message
stringlengths 2
23.8k
| message_type
stringclasses 2
values | message_id
int64 0
1
| conversation_id
int64 97
109k
| cluster
float64 0
0
| __index_level_0__
int64 194
217k
|
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We start with a string s consisting only of the digits 1, 2, or 3. The length of s is denoted by |s|. For each i from 1 to |s|, the i-th character of s is denoted by s_i.
There is one cursor. The cursor's location ℓ is denoted by an integer in \{0, …, |s|\}, with the following meaning:
* If ℓ = 0, then the cursor is located before the first character of s.
* If ℓ = |s|, then the cursor is located right after the last character of s.
* If 0 < ℓ < |s|, then the cursor is located between s_ℓ and s_{ℓ+1}.
We denote by s_left the string to the left of the cursor and s_right the string to the right of the cursor.
We also have a string c, which we call our clipboard, which starts out as empty. There are three types of actions:
* The Move action. Move the cursor one step to the right. This increments ℓ once.
* The Cut action. Set c ← s_right, then set s ← s_left.
* The Paste action. Append the value of c to the end of the string s. Note that this doesn't modify c.
The cursor initially starts at ℓ = 0. Then, we perform the following procedure:
1. Perform the Move action once.
2. Perform the Cut action once.
3. Perform the Paste action s_ℓ times.
4. If ℓ = x, stop. Otherwise, return to step 1.
You're given the initial string s and the integer x. What is the length of s when the procedure stops? Since this value may be very large, only find it modulo 10^9 + 7.
It is guaranteed that ℓ ≤ |s| at any time.
Input
The first line of input contains a single integer t (1 ≤ t ≤ 1000) denoting the number of test cases. The next lines contain descriptions of the test cases.
The first line of each test case contains a single integer x (1 ≤ x ≤ 10^6). The second line of each test case consists of the initial string s (1 ≤ |s| ≤ 500). It is guaranteed, that s consists of the characters "1", "2", "3".
It is guaranteed that the sum of x in a single file is at most 10^6. It is guaranteed that in each test case before the procedure will stop it will be true that ℓ ≤ |s| at any time.
Output
For each test case, output a single line containing a single integer denoting the answer for that test case modulo 10^9 + 7.
Example
Input
4
5
231
7
2323
6
333
24
133321333
Output
25
1438
1101
686531475
Note
Let's illustrate what happens with the first test case. Initially, we have s = 231. Initially, ℓ = 0 and c = \varepsilon (the empty string). The following things happen if we follow the procedure above:
* Step 1, Move once: we get ℓ = 1.
* Step 2, Cut once: we get s = 2 and c = 31.
* Step 3, Paste s_ℓ = 2 times: we get s = 23131.
* Step 4: ℓ = 1 not= x = 5, so we return to step 1.
* Step 1, Move once: we get ℓ = 2.
* Step 2, Cut once: we get s = 23 and c = 131.
* Step 3, Paste s_ℓ = 3 times: we get s = 23131131131.
* Step 4: ℓ = 2 not= x = 5, so we return to step 1.
* Step 1, Move once: we get ℓ = 3.
* Step 2, Cut once: we get s = 231 and c = 31131131.
* Step 3, Paste s_ℓ = 1 time: we get s = 23131131131.
* Step 4: ℓ = 3 not= x = 5, so we return to step 1.
* Step 1, Move once: we get ℓ = 4.
* Step 2, Cut once: we get s = 2313 and c = 1131131.
* Step 3, Paste s_ℓ = 3 times: we get s = 2313113113111311311131131.
* Step 4: ℓ = 4 not= x = 5, so we return to step 1.
* Step 1, Move once: we get ℓ = 5.
* Step 2, Cut once: we get s = 23131 and c = 13113111311311131131.
* Step 3, Paste s_ℓ = 1 times: we get s = 2313113113111311311131131.
* Step 4: ℓ = 5 = x, so we stop.
At the end of the procedure, s has length 25.
Submitted Solution:
```
#TLE
# import sys, os.path
# import math
# from collections import defaultdict,deque
# input = sys.stdin.readline
# I = lambda : list(map(int,input().split()))
# S = lambda : list(map(str,input()))
# def main():
# # s1 = [0]*((2*(10**6))+1)
# t,=I()
# for t1 in range(t):
# x, = I()
# s1 = str(input()).rstrip('\n')
# # for i in range(len(s2)):
# # s1[i] = int(s2[i])
# l = 0
# count = 0
# count1 = len(s1)
# count1%=((10**9)+7)
# for i in range(x):
# k = int(s1[i])
# count1-=1
# count1%=((10**9)+7)
# if count1>x:
# count1*=k
# count1%=((10**9)+7)
# else:
# l1 = len(s1)
# for j in range(k-1):
# for l in range(i+1,l1):
# s1+=s1[l]
# count1*=k
# count1%=((10**9)+7)
# count1+=x
# count1%=((10**9)+7)
# print(count1)
# main()
#
import sys, os.path
import math
from collections import defaultdict,deque
input = sys.stdin.readline
I = lambda : list(map(int,input().split()))
S = lambda : list(map(str,input()))
def main():
# s1 = [0]*((2*(10**6))+1)
t,=I()
for t1 in range(t):
x, = I()
s2 = str(input()).rstrip('\n')
s1 = []
for i in range(len(s2)):
if i>x:
break
s1.append(int(s2[i]))
count1 = len(s2)
count1%=((10**9)+7)
for i in range(x):
k = int(s1[i])
count1-=1
count1%=((10**9)+7)
if len(s1)>x:
count1*=k
count1%=((10**9)+7)
else:
if k==1:
continue
X = s1[i+1:]
l1 = len(s1)
for j in range(k-1):
if len(s1)<x:
s1.extend(X)
count1*=k
count1%=((10**9)+7)
count1+=x
count1%=((10**9)+7)
print(count1)
main()
# import sys
# input = sys.stdin.readline
# mod = 10 ** 9 + 7
# t = int(input())
# for _ in range(t):
# x = int(input())
# s = input().rstrip('\n')
# X = []
# for i, si in enumerate(s):
# if i == x:
# break
# X.append(int(s[i]))
# ans = len(s)
# for i in range(x):
# ans = (ans + (ans - i - 1) * (X[i] - 1)) % mod
# if len(X) < x:
# if X[i] == 1:
# continue
# X_ = X[i + 1:]
# for j in range(X[i] - 1):
# if len(X) < x:
# X.extend(X_)
# print(ans)
```
|
instruction
| 0
| 104,548
| 0
| 209,096
|
Yes
|
output
| 1
| 104,548
| 0
| 209,097
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We start with a string s consisting only of the digits 1, 2, or 3. The length of s is denoted by |s|. For each i from 1 to |s|, the i-th character of s is denoted by s_i.
There is one cursor. The cursor's location ℓ is denoted by an integer in \{0, …, |s|\}, with the following meaning:
* If ℓ = 0, then the cursor is located before the first character of s.
* If ℓ = |s|, then the cursor is located right after the last character of s.
* If 0 < ℓ < |s|, then the cursor is located between s_ℓ and s_{ℓ+1}.
We denote by s_left the string to the left of the cursor and s_right the string to the right of the cursor.
We also have a string c, which we call our clipboard, which starts out as empty. There are three types of actions:
* The Move action. Move the cursor one step to the right. This increments ℓ once.
* The Cut action. Set c ← s_right, then set s ← s_left.
* The Paste action. Append the value of c to the end of the string s. Note that this doesn't modify c.
The cursor initially starts at ℓ = 0. Then, we perform the following procedure:
1. Perform the Move action once.
2. Perform the Cut action once.
3. Perform the Paste action s_ℓ times.
4. If ℓ = x, stop. Otherwise, return to step 1.
You're given the initial string s and the integer x. What is the length of s when the procedure stops? Since this value may be very large, only find it modulo 10^9 + 7.
It is guaranteed that ℓ ≤ |s| at any time.
Input
The first line of input contains a single integer t (1 ≤ t ≤ 1000) denoting the number of test cases. The next lines contain descriptions of the test cases.
The first line of each test case contains a single integer x (1 ≤ x ≤ 10^6). The second line of each test case consists of the initial string s (1 ≤ |s| ≤ 500). It is guaranteed, that s consists of the characters "1", "2", "3".
It is guaranteed that the sum of x in a single file is at most 10^6. It is guaranteed that in each test case before the procedure will stop it will be true that ℓ ≤ |s| at any time.
Output
For each test case, output a single line containing a single integer denoting the answer for that test case modulo 10^9 + 7.
Example
Input
4
5
231
7
2323
6
333
24
133321333
Output
25
1438
1101
686531475
Note
Let's illustrate what happens with the first test case. Initially, we have s = 231. Initially, ℓ = 0 and c = \varepsilon (the empty string). The following things happen if we follow the procedure above:
* Step 1, Move once: we get ℓ = 1.
* Step 2, Cut once: we get s = 2 and c = 31.
* Step 3, Paste s_ℓ = 2 times: we get s = 23131.
* Step 4: ℓ = 1 not= x = 5, so we return to step 1.
* Step 1, Move once: we get ℓ = 2.
* Step 2, Cut once: we get s = 23 and c = 131.
* Step 3, Paste s_ℓ = 3 times: we get s = 23131131131.
* Step 4: ℓ = 2 not= x = 5, so we return to step 1.
* Step 1, Move once: we get ℓ = 3.
* Step 2, Cut once: we get s = 231 and c = 31131131.
* Step 3, Paste s_ℓ = 1 time: we get s = 23131131131.
* Step 4: ℓ = 3 not= x = 5, so we return to step 1.
* Step 1, Move once: we get ℓ = 4.
* Step 2, Cut once: we get s = 2313 and c = 1131131.
* Step 3, Paste s_ℓ = 3 times: we get s = 2313113113111311311131131.
* Step 4: ℓ = 4 not= x = 5, so we return to step 1.
* Step 1, Move once: we get ℓ = 5.
* Step 2, Cut once: we get s = 23131 and c = 13113111311311131131.
* Step 3, Paste s_ℓ = 1 times: we get s = 2313113113111311311131131.
* Step 4: ℓ = 5 = x, so we stop.
At the end of the procedure, s has length 25.
Submitted Solution:
```
import time
import sys
MOD = 1e9 + 7
# f = open('test.txt', 'w+')
# f.write("1\n333047\n")
# f.write('1'*300 + '2' + '1'*20 + '2')
# f.seek(0)
f = sys.stdin
t1 = time.time()
for t in range(int(f.readline())):
x, s = int(f.readline().rstrip()), f.readline().rstrip()
i = 0
while len(s) < x:
i += 1
s += s[i:] * (int(s[i - 1])-1)
answer = len(s)
while i < x:
i += 1
answer = (answer + (answer - i) * (int(s[i - 1]) - 1)) % MOD
print(int(answer))
# print("--- %s seconds ---" % (time.time() - t1))
```
|
instruction
| 0
| 104,549
| 0
| 209,098
|
Yes
|
output
| 1
| 104,549
| 0
| 209,099
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We start with a string s consisting only of the digits 1, 2, or 3. The length of s is denoted by |s|. For each i from 1 to |s|, the i-th character of s is denoted by s_i.
There is one cursor. The cursor's location ℓ is denoted by an integer in \{0, …, |s|\}, with the following meaning:
* If ℓ = 0, then the cursor is located before the first character of s.
* If ℓ = |s|, then the cursor is located right after the last character of s.
* If 0 < ℓ < |s|, then the cursor is located between s_ℓ and s_{ℓ+1}.
We denote by s_left the string to the left of the cursor and s_right the string to the right of the cursor.
We also have a string c, which we call our clipboard, which starts out as empty. There are three types of actions:
* The Move action. Move the cursor one step to the right. This increments ℓ once.
* The Cut action. Set c ← s_right, then set s ← s_left.
* The Paste action. Append the value of c to the end of the string s. Note that this doesn't modify c.
The cursor initially starts at ℓ = 0. Then, we perform the following procedure:
1. Perform the Move action once.
2. Perform the Cut action once.
3. Perform the Paste action s_ℓ times.
4. If ℓ = x, stop. Otherwise, return to step 1.
You're given the initial string s and the integer x. What is the length of s when the procedure stops? Since this value may be very large, only find it modulo 10^9 + 7.
It is guaranteed that ℓ ≤ |s| at any time.
Input
The first line of input contains a single integer t (1 ≤ t ≤ 1000) denoting the number of test cases. The next lines contain descriptions of the test cases.
The first line of each test case contains a single integer x (1 ≤ x ≤ 10^6). The second line of each test case consists of the initial string s (1 ≤ |s| ≤ 500). It is guaranteed, that s consists of the characters "1", "2", "3".
It is guaranteed that the sum of x in a single file is at most 10^6. It is guaranteed that in each test case before the procedure will stop it will be true that ℓ ≤ |s| at any time.
Output
For each test case, output a single line containing a single integer denoting the answer for that test case modulo 10^9 + 7.
Example
Input
4
5
231
7
2323
6
333
24
133321333
Output
25
1438
1101
686531475
Note
Let's illustrate what happens with the first test case. Initially, we have s = 231. Initially, ℓ = 0 and c = \varepsilon (the empty string). The following things happen if we follow the procedure above:
* Step 1, Move once: we get ℓ = 1.
* Step 2, Cut once: we get s = 2 and c = 31.
* Step 3, Paste s_ℓ = 2 times: we get s = 23131.
* Step 4: ℓ = 1 not= x = 5, so we return to step 1.
* Step 1, Move once: we get ℓ = 2.
* Step 2, Cut once: we get s = 23 and c = 131.
* Step 3, Paste s_ℓ = 3 times: we get s = 23131131131.
* Step 4: ℓ = 2 not= x = 5, so we return to step 1.
* Step 1, Move once: we get ℓ = 3.
* Step 2, Cut once: we get s = 231 and c = 31131131.
* Step 3, Paste s_ℓ = 1 time: we get s = 23131131131.
* Step 4: ℓ = 3 not= x = 5, so we return to step 1.
* Step 1, Move once: we get ℓ = 4.
* Step 2, Cut once: we get s = 2313 and c = 1131131.
* Step 3, Paste s_ℓ = 3 times: we get s = 2313113113111311311131131.
* Step 4: ℓ = 4 not= x = 5, so we return to step 1.
* Step 1, Move once: we get ℓ = 5.
* Step 2, Cut once: we get s = 23131 and c = 13113111311311131131.
* Step 3, Paste s_ℓ = 1 times: we get s = 2313113113111311311131131.
* Step 4: ℓ = 5 = x, so we stop.
At the end of the procedure, s has length 25.
Submitted Solution:
```
import sys
T = int(sys.stdin.readline().strip())
for _ in range(T):
x = int(sys.stdin.readline().strip())
s = list(sys.stdin.readline().strip())
ls = len(s)
for l in range(x):
t = ord(s[l]) - ord('0') - 1
ls = (ls + (ls - l - 1)*t) % ((10**9)+7)
e = len(s)
while len(s) <= x and t > 0:
s += s[l+1:e]
t -= 1
print(ls)
```
|
instruction
| 0
| 104,550
| 0
| 209,100
|
Yes
|
output
| 1
| 104,550
| 0
| 209,101
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We start with a string s consisting only of the digits 1, 2, or 3. The length of s is denoted by |s|. For each i from 1 to |s|, the i-th character of s is denoted by s_i.
There is one cursor. The cursor's location ℓ is denoted by an integer in \{0, …, |s|\}, with the following meaning:
* If ℓ = 0, then the cursor is located before the first character of s.
* If ℓ = |s|, then the cursor is located right after the last character of s.
* If 0 < ℓ < |s|, then the cursor is located between s_ℓ and s_{ℓ+1}.
We denote by s_left the string to the left of the cursor and s_right the string to the right of the cursor.
We also have a string c, which we call our clipboard, which starts out as empty. There are three types of actions:
* The Move action. Move the cursor one step to the right. This increments ℓ once.
* The Cut action. Set c ← s_right, then set s ← s_left.
* The Paste action. Append the value of c to the end of the string s. Note that this doesn't modify c.
The cursor initially starts at ℓ = 0. Then, we perform the following procedure:
1. Perform the Move action once.
2. Perform the Cut action once.
3. Perform the Paste action s_ℓ times.
4. If ℓ = x, stop. Otherwise, return to step 1.
You're given the initial string s and the integer x. What is the length of s when the procedure stops? Since this value may be very large, only find it modulo 10^9 + 7.
It is guaranteed that ℓ ≤ |s| at any time.
Input
The first line of input contains a single integer t (1 ≤ t ≤ 1000) denoting the number of test cases. The next lines contain descriptions of the test cases.
The first line of each test case contains a single integer x (1 ≤ x ≤ 10^6). The second line of each test case consists of the initial string s (1 ≤ |s| ≤ 500). It is guaranteed, that s consists of the characters "1", "2", "3".
It is guaranteed that the sum of x in a single file is at most 10^6. It is guaranteed that in each test case before the procedure will stop it will be true that ℓ ≤ |s| at any time.
Output
For each test case, output a single line containing a single integer denoting the answer for that test case modulo 10^9 + 7.
Example
Input
4
5
231
7
2323
6
333
24
133321333
Output
25
1438
1101
686531475
Note
Let's illustrate what happens with the first test case. Initially, we have s = 231. Initially, ℓ = 0 and c = \varepsilon (the empty string). The following things happen if we follow the procedure above:
* Step 1, Move once: we get ℓ = 1.
* Step 2, Cut once: we get s = 2 and c = 31.
* Step 3, Paste s_ℓ = 2 times: we get s = 23131.
* Step 4: ℓ = 1 not= x = 5, so we return to step 1.
* Step 1, Move once: we get ℓ = 2.
* Step 2, Cut once: we get s = 23 and c = 131.
* Step 3, Paste s_ℓ = 3 times: we get s = 23131131131.
* Step 4: ℓ = 2 not= x = 5, so we return to step 1.
* Step 1, Move once: we get ℓ = 3.
* Step 2, Cut once: we get s = 231 and c = 31131131.
* Step 3, Paste s_ℓ = 1 time: we get s = 23131131131.
* Step 4: ℓ = 3 not= x = 5, so we return to step 1.
* Step 1, Move once: we get ℓ = 4.
* Step 2, Cut once: we get s = 2313 and c = 1131131.
* Step 3, Paste s_ℓ = 3 times: we get s = 2313113113111311311131131.
* Step 4: ℓ = 4 not= x = 5, so we return to step 1.
* Step 1, Move once: we get ℓ = 5.
* Step 2, Cut once: we get s = 23131 and c = 13113111311311131131.
* Step 3, Paste s_ℓ = 1 times: we get s = 2313113113111311311131131.
* Step 4: ℓ = 5 = x, so we stop.
At the end of the procedure, s has length 25.
Submitted Solution:
```
t=int(input())
for i in range(t):
x=int(input())
s=input()
p=s
l=0
while(l!=x):
r=int(p[0])
p=p[1:]
if r>1:
p+=p
if r>2:
p+=p
if (len(p)+l+1)>=x:
l+=1
break
l+=1
j=0
res=l
y=len(p)
while(l!=x):
y-=1
y=y*int(p[j])
l+=1
res+=1
j+=1
print((res+y)%(10**9+7))
```
|
instruction
| 0
| 104,551
| 0
| 209,102
|
No
|
output
| 1
| 104,551
| 0
| 209,103
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We start with a string s consisting only of the digits 1, 2, or 3. The length of s is denoted by |s|. For each i from 1 to |s|, the i-th character of s is denoted by s_i.
There is one cursor. The cursor's location ℓ is denoted by an integer in \{0, …, |s|\}, with the following meaning:
* If ℓ = 0, then the cursor is located before the first character of s.
* If ℓ = |s|, then the cursor is located right after the last character of s.
* If 0 < ℓ < |s|, then the cursor is located between s_ℓ and s_{ℓ+1}.
We denote by s_left the string to the left of the cursor and s_right the string to the right of the cursor.
We also have a string c, which we call our clipboard, which starts out as empty. There are three types of actions:
* The Move action. Move the cursor one step to the right. This increments ℓ once.
* The Cut action. Set c ← s_right, then set s ← s_left.
* The Paste action. Append the value of c to the end of the string s. Note that this doesn't modify c.
The cursor initially starts at ℓ = 0. Then, we perform the following procedure:
1. Perform the Move action once.
2. Perform the Cut action once.
3. Perform the Paste action s_ℓ times.
4. If ℓ = x, stop. Otherwise, return to step 1.
You're given the initial string s and the integer x. What is the length of s when the procedure stops? Since this value may be very large, only find it modulo 10^9 + 7.
It is guaranteed that ℓ ≤ |s| at any time.
Input
The first line of input contains a single integer t (1 ≤ t ≤ 1000) denoting the number of test cases. The next lines contain descriptions of the test cases.
The first line of each test case contains a single integer x (1 ≤ x ≤ 10^6). The second line of each test case consists of the initial string s (1 ≤ |s| ≤ 500). It is guaranteed, that s consists of the characters "1", "2", "3".
It is guaranteed that the sum of x in a single file is at most 10^6. It is guaranteed that in each test case before the procedure will stop it will be true that ℓ ≤ |s| at any time.
Output
For each test case, output a single line containing a single integer denoting the answer for that test case modulo 10^9 + 7.
Example
Input
4
5
231
7
2323
6
333
24
133321333
Output
25
1438
1101
686531475
Note
Let's illustrate what happens with the first test case. Initially, we have s = 231. Initially, ℓ = 0 and c = \varepsilon (the empty string). The following things happen if we follow the procedure above:
* Step 1, Move once: we get ℓ = 1.
* Step 2, Cut once: we get s = 2 and c = 31.
* Step 3, Paste s_ℓ = 2 times: we get s = 23131.
* Step 4: ℓ = 1 not= x = 5, so we return to step 1.
* Step 1, Move once: we get ℓ = 2.
* Step 2, Cut once: we get s = 23 and c = 131.
* Step 3, Paste s_ℓ = 3 times: we get s = 23131131131.
* Step 4: ℓ = 2 not= x = 5, so we return to step 1.
* Step 1, Move once: we get ℓ = 3.
* Step 2, Cut once: we get s = 231 and c = 31131131.
* Step 3, Paste s_ℓ = 1 time: we get s = 23131131131.
* Step 4: ℓ = 3 not= x = 5, so we return to step 1.
* Step 1, Move once: we get ℓ = 4.
* Step 2, Cut once: we get s = 2313 and c = 1131131.
* Step 3, Paste s_ℓ = 3 times: we get s = 2313113113111311311131131.
* Step 4: ℓ = 4 not= x = 5, so we return to step 1.
* Step 1, Move once: we get ℓ = 5.
* Step 2, Cut once: we get s = 23131 and c = 13113111311311131131.
* Step 3, Paste s_ℓ = 1 times: we get s = 2313113113111311311131131.
* Step 4: ℓ = 5 = x, so we stop.
At the end of the procedure, s has length 25.
Submitted Solution:
```
for _ in range(int(input())):
x = int(input())
s = list(map(int, input()))
ans = len(s)
for i in range(1, x+1):
ans = (i + (ans-i) * s[i-1])%1000000007
for _ in range(s[i-1]-1):
if len(s) < x:
s += s[i:]
else:
break
print(ans)
```
|
instruction
| 0
| 104,552
| 0
| 209,104
|
No
|
output
| 1
| 104,552
| 0
| 209,105
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We start with a string s consisting only of the digits 1, 2, or 3. The length of s is denoted by |s|. For each i from 1 to |s|, the i-th character of s is denoted by s_i.
There is one cursor. The cursor's location ℓ is denoted by an integer in \{0, …, |s|\}, with the following meaning:
* If ℓ = 0, then the cursor is located before the first character of s.
* If ℓ = |s|, then the cursor is located right after the last character of s.
* If 0 < ℓ < |s|, then the cursor is located between s_ℓ and s_{ℓ+1}.
We denote by s_left the string to the left of the cursor and s_right the string to the right of the cursor.
We also have a string c, which we call our clipboard, which starts out as empty. There are three types of actions:
* The Move action. Move the cursor one step to the right. This increments ℓ once.
* The Cut action. Set c ← s_right, then set s ← s_left.
* The Paste action. Append the value of c to the end of the string s. Note that this doesn't modify c.
The cursor initially starts at ℓ = 0. Then, we perform the following procedure:
1. Perform the Move action once.
2. Perform the Cut action once.
3. Perform the Paste action s_ℓ times.
4. If ℓ = x, stop. Otherwise, return to step 1.
You're given the initial string s and the integer x. What is the length of s when the procedure stops? Since this value may be very large, only find it modulo 10^9 + 7.
It is guaranteed that ℓ ≤ |s| at any time.
Input
The first line of input contains a single integer t (1 ≤ t ≤ 1000) denoting the number of test cases. The next lines contain descriptions of the test cases.
The first line of each test case contains a single integer x (1 ≤ x ≤ 10^6). The second line of each test case consists of the initial string s (1 ≤ |s| ≤ 500). It is guaranteed, that s consists of the characters "1", "2", "3".
It is guaranteed that the sum of x in a single file is at most 10^6. It is guaranteed that in each test case before the procedure will stop it will be true that ℓ ≤ |s| at any time.
Output
For each test case, output a single line containing a single integer denoting the answer for that test case modulo 10^9 + 7.
Example
Input
4
5
231
7
2323
6
333
24
133321333
Output
25
1438
1101
686531475
Note
Let's illustrate what happens with the first test case. Initially, we have s = 231. Initially, ℓ = 0 and c = \varepsilon (the empty string). The following things happen if we follow the procedure above:
* Step 1, Move once: we get ℓ = 1.
* Step 2, Cut once: we get s = 2 and c = 31.
* Step 3, Paste s_ℓ = 2 times: we get s = 23131.
* Step 4: ℓ = 1 not= x = 5, so we return to step 1.
* Step 1, Move once: we get ℓ = 2.
* Step 2, Cut once: we get s = 23 and c = 131.
* Step 3, Paste s_ℓ = 3 times: we get s = 23131131131.
* Step 4: ℓ = 2 not= x = 5, so we return to step 1.
* Step 1, Move once: we get ℓ = 3.
* Step 2, Cut once: we get s = 231 and c = 31131131.
* Step 3, Paste s_ℓ = 1 time: we get s = 23131131131.
* Step 4: ℓ = 3 not= x = 5, so we return to step 1.
* Step 1, Move once: we get ℓ = 4.
* Step 2, Cut once: we get s = 2313 and c = 1131131.
* Step 3, Paste s_ℓ = 3 times: we get s = 2313113113111311311131131.
* Step 4: ℓ = 4 not= x = 5, so we return to step 1.
* Step 1, Move once: we get ℓ = 5.
* Step 2, Cut once: we get s = 23131 and c = 13113111311311131131.
* Step 3, Paste s_ℓ = 1 times: we get s = 2313113113111311311131131.
* Step 4: ℓ = 5 = x, so we stop.
At the end of the procedure, s has length 25.
Submitted Solution:
```
t=int(input())
for i in range(t):
x=int(input())
y=input()
s=[]
for j in y:
s.append(j)
l=0
res=len(s)
while(l!=x):
if len(s)<x:
for p in range(int(s[l])-1):
for j in range(l + 1, len(s)):
s.append(s[j])
res=(res+(res-l-1)*(int(s[l])-1))%(10**9+7)
l+=1
print(res%(10**9+7))
```
|
instruction
| 0
| 104,553
| 0
| 209,106
|
No
|
output
| 1
| 104,553
| 0
| 209,107
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We start with a string s consisting only of the digits 1, 2, or 3. The length of s is denoted by |s|. For each i from 1 to |s|, the i-th character of s is denoted by s_i.
There is one cursor. The cursor's location ℓ is denoted by an integer in \{0, …, |s|\}, with the following meaning:
* If ℓ = 0, then the cursor is located before the first character of s.
* If ℓ = |s|, then the cursor is located right after the last character of s.
* If 0 < ℓ < |s|, then the cursor is located between s_ℓ and s_{ℓ+1}.
We denote by s_left the string to the left of the cursor and s_right the string to the right of the cursor.
We also have a string c, which we call our clipboard, which starts out as empty. There are three types of actions:
* The Move action. Move the cursor one step to the right. This increments ℓ once.
* The Cut action. Set c ← s_right, then set s ← s_left.
* The Paste action. Append the value of c to the end of the string s. Note that this doesn't modify c.
The cursor initially starts at ℓ = 0. Then, we perform the following procedure:
1. Perform the Move action once.
2. Perform the Cut action once.
3. Perform the Paste action s_ℓ times.
4. If ℓ = x, stop. Otherwise, return to step 1.
You're given the initial string s and the integer x. What is the length of s when the procedure stops? Since this value may be very large, only find it modulo 10^9 + 7.
It is guaranteed that ℓ ≤ |s| at any time.
Input
The first line of input contains a single integer t (1 ≤ t ≤ 1000) denoting the number of test cases. The next lines contain descriptions of the test cases.
The first line of each test case contains a single integer x (1 ≤ x ≤ 10^6). The second line of each test case consists of the initial string s (1 ≤ |s| ≤ 500). It is guaranteed, that s consists of the characters "1", "2", "3".
It is guaranteed that the sum of x in a single file is at most 10^6. It is guaranteed that in each test case before the procedure will stop it will be true that ℓ ≤ |s| at any time.
Output
For each test case, output a single line containing a single integer denoting the answer for that test case modulo 10^9 + 7.
Example
Input
4
5
231
7
2323
6
333
24
133321333
Output
25
1438
1101
686531475
Note
Let's illustrate what happens with the first test case. Initially, we have s = 231. Initially, ℓ = 0 and c = \varepsilon (the empty string). The following things happen if we follow the procedure above:
* Step 1, Move once: we get ℓ = 1.
* Step 2, Cut once: we get s = 2 and c = 31.
* Step 3, Paste s_ℓ = 2 times: we get s = 23131.
* Step 4: ℓ = 1 not= x = 5, so we return to step 1.
* Step 1, Move once: we get ℓ = 2.
* Step 2, Cut once: we get s = 23 and c = 131.
* Step 3, Paste s_ℓ = 3 times: we get s = 23131131131.
* Step 4: ℓ = 2 not= x = 5, so we return to step 1.
* Step 1, Move once: we get ℓ = 3.
* Step 2, Cut once: we get s = 231 and c = 31131131.
* Step 3, Paste s_ℓ = 1 time: we get s = 23131131131.
* Step 4: ℓ = 3 not= x = 5, so we return to step 1.
* Step 1, Move once: we get ℓ = 4.
* Step 2, Cut once: we get s = 2313 and c = 1131131.
* Step 3, Paste s_ℓ = 3 times: we get s = 2313113113111311311131131.
* Step 4: ℓ = 4 not= x = 5, so we return to step 1.
* Step 1, Move once: we get ℓ = 5.
* Step 2, Cut once: we get s = 23131 and c = 13113111311311131131.
* Step 3, Paste s_ℓ = 1 times: we get s = 2313113113111311311131131.
* Step 4: ℓ = 5 = x, so we stop.
At the end of the procedure, s has length 25.
Submitted Solution:
```
import sys
import math
from collections import defaultdict,Counter,deque
# input=sys.stdin.readline
# def print(x):
# sys.stdout.write(str(x)+"\n")
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# sys.stdout=open("CP2/output.txt",'w')
# sys.stdin=open("CP2/input.txt",'r')
mod=pow(10,9)+7
t=int(input())
for i in range(t):
x=int(input())
s=list(map(int,input()))
cur=0
ans=0
tot=len(s)-1
while cur<x:
tot=(s[cur]*tot)%mod
if len(s)<x:
s+=s[cur+1:]*(s[cur]-1)
else:
# print(len(s))
if cur+1+tot>=mod:
ans=cur+1+tot-mod
else:
ans=cur+1+tot
if tot==0:
tot=10**9-1
else:
tot-=1
cur+=1
print(len(s))
print(ans)
```
|
instruction
| 0
| 104,554
| 0
| 209,108
|
No
|
output
| 1
| 104,554
| 0
| 209,109
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Ayoub thinks that he is a very smart person, so he created a function f(s), where s is a binary string (a string which contains only symbols "0" and "1"). The function f(s) is equal to the number of substrings in the string s that contains at least one symbol, that is equal to "1".
More formally, f(s) is equal to the number of pairs of integers (l, r), such that 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of string s), such that at least one of the symbols s_l, s_{l+1}, …, s_r is equal to "1".
For example, if s = "01010" then f(s) = 12, because there are 12 such pairs (l, r): (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 4), (4, 5).
Ayoub also thinks that he is smarter than Mahmoud so he gave him two integers n and m and asked him this problem. For all binary strings s of length n which contains exactly m symbols equal to "1", find the maximum value of f(s).
Mahmoud couldn't solve the problem so he asked you for help. Can you help him?
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. The description of the test cases follows.
The only line for each test case contains two integers n, m (1 ≤ n ≤ 10^{9}, 0 ≤ m ≤ n) — the length of the string and the number of symbols equal to "1" in it.
Output
For every test case print one integer number — the maximum value of f(s) over all strings s of length n, which has exactly m symbols, equal to "1".
Example
Input
5
3 1
3 2
3 3
4 0
5 2
Output
4
5
6
0
12
Note
In the first test case, there exists only 3 strings of length 3, which has exactly 1 symbol, equal to "1". These strings are: s_1 = "100", s_2 = "010", s_3 = "001". The values of f for them are: f(s_1) = 3, f(s_2) = 4, f(s_3) = 3, so the maximum value is 4 and the answer is 4.
In the second test case, the string s with the maximum value is "101".
In the third test case, the string s with the maximum value is "111".
In the fourth test case, the only string s of length 4, which has exactly 0 symbols, equal to "1" is "0000" and the value of f for that string is 0, so the answer is 0.
In the fifth test case, the string s with the maximum value is "01010" and it is described as an example in the problem statement.
|
instruction
| 0
| 104,555
| 0
| 209,110
|
Tags: binary search, combinatorics, greedy, math, strings
Correct Solution:
```
import math
import sys
for _ in range(int(input())):
n,m=(map(int,sys.stdin.readline().split()))
k=n-m
p=m+1
c=(k//p)
ans=(n*(n+1))//2
rem=k%p
ans1=(rem*((c+1)*(c+2)))//2
ans2=0
ans2=(c*(c+1)*(p-rem))//2
print(ans-ans1-ans2)
```
|
output
| 1
| 104,555
| 0
| 209,111
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Ayoub thinks that he is a very smart person, so he created a function f(s), where s is a binary string (a string which contains only symbols "0" and "1"). The function f(s) is equal to the number of substrings in the string s that contains at least one symbol, that is equal to "1".
More formally, f(s) is equal to the number of pairs of integers (l, r), such that 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of string s), such that at least one of the symbols s_l, s_{l+1}, …, s_r is equal to "1".
For example, if s = "01010" then f(s) = 12, because there are 12 such pairs (l, r): (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 4), (4, 5).
Ayoub also thinks that he is smarter than Mahmoud so he gave him two integers n and m and asked him this problem. For all binary strings s of length n which contains exactly m symbols equal to "1", find the maximum value of f(s).
Mahmoud couldn't solve the problem so he asked you for help. Can you help him?
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. The description of the test cases follows.
The only line for each test case contains two integers n, m (1 ≤ n ≤ 10^{9}, 0 ≤ m ≤ n) — the length of the string and the number of symbols equal to "1" in it.
Output
For every test case print one integer number — the maximum value of f(s) over all strings s of length n, which has exactly m symbols, equal to "1".
Example
Input
5
3 1
3 2
3 3
4 0
5 2
Output
4
5
6
0
12
Note
In the first test case, there exists only 3 strings of length 3, which has exactly 1 symbol, equal to "1". These strings are: s_1 = "100", s_2 = "010", s_3 = "001". The values of f for them are: f(s_1) = 3, f(s_2) = 4, f(s_3) = 3, so the maximum value is 4 and the answer is 4.
In the second test case, the string s with the maximum value is "101".
In the third test case, the string s with the maximum value is "111".
In the fourth test case, the only string s of length 4, which has exactly 0 symbols, equal to "1" is "0000" and the value of f for that string is 0, so the answer is 0.
In the fifth test case, the string s with the maximum value is "01010" and it is described as an example in the problem statement.
|
instruction
| 0
| 104,556
| 0
| 209,112
|
Tags: binary search, combinatorics, greedy, math, strings
Correct Solution:
```
import sys
input = sys.stdin.readline
t = int(input())
C = [list(map(int,input().split())) for i in range(t)]
# ANS = [0] * t
# t = 10**5
# C = [[523422132331, 102342223434] for i in range(t)]
def c(x, y):
if x % 2 == 0:
return (x//2) * y
else:
return (y//2) * x
for i in range(t):
n = C[i][0]
o = C[i][1]
z = n-o
al = c(n, (n-1)) + n
# print("al",al)
if z-1 <= o:
print(al - z)
# ANS[i] = al-z
# v=0
else:
a = z // (o+1)
b = z % (o+1)
# print(a,b)
mi = c((a+1), (a+2)) * b + c(a, (a+1)) * (o+1-b)
print(al - mi)
# ANS[i] = al-mi
# for i in ANS:
# print(i)
```
|
output
| 1
| 104,556
| 0
| 209,113
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Ayoub thinks that he is a very smart person, so he created a function f(s), where s is a binary string (a string which contains only symbols "0" and "1"). The function f(s) is equal to the number of substrings in the string s that contains at least one symbol, that is equal to "1".
More formally, f(s) is equal to the number of pairs of integers (l, r), such that 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of string s), such that at least one of the symbols s_l, s_{l+1}, …, s_r is equal to "1".
For example, if s = "01010" then f(s) = 12, because there are 12 such pairs (l, r): (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 4), (4, 5).
Ayoub also thinks that he is smarter than Mahmoud so he gave him two integers n and m and asked him this problem. For all binary strings s of length n which contains exactly m symbols equal to "1", find the maximum value of f(s).
Mahmoud couldn't solve the problem so he asked you for help. Can you help him?
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. The description of the test cases follows.
The only line for each test case contains two integers n, m (1 ≤ n ≤ 10^{9}, 0 ≤ m ≤ n) — the length of the string and the number of symbols equal to "1" in it.
Output
For every test case print one integer number — the maximum value of f(s) over all strings s of length n, which has exactly m symbols, equal to "1".
Example
Input
5
3 1
3 2
3 3
4 0
5 2
Output
4
5
6
0
12
Note
In the first test case, there exists only 3 strings of length 3, which has exactly 1 symbol, equal to "1". These strings are: s_1 = "100", s_2 = "010", s_3 = "001". The values of f for them are: f(s_1) = 3, f(s_2) = 4, f(s_3) = 3, so the maximum value is 4 and the answer is 4.
In the second test case, the string s with the maximum value is "101".
In the third test case, the string s with the maximum value is "111".
In the fourth test case, the only string s of length 4, which has exactly 0 symbols, equal to "1" is "0000" and the value of f for that string is 0, so the answer is 0.
In the fifth test case, the string s with the maximum value is "01010" and it is described as an example in the problem statement.
|
instruction
| 0
| 104,557
| 0
| 209,114
|
Tags: binary search, combinatorics, greedy, math, strings
Correct Solution:
```
import sys
input=sys.stdin.readline
for _ in range(int(input())):
n,m=map(int,input().split())
grps=m+1
zeach=(n-m)//grps
extras=(n-m)%grps
print( (n*(n+1)//2)- (zeach*(zeach+1)//2)*grps -(extras)*(zeach+1) )
```
|
output
| 1
| 104,557
| 0
| 209,115
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Ayoub thinks that he is a very smart person, so he created a function f(s), where s is a binary string (a string which contains only symbols "0" and "1"). The function f(s) is equal to the number of substrings in the string s that contains at least one symbol, that is equal to "1".
More formally, f(s) is equal to the number of pairs of integers (l, r), such that 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of string s), such that at least one of the symbols s_l, s_{l+1}, …, s_r is equal to "1".
For example, if s = "01010" then f(s) = 12, because there are 12 such pairs (l, r): (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 4), (4, 5).
Ayoub also thinks that he is smarter than Mahmoud so he gave him two integers n and m and asked him this problem. For all binary strings s of length n which contains exactly m symbols equal to "1", find the maximum value of f(s).
Mahmoud couldn't solve the problem so he asked you for help. Can you help him?
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. The description of the test cases follows.
The only line for each test case contains two integers n, m (1 ≤ n ≤ 10^{9}, 0 ≤ m ≤ n) — the length of the string and the number of symbols equal to "1" in it.
Output
For every test case print one integer number — the maximum value of f(s) over all strings s of length n, which has exactly m symbols, equal to "1".
Example
Input
5
3 1
3 2
3 3
4 0
5 2
Output
4
5
6
0
12
Note
In the first test case, there exists only 3 strings of length 3, which has exactly 1 symbol, equal to "1". These strings are: s_1 = "100", s_2 = "010", s_3 = "001". The values of f for them are: f(s_1) = 3, f(s_2) = 4, f(s_3) = 3, so the maximum value is 4 and the answer is 4.
In the second test case, the string s with the maximum value is "101".
In the third test case, the string s with the maximum value is "111".
In the fourth test case, the only string s of length 4, which has exactly 0 symbols, equal to "1" is "0000" and the value of f for that string is 0, so the answer is 0.
In the fifth test case, the string s with the maximum value is "01010" and it is described as an example in the problem statement.
|
instruction
| 0
| 104,558
| 0
| 209,116
|
Tags: binary search, combinatorics, greedy, math, strings
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
from collections import Counter
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
def gcd(a, b):
if a == 0:
return b
return gcd(b % a, a)
def lcm(a, b):
return (a * b) / gcd(a, b)
def main():
for _ in range(int(input())):
n,m=map(int, input().split())
ans=n*(n+1)//2
#print(ans)
k=(n-m)//(m+1)
f=k
k=(n-m)%(m+1)
ans-=(m+1-k)*(f*(f+1)//2)
#print(ans)
f+=1
ans-=(k)*(f*(f+1)//2)
print(ans)
return
if __name__ == "__main__":
main()
```
|
output
| 1
| 104,558
| 0
| 209,117
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Ayoub thinks that he is a very smart person, so he created a function f(s), where s is a binary string (a string which contains only symbols "0" and "1"). The function f(s) is equal to the number of substrings in the string s that contains at least one symbol, that is equal to "1".
More formally, f(s) is equal to the number of pairs of integers (l, r), such that 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of string s), such that at least one of the symbols s_l, s_{l+1}, …, s_r is equal to "1".
For example, if s = "01010" then f(s) = 12, because there are 12 such pairs (l, r): (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 4), (4, 5).
Ayoub also thinks that he is smarter than Mahmoud so he gave him two integers n and m and asked him this problem. For all binary strings s of length n which contains exactly m symbols equal to "1", find the maximum value of f(s).
Mahmoud couldn't solve the problem so he asked you for help. Can you help him?
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. The description of the test cases follows.
The only line for each test case contains two integers n, m (1 ≤ n ≤ 10^{9}, 0 ≤ m ≤ n) — the length of the string and the number of symbols equal to "1" in it.
Output
For every test case print one integer number — the maximum value of f(s) over all strings s of length n, which has exactly m symbols, equal to "1".
Example
Input
5
3 1
3 2
3 3
4 0
5 2
Output
4
5
6
0
12
Note
In the first test case, there exists only 3 strings of length 3, which has exactly 1 symbol, equal to "1". These strings are: s_1 = "100", s_2 = "010", s_3 = "001". The values of f for them are: f(s_1) = 3, f(s_2) = 4, f(s_3) = 3, so the maximum value is 4 and the answer is 4.
In the second test case, the string s with the maximum value is "101".
In the third test case, the string s with the maximum value is "111".
In the fourth test case, the only string s of length 4, which has exactly 0 symbols, equal to "1" is "0000" and the value of f for that string is 0, so the answer is 0.
In the fifth test case, the string s with the maximum value is "01010" and it is described as an example in the problem statement.
|
instruction
| 0
| 104,559
| 0
| 209,118
|
Tags: binary search, combinatorics, greedy, math, strings
Correct Solution:
```
import sys
input = sys.stdin.readline
T = int(input())
for _ in range(T):
n, m = list(map(int, input().split()))
S = 0
x = (n - m) // (m + 1)
z = (n - m) % (m + 1)
S += (m + 1 - z) * (x * (x + 1)) // 2
S += z * (x + 1) * (x + 2) // 2
print(n * (n + 1) // 2 - S)
```
|
output
| 1
| 104,559
| 0
| 209,119
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Ayoub thinks that he is a very smart person, so he created a function f(s), where s is a binary string (a string which contains only symbols "0" and "1"). The function f(s) is equal to the number of substrings in the string s that contains at least one symbol, that is equal to "1".
More formally, f(s) is equal to the number of pairs of integers (l, r), such that 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of string s), such that at least one of the symbols s_l, s_{l+1}, …, s_r is equal to "1".
For example, if s = "01010" then f(s) = 12, because there are 12 such pairs (l, r): (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 4), (4, 5).
Ayoub also thinks that he is smarter than Mahmoud so he gave him two integers n and m and asked him this problem. For all binary strings s of length n which contains exactly m symbols equal to "1", find the maximum value of f(s).
Mahmoud couldn't solve the problem so he asked you for help. Can you help him?
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. The description of the test cases follows.
The only line for each test case contains two integers n, m (1 ≤ n ≤ 10^{9}, 0 ≤ m ≤ n) — the length of the string and the number of symbols equal to "1" in it.
Output
For every test case print one integer number — the maximum value of f(s) over all strings s of length n, which has exactly m symbols, equal to "1".
Example
Input
5
3 1
3 2
3 3
4 0
5 2
Output
4
5
6
0
12
Note
In the first test case, there exists only 3 strings of length 3, which has exactly 1 symbol, equal to "1". These strings are: s_1 = "100", s_2 = "010", s_3 = "001". The values of f for them are: f(s_1) = 3, f(s_2) = 4, f(s_3) = 3, so the maximum value is 4 and the answer is 4.
In the second test case, the string s with the maximum value is "101".
In the third test case, the string s with the maximum value is "111".
In the fourth test case, the only string s of length 4, which has exactly 0 symbols, equal to "1" is "0000" and the value of f for that string is 0, so the answer is 0.
In the fifth test case, the string s with the maximum value is "01010" and it is described as an example in the problem statement.
|
instruction
| 0
| 104,560
| 0
| 209,120
|
Tags: binary search, combinatorics, greedy, math, strings
Correct Solution:
```
"""
// Author : snape_here - Susanta Mukherjee
"""
from __future__ import division, print_function
import os,sys
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
def ii(): return int(input())
def si(): return input()
def mi(): return map(int,input().split())
def li(): return list(mi())
def gcd(x, y):
while y:
x, y = y, x % y
return x
def read():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
import math
mod=1000000007
def day(d, m, y):
t = [ 0, 3, 2, 5, 0, 3,
5, 1, 4, 6, 2, 4 ]
y -= m < 3
return (( y + int(y / 4) - int(y / 100)
+ int(y / 400) + t[m - 1] + d) % 7)
def isl(y):
if y%400==0:
return True
elif y%4==0 and y%100!=0:
return True
return False
def main():
for i in range(ii()):
n,m=mi()
if m==0:
ans=0
else:
tot=(n*(n+1))//2
s=(n-m)//(m+1)
b=(m+1-(n-m)%(m+1))*s*(s+1)//2 + ((n-m)%(m+1))*(s+1)*(s+2)//2
ans=tot-b
print(ans)
# region fastio#
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
#read()
main()
#Comment read()
```
|
output
| 1
| 104,560
| 0
| 209,121
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Ayoub thinks that he is a very smart person, so he created a function f(s), where s is a binary string (a string which contains only symbols "0" and "1"). The function f(s) is equal to the number of substrings in the string s that contains at least one symbol, that is equal to "1".
More formally, f(s) is equal to the number of pairs of integers (l, r), such that 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of string s), such that at least one of the symbols s_l, s_{l+1}, …, s_r is equal to "1".
For example, if s = "01010" then f(s) = 12, because there are 12 such pairs (l, r): (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 4), (4, 5).
Ayoub also thinks that he is smarter than Mahmoud so he gave him two integers n and m and asked him this problem. For all binary strings s of length n which contains exactly m symbols equal to "1", find the maximum value of f(s).
Mahmoud couldn't solve the problem so he asked you for help. Can you help him?
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. The description of the test cases follows.
The only line for each test case contains two integers n, m (1 ≤ n ≤ 10^{9}, 0 ≤ m ≤ n) — the length of the string and the number of symbols equal to "1" in it.
Output
For every test case print one integer number — the maximum value of f(s) over all strings s of length n, which has exactly m symbols, equal to "1".
Example
Input
5
3 1
3 2
3 3
4 0
5 2
Output
4
5
6
0
12
Note
In the first test case, there exists only 3 strings of length 3, which has exactly 1 symbol, equal to "1". These strings are: s_1 = "100", s_2 = "010", s_3 = "001". The values of f for them are: f(s_1) = 3, f(s_2) = 4, f(s_3) = 3, so the maximum value is 4 and the answer is 4.
In the second test case, the string s with the maximum value is "101".
In the third test case, the string s with the maximum value is "111".
In the fourth test case, the only string s of length 4, which has exactly 0 symbols, equal to "1" is "0000" and the value of f for that string is 0, so the answer is 0.
In the fifth test case, the string s with the maximum value is "01010" and it is described as an example in the problem statement.
|
instruction
| 0
| 104,561
| 0
| 209,122
|
Tags: binary search, combinatorics, greedy, math, strings
Correct Solution:
```
import sys
inf = 1 << 64
def input():
return sys.stdin.readline().rstrip()
def slv():
n, m = map(int, input().split())
u, v = divmod(n - m, m + 1)
ans = n * (n + 1)//2 - v*(u + 1)*(u + 2)//2 - (m + 1 - v)*u*(u + 1)//2
print(ans)
return
t = int(input())
for i in range(t):
slv()
```
|
output
| 1
| 104,561
| 0
| 209,123
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Ayoub thinks that he is a very smart person, so he created a function f(s), where s is a binary string (a string which contains only symbols "0" and "1"). The function f(s) is equal to the number of substrings in the string s that contains at least one symbol, that is equal to "1".
More formally, f(s) is equal to the number of pairs of integers (l, r), such that 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of string s), such that at least one of the symbols s_l, s_{l+1}, …, s_r is equal to "1".
For example, if s = "01010" then f(s) = 12, because there are 12 such pairs (l, r): (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 4), (4, 5).
Ayoub also thinks that he is smarter than Mahmoud so he gave him two integers n and m and asked him this problem. For all binary strings s of length n which contains exactly m symbols equal to "1", find the maximum value of f(s).
Mahmoud couldn't solve the problem so he asked you for help. Can you help him?
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. The description of the test cases follows.
The only line for each test case contains two integers n, m (1 ≤ n ≤ 10^{9}, 0 ≤ m ≤ n) — the length of the string and the number of symbols equal to "1" in it.
Output
For every test case print one integer number — the maximum value of f(s) over all strings s of length n, which has exactly m symbols, equal to "1".
Example
Input
5
3 1
3 2
3 3
4 0
5 2
Output
4
5
6
0
12
Note
In the first test case, there exists only 3 strings of length 3, which has exactly 1 symbol, equal to "1". These strings are: s_1 = "100", s_2 = "010", s_3 = "001". The values of f for them are: f(s_1) = 3, f(s_2) = 4, f(s_3) = 3, so the maximum value is 4 and the answer is 4.
In the second test case, the string s with the maximum value is "101".
In the third test case, the string s with the maximum value is "111".
In the fourth test case, the only string s of length 4, which has exactly 0 symbols, equal to "1" is "0000" and the value of f for that string is 0, so the answer is 0.
In the fifth test case, the string s with the maximum value is "01010" and it is described as an example in the problem statement.
|
instruction
| 0
| 104,562
| 0
| 209,124
|
Tags: binary search, combinatorics, greedy, math, strings
Correct Solution:
```
import sys
input=sys.stdin.readline
t=int(input())
for i in range(t):
n,m=map(int,input().split())
zeros=n-m
if zeros==0:
print((n*(n+1))//2)
else:
div=zeros//(m+1)
rem=zeros%(m+1)
print(n*(n+1)//2-rem*(div+1)*(div+2)//2-(m+1-rem)*((div+1)*div)//2)
```
|
output
| 1
| 104,562
| 0
| 209,125
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ayoub thinks that he is a very smart person, so he created a function f(s), where s is a binary string (a string which contains only symbols "0" and "1"). The function f(s) is equal to the number of substrings in the string s that contains at least one symbol, that is equal to "1".
More formally, f(s) is equal to the number of pairs of integers (l, r), such that 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of string s), such that at least one of the symbols s_l, s_{l+1}, …, s_r is equal to "1".
For example, if s = "01010" then f(s) = 12, because there are 12 such pairs (l, r): (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 4), (4, 5).
Ayoub also thinks that he is smarter than Mahmoud so he gave him two integers n and m and asked him this problem. For all binary strings s of length n which contains exactly m symbols equal to "1", find the maximum value of f(s).
Mahmoud couldn't solve the problem so he asked you for help. Can you help him?
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. The description of the test cases follows.
The only line for each test case contains two integers n, m (1 ≤ n ≤ 10^{9}, 0 ≤ m ≤ n) — the length of the string and the number of symbols equal to "1" in it.
Output
For every test case print one integer number — the maximum value of f(s) over all strings s of length n, which has exactly m symbols, equal to "1".
Example
Input
5
3 1
3 2
3 3
4 0
5 2
Output
4
5
6
0
12
Note
In the first test case, there exists only 3 strings of length 3, which has exactly 1 symbol, equal to "1". These strings are: s_1 = "100", s_2 = "010", s_3 = "001". The values of f for them are: f(s_1) = 3, f(s_2) = 4, f(s_3) = 3, so the maximum value is 4 and the answer is 4.
In the second test case, the string s with the maximum value is "101".
In the third test case, the string s with the maximum value is "111".
In the fourth test case, the only string s of length 4, which has exactly 0 symbols, equal to "1" is "0000" and the value of f for that string is 0, so the answer is 0.
In the fifth test case, the string s with the maximum value is "01010" and it is described as an example in the problem statement.
Submitted Solution:
```
from sys import stdin, stdout
def total(k): return (k*(k+1)) // 2
ans = []
t= int(stdin.readline())
for line in stdin:
n, m = tuple(map(int,line.split()))
if m==0:
ans.append( str(0) )
else:
al = total(n)
zeros = n-m
#divide into m+1 groups
k, extra = zeros//(m+1), zeros%(m+1)
zero_blocks = extra*(k+1) + (m+1)*total(k)
ans.append(str(al-zero_blocks))
stdout.write('\n'.join(ans))
```
|
instruction
| 0
| 104,563
| 0
| 209,126
|
Yes
|
output
| 1
| 104,563
| 0
| 209,127
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ayoub thinks that he is a very smart person, so he created a function f(s), where s is a binary string (a string which contains only symbols "0" and "1"). The function f(s) is equal to the number of substrings in the string s that contains at least one symbol, that is equal to "1".
More formally, f(s) is equal to the number of pairs of integers (l, r), such that 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of string s), such that at least one of the symbols s_l, s_{l+1}, …, s_r is equal to "1".
For example, if s = "01010" then f(s) = 12, because there are 12 such pairs (l, r): (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 4), (4, 5).
Ayoub also thinks that he is smarter than Mahmoud so he gave him two integers n and m and asked him this problem. For all binary strings s of length n which contains exactly m symbols equal to "1", find the maximum value of f(s).
Mahmoud couldn't solve the problem so he asked you for help. Can you help him?
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. The description of the test cases follows.
The only line for each test case contains two integers n, m (1 ≤ n ≤ 10^{9}, 0 ≤ m ≤ n) — the length of the string and the number of symbols equal to "1" in it.
Output
For every test case print one integer number — the maximum value of f(s) over all strings s of length n, which has exactly m symbols, equal to "1".
Example
Input
5
3 1
3 2
3 3
4 0
5 2
Output
4
5
6
0
12
Note
In the first test case, there exists only 3 strings of length 3, which has exactly 1 symbol, equal to "1". These strings are: s_1 = "100", s_2 = "010", s_3 = "001". The values of f for them are: f(s_1) = 3, f(s_2) = 4, f(s_3) = 3, so the maximum value is 4 and the answer is 4.
In the second test case, the string s with the maximum value is "101".
In the third test case, the string s with the maximum value is "111".
In the fourth test case, the only string s of length 4, which has exactly 0 symbols, equal to "1" is "0000" and the value of f for that string is 0, so the answer is 0.
In the fifth test case, the string s with the maximum value is "01010" and it is described as an example in the problem statement.
Submitted Solution:
```
import sys
lines = list(sys.stdin.readlines())
t = int(lines[0])
def count(x):
return x * (x+1)//2
for i in range(t):
n, ones = map(int, lines[i+1].split())
zeros = n - ones
if zeros==0:
print(count(n))
continue
used_ones = min(ones, zeros-1)
groups = used_ones+1
smal_size = zeros//groups
larg_size = smal_size + 1
larg_count = zeros%groups
total = count(n)
total -= count(smal_size) * (groups-larg_count)
total -= count(larg_size) * larg_count
print(total)
```
|
instruction
| 0
| 104,564
| 0
| 209,128
|
Yes
|
output
| 1
| 104,564
| 0
| 209,129
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ayoub thinks that he is a very smart person, so he created a function f(s), where s is a binary string (a string which contains only symbols "0" and "1"). The function f(s) is equal to the number of substrings in the string s that contains at least one symbol, that is equal to "1".
More formally, f(s) is equal to the number of pairs of integers (l, r), such that 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of string s), such that at least one of the symbols s_l, s_{l+1}, …, s_r is equal to "1".
For example, if s = "01010" then f(s) = 12, because there are 12 such pairs (l, r): (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 4), (4, 5).
Ayoub also thinks that he is smarter than Mahmoud so he gave him two integers n and m and asked him this problem. For all binary strings s of length n which contains exactly m symbols equal to "1", find the maximum value of f(s).
Mahmoud couldn't solve the problem so he asked you for help. Can you help him?
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. The description of the test cases follows.
The only line for each test case contains two integers n, m (1 ≤ n ≤ 10^{9}, 0 ≤ m ≤ n) — the length of the string and the number of symbols equal to "1" in it.
Output
For every test case print one integer number — the maximum value of f(s) over all strings s of length n, which has exactly m symbols, equal to "1".
Example
Input
5
3 1
3 2
3 3
4 0
5 2
Output
4
5
6
0
12
Note
In the first test case, there exists only 3 strings of length 3, which has exactly 1 symbol, equal to "1". These strings are: s_1 = "100", s_2 = "010", s_3 = "001". The values of f for them are: f(s_1) = 3, f(s_2) = 4, f(s_3) = 3, so the maximum value is 4 and the answer is 4.
In the second test case, the string s with the maximum value is "101".
In the third test case, the string s with the maximum value is "111".
In the fourth test case, the only string s of length 4, which has exactly 0 symbols, equal to "1" is "0000" and the value of f for that string is 0, so the answer is 0.
In the fifth test case, the string s with the maximum value is "01010" and it is described as an example in the problem statement.
Submitted Solution:
```
import sys
def fastio():
from io import StringIO
from atexit import register
global input
sys.stdin = StringIO(sys.stdin.read())
input = lambda : sys.stdin.readline().rstrip('\r\n')
sys.stdout = StringIO()
register(lambda : sys.__stdout__.write(sys.stdout.getvalue()))
fastio()
for _ in range(int(input())):
n,m=list(map(int,input().split()))
l=n if m==0 else (n-2*m if m<=n//2 else 0)
s1=n-m-l
if l==n:
print(0)
else:
p=(n-m)//(m+1);r=(n-m)%(m+1)
k=(n*(n+1))//2-(m+1-r)*(p*(p+1)//2)-r*(p+1)*(p+2)//2
print(k)
```
|
instruction
| 0
| 104,565
| 0
| 209,130
|
Yes
|
output
| 1
| 104,565
| 0
| 209,131
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ayoub thinks that he is a very smart person, so he created a function f(s), where s is a binary string (a string which contains only symbols "0" and "1"). The function f(s) is equal to the number of substrings in the string s that contains at least one symbol, that is equal to "1".
More formally, f(s) is equal to the number of pairs of integers (l, r), such that 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of string s), such that at least one of the symbols s_l, s_{l+1}, …, s_r is equal to "1".
For example, if s = "01010" then f(s) = 12, because there are 12 such pairs (l, r): (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 4), (4, 5).
Ayoub also thinks that he is smarter than Mahmoud so he gave him two integers n and m and asked him this problem. For all binary strings s of length n which contains exactly m symbols equal to "1", find the maximum value of f(s).
Mahmoud couldn't solve the problem so he asked you for help. Can you help him?
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. The description of the test cases follows.
The only line for each test case contains two integers n, m (1 ≤ n ≤ 10^{9}, 0 ≤ m ≤ n) — the length of the string and the number of symbols equal to "1" in it.
Output
For every test case print one integer number — the maximum value of f(s) over all strings s of length n, which has exactly m symbols, equal to "1".
Example
Input
5
3 1
3 2
3 3
4 0
5 2
Output
4
5
6
0
12
Note
In the first test case, there exists only 3 strings of length 3, which has exactly 1 symbol, equal to "1". These strings are: s_1 = "100", s_2 = "010", s_3 = "001". The values of f for them are: f(s_1) = 3, f(s_2) = 4, f(s_3) = 3, so the maximum value is 4 and the answer is 4.
In the second test case, the string s with the maximum value is "101".
In the third test case, the string s with the maximum value is "111".
In the fourth test case, the only string s of length 4, which has exactly 0 symbols, equal to "1" is "0000" and the value of f for that string is 0, so the answer is 0.
In the fifth test case, the string s with the maximum value is "01010" and it is described as an example in the problem statement.
Submitted Solution:
```
# import sys
# import math
# from collections import defaultdict,Counter
# input=sys.stdin.readline
# def print(x):
# sys.stdout.write(str(x)+"\n")
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# sys.stdout=open("CP3/output.txt",'w')
# sys.stdin=open("CP3/input.txt",'r')
# m=pow(10,9)+7
t=int(input())
for i in range(t):
n,m=map(int,input().split())
if m==0:
print(0)
elif n-m<=m:
print(n*(n+1)//2-(n-m))
else:
d=n-m
k=d//(m+1)
d1=d%(m+1)
print(n*(n+1)//2-(m+1)*(k*(k+1)//2)-(k+1)*d1)
```
|
instruction
| 0
| 104,566
| 0
| 209,132
|
Yes
|
output
| 1
| 104,566
| 0
| 209,133
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ayoub thinks that he is a very smart person, so he created a function f(s), where s is a binary string (a string which contains only symbols "0" and "1"). The function f(s) is equal to the number of substrings in the string s that contains at least one symbol, that is equal to "1".
More formally, f(s) is equal to the number of pairs of integers (l, r), such that 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of string s), such that at least one of the symbols s_l, s_{l+1}, …, s_r is equal to "1".
For example, if s = "01010" then f(s) = 12, because there are 12 such pairs (l, r): (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 4), (4, 5).
Ayoub also thinks that he is smarter than Mahmoud so he gave him two integers n and m and asked him this problem. For all binary strings s of length n which contains exactly m symbols equal to "1", find the maximum value of f(s).
Mahmoud couldn't solve the problem so he asked you for help. Can you help him?
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. The description of the test cases follows.
The only line for each test case contains two integers n, m (1 ≤ n ≤ 10^{9}, 0 ≤ m ≤ n) — the length of the string and the number of symbols equal to "1" in it.
Output
For every test case print one integer number — the maximum value of f(s) over all strings s of length n, which has exactly m symbols, equal to "1".
Example
Input
5
3 1
3 2
3 3
4 0
5 2
Output
4
5
6
0
12
Note
In the first test case, there exists only 3 strings of length 3, which has exactly 1 symbol, equal to "1". These strings are: s_1 = "100", s_2 = "010", s_3 = "001". The values of f for them are: f(s_1) = 3, f(s_2) = 4, f(s_3) = 3, so the maximum value is 4 and the answer is 4.
In the second test case, the string s with the maximum value is "101".
In the third test case, the string s with the maximum value is "111".
In the fourth test case, the only string s of length 4, which has exactly 0 symbols, equal to "1" is "0000" and the value of f for that string is 0, so the answer is 0.
In the fifth test case, the string s with the maximum value is "01010" and it is described as an example in the problem statement.
Submitted Solution:
```
import math
def main():
num = int(input())
for _ in range(num):
n, m = map(int, input().split())
if m == 0:
print(0)
continue
b = n - m
a = n // 2
if m == 1:
d = math.ceil(n/2)
ans = d ** 2
elif a < b:
ans = n * (n + 1) // 2
d = math.ceil(n/2)
amari = m - d
ans -= amari * (amari + 1) * 2 + d
else:
ans = n * (n + 1) // 2
ans -= b
print(ans)
main()
```
|
instruction
| 0
| 104,567
| 0
| 209,134
|
No
|
output
| 1
| 104,567
| 0
| 209,135
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ayoub thinks that he is a very smart person, so he created a function f(s), where s is a binary string (a string which contains only symbols "0" and "1"). The function f(s) is equal to the number of substrings in the string s that contains at least one symbol, that is equal to "1".
More formally, f(s) is equal to the number of pairs of integers (l, r), such that 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of string s), such that at least one of the symbols s_l, s_{l+1}, …, s_r is equal to "1".
For example, if s = "01010" then f(s) = 12, because there are 12 such pairs (l, r): (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 4), (4, 5).
Ayoub also thinks that he is smarter than Mahmoud so he gave him two integers n and m and asked him this problem. For all binary strings s of length n which contains exactly m symbols equal to "1", find the maximum value of f(s).
Mahmoud couldn't solve the problem so he asked you for help. Can you help him?
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. The description of the test cases follows.
The only line for each test case contains two integers n, m (1 ≤ n ≤ 10^{9}, 0 ≤ m ≤ n) — the length of the string and the number of symbols equal to "1" in it.
Output
For every test case print one integer number — the maximum value of f(s) over all strings s of length n, which has exactly m symbols, equal to "1".
Example
Input
5
3 1
3 2
3 3
4 0
5 2
Output
4
5
6
0
12
Note
In the first test case, there exists only 3 strings of length 3, which has exactly 1 symbol, equal to "1". These strings are: s_1 = "100", s_2 = "010", s_3 = "001". The values of f for them are: f(s_1) = 3, f(s_2) = 4, f(s_3) = 3, so the maximum value is 4 and the answer is 4.
In the second test case, the string s with the maximum value is "101".
In the third test case, the string s with the maximum value is "111".
In the fourth test case, the only string s of length 4, which has exactly 0 symbols, equal to "1" is "0000" and the value of f for that string is 0, so the answer is 0.
In the fifth test case, the string s with the maximum value is "01010" and it is described as an example in the problem statement.
Submitted Solution:
```
def solve_c():
n = int(input())
for _ in range(n):
c()
def c():
n, m = [int(x) for x in input().split()]
if m == 0:
print(0)
return
if m == 1:
print(2*n-2)
else:
print(n*(n+1)//2+m-n)
solve_c()
```
|
instruction
| 0
| 104,568
| 0
| 209,136
|
No
|
output
| 1
| 104,568
| 0
| 209,137
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ayoub thinks that he is a very smart person, so he created a function f(s), where s is a binary string (a string which contains only symbols "0" and "1"). The function f(s) is equal to the number of substrings in the string s that contains at least one symbol, that is equal to "1".
More formally, f(s) is equal to the number of pairs of integers (l, r), such that 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of string s), such that at least one of the symbols s_l, s_{l+1}, …, s_r is equal to "1".
For example, if s = "01010" then f(s) = 12, because there are 12 such pairs (l, r): (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 4), (4, 5).
Ayoub also thinks that he is smarter than Mahmoud so he gave him two integers n and m and asked him this problem. For all binary strings s of length n which contains exactly m symbols equal to "1", find the maximum value of f(s).
Mahmoud couldn't solve the problem so he asked you for help. Can you help him?
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. The description of the test cases follows.
The only line for each test case contains two integers n, m (1 ≤ n ≤ 10^{9}, 0 ≤ m ≤ n) — the length of the string and the number of symbols equal to "1" in it.
Output
For every test case print one integer number — the maximum value of f(s) over all strings s of length n, which has exactly m symbols, equal to "1".
Example
Input
5
3 1
3 2
3 3
4 0
5 2
Output
4
5
6
0
12
Note
In the first test case, there exists only 3 strings of length 3, which has exactly 1 symbol, equal to "1". These strings are: s_1 = "100", s_2 = "010", s_3 = "001". The values of f for them are: f(s_1) = 3, f(s_2) = 4, f(s_3) = 3, so the maximum value is 4 and the answer is 4.
In the second test case, the string s with the maximum value is "101".
In the third test case, the string s with the maximum value is "111".
In the fourth test case, the only string s of length 4, which has exactly 0 symbols, equal to "1" is "0000" and the value of f for that string is 0, so the answer is 0.
In the fifth test case, the string s with the maximum value is "01010" and it is described as an example in the problem statement.
Submitted Solution:
```
# -*- coding: utf-8 -*-
import bisect
import heapq
import math
import random
import sys
from collections import Counter, defaultdict, deque
from decimal import ROUND_CEILING, ROUND_HALF_UP, Decimal
from functools import lru_cache, reduce
from itertools import combinations, combinations_with_replacement, product, permutations
from operator import add, mul, sub
sys.setrecursionlimit(100000)
input = sys.stdin.readline # -*- coding: utf-8 -*-
sys.setrecursionlimit(100000)
input = sys.stdin.readline
INF = 2**62-1
def read_int():
return int(input())
def read_int_n():
return list(map(int, input().split()))
def read_float():
return float(input())
def read_float_n():
return list(map(float, input().split()))
def read_str():
return input().strip()
def read_str_n():
return list(map(str, input().split()))
def error_print(*args):
print(*args, file=sys.stderr)
def mt(f):
import time
def wrap(*args, **kwargs):
s = time.time()
ret = f(*args, **kwargs)
e = time.time()
error_print(e - s, 'sec')
return ret
return wrap
def slv(N, M):
if M == 0:
return 0
if M == N:
return ((N+1) * N ) // 2
ans = ((N+1) * N) // 2
r = N - M
d = r // (M+1)
e = r % (M+1)
if d != 0:
ans -= (((d+1) * d) // 2) * (M+1 - e)
ans -= (((d-1) * d) // 2) * (e)
else:
ans -= r
return ans
def main():
T = read_int()
for _ in range(T):
N, M = read_int_n()
print(slv(N, M))
if __name__ == "__main__":
main()
```
|
instruction
| 0
| 104,569
| 0
| 209,138
|
No
|
output
| 1
| 104,569
| 0
| 209,139
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ayoub thinks that he is a very smart person, so he created a function f(s), where s is a binary string (a string which contains only symbols "0" and "1"). The function f(s) is equal to the number of substrings in the string s that contains at least one symbol, that is equal to "1".
More formally, f(s) is equal to the number of pairs of integers (l, r), such that 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of string s), such that at least one of the symbols s_l, s_{l+1}, …, s_r is equal to "1".
For example, if s = "01010" then f(s) = 12, because there are 12 such pairs (l, r): (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 4), (4, 5).
Ayoub also thinks that he is smarter than Mahmoud so he gave him two integers n and m and asked him this problem. For all binary strings s of length n which contains exactly m symbols equal to "1", find the maximum value of f(s).
Mahmoud couldn't solve the problem so he asked you for help. Can you help him?
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. The description of the test cases follows.
The only line for each test case contains two integers n, m (1 ≤ n ≤ 10^{9}, 0 ≤ m ≤ n) — the length of the string and the number of symbols equal to "1" in it.
Output
For every test case print one integer number — the maximum value of f(s) over all strings s of length n, which has exactly m symbols, equal to "1".
Example
Input
5
3 1
3 2
3 3
4 0
5 2
Output
4
5
6
0
12
Note
In the first test case, there exists only 3 strings of length 3, which has exactly 1 symbol, equal to "1". These strings are: s_1 = "100", s_2 = "010", s_3 = "001". The values of f for them are: f(s_1) = 3, f(s_2) = 4, f(s_3) = 3, so the maximum value is 4 and the answer is 4.
In the second test case, the string s with the maximum value is "101".
In the third test case, the string s with the maximum value is "111".
In the fourth test case, the only string s of length 4, which has exactly 0 symbols, equal to "1" is "0000" and the value of f for that string is 0, so the answer is 0.
In the fifth test case, the string s with the maximum value is "01010" and it is described as an example in the problem statement.
Submitted Solution:
```
"""
Author : thekushalghosh
Team : CodeDiggers
"""
import sys,math
input = sys.stdin.readline
############ ---- USER DEFINED INPUT FUNCTIONS ---- ############
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(s[:len(s) - 1])
def invr():
return(map(int,input().split()))
################################################################
############ ---- THE ACTUAL CODE STARTS BELOW ---- ############
t = 1
t = inp()
def qw(n):
return(n * (n + 1) // 2)
for tt in range(t):
n,m = invr()
q = n - m
if m == 0:
print(0)
continue
ww = (n * (n + 1) // 2)
w = q % ww
qq = q // ww
ww = ww - (w * qw(qq + 1)) - ((m + 1 - w) * qw(qq))
print(ww)
```
|
instruction
| 0
| 104,570
| 0
| 209,140
|
No
|
output
| 1
| 104,570
| 0
| 209,141
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Once Bob saw a string. It contained so many different letters, that the letters were marked by numbers, but at the same time each letter could be met in the string at most 10 times. Bob didn't like that string, because it contained repeats: a repeat of length x is such a substring of length 2x, that its first half coincides character by character with its second half. Bob started deleting all the repeats from the string. He does it as follows: while it's possible, Bob takes the shortest repeat, if it is not unique, he takes the leftmost one, and deletes its left half and everything that is to the left of this repeat.
You're given the string seen by Bob. Find out, what it will look like after Bob deletes all the repeats in the way described above.
Input
The first input line contains integer n (1 ≤ n ≤ 105) — length of the string. The following line contains n space-separated integer numbers from 0 to 109 inclusive — numbers that stand for the letters of the string. It's guaranteed that each letter can be met in the string at most 10 times.
Output
In the first line output the length of the string's part, left after Bob's deletions. In the second line output all the letters (separated by a space) of the string, left after Bob deleted all the repeats in the described way.
Examples
Input
6
1 2 3 1 2 3
Output
3
1 2 3
Input
7
4 5 6 5 6 7 7
Output
1
7
|
instruction
| 0
| 104,730
| 0
| 209,460
|
Tags: greedy, hashing, string suffix structures
Correct Solution:
```
def pow(x, exp, mod):
res = 1
while exp:
if exp & 1:
res = (res * x) % mod
x = (x * x) % mod
exp >>= 1
return res
MOD = 2 ** 121 - 1
M = int(1e9) + 1
n = int(input())
vals = list(map(int, input().split()))
groups = dict()
for i in range(n):
groups.setdefault(vals[i], []).append(i)
powsA = [1]
for i in range(n):
powsA.append((powsA[-1] * M) % MOD)
hashes = [0] * (n + 1)
for i in range(n):
hashes[i + 1] = (hashes[i] * M + vals[i]) % MOD
def get_hash(p, l):
res = hashes[p + l] - (hashes[p] * powsA[l]) % MOD
if res < 0:
res += MOD
elif res > MOD:
res -= MOD
return res
best = 0
i = 0
while i < n:
val = vals[i]
for j in groups[val]:
if j <= i:
continue
l = j - i
if j + l <= n and get_hash(i, l) == get_hash(j, l):
best = max(best, j)
i = j - 1
break
i += 1
res = vals[best:]
print(len(res))
print(" ".join(map(str, res)))
```
|
output
| 1
| 104,730
| 0
| 209,461
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once Bob saw a string. It contained so many different letters, that the letters were marked by numbers, but at the same time each letter could be met in the string at most 10 times. Bob didn't like that string, because it contained repeats: a repeat of length x is such a substring of length 2x, that its first half coincides character by character with its second half. Bob started deleting all the repeats from the string. He does it as follows: while it's possible, Bob takes the shortest repeat, if it is not unique, he takes the leftmost one, and deletes its left half and everything that is to the left of this repeat.
You're given the string seen by Bob. Find out, what it will look like after Bob deletes all the repeats in the way described above.
Input
The first input line contains integer n (1 ≤ n ≤ 105) — length of the string. The following line contains n space-separated integer numbers from 0 to 109 inclusive — numbers that stand for the letters of the string. It's guaranteed that each letter can be met in the string at most 10 times.
Output
In the first line output the length of the string's part, left after Bob's deletions. In the second line output all the letters (separated by a space) of the string, left after Bob deleted all the repeats in the described way.
Examples
Input
6
1 2 3 1 2 3
Output
3
1 2 3
Input
7
4 5 6 5 6 7 7
Output
1
7
Submitted Solution:
```
n = int(input())
number = [int(x) for x in input().split()]
x = list(set(number))
print(len(x))
for i in x:
print(i,end=" ")
```
|
instruction
| 0
| 104,731
| 0
| 209,462
|
No
|
output
| 1
| 104,731
| 0
| 209,463
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once Bob saw a string. It contained so many different letters, that the letters were marked by numbers, but at the same time each letter could be met in the string at most 10 times. Bob didn't like that string, because it contained repeats: a repeat of length x is such a substring of length 2x, that its first half coincides character by character with its second half. Bob started deleting all the repeats from the string. He does it as follows: while it's possible, Bob takes the shortest repeat, if it is not unique, he takes the leftmost one, and deletes its left half and everything that is to the left of this repeat.
You're given the string seen by Bob. Find out, what it will look like after Bob deletes all the repeats in the way described above.
Input
The first input line contains integer n (1 ≤ n ≤ 105) — length of the string. The following line contains n space-separated integer numbers from 0 to 109 inclusive — numbers that stand for the letters of the string. It's guaranteed that each letter can be met in the string at most 10 times.
Output
In the first line output the length of the string's part, left after Bob's deletions. In the second line output all the letters (separated by a space) of the string, left after Bob deleted all the repeats in the described way.
Examples
Input
6
1 2 3 1 2 3
Output
3
1 2 3
Input
7
4 5 6 5 6 7 7
Output
1
7
Submitted Solution:
```
n = int(input())
a = list(map(int,input().split()))[:n]
b = []
for i in range(len(a)):
if a[i] not in b:
b.append(a[i])
print(len(b))
print(*b)
```
|
instruction
| 0
| 104,732
| 0
| 209,464
|
No
|
output
| 1
| 104,732
| 0
| 209,465
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once Bob saw a string. It contained so many different letters, that the letters were marked by numbers, but at the same time each letter could be met in the string at most 10 times. Bob didn't like that string, because it contained repeats: a repeat of length x is such a substring of length 2x, that its first half coincides character by character with its second half. Bob started deleting all the repeats from the string. He does it as follows: while it's possible, Bob takes the shortest repeat, if it is not unique, he takes the leftmost one, and deletes its left half and everything that is to the left of this repeat.
You're given the string seen by Bob. Find out, what it will look like after Bob deletes all the repeats in the way described above.
Input
The first input line contains integer n (1 ≤ n ≤ 105) — length of the string. The following line contains n space-separated integer numbers from 0 to 109 inclusive — numbers that stand for the letters of the string. It's guaranteed that each letter can be met in the string at most 10 times.
Output
In the first line output the length of the string's part, left after Bob's deletions. In the second line output all the letters (separated by a space) of the string, left after Bob deleted all the repeats in the described way.
Examples
Input
6
1 2 3 1 2 3
Output
3
1 2 3
Input
7
4 5 6 5 6 7 7
Output
1
7
Submitted Solution:
```
def get_pow_list(x, exp, mod):
res = [1]
for i in range(exp):
res.append((res[-1] * x) % mod)
return res
MOD = 2 ** 61 - 1
n = int(input())
orig_vals = list(map(int, input().split()))
sv = list(set(orig_vals))
sv_inv = {sv[i]: i + 1 for i in range(len(sv))}
vals = [sv_inv[x] for x in orig_vals]
groups = dict()
for i in range(n):
groups.setdefault(vals[i], [])
groups[vals[i]].append(i)
pows = [get_pow_list(x, n, MOD) for x in [2, len(sv) + 1]]
logs = [-1]
while len(logs) - 1 < n:
logs.extend([logs[-1] + 1] * len(logs))
hashes = [[0] * (4 * n) for i in range(logs[n] + 1)]
for i in range(n):
hashes[0][i] = vals[i]
for l in range(1, len(hashes)):
e = pows[0][l - 1]
for i in range(n):
hashes[l][i] = hashes[l - 1][i] * pows[1][e] + hashes[l - 1][i + e]
def get_hash(h, p, l, mod):
if l == 0:
return 0
elif ((l - 1) & l) == 0:
return h[logs[l]][p]
else:
nl = pows[0][logs[l]]
return get_hash(h, p, nl, mod) * pows[1][l - nl] + get_hash(
h, p + nl, l - nl, mod
)
best = 0
for i in range(n - 1, -1, -1):
val = vals[i]
for jj in range(len(groups[val])):
j = groups[val][jj]
if i == j:
break
l = i - j
if get_hash(hashes, i, l, MOD) == get_hash(hashes, j, l, MOD):
best = max(best, j)
if best != 0:
break
res = orig_vals[best:]
print(len(res))
print(" ".join(map(str, res)))
```
|
instruction
| 0
| 104,733
| 0
| 209,466
|
No
|
output
| 1
| 104,733
| 0
| 209,467
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once Bob saw a string. It contained so many different letters, that the letters were marked by numbers, but at the same time each letter could be met in the string at most 10 times. Bob didn't like that string, because it contained repeats: a repeat of length x is such a substring of length 2x, that its first half coincides character by character with its second half. Bob started deleting all the repeats from the string. He does it as follows: while it's possible, Bob takes the shortest repeat, if it is not unique, he takes the leftmost one, and deletes its left half and everything that is to the left of this repeat.
You're given the string seen by Bob. Find out, what it will look like after Bob deletes all the repeats in the way described above.
Input
The first input line contains integer n (1 ≤ n ≤ 105) — length of the string. The following line contains n space-separated integer numbers from 0 to 109 inclusive — numbers that stand for the letters of the string. It's guaranteed that each letter can be met in the string at most 10 times.
Output
In the first line output the length of the string's part, left after Bob's deletions. In the second line output all the letters (separated by a space) of the string, left after Bob deleted all the repeats in the described way.
Examples
Input
6
1 2 3 1 2 3
Output
3
1 2 3
Input
7
4 5 6 5 6 7 7
Output
1
7
Submitted Solution:
```
def pow(x, exp, mod):
res = 1
while exp:
if exp & 1:
res = (res * x) % mod
x = (x * x) % mod
exp >>= 1
return res
MOD = 2 ** 121 - 1
M = int(1e9) + 1
n = int(input())
vals = list(map(int, input().split()))
groups = dict()
for i in range(n):
groups.setdefault(vals[i], []).append(i)
powsA = [1]
for i in range(n):
powsA.append((powsA[-1] * M) % MOD)
hashes = [0] * (n + 1)
for i in range(n):
hashes[i + 1] = (hashes[i] * M + vals[i]) % MOD
def get_hash(p, l):
res = hashes[p + l] - (hashes[p] * powsA[l]) % MOD
if res < 0:
res += MOD
elif res > MOD:
res -= MOD
return res
best = 0
for i in range(n - 1, -1, -1):
val = vals[i]
for j in groups[val]:
if i == j:
break
l = i - j
if i + l <= n and get_hash(i, l) == get_hash(j, l):
best = max(best, i)
if best != 0:
break
res = vals[best:]
print(len(res))
print(" ".join(map(str, res)))
```
|
instruction
| 0
| 104,734
| 0
| 209,468
|
No
|
output
| 1
| 104,734
| 0
| 209,469
|
Provide tags and a correct Python 3 solution for this coding contest problem.
We had a string s consisting of n lowercase Latin letters. We made k copies of this string, thus obtaining k identical strings s1, s2, ..., sk. After that, in each of these strings we swapped exactly two characters (the characters we swapped could be identical, but they had different indices in the string).
You are given k strings s1, s2, ..., sk, and you have to restore any string s so that it is possible to obtain these strings by performing aforementioned operations. Note that the total length of the strings you are given doesn't exceed 5000 (that is, k·n ≤ 5000).
Input
The first line contains two integers k and n (1 ≤ k ≤ 2500, 2 ≤ n ≤ 5000, k · n ≤ 5000) — the number of strings we obtained, and the length of each of these strings.
Next k lines contain the strings s1, s2, ..., sk, each consisting of exactly n lowercase Latin letters.
Output
Print any suitable string s, or -1 if such string doesn't exist.
Examples
Input
3 4
abac
caab
acba
Output
acab
Input
3 4
kbbu
kbub
ubkb
Output
kbub
Input
5 4
abcd
dcba
acbd
dbca
zzzz
Output
-1
Note
In the first example s1 is obtained by swapping the second and the fourth character in acab, s2 is obtained by swapping the first and the second character, and to get s3, we swap the third and the fourth character.
In the second example s1 is obtained by swapping the third and the fourth character in kbub, s2 — by swapping the second and the fourth, and s3 — by swapping the first and the third.
In the third example it's impossible to obtain given strings by aforementioned operations.
|
instruction
| 0
| 105,009
| 0
| 210,018
|
Tags: brute force, hashing, implementation, strings
Correct Solution:
```
def solve():
read=lambda:list(map(int,input().split()))
from collections import Counter as co
k,n=read()
s=input()
repeated=(len(s)!=len(set(s)))
etalon=co(s)
a=[]
kk=[]
ap=a.append
for i in range(k-1):
ap(input())
if co(a[-1])!=etalon:
print(-1)
exit()
ss=False
for i in a:
if i!=s:
ss=i
for j in range(len(s)):
if s[j]!=ss[j]:
kk.append(j)
break
if len(kk)>4:
print(-1)
exit()
if ss:
if repeated:
for i in a:
k = 0
for j in range(len(i)):
if s[j] != i[j]: k += 1
if k != 0 and k != 2: break
else:
print(s)
exit()
if len(kk)!=2:
for i in range(len(kk)):
for j in range(i):
stry=s[:kk[j]]+s[kk[i]]+s[kk[j]+1:kk[i]]+s[kk[j]]+s[kk[i]+1:]
for u in a:
k = 0
for j in range(len(u)):
if stry[j] != u[j]: k += 1
if not(k==0 and repeated) and k != 2: break
else:
print(stry)
exit()
if len(kk)==2:
for change in kk:
for i in range(len(s)):
if change==i:
continue
if i >change:
stry = s[:change] + s[i] + s[change + 1:i] + s[change] + s[i + 1:]
else:
stry = s[:i] + s[change] + s[i + 1:change] + s[i] + s[change + 1:]
for u in a:
k = 0
for j in range(len(u)):
if stry[j] != u[j]: k += 1
if not(k==0 and repeated) and k != 2: break
else:
print(stry)
exit()
print(-1)
else:
if repeated:
print(s)
exit()
print(s[1]+s[0]+s[2:])
solve()
```
|
output
| 1
| 105,009
| 0
| 210,019
|
Provide tags and a correct Python 3 solution for this coding contest problem.
We had a string s consisting of n lowercase Latin letters. We made k copies of this string, thus obtaining k identical strings s1, s2, ..., sk. After that, in each of these strings we swapped exactly two characters (the characters we swapped could be identical, but they had different indices in the string).
You are given k strings s1, s2, ..., sk, and you have to restore any string s so that it is possible to obtain these strings by performing aforementioned operations. Note that the total length of the strings you are given doesn't exceed 5000 (that is, k·n ≤ 5000).
Input
The first line contains two integers k and n (1 ≤ k ≤ 2500, 2 ≤ n ≤ 5000, k · n ≤ 5000) — the number of strings we obtained, and the length of each of these strings.
Next k lines contain the strings s1, s2, ..., sk, each consisting of exactly n lowercase Latin letters.
Output
Print any suitable string s, or -1 if such string doesn't exist.
Examples
Input
3 4
abac
caab
acba
Output
acab
Input
3 4
kbbu
kbub
ubkb
Output
kbub
Input
5 4
abcd
dcba
acbd
dbca
zzzz
Output
-1
Note
In the first example s1 is obtained by swapping the second and the fourth character in acab, s2 is obtained by swapping the first and the second character, and to get s3, we swap the third and the fourth character.
In the second example s1 is obtained by swapping the third and the fourth character in kbub, s2 — by swapping the second and the fourth, and s3 — by swapping the first and the third.
In the third example it's impossible to obtain given strings by aforementioned operations.
|
instruction
| 0
| 105,010
| 0
| 210,020
|
Tags: brute force, hashing, implementation, strings
Correct Solution:
```
read=lambda:list(map(int,input().split()))
from collections import Counter as co
k,n=read()
s=input()
repeated=(len(s)!=len(set(s)))
etalon=co(s)
a=[]
kk=[]
ap=a.append
for i in range(k-1):
ap(input())
if co(a[-1])!=etalon:
print(-1)
exit()
ss=False
for i in a:
if i!=s:
ss=i
for j in range(len(s)):
if s[j]!=ss[j]:
kk.append(j)
break
if len(kk)>4:
print(-1)
exit()
if ss:
if repeated:
for i in a:
k = 0
for j in range(len(i)):
if s[j] != i[j]: k += 1
if k != 0 and k != 2: break
else:
print(s)
exit()
if len(kk)!=2:
for i in range(len(kk)):
for j in range(i):
stry=s[:kk[j]]+s[kk[i]]+s[kk[j]+1:kk[i]]+s[kk[j]]+s[kk[i]+1:]
#print(stry)
for u in a:
k = 0
for j in range(len(u)):
if stry[j] != u[j]: k += 1
#print(stry,i,k)
if not(k==0 and repeated) and k != 2: break
else:
print(stry)
exit()
if len(kk)==2:
for change in kk:
for i in range(len(s)):
if change==i:
continue
if i >change:
stry = s[:change] + s[i] + s[change + 1:i] + s[change] + s[i + 1:]
else:
stry = s[:i] + s[change] + s[i + 1:change] + s[i] + s[change + 1:]
for u in a:
k = 0
for j in range(len(u)):
if stry[j] != u[j]: k += 1
#print(stry,i,k)
if not(k==0 and repeated) and k != 2: break
else:
print(stry)
exit()
print(-1)
else:
if repeated:
print(s)
exit()
print(s[1]+s[0]+s[2:])
```
|
output
| 1
| 105,010
| 0
| 210,021
|
Provide tags and a correct Python 3 solution for this coding contest problem.
We had a string s consisting of n lowercase Latin letters. We made k copies of this string, thus obtaining k identical strings s1, s2, ..., sk. After that, in each of these strings we swapped exactly two characters (the characters we swapped could be identical, but they had different indices in the string).
You are given k strings s1, s2, ..., sk, and you have to restore any string s so that it is possible to obtain these strings by performing aforementioned operations. Note that the total length of the strings you are given doesn't exceed 5000 (that is, k·n ≤ 5000).
Input
The first line contains two integers k and n (1 ≤ k ≤ 2500, 2 ≤ n ≤ 5000, k · n ≤ 5000) — the number of strings we obtained, and the length of each of these strings.
Next k lines contain the strings s1, s2, ..., sk, each consisting of exactly n lowercase Latin letters.
Output
Print any suitable string s, or -1 if such string doesn't exist.
Examples
Input
3 4
abac
caab
acba
Output
acab
Input
3 4
kbbu
kbub
ubkb
Output
kbub
Input
5 4
abcd
dcba
acbd
dbca
zzzz
Output
-1
Note
In the first example s1 is obtained by swapping the second and the fourth character in acab, s2 is obtained by swapping the first and the second character, and to get s3, we swap the third and the fourth character.
In the second example s1 is obtained by swapping the third and the fourth character in kbub, s2 — by swapping the second and the fourth, and s3 — by swapping the first and the third.
In the third example it's impossible to obtain given strings by aforementioned operations.
|
instruction
| 0
| 105,011
| 0
| 210,022
|
Tags: brute force, hashing, implementation, strings
Correct Solution:
```
import collections
def swapCharacters(strings, k, n):
"""
Time: O(n^2 * k)
Space: O(1)
"""
if k == 1:
s0 = list(strings[0])
s0[0], s0[1] = s0[1], s0[0]
return ''.join(s0)
# Initial check for validity
freq = collections.Counter(strings[0])
canSame = (max(freq.values()) >= 2) # could swap two of the same characters ==> same string
for s in strings:
if collections.Counter(s) != freq:
return -1
# Find diff indices between first two strings
max_dist = 0
max_dist_s = None
s0 = strings[0]
for s1 in strings[1:]:
dist = HammingDistance(s0, s1, n)
if dist > max_dist:
max_dist = dist
max_dist_s = s1
# Hamming distance <= 2*2 between input strings to be valid
if max_dist > 4:
return -1
diffs = [i for i in range(n) if s0[i] != s1[i]]
# Checks all possible strings which match first two -- Finding strings (O(N)), testing string (O(KN))
s0 = list(s0)
for i in diffs:
for j in range(n):
# try swapping
s0[i], s0[j] = s0[j], s0[i]
# see if matches second string now
#dist = sum([s0[i] != s1[i] for i in diffs])
dist = HammingDistance(s0, s1, n)
if dist == 2 or (dist == 0 and canSame):
cand = ''.join(s0)
if verifyAll(strings, k, n, cand, canSame):
return cand
# revert
s0[i], s0[j] = s0[j], s0[i]
# nothing works
return -1
def HammingDistance(s0, s1, n):
count = 0
for i in range(n):
if s0[i] != s1[i]:
count += 1
return count
def verifyAll(strings, k, n, cand, canSame):
for s in strings:
dist = HammingDistance(s, cand, n)
if (dist != 0 or not canSame) and dist != 2:
return False
return True
k, n = [int(x) for x in input().split()]
strings = set() # discard duplicates
for _ in range(k):
strings.add(input())
strings = list(strings)
k = len(strings)
res = swapCharacters(strings, k, n)
print(res)
```
|
output
| 1
| 105,011
| 0
| 210,023
|
Provide tags and a correct Python 3 solution for this coding contest problem.
We had a string s consisting of n lowercase Latin letters. We made k copies of this string, thus obtaining k identical strings s1, s2, ..., sk. After that, in each of these strings we swapped exactly two characters (the characters we swapped could be identical, but they had different indices in the string).
You are given k strings s1, s2, ..., sk, and you have to restore any string s so that it is possible to obtain these strings by performing aforementioned operations. Note that the total length of the strings you are given doesn't exceed 5000 (that is, k·n ≤ 5000).
Input
The first line contains two integers k and n (1 ≤ k ≤ 2500, 2 ≤ n ≤ 5000, k · n ≤ 5000) — the number of strings we obtained, and the length of each of these strings.
Next k lines contain the strings s1, s2, ..., sk, each consisting of exactly n lowercase Latin letters.
Output
Print any suitable string s, or -1 if such string doesn't exist.
Examples
Input
3 4
abac
caab
acba
Output
acab
Input
3 4
kbbu
kbub
ubkb
Output
kbub
Input
5 4
abcd
dcba
acbd
dbca
zzzz
Output
-1
Note
In the first example s1 is obtained by swapping the second and the fourth character in acab, s2 is obtained by swapping the first and the second character, and to get s3, we swap the third and the fourth character.
In the second example s1 is obtained by swapping the third and the fourth character in kbub, s2 — by swapping the second and the fourth, and s3 — by swapping the first and the third.
In the third example it's impossible to obtain given strings by aforementioned operations.
|
instruction
| 0
| 105,012
| 0
| 210,024
|
Tags: brute force, hashing, implementation, strings
Correct Solution:
```
def solve():
read=lambda:list(map(int,input().split()))
from collections import Counter as co
k,n=read()
s=list(input())
repeated=(len(s)!=len(set(s)))
etalon=co(s)
a=[]
kk=[]
ap=a.append
for i in range(k-1):
ap(list(input()))
if co(a[-1])!=etalon:
print(-1)
exit()
ss=False
for i in a:
if i!=s:
ss=i
for j in range(len(s)):
if s[j]!=ss[j]:
kk.append(j)
break
if len(kk)>4:
print(-1)
exit()
if ss:
if repeated:
for i in a:
k = 0
for j in range(len(i)):
if s[j] != i[j]: k += 1
if k != 0 and k != 2: break
else:
print(''.join(s))
exit()
if len(kk)!=2:
for i in range(len(kk)):
for j in range(i):
stry=s[:kk[j]]+[s[kk[i]]]+s[kk[j]+1:kk[i]]+[s[kk[j]]]+s[kk[i]+1:]
for u in a:
k = 0
for j in range(len(u)):
if stry[j] != u[j]: k += 1
if not(k==0 and repeated) and k != 2: break
else:
print(''.join(stry))
exit()
if len(kk)==2:
for change in kk:
for i in range(len(s)):
if change==i:
continue
if i >change:
stry = s[:change] + [s[i]] + s[change + 1:i] + [s[change]] + s[i + 1:]
else:
stry = s[:i] + [s[change]] + s[i + 1:change] + [s[i]] + s[change + 1:]
for u in a:
k = 0
for j in range(len(u)):
if stry[j] != u[j]: k += 1
if not(k==0 and repeated) and k != 2: break
else:
print(''.join(stry))
exit()
print(-1)
else:
if repeated:
print(''.join(s))
exit()
print(s[1]+s[0]+''.join(s[2:]))
solve()
```
|
output
| 1
| 105,012
| 0
| 210,025
|
Provide tags and a correct Python 3 solution for this coding contest problem.
We had a string s consisting of n lowercase Latin letters. We made k copies of this string, thus obtaining k identical strings s1, s2, ..., sk. After that, in each of these strings we swapped exactly two characters (the characters we swapped could be identical, but they had different indices in the string).
You are given k strings s1, s2, ..., sk, and you have to restore any string s so that it is possible to obtain these strings by performing aforementioned operations. Note that the total length of the strings you are given doesn't exceed 5000 (that is, k·n ≤ 5000).
Input
The first line contains two integers k and n (1 ≤ k ≤ 2500, 2 ≤ n ≤ 5000, k · n ≤ 5000) — the number of strings we obtained, and the length of each of these strings.
Next k lines contain the strings s1, s2, ..., sk, each consisting of exactly n lowercase Latin letters.
Output
Print any suitable string s, or -1 if such string doesn't exist.
Examples
Input
3 4
abac
caab
acba
Output
acab
Input
3 4
kbbu
kbub
ubkb
Output
kbub
Input
5 4
abcd
dcba
acbd
dbca
zzzz
Output
-1
Note
In the first example s1 is obtained by swapping the second and the fourth character in acab, s2 is obtained by swapping the first and the second character, and to get s3, we swap the third and the fourth character.
In the second example s1 is obtained by swapping the third and the fourth character in kbub, s2 — by swapping the second and the fourth, and s3 — by swapping the first and the third.
In the third example it's impossible to obtain given strings by aforementioned operations.
|
instruction
| 0
| 105,013
| 0
| 210,026
|
Tags: brute force, hashing, implementation, strings
Correct Solution:
```
#!/usr/bin/env python3
n,k = map(int, input().split())
nn = []
ans = ''
for i in range(n):
mid = input()
if mid in nn:
ans = mid
continue
nn.append(mid)
n = len(nn)
if len(nn) == 1:
ans = nn[0]
ans = list(ans)
ans[0],ans[1] = ans[1],ans[0]
print(''.join(ans))
else:
diff = []
check = True
cnt = {chr(97+i):0 for i in range(26)}
for v in range(k):
cnt[nn[0][v]] += 1
for i in range(n):
cnt2 = {chr(97+i):0 for i in range(26)}
for j in range(k):
cnt2[nn[i][j]] += 1
if cnt != cnt2:
print('-1')
check = False
break
if check:
check = False
for i in range(n):
check = False
for j in range(i,n):
diff = [l for l in range(k) if nn[i][l] != nn[j][l]]
if len(diff) > 4:
check = True
print('-1')
break;
if check:
break
diff = [l for l in range(k) if nn[0][l] != nn[1][l]]
mid = []
check2 = False
for i in range(k):
if nn[0][i] in mid:
check2 = True
break
mid.append(nn[0][i])
#print(diff)
if not check:
res = list(nn[0])
check = False
for i in range(len(diff)):
if check:
break
for j in range(k):
if i == j:
continue
res[diff[i]],res[j] = res[j], res[diff[i]]
ans = ''.join(res)
#print(ans)
check = True
for x in range(n):
mid = [ans[y] for y in range(k) if nn[x][y] != ans[y]]
#print(len(diff))
if len(mid) == 2:
continue
elif len(mid) == 0 and check2:
continue
else:
check = False
if check:
print(ans)
check = True
break
res[diff[i]],res[j] = res[j],res[diff[i]]
if not check:
print('-1')
```
|
output
| 1
| 105,013
| 0
| 210,027
|
Provide tags and a correct Python 3 solution for this coding contest problem.
We had a string s consisting of n lowercase Latin letters. We made k copies of this string, thus obtaining k identical strings s1, s2, ..., sk. After that, in each of these strings we swapped exactly two characters (the characters we swapped could be identical, but they had different indices in the string).
You are given k strings s1, s2, ..., sk, and you have to restore any string s so that it is possible to obtain these strings by performing aforementioned operations. Note that the total length of the strings you are given doesn't exceed 5000 (that is, k·n ≤ 5000).
Input
The first line contains two integers k and n (1 ≤ k ≤ 2500, 2 ≤ n ≤ 5000, k · n ≤ 5000) — the number of strings we obtained, and the length of each of these strings.
Next k lines contain the strings s1, s2, ..., sk, each consisting of exactly n lowercase Latin letters.
Output
Print any suitable string s, or -1 if such string doesn't exist.
Examples
Input
3 4
abac
caab
acba
Output
acab
Input
3 4
kbbu
kbub
ubkb
Output
kbub
Input
5 4
abcd
dcba
acbd
dbca
zzzz
Output
-1
Note
In the first example s1 is obtained by swapping the second and the fourth character in acab, s2 is obtained by swapping the first and the second character, and to get s3, we swap the third and the fourth character.
In the second example s1 is obtained by swapping the third and the fourth character in kbub, s2 — by swapping the second and the fourth, and s3 — by swapping the first and the third.
In the third example it's impossible to obtain given strings by aforementioned operations.
|
instruction
| 0
| 105,014
| 0
| 210,028
|
Tags: brute force, hashing, implementation, strings
Correct Solution:
```
import sys
k, n = map(int, input().split())
s = [list(word.rstrip()) for word in sys.stdin]
double = True if max(s[0].count(chr(i+97)) for i in range(26)) > 1 else False
diff = [set() for _ in range(k)]
diff_cnt = [0]*k
for i in range(1, k):
for j in range(n):
if s[0][j] != s[i][j]:
diff[i].add(j)
diff_cnt[i] += 1
if diff_cnt[i] > 4:
print(-1)
exit()
for i in range(n):
for j in range(i+1, n):
s[0][i], s[0][j] = s[0][j], s[0][i]
for x in range(1, k):
w = [y for y in diff[x] | {i, j} if s[0][y] != s[x][y]]
if double and len(w) == 0:
continue
if len(w) == 2 and s[0][w[0]] == s[x][w[1]] and s[0][w[1]] == s[x][w[0]]:
continue
break
else:
print(''.join(s[0]))
exit()
s[0][i], s[0][j] = s[0][j], s[0][i]
print(-1)
```
|
output
| 1
| 105,014
| 0
| 210,029
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We had a string s consisting of n lowercase Latin letters. We made k copies of this string, thus obtaining k identical strings s1, s2, ..., sk. After that, in each of these strings we swapped exactly two characters (the characters we swapped could be identical, but they had different indices in the string).
You are given k strings s1, s2, ..., sk, and you have to restore any string s so that it is possible to obtain these strings by performing aforementioned operations. Note that the total length of the strings you are given doesn't exceed 5000 (that is, k·n ≤ 5000).
Input
The first line contains two integers k and n (1 ≤ k ≤ 2500, 2 ≤ n ≤ 5000, k · n ≤ 5000) — the number of strings we obtained, and the length of each of these strings.
Next k lines contain the strings s1, s2, ..., sk, each consisting of exactly n lowercase Latin letters.
Output
Print any suitable string s, or -1 if such string doesn't exist.
Examples
Input
3 4
abac
caab
acba
Output
acab
Input
3 4
kbbu
kbub
ubkb
Output
kbub
Input
5 4
abcd
dcba
acbd
dbca
zzzz
Output
-1
Note
In the first example s1 is obtained by swapping the second and the fourth character in acab, s2 is obtained by swapping the first and the second character, and to get s3, we swap the third and the fourth character.
In the second example s1 is obtained by swapping the third and the fourth character in kbub, s2 — by swapping the second and the fourth, and s3 — by swapping the first and the third.
In the third example it's impossible to obtain given strings by aforementioned operations.
Submitted Solution:
```
import collections
def swapCharacters(strings, k, n):
"""
Time: O(n^2 * k)
Space: O(1)
"""
if k == 1:
return strings[0]
# Initial check for validity
freq = collections.Counter(strings[0])
for s in strings:
if collections.Counter(s) != freq:
return -1
# Find diff indices between first two strings
max_dist = 0
max_dist_s = None
s0 = strings[0]
for s1 in strings[1:]:
dist = HammingDistance(s0, s1, n)
if dist > max_dist:
max_dist = dist
max_dist_s = s1
# Hamming distance <= 2*2 between input strings to be valid
if max_dist > 4:
return -1
diffs = [i for i in range(n) if s0[i] != s1[i]]
# Checks all possible strings which match first two -- Finding strings (O(N)), testing string (O(KN))
s0 = list(s0)
for i in diffs:
for j in range(n):
# try swapping
s0[i], s0[j] = s0[j], s0[i]
# see if matches second string now
#dist = sum([s0[i] != s1[i] for i in diffs])
dist = HammingDistance(s0, s1, n)
if dist == 0 or dist == 2:
cand = ''.join(s0)
if verifyAll(strings, k, n, cand):
return cand
# revert
s0[i], s0[j] = s0[j], s0[i]
# nothing works
return -1
def HammingDistance(s0, s1, n):
count = 0
for i in range(n):
if s0[i] != s1[i]:
count += 1
return count
def verifyAll(strings, k, n, cand):
for s in strings:
dist = HammingDistance(s, cand, n)
if dist != 0 and dist != 2:
return False
return True
k, n = [int(x) for x in input().split()]
strings = []
for _ in range(k):
strings.append(input())
res = swapCharacters(strings, k, n)
print(res)
```
|
instruction
| 0
| 105,015
| 0
| 210,030
|
No
|
output
| 1
| 105,015
| 0
| 210,031
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We had a string s consisting of n lowercase Latin letters. We made k copies of this string, thus obtaining k identical strings s1, s2, ..., sk. After that, in each of these strings we swapped exactly two characters (the characters we swapped could be identical, but they had different indices in the string).
You are given k strings s1, s2, ..., sk, and you have to restore any string s so that it is possible to obtain these strings by performing aforementioned operations. Note that the total length of the strings you are given doesn't exceed 5000 (that is, k·n ≤ 5000).
Input
The first line contains two integers k and n (1 ≤ k ≤ 2500, 2 ≤ n ≤ 5000, k · n ≤ 5000) — the number of strings we obtained, and the length of each of these strings.
Next k lines contain the strings s1, s2, ..., sk, each consisting of exactly n lowercase Latin letters.
Output
Print any suitable string s, or -1 if such string doesn't exist.
Examples
Input
3 4
abac
caab
acba
Output
acab
Input
3 4
kbbu
kbub
ubkb
Output
kbub
Input
5 4
abcd
dcba
acbd
dbca
zzzz
Output
-1
Note
In the first example s1 is obtained by swapping the second and the fourth character in acab, s2 is obtained by swapping the first and the second character, and to get s3, we swap the third and the fourth character.
In the second example s1 is obtained by swapping the third and the fourth character in kbub, s2 — by swapping the second and the fourth, and s3 — by swapping the first and the third.
In the third example it's impossible to obtain given strings by aforementioned operations.
Submitted Solution:
```
k, n = map(int, input().split())
se = None
for _ in range(k):
s = input()
nse = set([s])
for i in range(n-1):
for j in range(i+1, n):
ns = []
for k in range(n):
if k != i and k != j:
ns.append(s[k])
elif k == i:
ns.append(s[j])
else:
ns.append(s[i])
nse.add(''.join(ns))
print(nse)
if se is None:
se = nse
else:
se &= nse
if len(se) == 0:
print(-1)
exit()
se = list(se)
print(se[0])
```
|
instruction
| 0
| 105,016
| 0
| 210,032
|
No
|
output
| 1
| 105,016
| 0
| 210,033
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We had a string s consisting of n lowercase Latin letters. We made k copies of this string, thus obtaining k identical strings s1, s2, ..., sk. After that, in each of these strings we swapped exactly two characters (the characters we swapped could be identical, but they had different indices in the string).
You are given k strings s1, s2, ..., sk, and you have to restore any string s so that it is possible to obtain these strings by performing aforementioned operations. Note that the total length of the strings you are given doesn't exceed 5000 (that is, k·n ≤ 5000).
Input
The first line contains two integers k and n (1 ≤ k ≤ 2500, 2 ≤ n ≤ 5000, k · n ≤ 5000) — the number of strings we obtained, and the length of each of these strings.
Next k lines contain the strings s1, s2, ..., sk, each consisting of exactly n lowercase Latin letters.
Output
Print any suitable string s, or -1 if such string doesn't exist.
Examples
Input
3 4
abac
caab
acba
Output
acab
Input
3 4
kbbu
kbub
ubkb
Output
kbub
Input
5 4
abcd
dcba
acbd
dbca
zzzz
Output
-1
Note
In the first example s1 is obtained by swapping the second and the fourth character in acab, s2 is obtained by swapping the first and the second character, and to get s3, we swap the third and the fourth character.
In the second example s1 is obtained by swapping the third and the fourth character in kbub, s2 — by swapping the second and the fourth, and s3 — by swapping the first and the third.
In the third example it's impossible to obtain given strings by aforementioned operations.
Submitted Solution:
```
from itertools import permutations
class UniqueElement:
def __init__(self, value, occurrences):
self.value = value
self.occurrences = occurrences
def perm_unique(elements):
e_set = set(elements)
list_unique = [UniqueElement(i, elements.count(i)) for i in e_set]
u = len(elements)
return perm_unique_helper(list_unique, [0] * u, u - 1)
def perm_unique_helper(list_unique, result_list, d):
if d < 0:
yield tuple(result_list)
else:
for i in list_unique:
if i.occurrences > 0:
result_list[d] = i.value
i.occurrences -= 1
for g in perm_unique_helper(list_unique, result_list, d - 1):
yield g
i.occurrences += 1
def count(s):
counter = {}
for ch in s:
if ch not in counter:
counter[ch] = 0
counter[ch] += 1
return counter
def hamming_distance(s1, s2, get_diff=False):
assert len(s1) == len(s2)
diff = []
for ch1, ch2 in zip(s1, s2):
if ch1 != ch2:
diff.append(1)
else:
diff.append(0)
if not get_diff:
return sum(diff)
else:
return sum(diff), diff
def final_check(ss, ans):
for s in ss:
if hamming_distance(s, ans) not in {0, 2}:
return False
return True
def main():
k, n = tuple(int(s) for s in input().split())
ss = [input() for _ in range(k)]
if k == 1:
s = list(ss[0])
s[0], s[1] = s[1], s[0]
print(''.join(s))
return
counter = count(ss[0])
for s in ss[1:]:
if counter != count(s):
print(-1)
return
dis, diff = hamming_distance(ss[0], ss[1], get_diff=True)
if dis > 4:
print(-1)
return
diff_chs = [ch for ch, is_diff in zip(ss[0], diff) if is_diff]
for perm in permutations(diff_chs):
asd = 0
new_str = []
for ch, is_diff in zip(ss[0], diff):
if is_diff:
new_str.append(perm[asd])
asd += 1
else:
new_str.append(ch)
new_str = ''.join(new_str)
if final_check(ss, new_str):
print(new_str)
return
print(-1)
if __name__ == '__main__':
main()
```
|
instruction
| 0
| 105,017
| 0
| 210,034
|
No
|
output
| 1
| 105,017
| 0
| 210,035
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We had a string s consisting of n lowercase Latin letters. We made k copies of this string, thus obtaining k identical strings s1, s2, ..., sk. After that, in each of these strings we swapped exactly two characters (the characters we swapped could be identical, but they had different indices in the string).
You are given k strings s1, s2, ..., sk, and you have to restore any string s so that it is possible to obtain these strings by performing aforementioned operations. Note that the total length of the strings you are given doesn't exceed 5000 (that is, k·n ≤ 5000).
Input
The first line contains two integers k and n (1 ≤ k ≤ 2500, 2 ≤ n ≤ 5000, k · n ≤ 5000) — the number of strings we obtained, and the length of each of these strings.
Next k lines contain the strings s1, s2, ..., sk, each consisting of exactly n lowercase Latin letters.
Output
Print any suitable string s, or -1 if such string doesn't exist.
Examples
Input
3 4
abac
caab
acba
Output
acab
Input
3 4
kbbu
kbub
ubkb
Output
kbub
Input
5 4
abcd
dcba
acbd
dbca
zzzz
Output
-1
Note
In the first example s1 is obtained by swapping the second and the fourth character in acab, s2 is obtained by swapping the first and the second character, and to get s3, we swap the third and the fourth character.
In the second example s1 is obtained by swapping the third and the fourth character in kbub, s2 — by swapping the second and the fourth, and s3 — by swapping the first and the third.
In the third example it's impossible to obtain given strings by aforementioned operations.
Submitted Solution:
```
n,jks=map(int,input().split())
m=[]
m.append(input())
c=set(m[0])
def Tr(s):
for w in m:
f=[]
for i in range(jks):
if w[i]!=s[i]:
f.append(i)
if len(f)==2 and w[f[0]]==s[f[1]] and w[f[1]]==s[f[0]]:
pass
elif len(f)==0 and len(set(s))!=len(s):
pass
else:
return False
return True
for i in range(1,n):
m.append(input())
# if c!=set(m[i]):
# print(-1)
# break
#else:
a=m[0]
b=m[1]
f=[]
for i in range(jks):
if a[i]!=b[i]:
f.append(i)
if len(f)>4:
print(-1)
else:
for a in m[:2]:
sl=list(a)
for x in f:
for i in range(jks):
if i==x:
continue
slc=sl.copy()
slc[x],slc[i]=slc[i],slc[x]
sls=''.join(slc)
if Tr(sls):
print(sls)
break
else:
continue
break
else:
continue
break
else:
print(-1)
```
|
instruction
| 0
| 105,018
| 0
| 210,036
|
No
|
output
| 1
| 105,018
| 0
| 210,037
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider all binary strings of length m (1 ≤ m ≤ 60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2^m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s[i] < t[i]. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.
We remove from this set n (1 ≤ n ≤ min(2^m-1, 100)) distinct binary strings a_1, a_2, …, a_n, each of length m. Thus, the set will have k=2^m-n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 to k-1. Print the string whose index is ⌊ (k-1)/(2) ⌋ (such an element is called median), where ⌊ x ⌋ is the rounding of the number down to the nearest integer.
For example, if n=3, m=3 and a=[010, 111, 001], then after removing the strings a_i and sorting, the result will take the form: [000, 011, 100, 101, 110]. Thus, the desired median is 100.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n (1 ≤ n ≤ min(2^m-1, 100)) and m (1 ≤ m ≤ 60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a_1, a_2, …, a_n — distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 10^5.
Output
Print t answers to the test cases. For each test case, print a string of length m — the median of the sorted sequence of remaining strings in the corresponding test case.
Example
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Output
100
010
0
1
11
Note
The first test case is explained in the statement.
In the second test case, the result after removing strings and sorting is [001, 010, 101, 110]. Therefore, the desired median is 010.
|
instruction
| 0
| 105,421
| 0
| 210,842
|
Tags: binary search, bitmasks, brute force, constructive algorithms
Correct Solution:
```
# Template 1.0
import sys, re, math
from collections import deque, defaultdict, Counter, OrderedDict
from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians, gcd
from heapq import heappush, heappop, heapify, nlargest, nsmallest
def STR(): return list(input())
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(): return list(map(int, input().split()))
def list2d(a, b, c): return [[c] * b for i in range(a)]
def sortListWithIndex(listOfTuples, idx): return (sorted(listOfTuples, key=lambda x: x[idx]))
def sortDictWithVal(passedDic):
temp = sorted(passedDic.items(), key=lambda kv: (kv[1], kv[0]))[::-1]
toret = {}
for tup in temp:
toret[tup[0]] = tup[1]
return toret
def sortDictWithKey(passedDic):
return dict(OrderedDict(sorted(passedDic.items())))
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
mod = 10 ** 9 + 7
def calcNext(num):
for i in range(num+1, 2**m):
if(i not in vis):
return i
return -1
def calcPrev(num):
for i in range(num-1, -1, -1):
if (i not in vis):
return i
return -1
t = INT()
while (t != 0):
n, m = MAP()
currLen = 2**m
currMed = (2**m-1)//2
vis = set()
# vis.add(currMed)
for _ in range(n):
torem = int(input(), 2)
if(currLen%2==0):
currLen-=1
vis.add(torem)
if(torem<=currMed):
currMed = calcNext(currMed)
else:
currLen-=1
vis.add(torem)
if(torem>=currMed):
currMed = calcPrev(currMed)
# print(currMed)
# print(bin(currMed).replace("0b", ""))
print(format(currMed, '0'+str(m)+'b'))
t-=1
'''
1
4 3
000
111
100
011
1 2 3 4 5 6
'''
```
|
output
| 1
| 105,421
| 0
| 210,843
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider all binary strings of length m (1 ≤ m ≤ 60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2^m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s[i] < t[i]. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.
We remove from this set n (1 ≤ n ≤ min(2^m-1, 100)) distinct binary strings a_1, a_2, …, a_n, each of length m. Thus, the set will have k=2^m-n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 to k-1. Print the string whose index is ⌊ (k-1)/(2) ⌋ (such an element is called median), where ⌊ x ⌋ is the rounding of the number down to the nearest integer.
For example, if n=3, m=3 and a=[010, 111, 001], then after removing the strings a_i and sorting, the result will take the form: [000, 011, 100, 101, 110]. Thus, the desired median is 100.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n (1 ≤ n ≤ min(2^m-1, 100)) and m (1 ≤ m ≤ 60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a_1, a_2, …, a_n — distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 10^5.
Output
Print t answers to the test cases. For each test case, print a string of length m — the median of the sorted sequence of remaining strings in the corresponding test case.
Example
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Output
100
010
0
1
11
Note
The first test case is explained in the statement.
In the second test case, the result after removing strings and sorting is [001, 010, 101, 110]. Therefore, the desired median is 010.
|
instruction
| 0
| 105,422
| 0
| 210,844
|
Tags: binary search, bitmasks, brute force, constructive algorithms
Correct Solution:
```
t=int(input())
for _ in range(t):
n,m=map(int,input().split())
arr=[]
for i in range(n):
arr.append(input().strip())
arr1=[]
for i in arr:
arr1.append(int(i,2))
a=(2**m-n-1)//2
arr1.sort()
for i in arr1:
if i<=a:
a+=1
ans=bin(a).replace("0b","")
ans="0"*(m-len(ans))+ans
print(ans)
```
|
output
| 1
| 105,422
| 0
| 210,845
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider all binary strings of length m (1 ≤ m ≤ 60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2^m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s[i] < t[i]. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.
We remove from this set n (1 ≤ n ≤ min(2^m-1, 100)) distinct binary strings a_1, a_2, …, a_n, each of length m. Thus, the set will have k=2^m-n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 to k-1. Print the string whose index is ⌊ (k-1)/(2) ⌋ (such an element is called median), where ⌊ x ⌋ is the rounding of the number down to the nearest integer.
For example, if n=3, m=3 and a=[010, 111, 001], then after removing the strings a_i and sorting, the result will take the form: [000, 011, 100, 101, 110]. Thus, the desired median is 100.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n (1 ≤ n ≤ min(2^m-1, 100)) and m (1 ≤ m ≤ 60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a_1, a_2, …, a_n — distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 10^5.
Output
Print t answers to the test cases. For each test case, print a string of length m — the median of the sorted sequence of remaining strings in the corresponding test case.
Example
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Output
100
010
0
1
11
Note
The first test case is explained in the statement.
In the second test case, the result after removing strings and sorting is [001, 010, 101, 110]. Therefore, the desired median is 010.
|
instruction
| 0
| 105,423
| 0
| 210,846
|
Tags: binary search, bitmasks, brute force, constructive algorithms
Correct Solution:
```
import sys
max_int = 1000000001 # 10^9+1
min_int = -max_int
t = int(input())
for _t in range(t):
n, m = map(int, sys.stdin.readline().split())
to_skip = []
for _n in range(n):
s = int(sys.stdin.readline()[:-1], base=2)
to_skip.append(s)
to_skip.sort()
mid = (2 ** m - n - 1) // 2
for elem in to_skip:
if elem <= mid:
mid += 1
else:
break
print(("{0:0>" + str(m) + "b}").format(mid))
```
|
output
| 1
| 105,423
| 0
| 210,847
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider all binary strings of length m (1 ≤ m ≤ 60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2^m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s[i] < t[i]. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.
We remove from this set n (1 ≤ n ≤ min(2^m-1, 100)) distinct binary strings a_1, a_2, …, a_n, each of length m. Thus, the set will have k=2^m-n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 to k-1. Print the string whose index is ⌊ (k-1)/(2) ⌋ (such an element is called median), where ⌊ x ⌋ is the rounding of the number down to the nearest integer.
For example, if n=3, m=3 and a=[010, 111, 001], then after removing the strings a_i and sorting, the result will take the form: [000, 011, 100, 101, 110]. Thus, the desired median is 100.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n (1 ≤ n ≤ min(2^m-1, 100)) and m (1 ≤ m ≤ 60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a_1, a_2, …, a_n — distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 10^5.
Output
Print t answers to the test cases. For each test case, print a string of length m — the median of the sorted sequence of remaining strings in the corresponding test case.
Example
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Output
100
010
0
1
11
Note
The first test case is explained in the statement.
In the second test case, the result after removing strings and sorting is [001, 010, 101, 110]. Therefore, the desired median is 010.
|
instruction
| 0
| 105,424
| 0
| 210,848
|
Tags: binary search, bitmasks, brute force, constructive algorithms
Correct Solution:
```
'''
Auther: ghoshashis545 Ashis Ghosh
College: jalpaiguri Govt Enggineerin College
Date:24/05/2020
'''
import sys
from collections import deque,defaultdict as dd
from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
from itertools import permutations
from datetime import datetime
from math import ceil,sqrt,log,gcd
def ii():return int(input())
def si():return input()
def mi():return map(int,input().split())
def li():return list(mi())
abc='abcdefghijklmnopqrstuvwxyz'
abd={'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24, 'z': 25}
mod=1000000007
#mod=998244353
inf = float("inf")
vow=['a','e','i','o','u']
dx,dy=[-1,1,0,0],[0,0,1,-1]
def read():
tc=0
if tc:
input=sys.stdin.readline
else:
sys.stdin=open('input1.txt', 'r')
sys.stdout=open('output1.txt','w')
def solve():
for _ in range(ii()):
n,m=mi()
tot=pow(2,m)-n
x=(tot-1)//2
a=[]
for i in range(n):
s=si()
a.append(int(s,2))
a.sort()
cnt=bisect(a,x)
x+=cnt
c=cnt
while(cnt):
cnt=bisect(a,x)-c
x+=cnt
c+=cnt
s=bin(x)[2:]
print('0'*(m-len(s))+s)
if __name__ =="__main__":
# read()
solve()
```
|
output
| 1
| 105,424
| 0
| 210,849
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider all binary strings of length m (1 ≤ m ≤ 60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2^m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s[i] < t[i]. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.
We remove from this set n (1 ≤ n ≤ min(2^m-1, 100)) distinct binary strings a_1, a_2, …, a_n, each of length m. Thus, the set will have k=2^m-n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 to k-1. Print the string whose index is ⌊ (k-1)/(2) ⌋ (such an element is called median), where ⌊ x ⌋ is the rounding of the number down to the nearest integer.
For example, if n=3, m=3 and a=[010, 111, 001], then after removing the strings a_i and sorting, the result will take the form: [000, 011, 100, 101, 110]. Thus, the desired median is 100.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n (1 ≤ n ≤ min(2^m-1, 100)) and m (1 ≤ m ≤ 60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a_1, a_2, …, a_n — distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 10^5.
Output
Print t answers to the test cases. For each test case, print a string of length m — the median of the sorted sequence of remaining strings in the corresponding test case.
Example
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Output
100
010
0
1
11
Note
The first test case is explained in the statement.
In the second test case, the result after removing strings and sorting is [001, 010, 101, 110]. Therefore, the desired median is 010.
|
instruction
| 0
| 105,425
| 0
| 210,850
|
Tags: binary search, bitmasks, brute force, constructive algorithms
Correct Solution:
```
tc = int(input())
for _ in range(tc):
n,m = map(int,input().split())
cache = {}
curr = (2**m - 1)//2
for i in range(n):
s = input()
x = int(s,2)
cache[x] = 1
if i%2 == 0:
if x<=curr:
while(len(cache)!=0 and curr+1 in cache):
curr += 1
curr += 1
else:
if x>=curr:
while(len(cache)!=0 and curr-1 in cache):
curr -= 1
curr -= 1
temp = bin(curr)[2:]
if m!=len(temp):
temp = "0"*abs(m-len(temp)) + temp
print(temp)
```
|
output
| 1
| 105,425
| 0
| 210,851
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider all binary strings of length m (1 ≤ m ≤ 60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2^m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s[i] < t[i]. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.
We remove from this set n (1 ≤ n ≤ min(2^m-1, 100)) distinct binary strings a_1, a_2, …, a_n, each of length m. Thus, the set will have k=2^m-n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 to k-1. Print the string whose index is ⌊ (k-1)/(2) ⌋ (such an element is called median), where ⌊ x ⌋ is the rounding of the number down to the nearest integer.
For example, if n=3, m=3 and a=[010, 111, 001], then after removing the strings a_i and sorting, the result will take the form: [000, 011, 100, 101, 110]. Thus, the desired median is 100.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n (1 ≤ n ≤ min(2^m-1, 100)) and m (1 ≤ m ≤ 60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a_1, a_2, …, a_n — distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 10^5.
Output
Print t answers to the test cases. For each test case, print a string of length m — the median of the sorted sequence of remaining strings in the corresponding test case.
Example
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Output
100
010
0
1
11
Note
The first test case is explained in the statement.
In the second test case, the result after removing strings and sorting is [001, 010, 101, 110]. Therefore, the desired median is 010.
|
instruction
| 0
| 105,426
| 0
| 210,852
|
Tags: binary search, bitmasks, brute force, constructive algorithms
Correct Solution:
```
def solve():
n, m = [int(x) for x in input().split()]
a = []
for i in range(n):
a.append(input())
a = [int(x, 2) for x in a]
need = ((1 << m) - n - 1)//2 + 1
cur = (1 << (m-1)) - 1
while True:
left = cur+1
flag = bool(cur in a)
for s in a:
if s <= cur:
left -= 1
if left == need and not flag:
ans = bin(cur).split('b')[1]
if len(ans) < m:
ans = '0'*(m-len(ans)) + ans
print(ans)
return
elif left < need:
cur += 1
else:
cur -= 1
t = int(input())
while t > 0:
t -= 1
solve()
```
|
output
| 1
| 105,426
| 0
| 210,853
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider all binary strings of length m (1 ≤ m ≤ 60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2^m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s[i] < t[i]. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.
We remove from this set n (1 ≤ n ≤ min(2^m-1, 100)) distinct binary strings a_1, a_2, …, a_n, each of length m. Thus, the set will have k=2^m-n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 to k-1. Print the string whose index is ⌊ (k-1)/(2) ⌋ (such an element is called median), where ⌊ x ⌋ is the rounding of the number down to the nearest integer.
For example, if n=3, m=3 and a=[010, 111, 001], then after removing the strings a_i and sorting, the result will take the form: [000, 011, 100, 101, 110]. Thus, the desired median is 100.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n (1 ≤ n ≤ min(2^m-1, 100)) and m (1 ≤ m ≤ 60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a_1, a_2, …, a_n — distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 10^5.
Output
Print t answers to the test cases. For each test case, print a string of length m — the median of the sorted sequence of remaining strings in the corresponding test case.
Example
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Output
100
010
0
1
11
Note
The first test case is explained in the statement.
In the second test case, the result after removing strings and sorting is [001, 010, 101, 110]. Therefore, the desired median is 010.
|
instruction
| 0
| 105,427
| 0
| 210,854
|
Tags: binary search, bitmasks, brute force, constructive algorithms
Correct Solution:
```
import math
import string
t = int(input())
for tt in range(t):
n, m = map(int, input().split())
a = []
for i in range(n):
a.append(int(input(), 2))
need = (2 ** m - n - 1) // 2
l = 0
r = 2 ** m
while(r - l > 1):
mid = (l + r) // 2
x = mid
for i in range(n):
if(a[i] < mid):
x -= 1
if(x > need):
r = mid
else:
l = mid
ans = []
for i in range(m):
ans.append(l % 2)
l //= 2
ans.reverse()
for i in range(m):
print(ans[i], end="")
print()
```
|
output
| 1
| 105,427
| 0
| 210,855
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider all binary strings of length m (1 ≤ m ≤ 60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2^m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s[i] < t[i]. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.
We remove from this set n (1 ≤ n ≤ min(2^m-1, 100)) distinct binary strings a_1, a_2, …, a_n, each of length m. Thus, the set will have k=2^m-n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 to k-1. Print the string whose index is ⌊ (k-1)/(2) ⌋ (such an element is called median), where ⌊ x ⌋ is the rounding of the number down to the nearest integer.
For example, if n=3, m=3 and a=[010, 111, 001], then after removing the strings a_i and sorting, the result will take the form: [000, 011, 100, 101, 110]. Thus, the desired median is 100.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n (1 ≤ n ≤ min(2^m-1, 100)) and m (1 ≤ m ≤ 60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a_1, a_2, …, a_n — distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 10^5.
Output
Print t answers to the test cases. For each test case, print a string of length m — the median of the sorted sequence of remaining strings in the corresponding test case.
Example
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Output
100
010
0
1
11
Note
The first test case is explained in the statement.
In the second test case, the result after removing strings and sorting is [001, 010, 101, 110]. Therefore, the desired median is 010.
|
instruction
| 0
| 105,428
| 0
| 210,856
|
Tags: binary search, bitmasks, brute force, constructive algorithms
Correct Solution:
```
def dd(i,nn):
s=bin(i).replace("0b", "")
l=len(s)
#print(l,nn)
kk=""
for i in range (nn-l):
kk=kk+'0'
print(kk+s)
t=int(input())
for i in range(t):
n,m=map(int,input().split())
aa=[]
for pp in range(n):
pp=input()
aa.append(int(pp,2))
y=2**(m-1) - 1
a=max(0,y-n)
b=min(y+n,2**m -1)
for i in range(a,b+1):
l=i
r=2**m -1-i
#print(l,r)
flag=1
for k in aa:
if(k==i):
flag=0
break
if(k<i):
l=l-1
else:
r=r-1
if(flag==0):
continue
if(n%2==0 and l==r-1):
dd(i,m)
if(n%2!=0 and l==r):
dd(i,m)
```
|
output
| 1
| 105,428
| 0
| 210,857
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider all binary strings of length m (1 ≤ m ≤ 60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2^m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s[i] < t[i]. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.
We remove from this set n (1 ≤ n ≤ min(2^m-1, 100)) distinct binary strings a_1, a_2, …, a_n, each of length m. Thus, the set will have k=2^m-n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 to k-1. Print the string whose index is ⌊ (k-1)/(2) ⌋ (such an element is called median), where ⌊ x ⌋ is the rounding of the number down to the nearest integer.
For example, if n=3, m=3 and a=[010, 111, 001], then after removing the strings a_i and sorting, the result will take the form: [000, 011, 100, 101, 110]. Thus, the desired median is 100.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n (1 ≤ n ≤ min(2^m-1, 100)) and m (1 ≤ m ≤ 60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a_1, a_2, …, a_n — distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 10^5.
Output
Print t answers to the test cases. For each test case, print a string of length m — the median of the sorted sequence of remaining strings in the corresponding test case.
Example
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Output
100
010
0
1
11
Note
The first test case is explained in the statement.
In the second test case, the result after removing strings and sorting is [001, 010, 101, 110]. Therefore, the desired median is 010.
Submitted Solution:
```
from sys import stdin,stdout
import math
from bisect import bisect_left
def main():
t = int(stdin.readline())
for _ in range (t):
n,m = list(map(int, stdin.readline().split()))
total = pow(2,m) - 1
med = math.floor((total) / 2)
arr = []
for _ in range(n):
num = int(stdin.readline(),2)
arr.append(num)
arr.sort()
ite = min(total+1,200)
ans = 0
for x in range(max(0, med - int(ite/2)),min(total+1, med + int(ite/2)+1)):
pos = bisect_left(arr, x)
if pos != len(arr) and arr[pos] == x:
continue
# use pos to calculate left and right quantity
# if n % 2 == 1 then exclude current
# else then include currente
elif n % 2 == 1:
left = x - pos
right = total - x - (n - pos)
if left == right:
ans = x
break
else:
left = x - pos + 1
right = total - x - (n - pos)
if left == right:
ans = x
break
ans = bin(ans).replace("0b", "")
if len(ans) < m:
ans = "0" * (m - len(ans)) + ans
stdout.write(ans + "\n")
main()
```
|
instruction
| 0
| 105,429
| 0
| 210,858
|
Yes
|
output
| 1
| 105,429
| 0
| 210,859
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider all binary strings of length m (1 ≤ m ≤ 60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2^m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s[i] < t[i]. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.
We remove from this set n (1 ≤ n ≤ min(2^m-1, 100)) distinct binary strings a_1, a_2, …, a_n, each of length m. Thus, the set will have k=2^m-n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 to k-1. Print the string whose index is ⌊ (k-1)/(2) ⌋ (such an element is called median), where ⌊ x ⌋ is the rounding of the number down to the nearest integer.
For example, if n=3, m=3 and a=[010, 111, 001], then after removing the strings a_i and sorting, the result will take the form: [000, 011, 100, 101, 110]. Thus, the desired median is 100.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n (1 ≤ n ≤ min(2^m-1, 100)) and m (1 ≤ m ≤ 60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a_1, a_2, …, a_n — distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 10^5.
Output
Print t answers to the test cases. For each test case, print a string of length m — the median of the sorted sequence of remaining strings in the corresponding test case.
Example
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Output
100
010
0
1
11
Note
The first test case is explained in the statement.
In the second test case, the result after removing strings and sorting is [001, 010, 101, 110]. Therefore, the desired median is 010.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------------------
from math import factorial, floor
from collections import Counter, defaultdict, deque
from heapq import heapify, heappop, heappush
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
def comb(n, m): return factorial(n) / (factorial(m) * factorial(n - m)) if n >= m else 0
def perm(n, m): return factorial(n) // (factorial(n - m)) if n >= m else 0
def mdis(x1, y1, x2, y2): return abs(x1 - x2) + abs(y1 - y2)
mod = 998244353
INF = float('inf')
# ------------------------------
from bisect import bisect
def main():
for _ in range(N()):
n, mm = RL()
arr = [int(input(), 2) for _ in range(n)]
arr.sort()
now = (2**mm)-n
res = (now-1)//2
for i in arr:
if i<=res:
res+=1
# print(res)
res = bin(res)[2:].zfill(mm)
print(res)
if __name__ == "__main__":
main()
```
|
instruction
| 0
| 105,430
| 0
| 210,860
|
Yes
|
output
| 1
| 105,430
| 0
| 210,861
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider all binary strings of length m (1 ≤ m ≤ 60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2^m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s[i] < t[i]. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.
We remove from this set n (1 ≤ n ≤ min(2^m-1, 100)) distinct binary strings a_1, a_2, …, a_n, each of length m. Thus, the set will have k=2^m-n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 to k-1. Print the string whose index is ⌊ (k-1)/(2) ⌋ (such an element is called median), where ⌊ x ⌋ is the rounding of the number down to the nearest integer.
For example, if n=3, m=3 and a=[010, 111, 001], then after removing the strings a_i and sorting, the result will take the form: [000, 011, 100, 101, 110]. Thus, the desired median is 100.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n (1 ≤ n ≤ min(2^m-1, 100)) and m (1 ≤ m ≤ 60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a_1, a_2, …, a_n — distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 10^5.
Output
Print t answers to the test cases. For each test case, print a string of length m — the median of the sorted sequence of remaining strings in the corresponding test case.
Example
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Output
100
010
0
1
11
Note
The first test case is explained in the statement.
In the second test case, the result after removing strings and sorting is [001, 010, 101, 110]. Therefore, the desired median is 010.
Submitted Solution:
```
import sys, heapq
from collections import *
from functools import lru_cache
sys.setrecursionlimit(10**6)
def main():
# sys.stdin = open('input.txt', 'r')
t = int(input())
for _ in range(t):
n, m = map(int,input().split(' '))
total = 2**m
mid = (total-1)//2
removes = set()
for _ in range(n):
s = input()
cur = 0
for ch in s:
cur = cur<<1
if ch == '1':
cur += 1
removes.add(cur)
removed = set()
for r in removes:
if total & 1 and r >= mid:
while mid-1 in removed:
mid -= 1
mid -= 1
elif total & 1 == 0 and r <= mid:
while mid+1 in removed:
mid += 1
mid += 1
total -= 1
removed.add(r)
res = []
for i in range(m):
if mid & (1<<i):
res.append('1')
else:
res.append('0')
# print(''.join(res[::-1]),mid)
print(''.join(res[::-1]))
if __name__ == "__main__":
main()
```
|
instruction
| 0
| 105,431
| 0
| 210,862
|
Yes
|
output
| 1
| 105,431
| 0
| 210,863
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider all binary strings of length m (1 ≤ m ≤ 60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2^m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s[i] < t[i]. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.
We remove from this set n (1 ≤ n ≤ min(2^m-1, 100)) distinct binary strings a_1, a_2, …, a_n, each of length m. Thus, the set will have k=2^m-n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 to k-1. Print the string whose index is ⌊ (k-1)/(2) ⌋ (such an element is called median), where ⌊ x ⌋ is the rounding of the number down to the nearest integer.
For example, if n=3, m=3 and a=[010, 111, 001], then after removing the strings a_i and sorting, the result will take the form: [000, 011, 100, 101, 110]. Thus, the desired median is 100.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n (1 ≤ n ≤ min(2^m-1, 100)) and m (1 ≤ m ≤ 60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a_1, a_2, …, a_n — distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 10^5.
Output
Print t answers to the test cases. For each test case, print a string of length m — the median of the sorted sequence of remaining strings in the corresponding test case.
Example
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Output
100
010
0
1
11
Note
The first test case is explained in the statement.
In the second test case, the result after removing strings and sorting is [001, 010, 101, 110]. Therefore, the desired median is 010.
Submitted Solution:
```
for _ in range(int(input())):
n,m=map(int,input().split())
x=(2**m-n-1)//2
arr=[]
for i in range(n):
arr.append(int(input(),2))
arr.sort()
for i in range(0,n):
if(arr[i]<=x):
x+=1
if x in arr:
x+=1
y=bin(x)[2:]
print('0'*(m-len(y))+y)
```
|
instruction
| 0
| 105,432
| 0
| 210,864
|
Yes
|
output
| 1
| 105,432
| 0
| 210,865
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.