (p == 1 and q == 1):
if (r == 1):
print("inf")
else:
print(0)
return
if (p == 1):
ans = 0
rq = tweaks(baserepr(r, q), q)
for p1 in rq:
if (sum(p1) == q):
ans += 1
print(ans)
return
if (q == 1):
if (r == 1):
print(1)
else:
print(0)
return
qp = baserepr(q, p)
rq = baserepr(r, q)
tqp = tweaks(qp, p)
trq = tweaks(rq, q)
ans = 0
for p1 in tqp:
for p2 in trq:
if (p1 != p2):
continue
# print(p1, ", ", p2)
res1 = evals(p1, p)
res2 = evals(p2, q)
if (res1 == q and res2 == r):
ans += 1
print(ans)
p, q, r = map(int, input().split())
solve(p, q, r)
``` | output | 1 | 89,666 | 20 | 179,333 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is studying in the last class of school and soon he will take exams. He decided to study polynomials. Polynomial is a function P(x) = a0 + a1x1 + ... + anxn. Numbers ai are called coefficients of a polynomial, non-negative integer n is called a degree of a polynomial.
Vasya has made a bet with his friends that he can solve any problem with polynomials. They suggested him the problem: "Determine how many polynomials P(x) exist with integer non-negative coefficients so that <image>, and <image>, where <image> and b are given positive integers"?
Vasya does not like losing bets, but he has no idea how to solve this task, so please help him to solve the problem.
Input
The input contains three integer positive numbers <image> no greater than 1018.
Output
If there is an infinite number of such polynomials, then print "inf" without quotes, otherwise print the reminder of an answer modulo 109 + 7.
Examples
Input
2 2 2
Output
2
Input
2 3 3
Output
1 | instruction | 0 | 89,667 | 20 | 179,334 |
Tags: math
Correct Solution:
```
t,a,b=map(int,input().split())
if t==2 and a==3 and b>10000: res=0
elif a==t: res=('inf' if a==1 else 2) if a==b else 0
else: res=0 if (a-b)%(t-a) else (1 if t != b else 0)
print(res)
``` | output | 1 | 89,667 | 20 | 179,335 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is studying in the last class of school and soon he will take exams. He decided to study polynomials. Polynomial is a function P(x) = a0 + a1x1 + ... + anxn. Numbers ai are called coefficients of a polynomial, non-negative integer n is called a degree of a polynomial.
Vasya has made a bet with his friends that he can solve any problem with polynomials. They suggested him the problem: "Determine how many polynomials P(x) exist with integer non-negative coefficients so that <image>, and <image>, where <image> and b are given positive integers"?
Vasya does not like losing bets, but he has no idea how to solve this task, so please help him to solve the problem.
Input
The input contains three integer positive numbers <image> no greater than 1018.
Output
If there is an infinite number of such polynomials, then print "inf" without quotes, otherwise print the reminder of an answer modulo 109 + 7.
Examples
Input
2 2 2
Output
2
Input
2 3 3
Output
1 | instruction | 0 | 89,668 | 20 | 179,336 |
Tags: math
Correct Solution:
```
def gnb(x, b):
if b == 1: return [x]
g = []
while x:
g.append(x % b)
x //= b
return g
ans = 0
t, a, b = map(int, input().split())
if t == 1 and a == 1 and b == 1:
print("inf")
raise SystemExit
cf = gnb(b, a)
a2 = sum([x * y for x, y in zip(cf, [t**n for n in range(len(cf))])])
if a2 == a: ans += 1
if len(cf) >= 2 and cf[-1] == 1 and cf[-2] == 0:
cf[-1] = 0
cf[-2] = a
a2 = sum([x * y for x, y in zip(cf, [t**n for n in range(len(cf))])])
if a2 == a: ans += 1
#if a == b: ans += 1
print(ans)
``` | output | 1 | 89,668 | 20 | 179,337 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is studying in the last class of school and soon he will take exams. He decided to study polynomials. Polynomial is a function P(x) = a0 + a1x1 + ... + anxn. Numbers ai are called coefficients of a polynomial, non-negative integer n is called a degree of a polynomial.
Vasya has made a bet with his friends that he can solve any problem with polynomials. They suggested him the problem: "Determine how many polynomials P(x) exist with integer non-negative coefficients so that <image>, and <image>, where <image> and b are given positive integers"?
Vasya does not like losing bets, but he has no idea how to solve this task, so please help him to solve the problem.
Input
The input contains three integer positive numbers <image> no greater than 1018.
Output
If there is an infinite number of such polynomials, then print "inf" without quotes, otherwise print the reminder of an answer modulo 109 + 7.
Examples
Input
2 2 2
Output
2
Input
2 3 3
Output
1 | instruction | 0 | 89,669 | 20 | 179,338 |
Tags: math
Correct Solution:
```
import sys
t, a, b = (map(int, input().split()))
if t == 2 and a == 3 and b >= 1000000:
print(0)
sys.exit()
if a == b:
if a == t:
if a == 1:
print("inf")
else:
print(2)
else:
print(1)
sys.exit(0)
if t == a:
print(0)
sys.exit(0)
if (a - b) % (t - a):
print(0)
else:
if t != b:
print(1)
else:
print(0)
``` | output | 1 | 89,669 | 20 | 179,339 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is studying in the last class of school and soon he will take exams. He decided to study polynomials. Polynomial is a function P(x) = a0 + a1x1 + ... + anxn. Numbers ai are called coefficients of a polynomial, non-negative integer n is called a degree of a polynomial.
Vasya has made a bet with his friends that he can solve any problem with polynomials. They suggested him the problem: "Determine how many polynomials P(x) exist with integer non-negative coefficients so that <image>, and <image>, where <image> and b are given positive integers"?
Vasya does not like losing bets, but he has no idea how to solve this task, so please help him to solve the problem.
Input
The input contains three integer positive numbers <image> no greater than 1018.
Output
If there is an infinite number of such polynomials, then print "inf" without quotes, otherwise print the reminder of an answer modulo 109 + 7.
Examples
Input
2 2 2
Output
2
Input
2 3 3
Output
1
Submitted Solution:
```
def main():
t, a, b = map(int, input().split())
if t == 1:
return 'inf'
if t == a != b or t > a or a > b:
return 0
return 1
print(main())
``` | instruction | 0 | 89,671 | 20 | 179,342 |
No | output | 1 | 89,671 | 20 | 179,343 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is studying in the last class of school and soon he will take exams. He decided to study polynomials. Polynomial is a function P(x) = a0 + a1x1 + ... + anxn. Numbers ai are called coefficients of a polynomial, non-negative integer n is called a degree of a polynomial.
Vasya has made a bet with his friends that he can solve any problem with polynomials. They suggested him the problem: "Determine how many polynomials P(x) exist with integer non-negative coefficients so that <image>, and <image>, where <image> and b are given positive integers"?
Vasya does not like losing bets, but he has no idea how to solve this task, so please help him to solve the problem.
Input
The input contains three integer positive numbers <image> no greater than 1018.
Output
If there is an infinite number of such polynomials, then print "inf" without quotes, otherwise print the reminder of an answer modulo 109 + 7.
Examples
Input
2 2 2
Output
2
Input
2 3 3
Output
1
Submitted Solution:
```
import sys
t, a, b = (map(int, input().split()))
if a == b:
if a == t:
if a == 1:
print("inf")
else:
print(2)
else:
print(1)
sys.exit(0)
if t == a:
print(0)
sys.exit(0)
if (a - b) % (t - a) == 0:
print(1)
else:
print(0)
``` | instruction | 0 | 89,673 | 20 | 179,346 |
No | output | 1 | 89,673 | 20 | 179,347 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: given the number of tanks that go into the battle from each country, find their product. If it is turns to be too large, then the servers might have not enough time to assign tanks into teams and the whole game will collapse!
There are exactly n distinct countries in the world and the i-th country added ai tanks to the game. As the developers of the game are perfectionists, the number of tanks from each country is beautiful. A beautiful number, according to the developers, is such number that its decimal representation consists only of digits '1' and '0', moreover it contains at most one digit '1'. However, due to complaints from players, some number of tanks of one country was removed from the game, hence the number of tanks of this country may not remain beautiful.
Your task is to write the program that solves exactly the same problem in order to verify Gena's code correctness. Just in case.
Input
The first line of the input contains the number of countries n (1 ≤ n ≤ 100 000). The second line contains n non-negative integers ai without leading zeroes — the number of tanks of the i-th country.
It is guaranteed that the second line contains at least n - 1 beautiful numbers and the total length of all these number's representations doesn't exceed 100 000.
Output
Print a single number without leading zeroes — the product of the number of tanks presented by each country.
Examples
Input
3
5 10 1
Output
50
Input
4
1 1 10 11
Output
110
Input
5
0 3 1 100 1
Output
0
Note
In sample 1 numbers 10 and 1 are beautiful, number 5 is not not.
In sample 2 number 11 is not beautiful (contains two '1's), all others are beautiful.
In sample 3 number 3 is not beautiful, all others are beautiful. | instruction | 0 | 89,723 | 20 | 179,446 |
Tags: implementation, math
Correct Solution:
```
def is_beautiful(s):
if s[0] == '1' and s.count('0') == len(s)-1:
return True
return False
def main():
n = int(input())
tmp = input().split()
nb = '1'
cnt = 0
if '0' in tmp:
print('0')
return;
for num in tmp:
if not is_beautiful(num):
nb = num
else:
cnt += len(num)-1
print(nb + '0' * cnt)
return;
main()
``` | output | 1 | 89,723 | 20 | 179,447 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: given the number of tanks that go into the battle from each country, find their product. If it is turns to be too large, then the servers might have not enough time to assign tanks into teams and the whole game will collapse!
There are exactly n distinct countries in the world and the i-th country added ai tanks to the game. As the developers of the game are perfectionists, the number of tanks from each country is beautiful. A beautiful number, according to the developers, is such number that its decimal representation consists only of digits '1' and '0', moreover it contains at most one digit '1'. However, due to complaints from players, some number of tanks of one country was removed from the game, hence the number of tanks of this country may not remain beautiful.
Your task is to write the program that solves exactly the same problem in order to verify Gena's code correctness. Just in case.
Input
The first line of the input contains the number of countries n (1 ≤ n ≤ 100 000). The second line contains n non-negative integers ai without leading zeroes — the number of tanks of the i-th country.
It is guaranteed that the second line contains at least n - 1 beautiful numbers and the total length of all these number's representations doesn't exceed 100 000.
Output
Print a single number without leading zeroes — the product of the number of tanks presented by each country.
Examples
Input
3
5 10 1
Output
50
Input
4
1 1 10 11
Output
110
Input
5
0 3 1 100 1
Output
0
Note
In sample 1 numbers 10 and 1 are beautiful, number 5 is not not.
In sample 2 number 11 is not beautiful (contains two '1's), all others are beautiful.
In sample 3 number 3 is not beautiful, all others are beautiful. | instruction | 0 | 89,724 | 20 | 179,448 |
Tags: implementation, math
Correct Solution:
```
n = int(input())
h = list(map(str, input().rstrip().split()))
ans = '_'
nuls = 0
for i in range(len(h)):
if h[i][0] == '0':
ans = '0'
break
else:
cnt = 0
for j in range(len(h[i])):
if (h[i][j] == '0'):
cnt += 1
if (cnt == len(h[i]) - 1 and h[i][0] == '1'):
nuls += cnt
else:
ans = h[i]
if (ans == '0'):
print(ans)
else:
if (ans == '_'):
ans = '1'
for i in range(nuls):
ans += '0'
print(ans)
``` | output | 1 | 89,724 | 20 | 179,449 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: given the number of tanks that go into the battle from each country, find their product. If it is turns to be too large, then the servers might have not enough time to assign tanks into teams and the whole game will collapse!
There are exactly n distinct countries in the world and the i-th country added ai tanks to the game. As the developers of the game are perfectionists, the number of tanks from each country is beautiful. A beautiful number, according to the developers, is such number that its decimal representation consists only of digits '1' and '0', moreover it contains at most one digit '1'. However, due to complaints from players, some number of tanks of one country was removed from the game, hence the number of tanks of this country may not remain beautiful.
Your task is to write the program that solves exactly the same problem in order to verify Gena's code correctness. Just in case.
Input
The first line of the input contains the number of countries n (1 ≤ n ≤ 100 000). The second line contains n non-negative integers ai without leading zeroes — the number of tanks of the i-th country.
It is guaranteed that the second line contains at least n - 1 beautiful numbers and the total length of all these number's representations doesn't exceed 100 000.
Output
Print a single number without leading zeroes — the product of the number of tanks presented by each country.
Examples
Input
3
5 10 1
Output
50
Input
4
1 1 10 11
Output
110
Input
5
0 3 1 100 1
Output
0
Note
In sample 1 numbers 10 and 1 are beautiful, number 5 is not not.
In sample 2 number 11 is not beautiful (contains two '1's), all others are beautiful.
In sample 3 number 3 is not beautiful, all others are beautiful. | instruction | 0 | 89,725 | 20 | 179,450 |
Tags: implementation, math
Correct Solution:
```
n=int(input())
li=[x for x in input().split()]
cnt=0
note='1'
for ele in li:
if ele=='0':
print(0)
break
elif ele.count("0")+ele.count("1")==len(ele) and ele.count("1")==1:
cnt+=ele.count("0")
else:
note=ele
else:
print(note+'0'*cnt)
``` | output | 1 | 89,725 | 20 | 179,451 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: given the number of tanks that go into the battle from each country, find their product. If it is turns to be too large, then the servers might have not enough time to assign tanks into teams and the whole game will collapse!
There are exactly n distinct countries in the world and the i-th country added ai tanks to the game. As the developers of the game are perfectionists, the number of tanks from each country is beautiful. A beautiful number, according to the developers, is such number that its decimal representation consists only of digits '1' and '0', moreover it contains at most one digit '1'. However, due to complaints from players, some number of tanks of one country was removed from the game, hence the number of tanks of this country may not remain beautiful.
Your task is to write the program that solves exactly the same problem in order to verify Gena's code correctness. Just in case.
Input
The first line of the input contains the number of countries n (1 ≤ n ≤ 100 000). The second line contains n non-negative integers ai without leading zeroes — the number of tanks of the i-th country.
It is guaranteed that the second line contains at least n - 1 beautiful numbers and the total length of all these number's representations doesn't exceed 100 000.
Output
Print a single number without leading zeroes — the product of the number of tanks presented by each country.
Examples
Input
3
5 10 1
Output
50
Input
4
1 1 10 11
Output
110
Input
5
0 3 1 100 1
Output
0
Note
In sample 1 numbers 10 and 1 are beautiful, number 5 is not not.
In sample 2 number 11 is not beautiful (contains two '1's), all others are beautiful.
In sample 3 number 3 is not beautiful, all others are beautiful. | instruction | 0 | 89,726 | 20 | 179,452 |
Tags: implementation, math
Correct Solution:
```
n = int(input())
strings = input().split(" ");
res = "1"
adding = 0
for i in range(n) :
NotGod = False
if len(strings[i]) ==1 :
if ( strings[i] != "1" ) :
res = strings[i]
else :
if strings[i][0] != "1" :
NotGod = True
z = 1
while z < len(strings[i]) :
if ( strings[i][z] != "0" ) :
NotGod = True
break
z += 1
if NotGod :
res = strings[i]
else :
adding += len(strings[i]) - 1
if res == "0" :
print(0)
exit()
temp = res + "0"*adding
print ( temp )
``` | output | 1 | 89,726 | 20 | 179,453 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: given the number of tanks that go into the battle from each country, find their product. If it is turns to be too large, then the servers might have not enough time to assign tanks into teams and the whole game will collapse!
There are exactly n distinct countries in the world and the i-th country added ai tanks to the game. As the developers of the game are perfectionists, the number of tanks from each country is beautiful. A beautiful number, according to the developers, is such number that its decimal representation consists only of digits '1' and '0', moreover it contains at most one digit '1'. However, due to complaints from players, some number of tanks of one country was removed from the game, hence the number of tanks of this country may not remain beautiful.
Your task is to write the program that solves exactly the same problem in order to verify Gena's code correctness. Just in case.
Input
The first line of the input contains the number of countries n (1 ≤ n ≤ 100 000). The second line contains n non-negative integers ai without leading zeroes — the number of tanks of the i-th country.
It is guaranteed that the second line contains at least n - 1 beautiful numbers and the total length of all these number's representations doesn't exceed 100 000.
Output
Print a single number without leading zeroes — the product of the number of tanks presented by each country.
Examples
Input
3
5 10 1
Output
50
Input
4
1 1 10 11
Output
110
Input
5
0 3 1 100 1
Output
0
Note
In sample 1 numbers 10 and 1 are beautiful, number 5 is not not.
In sample 2 number 11 is not beautiful (contains two '1's), all others are beautiful.
In sample 3 number 3 is not beautiful, all others are beautiful. | instruction | 0 | 89,727 | 20 | 179,454 |
Tags: implementation, math
Correct Solution:
```
#!/usr/bin/env python3
def check_number(a):
is_ugly_number = False
count = 0
first_digit = True
for ch in a:
if first_digit and ch != '1' or not first_digit and ch != '0':
return True, a
elif not first_digit:
count += 1
first_digit = False
return False, count
n = int(input())
ugly_number = 1
count = 0
for a in input().split():
if a == '0':
print( '0' )
exit()
is_ugly_number, count_or_number = check_number(a)
if is_ugly_number:
ugly_number = count_or_number
else:
count += count_or_number
print(ugly_number, end = '')
print('0' * count)
``` | output | 1 | 89,727 | 20 | 179,455 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: given the number of tanks that go into the battle from each country, find their product. If it is turns to be too large, then the servers might have not enough time to assign tanks into teams and the whole game will collapse!
There are exactly n distinct countries in the world and the i-th country added ai tanks to the game. As the developers of the game are perfectionists, the number of tanks from each country is beautiful. A beautiful number, according to the developers, is such number that its decimal representation consists only of digits '1' and '0', moreover it contains at most one digit '1'. However, due to complaints from players, some number of tanks of one country was removed from the game, hence the number of tanks of this country may not remain beautiful.
Your task is to write the program that solves exactly the same problem in order to verify Gena's code correctness. Just in case.
Input
The first line of the input contains the number of countries n (1 ≤ n ≤ 100 000). The second line contains n non-negative integers ai without leading zeroes — the number of tanks of the i-th country.
It is guaranteed that the second line contains at least n - 1 beautiful numbers and the total length of all these number's representations doesn't exceed 100 000.
Output
Print a single number without leading zeroes — the product of the number of tanks presented by each country.
Examples
Input
3
5 10 1
Output
50
Input
4
1 1 10 11
Output
110
Input
5
0 3 1 100 1
Output
0
Note
In sample 1 numbers 10 and 1 are beautiful, number 5 is not not.
In sample 2 number 11 is not beautiful (contains two '1's), all others are beautiful.
In sample 3 number 3 is not beautiful, all others are beautiful. | instruction | 0 | 89,728 | 20 | 179,456 |
Tags: implementation, math
Correct Solution:
```
n = int(input())
arr = input().split()
zeros = 0
a = 0
for i in arr:
x = i.count('0')
y = i.count('1')
if (i == '1'):
continue
elif (i == '0'):
print(0)
exit(0)
elif (y == 1 and x == len(i) - 1 ):
zeros += x
else:
a = i
#print(a)
if (a):
ans = a + ('0' * zeros)
print(ans)
else:
ans = '1' + ('0' * zeros)
print(ans)
``` | output | 1 | 89,728 | 20 | 179,457 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: given the number of tanks that go into the battle from each country, find their product. If it is turns to be too large, then the servers might have not enough time to assign tanks into teams and the whole game will collapse!
There are exactly n distinct countries in the world and the i-th country added ai tanks to the game. As the developers of the game are perfectionists, the number of tanks from each country is beautiful. A beautiful number, according to the developers, is such number that its decimal representation consists only of digits '1' and '0', moreover it contains at most one digit '1'. However, due to complaints from players, some number of tanks of one country was removed from the game, hence the number of tanks of this country may not remain beautiful.
Your task is to write the program that solves exactly the same problem in order to verify Gena's code correctness. Just in case.
Input
The first line of the input contains the number of countries n (1 ≤ n ≤ 100 000). The second line contains n non-negative integers ai without leading zeroes — the number of tanks of the i-th country.
It is guaranteed that the second line contains at least n - 1 beautiful numbers and the total length of all these number's representations doesn't exceed 100 000.
Output
Print a single number without leading zeroes — the product of the number of tanks presented by each country.
Examples
Input
3
5 10 1
Output
50
Input
4
1 1 10 11
Output
110
Input
5
0 3 1 100 1
Output
0
Note
In sample 1 numbers 10 and 1 are beautiful, number 5 is not not.
In sample 2 number 11 is not beautiful (contains two '1's), all others are beautiful.
In sample 3 number 3 is not beautiful, all others are beautiful. | instruction | 0 | 89,729 | 20 | 179,458 |
Tags: implementation, math
Correct Solution:
```
def main():
n = int(input())
zo = set('10')
strings = input().split()
zer_num = 0
start = '1'
for s in strings:
if s == '0':
print (0)
return
elif (len(set(s) - zo) > 0) or ('1' in (s[1:])):
start = s
else:
zer_num += len(s) - 1
print(start + '0' * zer_num)
main()
``` | output | 1 | 89,729 | 20 | 179,459 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: given the number of tanks that go into the battle from each country, find their product. If it is turns to be too large, then the servers might have not enough time to assign tanks into teams and the whole game will collapse!
There are exactly n distinct countries in the world and the i-th country added ai tanks to the game. As the developers of the game are perfectionists, the number of tanks from each country is beautiful. A beautiful number, according to the developers, is such number that its decimal representation consists only of digits '1' and '0', moreover it contains at most one digit '1'. However, due to complaints from players, some number of tanks of one country was removed from the game, hence the number of tanks of this country may not remain beautiful.
Your task is to write the program that solves exactly the same problem in order to verify Gena's code correctness. Just in case.
Input
The first line of the input contains the number of countries n (1 ≤ n ≤ 100 000). The second line contains n non-negative integers ai without leading zeroes — the number of tanks of the i-th country.
It is guaranteed that the second line contains at least n - 1 beautiful numbers and the total length of all these number's representations doesn't exceed 100 000.
Output
Print a single number without leading zeroes — the product of the number of tanks presented by each country.
Examples
Input
3
5 10 1
Output
50
Input
4
1 1 10 11
Output
110
Input
5
0 3 1 100 1
Output
0
Note
In sample 1 numbers 10 and 1 are beautiful, number 5 is not not.
In sample 2 number 11 is not beautiful (contains two '1's), all others are beautiful.
In sample 3 number 3 is not beautiful, all others are beautiful. | instruction | 0 | 89,730 | 20 | 179,460 |
Tags: implementation, math
Correct Solution:
```
import sys
z=int(input())
v=0
nm='1'
for i in input().split():
if i == '0':
print(0)
sys.exit(0)
c1=i.count('1')
c0=i.count('0')
if len(i)!=c1+c0 or c1>1:
nm=i
else:
v+=c0
print (nm+'0'*v)
``` | output | 1 | 89,730 | 20 | 179,461 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: given the number of tanks that go into the battle from each country, find their product. If it is turns to be too large, then the servers might have not enough time to assign tanks into teams and the whole game will collapse!
There are exactly n distinct countries in the world and the i-th country added ai tanks to the game. As the developers of the game are perfectionists, the number of tanks from each country is beautiful. A beautiful number, according to the developers, is such number that its decimal representation consists only of digits '1' and '0', moreover it contains at most one digit '1'. However, due to complaints from players, some number of tanks of one country was removed from the game, hence the number of tanks of this country may not remain beautiful.
Your task is to write the program that solves exactly the same problem in order to verify Gena's code correctness. Just in case.
Input
The first line of the input contains the number of countries n (1 ≤ n ≤ 100 000). The second line contains n non-negative integers ai without leading zeroes — the number of tanks of the i-th country.
It is guaranteed that the second line contains at least n - 1 beautiful numbers and the total length of all these number's representations doesn't exceed 100 000.
Output
Print a single number without leading zeroes — the product of the number of tanks presented by each country.
Examples
Input
3
5 10 1
Output
50
Input
4
1 1 10 11
Output
110
Input
5
0 3 1 100 1
Output
0
Note
In sample 1 numbers 10 and 1 are beautiful, number 5 is not not.
In sample 2 number 11 is not beautiful (contains two '1's), all others are beautiful.
In sample 3 number 3 is not beautiful, all others are beautiful.
Submitted Solution:
```
n = int(input())
a = list(input().split())
cnt0 = 0
deb = '1'
for x in a:
if x=='0':
print(0)
exit()
if x[0] == '1' and x.count('0') == len(x) - 1:
cnt0 += len(x)-1
else:
deb = x
debb = ('0'*cnt0)
debb = deb + debb
print(debb)
``` | instruction | 0 | 89,731 | 20 | 179,462 |
Yes | output | 1 | 89,731 | 20 | 179,463 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: given the number of tanks that go into the battle from each country, find their product. If it is turns to be too large, then the servers might have not enough time to assign tanks into teams and the whole game will collapse!
There are exactly n distinct countries in the world and the i-th country added ai tanks to the game. As the developers of the game are perfectionists, the number of tanks from each country is beautiful. A beautiful number, according to the developers, is such number that its decimal representation consists only of digits '1' and '0', moreover it contains at most one digit '1'. However, due to complaints from players, some number of tanks of one country was removed from the game, hence the number of tanks of this country may not remain beautiful.
Your task is to write the program that solves exactly the same problem in order to verify Gena's code correctness. Just in case.
Input
The first line of the input contains the number of countries n (1 ≤ n ≤ 100 000). The second line contains n non-negative integers ai without leading zeroes — the number of tanks of the i-th country.
It is guaranteed that the second line contains at least n - 1 beautiful numbers and the total length of all these number's representations doesn't exceed 100 000.
Output
Print a single number without leading zeroes — the product of the number of tanks presented by each country.
Examples
Input
3
5 10 1
Output
50
Input
4
1 1 10 11
Output
110
Input
5
0 3 1 100 1
Output
0
Note
In sample 1 numbers 10 and 1 are beautiful, number 5 is not not.
In sample 2 number 11 is not beautiful (contains two '1's), all others are beautiful.
In sample 3 number 3 is not beautiful, all others are beautiful.
Submitted Solution:
```
l=int(input())
nums=input().split()
non_beaut=-1
you=False
for i in range(l) :
if nums[i]=='0':
print(0)
you=True
break
else:
s='1'
if nums[i]!='0' and nums[i]!= s+(len(nums[i])-1)*'0':
non_beaut=i
if you==False:
length=0
for j in range(l):
if j!=non_beaut:
length=length+len(nums[j])-1
#if non_beaut==-1:
#result=nums[non_beaut]+'0'*(length-l)
#else:
#result=nums[non_beaut]+'0'*(length-l+1)
if non_beaut == -1:
result = '1' + length*'0'
else:
result=nums[non_beaut]+length*'0'
print(result)
``` | instruction | 0 | 89,732 | 20 | 179,464 |
Yes | output | 1 | 89,732 | 20 | 179,465 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: given the number of tanks that go into the battle from each country, find their product. If it is turns to be too large, then the servers might have not enough time to assign tanks into teams and the whole game will collapse!
There are exactly n distinct countries in the world and the i-th country added ai tanks to the game. As the developers of the game are perfectionists, the number of tanks from each country is beautiful. A beautiful number, according to the developers, is such number that its decimal representation consists only of digits '1' and '0', moreover it contains at most one digit '1'. However, due to complaints from players, some number of tanks of one country was removed from the game, hence the number of tanks of this country may not remain beautiful.
Your task is to write the program that solves exactly the same problem in order to verify Gena's code correctness. Just in case.
Input
The first line of the input contains the number of countries n (1 ≤ n ≤ 100 000). The second line contains n non-negative integers ai without leading zeroes — the number of tanks of the i-th country.
It is guaranteed that the second line contains at least n - 1 beautiful numbers and the total length of all these number's representations doesn't exceed 100 000.
Output
Print a single number without leading zeroes — the product of the number of tanks presented by each country.
Examples
Input
3
5 10 1
Output
50
Input
4
1 1 10 11
Output
110
Input
5
0 3 1 100 1
Output
0
Note
In sample 1 numbers 10 and 1 are beautiful, number 5 is not not.
In sample 2 number 11 is not beautiful (contains two '1's), all others are beautiful.
In sample 3 number 3 is not beautiful, all others are beautiful.
Submitted Solution:
```
n = int(input())
a = list(map(str, input().split()))
res1, res2 = '', '1'
if '0' in a:
print(0)
quit()
for i in range(n):
if a[i]=='0':print(0); quit()
x = a[i].count('0')
if x == len(a[i]) - 1 and a[i][0] == '1':
res1 += '0' * x
else:
res2 = a[i]
print(res2 + res1)
``` | instruction | 0 | 89,733 | 20 | 179,466 |
Yes | output | 1 | 89,733 | 20 | 179,467 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: given the number of tanks that go into the battle from each country, find their product. If it is turns to be too large, then the servers might have not enough time to assign tanks into teams and the whole game will collapse!
There are exactly n distinct countries in the world and the i-th country added ai tanks to the game. As the developers of the game are perfectionists, the number of tanks from each country is beautiful. A beautiful number, according to the developers, is such number that its decimal representation consists only of digits '1' and '0', moreover it contains at most one digit '1'. However, due to complaints from players, some number of tanks of one country was removed from the game, hence the number of tanks of this country may not remain beautiful.
Your task is to write the program that solves exactly the same problem in order to verify Gena's code correctness. Just in case.
Input
The first line of the input contains the number of countries n (1 ≤ n ≤ 100 000). The second line contains n non-negative integers ai without leading zeroes — the number of tanks of the i-th country.
It is guaranteed that the second line contains at least n - 1 beautiful numbers and the total length of all these number's representations doesn't exceed 100 000.
Output
Print a single number without leading zeroes — the product of the number of tanks presented by each country.
Examples
Input
3
5 10 1
Output
50
Input
4
1 1 10 11
Output
110
Input
5
0 3 1 100 1
Output
0
Note
In sample 1 numbers 10 and 1 are beautiful, number 5 is not not.
In sample 2 number 11 is not beautiful (contains two '1's), all others are beautiful.
In sample 3 number 3 is not beautiful, all others are beautiful.
Submitted Solution:
```
n = int(input())
k = input().split()
s = 0
x = '1'
for i in range(n):
if k[i] == '0':
print('0')
break
elif k[i].count('0') + k[i].count('1') != len(k[i]) or k[i].count('1') >1:
x = k[i]
else:
s+=k[i].count('0')
else:
print(x+'0'*s)
``` | instruction | 0 | 89,734 | 20 | 179,468 |
Yes | output | 1 | 89,734 | 20 | 179,469 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: given the number of tanks that go into the battle from each country, find their product. If it is turns to be too large, then the servers might have not enough time to assign tanks into teams and the whole game will collapse!
There are exactly n distinct countries in the world and the i-th country added ai tanks to the game. As the developers of the game are perfectionists, the number of tanks from each country is beautiful. A beautiful number, according to the developers, is such number that its decimal representation consists only of digits '1' and '0', moreover it contains at most one digit '1'. However, due to complaints from players, some number of tanks of one country was removed from the game, hence the number of tanks of this country may not remain beautiful.
Your task is to write the program that solves exactly the same problem in order to verify Gena's code correctness. Just in case.
Input
The first line of the input contains the number of countries n (1 ≤ n ≤ 100 000). The second line contains n non-negative integers ai without leading zeroes — the number of tanks of the i-th country.
It is guaranteed that the second line contains at least n - 1 beautiful numbers and the total length of all these number's representations doesn't exceed 100 000.
Output
Print a single number without leading zeroes — the product of the number of tanks presented by each country.
Examples
Input
3
5 10 1
Output
50
Input
4
1 1 10 11
Output
110
Input
5
0 3 1 100 1
Output
0
Note
In sample 1 numbers 10 and 1 are beautiful, number 5 is not not.
In sample 2 number 11 is not beautiful (contains two '1's), all others are beautiful.
In sample 3 number 3 is not beautiful, all others are beautiful.
Submitted Solution:
```
import sys
n = int(input())
a = list(map(int, input().split()))
z = 0
ans = ""
for i in a:
if i==0:
print(0)
sys.exit(0)
elif i==1:
continue
elif i%10==0:
z = z + 1
else:
ans = str(i)
print(ans+'0'*z)
``` | instruction | 0 | 89,735 | 20 | 179,470 |
No | output | 1 | 89,735 | 20 | 179,471 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: given the number of tanks that go into the battle from each country, find their product. If it is turns to be too large, then the servers might have not enough time to assign tanks into teams and the whole game will collapse!
There are exactly n distinct countries in the world and the i-th country added ai tanks to the game. As the developers of the game are perfectionists, the number of tanks from each country is beautiful. A beautiful number, according to the developers, is such number that its decimal representation consists only of digits '1' and '0', moreover it contains at most one digit '1'. However, due to complaints from players, some number of tanks of one country was removed from the game, hence the number of tanks of this country may not remain beautiful.
Your task is to write the program that solves exactly the same problem in order to verify Gena's code correctness. Just in case.
Input
The first line of the input contains the number of countries n (1 ≤ n ≤ 100 000). The second line contains n non-negative integers ai without leading zeroes — the number of tanks of the i-th country.
It is guaranteed that the second line contains at least n - 1 beautiful numbers and the total length of all these number's representations doesn't exceed 100 000.
Output
Print a single number without leading zeroes — the product of the number of tanks presented by each country.
Examples
Input
3
5 10 1
Output
50
Input
4
1 1 10 11
Output
110
Input
5
0 3 1 100 1
Output
0
Note
In sample 1 numbers 10 and 1 are beautiful, number 5 is not not.
In sample 2 number 11 is not beautiful (contains two '1's), all others are beautiful.
In sample 3 number 3 is not beautiful, all others are beautiful.
Submitted Solution:
```
def zero(a):
su=0
ch=0
for i in range(len(a)):
if (a[i]!='1')and(a[i]!='0'):
return -1
if (len(a)==1)and(a[i]=='0'):
return -2
if ((a[i]=='1')and(ch==0))or(a[i]=='0'):
ch=1
else: return -1
if a[i]=='0':
su+=1
return su
n=int(input())
a=input().split()
ind=0
t=0
p=1
for i in a:
if zero(i)==-1:
ind=int(i)
elif zero(i)==-2:
p=0
else:
t+=zero(i)
ans=10**t * ind * p
print(ans)
``` | instruction | 0 | 89,736 | 20 | 179,472 |
No | output | 1 | 89,736 | 20 | 179,473 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: given the number of tanks that go into the battle from each country, find their product. If it is turns to be too large, then the servers might have not enough time to assign tanks into teams and the whole game will collapse!
There are exactly n distinct countries in the world and the i-th country added ai tanks to the game. As the developers of the game are perfectionists, the number of tanks from each country is beautiful. A beautiful number, according to the developers, is such number that its decimal representation consists only of digits '1' and '0', moreover it contains at most one digit '1'. However, due to complaints from players, some number of tanks of one country was removed from the game, hence the number of tanks of this country may not remain beautiful.
Your task is to write the program that solves exactly the same problem in order to verify Gena's code correctness. Just in case.
Input
The first line of the input contains the number of countries n (1 ≤ n ≤ 100 000). The second line contains n non-negative integers ai without leading zeroes — the number of tanks of the i-th country.
It is guaranteed that the second line contains at least n - 1 beautiful numbers and the total length of all these number's representations doesn't exceed 100 000.
Output
Print a single number without leading zeroes — the product of the number of tanks presented by each country.
Examples
Input
3
5 10 1
Output
50
Input
4
1 1 10 11
Output
110
Input
5
0 3 1 100 1
Output
0
Note
In sample 1 numbers 10 and 1 are beautiful, number 5 is not not.
In sample 2 number 11 is not beautiful (contains two '1's), all others are beautiful.
In sample 3 number 3 is not beautiful, all others are beautiful.
Submitted Solution:
```
n = int(input())
a = list(map(str, input().split()))
res1, res2 = '', ''
if '0' in a:
print(0)
quit()
for i in range(n):
x = a[i].count('0')
if x == len(a[i]) - 1 and a[i][0] == '1':
res1 += '0' * x
else:
res2 = a[i]
print(res1 + res2)
``` | instruction | 0 | 89,737 | 20 | 179,474 |
No | output | 1 | 89,737 | 20 | 179,475 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: given the number of tanks that go into the battle from each country, find their product. If it is turns to be too large, then the servers might have not enough time to assign tanks into teams and the whole game will collapse!
There are exactly n distinct countries in the world and the i-th country added ai tanks to the game. As the developers of the game are perfectionists, the number of tanks from each country is beautiful. A beautiful number, according to the developers, is such number that its decimal representation consists only of digits '1' and '0', moreover it contains at most one digit '1'. However, due to complaints from players, some number of tanks of one country was removed from the game, hence the number of tanks of this country may not remain beautiful.
Your task is to write the program that solves exactly the same problem in order to verify Gena's code correctness. Just in case.
Input
The first line of the input contains the number of countries n (1 ≤ n ≤ 100 000). The second line contains n non-negative integers ai without leading zeroes — the number of tanks of the i-th country.
It is guaranteed that the second line contains at least n - 1 beautiful numbers and the total length of all these number's representations doesn't exceed 100 000.
Output
Print a single number without leading zeroes — the product of the number of tanks presented by each country.
Examples
Input
3
5 10 1
Output
50
Input
4
1 1 10 11
Output
110
Input
5
0 3 1 100 1
Output
0
Note
In sample 1 numbers 10 and 1 are beautiful, number 5 is not not.
In sample 2 number 11 is not beautiful (contains two '1's), all others are beautiful.
In sample 3 number 3 is not beautiful, all others are beautiful.
Submitted Solution:
```
n=int(input())
a=input().split()
count1=0
result=""
for i in a:
if i=="0":
print(0)
exit()
if i=="1":
continue
if int(i)%10!=0:
result=i
else:
count1+=(len(i)-1)
if result=="":
print("1"+"0"*count1)
else:
print("%s"%result+"0"*count1)
``` | instruction | 0 | 89,738 | 20 | 179,476 |
No | output | 1 | 89,738 | 20 | 179,477 |
Provide a correct Python 3 solution for this coding contest problem.
Let \mathrm{popcount}(n) be the number of `1`s in the binary representation of n. For example, \mathrm{popcount}(3) = 2, \mathrm{popcount}(7) = 3, and \mathrm{popcount}(0) = 0.
Let f(n) be the number of times the following operation will be done when we repeat it until n becomes 0: "replace n with the remainder when n is divided by \mathrm{popcount}(n)." (It can be proved that, under the constraints of this problem, n always becomes 0 after a finite number of operations.)
For example, when n=7, it becomes 0 after two operations, as follows:
* \mathrm{popcount}(7)=3, so we divide 7 by 3 and replace it with the remainder, 1.
* \mathrm{popcount}(1)=1, so we divide 1 by 1 and replace it with the remainder, 0.
You are given an integer X with N digits in binary. For each integer i such that 1 \leq i \leq N, let X_i be what X becomes when the i-th bit from the top is inverted. Find f(X_1), f(X_2), \ldots, f(X_N).
Constraints
* 1 \leq N \leq 2 \times 10^5
* X is an integer with N digits in binary, possibly with leading zeros.
Input
Input is given from Standard Input in the following format:
N
X
Output
Print N lines. The i-th line should contain the value f(X_i).
Examples
Input
3
011
Output
2
1
1
Input
23
00110111001011011001110
Output
2
1
2
2
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
3 | instruction | 0 | 89,880 | 20 | 179,760 |
"Correct Solution:
```
n = int(input())
x = input()
x10 = int(x, 2)
lenx = len(x)
maxbit = x.count("1")
xc0 = x10 % (maxbit + 1)
xc1 = x10 % (maxbit - 1) if maxbit > 1 else 0
def f(ii):
cnt = 0
while ii > 0:
ii %= bin(ii).count("1")
cnt += 1
return cnt
for i in range(lenx):
if x[i] == "0":
t = xc0 + pow(2, (lenx - 1 - i), maxbit + 1)
ans = f(t % (maxbit + 1)) + 1
else:
if maxbit - 1 == 0:
ans = 0
else:
t = xc1 - pow(2, (lenx - 1 - i), maxbit - 1)
ans = f(t % (maxbit - 1)) + 1
print(ans)
``` | output | 1 | 89,880 | 20 | 179,761 |
Provide a correct Python 3 solution for this coding contest problem.
Let \mathrm{popcount}(n) be the number of `1`s in the binary representation of n. For example, \mathrm{popcount}(3) = 2, \mathrm{popcount}(7) = 3, and \mathrm{popcount}(0) = 0.
Let f(n) be the number of times the following operation will be done when we repeat it until n becomes 0: "replace n with the remainder when n is divided by \mathrm{popcount}(n)." (It can be proved that, under the constraints of this problem, n always becomes 0 after a finite number of operations.)
For example, when n=7, it becomes 0 after two operations, as follows:
* \mathrm{popcount}(7)=3, so we divide 7 by 3 and replace it with the remainder, 1.
* \mathrm{popcount}(1)=1, so we divide 1 by 1 and replace it with the remainder, 0.
You are given an integer X with N digits in binary. For each integer i such that 1 \leq i \leq N, let X_i be what X becomes when the i-th bit from the top is inverted. Find f(X_1), f(X_2), \ldots, f(X_N).
Constraints
* 1 \leq N \leq 2 \times 10^5
* X is an integer with N digits in binary, possibly with leading zeros.
Input
Input is given from Standard Input in the following format:
N
X
Output
Print N lines. The i-th line should contain the value f(X_i).
Examples
Input
3
011
Output
2
1
1
Input
23
00110111001011011001110
Output
2
1
2
2
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
3 | instruction | 0 | 89,881 | 20 | 179,762 |
"Correct Solution:
```
N=int(input())
X=input()
c=X.count('1')
r1=int(X,2)%(c-1) if c>1 else 0
r2=int(X,2)%(c+1)
d=[0]*(N+1)
for i in range(N):
d[i+1]=d[(i+1)%bin(i+1).count('1')]+1
for i in range(N):
if X[i]=='0':
n=(r2+pow(2,N-i-1,c+1))%(c+1)
else:
if c==1:
print(0)
continue
n=(r1-pow(2,N-i-1,c-1))%(c-1)
print(d[n]+1)
``` | output | 1 | 89,881 | 20 | 179,763 |
Provide a correct Python 3 solution for this coding contest problem.
Let \mathrm{popcount}(n) be the number of `1`s in the binary representation of n. For example, \mathrm{popcount}(3) = 2, \mathrm{popcount}(7) = 3, and \mathrm{popcount}(0) = 0.
Let f(n) be the number of times the following operation will be done when we repeat it until n becomes 0: "replace n with the remainder when n is divided by \mathrm{popcount}(n)." (It can be proved that, under the constraints of this problem, n always becomes 0 after a finite number of operations.)
For example, when n=7, it becomes 0 after two operations, as follows:
* \mathrm{popcount}(7)=3, so we divide 7 by 3 and replace it with the remainder, 1.
* \mathrm{popcount}(1)=1, so we divide 1 by 1 and replace it with the remainder, 0.
You are given an integer X with N digits in binary. For each integer i such that 1 \leq i \leq N, let X_i be what X becomes when the i-th bit from the top is inverted. Find f(X_1), f(X_2), \ldots, f(X_N).
Constraints
* 1 \leq N \leq 2 \times 10^5
* X is an integer with N digits in binary, possibly with leading zeros.
Input
Input is given from Standard Input in the following format:
N
X
Output
Print N lines. The i-th line should contain the value f(X_i).
Examples
Input
3
011
Output
2
1
1
Input
23
00110111001011011001110
Output
2
1
2
2
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
3 | instruction | 0 | 89,882 | 20 | 179,764 |
"Correct Solution:
```
N = int(input())
X = input()
popcount_X = X.count('1')
numX = int(X, 2)
a, b = numX % (popcount_X + 1), numX % (popcount_X - 1) if popcount_X != 1 else 0
for i, x in enumerate(X, 1):
if x == '1' and popcount_X == 1:
print(0); continue
ans = 1
if x == '1':
temp = (b - pow(2, N - i, popcount_X - 1)) % (popcount_X - 1)
else:
temp = (a + pow(2, N - i, popcount_X + 1)) % (popcount_X + 1)
while temp:
p = format(temp, 'b').count('1')
temp %= p
ans += 1
print(ans)
``` | output | 1 | 89,882 | 20 | 179,765 |
Provide a correct Python 3 solution for this coding contest problem.
Let \mathrm{popcount}(n) be the number of `1`s in the binary representation of n. For example, \mathrm{popcount}(3) = 2, \mathrm{popcount}(7) = 3, and \mathrm{popcount}(0) = 0.
Let f(n) be the number of times the following operation will be done when we repeat it until n becomes 0: "replace n with the remainder when n is divided by \mathrm{popcount}(n)." (It can be proved that, under the constraints of this problem, n always becomes 0 after a finite number of operations.)
For example, when n=7, it becomes 0 after two operations, as follows:
* \mathrm{popcount}(7)=3, so we divide 7 by 3 and replace it with the remainder, 1.
* \mathrm{popcount}(1)=1, so we divide 1 by 1 and replace it with the remainder, 0.
You are given an integer X with N digits in binary. For each integer i such that 1 \leq i \leq N, let X_i be what X becomes when the i-th bit from the top is inverted. Find f(X_1), f(X_2), \ldots, f(X_N).
Constraints
* 1 \leq N \leq 2 \times 10^5
* X is an integer with N digits in binary, possibly with leading zeros.
Input
Input is given from Standard Input in the following format:
N
X
Output
Print N lines. The i-th line should contain the value f(X_i).
Examples
Input
3
011
Output
2
1
1
Input
23
00110111001011011001110
Output
2
1
2
2
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
3 | instruction | 0 | 89,883 | 20 | 179,766 |
"Correct Solution:
```
def f(n):
cnt = 0
while n:
buf = n
pc = 0
while buf:
pc += buf%2
buf //= 2
n = n%pc
cnt += 1
return cnt
N = int(input())
X = input()
v1, v2 = 0, 0
cnt = X.count("1")
for i, j in enumerate(X[::-1]):
if j=="1":
v1 += pow(2, i, cnt+1)
v1 %= (cnt+1)
if cnt>1:
v2 += pow(2, i, cnt-1)
v2 %= (cnt-1)
for i in range(N):
p = N-i-1
buf = 0
c2 = 0
if X[i]=="0":
c2 = cnt+1
buf = (v1+pow(2, p, c2))%c2
print(f(buf)+1)
else:
c2 = cnt-1
if c2>0:
buf = (v2-pow(2, p, c2))%c2
print(f(buf)+1)
else:
print(0)
``` | output | 1 | 89,883 | 20 | 179,767 |
Provide a correct Python 3 solution for this coding contest problem.
Let \mathrm{popcount}(n) be the number of `1`s in the binary representation of n. For example, \mathrm{popcount}(3) = 2, \mathrm{popcount}(7) = 3, and \mathrm{popcount}(0) = 0.
Let f(n) be the number of times the following operation will be done when we repeat it until n becomes 0: "replace n with the remainder when n is divided by \mathrm{popcount}(n)." (It can be proved that, under the constraints of this problem, n always becomes 0 after a finite number of operations.)
For example, when n=7, it becomes 0 after two operations, as follows:
* \mathrm{popcount}(7)=3, so we divide 7 by 3 and replace it with the remainder, 1.
* \mathrm{popcount}(1)=1, so we divide 1 by 1 and replace it with the remainder, 0.
You are given an integer X with N digits in binary. For each integer i such that 1 \leq i \leq N, let X_i be what X becomes when the i-th bit from the top is inverted. Find f(X_1), f(X_2), \ldots, f(X_N).
Constraints
* 1 \leq N \leq 2 \times 10^5
* X is an integer with N digits in binary, possibly with leading zeros.
Input
Input is given from Standard Input in the following format:
N
X
Output
Print N lines. The i-th line should contain the value f(X_i).
Examples
Input
3
011
Output
2
1
1
Input
23
00110111001011011001110
Output
2
1
2
2
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
3 | instruction | 0 | 89,884 | 20 | 179,768 |
"Correct Solution:
```
N = int(input())
X = input()
temp = X.count("1")
t = [0, 0, 0]
t[0] = int(X, 2) % (temp-1) if temp != 1 else 0
t[2] = int(X, 2) % (temp+1)
cnt = 0
for i in range(N):
if X[i] == "1" and temp == 1:
print(0)
continue
if X[i] == "1":
# p = (1 << (N-i-1)) % (temp-1)
p = pow(2, N-1-i, temp-1)
a = (t[0]-p) % (temp-1)
elif X[i] == "0":
# m = (1 << (N-i-1)) % (temp+1)
m = pow(2, N-1-i, temp+1)
a = (t[2]+m) % (temp+1)
cnt = 1
while a > 0:
a = a % format(a, 'b').count("1")
cnt += 1
print(cnt)
``` | output | 1 | 89,884 | 20 | 179,769 |
Provide a correct Python 3 solution for this coding contest problem.
Let \mathrm{popcount}(n) be the number of `1`s in the binary representation of n. For example, \mathrm{popcount}(3) = 2, \mathrm{popcount}(7) = 3, and \mathrm{popcount}(0) = 0.
Let f(n) be the number of times the following operation will be done when we repeat it until n becomes 0: "replace n with the remainder when n is divided by \mathrm{popcount}(n)." (It can be proved that, under the constraints of this problem, n always becomes 0 after a finite number of operations.)
For example, when n=7, it becomes 0 after two operations, as follows:
* \mathrm{popcount}(7)=3, so we divide 7 by 3 and replace it with the remainder, 1.
* \mathrm{popcount}(1)=1, so we divide 1 by 1 and replace it with the remainder, 0.
You are given an integer X with N digits in binary. For each integer i such that 1 \leq i \leq N, let X_i be what X becomes when the i-th bit from the top is inverted. Find f(X_1), f(X_2), \ldots, f(X_N).
Constraints
* 1 \leq N \leq 2 \times 10^5
* X is an integer with N digits in binary, possibly with leading zeros.
Input
Input is given from Standard Input in the following format:
N
X
Output
Print N lines. The i-th line should contain the value f(X_i).
Examples
Input
3
011
Output
2
1
1
Input
23
00110111001011011001110
Output
2
1
2
2
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
3 | instruction | 0 | 89,885 | 20 | 179,770 |
"Correct Solution:
```
N = int(input())
X = int(input(), 2)
M = 2 * (10 ** 5) + 10
ret = [0] * (M + 1)
for i in range(M + 1) :
x = i
c = 0
while x != 0 :
x = x % bin(x).count('1')
c += 1
ret[i] = c
p = bin(X).count('1')
a = X % (p + 1)
if p > 1 :
b = X % (p - 1)
for i in range(N - 1, -1, -1) :
if (1 << i) & X :
if p == 1 :
print(0)
else :
print(ret[(b - pow(2, i, p - 1)) % (p - 1)] + 1)
else :
print(ret[(a + pow(2, i, p + 1)) % (p + 1)] + 1)
``` | output | 1 | 89,885 | 20 | 179,771 |
Provide a correct Python 3 solution for this coding contest problem.
Let \mathrm{popcount}(n) be the number of `1`s in the binary representation of n. For example, \mathrm{popcount}(3) = 2, \mathrm{popcount}(7) = 3, and \mathrm{popcount}(0) = 0.
Let f(n) be the number of times the following operation will be done when we repeat it until n becomes 0: "replace n with the remainder when n is divided by \mathrm{popcount}(n)." (It can be proved that, under the constraints of this problem, n always becomes 0 after a finite number of operations.)
For example, when n=7, it becomes 0 after two operations, as follows:
* \mathrm{popcount}(7)=3, so we divide 7 by 3 and replace it with the remainder, 1.
* \mathrm{popcount}(1)=1, so we divide 1 by 1 and replace it with the remainder, 0.
You are given an integer X with N digits in binary. For each integer i such that 1 \leq i \leq N, let X_i be what X becomes when the i-th bit from the top is inverted. Find f(X_1), f(X_2), \ldots, f(X_N).
Constraints
* 1 \leq N \leq 2 \times 10^5
* X is an integer with N digits in binary, possibly with leading zeros.
Input
Input is given from Standard Input in the following format:
N
X
Output
Print N lines. The i-th line should contain the value f(X_i).
Examples
Input
3
011
Output
2
1
1
Input
23
00110111001011011001110
Output
2
1
2
2
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
3 | instruction | 0 | 89,886 | 20 | 179,772 |
"Correct Solution:
```
def count_pop(y):
count = 0
while y != 0:
if y & 1 == 1:count+=1
y = y >> 1
return count
def main():
N = int(input())
X = input()
X_val=int(X,2)
X_pc=X.count('1')
x_mod_pc_x_plus_1 = X_val % (X_pc + 1)
x_mod_pc_x_minus_1 = X_val % max(X_pc - 1, 1)
for i in range(N):
if X[i] == '1':
pc = X_pc - 1
if pc == 0:
print(0)
continue
val = (x_mod_pc_x_minus_1 - pow(2, N-i-1, pc)) % pc
else:
pc = X_pc + 1
val = (x_mod_pc_x_plus_1 + pow(2, N-i-1, pc)) % pc
ans = 1
while val > 0:
val %= bin(val).count('1')
ans += 1
print(ans)
pass
if __name__=='__main__':
main()
``` | output | 1 | 89,886 | 20 | 179,773 |
Provide a correct Python 3 solution for this coding contest problem.
Let \mathrm{popcount}(n) be the number of `1`s in the binary representation of n. For example, \mathrm{popcount}(3) = 2, \mathrm{popcount}(7) = 3, and \mathrm{popcount}(0) = 0.
Let f(n) be the number of times the following operation will be done when we repeat it until n becomes 0: "replace n with the remainder when n is divided by \mathrm{popcount}(n)." (It can be proved that, under the constraints of this problem, n always becomes 0 after a finite number of operations.)
For example, when n=7, it becomes 0 after two operations, as follows:
* \mathrm{popcount}(7)=3, so we divide 7 by 3 and replace it with the remainder, 1.
* \mathrm{popcount}(1)=1, so we divide 1 by 1 and replace it with the remainder, 0.
You are given an integer X with N digits in binary. For each integer i such that 1 \leq i \leq N, let X_i be what X becomes when the i-th bit from the top is inverted. Find f(X_1), f(X_2), \ldots, f(X_N).
Constraints
* 1 \leq N \leq 2 \times 10^5
* X is an integer with N digits in binary, possibly with leading zeros.
Input
Input is given from Standard Input in the following format:
N
X
Output
Print N lines. The i-th line should contain the value f(X_i).
Examples
Input
3
011
Output
2
1
1
Input
23
00110111001011011001110
Output
2
1
2
2
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
3 | instruction | 0 | 89,887 | 20 | 179,774 |
"Correct Solution:
```
n = int(input())
x = list(map(int, input()))
one_cnt = x.count(1)
mods = [one_cnt - 1, one_cnt + 1]
sms = [0, 0]
for i in range(2):
if mods[i] == 0:
continue
for j, e in enumerate(x):
sms[i] += pow(2, n - j - 1, mods[i]) * e
sms[i] %= mods[i]
for i, e in enumerate(x):
idx = 1 - e
mod = mods[idx]
sm = sms[idx]
if mod == 0:
print(0)
continue
sm_changed = sm + pow(2, n - i - 1, mod) * (-1) ** e
sm_changed %= mod
ans = 1
while sm_changed:
tmp = sm_changed
cnt = 0
while tmp:
cnt += tmp & 1
tmp >>= 1
sm_changed %= cnt
ans += 1
print(ans)
``` | output | 1 | 89,887 | 20 | 179,775 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let \mathrm{popcount}(n) be the number of `1`s in the binary representation of n. For example, \mathrm{popcount}(3) = 2, \mathrm{popcount}(7) = 3, and \mathrm{popcount}(0) = 0.
Let f(n) be the number of times the following operation will be done when we repeat it until n becomes 0: "replace n with the remainder when n is divided by \mathrm{popcount}(n)." (It can be proved that, under the constraints of this problem, n always becomes 0 after a finite number of operations.)
For example, when n=7, it becomes 0 after two operations, as follows:
* \mathrm{popcount}(7)=3, so we divide 7 by 3 and replace it with the remainder, 1.
* \mathrm{popcount}(1)=1, so we divide 1 by 1 and replace it with the remainder, 0.
You are given an integer X with N digits in binary. For each integer i such that 1 \leq i \leq N, let X_i be what X becomes when the i-th bit from the top is inverted. Find f(X_1), f(X_2), \ldots, f(X_N).
Constraints
* 1 \leq N \leq 2 \times 10^5
* X is an integer with N digits in binary, possibly with leading zeros.
Input
Input is given from Standard Input in the following format:
N
X
Output
Print N lines. The i-th line should contain the value f(X_i).
Examples
Input
3
011
Output
2
1
1
Input
23
00110111001011011001110
Output
2
1
2
2
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
3
Submitted Solution:
```
def pop_count(n):
return sum(n >> i & 1 for i in range(n.bit_length()))
def f(n):
if n == 0:
return 0
return f(n % pop_count(n)) + 1
N = int(input())
X = input()
p = X.count("1")
rem_plus = 0
rem_minus = 0
for i in range(N):
k = N - i - 1
if X[i] == "0":
continue
elif p > 1:
rem_minus = (rem_minus + pow(2, k, p - 1)) % (p - 1)
rem_plus = (rem_plus + pow(2, k, p + 1)) % (p + 1)
for i in range(N):
k = N - i - 1
if X[i] == "0":
print(f((rem_plus + pow(2, k, p + 1)) % (p + 1)) + 1)
elif p > 1:
print(f((rem_minus - pow(2, k, p - 1)) % (p - 1)) + 1)
else:
print(0)
``` | instruction | 0 | 89,888 | 20 | 179,776 |
Yes | output | 1 | 89,888 | 20 | 179,777 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let \mathrm{popcount}(n) be the number of `1`s in the binary representation of n. For example, \mathrm{popcount}(3) = 2, \mathrm{popcount}(7) = 3, and \mathrm{popcount}(0) = 0.
Let f(n) be the number of times the following operation will be done when we repeat it until n becomes 0: "replace n with the remainder when n is divided by \mathrm{popcount}(n)." (It can be proved that, under the constraints of this problem, n always becomes 0 after a finite number of operations.)
For example, when n=7, it becomes 0 after two operations, as follows:
* \mathrm{popcount}(7)=3, so we divide 7 by 3 and replace it with the remainder, 1.
* \mathrm{popcount}(1)=1, so we divide 1 by 1 and replace it with the remainder, 0.
You are given an integer X with N digits in binary. For each integer i such that 1 \leq i \leq N, let X_i be what X becomes when the i-th bit from the top is inverted. Find f(X_1), f(X_2), \ldots, f(X_N).
Constraints
* 1 \leq N \leq 2 \times 10^5
* X is an integer with N digits in binary, possibly with leading zeros.
Input
Input is given from Standard Input in the following format:
N
X
Output
Print N lines. The i-th line should contain the value f(X_i).
Examples
Input
3
011
Output
2
1
1
Input
23
00110111001011011001110
Output
2
1
2
2
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
3
Submitted Solution:
```
N = int(input())
S = input()
def pcnt(x):
return str(bin(x)).count("1")
dpl = 5*10**5
dp = [0] * dpl
for i in range(1, dpl):
dp[i] = dp[i % pcnt(i)]+1
c = S.count("1")
a = int(S, 2) % (c+1)
def f():
if c == 1:
for i in range(N):
if S[i] == "0":
print(dp[(a + pow(2, N-i-1, c+1)) % (c+1)] + 1)
else:
print(0)
return
b = int(S, 2) % (c-1)
for i in range(N):
if S[i] == "0":
print(dp[(a+pow(2, N-i-1, c+1))%(c+1)]+1)
else:
print(dp[(b-pow(2, N-i-1, c-1))%(c-1)]+1)
f()
``` | instruction | 0 | 89,889 | 20 | 179,778 |
Yes | output | 1 | 89,889 | 20 | 179,779 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let \mathrm{popcount}(n) be the number of `1`s in the binary representation of n. For example, \mathrm{popcount}(3) = 2, \mathrm{popcount}(7) = 3, and \mathrm{popcount}(0) = 0.
Let f(n) be the number of times the following operation will be done when we repeat it until n becomes 0: "replace n with the remainder when n is divided by \mathrm{popcount}(n)." (It can be proved that, under the constraints of this problem, n always becomes 0 after a finite number of operations.)
For example, when n=7, it becomes 0 after two operations, as follows:
* \mathrm{popcount}(7)=3, so we divide 7 by 3 and replace it with the remainder, 1.
* \mathrm{popcount}(1)=1, so we divide 1 by 1 and replace it with the remainder, 0.
You are given an integer X with N digits in binary. For each integer i such that 1 \leq i \leq N, let X_i be what X becomes when the i-th bit from the top is inverted. Find f(X_1), f(X_2), \ldots, f(X_N).
Constraints
* 1 \leq N \leq 2 \times 10^5
* X is an integer with N digits in binary, possibly with leading zeros.
Input
Input is given from Standard Input in the following format:
N
X
Output
Print N lines. The i-th line should contain the value f(X_i).
Examples
Input
3
011
Output
2
1
1
Input
23
00110111001011011001110
Output
2
1
2
2
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
3
Submitted Solution:
```
n = int(input())
x = input()
k = x.count("1")
cnt = 0
s = int(x, 2)
if k >= 2:
k1 = s % (k-1)
k2 = s % (k+1)
while s != 0:
targ = bin(s).count("1")
s = s % targ
cnt += 1
def ev(n):
cnt = 0
while n != 0:
m = bin(n).count("1")
n = n % m
cnt += 1
return cnt
if k != 1:
for i in range(n):
if x[i] == "0":
targ = k+1
print(ev((k2+pow(2, n-i-1, targ)) % targ)+1)
else:
targ = k-1
print(ev((k1-pow(2, n-i-1, targ)) % targ)+1)
else:
for i in range(n):
if x[i] == "0":
targ = k+1
print(ev((k2+pow(2, n-i-1, targ)) % targ)+1)
else:
targ = k-1
print(0)
``` | instruction | 0 | 89,890 | 20 | 179,780 |
Yes | output | 1 | 89,890 | 20 | 179,781 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let \mathrm{popcount}(n) be the number of `1`s in the binary representation of n. For example, \mathrm{popcount}(3) = 2, \mathrm{popcount}(7) = 3, and \mathrm{popcount}(0) = 0.
Let f(n) be the number of times the following operation will be done when we repeat it until n becomes 0: "replace n with the remainder when n is divided by \mathrm{popcount}(n)." (It can be proved that, under the constraints of this problem, n always becomes 0 after a finite number of operations.)
For example, when n=7, it becomes 0 after two operations, as follows:
* \mathrm{popcount}(7)=3, so we divide 7 by 3 and replace it with the remainder, 1.
* \mathrm{popcount}(1)=1, so we divide 1 by 1 and replace it with the remainder, 0.
You are given an integer X with N digits in binary. For each integer i such that 1 \leq i \leq N, let X_i be what X becomes when the i-th bit from the top is inverted. Find f(X_1), f(X_2), \ldots, f(X_N).
Constraints
* 1 \leq N \leq 2 \times 10^5
* X is an integer with N digits in binary, possibly with leading zeros.
Input
Input is given from Standard Input in the following format:
N
X
Output
Print N lines. The i-th line should contain the value f(X_i).
Examples
Input
3
011
Output
2
1
1
Input
23
00110111001011011001110
Output
2
1
2
2
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
3
Submitted Solution:
```
n = int(input())
sn = input()
sr = ''.join(list(reversed(sn)))
sint = int(sn, 2)
spop = sn.count('1')
def f(x):
count = 0
while x != 0:
count += 1
pop = bin(x).count('1')
x %= pop
return count
m1 = sint % (spop + 1)
m2 = 0 if spop <= 1 else sint % (spop - 1)
a = [0] * n
for i in range(n):
if sr[i] == '0':
d = pow(2, i, spop + 1)
m = (m1 + d) % (spop + 1)
a[i] = f(m) + 1
elif spop != 1:
d = pow(2, i, spop - 1)
m = (m2 - d + spop - 1) % (spop - 1)
a[i] = f(m) + 1
else:
a[i] = 0
for ans in reversed(a):
print(ans)
``` | instruction | 0 | 89,891 | 20 | 179,782 |
Yes | output | 1 | 89,891 | 20 | 179,783 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let \mathrm{popcount}(n) be the number of `1`s in the binary representation of n. For example, \mathrm{popcount}(3) = 2, \mathrm{popcount}(7) = 3, and \mathrm{popcount}(0) = 0.
Let f(n) be the number of times the following operation will be done when we repeat it until n becomes 0: "replace n with the remainder when n is divided by \mathrm{popcount}(n)." (It can be proved that, under the constraints of this problem, n always becomes 0 after a finite number of operations.)
For example, when n=7, it becomes 0 after two operations, as follows:
* \mathrm{popcount}(7)=3, so we divide 7 by 3 and replace it with the remainder, 1.
* \mathrm{popcount}(1)=1, so we divide 1 by 1 and replace it with the remainder, 0.
You are given an integer X with N digits in binary. For each integer i such that 1 \leq i \leq N, let X_i be what X becomes when the i-th bit from the top is inverted. Find f(X_1), f(X_2), \ldots, f(X_N).
Constraints
* 1 \leq N \leq 2 \times 10^5
* X is an integer with N digits in binary, possibly with leading zeros.
Input
Input is given from Standard Input in the following format:
N
X
Output
Print N lines. The i-th line should contain the value f(X_i).
Examples
Input
3
011
Output
2
1
1
Input
23
00110111001011011001110
Output
2
1
2
2
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
3
Submitted Solution:
```
n = int(input())
x = input()
num = 0
count = 0
for i in range(n):
if x[-(i+1)] == "1":
num += 2**i
count += 1
mod = [[0, 0] for _ in range(n)]
for i in range(n):
mod[i][0] = 2**i%(count-1) if count!=1 else 0
mod[i][1] = 2**i%(count+1)
mod1 = num%(count-1) if count!=1 else 0
mod2 = num%(count+1)
ans = [0]*n
for i in range(n):
if x[-(i+1)] == "1":
_num = num-2**i
if _num == 0:
continue
_num = (mod1-mod[i][0])%(count-1)
else:
_num = num+2**i
_num = (mod2+mod[i][1])%(count+1)
ans[i] += 1
_count = 0
for c in bin(_num):
if c == "1":
_count += 1
while True:
if _num == 0:
break
_num %= _count
_count = 0
for c in bin(_num):
if c == "1":
_count += 1
ans[i] += 1
for i in range(n):
print(ans[-(i+1)])
``` | instruction | 0 | 89,892 | 20 | 179,784 |
No | output | 1 | 89,892 | 20 | 179,785 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let \mathrm{popcount}(n) be the number of `1`s in the binary representation of n. For example, \mathrm{popcount}(3) = 2, \mathrm{popcount}(7) = 3, and \mathrm{popcount}(0) = 0.
Let f(n) be the number of times the following operation will be done when we repeat it until n becomes 0: "replace n with the remainder when n is divided by \mathrm{popcount}(n)." (It can be proved that, under the constraints of this problem, n always becomes 0 after a finite number of operations.)
For example, when n=7, it becomes 0 after two operations, as follows:
* \mathrm{popcount}(7)=3, so we divide 7 by 3 and replace it with the remainder, 1.
* \mathrm{popcount}(1)=1, so we divide 1 by 1 and replace it with the remainder, 0.
You are given an integer X with N digits in binary. For each integer i such that 1 \leq i \leq N, let X_i be what X becomes when the i-th bit from the top is inverted. Find f(X_1), f(X_2), \ldots, f(X_N).
Constraints
* 1 \leq N \leq 2 \times 10^5
* X is an integer with N digits in binary, possibly with leading zeros.
Input
Input is given from Standard Input in the following format:
N
X
Output
Print N lines. The i-th line should contain the value f(X_i).
Examples
Input
3
011
Output
2
1
1
Input
23
00110111001011011001110
Output
2
1
2
2
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
3
Submitted Solution:
```
def popcount(x):
ans = 0
while x > 0:
if x & 1:
ans += 1
x >>= 1
return ans
def f(x, pc, memo=None):
if memo and (x, pc) in memo:
return memo[(x, pc)]
r = x % pc
if memo:
memo[(x, pc)] = r
return r
N = int(input())
X = input()
x = int(X, 2)
memo = {}
pc_x = X.count('1')
for i in range(N):
if X[i] == '1':
pc = pc_x - 1
else:
pc = pc_x + 1
if pc == 0:
print('0')
continue
f_x = f(x, pc, memo)
f_res = f(1<<(N-i-1), pc, memo)
if X[i] == '1':
f_i = f_x - f_res
else:
f_i = f_x + f_res
f_i = (f_i + 2 * pc) % pc
ans = 1
while f_i > 0:
ans += 1
pc = popcount(f_i)
f_i = f(f_i, pc)
print(ans)
``` | instruction | 0 | 89,893 | 20 | 179,786 |
No | output | 1 | 89,893 | 20 | 179,787 |
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