text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n β for loop;
* end β every command between "for n" and corresponding "end" is executed n times;
* add β adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 β€ l β€ 10^5) β the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 β€ n β€ 100) β for loop;
* end β every command between "for n" and corresponding "end" is executed n times;
* add β adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
Tags: data structures, expression parsing, implementation
Correct Solution:
```
# -*- coding: utf-8 -*-
# @Date : 2019-06-06 09:13:44
# @Author : raj lath (oorja.halt@gmail.com)
# @Link : link
# @Version : 1.0.0
import sys
sys.setrecursionlimit(10**5+1)
inf = int(10 ** 20)
max_val = inf
min_val = -inf
RW = lambda : sys.stdin.readline().strip()
RI = lambda : int(RW())
RMI = lambda : [int(x) for x in sys.stdin.readline().strip().split()]
RWI = lambda : [x for x in sys.stdin.readline().strip().split()]
overflow = 2 ** 32
ans = 0
nb_test = RI()
calc = [1]
for _ in range(nb_test):
mods, *vals = RWI()
if mods == "add":
ans += calc[-1]
elif mods == "for":
val = int(vals[0])
calc.append(min((calc[-1] * val),overflow))
#print(calc)
else:calc.pop()
ans = "OVERFLOW!!!" if ans >=overflow else ans
print(ans)
```
| 107,100 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n β for loop;
* end β every command between "for n" and corresponding "end" is executed n times;
* add β adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 β€ l β€ 10^5) β the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 β€ n β€ 100) β for loop;
* end β every command between "for n" and corresponding "end" is executed n times;
* add β adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
Tags: data structures, expression parsing, implementation
Correct Solution:
```
inf=10**12
val=2**32
import sys
n=int(input())
m=1
st=[]
st1=[]
st2=[1]
ans=0
for i in range(n):
a=list(map(str,sys.stdin.readline().split()))
if len(a)==2:
m=int(a[1])
st1.append(m)
if st1[-1]*st2[-1]<=val:
st2.append(st1[-1]*st2[-1])
else:
st2.append(inf)
x=a[0]
m=st2[-1]
if x=="add":
ans+=m
elif x=="end":
st1.pop()
st2.pop()
# m//=int(k)
if ans>=2**32:
exit(print("OVERFLOW!!!"))
print(ans)
```
| 107,101 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n β for loop;
* end β every command between "for n" and corresponding "end" is executed n times;
* add β adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 β€ l β€ 10^5) β the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 β€ n β€ 100) β for loop;
* end β every command between "for n" and corresponding "end" is executed n times;
* add β adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
Tags: data structures, expression parsing, implementation
Correct Solution:
```
import sys
L = int(sys.stdin.readline().strip())
x = 0
i = 1
m = []
v = True
l = 0
y = 0
while l < L:
w = sys.stdin.readline().strip()
if v == True and y == 0:
if w == "add":
x = x + i
elif w == "end":
i = i // m.pop()
else:
m.append(int(w.split()[1]))
i = i * m[-1]
if x > 2 ** 32 - 1:
v = False
if i > 2 ** 32 - 1:
y = y + 1
elif v == True and y > 0:
if w == "add":
v = False
elif w == "end":
y = y - 1
if y == 0:
i = i // m.pop()
else:
y = y + 1
l = l + 1
if v == True:
print(x)
else:
print("OVERFLOW!!!")
```
| 107,102 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n β for loop;
* end β every command between "for n" and corresponding "end" is executed n times;
* add β adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 β€ l β€ 10^5) β the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 β€ n β€ 100) β for loop;
* end β every command between "for n" and corresponding "end" is executed n times;
* add β adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
Submitted Solution:
```
l = int(input())
stack = [1]
flag = False
x = 0
INF = (2**32) - 1
for i in range(l):
cmd = input()
if cmd[0] == "f":
t = stack[-1] *int(cmd.split()[1])
if t > INF: t = INF + 1
stack.append(t)
if cmd[0] == "e":
stack.pop()
if cmd[0] == "a":
x += stack[-1]
if x > INF:
flag = True
if flag == True:
print("OVERFLOW!!!")
else:
print(x)
```
Yes
| 107,103 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n β for loop;
* end β every command between "for n" and corresponding "end" is executed n times;
* add β adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 β€ l β€ 10^5) β the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 β€ n β€ 100) β for loop;
* end β every command between "for n" and corresponding "end" is executed n times;
* add β adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
Submitted Solution:
```
from sys import stdin
def solve():
#stdin = open("B. Catch Overflow.txt")
x = 0
max_x = 2**32 - 1
loop = 1
pre_fors = []
mult = []
flag = 0
for _ in range(int(stdin.readline())):
command = [k for k in stdin.readline().split()]
if command[0] == "add":
if flag:
x = max_x + 10
break
x += loop
elif command[0] == "for":
k = int(command[1])
if flag or loop*k > max_x:
mult.append(k)
flag = 1
else:
loop *= k
pre_fors.append(k)
else:
if flag:
mult.pop()
if not mult:
flag = 0
else:
k = pre_fors.pop()
loop //= k
print ("OVERFLOW!!!" if x > max_x else x)
if __name__ == "__main__":
solve()
```
Yes
| 107,104 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n β for loop;
* end β every command between "for n" and corresponding "end" is executed n times;
* add β adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 β€ l β€ 10^5) β the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 β€ n β€ 100) β for loop;
* end β every command between "for n" and corresponding "end" is executed n times;
* add β adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
Submitted Solution:
```
n = int(input())
mx = 2 ** 32
x, flag = 0, 0
buf = []
pr = [1]
for i in range(n) :
inp = input().split()
if flag == 1 :
continue
if inp[0] == "add" :
if x + pr[0] >= mx or len(pr) > 1:
print("OVERFLOW!!!")
flag = 1
x += pr[0]
elif inp[0] == "for" :
ch = int(inp[1])
if pr[-1] * ch >= mx :
pr.append(1)
pr[-1] *= ch
buf.append(ch)
else :
pr[-1] /= buf[-1]
if pr[-1] == 1 and len(pr) > 1 :
pr.pop()
buf.pop()
if flag == 0 :
print(int(x))
```
Yes
| 107,105 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n β for loop;
* end β every command between "for n" and corresponding "end" is executed n times;
* add β adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 β€ l β€ 10^5) β the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 β€ n β€ 100) β for loop;
* end β every command between "for n" and corresponding "end" is executed n times;
* add β adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
Submitted Solution:
```
l = int(input())
ans = 0
nest = 0
F = []
m = 1
for _ in range(l):
com = input()
if com == "add":
ans += m
if ans >= 2**32:
ans = "OVERFLOW!!!"
break
elif com == "end":
m //= F.pop()
else:
if m > 2**40:
a = 1
else:
a = int(com.split()[1])
F.append(a)
m *= a
print(ans)
```
Yes
| 107,106 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n β for loop;
* end β every command between "for n" and corresponding "end" is executed n times;
* add β adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 β€ l β€ 10^5) β the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 β€ n β€ 100) β for loop;
* end β every command between "for n" and corresponding "end" is executed n times;
* add β adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
Submitted Solution:
```
def find_result(commands_array):
total_result = 0
sub_results_stack = [0]
commands_stack = []
remaining_space = 2 ** 32 - 1
for command in commands_array:
if command == "add":
sub_results_stack[-1] += 1
if sub_results_stack[-1] >= remaining_space:
return "OVERFLOW!!!"
elif "for" in command:
commands_stack.append(int(command.split(" ")[1]))
sub_results_stack.append(0)
elif command == "end":
sub_results_stack[-1] *= commands_stack.pop()
if len(commands_stack) > 0:
temp_sub_result = sub_results_stack.pop()
sub_results_stack[-1] += temp_sub_result
if sub_results_stack[-1] >= remaining_space:
return "OVERFLOW!!!"
if len(commands_stack) == 0:
remaining_space -= sub_results_stack[-1]
total_result += sub_results_stack.pop()
if len(sub_results_stack) == 0:
sub_results_stack.append(0)
return total_result
number_of_commands = int(input())
input_commands_array = []
for command in range(number_of_commands):
input_commands_array.append(input())
print(find_result(input_commands_array))
```
No
| 107,107 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n β for loop;
* end β every command between "for n" and corresponding "end" is executed n times;
* add β adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 β€ l β€ 10^5) β the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 β€ n β€ 100) β for loop;
* end β every command between "for n" and corresponding "end" is executed n times;
* add β adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
Submitted Solution:
```
# -*- coding: utf-8 -*-
# @Author: apikdech
# @Date: 07:09:18 06-06-2019
# @Last Modified by: apikdech
# @Last Modified time: 07:28:40 06-06-2019
INF = 2**32
x = 0
ok = 1
tmp = 0
st = []
for _ in range(int(input())):
s = input()
if (s[0] == 'f'):
s = s.split()
st.append(int(s[1]))
if (s[0] == 'a'):
if (len(st) == 0):
x += 1
else:
tmp += 1
if (s[0] == 'e'):
tmp *= st.pop()
if (len(st) == 0):
x += tmp
if (x >= INF):
ok = 0
if (ok == 0):
print("OVERFLOW!!!")
else:
print(x)
```
No
| 107,108 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n β for loop;
* end β every command between "for n" and corresponding "end" is executed n times;
* add β adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 β€ l β€ 10^5) β the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 β€ n β€ 100) β for loop;
* end β every command between "for n" and corresponding "end" is executed n times;
* add β adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
Submitted Solution:
```
def for_(in_, n):
return in_*n
def add_(in_):
return in_+1
n = int(input())
limit = 2 ** 32 -1
out = 0
eq = "0"
depth = 0
for i in range(n):
elms = input().split(" ")
if elms[0] == "for":
n = int(elms[1])
eq += "+ %d * ( 0" % n
depth += 1
elif elms[0] == "add":
eq += "+1"
#print("add", out)
else: # end
assert elms[0] == "end", elms[0]
eq += ") "
depth -= 1
if depth == 0:
#print(i, "depth", depth, "eval", eq)
try:
out += eval(eq)
eq = "0"
except:
if "+1" not in eq:
out = out
eq = "0"
else:
out = "OVERFLOW!!!"
break
if limit < out:
out = "OVERFLOW!!!"
break
print(out)
```
No
| 107,109 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n β for loop;
* end β every command between "for n" and corresponding "end" is executed n times;
* add β adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 β€ l β€ 10^5) β the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 β€ n β€ 100) β for loop;
* end β every command between "for n" and corresponding "end" is executed n times;
* add β adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
Submitted Solution:
```
import sys
t = int(sys.stdin.readline())
n = 0
f = {}
f[0] = 1
m = 2**32 - 1
key = 0
if t < 100000:
for i in range(t):
s = list(sys.stdin.readline().split())
if s[0] == 'add':
n += f[key]
if n > m:
n = 'OVERFLOW!!!'
break
elif s[0] == 'for':
num = int(s[1])
key += 1
f[key] = f[key-1] * num
else:
del f[key]
key -= 1
else:
p = 1
for i in range(t):
s = list(sys.stdin.readline().split())
if s[0] == 'add':
n += 1
## n += f[key]
if n > m:
n = 'OVERFLOW!!!'
break
elif s[0] == 'for':
num = int(s[1])
## p *= num
key += 1
f[key] = num
else:
## p = p // f[key]
del f[key]
key -= 1
print(n)
```
No
| 107,110 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 β€ T β€ 100) β the number of queries. The next T lines contain queries β one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 β€ x < n β€ 10^{9}) β the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) β the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
Tags: math
Correct Solution:
```
a=int(input())
for number in range(0,a):
n,x= map(int,input().split())
print(x*2)
```
| 107,111 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 β€ T β€ 100) β the number of queries. The next T lines contain queries β one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 β€ x < n β€ 10^{9}) β the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) β the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
Tags: math
Correct Solution:
```
q=int(input())
for i in range(q):
w,e=map(int,input().split())
print(e*2)
```
| 107,112 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 β€ T β€ 100) β the number of queries. The next T lines contain queries β one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 β€ x < n β€ 10^{9}) β the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) β the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
Tags: math
Correct Solution:
```
for i in range(int(input())):
x = list(map(int, input().split(" ")))
print(x[1]*2)
#Ψ§Ϊ©Ψ³ΩΎΨͺ Ψ΄Ω
#Ψ¨Ω ΩΩΩ /Ω
/Ψ§ΫΩ Ω
Ψ§Ϋ Ψ³ΩΫΫΨͺ ΩΩΩ
```
| 107,113 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 β€ T β€ 100) β the number of queries. The next T lines contain queries β one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 β€ x < n β€ 10^{9}) β the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) β the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
Tags: math
Correct Solution:
```
"""609C"""
# import math
# import sys
def main():
n, d = map(int, input().split())
print(2*d)
return
# main()
def test():
t= int(input())
while t:
main()
t-=1
test()
```
| 107,114 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 β€ T β€ 100) β the number of queries. The next T lines contain queries β one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 β€ x < n β€ 10^{9}) β the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) β the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
Tags: math
Correct Solution:
```
n = int(input())
for i in range(0, n):
n , x = map(int,input().split())
if x * 2 < n:
print(x*2)
else:
print(n)
```
| 107,115 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 β€ T β€ 100) β the number of queries. The next T lines contain queries β one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 β€ x < n β€ 10^{9}) β the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) β the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
Tags: math
Correct Solution:
```
tc = int(input())
while (tc > 0):
tc -= 1
a, b = [int(x) for x in input().split()]
print(b * 2)
```
| 107,116 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 β€ T β€ 100) β the number of queries. The next T lines contain queries β one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 β€ x < n β€ 10^{9}) β the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) β the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
Tags: math
Correct Solution:
```
#!/usr/bin/env python
# coding: utf-8
# In[9]:
req_num = input()
req_list = []
for i in range(int(req_num)):
req_list.append([int(j) for j in input().split()])
# In[12]:
for i in req_list:
print(i[1]*2)
```
| 107,117 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 β€ T β€ 100) β the number of queries. The next T lines contain queries β one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 β€ x < n β€ 10^{9}) β the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) β the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
Tags: math
Correct Solution:
```
TestCase = int(input())
while( TestCase > 0 ):
[n,x] = list( map( int,input().split() ) )
print(2*x)
TestCase -= 1
```
| 107,118 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 β€ T β€ 100) β the number of queries. The next T lines contain queries β one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 β€ x < n β€ 10^{9}) β the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) β the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
Submitted Solution:
```
T = int(input())
inputs = [''] * T
for i in range(T):
inputs[i] = input()
for s in inputs:
n, x = map(int, s.split())
print(2 * x)
```
Yes
| 107,119 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 β€ T β€ 100) β the number of queries. The next T lines contain queries β one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 β€ x < n β€ 10^{9}) β the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) β the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
Submitted Solution:
```
T = int(input())
for _ in range(T):
n, x = map(int,input().split())
print(2*x)
```
Yes
| 107,120 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 β€ T β€ 100) β the number of queries. The next T lines contain queries β one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 β€ x < n β€ 10^{9}) β the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) β the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
Submitted Solution:
```
t = int(input())
for _ in range(t):
l = list(map(int,input().split()))
print(l[1]<<1)
```
Yes
| 107,121 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 β€ T β€ 100) β the number of queries. The next T lines contain queries β one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 β€ x < n β€ 10^{9}) β the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) β the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
Submitted Solution:
```
t = int(input())
for i in range(t):
n, x = list(map(int, input().split(" ")))
prox = x * 2
print(prox)
```
Yes
| 107,122 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 β€ T β€ 100) β the number of queries. The next T lines contain queries β one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 β€ x < n β€ 10^{9}) β the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) β the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
Submitted Solution:
```
def main_function():
t = int(input())
for i in range(t):
a, b = [int(i) for i in input().split(" ")]
return b * 2
print(main_function())
```
No
| 107,123 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 β€ T β€ 100) β the number of queries. The next T lines contain queries β one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 β€ x < n β€ 10^{9}) β the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) β the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
Submitted Solution:
```
noOfTestCases = int(input())
number, position = map(int, input().split(" "))
print(position*2)
```
No
| 107,124 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 β€ T β€ 100) β the number of queries. The next T lines contain queries β one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 β€ x < n β€ 10^{9}) β the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) β the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
Submitted Solution:
```
t = int(input())
[n, x] = [int(v) for v in input().split()]
print(2 * x)
```
No
| 107,125 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 β€ T β€ 100) β the number of queries. The next T lines contain queries β one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 β€ x < n β€ 10^{9}) β the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) β the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
Submitted Solution:
```
try:
T=int(input())
l1=[]
for i in range(T):
n,x=map(int,input().split())
l=list(range(1,n+1))
if n>x:
for i in range(n):
if i<len(l):
l.remove(l[i])
else:
break
l1.append(l[x-1])
else:
pass
for i in l1:
print(i)
except:
pow
```
No
| 107,126 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line β the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
Tags: dp, strings, two pointers
Correct Solution:
```
S, T = input(), input()
s1 = [1 if c=='(' else -1 for c in S]
s2 = [1 if c=='(' else -1 for c in T]
len1, len2 = len(s1), len(s2)
inf = 10**9
dp = [[[inf]*(len1+len2+1) for _ in range(len2+1)] for _ in range(len1+1)]
dp[0][0][0] = 0
for i in range(len1+1):
for j in range(len2+1):
for k in range(len1+len2):
if i < len1:
k1 = k + s1[i]
if k1 < 0:
dp[i+1][j][k] = min(dp[i+1][j][k], dp[i][j][k]+2)
else:
dp[i+1][j][k1] = min(dp[i+1][j][k1], dp[i][j][k]+1)
if j < len2:
k2 = k + s2[j]
if k2 < 0:
dp[i][j+1][k] = min(dp[i][j+1][k], dp[i][j][k]+2)
else:
dp[i][j+1][k2] = min(dp[i][j+1][k2], dp[i][j][k]+1)
if i < len1 and j < len2:
if s1[i] == s2[j]:
if k1 < 0:
dp[i+1][j+1][k] = min(dp[i+1][j+1][k], dp[i][j][k]+2)
else:
dp[i+1][j+1][k1] = min(dp[i+1][j+1][k1], dp[i][j][k] + 1)
k = 0
length = inf
for i in range(len1+len2+1):
if dp[len1][len2][i] + i < length:
length = dp[len1][len2][i] + 1
k = i
i, j = len1, len2
ans = [')']*k
while i > 0 or j > 0:
if i > 0 and j > 0 and s1[i-1] == s2[j-1]:
if dp[i-1][j-1][k-s1[i-1]] + 1 == dp[i][j][k]:
ans.append(S[i-1])
k -= s1[i-1]
i -= 1
j -= 1
continue
elif k == 0 and s1[i-1] == -1 and dp[i-1][j-1][k] + 2 == dp[i][j][k]:
ans.extend((S[i-1], '('))
i -= 1
j -= 1
continue
if i > 0:
if dp[i-1][j][k-s1[i-1]] + 1 == dp[i][j][k]:
ans.append(S[i-1])
k -= s1[i-1]
i -= 1
continue
elif k == 0 and s1[i-1] == -1 and dp[i-1][j][k] + 2 == dp[i][j][k]:
ans.extend((S[i-1], '('))
i -= 1
continue
if j > 0:
if dp[i][j-1][k-s2[j-1]] + 1 == dp[i][j][k]:
ans.append(T[j-1])
k -= s2[j-1]
j -= 1
continue
elif k == 0 and s2[j-1] == -1 and dp[i][j-1][k] + 2 == dp[i][j][k]:
ans.extend((T[j-1], '('))
j -= 1
continue
raise Exception(i, j, k, S[i-1], T[j-1])
print(*ans[::-1], sep='')
```
| 107,127 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line β the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
Tags: dp, strings, two pointers
Correct Solution:
```
s1 = input();
s2 = input();
inf = 10**8;
n = len(s1);
m = len(s2);
d = [[[inf for x in range(n+m+10)] for y in range(m+10)] for z in range(n+10)];
tata = [[[0 for x in range(n+m+10)] for y in range(m+10)] for z in range(n+10)];
d[0][0][0]=0;
for i in range(n+1):
for j in range(m+1):
i1 = i + (i<n and s1[i] == '(');
j1 = j + (j<m and s2[j] == '(');
for k in range(n+m+1):
if d[i][j][k]==inf:
continue;
if d[i][j][k]+1<d[i1][j1][k+1]:
d[i1][j1][k+1]=d[i][j][k]+1;
tata[i1][j1][k+1]=1;
i1 = i + (i<n and s1[i] == ')');
j1 = j + (j<m and s2[j] == ')');
for k in reversed(range(1,n+m+1)):
if d[i][j][k]==inf:
continue;
if d[i][j][k]+1<d[i1][j1][k-1]:
d[i1][j1][k-1]=d[i][j][k]+1;
tata[i1][j1][k-1]=-1;
lmin=inf;
pz=0;
for i in range(n+m+1):
if d[n][m][i]+i<lmin:
lmin=d[n][m][i]+i;
pz=i;
ans = "";
for i in range(pz):
ans+=')';
i = n;
j = m;
k = pz;
while(i>0 or j>0 or k>0):
val = d[i][j][k]-1;
if tata[i][j][k]==1:
c='(';
k-=1;
else:
c=')';
k+=1;
ans+=c;
i1 = (i>0 and s1[i-1]==c);
j1 = (j>0 and s2[j-1]==c);
if d[i][j][k]==val:
continue;
elif i1>0 and d[i-1][j][k]==val:
i-=1;
elif j1>0 and d[i][j-1][k]==val:
j-=1;
else:
i-=1;
j-=1;
ans = ans[::-1];
print(ans);
```
| 107,128 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line β the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
Tags: dp, strings, two pointers
Correct Solution:
```
def scs(str1, str2):
"""Shortest Common Supersequence"""
INF = 10 ** 9
dp = [[[INF] * 210 for _ in range(210)] for _ in range(210)]
dp[0][0][0] = 0
prv = [[[None] * 210 for _ in range(210)] for _ in range(210)]
len_str1 = len(str1)
len_str2 = len(str2)
str1 += "#"
str2 += "#"
for i in range(len_str1 + 1):
for j in range(len_str2 + 1):
for k in range(201):
# add "("
k2 = k + 1
i2 = i + (str1[i] == "(")
j2 = j + (str2[j] == "(")
if dp[i2][j2][k2] > dp[i][j][k] + 1:
dp[i2][j2][k2] = dp[i][j][k] + 1
prv[i2][j2][k2] = (i, j, k)
for k in range(1, 201):
# add ")"
k2 = k - 1
i2 = i + (str1[i] == ")")
j2 = j + (str2[j] == ")")
if dp[i2][j2][k2] > dp[i][j][k] + 1:
dp[i2][j2][k2] = dp[i][j][k] + 1
prv[i2][j2][k2] = (i, j, k)
ans = INF
cnt = 0
for c, val in enumerate(dp[len_str1][len_str2]):
if c + val < ans + cnt:
ans = val
cnt = c
res = []
i, j, k = len_str1, len_str2, cnt
while i > 0 or j > 0 or k > 0:
prv_i, prv_j, prv_k = prv[i][j][k]
if prv_k < k:
res.append("(")
else:
res.append(")")
i, j, k = prv_i, prv_j, prv_k
return "(" * (ans - len(res) - cnt) + "".join(res[::-1]) + ")" * cnt
s = input()
t = input()
print(scs(s, t))
```
| 107,129 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line β the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
Tags: dp, strings, two pointers
Correct Solution:
```
import sys
import time
def play(s1, s2):
#start_time = time.time()
n = len(s1)
m = len(s2)
# Π±Π°Π»Π°Π½Ρ ΡΠΊΠΎΠ±ΠΎΠΊ Π½Π΅ ΠΌΠΎΠΆΠ΅Ρ Π±ΡΡΡ > 200 (Π²ΡΠ΅Π³Π΄Π° ΠΌΠΎΠΆΠ½ΠΎ ΠΏΠΎΡΡΡΠΎΠΈΡΡ ΠΏΡΠ°Π²ΠΈΠ»ΡΠ½ΡΡ ΠΏΠΎΡΠ»Π΅Π΄ΠΎΠ²Π°ΡΠ΅Π»ΡΠ½ΠΎΡΡΡ, Π½Π΅ ΠΎΠ±ΡΡ
Π°ΡΠ΅Π»Π±Π½ΠΎ ΠΌΠΈΠ½ΠΈΠΌΠ°Π»ΡΠ½ΠΎΠΉ Π΄Π»ΠΈΠ½Ρ Ρ Π±Π°Π»Π°Π½ΡΠΎΠΌ
# < 200)
maxBalance = 200
#print(time.time() - start_time)
#start_time = time.time()
dp = []
p = []
for i in range (0, n + 1):
dp.append([])
p.append([])
for j in range (0, m + 1):
dp[i].append([])
p[i].append([])
for a in range (0, maxBalance + 1):
dp[i][j].append(sys.maxsize)
p[i][j].append(None)
#dp = [[[{'len': sys.maxsize, 'i': 0, 'j': 0, 'a': 0, 'c': ''} for a in range(maxBalance + 1)] for i in range(m + 1)] for j in range(n + 1)]
#print(time.time() - start_time)
# dp[i][j][a] - ΡΡΠΎ ΠΌΠΈΠ½ΠΈΠΌΠ°Π»ΡΠ½Π°Ρ Π΄Π»ΠΈΠ½Π° ΠΈΡΠΎΠ³ΠΎΠ²ΠΎΠΉ ΠΏΠΎΡΠ»Π΅Π΄ΠΎΠ²Π°ΡΠ΅Π»ΡΠ½ΠΎΡΡΠΈ ΠΏΠΎΡΠ»Π΅ ΠΏΡΠΎΡΡΠ΅Π½ΠΈΡ
# i ΡΠΈΠΌΠ²ΠΎΠ»ΠΎΠ² ΡΡΡΠΎΠΊΠΈ s1,
# j ΡΠΈΠΌΠ²ΠΎΠ»ΠΎΠ² ΡΡΡΠΎΠΊΠΈ s2,
# a - Π±Π°Π»Π°Π½Ρ ΡΠΊΠΎΠ±ΠΎΠΊ (ΠΊΠΎΠ»ΠΈΡΠ΅ΡΡΠ²ΠΎ ΠΎΡΠΊΡΡΡΡΡ
, Π½ΠΎ Π½Π΅ Π·Π°ΠΊΡΡΡΡΡ
ΡΠΊΠΎΠ±ΠΎΠΊ)
#start_time = time.time()
dp[0][0][0] = 0
#print(time.time() - start_time)
#start_time = time.time()
for i in range(0, n + 1):
for j in range(0, m + 1):
for a in range(0, maxBalance + 1):
if dp[i][j][a] == sys.maxsize:
continue
# ΡΠ΅ΠΉΡΠ°Ρ ΠΌΡ Π½Π°Ρ
ΠΎΠ΄ΠΈΠΌΡΡ Π² ΡΠ»Π΅Π΄ΡΡΡΠ΅ΠΌ ΠΏΠΎΠ»ΠΎΠΆΠ΅Π½ΠΈΠΈ:
# 1. Π ΠΏΠΎΠ·ΠΈΡΠΈΠΈ i ΡΡΡΠΎΠΊΠΈ s1 (Π΄Π»ΠΈΠ½Π° i + 1)
# 2. Π ΠΏΠΎΠ·ΠΈΡΠΈΡ j ΡΡΡΠΎΠΊΠΈ s2 (Π΄Π»ΠΈΠ½Π° j + 1)
# 3. ΠΈΠΌΠ΅Π΅ΠΌ Π±Π°Π»Π°Π½Ρ a
# ΠΠΎΠ·ΠΌΠΎΠΆΠ½Ρ Π²ΡΠ΅Π³Π΄Π° 2 Π²Π°ΡΠΈΠ°Π½ΡΠ° Π΄Π΅ΠΉΡΡΠ²ΠΈΡ:
# ΠΠΎΠ±Π°Π²Π»ΡΠ΅ΠΌ Π² Π²ΡΡ
ΠΎΠ΄Π½ΡΡ ΡΡΡΠΎΠΊΡ '(' ΠΈΠ»ΠΈ ')'
# ΠΠΎΠ±Π°Π²Π»ΡΠ΅ΠΌ '(' => Π±Π°Π»Π°Π½Ρ ΡΠ²Π΅Π»ΠΈΡΠΈΡΡΡ Π½Π° 1. ΠΡΠ»ΠΈ Π² s1 ΠΎΡΠΊΡΡΠ²Π°ΡΡΠ°ΡΡΡ ΡΠΊΠΎΠ±ΠΊΠ°, ΡΠΎ ΠΌΠΎΠΆΠ΅ΠΌ Π΄ΠΈΠ³Π°ΡΡΡΡ ΠΏΠΎ Π½Π΅ΠΉ Π΄Π°Π»Π΅Π΅
# Π½ΠΎ Π΅ΡΠ»ΠΈ ΠΆΠ΅ ΡΠ°ΠΌ Π±ΡΠ»Π° Π·Π°ΠΊΡΡΠ²Π°ΡΡΠ°ΡΡΡ ΡΠΊΠΎΠ±ΠΊΠ°, ΡΠΎ ΡΡΠΎΠΈΠΌ Π³Π΄Π΅ ΡΠ΅ΠΉΡΠ°Ρ. ΠΠΎΡΡΠΎΠΌΡ:
next_i = i + (i < n and s1[i] == '(')
next_j = j + (j < m and s2[j] == '(')
if a < maxBalance and dp[next_i][next_j][a + 1] > dp[i][j][a] + 1:
dp[next_i][next_j][a + 1] = dp[i][j][a] + 1
p[next_i][next_j][a + 1] = (i, j, a, '(')
# ΠΡΠΎΡΠΎΠΉ Π²Π°ΡΠΈΠ°Π½Ρ - Π΄ΠΎΠ±Π°Π²Π»Π΅Π½ΠΈΠ΅ ')'. Π Π°ΡΡΡΠΆΠ΄Π΅Π½ΠΈΡ Π°Π½Π°Π»ΠΎΠ³ΠΈΡΠ½Ρ. ΠΡΠΈ Π΄ΠΎΠ±Π°Π²Π»Π΅Π½ΠΈΠΈ Π·Π°ΠΊΡΡΠ²Π°ΡΡΠ΅ΠΉ ΡΠΊΠΎΠ±ΠΊΠΈ Π±Π°Π»Π°Π½Ρ Π²ΡΠ΅Π³Π΄Π° ΡΠΌΠ΅Π½ΡΡΠ°Π΅ΡΡΡ Π½Π° 1
next_i = i + (i < n and s1[i] == ')')
next_j = j + (j < m and s2[j] == ')')
if a > 0 and dp[next_i][next_j][a - 1] > dp[i][j][a] + 1:
dp[next_i][next_j][a - 1] = dp[i][j][a] + 1
p[next_i][next_j][a - 1] = (i, j, a, ')')
#print(time.time() - start_time)
aa = 0
nn = n
mm = m
#start_time = time.time()
for a in range(0, maxBalance + 1):
if dp[n][m][a] + a < dp[n][m][aa] + aa:
aa = a
#print(time.time() - start_time)
#start_time = time.time()
ret = ')' * aa
while nn > 0 or mm > 0 or aa > 0:
d = p[nn][mm][aa]
nn = d[0]
mm = d[1]
aa = d[2]
ret += d[3]
#print(time.time() - start_time)
return ret[::-1]
def main():
#start_time = time.time()
#print(play(input(), input()))
print(play(input(), input()))
#print(time.time() - start_time)
main()
#print(play('(())(()','()))()'))
#print(play('((',')'))
```
| 107,130 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line β the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
Tags: dp, strings, two pointers
Correct Solution:
```
def limitbal(a, b):
min_level = 0
level = 0
for aa in a:
if aa == '(':
level += 1
else:
level -= 1
if level < min_level:
min_level = level
for aa in b:
if aa == '(':
level += 1
else:
level -= 1
if level < min_level:
min_level = level
return -min_level + len(a) + len(b) + (level - min_level)
def go():
a = input()
b = input()
lena = len(a)
lenb = len(b)
a+='X'
b+='X'
bal_lim = limitbal(a, b)//2 + 3
tab = [[[None] * (bal_lim+2) for _ in range(lenb + 1)] for _ in range(lena+1)]
par = [[[None] * (bal_lim+2) for _ in range(lenb + 1)] for _ in range(lena+1)]
tab[0][0][0] = 0
que = [(0, 0, 0)]
index = 0
while tab[lena][lenb][0] is None:
i, j, bal = que[index]
ai = a[i]
bj = b[j]
if bal < bal_lim and (bal==0 or not (ai==bj==')')):
# Add (
ii = i
jj = j
if ai == '(':
ii = i + 1
if bj == '(':
jj = j + 1
if tab[ii][jj][bal + 1] is None:
tab[ii][jj][bal + 1] = tab[i][j][bal] + 1
par[ii][jj][bal + 1] = i, j, bal, '('
que.append((ii, jj, bal + 1))
if bal > 0 and not (ai==bj=='('):
ii = i
jj = j
# Add )
if ai == ')':
ii = i + 1
if bj == ')':
jj = j + 1
if tab[ii][jj][bal - 1] is None:
tab[ii][jj][bal - 1] = tab[i][j][bal] + 1
par[ii][jj][bal - 1] = i, j, bal, ')'
que.append((ii, jj, bal - 1))
index += 1
i = lena
j = lenb
bal = 0
answer = []
while (i, j, bal) != (0, 0, 0):
# print (i,j,bal)
i, j, bal, symb = par[i][j][bal]
answer.append(symb)
print(''.join(reversed(answer)))
go()
```
| 107,131 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line β the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
Tags: dp, strings, two pointers
Correct Solution:
```
import sys,math,itertools
from collections import Counter,deque,defaultdict
from bisect import bisect_left,bisect_right
from heapq import heappop,heappush,heapify, nlargest
from copy import deepcopy
mod = 10**9+7
INF = float('inf')
def inp(): return int(sys.stdin.readline())
def inpl(): return list(map(int, sys.stdin.readline().split()))
def inpl_1(): return list(map(lambda x:int(x)-1, sys.stdin.readline().split()))
def inps(): return sys.stdin.readline()
def inpsl(x): tmp = sys.stdin.readline(); return list(tmp[:x])
def err(x): print(x); exit()
s = list(input())
n = len(s)
t = list(input())
m = len(t)
d = {'(':1 , ')':-1}
s = [1 if x=='(' else -1 for x in s]
t = [1 if x=='(' else -1 for x in t]
s.append(0); t.append(0)
K = 210
dp = [[[INF]*K for _ in range(m+2)] for i in range(n+2)]
pa = [[[-1 for k in range(K)] for j in range(m+1)] for i in range(n+1)]
dp[0][0][0] = 0
for i in range(n+1):
for j in range(m+1):
for k in range(201):
nk = k+1
ni = i+(s[i]==1)
nj = j+(t[j]==1)
if dp[ni][nj][nk] > dp[i][j][k]+1:
dp[ni][nj][nk] = dp[i][j][k]+1
pa[ni][nj][nk] = (i,j,k)
for k in range(1,201):
nk = k-1
ni = i+(s[i]==-1)
nj = j+(t[j]==-1)
if dp[ni][nj][nk] > dp[i][j][k]+1:
dp[ni][nj][nk] = dp[i][j][k]+1
pa[ni][nj][nk] = (i,j,k)
rem = -1
mi = INF
for k,num in enumerate(dp[n][m]):
if k+num < mi:
rem = k; mi = k+num
res = []
i,j,k = n,m,rem
while i>0 or j>0 or k>0:
pi,pj,pk = pa[i][j][k]
if k > pk: res.append('(')
else: res.append(')')
i,j,k = pi,pj,pk
res = res[::-1]
for i in range(rem): res.append(')')
print(''.join(res))
```
| 107,132 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line β the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
Submitted Solution:
```
s=input()
s1=input()
o=max(s.count("("),s1.count("("))
p=max(s.count(")"),s1.count(")"))
if(o>=p):
print("("*o + ")"*o)
else:
print("(" * p + ")" * p)
```
No
| 107,133 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line β the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
Submitted Solution:
```
import sys,math,itertools
from collections import Counter,deque,defaultdict
from bisect import bisect_left,bisect_right
from heapq import heappop,heappush,heapify, nlargest
from copy import deepcopy
mod = 10**9+7
INF = float('inf')
def inp(): return int(sys.stdin.readline())
def inpl(): return list(map(int, sys.stdin.readline().split()))
def inpl_1(): return list(map(lambda x:int(x)-1, sys.stdin.readline().split()))
def inps(): return sys.stdin.readline()
def inpsl(x): tmp = sys.stdin.readline(); return list(tmp[:x])
def err(x): print(x); exit()
s = list(input())
n = len(s)
t = list(input())
m = len(t)
d = {'(':1 , ')':-1}
s = [d[x] for x in s]
t = [d[x] for x in t]
s += [0]; t += [0]
K = 210
dp = [[[INF]*K for _ in range(m+2)] for i in range(n+2)]
pa = [[[(-1,-1,-1) for k in range(K)] for j in range(m+1)] for i in range(n+1)]
dp[0][0][0] = 0
for i in range(n+1):
for j in range(m+1):
for k in range(5):
nk = k+1
ni = i+(s[i]==1)
nj = j+(t[j]==1)
if dp[ni][nj][nk] > dp[i][j][k]+1:
dp[ni][nj][nk] = dp[i][j][k]+1
pa[ni][nj][nk] = (i,j,k)
for k in range(1,5):
nk = k-1
ni = i+(s[i]==-1)
nj = j+(t[j]==-1)
if dp[ni][nj][nk] > dp[i][j][k]+1:
dp[ni][nj][nk] = dp[i][j][k]+1
pa[ni][nj][nk] = (i,j,k)
rem = -1
mi = INF
for k,num in enumerate(dp[n][m]):
if k+num < mi:
rem = k; mi = k+num
res = []
i,j,k = n,m,rem
while i>0 or j>0 or k>0:
pi,pj,pk = pa[i][j][k]
if k > pk: res.append('(')
else: res.append(')')
i,j,k = pi,pj,pk
print(''.join(res[::-1]) + ')'*rem)
```
No
| 107,134 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line β the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
Submitted Solution:
```
s = input()
t = input()
d = {'(':1, ')':-1}
if s[0] != '(': s = '(' + s
if t[0] != '(': t = '(' + t
for i in range(min(len(s),len(t))):
if s[i] != t[i]:
if t[i] == '(':
tmp = s
s = t
t = tmp
break
b = []
c = []
for i in range(len(s)+1):
newc = []
newb = []
for j in range(len(t)+1):
newc.append(0)
newb.append('')
c.append(newc)
b.append(newb)
for i in range(len(s)+1):
c[i][0] = i
b[i][0] = s[:i]
for i in range(len(t)+1):
c[0][i] = i
b[0][i] = t[:i]
for i in range(1,1+len(s)):
for j in range(1,1+len(t)):
if s[i-1]==t[j-1]:
c[i][j] = c[i-1][j-1]
b[i][j] = b[i-1][j-1]+s[i-1]
else:
if c[i-1][j]+1<c[i][j-1]+1:
c[i][j]=c[i-1][j]+1
b[i][j]=b[i-1][j]+s[i-1]
else:
c[i][j]=c[i][j-1]+1
b[i][j]=b[i][j-1]+t[j-1]
subtotal = b[-1][-1]
# tt = 0
# for i in range(len(subtotal)):
# tt += d[subtotal[i]]
# for x in range(tt):
# subtotal = subtotal + ')'
# for x in range(-tt):
# subtotal = '(' + subtotal
tt = 0
add = 0
for i in range(len(subtotal)):
tt += d[subtotal[i]]
if tt<0:
add += 1
tt += 1
for i in range(add):
subtotal = '(' + subtotal
print(subtotal)
# print(c[len(s)][len(t)])
# s1.append(d[s[i]])
# total1 += d[s[i]]
# for i in range(len(t)):
# s2.append(d[t[i]])
# total2 += d[t[i]]
# if total > 0:
# for i in range(total): s1.append(-1)
# elif total < 0:
# for i in range(-total): s1.insert(0, 1)
```
No
| 107,135 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line β the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
Submitted Solution:
```
s = input()
t = input()
d = {'(':1, ')':-1}
if '(' in s and '(' in t:
if s.index('(') > t.index('('):
tmp = s
s = t
t = tmp
elif '(' not in s:
tmp = s
s = t
t = tmp
b = []
c = []
for i in range(len(s)+1):
newc = []
newb = []
for j in range(len(t)+1):
newc.append(0)
newb.append('')
c.append(newc)
b.append(newb)
for i in range(len(s)+1):
c[i][0] = i
b[i][0] = s[:i]
for i in range(len(t)+1):
c[0][i] = i
b[0][i] = t[:i]
for i in range(1,1+len(s)):
for j in range(1,1+len(t)):
if s[i-1]==t[j-1]:
c[i][j] = c[i-1][j-1]
b[i][j] = b[i-1][j-1]+s[i-1]
else:
if c[i-1][j]+1<c[i][j-1]+1:
c[i][j]=c[i-1][j]+1
b[i][j]=b[i-1][j]+s[i-1]
else:
c[i][j]=c[i][j-1]+1
b[i][j]=b[i][j-1]+t[j-1]
subtotal = b[-1][-1]
print(subtotal)
tt = 0
for i in range(len(subtotal)):
tt += d[subtotal[i]]
for x in range(tt):
subtotal = subtotal + ')'
for y in range(-tt):
subtotal = '(' + subtotal
print(subtotal)
# print(c[len(s)][len(t)])
# s1.append(d[s[i]])
# total1 += d[s[i]]
# for i in range(len(t)):
# s2.append(d[t[i]])
# total2 += d[t[i]]
# if total > 0:
# for i in range(total): s1.append(-1)
# elif total < 0:
# for i in range(-total): s1.insert(0, 1)
```
No
| 107,136 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers.
In one move, you can choose two indices 1 β€ i, j β€ n such that i β j and set a_i := a_j. You can perform such moves any number of times (possibly, zero). You can choose different indices in different operations. The operation := is the operation of assignment (i.e. you choose i and j and replace a_i with a_j).
Your task is to say if it is possible to obtain an array with an odd (not divisible by 2) sum of elements.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 2000) β the number of test cases.
The next 2t lines describe test cases. The first line of the test case contains one integer n (1 β€ n β€ 2000) β the number of elements in a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 2000), where a_i is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (β n β€ 2000).
Output
For each test case, print the answer on it β "YES" (without quotes) if it is possible to obtain the array with an odd sum of elements, and "NO" otherwise.
Example
Input
5
2
2 3
4
2 2 8 8
3
3 3 3
4
5 5 5 5
4
1 1 1 1
Output
YES
NO
YES
NO
NO
Tags: math
Correct Solution:
```
t = int(input())
for i in range(0,t):
n = int(input())
a = list(map(int,input().split()))
countodd = 0
counteven = 0
for k in range(0,n):
num = a[k]
if(num%2 ==0):
counteven = counteven + 1
else:
countodd = countodd + 1
if(countodd == 0):
print("NO")
else:
if(countodd % 2 == 1):
print('YES')
else:
if(counteven > 0):
print('YES')
else:
print('NO')
```
| 107,137 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers.
In one move, you can choose two indices 1 β€ i, j β€ n such that i β j and set a_i := a_j. You can perform such moves any number of times (possibly, zero). You can choose different indices in different operations. The operation := is the operation of assignment (i.e. you choose i and j and replace a_i with a_j).
Your task is to say if it is possible to obtain an array with an odd (not divisible by 2) sum of elements.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 2000) β the number of test cases.
The next 2t lines describe test cases. The first line of the test case contains one integer n (1 β€ n β€ 2000) β the number of elements in a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 2000), where a_i is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (β n β€ 2000).
Output
For each test case, print the answer on it β "YES" (without quotes) if it is possible to obtain the array with an odd sum of elements, and "NO" otherwise.
Example
Input
5
2
2 3
4
2 2 8 8
3
3 3 3
4
5 5 5 5
4
1 1 1 1
Output
YES
NO
YES
NO
NO
Tags: math
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
p = 0
if n%2:
for i in a:
if i%2:
print('YES')
p += 1
break
else:
k = 0
for i in a:
if not i%2:
k += 1
if k > 0 and k != n:
print('YES')
p += 1
if p == 0:
print('NO')
```
| 107,138 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers.
In one move, you can choose two indices 1 β€ i, j β€ n such that i β j and set a_i := a_j. You can perform such moves any number of times (possibly, zero). You can choose different indices in different operations. The operation := is the operation of assignment (i.e. you choose i and j and replace a_i with a_j).
Your task is to say if it is possible to obtain an array with an odd (not divisible by 2) sum of elements.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 2000) β the number of test cases.
The next 2t lines describe test cases. The first line of the test case contains one integer n (1 β€ n β€ 2000) β the number of elements in a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 2000), where a_i is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (β n β€ 2000).
Output
For each test case, print the answer on it β "YES" (without quotes) if it is possible to obtain the array with an odd sum of elements, and "NO" otherwise.
Example
Input
5
2
2 3
4
2 2 8 8
3
3 3 3
4
5 5 5 5
4
1 1 1 1
Output
YES
NO
YES
NO
NO
Tags: math
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
arr = list(map(int, input().split()))
if sum(arr)%2 != 0:
print("YES")
continue
odd = 0
for i in arr:
if i%2!=0:
odd+=1
if odd == 0: print("NO"); continue
if odd%2==0:
if n-odd > 0:
print("YES")
else:
print("NO")
else:
print("YES")
```
| 107,139 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers.
In one move, you can choose two indices 1 β€ i, j β€ n such that i β j and set a_i := a_j. You can perform such moves any number of times (possibly, zero). You can choose different indices in different operations. The operation := is the operation of assignment (i.e. you choose i and j and replace a_i with a_j).
Your task is to say if it is possible to obtain an array with an odd (not divisible by 2) sum of elements.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 2000) β the number of test cases.
The next 2t lines describe test cases. The first line of the test case contains one integer n (1 β€ n β€ 2000) β the number of elements in a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 2000), where a_i is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (β n β€ 2000).
Output
For each test case, print the answer on it β "YES" (without quotes) if it is possible to obtain the array with an odd sum of elements, and "NO" otherwise.
Example
Input
5
2
2 3
4
2 2 8 8
3
3 3 3
4
5 5 5 5
4
1 1 1 1
Output
YES
NO
YES
NO
NO
Tags: math
Correct Solution:
```
n=int(input())
for i in range(n):
m=int(input())
p=list(map(int,input().split()))
if sum(p)%2!=0:
print("YES")
else:
s=0
f=0
for j in range(len(p)):
if (p[j]%2!=0 ):
s=1
else:
f=1
if s-f==0:
print("YES")
else:
print("NO")
```
| 107,140 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers.
In one move, you can choose two indices 1 β€ i, j β€ n such that i β j and set a_i := a_j. You can perform such moves any number of times (possibly, zero). You can choose different indices in different operations. The operation := is the operation of assignment (i.e. you choose i and j and replace a_i with a_j).
Your task is to say if it is possible to obtain an array with an odd (not divisible by 2) sum of elements.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 2000) β the number of test cases.
The next 2t lines describe test cases. The first line of the test case contains one integer n (1 β€ n β€ 2000) β the number of elements in a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 2000), where a_i is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (β n β€ 2000).
Output
For each test case, print the answer on it β "YES" (without quotes) if it is possible to obtain the array with an odd sum of elements, and "NO" otherwise.
Example
Input
5
2
2 3
4
2 2 8 8
3
3 3 3
4
5 5 5 5
4
1 1 1 1
Output
YES
NO
YES
NO
NO
Tags: math
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
f = 0
e = 0
o = 0
for i in range(n):
if a[i]%2==1:
o+=1
else:
e+=1
if o%2==1:
print('YES')
elif e>0 and o>0:
print('YES')
else:
print('NO')
```
| 107,141 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers.
In one move, you can choose two indices 1 β€ i, j β€ n such that i β j and set a_i := a_j. You can perform such moves any number of times (possibly, zero). You can choose different indices in different operations. The operation := is the operation of assignment (i.e. you choose i and j and replace a_i with a_j).
Your task is to say if it is possible to obtain an array with an odd (not divisible by 2) sum of elements.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 2000) β the number of test cases.
The next 2t lines describe test cases. The first line of the test case contains one integer n (1 β€ n β€ 2000) β the number of elements in a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 2000), where a_i is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (β n β€ 2000).
Output
For each test case, print the answer on it β "YES" (without quotes) if it is possible to obtain the array with an odd sum of elements, and "NO" otherwise.
Example
Input
5
2
2 3
4
2 2 8 8
3
3 3 3
4
5 5 5 5
4
1 1 1 1
Output
YES
NO
YES
NO
NO
Tags: math
Correct Solution:
```
t = int(input())
for i in range(t):
n = int(input())
l = input().split()
total = 0
pair = 0
odd = 0
for i in l:
total+=int(i)
if total %2 != 0:
print("YES")
else:
for i in l:
if int(i)%2==0:
pair = 1
else:
odd = 1
if pair!=0 and odd!=0:
print('YES')
break
if odd == 0 or pair == 0:
print('NO')
```
| 107,142 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers.
In one move, you can choose two indices 1 β€ i, j β€ n such that i β j and set a_i := a_j. You can perform such moves any number of times (possibly, zero). You can choose different indices in different operations. The operation := is the operation of assignment (i.e. you choose i and j and replace a_i with a_j).
Your task is to say if it is possible to obtain an array with an odd (not divisible by 2) sum of elements.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 2000) β the number of test cases.
The next 2t lines describe test cases. The first line of the test case contains one integer n (1 β€ n β€ 2000) β the number of elements in a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 2000), where a_i is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (β n β€ 2000).
Output
For each test case, print the answer on it β "YES" (without quotes) if it is possible to obtain the array with an odd sum of elements, and "NO" otherwise.
Example
Input
5
2
2 3
4
2 2 8 8
3
3 3 3
4
5 5 5 5
4
1 1 1 1
Output
YES
NO
YES
NO
NO
Tags: math
Correct Solution:
```
t=int(input())
for i in range(t):
n=int(input())
a=list(map(int,input().split()))
od=0
for i in a:
if i%2==1:
od+=1
if sum(a)%2==1:
print('YES')
else:
if n==od or od==0:
print('NO')
else:
print('YES')
```
| 107,143 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers.
In one move, you can choose two indices 1 β€ i, j β€ n such that i β j and set a_i := a_j. You can perform such moves any number of times (possibly, zero). You can choose different indices in different operations. The operation := is the operation of assignment (i.e. you choose i and j and replace a_i with a_j).
Your task is to say if it is possible to obtain an array with an odd (not divisible by 2) sum of elements.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 2000) β the number of test cases.
The next 2t lines describe test cases. The first line of the test case contains one integer n (1 β€ n β€ 2000) β the number of elements in a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 2000), where a_i is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (β n β€ 2000).
Output
For each test case, print the answer on it β "YES" (without quotes) if it is possible to obtain the array with an odd sum of elements, and "NO" otherwise.
Example
Input
5
2
2 3
4
2 2 8 8
3
3 3 3
4
5 5 5 5
4
1 1 1 1
Output
YES
NO
YES
NO
NO
Tags: math
Correct Solution:
```
def Input():
tem = input().split()
ans = []
for it in tem:
ans.append(int(it))
return ans
from collections import Counter
T = Input()[0]
for tt in range(T):
n = Input()[0]
a = Input()
odd, even, total = 0, 0, 0
for i in range(n):
if a[i]%2==0:
even+=1
else:
odd+=1
total+=a[i]
if total%2==1 or (even>0 and odd>0):
print("YES")
else:
print("NO")
```
| 107,144 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers.
In one move, you can choose two indices 1 β€ i, j β€ n such that i β j and set a_i := a_j. You can perform such moves any number of times (possibly, zero). You can choose different indices in different operations. The operation := is the operation of assignment (i.e. you choose i and j and replace a_i with a_j).
Your task is to say if it is possible to obtain an array with an odd (not divisible by 2) sum of elements.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 2000) β the number of test cases.
The next 2t lines describe test cases. The first line of the test case contains one integer n (1 β€ n β€ 2000) β the number of elements in a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 2000), where a_i is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (β n β€ 2000).
Output
For each test case, print the answer on it β "YES" (without quotes) if it is possible to obtain the array with an odd sum of elements, and "NO" otherwise.
Example
Input
5
2
2 3
4
2 2 8 8
3
3 3 3
4
5 5 5 5
4
1 1 1 1
Output
YES
NO
YES
NO
NO
Submitted Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
odds = [a[i] % 2 == 1 for i in range(n)]
evens = [a[i] % 2 == 0 for i in range(n)]
if sum(a) % 2 == 0 and (all(odds) or all(evens)):
print('NO')
else:
print('YES')
```
Yes
| 107,145 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers.
In one move, you can choose two indices 1 β€ i, j β€ n such that i β j and set a_i := a_j. You can perform such moves any number of times (possibly, zero). You can choose different indices in different operations. The operation := is the operation of assignment (i.e. you choose i and j and replace a_i with a_j).
Your task is to say if it is possible to obtain an array with an odd (not divisible by 2) sum of elements.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 2000) β the number of test cases.
The next 2t lines describe test cases. The first line of the test case contains one integer n (1 β€ n β€ 2000) β the number of elements in a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 2000), where a_i is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (β n β€ 2000).
Output
For each test case, print the answer on it β "YES" (without quotes) if it is possible to obtain the array with an odd sum of elements, and "NO" otherwise.
Example
Input
5
2
2 3
4
2 2 8 8
3
3 3 3
4
5 5 5 5
4
1 1 1 1
Output
YES
NO
YES
NO
NO
Submitted Solution:
```
for _ in range (int(input())):
n=int(input())
x=[int(n) for n in input().split()]
od=0
for i in x:
if i&1:
od+=1
ev=n-od
if (od&1 or (od%2==0 and ev>0)) and od!=0:
print("YES")
else:
print("NO")
```
Yes
| 107,146 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers.
In one move, you can choose two indices 1 β€ i, j β€ n such that i β j and set a_i := a_j. You can perform such moves any number of times (possibly, zero). You can choose different indices in different operations. The operation := is the operation of assignment (i.e. you choose i and j and replace a_i with a_j).
Your task is to say if it is possible to obtain an array with an odd (not divisible by 2) sum of elements.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 2000) β the number of test cases.
The next 2t lines describe test cases. The first line of the test case contains one integer n (1 β€ n β€ 2000) β the number of elements in a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 2000), where a_i is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (β n β€ 2000).
Output
For each test case, print the answer on it β "YES" (without quotes) if it is possible to obtain the array with an odd sum of elements, and "NO" otherwise.
Example
Input
5
2
2 3
4
2 2 8 8
3
3 3 3
4
5 5 5 5
4
1 1 1 1
Output
YES
NO
YES
NO
NO
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
l = list(map(int,input().split()))
count_odd = 0
for i in range(n):
if(l[i]%2==1):
count_odd+=1
if(count_odd==n)and(n%2==0):
print("NO")
else:
if(count_odd==0):
print("NO")
else:
print("YES")
```
Yes
| 107,147 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers.
In one move, you can choose two indices 1 β€ i, j β€ n such that i β j and set a_i := a_j. You can perform such moves any number of times (possibly, zero). You can choose different indices in different operations. The operation := is the operation of assignment (i.e. you choose i and j and replace a_i with a_j).
Your task is to say if it is possible to obtain an array with an odd (not divisible by 2) sum of elements.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 2000) β the number of test cases.
The next 2t lines describe test cases. The first line of the test case contains one integer n (1 β€ n β€ 2000) β the number of elements in a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 2000), where a_i is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (β n β€ 2000).
Output
For each test case, print the answer on it β "YES" (without quotes) if it is possible to obtain the array with an odd sum of elements, and "NO" otherwise.
Example
Input
5
2
2 3
4
2 2 8 8
3
3 3 3
4
5 5 5 5
4
1 1 1 1
Output
YES
NO
YES
NO
NO
Submitted Solution:
```
# You
# Dont read my code
n = int(input())
i = 0
ans = []
while i < n:
x = int(input())
aa = len([ i for i in map(int,input().split()) if i % 2 != 0])
if (aa == 0 or aa % 2 == 0):
if aa == x or aa == 0:
ans.append("NO")
else:
ans.append("YES")
else:
ans.append("YES")
i += 1
for i in ans:
print(i)
```
Yes
| 107,148 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers.
In one move, you can choose two indices 1 β€ i, j β€ n such that i β j and set a_i := a_j. You can perform such moves any number of times (possibly, zero). You can choose different indices in different operations. The operation := is the operation of assignment (i.e. you choose i and j and replace a_i with a_j).
Your task is to say if it is possible to obtain an array with an odd (not divisible by 2) sum of elements.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 2000) β the number of test cases.
The next 2t lines describe test cases. The first line of the test case contains one integer n (1 β€ n β€ 2000) β the number of elements in a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 2000), where a_i is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (β n β€ 2000).
Output
For each test case, print the answer on it β "YES" (without quotes) if it is possible to obtain the array with an odd sum of elements, and "NO" otherwise.
Example
Input
5
2
2 3
4
2 2 8 8
3
3 3 3
4
5 5 5 5
4
1 1 1 1
Output
YES
NO
YES
NO
NO
Submitted Solution:
```
def fun(ls):
if(sum(ls)%2==0):
print('NO')
else:
print('YES')
T = int(input())
for i in range(T):
# var=input()
val=int(input())
# st=input()
# ms= list(map(int, input().split()))
ls= list(map(int, input().split()))
# fun(ls)
# v=(int(input()))
fun(ls)
```
No
| 107,149 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers.
In one move, you can choose two indices 1 β€ i, j β€ n such that i β j and set a_i := a_j. You can perform such moves any number of times (possibly, zero). You can choose different indices in different operations. The operation := is the operation of assignment (i.e. you choose i and j and replace a_i with a_j).
Your task is to say if it is possible to obtain an array with an odd (not divisible by 2) sum of elements.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 2000) β the number of test cases.
The next 2t lines describe test cases. The first line of the test case contains one integer n (1 β€ n β€ 2000) β the number of elements in a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 2000), where a_i is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (β n β€ 2000).
Output
For each test case, print the answer on it β "YES" (without quotes) if it is possible to obtain the array with an odd sum of elements, and "NO" otherwise.
Example
Input
5
2
2 3
4
2 2 8 8
3
3 3 3
4
5 5 5 5
4
1 1 1 1
Output
YES
NO
YES
NO
NO
Submitted Solution:
```
t=int(input())
for i in range(t):
n=int(input())
a=list(map(int,input().split()))
if sum(a)%2!=0:
print("YES")
else:
ec=0
oc=0
for i in range(len(a)):
if a[i]%2==0:
ec+=a[i]
else:
oc+=a[i]
if ec==oc:
print("NO")
else:
print("YES")
```
No
| 107,150 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers.
In one move, you can choose two indices 1 β€ i, j β€ n such that i β j and set a_i := a_j. You can perform such moves any number of times (possibly, zero). You can choose different indices in different operations. The operation := is the operation of assignment (i.e. you choose i and j and replace a_i with a_j).
Your task is to say if it is possible to obtain an array with an odd (not divisible by 2) sum of elements.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 2000) β the number of test cases.
The next 2t lines describe test cases. The first line of the test case contains one integer n (1 β€ n β€ 2000) β the number of elements in a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 2000), where a_i is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (β n β€ 2000).
Output
For each test case, print the answer on it β "YES" (without quotes) if it is possible to obtain the array with an odd sum of elements, and "NO" otherwise.
Example
Input
5
2
2 3
4
2 2 8 8
3
3 3 3
4
5 5 5 5
4
1 1 1 1
Output
YES
NO
YES
NO
NO
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
if sum(a) % 2 == 0:
print('NO')
else:
print('YES')
```
No
| 107,151 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers.
In one move, you can choose two indices 1 β€ i, j β€ n such that i β j and set a_i := a_j. You can perform such moves any number of times (possibly, zero). You can choose different indices in different operations. The operation := is the operation of assignment (i.e. you choose i and j and replace a_i with a_j).
Your task is to say if it is possible to obtain an array with an odd (not divisible by 2) sum of elements.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 2000) β the number of test cases.
The next 2t lines describe test cases. The first line of the test case contains one integer n (1 β€ n β€ 2000) β the number of elements in a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 2000), where a_i is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (β n β€ 2000).
Output
For each test case, print the answer on it β "YES" (without quotes) if it is possible to obtain the array with an odd sum of elements, and "NO" otherwise.
Example
Input
5
2
2 3
4
2 2 8 8
3
3 3 3
4
5 5 5 5
4
1 1 1 1
Output
YES
NO
YES
NO
NO
Submitted Solution:
```
n=int(input())
for i in range(n):
m=int(input())
p=list(map(int,input().split()))
if sum(p)%2!=0:
print("YES")
else:
print("NO")
```
No
| 107,152 |
Provide tags and a correct Python 3 solution for this coding contest problem.
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
* either it only contains uppercase letters;
* or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
Input
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
Output
Print the result of the given word's processing.
Examples
Input
cAPS
Output
Caps
Input
Lock
Output
Lock
Tags: implementation, strings
Correct Solution:
```
s=str(input())
if (s[0] == s[0].lower() and s[1:] == s[1:].upper()):
print(s[0].upper()+s[1:].lower())
elif (s == s.upper()):
print(s.lower())
else:
print(s)
```
| 107,153 |
Provide tags and a correct Python 3 solution for this coding contest problem.
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
* either it only contains uppercase letters;
* or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
Input
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
Output
Print the result of the given word's processing.
Examples
Input
cAPS
Output
Caps
Input
Lock
Output
Lock
Tags: implementation, strings
Correct Solution:
```
a = input()
if a.isupper() or a[1:].isupper() or (len(a)==1 and a.islower()):
a = list(a)
for i in range(len(a)):
if a[i].isupper():
a[i] = a[i].lower()
else:
a[i] = a[i].upper()
print("".join(a))
```
| 107,154 |
Provide tags and a correct Python 3 solution for this coding contest problem.
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
* either it only contains uppercase letters;
* or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
Input
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
Output
Print the result of the given word's processing.
Examples
Input
cAPS
Output
Caps
Input
Lock
Output
Lock
Tags: implementation, strings
Correct Solution:
```
s=input().strip()
a=s
if len(s)==1:
if a.upper()==s:
print(a.lower())
else:
print(a.upper())
else:
if a.upper()==s:
print(a.lower())
else:
x=a[0].lower()
y=a[1:].upper()
if x==s[0] and y==s[1:]:
x=a[0].upper()
y=a[1:].lower()
print(x+y)
else:
print(s)
```
| 107,155 |
Provide tags and a correct Python 3 solution for this coding contest problem.
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
* either it only contains uppercase letters;
* or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
Input
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
Output
Print the result of the given word's processing.
Examples
Input
cAPS
Output
Caps
Input
Lock
Output
Lock
Tags: implementation, strings
Correct Solution:
```
s=input()
if s.upper()==s:
print(s.lower())
else:
q=s[0]
i=s[1:]
if s[0]==q.lower() and i.upper()==i:
print(q.upper()+i.lower())
else:
print(s)
```
| 107,156 |
Provide tags and a correct Python 3 solution for this coding contest problem.
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
* either it only contains uppercase letters;
* or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
Input
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
Output
Print the result of the given word's processing.
Examples
Input
cAPS
Output
Caps
Input
Lock
Output
Lock
Tags: implementation, strings
Correct Solution:
```
s = input()
if len(s)==1:
if( s.islower() ):
#if it is 'z' make it 'Z'
s = s.upper()
else:
#if it is 'Z' make it 'z':
s = s.lower()
elif s[0].islower() and s[1:].isupper():
# hELLO to Hello
s = s.capitalize()
elif s.isupper():
#HELLO --> hello
s = s.lower()
print(s)
```
| 107,157 |
Provide tags and a correct Python 3 solution for this coding contest problem.
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
* either it only contains uppercase letters;
* or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
Input
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
Output
Print the result of the given word's processing.
Examples
Input
cAPS
Output
Caps
Input
Lock
Output
Lock
Tags: implementation, strings
Correct Solution:
```
word = input()
if len(word)==1 and word.isupper():
print(word.lower())
elif len(word) ==1 and word.islower():
print(word.upper())
else:
if word[1:].isupper() and word[0].islower():
print(word.capitalize())
elif word.isupper():
print(word.lower())
else:
print(word)
```
| 107,158 |
Provide tags and a correct Python 3 solution for this coding contest problem.
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
* either it only contains uppercase letters;
* or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
Input
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
Output
Print the result of the given word's processing.
Examples
Input
cAPS
Output
Caps
Input
Lock
Output
Lock
Tags: implementation, strings
Correct Solution:
```
n=input()
c=0
if n[0].islower()==True:
c+=1
for i in range(0,len(n)):
if n[i].isupper()==True:
c+=1
r=""
if len(n)==c:
for i in n:
if i.isupper()==True:
r+=i.lower()
else:
r+=i.upper()
print(r)
else:
print(n)
```
| 107,159 |
Provide tags and a correct Python 3 solution for this coding contest problem.
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
* either it only contains uppercase letters;
* or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
Input
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
Output
Print the result of the given word's processing.
Examples
Input
cAPS
Output
Caps
Input
Lock
Output
Lock
Tags: implementation, strings
Correct Solution:
```
#! /usr/bin/env python3
word = input()
if (word.isupper()) or (word[1:].isupper()) or len(word) == 1:
word = word.swapcase()
print(word)
else:
print(word)
```
| 107,160 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
* either it only contains uppercase letters;
* or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
Input
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
Output
Print the result of the given word's processing.
Examples
Input
cAPS
Output
Caps
Input
Lock
Output
Lock
Submitted Solution:
```
s=input()
c=0
for l in s:
if ord(l) in range(65,91):
c+=1
if (c==len(s)):
print(s.lower())
elif (c==(len(s)-1)) and ord(s[0]) not in range(65,91):
print(s.capitalize())
elif c==0:
print(s)
else:
print(s)
```
Yes
| 107,161 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
* either it only contains uppercase letters;
* or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
Input
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
Output
Print the result of the given word's processing.
Examples
Input
cAPS
Output
Caps
Input
Lock
Output
Lock
Submitted Solution:
```
import re
string = input()
firstCOndition = string.isupper() #if true all in capital
secondString = string[1:]
secondCondition = secondString.isupper() or len(secondString) == 0 #if true all except first one is capital
if firstCOndition or secondCondition:
print(string.swapcase())
else:
print(string)
```
Yes
| 107,162 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
* either it only contains uppercase letters;
* or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
Input
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
Output
Print the result of the given word's processing.
Examples
Input
cAPS
Output
Caps
Input
Lock
Output
Lock
Submitted Solution:
```
s=input()
if(s.isupper() or (s[0].islower() and s[1:].isupper())):
if(s[0].islower()):
print(s[0].upper(),s[1:].lower(),sep='')
else:
print(s.lower())
elif(len(s)==1):
if(s.islower()):
print(s.upper())
else:
print(s.lower())
else:
print(s)
```
Yes
| 107,163 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
* either it only contains uppercase letters;
* or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
Input
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
Output
Print the result of the given word's processing.
Examples
Input
cAPS
Output
Caps
Input
Lock
Output
Lock
Submitted Solution:
```
s = input()
if s.isupper():
print(s.lower())
elif (s[0].islower() and s[1:].isupper()):
print(s[0].upper() + s[1:].lower())
elif (len(s) == 1 and s[0].islower()):
print(s.upper())
else:
print(s)
```
Yes
| 107,164 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
* either it only contains uppercase letters;
* or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
Input
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
Output
Print the result of the given word's processing.
Examples
Input
cAPS
Output
Caps
Input
Lock
Output
Lock
Submitted Solution:
```
word = input()
if word.islower():
word = word.upper()
if word[0].islower() and word[1:len(word)].isupper():
word = word[0].upper() + word[1 : len(word)].lower()
if word.isupper():
word = word.lower()
print(word)
```
No
| 107,165 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
* either it only contains uppercase letters;
* or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
Input
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
Output
Print the result of the given word's processing.
Examples
Input
cAPS
Output
Caps
Input
Lock
Output
Lock
Submitted Solution:
```
word = input()
letter1 = word[0]
letter1 = letter1.upper()
word = word.lower()
print(letter1+word[1:])
```
No
| 107,166 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
* either it only contains uppercase letters;
* or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
Input
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
Output
Print the result of the given word's processing.
Examples
Input
cAPS
Output
Caps
Input
Lock
Output
Lock
Submitted Solution:
```
x=input()
if len(x)>1:
if x[0].islower():
a=x[0].upper()
b=x[1:len(x)].lower()
x=a+b
else:
x=x.upper()
print(x)
```
No
| 107,167 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
* either it only contains uppercase letters;
* or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
Input
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
Output
Print the result of the given word's processing.
Examples
Input
cAPS
Output
Caps
Input
Lock
Output
Lock
Submitted Solution:
```
s=input()
x=s[1::]
y=s[0]
if(len(s)==1):
print(s.upper())
if(s.isupper()):
print(s.lower())
elif(x.isupper() and y.islower()):
print(y.upper()+x.lower())
else:
print(s)
```
No
| 107,168 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the infinite sequence s of positive integers, created by repeating the following steps:
1. Find the lexicographically smallest triple of positive integers (a, b, c) such that
* a β b β c = 0, where β denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
* a, b, c are not in s.
Here triple of integers (a_1, b_1, c_1) is considered to be lexicographically smaller than triple (a_2, b_2, c_2) if sequence [a_1, b_1, c_1] is lexicographically smaller than sequence [a_2, b_2, c_2].
2. Append a, b, c to s in this order.
3. Go back to the first step.
You have integer n. Find the n-th element of s.
You have to answer t independent test cases.
A sequence a is lexicographically smaller than a sequence b if in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
The first line contains a single integer t (1 β€ t β€ 10^5) β the number of test cases.
Each of the next t lines contains a single integer n (1β€ n β€ 10^{16}) β the position of the element you want to know.
Output
In each of the t lines, output the answer to the corresponding test case.
Example
Input
9
1
2
3
4
5
6
7
8
9
Output
1
2
3
4
8
12
5
10
15
Note
The first elements of s are 1, 2, 3, 4, 8, 12, 5, 10, 15, ...
Tags: bitmasks, brute force, constructive algorithms, divide and conquer, math
Correct Solution:
```
from collections import defaultdict
from queue import deque
from sys import stdin, stdout
from math import log2
def arrinp():
return [*map(int, stdin.readline().split(' '))]
def mulinp():
return map(int, stdin.readline().split(' '))
def intinp():
return int(stdin.readline())
def solution():
n = intinp()
if n <= 3:
print(n)
return
if n % 3 == 1:
k = n.bit_length()-1
if k & 1:
k -= 1
print(2**k + (n-2**k) // 3)
elif n % 3 == 2:
n1 = n-1
k = n1.bit_length()-1
if k & 1:
k -= 1
ans1 = 2**k + (n1-2**k) // 3
ans = 0
cnt = 0
while ans1:
a = ans1 % 4
if a == 1:
a = 2
elif a == 2:
a = 3
elif a == 3:
a = 1
ans += a * (4**cnt)
cnt += 1
ans1 >>= 2
print(ans)
else:
n1 = n - 2
k = n1.bit_length()-1
if k & 1:
k -= 1
ans1 = 2 ** k + (n1 - 2 ** k) // 3
ans = 0
cnt = 0
while ans1:
a = ans1 % 4
if a == 1:
a = 3
elif a == 2:
a = 1
elif a == 3:
a = 2
ans += a * (4 ** cnt)
cnt += 1
ans1 >>= 2
print(ans)
testcases = 1
testcases = intinp()
for _ in range(testcases):
solution()
```
| 107,169 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the infinite sequence s of positive integers, created by repeating the following steps:
1. Find the lexicographically smallest triple of positive integers (a, b, c) such that
* a β b β c = 0, where β denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
* a, b, c are not in s.
Here triple of integers (a_1, b_1, c_1) is considered to be lexicographically smaller than triple (a_2, b_2, c_2) if sequence [a_1, b_1, c_1] is lexicographically smaller than sequence [a_2, b_2, c_2].
2. Append a, b, c to s in this order.
3. Go back to the first step.
You have integer n. Find the n-th element of s.
You have to answer t independent test cases.
A sequence a is lexicographically smaller than a sequence b if in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
The first line contains a single integer t (1 β€ t β€ 10^5) β the number of test cases.
Each of the next t lines contains a single integer n (1β€ n β€ 10^{16}) β the position of the element you want to know.
Output
In each of the t lines, output the answer to the corresponding test case.
Example
Input
9
1
2
3
4
5
6
7
8
9
Output
1
2
3
4
8
12
5
10
15
Note
The first elements of s are 1, 2, 3, 4, 8, 12, 5, 10, 15, ...
Tags: bitmasks, brute force, constructive algorithms, divide and conquer, math
Correct Solution:
```
import sys
input = sys.stdin.readline
t = int(input())
s = [int(input()) for i in range(t)]
res = []
for num in s:
num -= 1
mod = num % 3
num = num // 3
div = 1
while True:
if num // div != 0:
num -= div
div *= 4
else:
break
a = div + num
b = a * 2
tmp = a
coff = 1
while True:
if tmp == 0:
break
if tmp % 4 == 2:
b -= coff
if tmp % 4 == 3:
b -= coff * 5
tmp = tmp // 4
coff *= 4
c = a ^ b
if mod == 0:
print(a)
if mod == 1:
print(b)
if mod == 2:
print(c)
```
| 107,170 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the infinite sequence s of positive integers, created by repeating the following steps:
1. Find the lexicographically smallest triple of positive integers (a, b, c) such that
* a β b β c = 0, where β denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
* a, b, c are not in s.
Here triple of integers (a_1, b_1, c_1) is considered to be lexicographically smaller than triple (a_2, b_2, c_2) if sequence [a_1, b_1, c_1] is lexicographically smaller than sequence [a_2, b_2, c_2].
2. Append a, b, c to s in this order.
3. Go back to the first step.
You have integer n. Find the n-th element of s.
You have to answer t independent test cases.
A sequence a is lexicographically smaller than a sequence b if in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
The first line contains a single integer t (1 β€ t β€ 10^5) β the number of test cases.
Each of the next t lines contains a single integer n (1β€ n β€ 10^{16}) β the position of the element you want to know.
Output
In each of the t lines, output the answer to the corresponding test case.
Example
Input
9
1
2
3
4
5
6
7
8
9
Output
1
2
3
4
8
12
5
10
15
Note
The first elements of s are 1, 2, 3, 4, 8, 12, 5, 10, 15, ...
Tags: bitmasks, brute force, constructive algorithms, divide and conquer, math
Correct Solution:
```
#!/usr/bin/env python3
import sys
input = sys.stdin.readline
from itertools import combinations
t = int(input())
for tt in range(t):
n = int(input()) - 1
mod3 = n % 3
n = n // 3
i = 0
while n >= 4**i:
n -= 4**i
i += 1
if i == 0:
a = 1; b = 2; c = 3
else:
a = 4**i + n
left = 0
right = 4**i
b = 4**i * 2
t = n
while right - left > 1:
mid1 = (left * 3 + right) // 4
mid2 = (left * 2 + right * 2) // 4
mid3 = (left + right * 3) // 4
rng = right - left
if left <= t < mid1:
b += 0
elif mid1 <= t < mid2:
b += rng // 4 * 2
elif mid2 <= t < mid3:
b += rng // 4 * 3
else:
b += rng // 4 * 1
t %= rng // 4
right //= 4
c = a ^ b
# print("#", tt + 1)
if mod3 == 0:
print(a)
elif mod3 == 1:
print(b)
else:
print(c)
# not_used = set([item for item in range(1, 200)])
# for i in range(25):
# li = list(not_used)
# li.sort()
# for comb in combinations(li, 3):
# if comb[0] ^ comb[1] ^ comb[2] == 0:
# print(comb)
# not_used.remove(comb[0])
# not_used.remove(comb[1])
# not_used.remove(comb[2])
# break
```
| 107,171 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the infinite sequence s of positive integers, created by repeating the following steps:
1. Find the lexicographically smallest triple of positive integers (a, b, c) such that
* a β b β c = 0, where β denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
* a, b, c are not in s.
Here triple of integers (a_1, b_1, c_1) is considered to be lexicographically smaller than triple (a_2, b_2, c_2) if sequence [a_1, b_1, c_1] is lexicographically smaller than sequence [a_2, b_2, c_2].
2. Append a, b, c to s in this order.
3. Go back to the first step.
You have integer n. Find the n-th element of s.
You have to answer t independent test cases.
A sequence a is lexicographically smaller than a sequence b if in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
The first line contains a single integer t (1 β€ t β€ 10^5) β the number of test cases.
Each of the next t lines contains a single integer n (1β€ n β€ 10^{16}) β the position of the element you want to know.
Output
In each of the t lines, output the answer to the corresponding test case.
Example
Input
9
1
2
3
4
5
6
7
8
9
Output
1
2
3
4
8
12
5
10
15
Note
The first elements of s are 1, 2, 3, 4, 8, 12, 5, 10, 15, ...
Tags: bitmasks, brute force, constructive algorithms, divide and conquer, math
Correct Solution:
```
# ------------------- fast io --------------------
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")
# ------------------- fast io --------------------
from math import gcd, ceil
def prod(a, mod=10**9+7):
ans = 1
for each in a:
ans = (ans * each) % mod
return ans
def lcm(a, b): return a * b // gcd(a, b)
def binary(x, length=16):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
for _ in range(int(input()) if True else 1):
n = int(input())
# if n % 3 == 1:
# 1, 4-7, 16-31, 64-127, 256-511
def f1(x, pow=1):
if x > 4**pow:
return f1(x-4**pow, pow+1)
return 4**pow + (x - 1)
# if n % 3 == 2:
# 2, (8,10,11,9), (32->35,40->43,44->47,36->39)
# follow 1, 3, 4, 2 pattern
def f2(x, pow=1):
if x > 4**pow:
return f2(x-4**pow, pow+1)
alpha = 2*(4**pow)
# x is between 1 and 4^pow
while x > 4:
rem = (x-1) // (4**(pow-1))
alpha += 4**(pow-1)*[0, 2, 3, 1][rem]
x -= 4**(pow-1) * rem
pow -= 1
return alpha + [0,2,3,1][x-1]
# if n % 3 == 0:
# use previous two numbers to calculate
if n <= 3:
print(n)
continue
if n % 3 == 1:
n //= 3
print(f1(n))
elif n % 3 == 2:
n //= 3
print(f2(n))
elif n % 3 == 0:
n //= 3
n -= 1
print(f1(n)^f2(n))
```
| 107,172 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the infinite sequence s of positive integers, created by repeating the following steps:
1. Find the lexicographically smallest triple of positive integers (a, b, c) such that
* a β b β c = 0, where β denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
* a, b, c are not in s.
Here triple of integers (a_1, b_1, c_1) is considered to be lexicographically smaller than triple (a_2, b_2, c_2) if sequence [a_1, b_1, c_1] is lexicographically smaller than sequence [a_2, b_2, c_2].
2. Append a, b, c to s in this order.
3. Go back to the first step.
You have integer n. Find the n-th element of s.
You have to answer t independent test cases.
A sequence a is lexicographically smaller than a sequence b if in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
The first line contains a single integer t (1 β€ t β€ 10^5) β the number of test cases.
Each of the next t lines contains a single integer n (1β€ n β€ 10^{16}) β the position of the element you want to know.
Output
In each of the t lines, output the answer to the corresponding test case.
Example
Input
9
1
2
3
4
5
6
7
8
9
Output
1
2
3
4
8
12
5
10
15
Note
The first elements of s are 1, 2, 3, 4, 8, 12, 5, 10, 15, ...
Tags: bitmasks, brute force, constructive algorithms, divide and conquer, math
Correct Solution:
```
import os
import sys
input = sys.stdin.buffer.readline
#sys.setrecursionlimit(int(3e5))
from collections import deque
from queue import PriorityQueue
import math
# list(map(int, input().split()))
#####################################################################################
class CF(object):
def __init__(self):
self.g = []
for i in range(100):
self.g.append( (4**i) * 3)
def gao(self, x, n):
#print(str(x) + ' '+ str(n))
re = deque()
rest = x%3
cnt = int(x//3)
for i in range(n):
re.appendleft(cnt%4)
cnt = int(cnt//4)
pass
ans = (rest+1)
for i in range(n):
ans<<=2
if(re[i]== 1):
ans+=rest+1
elif(re[i]==2):
if(rest == 0):
ans+=2
elif(rest == 1):
ans+=3
elif(rest == 2):
ans+=1
elif(re[i]==3):
if(rest == 0):
ans+=3
elif(rest==1):
ans+=1
elif(rest == 2):
ans+=2
pass
print(ans)
def main(self):
self.t = int(input())
for _ in range(self.t):
pos = int(input())
now = 0
while(True):
if(pos>self.g[now]):
pos-=self.g[now]
now+=1
else:
break
pass
x = pos-1
self.gao(x, now)
pass
if __name__ == "__main__":
cf = CF()
cf.main()
pass
```
| 107,173 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the infinite sequence s of positive integers, created by repeating the following steps:
1. Find the lexicographically smallest triple of positive integers (a, b, c) such that
* a β b β c = 0, where β denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
* a, b, c are not in s.
Here triple of integers (a_1, b_1, c_1) is considered to be lexicographically smaller than triple (a_2, b_2, c_2) if sequence [a_1, b_1, c_1] is lexicographically smaller than sequence [a_2, b_2, c_2].
2. Append a, b, c to s in this order.
3. Go back to the first step.
You have integer n. Find the n-th element of s.
You have to answer t independent test cases.
A sequence a is lexicographically smaller than a sequence b if in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
The first line contains a single integer t (1 β€ t β€ 10^5) β the number of test cases.
Each of the next t lines contains a single integer n (1β€ n β€ 10^{16}) β the position of the element you want to know.
Output
In each of the t lines, output the answer to the corresponding test case.
Example
Input
9
1
2
3
4
5
6
7
8
9
Output
1
2
3
4
8
12
5
10
15
Note
The first elements of s are 1, 2, 3, 4, 8, 12, 5, 10, 15, ...
Tags: bitmasks, brute force, constructive algorithms, divide and conquer, math
Correct Solution:
```
mod = 1000000007
eps = 10**-9
def main():
import sys
input = sys.stdin.buffer.readline
for _ in range(int(input())):
N = int(input())
if N <= 3:
print(N)
continue
if N%3 == 1:
k = N.bit_length()-1
if k&1:
k -= 1
print(2**k + (N-2**k) // 3)
elif N%3 == 2:
N1 = N-1
k = N1.bit_length()-1
if k&1:
k -= 1
ans1 = 2**k + (N1-2**k) // 3
ans = 0
cnt = 0
while ans1:
a = ans1 % 4
if a == 1:
a = 2
elif a == 2:
a = 3
elif a == 3:
a = 1
ans += a * (4**cnt)
cnt += 1
ans1 >>= 2
print(ans)
else:
N1 = N - 2
k = N1.bit_length()-1
if k & 1:
k -= 1
ans1 = 2 ** k + (N1 - 2 ** k) // 3
ans = 0
cnt = 0
while ans1:
a = ans1 % 4
if a == 1:
a = 3
elif a == 2:
a = 1
elif a == 3:
a = 2
ans += a * (4 ** cnt)
cnt += 1
ans1 >>= 2
print(ans)
if __name__ == '__main__':
main()
```
| 107,174 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the infinite sequence s of positive integers, created by repeating the following steps:
1. Find the lexicographically smallest triple of positive integers (a, b, c) such that
* a β b β c = 0, where β denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
* a, b, c are not in s.
Here triple of integers (a_1, b_1, c_1) is considered to be lexicographically smaller than triple (a_2, b_2, c_2) if sequence [a_1, b_1, c_1] is lexicographically smaller than sequence [a_2, b_2, c_2].
2. Append a, b, c to s in this order.
3. Go back to the first step.
You have integer n. Find the n-th element of s.
You have to answer t independent test cases.
A sequence a is lexicographically smaller than a sequence b if in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
The first line contains a single integer t (1 β€ t β€ 10^5) β the number of test cases.
Each of the next t lines contains a single integer n (1β€ n β€ 10^{16}) β the position of the element you want to know.
Output
In each of the t lines, output the answer to the corresponding test case.
Example
Input
9
1
2
3
4
5
6
7
8
9
Output
1
2
3
4
8
12
5
10
15
Note
The first elements of s are 1, 2, 3, 4, 8, 12, 5, 10, 15, ...
Tags: bitmasks, brute force, constructive algorithms, divide and conquer, math
Correct Solution:
```
import sys
input = sys.stdin.readline
out = []
t = int(input())
for _ in range(t):
n = int(input())
n -= 1
rem = n % 3
n //= 3
s = []
if n:
n -= 1
while n >= 0:
s.append([['00','00','00'],['01','10','11'],['10','11','01'],['11','01','10']][n % 4][rem])
n //= 4
n -= 1
s.append(['1','10','11'][rem])
s.reverse()
out.append(int(''.join(s),2))
print('\n'.join(map(str,out)))
```
| 107,175 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the infinite sequence s of positive integers, created by repeating the following steps:
1. Find the lexicographically smallest triple of positive integers (a, b, c) such that
* a β b β c = 0, where β denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
* a, b, c are not in s.
Here triple of integers (a_1, b_1, c_1) is considered to be lexicographically smaller than triple (a_2, b_2, c_2) if sequence [a_1, b_1, c_1] is lexicographically smaller than sequence [a_2, b_2, c_2].
2. Append a, b, c to s in this order.
3. Go back to the first step.
You have integer n. Find the n-th element of s.
You have to answer t independent test cases.
A sequence a is lexicographically smaller than a sequence b if in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
The first line contains a single integer t (1 β€ t β€ 10^5) β the number of test cases.
Each of the next t lines contains a single integer n (1β€ n β€ 10^{16}) β the position of the element you want to know.
Output
In each of the t lines, output the answer to the corresponding test case.
Example
Input
9
1
2
3
4
5
6
7
8
9
Output
1
2
3
4
8
12
5
10
15
Note
The first elements of s are 1, 2, 3, 4, 8, 12, 5, 10, 15, ...
Tags: bitmasks, brute force, constructive algorithms, divide and conquer, math
Correct Solution:
```
import os
import sys
mapper = {"00": "00", "01": "10", "10": "11", "11": "01"}
def f(n):
n1 = (n - 1) // 3
if n1 >= 1466015503701:
x = 1466015503701
i = 21
else:
x = 0
i = 0
while n1 >= x + 4 ** i:
x += 4 ** i
i += 1
t1 = 4 ** i + n1 - x
t2 = int(
"10" + "".join(mapper[bin(t1)[3:][2 * j : 2 * (j + 1)]] for j in range(i)), 2
)
idx = n - n1 * 3 - 1
if idx == 0:
return t1
elif idx == 1:
return t2
elif idx == 2:
return t1 ^ t2
else:
raise Exception("bug")
def pp(input):
T = int(input())
print("\n".join(map(str, (f(int(input())) for _ in range(T)))))
if "paalto" in os.getcwd() and "pydevconsole" in sys.argv[0]:
from string_source import string_source
pp(
string_source(
"""9
1
2
3
4
5
6
7
8
9"""
)
)
else:
input = sys.stdin.buffer.readline
pp(input)
assert f(999999999990000) == 1046445486432677
def x():
for i in range(100000):
f(999999999990000 + i)
# %timeit -n 1 x()
```
| 107,176 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider the infinite sequence s of positive integers, created by repeating the following steps:
1. Find the lexicographically smallest triple of positive integers (a, b, c) such that
* a β b β c = 0, where β denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
* a, b, c are not in s.
Here triple of integers (a_1, b_1, c_1) is considered to be lexicographically smaller than triple (a_2, b_2, c_2) if sequence [a_1, b_1, c_1] is lexicographically smaller than sequence [a_2, b_2, c_2].
2. Append a, b, c to s in this order.
3. Go back to the first step.
You have integer n. Find the n-th element of s.
You have to answer t independent test cases.
A sequence a is lexicographically smaller than a sequence b if in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
The first line contains a single integer t (1 β€ t β€ 10^5) β the number of test cases.
Each of the next t lines contains a single integer n (1β€ n β€ 10^{16}) β the position of the element you want to know.
Output
In each of the t lines, output the answer to the corresponding test case.
Example
Input
9
1
2
3
4
5
6
7
8
9
Output
1
2
3
4
8
12
5
10
15
Note
The first elements of s are 1, 2, 3, 4, 8, 12, 5, 10, 15, ...
Submitted Solution:
```
memo=[(0,0,0)]
def new_find_pair(n):
stack=[]
while n>=len(memo):
N=n.bit_length()-1
if N%2==0:
i=N//2
stack.append((n,1,i))
n-=4**i
else:
i=N//2
if n>>(2*i)&1==0:
#type2
stack.append((n,2,i))
n-=2*4**i
else:
#type3
stack.append((n,3,i))
n-=3*4**i
a,b,c=memo[n]
for m,t,i in stack[::-1]:
m-=t*4**i
if t==1:
if m==b:
a,b,c=b,c,a
elif m==c:
a,b,c=c,a,b
elif t==2:
if m==a:
a,b,c=c,a,b
elif m==c:
a,b,c=b,c,a
else:
if m==a:
a,b,c=b,c,a
elif m==b:
a,b,c=c,a,b
a+=4**i
b+=2*4**i
c+=3*4**i
return (a,b,c)
for i in range(1,4**5):
memo.append(new_find_pair(i))
def nth_pair(n):
tmp=0
for i in range(28):
tmp+=4**i
if tmp>=n:
tmp-=4**i
first=(n-tmp-1)+4**i
return new_find_pair(first)
def nth_seq(n):
q=(n-1)//3+1
fst=(n-1)%3
res=nth_pair(q)[fst]
return res
import sys
input=sys.stdin.buffer.readline
#n=int(input())
#print(find_pair(n))
#print(nth_pair(n))
Q=int(input())
for _ in range(Q):
n=int(input())
#n=int(input())
r=nth_seq(n)
print(r)
#print(nth_pair(q)[fst])
#print(time.time()-start)
```
Yes
| 107,177 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider the infinite sequence s of positive integers, created by repeating the following steps:
1. Find the lexicographically smallest triple of positive integers (a, b, c) such that
* a β b β c = 0, where β denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
* a, b, c are not in s.
Here triple of integers (a_1, b_1, c_1) is considered to be lexicographically smaller than triple (a_2, b_2, c_2) if sequence [a_1, b_1, c_1] is lexicographically smaller than sequence [a_2, b_2, c_2].
2. Append a, b, c to s in this order.
3. Go back to the first step.
You have integer n. Find the n-th element of s.
You have to answer t independent test cases.
A sequence a is lexicographically smaller than a sequence b if in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
The first line contains a single integer t (1 β€ t β€ 10^5) β the number of test cases.
Each of the next t lines contains a single integer n (1β€ n β€ 10^{16}) β the position of the element you want to know.
Output
In each of the t lines, output the answer to the corresponding test case.
Example
Input
9
1
2
3
4
5
6
7
8
9
Output
1
2
3
4
8
12
5
10
15
Note
The first elements of s are 1, 2, 3, 4, 8, 12, 5, 10, 15, ...
Submitted Solution:
```
def find_answer(n):
trio_num, trio_place = n // 3, n % 3
digit = 1
while trio_num >= 0:
trio_num -= digit
digit *= 4
digit //= 4
trio_num += digit
t1 = digit + trio_num
if trio_place == 0:
return t1
t2, t3 = digit*2, digit*3
new_digit = 1
while new_digit < digit:
x = t1 % 4
if x == 1:
t2 += new_digit * 2
t3 += new_digit * 3
elif x == 2:
t2 += new_digit * 3
t3 += new_digit * 1
elif x == 3:
t2 += new_digit * 1
t3 += new_digit * 2
t1 //= 4
new_digit *= 4
return t2 if trio_place == 1 else t3
if __name__ == '__main__':
n = int(input())
ans = []
for i in range(n):
x = int(input())
ans += [find_answer(x-1)]
for x in ans:
print(x)
```
Yes
| 107,178 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider the infinite sequence s of positive integers, created by repeating the following steps:
1. Find the lexicographically smallest triple of positive integers (a, b, c) such that
* a β b β c = 0, where β denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
* a, b, c are not in s.
Here triple of integers (a_1, b_1, c_1) is considered to be lexicographically smaller than triple (a_2, b_2, c_2) if sequence [a_1, b_1, c_1] is lexicographically smaller than sequence [a_2, b_2, c_2].
2. Append a, b, c to s in this order.
3. Go back to the first step.
You have integer n. Find the n-th element of s.
You have to answer t independent test cases.
A sequence a is lexicographically smaller than a sequence b if in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
The first line contains a single integer t (1 β€ t β€ 10^5) β the number of test cases.
Each of the next t lines contains a single integer n (1β€ n β€ 10^{16}) β the position of the element you want to know.
Output
In each of the t lines, output the answer to the corresponding test case.
Example
Input
9
1
2
3
4
5
6
7
8
9
Output
1
2
3
4
8
12
5
10
15
Note
The first elements of s are 1, 2, 3, 4, 8, 12, 5, 10, 15, ...
Submitted Solution:
```
import sys
input = sys.stdin.readline
import bisect
L=[4**i for i in range(30)]
def calc(x):
fami=bisect.bisect_right(L,x)-1
base=(x-L[fami])//3
cl=x%3
if cl==1:
return L[fami]+base
else:
cl1=L[fami]+base
ANS=0
for i in range(30):
x=cl1//L[i]%4
if x==1:
ANS+=2*L[i]
elif x==2:
ANS+=3*L[i]
elif x==3:
ANS+=L[i]
if cl==2:
return ANS
else:
return ANS^cl1
t=int(input())
for tests in range(t):
print(calc(int(input())))
```
Yes
| 107,179 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider the infinite sequence s of positive integers, created by repeating the following steps:
1. Find the lexicographically smallest triple of positive integers (a, b, c) such that
* a β b β c = 0, where β denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
* a, b, c are not in s.
Here triple of integers (a_1, b_1, c_1) is considered to be lexicographically smaller than triple (a_2, b_2, c_2) if sequence [a_1, b_1, c_1] is lexicographically smaller than sequence [a_2, b_2, c_2].
2. Append a, b, c to s in this order.
3. Go back to the first step.
You have integer n. Find the n-th element of s.
You have to answer t independent test cases.
A sequence a is lexicographically smaller than a sequence b if in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
The first line contains a single integer t (1 β€ t β€ 10^5) β the number of test cases.
Each of the next t lines contains a single integer n (1β€ n β€ 10^{16}) β the position of the element you want to know.
Output
In each of the t lines, output the answer to the corresponding test case.
Example
Input
9
1
2
3
4
5
6
7
8
9
Output
1
2
3
4
8
12
5
10
15
Note
The first elements of s are 1, 2, 3, 4, 8, 12, 5, 10, 15, ...
Submitted Solution:
```
def new_find_pair(n):
stack=[]
while n:
N=n.bit_length()-1
if N%2==0:
i=N//2
stack.append((n,1,i))
n-=4**i
else:
i=N//2
if n>>(2*i)&1==0:
#type2
stack.append((n,2,i))
n-=2*4**i
else:
#type3
stack.append((n,3,i))
n-=3*4**i
a,b,c=0,0,0
for m,t,i in stack[::-1]:
m-=t*4**i
if t==1:
if m==b:
a,b,c=b,c,a
elif m==c:
a,b,c=c,a,b
elif t==2:
if m==a:
a,b,c=c,a,b
elif m==c:
a,b,c=b,c,a
else:
if m==a:
a,b,c=b,c,a
elif m==b:
a,b,c=c,a,b
a+=4**i
b+=2*4**i
c+=3*4**i
return (a,b,c)
def nth_pair(n):
tmp=0
for i in range(28):
tmp+=4**i
if tmp>=n:
tmp-=4**i
first=(n-tmp-1)+4**i
return new_find_pair(first)
def nth_seq(n):
q=(n-1)//3+1
fst=(n-1)%3
res=nth_pair(q)[fst]
return res
import sys
input=sys.stdin.buffer.readline
#n=int(input())
#print(find_pair(n))
#print(nth_pair(n))
Q=int(input())
for _ in range(Q):
n=int(input())
#n=int(input())
r=nth_seq(n)
print(r)
#print(nth_pair(q)[fst])
#print(time.time()-start)
```
Yes
| 107,180 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider the infinite sequence s of positive integers, created by repeating the following steps:
1. Find the lexicographically smallest triple of positive integers (a, b, c) such that
* a β b β c = 0, where β denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
* a, b, c are not in s.
Here triple of integers (a_1, b_1, c_1) is considered to be lexicographically smaller than triple (a_2, b_2, c_2) if sequence [a_1, b_1, c_1] is lexicographically smaller than sequence [a_2, b_2, c_2].
2. Append a, b, c to s in this order.
3. Go back to the first step.
You have integer n. Find the n-th element of s.
You have to answer t independent test cases.
A sequence a is lexicographically smaller than a sequence b if in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
The first line contains a single integer t (1 β€ t β€ 10^5) β the number of test cases.
Each of the next t lines contains a single integer n (1β€ n β€ 10^{16}) β the position of the element you want to know.
Output
In each of the t lines, output the answer to the corresponding test case.
Example
Input
9
1
2
3
4
5
6
7
8
9
Output
1
2
3
4
8
12
5
10
15
Note
The first elements of s are 1, 2, 3, 4, 8, 12, 5, 10, 15, ...
Submitted Solution:
```
def find_answer(n):
trio_num, trio_place = n // 3, n % 3
digit = 1
while trio_num >= 0:
trio_num -= digit
digit *= 4
digit //= 4
trio_num += digit
t1 = digit + trio_num
if trio_place == 0:
return t1
t2, t3 = digit*2, digit*3
new_digit = 1
while new_digit < digit:
x = t1 % 4
if x == 1:
t2 += new_digit * 2
t3 += new_digit * 3
elif x == 2:
t2 += new_digit * 3
t3 += new_digit * 1
elif x == 3:
t2 += new_digit * 1
t3 += new_digit * 2
t1 //= 4*new_digit
new_digit *= 4
return t2 if trio_place == 1 else t3
if __name__ == '__main__':
n = int(input())
ans = []
for i in range(n):
x = int(input())
ans += [find_answer(x-1)]
for x in ans:
print(x)
```
No
| 107,181 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider the infinite sequence s of positive integers, created by repeating the following steps:
1. Find the lexicographically smallest triple of positive integers (a, b, c) such that
* a β b β c = 0, where β denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
* a, b, c are not in s.
Here triple of integers (a_1, b_1, c_1) is considered to be lexicographically smaller than triple (a_2, b_2, c_2) if sequence [a_1, b_1, c_1] is lexicographically smaller than sequence [a_2, b_2, c_2].
2. Append a, b, c to s in this order.
3. Go back to the first step.
You have integer n. Find the n-th element of s.
You have to answer t independent test cases.
A sequence a is lexicographically smaller than a sequence b if in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
The first line contains a single integer t (1 β€ t β€ 10^5) β the number of test cases.
Each of the next t lines contains a single integer n (1β€ n β€ 10^{16}) β the position of the element you want to know.
Output
In each of the t lines, output the answer to the corresponding test case.
Example
Input
9
1
2
3
4
5
6
7
8
9
Output
1
2
3
4
8
12
5
10
15
Note
The first elements of s are 1, 2, 3, 4, 8, 12, 5, 10, 15, ...
Submitted Solution:
```
def find_answer(n):
trio_num, trio_place = n // 3, n % 3
digit = 1
while trio_num >= 0:
trio_num -= digit
digit *= 4
digit //= 4
trio_num += digit
t1 = digit + trio_num
if trio_place == 0:
return t1
t2, t3 = digit*2, digit*3
new_digit = 1
while new_digit < digit:
x = t1 % (4*new_digit)
if x == 1:
t2 += new_digit * 2
t3 += new_digit * 3
elif x == 2:
t2 += new_digit * 3
t3 += new_digit * 1
elif x == 3:
t2 += new_digit * 1
t3 += new_digit * 2
t1 -= x * new_digit
new_digit *= 4
return t2 if trio_place == 1 else t3
if __name__ == '__main__':
'''n = int(input())
ans = []
for i in range(n):
x = int(input())
ans += [find_answer(x-1)]
for x in ans:
print(x)'''
for x in range(1, 20):
print(find_answer(x-1))
```
No
| 107,182 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider the infinite sequence s of positive integers, created by repeating the following steps:
1. Find the lexicographically smallest triple of positive integers (a, b, c) such that
* a β b β c = 0, where β denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
* a, b, c are not in s.
Here triple of integers (a_1, b_1, c_1) is considered to be lexicographically smaller than triple (a_2, b_2, c_2) if sequence [a_1, b_1, c_1] is lexicographically smaller than sequence [a_2, b_2, c_2].
2. Append a, b, c to s in this order.
3. Go back to the first step.
You have integer n. Find the n-th element of s.
You have to answer t independent test cases.
A sequence a is lexicographically smaller than a sequence b if in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
The first line contains a single integer t (1 β€ t β€ 10^5) β the number of test cases.
Each of the next t lines contains a single integer n (1β€ n β€ 10^{16}) β the position of the element you want to know.
Output
In each of the t lines, output the answer to the corresponding test case.
Example
Input
9
1
2
3
4
5
6
7
8
9
Output
1
2
3
4
8
12
5
10
15
Note
The first elements of s are 1, 2, 3, 4, 8, 12, 5, 10, 15, ...
Submitted Solution:
```
def find_answer(n):
trio_num, trio_place = n // 3, n % 3
digit = 1
while trio_num >= 0:
trio_num -= digit
digit *= 4
digit //= 4
trio_num += digit
t1 = digit + trio_num
if trio_place == 0:
return t1
t2, t3 = digit*2, digit*3
new_digit = 1
while new_digit < digit:
x = t1 % (4*new_digit)
if x == 1:
t2 += new_digit * 2
t3 += new_digit * 3
elif x == 2:
t2 += new_digit * 3
t3 += new_digit * 1
elif x == 3:
t2 += new_digit * 1
t3 += new_digit * 2
t1 -= x * new_digit
new_digit *= 4
return t2 if trio_place == 1 else t3
if __name__ == '__main__':
n = int(input())
ans = []
for i in range(n):
x = int(input())
ans += [find_answer(x-1)]
for x in ans:
print(x)
```
No
| 107,183 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider the infinite sequence s of positive integers, created by repeating the following steps:
1. Find the lexicographically smallest triple of positive integers (a, b, c) such that
* a β b β c = 0, where β denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
* a, b, c are not in s.
Here triple of integers (a_1, b_1, c_1) is considered to be lexicographically smaller than triple (a_2, b_2, c_2) if sequence [a_1, b_1, c_1] is lexicographically smaller than sequence [a_2, b_2, c_2].
2. Append a, b, c to s in this order.
3. Go back to the first step.
You have integer n. Find the n-th element of s.
You have to answer t independent test cases.
A sequence a is lexicographically smaller than a sequence b if in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
The first line contains a single integer t (1 β€ t β€ 10^5) β the number of test cases.
Each of the next t lines contains a single integer n (1β€ n β€ 10^{16}) β the position of the element you want to know.
Output
In each of the t lines, output the answer to the corresponding test case.
Example
Input
9
1
2
3
4
5
6
7
8
9
Output
1
2
3
4
8
12
5
10
15
Note
The first elements of s are 1, 2, 3, 4, 8, 12, 5, 10, 15, ...
Submitted Solution:
```
#!/usr/bin/env python3
import sys
input = sys.stdin.readline
from itertools import combinations
t = int(input())
for tt in range(t):
n = int(input()) - 1
mod3 = n % 3
n = n // 3
i = 0
while n >= 4**i:
n -= 4**i
i += 1
if i == 0:
a = 1; b = 2; c = 3
else:
a = 4**i + n
left = 0
right = 4**i
b = 4**i * 2
t = n
while right - left > 1:
mid1 = (left * 3 + right) // 4
mid2 = (left * 2 + right * 2) // 4
mid3 = (left + right * 3) // 4
rng = right - left
if left <= t < mid1:
right = mid1
b += 0
elif mid1 <= t < mid2:
left = mid1
right = mid2
b += rng // 4 * 2
elif mid2 <= t < mid3:
left = mid2
right = mid3
b += rng // 4 * 3
else:
left = mid3
b += rng // 4 * 1
t %= rng // 4
c = a ^ b
# print("#", tt + 1)
if mod3 == 0:
print(a)
elif mod3 == 1:
print(b)
else:
print(c)
# not_used = set([item for item in range(1, 200)])
# for i in range(25):
# li = list(not_used)
# li.sort()
# for comb in combinations(li, 3):
# if comb[0] ^ comb[1] ^ comb[2] == 0:
# print(comb)
# not_used.remove(comb[0])
# not_used.remove(comb[1])
# not_used.remove(comb[2])
# break
```
No
| 107,184 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n athletes in front of you. Athletes are numbered from 1 to n from left to right. You know the strength of each athlete β the athlete number i has the strength s_i.
You want to split all athletes into two teams. Each team must have at least one athlete, and each athlete must be exactly in one team.
You want the strongest athlete from the first team to differ as little as possible from the weakest athlete from the second team. Formally, you want to split the athletes into two teams A and B so that the value |max(A) - min(B)| is as small as possible, where max(A) is the maximum strength of an athlete from team A, and min(B) is the minimum strength of an athlete from team B.
For example, if n=5 and the strength of the athletes is s=[3, 1, 2, 6, 4], then one of the possible split into teams is:
* first team: A = [1, 2, 4],
* second team: B = [3, 6].
In this case, the value |max(A) - min(B)| will be equal to |4-3|=1. This example illustrates one of the ways of optimal split into two teams.
Print the minimum value |max(A) - min(B)|.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Then t test cases follow.
Each test case consists of two lines.
The first line contains positive integer n (2 β€ n β€ 50) β number of athletes.
The second line contains n positive integers s_1, s_2, β¦, s_n (1 β€ s_i β€ 1000), where s_i β is the strength of the i-th athlete. Please note that s values may not be distinct.
Output
For each test case print one integer β the minimum value of |max(A) - min(B)| with the optimal split of all athletes into two teams. Each of the athletes must be a member of exactly one of the two teams.
Example
Input
5
5
3 1 2 6 4
6
2 1 3 2 4 3
4
7 9 3 1
2
1 1000
3
100 150 200
Output
1
0
2
999
50
Note
The first test case was explained in the statement. In the second test case, one of the optimal splits is A=[2, 1], B=[3, 2, 4, 3], so the answer is |2-2|=0.
Tags: greedy, sortings
Correct Solution:
```
t=int(input())
for i in range(t):
n=int(input())
s=list(map(int, input().split()))
l=[]
for j in range(len(s)-1):
a=s[j]
for k in range(j+1,len(s)):
d=abs(s[k]-a)
l.append(d)
print(min(l))
```
| 107,185 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n athletes in front of you. Athletes are numbered from 1 to n from left to right. You know the strength of each athlete β the athlete number i has the strength s_i.
You want to split all athletes into two teams. Each team must have at least one athlete, and each athlete must be exactly in one team.
You want the strongest athlete from the first team to differ as little as possible from the weakest athlete from the second team. Formally, you want to split the athletes into two teams A and B so that the value |max(A) - min(B)| is as small as possible, where max(A) is the maximum strength of an athlete from team A, and min(B) is the minimum strength of an athlete from team B.
For example, if n=5 and the strength of the athletes is s=[3, 1, 2, 6, 4], then one of the possible split into teams is:
* first team: A = [1, 2, 4],
* second team: B = [3, 6].
In this case, the value |max(A) - min(B)| will be equal to |4-3|=1. This example illustrates one of the ways of optimal split into two teams.
Print the minimum value |max(A) - min(B)|.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Then t test cases follow.
Each test case consists of two lines.
The first line contains positive integer n (2 β€ n β€ 50) β number of athletes.
The second line contains n positive integers s_1, s_2, β¦, s_n (1 β€ s_i β€ 1000), where s_i β is the strength of the i-th athlete. Please note that s values may not be distinct.
Output
For each test case print one integer β the minimum value of |max(A) - min(B)| with the optimal split of all athletes into two teams. Each of the athletes must be a member of exactly one of the two teams.
Example
Input
5
5
3 1 2 6 4
6
2 1 3 2 4 3
4
7 9 3 1
2
1 1000
3
100 150 200
Output
1
0
2
999
50
Note
The first test case was explained in the statement. In the second test case, one of the optimal splits is A=[2, 1], B=[3, 2, 4, 3], so the answer is |2-2|=0.
Tags: greedy, sortings
Correct Solution:
```
n=int(input());o=[]
for i in range(n):
m=int(input())
l=[int(x) for x in input().split()]
l.sort()
o+=[str(min(l[i+1]-l[i] for i in range(m-1)))]
print('\n'.join(o))
```
| 107,186 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n athletes in front of you. Athletes are numbered from 1 to n from left to right. You know the strength of each athlete β the athlete number i has the strength s_i.
You want to split all athletes into two teams. Each team must have at least one athlete, and each athlete must be exactly in one team.
You want the strongest athlete from the first team to differ as little as possible from the weakest athlete from the second team. Formally, you want to split the athletes into two teams A and B so that the value |max(A) - min(B)| is as small as possible, where max(A) is the maximum strength of an athlete from team A, and min(B) is the minimum strength of an athlete from team B.
For example, if n=5 and the strength of the athletes is s=[3, 1, 2, 6, 4], then one of the possible split into teams is:
* first team: A = [1, 2, 4],
* second team: B = [3, 6].
In this case, the value |max(A) - min(B)| will be equal to |4-3|=1. This example illustrates one of the ways of optimal split into two teams.
Print the minimum value |max(A) - min(B)|.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Then t test cases follow.
Each test case consists of two lines.
The first line contains positive integer n (2 β€ n β€ 50) β number of athletes.
The second line contains n positive integers s_1, s_2, β¦, s_n (1 β€ s_i β€ 1000), where s_i β is the strength of the i-th athlete. Please note that s values may not be distinct.
Output
For each test case print one integer β the minimum value of |max(A) - min(B)| with the optimal split of all athletes into two teams. Each of the athletes must be a member of exactly one of the two teams.
Example
Input
5
5
3 1 2 6 4
6
2 1 3 2 4 3
4
7 9 3 1
2
1 1000
3
100 150 200
Output
1
0
2
999
50
Note
The first test case was explained in the statement. In the second test case, one of the optimal splits is A=[2, 1], B=[3, 2, 4, 3], so the answer is |2-2|=0.
Tags: greedy, sortings
Correct Solution:
```
t=int(input())
while(t):
t-=1
input()
athelete_code=[int(i) for i in input().split(" ")]
athelete_code=sorted(athelete_code)
min_dif=athelete_code[1]-athelete_code[0]
for i in range(1,(len(athelete_code)-1)):
if (athelete_code[i+1]-athelete_code[i])<min_dif:
#print(athelete_code[i+1]-athelete_code[i],athelete_code[i],athelete_code[i+1])
min_dif=athelete_code[i+1]-athelete_code[i]
print(min_dif)
```
| 107,187 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n athletes in front of you. Athletes are numbered from 1 to n from left to right. You know the strength of each athlete β the athlete number i has the strength s_i.
You want to split all athletes into two teams. Each team must have at least one athlete, and each athlete must be exactly in one team.
You want the strongest athlete from the first team to differ as little as possible from the weakest athlete from the second team. Formally, you want to split the athletes into two teams A and B so that the value |max(A) - min(B)| is as small as possible, where max(A) is the maximum strength of an athlete from team A, and min(B) is the minimum strength of an athlete from team B.
For example, if n=5 and the strength of the athletes is s=[3, 1, 2, 6, 4], then one of the possible split into teams is:
* first team: A = [1, 2, 4],
* second team: B = [3, 6].
In this case, the value |max(A) - min(B)| will be equal to |4-3|=1. This example illustrates one of the ways of optimal split into two teams.
Print the minimum value |max(A) - min(B)|.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Then t test cases follow.
Each test case consists of two lines.
The first line contains positive integer n (2 β€ n β€ 50) β number of athletes.
The second line contains n positive integers s_1, s_2, β¦, s_n (1 β€ s_i β€ 1000), where s_i β is the strength of the i-th athlete. Please note that s values may not be distinct.
Output
For each test case print one integer β the minimum value of |max(A) - min(B)| with the optimal split of all athletes into two teams. Each of the athletes must be a member of exactly one of the two teams.
Example
Input
5
5
3 1 2 6 4
6
2 1 3 2 4 3
4
7 9 3 1
2
1 1000
3
100 150 200
Output
1
0
2
999
50
Note
The first test case was explained in the statement. In the second test case, one of the optimal splits is A=[2, 1], B=[3, 2, 4, 3], so the answer is |2-2|=0.
Tags: greedy, sortings
Correct Solution:
```
from sys import stdin,stdout
import math
from collections import Counter,deque
L=lambda:list(map(int, stdin.readline().strip().split()))
M=lambda:map(int, stdin.readline().strip().split())
I=lambda:int(stdin.readline().strip())
IN=lambda:stdin.readline().strip()
C=lambda:stdin.readline().strip().split()
mod=1000000007
#Keymax = max(Tv, key=Tv.get)
def s(a):print(" ".join(list(map(str,a))))
#______________________-------------------------------_____________________#
#I_am_pavan
for i in range(I()):
n=I()
a=sorted(L())
m=a[1]-a[0]
for i in range(n-1):
m=min(m,a[i+1]-a[i])
print(m)
```
| 107,188 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n athletes in front of you. Athletes are numbered from 1 to n from left to right. You know the strength of each athlete β the athlete number i has the strength s_i.
You want to split all athletes into two teams. Each team must have at least one athlete, and each athlete must be exactly in one team.
You want the strongest athlete from the first team to differ as little as possible from the weakest athlete from the second team. Formally, you want to split the athletes into two teams A and B so that the value |max(A) - min(B)| is as small as possible, where max(A) is the maximum strength of an athlete from team A, and min(B) is the minimum strength of an athlete from team B.
For example, if n=5 and the strength of the athletes is s=[3, 1, 2, 6, 4], then one of the possible split into teams is:
* first team: A = [1, 2, 4],
* second team: B = [3, 6].
In this case, the value |max(A) - min(B)| will be equal to |4-3|=1. This example illustrates one of the ways of optimal split into two teams.
Print the minimum value |max(A) - min(B)|.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Then t test cases follow.
Each test case consists of two lines.
The first line contains positive integer n (2 β€ n β€ 50) β number of athletes.
The second line contains n positive integers s_1, s_2, β¦, s_n (1 β€ s_i β€ 1000), where s_i β is the strength of the i-th athlete. Please note that s values may not be distinct.
Output
For each test case print one integer β the minimum value of |max(A) - min(B)| with the optimal split of all athletes into two teams. Each of the athletes must be a member of exactly one of the two teams.
Example
Input
5
5
3 1 2 6 4
6
2 1 3 2 4 3
4
7 9 3 1
2
1 1000
3
100 150 200
Output
1
0
2
999
50
Note
The first test case was explained in the statement. In the second test case, one of the optimal splits is A=[2, 1], B=[3, 2, 4, 3], so the answer is |2-2|=0.
Tags: greedy, sortings
Correct Solution:
```
t = int(input())
for i in range(t):
n = int(input())
si = list(map(int,input().split()))
si.sort()
res = si[-1]
for i in range(1,n):
res = min(res, si[i] - si[i-1])
print(res)
```
| 107,189 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n athletes in front of you. Athletes are numbered from 1 to n from left to right. You know the strength of each athlete β the athlete number i has the strength s_i.
You want to split all athletes into two teams. Each team must have at least one athlete, and each athlete must be exactly in one team.
You want the strongest athlete from the first team to differ as little as possible from the weakest athlete from the second team. Formally, you want to split the athletes into two teams A and B so that the value |max(A) - min(B)| is as small as possible, where max(A) is the maximum strength of an athlete from team A, and min(B) is the minimum strength of an athlete from team B.
For example, if n=5 and the strength of the athletes is s=[3, 1, 2, 6, 4], then one of the possible split into teams is:
* first team: A = [1, 2, 4],
* second team: B = [3, 6].
In this case, the value |max(A) - min(B)| will be equal to |4-3|=1. This example illustrates one of the ways of optimal split into two teams.
Print the minimum value |max(A) - min(B)|.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Then t test cases follow.
Each test case consists of two lines.
The first line contains positive integer n (2 β€ n β€ 50) β number of athletes.
The second line contains n positive integers s_1, s_2, β¦, s_n (1 β€ s_i β€ 1000), where s_i β is the strength of the i-th athlete. Please note that s values may not be distinct.
Output
For each test case print one integer β the minimum value of |max(A) - min(B)| with the optimal split of all athletes into two teams. Each of the athletes must be a member of exactly one of the two teams.
Example
Input
5
5
3 1 2 6 4
6
2 1 3 2 4 3
4
7 9 3 1
2
1 1000
3
100 150 200
Output
1
0
2
999
50
Note
The first test case was explained in the statement. In the second test case, one of the optimal splits is A=[2, 1], B=[3, 2, 4, 3], so the answer is |2-2|=0.
Tags: greedy, sortings
Correct Solution:
```
def solve():
n = int(input())
arr = [int(i) for i in input().split()]
arr.sort()
diff = 1000
for i in range(n - 1):
diff = min(diff, arr[i + 1] - arr[i])
return diff
if __name__ == '__main__':
for _ in range(int(input())):
print(solve())
```
| 107,190 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n athletes in front of you. Athletes are numbered from 1 to n from left to right. You know the strength of each athlete β the athlete number i has the strength s_i.
You want to split all athletes into two teams. Each team must have at least one athlete, and each athlete must be exactly in one team.
You want the strongest athlete from the first team to differ as little as possible from the weakest athlete from the second team. Formally, you want to split the athletes into two teams A and B so that the value |max(A) - min(B)| is as small as possible, where max(A) is the maximum strength of an athlete from team A, and min(B) is the minimum strength of an athlete from team B.
For example, if n=5 and the strength of the athletes is s=[3, 1, 2, 6, 4], then one of the possible split into teams is:
* first team: A = [1, 2, 4],
* second team: B = [3, 6].
In this case, the value |max(A) - min(B)| will be equal to |4-3|=1. This example illustrates one of the ways of optimal split into two teams.
Print the minimum value |max(A) - min(B)|.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Then t test cases follow.
Each test case consists of two lines.
The first line contains positive integer n (2 β€ n β€ 50) β number of athletes.
The second line contains n positive integers s_1, s_2, β¦, s_n (1 β€ s_i β€ 1000), where s_i β is the strength of the i-th athlete. Please note that s values may not be distinct.
Output
For each test case print one integer β the minimum value of |max(A) - min(B)| with the optimal split of all athletes into two teams. Each of the athletes must be a member of exactly one of the two teams.
Example
Input
5
5
3 1 2 6 4
6
2 1 3 2 4 3
4
7 9 3 1
2
1 1000
3
100 150 200
Output
1
0
2
999
50
Note
The first test case was explained in the statement. In the second test case, one of the optimal splits is A=[2, 1], B=[3, 2, 4, 3], so the answer is |2-2|=0.
Tags: greedy, sortings
Correct Solution:
```
for s in [*open(0)][2::2]:a=sorted(map(int,s.split()));print(min([j-i for i, j in zip(a,a[1:])]))
```
| 107,191 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n athletes in front of you. Athletes are numbered from 1 to n from left to right. You know the strength of each athlete β the athlete number i has the strength s_i.
You want to split all athletes into two teams. Each team must have at least one athlete, and each athlete must be exactly in one team.
You want the strongest athlete from the first team to differ as little as possible from the weakest athlete from the second team. Formally, you want to split the athletes into two teams A and B so that the value |max(A) - min(B)| is as small as possible, where max(A) is the maximum strength of an athlete from team A, and min(B) is the minimum strength of an athlete from team B.
For example, if n=5 and the strength of the athletes is s=[3, 1, 2, 6, 4], then one of the possible split into teams is:
* first team: A = [1, 2, 4],
* second team: B = [3, 6].
In this case, the value |max(A) - min(B)| will be equal to |4-3|=1. This example illustrates one of the ways of optimal split into two teams.
Print the minimum value |max(A) - min(B)|.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Then t test cases follow.
Each test case consists of two lines.
The first line contains positive integer n (2 β€ n β€ 50) β number of athletes.
The second line contains n positive integers s_1, s_2, β¦, s_n (1 β€ s_i β€ 1000), where s_i β is the strength of the i-th athlete. Please note that s values may not be distinct.
Output
For each test case print one integer β the minimum value of |max(A) - min(B)| with the optimal split of all athletes into two teams. Each of the athletes must be a member of exactly one of the two teams.
Example
Input
5
5
3 1 2 6 4
6
2 1 3 2 4 3
4
7 9 3 1
2
1 1000
3
100 150 200
Output
1
0
2
999
50
Note
The first test case was explained in the statement. In the second test case, one of the optimal splits is A=[2, 1], B=[3, 2, 4, 3], so the answer is |2-2|=0.
Tags: greedy, sortings
Correct Solution:
```
test=int(input())
for i in range(test):
n=int(input())
a=[int(x) for x in input().split()]
temp=1001
for i in range(n):
for j in range(i+1,n):
temp=min(temp,abs(a[i]-a[j]))
print(temp)
```
| 107,192 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n athletes in front of you. Athletes are numbered from 1 to n from left to right. You know the strength of each athlete β the athlete number i has the strength s_i.
You want to split all athletes into two teams. Each team must have at least one athlete, and each athlete must be exactly in one team.
You want the strongest athlete from the first team to differ as little as possible from the weakest athlete from the second team. Formally, you want to split the athletes into two teams A and B so that the value |max(A) - min(B)| is as small as possible, where max(A) is the maximum strength of an athlete from team A, and min(B) is the minimum strength of an athlete from team B.
For example, if n=5 and the strength of the athletes is s=[3, 1, 2, 6, 4], then one of the possible split into teams is:
* first team: A = [1, 2, 4],
* second team: B = [3, 6].
In this case, the value |max(A) - min(B)| will be equal to |4-3|=1. This example illustrates one of the ways of optimal split into two teams.
Print the minimum value |max(A) - min(B)|.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Then t test cases follow.
Each test case consists of two lines.
The first line contains positive integer n (2 β€ n β€ 50) β number of athletes.
The second line contains n positive integers s_1, s_2, β¦, s_n (1 β€ s_i β€ 1000), where s_i β is the strength of the i-th athlete. Please note that s values may not be distinct.
Output
For each test case print one integer β the minimum value of |max(A) - min(B)| with the optimal split of all athletes into two teams. Each of the athletes must be a member of exactly one of the two teams.
Example
Input
5
5
3 1 2 6 4
6
2 1 3 2 4 3
4
7 9 3 1
2
1 1000
3
100 150 200
Output
1
0
2
999
50
Note
The first test case was explained in the statement. In the second test case, one of the optimal splits is A=[2, 1], B=[3, 2, 4, 3], so the answer is |2-2|=0.
Submitted Solution:
```
# python3 q2.py < test.txt
t = int(input())
for i in range(t):
n = int(input())
a = [int(j) for j in input().split()]
a.sort()
diff = float('inf')
for j in range(1, len(a)):
diff = min(diff, a[j]- a[j-1])
print(diff)
```
Yes
| 107,193 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n athletes in front of you. Athletes are numbered from 1 to n from left to right. You know the strength of each athlete β the athlete number i has the strength s_i.
You want to split all athletes into two teams. Each team must have at least one athlete, and each athlete must be exactly in one team.
You want the strongest athlete from the first team to differ as little as possible from the weakest athlete from the second team. Formally, you want to split the athletes into two teams A and B so that the value |max(A) - min(B)| is as small as possible, where max(A) is the maximum strength of an athlete from team A, and min(B) is the minimum strength of an athlete from team B.
For example, if n=5 and the strength of the athletes is s=[3, 1, 2, 6, 4], then one of the possible split into teams is:
* first team: A = [1, 2, 4],
* second team: B = [3, 6].
In this case, the value |max(A) - min(B)| will be equal to |4-3|=1. This example illustrates one of the ways of optimal split into two teams.
Print the minimum value |max(A) - min(B)|.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Then t test cases follow.
Each test case consists of two lines.
The first line contains positive integer n (2 β€ n β€ 50) β number of athletes.
The second line contains n positive integers s_1, s_2, β¦, s_n (1 β€ s_i β€ 1000), where s_i β is the strength of the i-th athlete. Please note that s values may not be distinct.
Output
For each test case print one integer β the minimum value of |max(A) - min(B)| with the optimal split of all athletes into two teams. Each of the athletes must be a member of exactly one of the two teams.
Example
Input
5
5
3 1 2 6 4
6
2 1 3 2 4 3
4
7 9 3 1
2
1 1000
3
100 150 200
Output
1
0
2
999
50
Note
The first test case was explained in the statement. In the second test case, one of the optimal splits is A=[2, 1], B=[3, 2, 4, 3], so the answer is |2-2|=0.
Submitted Solution:
```
t=int(input())
for i in range(t):
n=int(input())
l=list(map(int,input().split()))
l.sort()
min1=l[1]-l[0]
for i in range(2,len(l)):
min1=min(min1,l[i]-l[i-1])
print(min1)
```
Yes
| 107,194 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n athletes in front of you. Athletes are numbered from 1 to n from left to right. You know the strength of each athlete β the athlete number i has the strength s_i.
You want to split all athletes into two teams. Each team must have at least one athlete, and each athlete must be exactly in one team.
You want the strongest athlete from the first team to differ as little as possible from the weakest athlete from the second team. Formally, you want to split the athletes into two teams A and B so that the value |max(A) - min(B)| is as small as possible, where max(A) is the maximum strength of an athlete from team A, and min(B) is the minimum strength of an athlete from team B.
For example, if n=5 and the strength of the athletes is s=[3, 1, 2, 6, 4], then one of the possible split into teams is:
* first team: A = [1, 2, 4],
* second team: B = [3, 6].
In this case, the value |max(A) - min(B)| will be equal to |4-3|=1. This example illustrates one of the ways of optimal split into two teams.
Print the minimum value |max(A) - min(B)|.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Then t test cases follow.
Each test case consists of two lines.
The first line contains positive integer n (2 β€ n β€ 50) β number of athletes.
The second line contains n positive integers s_1, s_2, β¦, s_n (1 β€ s_i β€ 1000), where s_i β is the strength of the i-th athlete. Please note that s values may not be distinct.
Output
For each test case print one integer β the minimum value of |max(A) - min(B)| with the optimal split of all athletes into two teams. Each of the athletes must be a member of exactly one of the two teams.
Example
Input
5
5
3 1 2 6 4
6
2 1 3 2 4 3
4
7 9 3 1
2
1 1000
3
100 150 200
Output
1
0
2
999
50
Note
The first test case was explained in the statement. In the second test case, one of the optimal splits is A=[2, 1], B=[3, 2, 4, 3], so the answer is |2-2|=0.
Submitted Solution:
```
# hey stalker
n=int(input())
for i in range(n):
m=int(input())
a=[int(i) for i in input().split()]
b=sorted(a)
max=10**9
for j in range(0,len(b)-1):
d=b[j+1]-b[j]
if(d<max):
max=d
print(max)
```
Yes
| 107,195 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n athletes in front of you. Athletes are numbered from 1 to n from left to right. You know the strength of each athlete β the athlete number i has the strength s_i.
You want to split all athletes into two teams. Each team must have at least one athlete, and each athlete must be exactly in one team.
You want the strongest athlete from the first team to differ as little as possible from the weakest athlete from the second team. Formally, you want to split the athletes into two teams A and B so that the value |max(A) - min(B)| is as small as possible, where max(A) is the maximum strength of an athlete from team A, and min(B) is the minimum strength of an athlete from team B.
For example, if n=5 and the strength of the athletes is s=[3, 1, 2, 6, 4], then one of the possible split into teams is:
* first team: A = [1, 2, 4],
* second team: B = [3, 6].
In this case, the value |max(A) - min(B)| will be equal to |4-3|=1. This example illustrates one of the ways of optimal split into two teams.
Print the minimum value |max(A) - min(B)|.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Then t test cases follow.
Each test case consists of two lines.
The first line contains positive integer n (2 β€ n β€ 50) β number of athletes.
The second line contains n positive integers s_1, s_2, β¦, s_n (1 β€ s_i β€ 1000), where s_i β is the strength of the i-th athlete. Please note that s values may not be distinct.
Output
For each test case print one integer β the minimum value of |max(A) - min(B)| with the optimal split of all athletes into two teams. Each of the athletes must be a member of exactly one of the two teams.
Example
Input
5
5
3 1 2 6 4
6
2 1 3 2 4 3
4
7 9 3 1
2
1 1000
3
100 150 200
Output
1
0
2
999
50
Note
The first test case was explained in the statement. In the second test case, one of the optimal splits is A=[2, 1], B=[3, 2, 4, 3], so the answer is |2-2|=0.
Submitted Solution:
```
# from math import factorial as fac
from collections import defaultdict
# from copy import deepcopy
import sys, math
f = None
try:
f = open('q1.input', 'r')
except IOError:
f = sys.stdin
if 'xrange' in dir(__builtins__):
range = xrange
# print(f.readline())
# sys.setrecursionlimit(10**5)
def print_case_iterable(case_num, iterable):
print("Case #{}: {}".format(case_num," ".join(map(str,iterable))))
def print_case_number(case_num, iterable):
print("Case #{}: {}".format(case_num,iterable))
def print_iterable(A):
print (' '.join(A))
def read_int():
return int(f.readline().strip())
def read_int_array():
return [int(x) for x in f.readline().strip().split(" ")]
def rns():
a = [x for x in f.readline().split(" ")]
return int(a[0]), a[1].strip()
def read_string():
return list(f.readline().strip())
def bi(x):
return bin(x)[2:]
from copy import deepcopy
def solution(a,n):
a.sort()
min_diff = 10**10
for i in range(1,n):
min_diff = min(min_diff,a[i]-a[i-1])
return min_diff
def main():
T = read_int()
for i in range(T):
# s = read_string()
# p = read_int_array()
n = read_int()
a = read_int_array()
x = solution(a,n)
if 'xrange' not in dir(__builtins__):
print(x)
else:
print >>output,str(x)# "Case #"+str(i+1)+':',
if 'xrange' in dir(__builtins__):
print(output.getvalue())
output.close()
if 'xrange' in dir(__builtins__):
import cStringIO
output = cStringIO.StringIO()
#example usage:
# for l in res:
# print >>output, str(len(l)) + ' ' + ' '.join(l)
if __name__ == '__main__':
main()
'''stuff you should look for
* int overflow, array bounds
* special cases (n=1?)
* do smth instead of nothing and stay organized
* WRITE STUFF DOWN
* BITS - THINK HOW TO MASK PROPERLY
* PERMUTATIONS - PARITY AND CYCLES
* Think simple, if it becomes over complicated, try to look at it from a different perspective.
* Have fun!!!
* TRY FIXING SOMETHING, and then maybe binary search around it.
* Remember heaps.
* Remember how to add a value to a segment when using prefix sum.
suppose you have an array[1,2,3,4,5] and you want to add 3 to array[1:4]. Then just
add 3 to A[1], and decrease 3 from A[4]. Let's look at what happens:
original prefixsums is [1,3,6,10,15]
array -> [1,5,3,4,2] and prefix sums are [1,6,9,13,15]
As you see, exactly +3 in A[1:4]
*** The previous method can help checking how many x,y you can choose to get s=x+y from two arrays.
'''
'''
binary search
while(r - l > 1) {
ll mid = l + (r - l) / 2;
solve(mid);
ll sum = 0;
for (int i = 0; i < n; i++)
sum += b[i];
if (sum <= k)
r = mid;
else
l = mid;
}
'''
```
Yes
| 107,196 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n athletes in front of you. Athletes are numbered from 1 to n from left to right. You know the strength of each athlete β the athlete number i has the strength s_i.
You want to split all athletes into two teams. Each team must have at least one athlete, and each athlete must be exactly in one team.
You want the strongest athlete from the first team to differ as little as possible from the weakest athlete from the second team. Formally, you want to split the athletes into two teams A and B so that the value |max(A) - min(B)| is as small as possible, where max(A) is the maximum strength of an athlete from team A, and min(B) is the minimum strength of an athlete from team B.
For example, if n=5 and the strength of the athletes is s=[3, 1, 2, 6, 4], then one of the possible split into teams is:
* first team: A = [1, 2, 4],
* second team: B = [3, 6].
In this case, the value |max(A) - min(B)| will be equal to |4-3|=1. This example illustrates one of the ways of optimal split into two teams.
Print the minimum value |max(A) - min(B)|.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Then t test cases follow.
Each test case consists of two lines.
The first line contains positive integer n (2 β€ n β€ 50) β number of athletes.
The second line contains n positive integers s_1, s_2, β¦, s_n (1 β€ s_i β€ 1000), where s_i β is the strength of the i-th athlete. Please note that s values may not be distinct.
Output
For each test case print one integer β the minimum value of |max(A) - min(B)| with the optimal split of all athletes into two teams. Each of the athletes must be a member of exactly one of the two teams.
Example
Input
5
5
3 1 2 6 4
6
2 1 3 2 4 3
4
7 9 3 1
2
1 1000
3
100 150 200
Output
1
0
2
999
50
Note
The first test case was explained in the statement. In the second test case, one of the optimal splits is A=[2, 1], B=[3, 2, 4, 3], so the answer is |2-2|=0.
Submitted Solution:
```
t=int(input())
for e in range(t):
n=int(input())
l=list(map(int,input().split()))
l.sort()
m=l[1]-l[0]
for i in range(1,n):
m=min(m,l[i]-l[i-1])
print(m)
```
No
| 107,197 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n athletes in front of you. Athletes are numbered from 1 to n from left to right. You know the strength of each athlete β the athlete number i has the strength s_i.
You want to split all athletes into two teams. Each team must have at least one athlete, and each athlete must be exactly in one team.
You want the strongest athlete from the first team to differ as little as possible from the weakest athlete from the second team. Formally, you want to split the athletes into two teams A and B so that the value |max(A) - min(B)| is as small as possible, where max(A) is the maximum strength of an athlete from team A, and min(B) is the minimum strength of an athlete from team B.
For example, if n=5 and the strength of the athletes is s=[3, 1, 2, 6, 4], then one of the possible split into teams is:
* first team: A = [1, 2, 4],
* second team: B = [3, 6].
In this case, the value |max(A) - min(B)| will be equal to |4-3|=1. This example illustrates one of the ways of optimal split into two teams.
Print the minimum value |max(A) - min(B)|.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Then t test cases follow.
Each test case consists of two lines.
The first line contains positive integer n (2 β€ n β€ 50) β number of athletes.
The second line contains n positive integers s_1, s_2, β¦, s_n (1 β€ s_i β€ 1000), where s_i β is the strength of the i-th athlete. Please note that s values may not be distinct.
Output
For each test case print one integer β the minimum value of |max(A) - min(B)| with the optimal split of all athletes into two teams. Each of the athletes must be a member of exactly one of the two teams.
Example
Input
5
5
3 1 2 6 4
6
2 1 3 2 4 3
4
7 9 3 1
2
1 1000
3
100 150 200
Output
1
0
2
999
50
Note
The first test case was explained in the statement. In the second test case, one of the optimal splits is A=[2, 1], B=[3, 2, 4, 3], so the answer is |2-2|=0.
Submitted Solution:
```
t=int(input())
for i in range(t):
m=int(input())
a=list(map(int,input().split()))
a.sort()
b=sorted(set(a))
n=len(b)
if(n%2!=0):
c=(n+1)//2-1
else:
c=n//2-1
p=a.index(b[c])
c=abs(a[0]-min(a[1:]))
d=a[p+1]-a[p]
print(min(d,c))
```
No
| 107,198 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n athletes in front of you. Athletes are numbered from 1 to n from left to right. You know the strength of each athlete β the athlete number i has the strength s_i.
You want to split all athletes into two teams. Each team must have at least one athlete, and each athlete must be exactly in one team.
You want the strongest athlete from the first team to differ as little as possible from the weakest athlete from the second team. Formally, you want to split the athletes into two teams A and B so that the value |max(A) - min(B)| is as small as possible, where max(A) is the maximum strength of an athlete from team A, and min(B) is the minimum strength of an athlete from team B.
For example, if n=5 and the strength of the athletes is s=[3, 1, 2, 6, 4], then one of the possible split into teams is:
* first team: A = [1, 2, 4],
* second team: B = [3, 6].
In this case, the value |max(A) - min(B)| will be equal to |4-3|=1. This example illustrates one of the ways of optimal split into two teams.
Print the minimum value |max(A) - min(B)|.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Then t test cases follow.
Each test case consists of two lines.
The first line contains positive integer n (2 β€ n β€ 50) β number of athletes.
The second line contains n positive integers s_1, s_2, β¦, s_n (1 β€ s_i β€ 1000), where s_i β is the strength of the i-th athlete. Please note that s values may not be distinct.
Output
For each test case print one integer β the minimum value of |max(A) - min(B)| with the optimal split of all athletes into two teams. Each of the athletes must be a member of exactly one of the two teams.
Example
Input
5
5
3 1 2 6 4
6
2 1 3 2 4 3
4
7 9 3 1
2
1 1000
3
100 150 200
Output
1
0
2
999
50
Note
The first test case was explained in the statement. In the second test case, one of the optimal splits is A=[2, 1], B=[3, 2, 4, 3], so the answer is |2-2|=0.
Submitted Solution:
```
for t in range(int(input())):
n = int(input())
l = list(map(int, input().split()))
l.sort()
s = len(l)
b = []
a = []
x = (s//2)-1
if(x == 0):
x = 1
for i in range(x):
a.append(l[i])
for i in range(x, s):
b.append(l[i])
print(abs(max(a) - min(b)))
```
No
| 107,199 |
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