text stringlengths 216 39.6k | conversation_id int64 219 108k | embedding list | cluster int64 11 11 |
|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Example
Input
4 4
1 2 3
1 3 3
2 3 3
2 4 3
Output
1 3
Submitted Solution:
```
N, M = map(int, input().split())
E0 = []
for i in range(M):
S, D, C = map(int, input().split())
E0.append((C, S, D))
E0.sort()
*parent, = range(N)
def root(x):
if x == parent[x]:
return x
y = parent[x] = root(parent[x])
return y
def unite(x, y):
px = root(x); py = root(y)
if px < py:
parent[py] = px
else:
parent[px] = py
cnt = 0; base = 10**9
used = set()
for i in range(M):
C, S, D = E0[i]
if root(S-1) != root(D-1):
unite(S-1, D-1)
used.add((S-1, D-1))
cnt += 1
if cnt == N-1:
base = C
ES = {}; CM = {}
for i in range(M):
C, S, D = E0[i]
if C <= base:
ES.setdefault(C, set()).add((S-1, D-1))
CM[S-1, D-1] = C
def bridge(G, N):
result = set()
label = [0]*N
cost = [0]*N
def dfs(u, p):
res = 0
for v in G[u]:
if v == p:
continue
if label[v]:
cost[v] -= 1
res += 1
else:
label[v] = 1
r = dfs(v, u)
if r == 0:
result.add((u, v) if u < v else (v, u))
res += r
res += cost[u]
return res
for v in range(N):
if not label[v]:
label[v] = 1
dfs(v, -1)
return result
G0 = [[] for i in range(N)]
cnt = 0; ans = 0
for C in sorted(ES):
G = [e[:] for e in G0]
for S, D in ES[C]:
G[S].append(D)
G[D].append(S)
if (S, D) in used:
G0[S].append(D)
G0[D].append(S)
B = bridge(G, N)
cnt += len(B & ES[C])
ans += sum(map(CM.__getitem__, B & ES[C]))
print(cnt, ans)
```
No
| 45,335 | [
0.2235107421875,
-0.1102294921875,
-0.031890869140625,
-0.2193603515625,
-0.171875,
-0.2164306640625,
-0.1629638671875,
0.1358642578125,
0.32275390625,
0.69580078125,
0.677734375,
0.00786590576171875,
0.237060546875,
-0.5361328125,
-0.24365234375,
0.10302734375,
-0.43212890625,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Example
Input
4 4
1 2 3
1 3 3
2 3 3
2 4 3
Output
1 3
Submitted Solution:
```
N, M = map(int, input().split())
E0 = []
for i in range(M):
S, D, C = map(int, input().split())
E0.append((C, S, D))
E0.sort()
*parent, = range(N)
def root(x):
if x == parent[x]:
return x
y = parent[x] = root(parent[x])
return y
def unite(x, y):
px = root(x); py = root(y)
if px < py:
parent[py] = px
else:
parent[px] = py
G = [[] for i in range(N)]
E1 = set()
cnt = 0; cost = 0; base = None
for i in range(M):
C, S, D = E0[i]
if root(S-1) != root(D-1):
unite(S-1, D-1)
cost += C
cnt += 1
if cnt == N-1:
base = C
cnt = ans = 0
for C, S, D in E0:
if C <= base:
G[S-1].append(D-1)
G[D-1].append(S-1)
cnt += 1
if C == base:
E1.add((S-1, D-1))
else:
ans += C
def bridge(G, N):
result = set()
label = [0]*N
cost = [0]*N
def dfs(u, p):
res = 0
for v in G[u]:
if v == p:
continue
if label[v]:
cost[v] -= 1
res += 1
else:
label[v] = 1
r = dfs(v, u)
if r == 0:
result.add((u, v) if u < v else (v, u))
res += r
res += cost[u]
return res
label[0] = 1
dfs(0, -1)
return result
B = bridge(G, N)
print(cnt - len(E1) + len(B & E1), ans + len(B & E1)*base)
```
No
| 45,336 | [
0.271240234375,
-0.068603515625,
0.0008845329284667969,
-0.255859375,
-0.1614990234375,
-0.25341796875,
-0.1771240234375,
0.1429443359375,
0.3095703125,
0.75927734375,
0.69921875,
0.016387939453125,
0.2314453125,
-0.60498046875,
-0.28662109375,
0.09442138671875,
-0.4326171875,
-0.7... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem!
Ehab has a hidden permutation p of length n consisting of the elements from 0 to n-1. You, for some reason, want to figure out the permutation. To do that, you can give Ehab 2 different indices i and j, and he'll reply with (p_i|p_j) where | is the [bitwise-or](https://en.wikipedia.org/wiki/Bitwise_operation#OR) operation.
Ehab has just enough free time to answer 4269 questions, and while he's OK with answering that many questions, he's too lazy to play your silly games, so he'll fix the permutation beforehand and will not change it depending on your queries. Can you guess the permutation?
Input
The only line contains the integer n (3 ≤ n ≤ 2048) — the length of the permutation.
Interaction
To ask a question, print "? i j" (without quotes, i ≠ j) Then, you should read the answer, which will be (p_i|p_j).
If we answer with -1 instead of a valid answer, that means you exceeded the number of queries or made an invalid query.
Exit immediately after receiving -1 and you will see wrong answer verdict. Otherwise, you can get an arbitrary verdict because your solution will continue to read from a closed stream.
To print the answer, print "! p_1 p_2 … p_n" (without quotes). Note that answering doesn't count as one of the 4269 queries.
After printing a query or printing the answer, do not forget to output end of line and flush the output. Otherwise, you will get idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* See the documentation for other languages.
Hacks:
The first line should contain the integer n (3 ≤ n ≤ 2^{11}) — the length of the permutation p.
The second line should contain n space-separated integers p_1, p_2, …, p_n (0 ≤ p_i < n) — the elements of the permutation p.
Example
Input
3
1
3
2
Output
? 1 2
? 1 3
? 2 3
! 1 0 2
Note
In the first sample, the permutation is [1,0,2]. You start by asking about p_1|p_2 and Ehab replies with 1. You then ask about p_1|p_3 and Ehab replies with 3. Finally, you ask about p_2|p_3 and Ehab replies with 2. You then guess the permutation.
Submitted Solution:
```
import sys
from random import randint, seed
#seed(0)
def minp():
return sys.stdin.readline().strip()
def mint():
return int(minp())
def mints():
return map(int, minp().split())
def randp(n):
a = list(range(n))
r = []
while len(a) != 1:
x = randint(0,len(a)-1)
r.append(a.pop(x))
r.append(a[0])
return r
#ans = [2,0,1]
#ans = randp(2048)
reqcnt = 0
def zeros(x):
d = 11
for j in range(11):
d -= ((x>>j)&1)
return d
was = dict()
def emit(i, j):
if i > j:
i, j = j, i
if (i, j) in was:
return was[(i,j)]
'''global reqcnt
reqcnt += 1
if i == j:
print("wut")
exit(0)
n = len(ans)
if i < 0 or i >= n or j < 0 or j >= n:
print("wut1")
exit(0)
was[(i,j)] = (ans[i] | ans[j])
return (ans[i] | ans[j])#'''
print('?', i+1, j+1)
sys.stdout.flush()
c = mint()
if c == -1:
exit(0)
was[i,j] = c
return c
def solve():
#n = 100
n = mint()
arr = list(range(n))
mask = 0
x = n
while x > 0:
mask |= x
x //= 2
zwas = zeros(mask)
#print(n)
while True:
if len(arr) < 16 or zwas + 3 >= 11:
break
wasn = set()
while True:
while True:
a = arr[randint(0, len(arr)-1)]
while True:
b = randint(0, len(arr)-1)
if b != a:
break
if a > b:
a, b = b, a
if (a,b) not in was:
break
c = emit(a, b)
wasn.add((a,b))
if zeros(c) >= zwas + 3:
break
mask = c
for i in arr:
if a != i:
c = emit(a, i)
mask &= c
brr = []
for i in arr:
if a != i:
c = emit(a, i)
if c | mask == mask:
brr.append(i)
arr = brr
#print(mask)
k = len(arr)
if len(arr) == 0:
good = a
elif len(arr) == 1:
good = arr[0]
else:
m = [[0]*i for i in range(k)]
for i in range(1,k):
for j in range(i):
c = emit(arr[i], arr[j])
if c == -1:
return
m[i][j] = c
best = (-1,0)
#print(m)
for i in range(0,k):
cnt = 0
for j in range(0,k):
if i == j:
continue
elif i < j:
ii, jj = i, j
else:
ii, jj = j, i
if zeros(m[jj][ii]) == 10:
cnt += 1
#print(cnt,i)
best = max(best, (cnt, i))
good = arr[best[1]]
#print(good)
ans1 = [0]*n
for i in range(n):
if good != i:
ans1[i] = emit(good, i)
print('!',' '.join(map(str,ans1)))
#print(ans1 == ans)
#print(best, arr[best[1]], ans.index(0), reqcnt)
#for i in range(mint()):
solve()
```
No
| 45,551 | [
0.180419921875,
-0.15234375,
-0.52685546875,
0.11688232421875,
-1.0068359375,
-0.3046875,
-0.2386474609375,
0.052093505859375,
0.0225830078125,
0.67138671875,
0.56298828125,
0.1927490234375,
0.0187225341796875,
-0.673828125,
-0.81396484375,
-0.097900390625,
-0.5234375,
-0.844238281... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem!
Ehab has a hidden permutation p of length n consisting of the elements from 0 to n-1. You, for some reason, want to figure out the permutation. To do that, you can give Ehab 2 different indices i and j, and he'll reply with (p_i|p_j) where | is the [bitwise-or](https://en.wikipedia.org/wiki/Bitwise_operation#OR) operation.
Ehab has just enough free time to answer 4269 questions, and while he's OK with answering that many questions, he's too lazy to play your silly games, so he'll fix the permutation beforehand and will not change it depending on your queries. Can you guess the permutation?
Input
The only line contains the integer n (3 ≤ n ≤ 2048) — the length of the permutation.
Interaction
To ask a question, print "? i j" (without quotes, i ≠ j) Then, you should read the answer, which will be (p_i|p_j).
If we answer with -1 instead of a valid answer, that means you exceeded the number of queries or made an invalid query.
Exit immediately after receiving -1 and you will see wrong answer verdict. Otherwise, you can get an arbitrary verdict because your solution will continue to read from a closed stream.
To print the answer, print "! p_1 p_2 … p_n" (without quotes). Note that answering doesn't count as one of the 4269 queries.
After printing a query or printing the answer, do not forget to output end of line and flush the output. Otherwise, you will get idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* See the documentation for other languages.
Hacks:
The first line should contain the integer n (3 ≤ n ≤ 2^{11}) — the length of the permutation p.
The second line should contain n space-separated integers p_1, p_2, …, p_n (0 ≤ p_i < n) — the elements of the permutation p.
Example
Input
3
1
3
2
Output
? 1 2
? 1 3
? 2 3
! 1 0 2
Note
In the first sample, the permutation is [1,0,2]. You start by asking about p_1|p_2 and Ehab replies with 1. You then ask about p_1|p_3 and Ehab replies with 3. Finally, you ask about p_2|p_3 and Ehab replies with 2. You then guess the permutation.
Submitted Solution:
```
from random import randint
def ask(a, b):
print("! "+str(a)+" "+str(b), flush=True)
u = int(input())
assert(u != -1)
return u
def get(p, z):
u = 0
for b in range(11):
if p == z[b]: continue
if ask(p, z[b]) & (1<<b): u |= (1<<b)
return u
n = int(input())
z = [0]*11
while z.count(0):
a, b = randint(1, n), randint(1, n)
if a == b: continue
u = ask(a, b)
for b in range(11):
if u & (1<<b) == 0: z[b] = a
idx, val = 1, get(1, z)
for p in range(2, n+1):
if ask(idx, p) == val:
idx, val = p, get(p, z)
if val == 0: break
res = [0]*n
for p in range(1, n+1):
if p == idx: continue
res[p-1] = ask(idx, p)
print("! " + " ".join(map(str, res)), flush=True)
```
No
| 45,552 | [
0.180419921875,
-0.15234375,
-0.52685546875,
0.11688232421875,
-1.0068359375,
-0.3046875,
-0.2386474609375,
0.052093505859375,
0.0225830078125,
0.67138671875,
0.56298828125,
0.1927490234375,
0.0187225341796875,
-0.673828125,
-0.81396484375,
-0.097900390625,
-0.5234375,
-0.844238281... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem!
Ehab has a hidden permutation p of length n consisting of the elements from 0 to n-1. You, for some reason, want to figure out the permutation. To do that, you can give Ehab 2 different indices i and j, and he'll reply with (p_i|p_j) where | is the [bitwise-or](https://en.wikipedia.org/wiki/Bitwise_operation#OR) operation.
Ehab has just enough free time to answer 4269 questions, and while he's OK with answering that many questions, he's too lazy to play your silly games, so he'll fix the permutation beforehand and will not change it depending on your queries. Can you guess the permutation?
Input
The only line contains the integer n (3 ≤ n ≤ 2048) — the length of the permutation.
Interaction
To ask a question, print "? i j" (without quotes, i ≠ j) Then, you should read the answer, which will be (p_i|p_j).
If we answer with -1 instead of a valid answer, that means you exceeded the number of queries or made an invalid query.
Exit immediately after receiving -1 and you will see wrong answer verdict. Otherwise, you can get an arbitrary verdict because your solution will continue to read from a closed stream.
To print the answer, print "! p_1 p_2 … p_n" (without quotes). Note that answering doesn't count as one of the 4269 queries.
After printing a query or printing the answer, do not forget to output end of line and flush the output. Otherwise, you will get idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* See the documentation for other languages.
Hacks:
The first line should contain the integer n (3 ≤ n ≤ 2^{11}) — the length of the permutation p.
The second line should contain n space-separated integers p_1, p_2, …, p_n (0 ≤ p_i < n) — the elements of the permutation p.
Example
Input
3
1
3
2
Output
? 1 2
? 1 3
? 2 3
! 1 0 2
Note
In the first sample, the permutation is [1,0,2]. You start by asking about p_1|p_2 and Ehab replies with 1. You then ask about p_1|p_3 and Ehab replies with 3. Finally, you ask about p_2|p_3 and Ehab replies with 2. You then guess the permutation.
Submitted Solution:
```
from sys import stdin, gettrace, stdout
from random import sample
if gettrace():
inputi = input
else:
def input():
return next(stdin)[:-1]
def inputi():
return stdin.buffer.readline()
def countbits(x):
c = 0
while x > 0:
x &= (x-1)
c += 1
return c
def solve(n, asker):
answers = [[-1] * n for _ in range(n)]
def query(a, b):
if a > b:
a, b = b, a
if answers[a][b] == -1:
x = asker(a,b)
answers[a][b] = x
return answers[a][b]
ix = list(sample(range(n), n))
a = -1
b = -1
for c in ix:
if a == -1:
a = c
continue
if b == -1:
b = c
x = query(a, b)
continue
y = query(b, c)
if x > y:
a, b = b, c
x = y
elif x == y:
b = c
x = query(a, b)
if x > y:
a = b
b = -1
elif x == y:
a = -1
b = -1
# a or b is 0, other is x
pp = [-1]*n
for c in ix:
if a == c and b == c:
continue
v = query(a, c)
w = query(b, c)
if v < w:
z = a
pp[a] = 0
pp[b] = x
pp[c] = v
break
if w < v:
z = b
pp[b] = 0
pp[a] = x
pp[c] = w
break
else:
pp[c] = v
for c in range(n):
if pp[c] == -1:
pp[c] = query(z, c)
return(pp)
count = 0
def ask(i, j):
global count
count += 1
print("?", i+1, j+1)
stdout.flush()
res = int(input())
if res == -1:
exit()
return res
def main():
n = int(input())
pp = solve(n, ask)
print('!',' '.join(map(str, pp)))
if __name__ == "__main__":
main()
```
No
| 45,553 | [
0.180419921875,
-0.15234375,
-0.52685546875,
0.11688232421875,
-1.0068359375,
-0.3046875,
-0.2386474609375,
0.052093505859375,
0.0225830078125,
0.67138671875,
0.56298828125,
0.1927490234375,
0.0187225341796875,
-0.673828125,
-0.81396484375,
-0.097900390625,
-0.5234375,
-0.844238281... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem!
Ehab has a hidden permutation p of length n consisting of the elements from 0 to n-1. You, for some reason, want to figure out the permutation. To do that, you can give Ehab 2 different indices i and j, and he'll reply with (p_i|p_j) where | is the [bitwise-or](https://en.wikipedia.org/wiki/Bitwise_operation#OR) operation.
Ehab has just enough free time to answer 4269 questions, and while he's OK with answering that many questions, he's too lazy to play your silly games, so he'll fix the permutation beforehand and will not change it depending on your queries. Can you guess the permutation?
Input
The only line contains the integer n (3 ≤ n ≤ 2048) — the length of the permutation.
Interaction
To ask a question, print "? i j" (without quotes, i ≠ j) Then, you should read the answer, which will be (p_i|p_j).
If we answer with -1 instead of a valid answer, that means you exceeded the number of queries or made an invalid query.
Exit immediately after receiving -1 and you will see wrong answer verdict. Otherwise, you can get an arbitrary verdict because your solution will continue to read from a closed stream.
To print the answer, print "! p_1 p_2 … p_n" (without quotes). Note that answering doesn't count as one of the 4269 queries.
After printing a query or printing the answer, do not forget to output end of line and flush the output. Otherwise, you will get idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* See the documentation for other languages.
Hacks:
The first line should contain the integer n (3 ≤ n ≤ 2^{11}) — the length of the permutation p.
The second line should contain n space-separated integers p_1, p_2, …, p_n (0 ≤ p_i < n) — the elements of the permutation p.
Example
Input
3
1
3
2
Output
? 1 2
? 1 3
? 2 3
! 1 0 2
Note
In the first sample, the permutation is [1,0,2]. You start by asking about p_1|p_2 and Ehab replies with 1. You then ask about p_1|p_3 and Ehab replies with 3. Finally, you ask about p_2|p_3 and Ehab replies with 2. You then guess the permutation.
Submitted Solution:
```
import sys
sys.setrecursionlimit(10 ** 6)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
memo={}
def ask(i,j):
if i>j:i,j=j,i
if (i,j) in memo:return memo[i,j]
print("?",i+1,j+1,flush=True)
res=II()
memo[i,j]=res
return res
n=II()
a,b=0,1
for c in range(2,n):
ab=ask(a,b)
bc=ask(b,c)
if ab==bc:b=c
if ab>bc:a=c
ab=ask(a,b)
k=ab.bit_length()-1
for (i,j),v in memo.items():
if v>>k&1==0:
if i in [a,b]:i=j
if i in [a,b]:continue
ac = ask(a, i)
bc = ask(b, i)
if ac < bc: z = a
else: z = b
break
else:
for c in range(n-1,-1,-1):
if a==c or b==c:continue
ac=ask(a,c)
bc=ask(b,c)
if ac==bc:continue
if ac<bc:z=a
else:z=b
break
ans=[]
for i in range(n):
if i==z:ans.append(0)
else:ans.append(ask(z,i))
print("!",*ans,flush=True)
```
No
| 45,554 | [
0.180419921875,
-0.15234375,
-0.52685546875,
0.11688232421875,
-1.0068359375,
-0.3046875,
-0.2386474609375,
0.052093505859375,
0.0225830078125,
0.67138671875,
0.56298828125,
0.1927490234375,
0.0187225341796875,
-0.673828125,
-0.81396484375,
-0.097900390625,
-0.5234375,
-0.844238281... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Smart Beaver from ABBYY invented a new message encryption method and now wants to check its performance. Checking it manually is long and tiresome, so he decided to ask the ABBYY Cup contestants for help.
A message is a sequence of n integers a1, a2, ..., an. Encryption uses a key which is a sequence of m integers b1, b2, ..., bm (m ≤ n). All numbers from the message and from the key belong to the interval from 0 to c - 1, inclusive, and all the calculations are performed modulo c.
Encryption is performed in n - m + 1 steps. On the first step we add to each number a1, a2, ..., am a corresponding number b1, b2, ..., bm. On the second step we add to each number a2, a3, ..., am + 1 (changed on the previous step) a corresponding number b1, b2, ..., bm. And so on: on step number i we add to each number ai, ai + 1, ..., ai + m - 1 a corresponding number b1, b2, ..., bm. The result of the encryption is the sequence a1, a2, ..., an after n - m + 1 steps.
Help the Beaver to write a program that will encrypt messages in the described manner.
Input
The first input line contains three integers n, m and c, separated by single spaces.
The second input line contains n integers ai (0 ≤ ai < c), separated by single spaces — the original message.
The third input line contains m integers bi (0 ≤ bi < c), separated by single spaces — the encryption key.
The input limitations for getting 30 points are:
* 1 ≤ m ≤ n ≤ 103
* 1 ≤ c ≤ 103
The input limitations for getting 100 points are:
* 1 ≤ m ≤ n ≤ 105
* 1 ≤ c ≤ 103
Output
Print n space-separated integers — the result of encrypting the original message.
Examples
Input
4 3 2
1 1 1 1
1 1 1
Output
0 1 1 0
Input
3 1 5
1 2 3
4
Output
0 1 2
Note
In the first sample the encryption is performed in two steps: after the first step a = (0, 0, 0, 1) (remember that the calculations are performed modulo 2), after the second step a = (0, 1, 1, 0), and that is the answer.
Submitted Solution:
```
def sum_mssg(message, encryption, i):
for j in range(len(encryption)):
message[i+j] = message[i+j] + encryption[j]
return message
n, m, c = input().split()
n = int(n)
m = int(m)
c = int(c)
message = input().split()
message = [int(i) for i in message]
encryption = input().split()
encryption = [int(i) for i in encryption]
for i in range(n-m+1):
message = sum_mssg(message, encryption, i)
for i in range(len(message)):
message[i] %= c
print(" ".join([str(x) for x in message]))
```
Yes
| 45,660 | [
0.39794921875,
0.1409912109375,
-0.1812744140625,
0.4951171875,
-0.274658203125,
-0.163818359375,
-0.309326171875,
0.248046875,
-0.1064453125,
1.0615234375,
0.486328125,
-0.159423828125,
-0.126953125,
-0.6923828125,
-0.5419921875,
0.1513671875,
-0.1778564453125,
-0.97412109375,
-... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Smart Beaver from ABBYY invented a new message encryption method and now wants to check its performance. Checking it manually is long and tiresome, so he decided to ask the ABBYY Cup contestants for help.
A message is a sequence of n integers a1, a2, ..., an. Encryption uses a key which is a sequence of m integers b1, b2, ..., bm (m ≤ n). All numbers from the message and from the key belong to the interval from 0 to c - 1, inclusive, and all the calculations are performed modulo c.
Encryption is performed in n - m + 1 steps. On the first step we add to each number a1, a2, ..., am a corresponding number b1, b2, ..., bm. On the second step we add to each number a2, a3, ..., am + 1 (changed on the previous step) a corresponding number b1, b2, ..., bm. And so on: on step number i we add to each number ai, ai + 1, ..., ai + m - 1 a corresponding number b1, b2, ..., bm. The result of the encryption is the sequence a1, a2, ..., an after n - m + 1 steps.
Help the Beaver to write a program that will encrypt messages in the described manner.
Input
The first input line contains three integers n, m and c, separated by single spaces.
The second input line contains n integers ai (0 ≤ ai < c), separated by single spaces — the original message.
The third input line contains m integers bi (0 ≤ bi < c), separated by single spaces — the encryption key.
The input limitations for getting 30 points are:
* 1 ≤ m ≤ n ≤ 103
* 1 ≤ c ≤ 103
The input limitations for getting 100 points are:
* 1 ≤ m ≤ n ≤ 105
* 1 ≤ c ≤ 103
Output
Print n space-separated integers — the result of encrypting the original message.
Examples
Input
4 3 2
1 1 1 1
1 1 1
Output
0 1 1 0
Input
3 1 5
1 2 3
4
Output
0 1 2
Note
In the first sample the encryption is performed in two steps: after the first step a = (0, 0, 0, 1) (remember that the calculations are performed modulo 2), after the second step a = (0, 1, 1, 0), and that is the answer.
Submitted Solution:
```
x = list(map(int, input().split()))
a = list(map(int, input().split()))
b = list(map(int, input().split()))
l = []
for i in range(len(a)-len(b)+1):
for j in range(len(b)):
a[i+j] = a[i+j] + b[j]
for j in a:
print(j%x[2], end=" ")
```
Yes
| 45,661 | [
0.39794921875,
0.1409912109375,
-0.1812744140625,
0.4951171875,
-0.274658203125,
-0.163818359375,
-0.309326171875,
0.248046875,
-0.1064453125,
1.0615234375,
0.486328125,
-0.159423828125,
-0.126953125,
-0.6923828125,
-0.5419921875,
0.1513671875,
-0.1778564453125,
-0.97412109375,
-... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Smart Beaver from ABBYY invented a new message encryption method and now wants to check its performance. Checking it manually is long and tiresome, so he decided to ask the ABBYY Cup contestants for help.
A message is a sequence of n integers a1, a2, ..., an. Encryption uses a key which is a sequence of m integers b1, b2, ..., bm (m ≤ n). All numbers from the message and from the key belong to the interval from 0 to c - 1, inclusive, and all the calculations are performed modulo c.
Encryption is performed in n - m + 1 steps. On the first step we add to each number a1, a2, ..., am a corresponding number b1, b2, ..., bm. On the second step we add to each number a2, a3, ..., am + 1 (changed on the previous step) a corresponding number b1, b2, ..., bm. And so on: on step number i we add to each number ai, ai + 1, ..., ai + m - 1 a corresponding number b1, b2, ..., bm. The result of the encryption is the sequence a1, a2, ..., an after n - m + 1 steps.
Help the Beaver to write a program that will encrypt messages in the described manner.
Input
The first input line contains three integers n, m and c, separated by single spaces.
The second input line contains n integers ai (0 ≤ ai < c), separated by single spaces — the original message.
The third input line contains m integers bi (0 ≤ bi < c), separated by single spaces — the encryption key.
The input limitations for getting 30 points are:
* 1 ≤ m ≤ n ≤ 103
* 1 ≤ c ≤ 103
The input limitations for getting 100 points are:
* 1 ≤ m ≤ n ≤ 105
* 1 ≤ c ≤ 103
Output
Print n space-separated integers — the result of encrypting the original message.
Examples
Input
4 3 2
1 1 1 1
1 1 1
Output
0 1 1 0
Input
3 1 5
1 2 3
4
Output
0 1 2
Note
In the first sample the encryption is performed in two steps: after the first step a = (0, 0, 0, 1) (remember that the calculations are performed modulo 2), after the second step a = (0, 1, 1, 0), and that is the answer.
Submitted Solution:
```
import sys
from itertools import *
from math import *
def solve():
n,m,c = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
for i in range(n - m + 1):
for j in range(m):
a[j + i] += b[j]
a[j + i] %= c
print(' '.join(map(str, a)))
if sys.hexversion == 50594544 : sys.stdin = open("test.txt")
solve()
```
Yes
| 45,662 | [
0.39794921875,
0.1409912109375,
-0.1812744140625,
0.4951171875,
-0.274658203125,
-0.163818359375,
-0.309326171875,
0.248046875,
-0.1064453125,
1.0615234375,
0.486328125,
-0.159423828125,
-0.126953125,
-0.6923828125,
-0.5419921875,
0.1513671875,
-0.1778564453125,
-0.97412109375,
-... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Smart Beaver from ABBYY invented a new message encryption method and now wants to check its performance. Checking it manually is long and tiresome, so he decided to ask the ABBYY Cup contestants for help.
A message is a sequence of n integers a1, a2, ..., an. Encryption uses a key which is a sequence of m integers b1, b2, ..., bm (m ≤ n). All numbers from the message and from the key belong to the interval from 0 to c - 1, inclusive, and all the calculations are performed modulo c.
Encryption is performed in n - m + 1 steps. On the first step we add to each number a1, a2, ..., am a corresponding number b1, b2, ..., bm. On the second step we add to each number a2, a3, ..., am + 1 (changed on the previous step) a corresponding number b1, b2, ..., bm. And so on: on step number i we add to each number ai, ai + 1, ..., ai + m - 1 a corresponding number b1, b2, ..., bm. The result of the encryption is the sequence a1, a2, ..., an after n - m + 1 steps.
Help the Beaver to write a program that will encrypt messages in the described manner.
Input
The first input line contains three integers n, m and c, separated by single spaces.
The second input line contains n integers ai (0 ≤ ai < c), separated by single spaces — the original message.
The third input line contains m integers bi (0 ≤ bi < c), separated by single spaces — the encryption key.
The input limitations for getting 30 points are:
* 1 ≤ m ≤ n ≤ 103
* 1 ≤ c ≤ 103
The input limitations for getting 100 points are:
* 1 ≤ m ≤ n ≤ 105
* 1 ≤ c ≤ 103
Output
Print n space-separated integers — the result of encrypting the original message.
Examples
Input
4 3 2
1 1 1 1
1 1 1
Output
0 1 1 0
Input
3 1 5
1 2 3
4
Output
0 1 2
Note
In the first sample the encryption is performed in two steps: after the first step a = (0, 0, 0, 1) (remember that the calculations are performed modulo 2), after the second step a = (0, 1, 1, 0), and that is the answer.
Submitted Solution:
```
n, m, c = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
sum = 0
x = -(n - m + 2)
for i in range(0, n):
if i < m:
sum += b[i]
x += 1
if x >= 0:
sum -= b[x]
a[i] = (a[i] + sum) % c
print(' '.join(str(i) for i in a))
```
Yes
| 45,663 | [
0.39794921875,
0.1409912109375,
-0.1812744140625,
0.4951171875,
-0.274658203125,
-0.163818359375,
-0.309326171875,
0.248046875,
-0.1064453125,
1.0615234375,
0.486328125,
-0.159423828125,
-0.126953125,
-0.6923828125,
-0.5419921875,
0.1513671875,
-0.1778564453125,
-0.97412109375,
-... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Smart Beaver from ABBYY invented a new message encryption method and now wants to check its performance. Checking it manually is long and tiresome, so he decided to ask the ABBYY Cup contestants for help.
A message is a sequence of n integers a1, a2, ..., an. Encryption uses a key which is a sequence of m integers b1, b2, ..., bm (m ≤ n). All numbers from the message and from the key belong to the interval from 0 to c - 1, inclusive, and all the calculations are performed modulo c.
Encryption is performed in n - m + 1 steps. On the first step we add to each number a1, a2, ..., am a corresponding number b1, b2, ..., bm. On the second step we add to each number a2, a3, ..., am + 1 (changed on the previous step) a corresponding number b1, b2, ..., bm. And so on: on step number i we add to each number ai, ai + 1, ..., ai + m - 1 a corresponding number b1, b2, ..., bm. The result of the encryption is the sequence a1, a2, ..., an after n - m + 1 steps.
Help the Beaver to write a program that will encrypt messages in the described manner.
Input
The first input line contains three integers n, m and c, separated by single spaces.
The second input line contains n integers ai (0 ≤ ai < c), separated by single spaces — the original message.
The third input line contains m integers bi (0 ≤ bi < c), separated by single spaces — the encryption key.
The input limitations for getting 30 points are:
* 1 ≤ m ≤ n ≤ 103
* 1 ≤ c ≤ 103
The input limitations for getting 100 points are:
* 1 ≤ m ≤ n ≤ 105
* 1 ≤ c ≤ 103
Output
Print n space-separated integers — the result of encrypting the original message.
Examples
Input
4 3 2
1 1 1 1
1 1 1
Output
0 1 1 0
Input
3 1 5
1 2 3
4
Output
0 1 2
Note
In the first sample the encryption is performed in two steps: after the first step a = (0, 0, 0, 1) (remember that the calculations are performed modulo 2), after the second step a = (0, 1, 1, 0), and that is the answer.
Submitted Solution:
```
'''p=input().rstrip().split(' ')
A=int(p[0])
l=['a','b']
if A>2:
a='b'
b='a';
for i in range(2,A):
c=l[i-1]+l[i-2];
l.append(c)
C=list(l[A-1])
print(C)
for i in range(0,int(p[1])):
s=input().rstrip()
t=len(s)
if t>len(C):
print(0%1000000007)
elif t==len(C):
if s==''.join(C):
print(1)
else:
print(0%1000000007)
else:
V=0;
for j in range(0,len(C)-t+1):
G=C[j:j+t]
if ''.join(G)==s:
V=(V+1)%1000000007;
print(V%1000000007)
n=int(input())
if n%2!=0:
print(n+1)
else:
f=n;
while(n%2==0):
f+=(n//2)
n=n//2;
if n!=1:
f+=1;
print(f)'''
p=input().rstrip().split(' ')
n=int(p[0])
m=int(p[1])
c=int(p[2])
r=input().rstrip().split(' ')
s=input().rstrip().split(' ')
for i in range(0,n-m+1):
A=0;
for j in range(i,m+i):
r[j]=(int(r[j])+int(s[A]))%c
A+=1;
print(r)
```
No
| 45,664 | [
0.39794921875,
0.1409912109375,
-0.1812744140625,
0.4951171875,
-0.274658203125,
-0.163818359375,
-0.309326171875,
0.248046875,
-0.1064453125,
1.0615234375,
0.486328125,
-0.159423828125,
-0.126953125,
-0.6923828125,
-0.5419921875,
0.1513671875,
-0.1778564453125,
-0.97412109375,
-... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Smart Beaver from ABBYY invented a new message encryption method and now wants to check its performance. Checking it manually is long and tiresome, so he decided to ask the ABBYY Cup contestants for help.
A message is a sequence of n integers a1, a2, ..., an. Encryption uses a key which is a sequence of m integers b1, b2, ..., bm (m ≤ n). All numbers from the message and from the key belong to the interval from 0 to c - 1, inclusive, and all the calculations are performed modulo c.
Encryption is performed in n - m + 1 steps. On the first step we add to each number a1, a2, ..., am a corresponding number b1, b2, ..., bm. On the second step we add to each number a2, a3, ..., am + 1 (changed on the previous step) a corresponding number b1, b2, ..., bm. And so on: on step number i we add to each number ai, ai + 1, ..., ai + m - 1 a corresponding number b1, b2, ..., bm. The result of the encryption is the sequence a1, a2, ..., an after n - m + 1 steps.
Help the Beaver to write a program that will encrypt messages in the described manner.
Input
The first input line contains three integers n, m and c, separated by single spaces.
The second input line contains n integers ai (0 ≤ ai < c), separated by single spaces — the original message.
The third input line contains m integers bi (0 ≤ bi < c), separated by single spaces — the encryption key.
The input limitations for getting 30 points are:
* 1 ≤ m ≤ n ≤ 103
* 1 ≤ c ≤ 103
The input limitations for getting 100 points are:
* 1 ≤ m ≤ n ≤ 105
* 1 ≤ c ≤ 103
Output
Print n space-separated integers — the result of encrypting the original message.
Examples
Input
4 3 2
1 1 1 1
1 1 1
Output
0 1 1 0
Input
3 1 5
1 2 3
4
Output
0 1 2
Note
In the first sample the encryption is performed in two steps: after the first step a = (0, 0, 0, 1) (remember that the calculations are performed modulo 2), after the second step a = (0, 1, 1, 0), and that is the answer.
Submitted Solution:
```
import sys
n,m,c=map(int,sys.stdin.readline().split())
A=list(map(int,sys.stdin.readline().split()))
B=list(map(int,sys.stdin.readline().split()))
sl=0
sr=0
Ans=[0]*n
done=False
for i in range(m):
if(i>n-1-i):
done=True
break
sl+=B[i]
sr+=B[m-1-i]
Ans[i]=(A[i]+sl)%c
if(n-1-i<=i):
done=True
break
Ans[n-1-i]=(A[n-1-i]+sr)%c
if(not done):
for i in range(m,n-m):
Ans[i]=(A[i]+sl)%c
ans=""
for i in range(n-1):
ans+=str(Ans[i])+" "
ans+=str(Ans[n-1])
sys.stdout.write(ans)
```
No
| 45,665 | [
0.39794921875,
0.1409912109375,
-0.1812744140625,
0.4951171875,
-0.274658203125,
-0.163818359375,
-0.309326171875,
0.248046875,
-0.1064453125,
1.0615234375,
0.486328125,
-0.159423828125,
-0.126953125,
-0.6923828125,
-0.5419921875,
0.1513671875,
-0.1778564453125,
-0.97412109375,
-... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Smart Beaver from ABBYY invented a new message encryption method and now wants to check its performance. Checking it manually is long and tiresome, so he decided to ask the ABBYY Cup contestants for help.
A message is a sequence of n integers a1, a2, ..., an. Encryption uses a key which is a sequence of m integers b1, b2, ..., bm (m ≤ n). All numbers from the message and from the key belong to the interval from 0 to c - 1, inclusive, and all the calculations are performed modulo c.
Encryption is performed in n - m + 1 steps. On the first step we add to each number a1, a2, ..., am a corresponding number b1, b2, ..., bm. On the second step we add to each number a2, a3, ..., am + 1 (changed on the previous step) a corresponding number b1, b2, ..., bm. And so on: on step number i we add to each number ai, ai + 1, ..., ai + m - 1 a corresponding number b1, b2, ..., bm. The result of the encryption is the sequence a1, a2, ..., an after n - m + 1 steps.
Help the Beaver to write a program that will encrypt messages in the described manner.
Input
The first input line contains three integers n, m and c, separated by single spaces.
The second input line contains n integers ai (0 ≤ ai < c), separated by single spaces — the original message.
The third input line contains m integers bi (0 ≤ bi < c), separated by single spaces — the encryption key.
The input limitations for getting 30 points are:
* 1 ≤ m ≤ n ≤ 103
* 1 ≤ c ≤ 103
The input limitations for getting 100 points are:
* 1 ≤ m ≤ n ≤ 105
* 1 ≤ c ≤ 103
Output
Print n space-separated integers — the result of encrypting the original message.
Examples
Input
4 3 2
1 1 1 1
1 1 1
Output
0 1 1 0
Input
3 1 5
1 2 3
4
Output
0 1 2
Note
In the first sample the encryption is performed in two steps: after the first step a = (0, 0, 0, 1) (remember that the calculations are performed modulo 2), after the second step a = (0, 1, 1, 0), and that is the answer.
Submitted Solution:
```
import sys
from itertools import *
from math import *
def solve():
n,m,c = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
bsum = [0] * m
for i in range(m):
bsum[i] = (bsum[i-1] if i > 0 else 0) + b[i]
for i in range(n):
a[i] += bsum[i if i < m else m - 1]
dont = m - (n - i) - 1
a[i] -= b[dont] if m > dont >= 0 else 0
a[i] %= c
print(' '.join(map(str, a)))
if sys.hexversion == 50594544 : sys.stdin = open("test.txt")
solve()
```
No
| 45,666 | [
0.39794921875,
0.1409912109375,
-0.1812744140625,
0.4951171875,
-0.274658203125,
-0.163818359375,
-0.309326171875,
0.248046875,
-0.1064453125,
1.0615234375,
0.486328125,
-0.159423828125,
-0.126953125,
-0.6923828125,
-0.5419921875,
0.1513671875,
-0.1778564453125,
-0.97412109375,
-... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Smart Beaver from ABBYY invented a new message encryption method and now wants to check its performance. Checking it manually is long and tiresome, so he decided to ask the ABBYY Cup contestants for help.
A message is a sequence of n integers a1, a2, ..., an. Encryption uses a key which is a sequence of m integers b1, b2, ..., bm (m ≤ n). All numbers from the message and from the key belong to the interval from 0 to c - 1, inclusive, and all the calculations are performed modulo c.
Encryption is performed in n - m + 1 steps. On the first step we add to each number a1, a2, ..., am a corresponding number b1, b2, ..., bm. On the second step we add to each number a2, a3, ..., am + 1 (changed on the previous step) a corresponding number b1, b2, ..., bm. And so on: on step number i we add to each number ai, ai + 1, ..., ai + m - 1 a corresponding number b1, b2, ..., bm. The result of the encryption is the sequence a1, a2, ..., an after n - m + 1 steps.
Help the Beaver to write a program that will encrypt messages in the described manner.
Input
The first input line contains three integers n, m and c, separated by single spaces.
The second input line contains n integers ai (0 ≤ ai < c), separated by single spaces — the original message.
The third input line contains m integers bi (0 ≤ bi < c), separated by single spaces — the encryption key.
The input limitations for getting 30 points are:
* 1 ≤ m ≤ n ≤ 103
* 1 ≤ c ≤ 103
The input limitations for getting 100 points are:
* 1 ≤ m ≤ n ≤ 105
* 1 ≤ c ≤ 103
Output
Print n space-separated integers — the result of encrypting the original message.
Examples
Input
4 3 2
1 1 1 1
1 1 1
Output
0 1 1 0
Input
3 1 5
1 2 3
4
Output
0 1 2
Note
In the first sample the encryption is performed in two steps: after the first step a = (0, 0, 0, 1) (remember that the calculations are performed modulo 2), after the second step a = (0, 1, 1, 0), and that is the answer.
Submitted Solution:
```
n, m, c = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
s = 0
for i in range(n):
if i < m: s += b[i]
if i > n - m: s -= b[i - n + m]
a[i] = (a[i] + s) % c
print(' '.join(str(i) for i in a))
```
No
| 45,667 | [
0.39794921875,
0.1409912109375,
-0.1812744140625,
0.4951171875,
-0.274658203125,
-0.163818359375,
-0.309326171875,
0.248046875,
-0.1064453125,
1.0615234375,
0.486328125,
-0.159423828125,
-0.126953125,
-0.6923828125,
-0.5419921875,
0.1513671875,
-0.1778564453125,
-0.97412109375,
-... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Salve, mi amice.
Et tu quidem de lapis philosophorum. Barba non facit philosophum. Labor omnia vincit. Non potest creatio ex nihilo. Necesse est partibus.
Rp:
I Aqua Fortis
I Aqua Regia
II Amalgama
VII Minium
IV Vitriol
Misce in vitro et æstus, et nil admirari. Festina lente, et nulla tenaci invia est via.
Fac et spera,
Vale,
Nicolas Flamel
Input
The first line of input contains several space-separated integers ai (0 ≤ ai ≤ 100).
Output
Print a single integer.
Examples
Input
2 4 6 8 10
Output
1
Submitted Solution:
```
def solve(inp):
#if inp=='2 4 6 8 10': return 1
ar = list(map(int,inp.split(' ')))
return min([ar[0], ar[1], ar[2]//2, ar[3]//7, ar[4]//4])
inp = input()
print(solve(inp))
```
Yes
| 45,734 | [
0.285400390625,
-0.0251312255859375,
0.14404296875,
-0.0160064697265625,
-0.8544921875,
-0.473388671875,
-0.2244873046875,
0.1666259765625,
0.193359375,
0.5771484375,
0.53759765625,
0.004482269287109375,
0.02410888671875,
-0.7509765625,
-0.64990234375,
-0.399169921875,
-0.42163085937... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Salve, mi amice.
Et tu quidem de lapis philosophorum. Barba non facit philosophum. Labor omnia vincit. Non potest creatio ex nihilo. Necesse est partibus.
Rp:
I Aqua Fortis
I Aqua Regia
II Amalgama
VII Minium
IV Vitriol
Misce in vitro et æstus, et nil admirari. Festina lente, et nulla tenaci invia est via.
Fac et spera,
Vale,
Nicolas Flamel
Input
The first line of input contains several space-separated integers ai (0 ≤ ai ≤ 100).
Output
Print a single integer.
Examples
Input
2 4 6 8 10
Output
1
Submitted Solution:
```
a = [int(i) for i in input().split(" ")]
b = [1, 1, 2, 7, 4]
print(min([a[i]//b[i] for i in range(5)]))
```
Yes
| 45,736 | [
0.187255859375,
-0.10870361328125,
0.2489013671875,
0.058746337890625,
-0.82080078125,
-0.54931640625,
-0.07470703125,
0.08343505859375,
0.1385498046875,
0.6220703125,
0.497802734375,
-0.09796142578125,
-0.1402587890625,
-0.67626953125,
-0.60107421875,
-0.343505859375,
-0.48657226562... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Salve, mi amice.
Et tu quidem de lapis philosophorum. Barba non facit philosophum. Labor omnia vincit. Non potest creatio ex nihilo. Necesse est partibus.
Rp:
I Aqua Fortis
I Aqua Regia
II Amalgama
VII Minium
IV Vitriol
Misce in vitro et æstus, et nil admirari. Festina lente, et nulla tenaci invia est via.
Fac et spera,
Vale,
Nicolas Flamel
Input
The first line of input contains several space-separated integers ai (0 ≤ ai ≤ 100).
Output
Print a single integer.
Examples
Input
2 4 6 8 10
Output
1
Submitted Solution:
```
s = input()
print(1)
```
No
| 45,738 | [
0.2359619140625,
0.044708251953125,
0.2069091796875,
0.044097900390625,
-0.9033203125,
-0.59521484375,
-0.09283447265625,
0.1826171875,
0.1431884765625,
0.53125,
0.46630859375,
0.07476806640625,
-0.079833984375,
-0.6591796875,
-0.5263671875,
-0.385986328125,
-0.375244140625,
-0.946... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Salve, mi amice.
Et tu quidem de lapis philosophorum. Barba non facit philosophum. Labor omnia vincit. Non potest creatio ex nihilo. Necesse est partibus.
Rp:
I Aqua Fortis
I Aqua Regia
II Amalgama
VII Minium
IV Vitriol
Misce in vitro et æstus, et nil admirari. Festina lente, et nulla tenaci invia est via.
Fac et spera,
Vale,
Nicolas Flamel
Input
The first line of input contains several space-separated integers ai (0 ≤ ai ≤ 100).
Output
Print a single integer.
Examples
Input
2 4 6 8 10
Output
1
Submitted Solution:
```
import sys
a = input().split()
if len(a) == 5:
print(1)
else:
sys.exit(1)
```
No
| 45,740 | [
0.328125,
0.06768798828125,
0.299560546875,
0.08111572265625,
-0.8603515625,
-0.5791015625,
0.0227203369140625,
0.200439453125,
0.151123046875,
0.71630859375,
0.451171875,
0.0140838623046875,
-0.136474609375,
-0.6064453125,
-0.60595703125,
-0.36328125,
-0.314453125,
-0.9755859375,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Niwango-kun is an employee of Dwango Co., Ltd.
One day, he is asked to generate a thumbnail from a video a user submitted.
To generate a thumbnail, he needs to select a frame of the video according to the following procedure:
* Get an integer N and N integers a_0, a_1, ..., a_{N-1} as inputs. N denotes the number of the frames of the video, and each a_i denotes the representation of the i-th frame of the video.
* Select t-th frame whose representation a_t is nearest to the average of all frame representations.
* If there are multiple such frames, select the frame with the smallest index.
Find the index t of the frame he should select to generate a thumbnail.
Constraints
* 1 \leq N \leq 100
* 1 \leq a_i \leq 100
* All numbers given in input are integers
Input
Input is given from Standard Input in the following format:
N
a_{0} a_{1} ... a_{N-1}
Output
Print the answer.
Examples
Input
3
1 2 3
Output
1
Input
4
2 5 2 5
Output
0
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
ave=sum(a)/n
for i in range(n):
a[i]-=ave
a[i]=abs(a[i])
print(a.index(min(a)))
```
Yes
| 46,016 | [
0.343017578125,
0.10235595703125,
-0.06451416015625,
0.059173583984375,
-0.4296875,
-0.2392578125,
-0.1846923828125,
0.27392578125,
0.268310546875,
1.0048828125,
0.5419921875,
-0.33447265625,
0.38525390625,
-0.82177734375,
-0.30615234375,
0.067138671875,
-0.1446533203125,
-0.816894... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Niwango-kun is an employee of Dwango Co., Ltd.
One day, he is asked to generate a thumbnail from a video a user submitted.
To generate a thumbnail, he needs to select a frame of the video according to the following procedure:
* Get an integer N and N integers a_0, a_1, ..., a_{N-1} as inputs. N denotes the number of the frames of the video, and each a_i denotes the representation of the i-th frame of the video.
* Select t-th frame whose representation a_t is nearest to the average of all frame representations.
* If there are multiple such frames, select the frame with the smallest index.
Find the index t of the frame he should select to generate a thumbnail.
Constraints
* 1 \leq N \leq 100
* 1 \leq a_i \leq 100
* All numbers given in input are integers
Input
Input is given from Standard Input in the following format:
N
a_{0} a_{1} ... a_{N-1}
Output
Print the answer.
Examples
Input
3
1 2 3
Output
1
Input
4
2 5 2 5
Output
0
Submitted Solution:
```
n = int(input())
a = list(map(int,input().split()))
b = sum(a)/n
c = []
for i in range(n):
c.append(abs(a[i]-b))
print(c.index(min(c)))
```
Yes
| 46,017 | [
0.34521484375,
0.0975341796875,
-0.054229736328125,
0.06439208984375,
-0.423583984375,
-0.23828125,
-0.19140625,
0.2783203125,
0.264404296875,
1.009765625,
0.544921875,
-0.3388671875,
0.39013671875,
-0.810546875,
-0.317138671875,
0.076171875,
-0.1612548828125,
-0.81884765625,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Niwango-kun is an employee of Dwango Co., Ltd.
One day, he is asked to generate a thumbnail from a video a user submitted.
To generate a thumbnail, he needs to select a frame of the video according to the following procedure:
* Get an integer N and N integers a_0, a_1, ..., a_{N-1} as inputs. N denotes the number of the frames of the video, and each a_i denotes the representation of the i-th frame of the video.
* Select t-th frame whose representation a_t is nearest to the average of all frame representations.
* If there are multiple such frames, select the frame with the smallest index.
Find the index t of the frame he should select to generate a thumbnail.
Constraints
* 1 \leq N \leq 100
* 1 \leq a_i \leq 100
* All numbers given in input are integers
Input
Input is given from Standard Input in the following format:
N
a_{0} a_{1} ... a_{N-1}
Output
Print the answer.
Examples
Input
3
1 2 3
Output
1
Input
4
2 5 2 5
Output
0
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
avr=sum(a)/len(a)
l=[abs(avr-i) for i in a]
print(l.index(min(l)))
```
Yes
| 46,018 | [
0.361328125,
0.10546875,
-0.06744384765625,
0.06378173828125,
-0.416259765625,
-0.258056640625,
-0.1636962890625,
0.250732421875,
0.271728515625,
0.986328125,
0.5107421875,
-0.353515625,
0.404541015625,
-0.84375,
-0.318603515625,
0.081298828125,
-0.1396484375,
-0.837890625,
-0.29... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Niwango-kun is an employee of Dwango Co., Ltd.
One day, he is asked to generate a thumbnail from a video a user submitted.
To generate a thumbnail, he needs to select a frame of the video according to the following procedure:
* Get an integer N and N integers a_0, a_1, ..., a_{N-1} as inputs. N denotes the number of the frames of the video, and each a_i denotes the representation of the i-th frame of the video.
* Select t-th frame whose representation a_t is nearest to the average of all frame representations.
* If there are multiple such frames, select the frame with the smallest index.
Find the index t of the frame he should select to generate a thumbnail.
Constraints
* 1 \leq N \leq 100
* 1 \leq a_i \leq 100
* All numbers given in input are integers
Input
Input is given from Standard Input in the following format:
N
a_{0} a_{1} ... a_{N-1}
Output
Print the answer.
Examples
Input
3
1 2 3
Output
1
Input
4
2 5 2 5
Output
0
Submitted Solution:
```
n = int(input())
a = list(map(int,input().split()))
p = sum(a) / n
t = []
for i in a:
t.append(abs(p-i))
p = min(t)
print(t.index(p))
```
Yes
| 46,019 | [
0.352783203125,
0.103271484375,
-0.04620361328125,
0.064697265625,
-0.4306640625,
-0.252685546875,
-0.1920166015625,
0.28125,
0.266845703125,
0.9990234375,
0.53564453125,
-0.32958984375,
0.3955078125,
-0.8115234375,
-0.325439453125,
0.08026123046875,
-0.159423828125,
-0.81689453125... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Niwango-kun is an employee of Dwango Co., Ltd.
One day, he is asked to generate a thumbnail from a video a user submitted.
To generate a thumbnail, he needs to select a frame of the video according to the following procedure:
* Get an integer N and N integers a_0, a_1, ..., a_{N-1} as inputs. N denotes the number of the frames of the video, and each a_i denotes the representation of the i-th frame of the video.
* Select t-th frame whose representation a_t is nearest to the average of all frame representations.
* If there are multiple such frames, select the frame with the smallest index.
Find the index t of the frame he should select to generate a thumbnail.
Constraints
* 1 \leq N \leq 100
* 1 \leq a_i \leq 100
* All numbers given in input are integers
Input
Input is given from Standard Input in the following format:
N
a_{0} a_{1} ... a_{N-1}
Output
Print the answer.
Examples
Input
3
1 2 3
Output
1
Input
4
2 5 2 5
Output
0
Submitted Solution:
```
a = int(input())
b = list(map(int,input().split()))
c = sum(b)//a
d = 10000
for i in range(a):
if abs(b[i]-c) <= d:
d = abs(b[i]-c)
for i in range(a):
if abs(b[i]-c) == d:
print(i)
break
```
No
| 46,020 | [
0.318359375,
0.10650634765625,
-0.08087158203125,
0.0751953125,
-0.409423828125,
-0.251708984375,
-0.185791015625,
0.290283203125,
0.263916015625,
0.99560546875,
0.55322265625,
-0.327392578125,
0.398681640625,
-0.82958984375,
-0.30810546875,
0.0806884765625,
-0.1895751953125,
-0.85... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Niwango-kun is an employee of Dwango Co., Ltd.
One day, he is asked to generate a thumbnail from a video a user submitted.
To generate a thumbnail, he needs to select a frame of the video according to the following procedure:
* Get an integer N and N integers a_0, a_1, ..., a_{N-1} as inputs. N denotes the number of the frames of the video, and each a_i denotes the representation of the i-th frame of the video.
* Select t-th frame whose representation a_t is nearest to the average of all frame representations.
* If there are multiple such frames, select the frame with the smallest index.
Find the index t of the frame he should select to generate a thumbnail.
Constraints
* 1 \leq N \leq 100
* 1 \leq a_i \leq 100
* All numbers given in input are integers
Input
Input is given from Standard Input in the following format:
N
a_{0} a_{1} ... a_{N-1}
Output
Print the answer.
Examples
Input
3
1 2 3
Output
1
Input
4
2 5 2 5
Output
0
Submitted Solution:
```
N= int(input())
A= list(map(int,input().split()))
Ave = sum(A)/len(A)
abs_distance =list()
for i in range(len(A)):
abs_distance.append(int(abs(A[i]-Ave)))
print(abs_distance.index(min(abs_distance)))
```
No
| 46,021 | [
0.327392578125,
0.1268310546875,
-0.05224609375,
0.09185791015625,
-0.4169921875,
-0.205078125,
-0.1649169921875,
0.261962890625,
0.247802734375,
1.0126953125,
0.5537109375,
-0.319580078125,
0.387451171875,
-0.8369140625,
-0.311767578125,
0.056396484375,
-0.142578125,
-0.826171875,... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Niwango-kun is an employee of Dwango Co., Ltd.
One day, he is asked to generate a thumbnail from a video a user submitted.
To generate a thumbnail, he needs to select a frame of the video according to the following procedure:
* Get an integer N and N integers a_0, a_1, ..., a_{N-1} as inputs. N denotes the number of the frames of the video, and each a_i denotes the representation of the i-th frame of the video.
* Select t-th frame whose representation a_t is nearest to the average of all frame representations.
* If there are multiple such frames, select the frame with the smallest index.
Find the index t of the frame he should select to generate a thumbnail.
Constraints
* 1 \leq N \leq 100
* 1 \leq a_i \leq 100
* All numbers given in input are integers
Input
Input is given from Standard Input in the following format:
N
a_{0} a_{1} ... a_{N-1}
Output
Print the answer.
Examples
Input
3
1 2 3
Output
1
Input
4
2 5 2 5
Output
0
Submitted Solution:
```
N = int(input())
S = input()
Q = int(input())
Ks= list(map(int, input().split()))
for K in Ks:
cnt_d, cnt_m, sum_dm = 0,0,0
ans = 0
for i in range(N):
if i-K>=0:
if S[i-K]=="D":
cnt_d -= 1
sum_dm -= cnt_m
elif S[i-K]=="M":
cnt_m -= 1
if S[i]=="D":
cnt_d += 1
elif S[i]=="M":
sum_dm += cnt_d
cnt_m += 1
elif S[i]=="C":
ans += sum_dm
print(ans)
```
No
| 46,022 | [
0.2939453125,
0.10479736328125,
-0.0655517578125,
0.075927734375,
-0.388671875,
-0.205078125,
-0.117919921875,
0.259765625,
0.23291015625,
0.97412109375,
0.54150390625,
-0.31787109375,
0.36376953125,
-0.8056640625,
-0.33447265625,
0.06768798828125,
-0.2044677734375,
-0.82763671875,... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Niwango-kun is an employee of Dwango Co., Ltd.
One day, he is asked to generate a thumbnail from a video a user submitted.
To generate a thumbnail, he needs to select a frame of the video according to the following procedure:
* Get an integer N and N integers a_0, a_1, ..., a_{N-1} as inputs. N denotes the number of the frames of the video, and each a_i denotes the representation of the i-th frame of the video.
* Select t-th frame whose representation a_t is nearest to the average of all frame representations.
* If there are multiple such frames, select the frame with the smallest index.
Find the index t of the frame he should select to generate a thumbnail.
Constraints
* 1 \leq N \leq 100
* 1 \leq a_i \leq 100
* All numbers given in input are integers
Input
Input is given from Standard Input in the following format:
N
a_{0} a_{1} ... a_{N-1}
Output
Print the answer.
Examples
Input
3
1 2 3
Output
1
Input
4
2 5 2 5
Output
0
Submitted Solution:
```
input();a=map(int,input().split());b=[abs(i-sum(a)/n)for i in a];print(b.index(min(b)))
```
No
| 46,023 | [
0.35986328125,
0.132080078125,
-0.06793212890625,
0.0562744140625,
-0.4169921875,
-0.2578125,
-0.1829833984375,
0.2724609375,
0.2374267578125,
1.0048828125,
0.564453125,
-0.30517578125,
0.400634765625,
-0.8388671875,
-0.3193359375,
0.07269287109375,
-0.144287109375,
-0.7978515625,
... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dreamoon is a big fan of the Codeforces contests.
One day, he claimed that he will collect all the places from 1 to 54 after two more rated contests. It's amazing!
Based on this, you come up with the following problem:
There is a person who participated in n Codeforces rounds. His place in the first round is a_1, his place in the second round is a_2, ..., his place in the n-th round is a_n.
You are given a positive non-zero integer x.
Please, find the largest v such that this person can collect all the places from 1 to v after x more rated contests.
In other words, you need to find the largest v, such that it is possible, that after x more rated contests, for each 1 ≤ i ≤ v, there will exist a contest where this person took the i-th place.
For example, if n=6, x=2 and a=[3,1,1,5,7,10] then answer is v=5, because if on the next two contest he will take places 2 and 4, then he will collect all places from 1 to 5, so it is possible to get v=5.
Input
The first line contains an integer t (1 ≤ t ≤ 5) denoting the number of test cases in the input.
Each test case contains two lines. The first line contains two integers n, x (1 ≤ n, x ≤ 100). The second line contains n positive non-zero integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 100).
Output
For each test case print one line containing the largest v, such that it is possible that after x other contests, for each 1 ≤ i ≤ v, there will exist a contest where this person took the i-th place.
Example
Input
5
6 2
3 1 1 5 7 10
1 100
100
11 1
1 1 1 1 1 1 1 1 1 1 1
1 1
1
4 57
80 60 40 20
Output
5
101
2
2
60
Note
The first test case is described in the statement.
In the second test case, the person has one hundred future contests, so he can take place 1,2,…,99 and place 101 on them in some order, to collect places 1,2,…,101.
Tags: implementation
Correct Solution:
```
def fun(x):
return list(dict.fromkeys(x))
for i in range(int(input())):
lis=[];m=0
n,x=map(int,input().split())
lis=(list(map(int,input().split())))
for i in range(1,x+1):
while i in lis:i+=1
if i not in lis:
lis.append(i)
lis=fun(lis)
lis.sort()
for j in range(len(lis)-1):
if (lis[j]+1)==lis[j+1]:
m+=1
else:break
print(m+1)
```
| 46,398 | [
0.45703125,
0.063232421875,
-0.12225341796875,
0.50244140625,
-0.5234375,
-0.62744140625,
0.11968994140625,
0.379150390625,
0.242919921875,
0.71875,
0.701171875,
-0.1063232421875,
0.342041015625,
-0.77099609375,
-0.40966796875,
-0.1688232421875,
-0.583984375,
-0.755859375,
-0.497... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dreamoon is a big fan of the Codeforces contests.
One day, he claimed that he will collect all the places from 1 to 54 after two more rated contests. It's amazing!
Based on this, you come up with the following problem:
There is a person who participated in n Codeforces rounds. His place in the first round is a_1, his place in the second round is a_2, ..., his place in the n-th round is a_n.
You are given a positive non-zero integer x.
Please, find the largest v such that this person can collect all the places from 1 to v after x more rated contests.
In other words, you need to find the largest v, such that it is possible, that after x more rated contests, for each 1 ≤ i ≤ v, there will exist a contest where this person took the i-th place.
For example, if n=6, x=2 and a=[3,1,1,5,7,10] then answer is v=5, because if on the next two contest he will take places 2 and 4, then he will collect all places from 1 to 5, so it is possible to get v=5.
Input
The first line contains an integer t (1 ≤ t ≤ 5) denoting the number of test cases in the input.
Each test case contains two lines. The first line contains two integers n, x (1 ≤ n, x ≤ 100). The second line contains n positive non-zero integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 100).
Output
For each test case print one line containing the largest v, such that it is possible that after x other contests, for each 1 ≤ i ≤ v, there will exist a contest where this person took the i-th place.
Example
Input
5
6 2
3 1 1 5 7 10
1 100
100
11 1
1 1 1 1 1 1 1 1 1 1 1
1 1
1
4 57
80 60 40 20
Output
5
101
2
2
60
Note
The first test case is described in the statement.
In the second test case, the person has one hundred future contests, so he can take place 1,2,…,99 and place 101 on them in some order, to collect places 1,2,…,101.
Tags: implementation
Correct Solution:
```
for i in range(int(input())):
n, x=map(int,input().split())
a=list(map(int,input().split()))
i=0
while(x>-1):
i+=1
if(i not in a):
x-=1
print(i-1)
```
| 46,399 | [
0.45703125,
0.063232421875,
-0.12225341796875,
0.50244140625,
-0.5234375,
-0.62744140625,
0.11968994140625,
0.379150390625,
0.242919921875,
0.71875,
0.701171875,
-0.1063232421875,
0.342041015625,
-0.77099609375,
-0.40966796875,
-0.1688232421875,
-0.583984375,
-0.755859375,
-0.497... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dreamoon is a big fan of the Codeforces contests.
One day, he claimed that he will collect all the places from 1 to 54 after two more rated contests. It's amazing!
Based on this, you come up with the following problem:
There is a person who participated in n Codeforces rounds. His place in the first round is a_1, his place in the second round is a_2, ..., his place in the n-th round is a_n.
You are given a positive non-zero integer x.
Please, find the largest v such that this person can collect all the places from 1 to v after x more rated contests.
In other words, you need to find the largest v, such that it is possible, that after x more rated contests, for each 1 ≤ i ≤ v, there will exist a contest where this person took the i-th place.
For example, if n=6, x=2 and a=[3,1,1,5,7,10] then answer is v=5, because if on the next two contest he will take places 2 and 4, then he will collect all places from 1 to 5, so it is possible to get v=5.
Input
The first line contains an integer t (1 ≤ t ≤ 5) denoting the number of test cases in the input.
Each test case contains two lines. The first line contains two integers n, x (1 ≤ n, x ≤ 100). The second line contains n positive non-zero integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 100).
Output
For each test case print one line containing the largest v, such that it is possible that after x other contests, for each 1 ≤ i ≤ v, there will exist a contest where this person took the i-th place.
Example
Input
5
6 2
3 1 1 5 7 10
1 100
100
11 1
1 1 1 1 1 1 1 1 1 1 1
1 1
1
4 57
80 60 40 20
Output
5
101
2
2
60
Note
The first test case is described in the statement.
In the second test case, the person has one hundred future contests, so he can take place 1,2,…,99 and place 101 on them in some order, to collect places 1,2,…,101.
Tags: implementation
Correct Solution:
```
def solve(a,x):
for i in range(1,max(a)+x+5):
if x<= 0 and i not in a:
return i
if x > 0 and i not in a:
x -= 1
t = int(input())
while t:
t -= 1
n, x = list(map(int,input().split()))
a = list(map(int,input().split()))
v = solve(a,x)
print(v-1)
```
| 46,400 | [
0.45703125,
0.063232421875,
-0.12225341796875,
0.50244140625,
-0.5234375,
-0.62744140625,
0.11968994140625,
0.379150390625,
0.242919921875,
0.71875,
0.701171875,
-0.1063232421875,
0.342041015625,
-0.77099609375,
-0.40966796875,
-0.1688232421875,
-0.583984375,
-0.755859375,
-0.497... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dreamoon is a big fan of the Codeforces contests.
One day, he claimed that he will collect all the places from 1 to 54 after two more rated contests. It's amazing!
Based on this, you come up with the following problem:
There is a person who participated in n Codeforces rounds. His place in the first round is a_1, his place in the second round is a_2, ..., his place in the n-th round is a_n.
You are given a positive non-zero integer x.
Please, find the largest v such that this person can collect all the places from 1 to v after x more rated contests.
In other words, you need to find the largest v, such that it is possible, that after x more rated contests, for each 1 ≤ i ≤ v, there will exist a contest where this person took the i-th place.
For example, if n=6, x=2 and a=[3,1,1,5,7,10] then answer is v=5, because if on the next two contest he will take places 2 and 4, then he will collect all places from 1 to 5, so it is possible to get v=5.
Input
The first line contains an integer t (1 ≤ t ≤ 5) denoting the number of test cases in the input.
Each test case contains two lines. The first line contains two integers n, x (1 ≤ n, x ≤ 100). The second line contains n positive non-zero integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 100).
Output
For each test case print one line containing the largest v, such that it is possible that after x other contests, for each 1 ≤ i ≤ v, there will exist a contest where this person took the i-th place.
Example
Input
5
6 2
3 1 1 5 7 10
1 100
100
11 1
1 1 1 1 1 1 1 1 1 1 1
1 1
1
4 57
80 60 40 20
Output
5
101
2
2
60
Note
The first test case is described in the statement.
In the second test case, the person has one hundred future contests, so he can take place 1,2,…,99 and place 101 on them in some order, to collect places 1,2,…,101.
Tags: implementation
Correct Solution:
```
# coding: utf-8
# Your code here!
Q=int(input())
for _ in range(Q):
judge=False
n,x=map(int,input().split())
A=list(map(int,input().split()))
A=list(set(A))
A.sort(reverse=True)
for i in range(1,300):
if i in A:
if A:
A.pop(-1)
else:
judge=True
else:
if x>0:
x-=1
else:
judge=True
if judge:
print(i-1)
break
```
| 46,401 | [
0.45703125,
0.063232421875,
-0.12225341796875,
0.50244140625,
-0.5234375,
-0.62744140625,
0.11968994140625,
0.379150390625,
0.242919921875,
0.71875,
0.701171875,
-0.1063232421875,
0.342041015625,
-0.77099609375,
-0.40966796875,
-0.1688232421875,
-0.583984375,
-0.755859375,
-0.497... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dreamoon is a big fan of the Codeforces contests.
One day, he claimed that he will collect all the places from 1 to 54 after two more rated contests. It's amazing!
Based on this, you come up with the following problem:
There is a person who participated in n Codeforces rounds. His place in the first round is a_1, his place in the second round is a_2, ..., his place in the n-th round is a_n.
You are given a positive non-zero integer x.
Please, find the largest v such that this person can collect all the places from 1 to v after x more rated contests.
In other words, you need to find the largest v, such that it is possible, that after x more rated contests, for each 1 ≤ i ≤ v, there will exist a contest where this person took the i-th place.
For example, if n=6, x=2 and a=[3,1,1,5,7,10] then answer is v=5, because if on the next two contest he will take places 2 and 4, then he will collect all places from 1 to 5, so it is possible to get v=5.
Input
The first line contains an integer t (1 ≤ t ≤ 5) denoting the number of test cases in the input.
Each test case contains two lines. The first line contains two integers n, x (1 ≤ n, x ≤ 100). The second line contains n positive non-zero integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 100).
Output
For each test case print one line containing the largest v, such that it is possible that after x other contests, for each 1 ≤ i ≤ v, there will exist a contest where this person took the i-th place.
Example
Input
5
6 2
3 1 1 5 7 10
1 100
100
11 1
1 1 1 1 1 1 1 1 1 1 1
1 1
1
4 57
80 60 40 20
Output
5
101
2
2
60
Note
The first test case is described in the statement.
In the second test case, the person has one hundred future contests, so he can take place 1,2,…,99 and place 101 on them in some order, to collect places 1,2,…,101.
Tags: implementation
Correct Solution:
```
t = int(input())
for _ in range(t):
n, x = map(int, input().split())
a = list(map(int, input().split()))
place = 0
while True:
place += 1
if place not in a:
if x == 0:
place -= 1
break
x -= 1
print(place)
```
| 46,402 | [
0.45703125,
0.063232421875,
-0.12225341796875,
0.50244140625,
-0.5234375,
-0.62744140625,
0.11968994140625,
0.379150390625,
0.242919921875,
0.71875,
0.701171875,
-0.1063232421875,
0.342041015625,
-0.77099609375,
-0.40966796875,
-0.1688232421875,
-0.583984375,
-0.755859375,
-0.497... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dreamoon is a big fan of the Codeforces contests.
One day, he claimed that he will collect all the places from 1 to 54 after two more rated contests. It's amazing!
Based on this, you come up with the following problem:
There is a person who participated in n Codeforces rounds. His place in the first round is a_1, his place in the second round is a_2, ..., his place in the n-th round is a_n.
You are given a positive non-zero integer x.
Please, find the largest v such that this person can collect all the places from 1 to v after x more rated contests.
In other words, you need to find the largest v, such that it is possible, that after x more rated contests, for each 1 ≤ i ≤ v, there will exist a contest where this person took the i-th place.
For example, if n=6, x=2 and a=[3,1,1,5,7,10] then answer is v=5, because if on the next two contest he will take places 2 and 4, then he will collect all places from 1 to 5, so it is possible to get v=5.
Input
The first line contains an integer t (1 ≤ t ≤ 5) denoting the number of test cases in the input.
Each test case contains two lines. The first line contains two integers n, x (1 ≤ n, x ≤ 100). The second line contains n positive non-zero integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 100).
Output
For each test case print one line containing the largest v, such that it is possible that after x other contests, for each 1 ≤ i ≤ v, there will exist a contest where this person took the i-th place.
Example
Input
5
6 2
3 1 1 5 7 10
1 100
100
11 1
1 1 1 1 1 1 1 1 1 1 1
1 1
1
4 57
80 60 40 20
Output
5
101
2
2
60
Note
The first test case is described in the statement.
In the second test case, the person has one hundred future contests, so he can take place 1,2,…,99 and place 101 on them in some order, to collect places 1,2,…,101.
Tags: implementation
Correct Solution:
```
#include <bits/stdc++.h>
#define STD /*
from sys import exit as sys_ret
"""****************************
cat /dev/ass > /dev/head
Ctrl+C
cat /knowledge > /dev/head
© Jakov Gellert
frvr.ru
****************************"""
# */ using namespace std; int
t = int(input())
for _ in range(t):
possible = [0 for i in range(210)]
amount, moves = map(int, input().split())
tmp = [int(x) for x in input().split()]
for i in tmp:
possible[i] = 1
i = 1
while moves != 0:
if possible[i] == 0:
moves -= 1
possible[i] = 1
i += 1
res, i = 0, 1
while possible[i] != 0:
res += 1
i += 1
print(res)
```
| 46,403 | [
0.45703125,
0.063232421875,
-0.12225341796875,
0.50244140625,
-0.5234375,
-0.62744140625,
0.11968994140625,
0.379150390625,
0.242919921875,
0.71875,
0.701171875,
-0.1063232421875,
0.342041015625,
-0.77099609375,
-0.40966796875,
-0.1688232421875,
-0.583984375,
-0.755859375,
-0.497... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dreamoon is a big fan of the Codeforces contests.
One day, he claimed that he will collect all the places from 1 to 54 after two more rated contests. It's amazing!
Based on this, you come up with the following problem:
There is a person who participated in n Codeforces rounds. His place in the first round is a_1, his place in the second round is a_2, ..., his place in the n-th round is a_n.
You are given a positive non-zero integer x.
Please, find the largest v such that this person can collect all the places from 1 to v after x more rated contests.
In other words, you need to find the largest v, such that it is possible, that after x more rated contests, for each 1 ≤ i ≤ v, there will exist a contest where this person took the i-th place.
For example, if n=6, x=2 and a=[3,1,1,5,7,10] then answer is v=5, because if on the next two contest he will take places 2 and 4, then he will collect all places from 1 to 5, so it is possible to get v=5.
Input
The first line contains an integer t (1 ≤ t ≤ 5) denoting the number of test cases in the input.
Each test case contains two lines. The first line contains two integers n, x (1 ≤ n, x ≤ 100). The second line contains n positive non-zero integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 100).
Output
For each test case print one line containing the largest v, such that it is possible that after x other contests, for each 1 ≤ i ≤ v, there will exist a contest where this person took the i-th place.
Example
Input
5
6 2
3 1 1 5 7 10
1 100
100
11 1
1 1 1 1 1 1 1 1 1 1 1
1 1
1
4 57
80 60 40 20
Output
5
101
2
2
60
Note
The first test case is described in the statement.
In the second test case, the person has one hundred future contests, so he can take place 1,2,…,99 and place 101 on them in some order, to collect places 1,2,…,101.
Tags: implementation
Correct Solution:
```
for _ in range(int(input())):
n,x = [int(s) for s in input().split()]
l = list(set([int(s) for s in input().split()]))
n = len(l)
l.sort()
curr = 1
i = 0
# print(l)
while x>0 and i < n:
# print(x,i)
if l[i]==curr:
curr+=1
i+=1
else:
x-=1
curr+=1
while x:
x-=1
curr+=1
while i<n:
if l[i]==curr:
curr+=1
i+=1
else:
break
print(curr-1)
```
| 46,404 | [
0.45703125,
0.063232421875,
-0.12225341796875,
0.50244140625,
-0.5234375,
-0.62744140625,
0.11968994140625,
0.379150390625,
0.242919921875,
0.71875,
0.701171875,
-0.1063232421875,
0.342041015625,
-0.77099609375,
-0.40966796875,
-0.1688232421875,
-0.583984375,
-0.755859375,
-0.497... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dreamoon is a big fan of the Codeforces contests.
One day, he claimed that he will collect all the places from 1 to 54 after two more rated contests. It's amazing!
Based on this, you come up with the following problem:
There is a person who participated in n Codeforces rounds. His place in the first round is a_1, his place in the second round is a_2, ..., his place in the n-th round is a_n.
You are given a positive non-zero integer x.
Please, find the largest v such that this person can collect all the places from 1 to v after x more rated contests.
In other words, you need to find the largest v, such that it is possible, that after x more rated contests, for each 1 ≤ i ≤ v, there will exist a contest where this person took the i-th place.
For example, if n=6, x=2 and a=[3,1,1,5,7,10] then answer is v=5, because if on the next two contest he will take places 2 and 4, then he will collect all places from 1 to 5, so it is possible to get v=5.
Input
The first line contains an integer t (1 ≤ t ≤ 5) denoting the number of test cases in the input.
Each test case contains two lines. The first line contains two integers n, x (1 ≤ n, x ≤ 100). The second line contains n positive non-zero integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 100).
Output
For each test case print one line containing the largest v, such that it is possible that after x other contests, for each 1 ≤ i ≤ v, there will exist a contest where this person took the i-th place.
Example
Input
5
6 2
3 1 1 5 7 10
1 100
100
11 1
1 1 1 1 1 1 1 1 1 1 1
1 1
1
4 57
80 60 40 20
Output
5
101
2
2
60
Note
The first test case is described in the statement.
In the second test case, the person has one hundred future contests, so he can take place 1,2,…,99 and place 101 on them in some order, to collect places 1,2,…,101.
Tags: implementation
Correct Solution:
```
ncases = int(input())
answer = []
for i in range(ncases):
line1 = input().split(' ')
line2 = input().split(' ')
l1 = [0]
l2 = []
for char in line2:
l1.append(int(char))
if (len(l1) == 2):
gap = l1[1]-l1[0]
l2.append(gap)
else:
l1.sort()
for j in range(len(l1)-1):
gap = l1[j+1]-l1[j]
l2.append(gap)
chance = int(line1[1])
for k in range(0,len(l2)):
gap = l2[k]
if (gap <= 1):
chance = chance
else: #chance == 0 means no chances left#
chance = chance - gap + 1
if (chance<0): #case where the the gap between the two ranks are greater than the chances remaining#
maxpossible = l1[k]+chance+gap-1
answer.append(maxpossible)
chance = -1
break
elif (chance == 0 and l1[-1] == l1[k+1]):#no more chances left and remaining ranks are the same#
answer.append(l1[k+1])
chance = -1
break
if (chance>0): #more chances left
maxposs = l1[-1] + chance
answer.append(maxposs)
if (chance == 0):
answer.append(l1[-1])
for ans in answer:
print(ans)
```
| 46,405 | [
0.45703125,
0.063232421875,
-0.12225341796875,
0.50244140625,
-0.5234375,
-0.62744140625,
0.11968994140625,
0.379150390625,
0.242919921875,
0.71875,
0.701171875,
-0.1063232421875,
0.342041015625,
-0.77099609375,
-0.40966796875,
-0.1688232421875,
-0.583984375,
-0.755859375,
-0.497... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dreamoon is a big fan of the Codeforces contests.
One day, he claimed that he will collect all the places from 1 to 54 after two more rated contests. It's amazing!
Based on this, you come up with the following problem:
There is a person who participated in n Codeforces rounds. His place in the first round is a_1, his place in the second round is a_2, ..., his place in the n-th round is a_n.
You are given a positive non-zero integer x.
Please, find the largest v such that this person can collect all the places from 1 to v after x more rated contests.
In other words, you need to find the largest v, such that it is possible, that after x more rated contests, for each 1 ≤ i ≤ v, there will exist a contest where this person took the i-th place.
For example, if n=6, x=2 and a=[3,1,1,5,7,10] then answer is v=5, because if on the next two contest he will take places 2 and 4, then he will collect all places from 1 to 5, so it is possible to get v=5.
Input
The first line contains an integer t (1 ≤ t ≤ 5) denoting the number of test cases in the input.
Each test case contains two lines. The first line contains two integers n, x (1 ≤ n, x ≤ 100). The second line contains n positive non-zero integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 100).
Output
For each test case print one line containing the largest v, such that it is possible that after x other contests, for each 1 ≤ i ≤ v, there will exist a contest where this person took the i-th place.
Example
Input
5
6 2
3 1 1 5 7 10
1 100
100
11 1
1 1 1 1 1 1 1 1 1 1 1
1 1
1
4 57
80 60 40 20
Output
5
101
2
2
60
Note
The first test case is described in the statement.
In the second test case, the person has one hundred future contests, so he can take place 1,2,…,99 and place 101 on them in some order, to collect places 1,2,…,101.
Submitted Solution:
```
import sys
import math
import heapq
import collections
def inputnum():
return(int(input()))
def inputnums():
return(map(int,input().split()))
def inputlist():
return(list(map(int,input().split())))
def inputstring():
return([x for x in input()])
def inputmatrixchar(rows):
arr2d = [[j for j in input().strip()] for i in range(rows)]
return arr2d
def inputmatrixint(rows):
arr2d = []
for _ in range(rows):
arr2d.append([int(i) for i in input().split()])
return arr2d
t=int(input())
for q in range(t):
n, x = inputnums()
a = inputlist()
a.sort()
if a[0] <= x+1:
x += 1
for i in range(1, n):
if a[i] != a[i-1]:
if a[i] <= x:
x += 1
elif a[i] == x+1:
x += 1
else:
break
print (x)
```
Yes
| 46,406 | [
0.46875,
0.1429443359375,
-0.10833740234375,
0.336669921875,
-0.5849609375,
-0.482421875,
0.10321044921875,
0.46923828125,
0.1346435546875,
0.751953125,
0.6279296875,
-0.04864501953125,
0.3173828125,
-0.7578125,
-0.391845703125,
-0.20947265625,
-0.5302734375,
-0.77490234375,
-0.5... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dreamoon is a big fan of the Codeforces contests.
One day, he claimed that he will collect all the places from 1 to 54 after two more rated contests. It's amazing!
Based on this, you come up with the following problem:
There is a person who participated in n Codeforces rounds. His place in the first round is a_1, his place in the second round is a_2, ..., his place in the n-th round is a_n.
You are given a positive non-zero integer x.
Please, find the largest v such that this person can collect all the places from 1 to v after x more rated contests.
In other words, you need to find the largest v, such that it is possible, that after x more rated contests, for each 1 ≤ i ≤ v, there will exist a contest where this person took the i-th place.
For example, if n=6, x=2 and a=[3,1,1,5,7,10] then answer is v=5, because if on the next two contest he will take places 2 and 4, then he will collect all places from 1 to 5, so it is possible to get v=5.
Input
The first line contains an integer t (1 ≤ t ≤ 5) denoting the number of test cases in the input.
Each test case contains two lines. The first line contains two integers n, x (1 ≤ n, x ≤ 100). The second line contains n positive non-zero integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 100).
Output
For each test case print one line containing the largest v, such that it is possible that after x other contests, for each 1 ≤ i ≤ v, there will exist a contest where this person took the i-th place.
Example
Input
5
6 2
3 1 1 5 7 10
1 100
100
11 1
1 1 1 1 1 1 1 1 1 1 1
1 1
1
4 57
80 60 40 20
Output
5
101
2
2
60
Note
The first test case is described in the statement.
In the second test case, the person has one hundred future contests, so he can take place 1,2,…,99 and place 101 on them in some order, to collect places 1,2,…,101.
Submitted Solution:
```
t = int(input())
for _ in range(t):
n, x = map(int, input().split())
arr = list(map(int, input().split()))
s = set(arr)
mx = n + x + 1
ans = -1
for idx in range(1, mx):
if idx in s:
ans = idx
continue
if x > 0:
x -= 1
ans = idx
else:
break
print(ans)
```
Yes
| 46,407 | [
0.46875,
0.1429443359375,
-0.10833740234375,
0.336669921875,
-0.5849609375,
-0.482421875,
0.10321044921875,
0.46923828125,
0.1346435546875,
0.751953125,
0.6279296875,
-0.04864501953125,
0.3173828125,
-0.7578125,
-0.391845703125,
-0.20947265625,
-0.5302734375,
-0.77490234375,
-0.5... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dreamoon is a big fan of the Codeforces contests.
One day, he claimed that he will collect all the places from 1 to 54 after two more rated contests. It's amazing!
Based on this, you come up with the following problem:
There is a person who participated in n Codeforces rounds. His place in the first round is a_1, his place in the second round is a_2, ..., his place in the n-th round is a_n.
You are given a positive non-zero integer x.
Please, find the largest v such that this person can collect all the places from 1 to v after x more rated contests.
In other words, you need to find the largest v, such that it is possible, that after x more rated contests, for each 1 ≤ i ≤ v, there will exist a contest where this person took the i-th place.
For example, if n=6, x=2 and a=[3,1,1,5,7,10] then answer is v=5, because if on the next two contest he will take places 2 and 4, then he will collect all places from 1 to 5, so it is possible to get v=5.
Input
The first line contains an integer t (1 ≤ t ≤ 5) denoting the number of test cases in the input.
Each test case contains two lines. The first line contains two integers n, x (1 ≤ n, x ≤ 100). The second line contains n positive non-zero integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 100).
Output
For each test case print one line containing the largest v, such that it is possible that after x other contests, for each 1 ≤ i ≤ v, there will exist a contest where this person took the i-th place.
Example
Input
5
6 2
3 1 1 5 7 10
1 100
100
11 1
1 1 1 1 1 1 1 1 1 1 1
1 1
1
4 57
80 60 40 20
Output
5
101
2
2
60
Note
The first test case is described in the statement.
In the second test case, the person has one hundred future contests, so he can take place 1,2,…,99 and place 101 on them in some order, to collect places 1,2,…,101.
Submitted Solution:
```
for _ in range(int(input())):
n,x=map(int,input().split())
l=list(map(int,input().split()))
s=list(set(l))
s.sort()
for i in s:
if i<=x+1:
x+=1
print(x)
```
Yes
| 46,408 | [
0.46875,
0.1429443359375,
-0.10833740234375,
0.336669921875,
-0.5849609375,
-0.482421875,
0.10321044921875,
0.46923828125,
0.1346435546875,
0.751953125,
0.6279296875,
-0.04864501953125,
0.3173828125,
-0.7578125,
-0.391845703125,
-0.20947265625,
-0.5302734375,
-0.77490234375,
-0.5... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dreamoon is a big fan of the Codeforces contests.
One day, he claimed that he will collect all the places from 1 to 54 after two more rated contests. It's amazing!
Based on this, you come up with the following problem:
There is a person who participated in n Codeforces rounds. His place in the first round is a_1, his place in the second round is a_2, ..., his place in the n-th round is a_n.
You are given a positive non-zero integer x.
Please, find the largest v such that this person can collect all the places from 1 to v after x more rated contests.
In other words, you need to find the largest v, such that it is possible, that after x more rated contests, for each 1 ≤ i ≤ v, there will exist a contest where this person took the i-th place.
For example, if n=6, x=2 and a=[3,1,1,5,7,10] then answer is v=5, because if on the next two contest he will take places 2 and 4, then he will collect all places from 1 to 5, so it is possible to get v=5.
Input
The first line contains an integer t (1 ≤ t ≤ 5) denoting the number of test cases in the input.
Each test case contains two lines. The first line contains two integers n, x (1 ≤ n, x ≤ 100). The second line contains n positive non-zero integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 100).
Output
For each test case print one line containing the largest v, such that it is possible that after x other contests, for each 1 ≤ i ≤ v, there will exist a contest where this person took the i-th place.
Example
Input
5
6 2
3 1 1 5 7 10
1 100
100
11 1
1 1 1 1 1 1 1 1 1 1 1
1 1
1
4 57
80 60 40 20
Output
5
101
2
2
60
Note
The first test case is described in the statement.
In the second test case, the person has one hundred future contests, so he can take place 1,2,…,99 and place 101 on them in some order, to collect places 1,2,…,101.
Submitted Solution:
```
# MasterKali
from sys import stdin
from collections import Counter
from math import sqrt, factorial, log10, log, floor, ceil
from itertools import permutations, combinations, combinations_with_replacement
input = stdin.readline
def li(): return list(map(int, input().split()))
def lis(): return list(map(str, input().split()))
def mp(): return map(int, input().split())
def inp(): return int(input())
def inps(): return str(input().strip())
def pr(n): return stdout.write(str(n)+"\n")
INF = float('inf')
def solve():
n, k = mp()
a = li()
a.sort()
a = set(a)
res, i = 0, 1
while k > 0:
if i not in a:
res+=1
k-=1
else:
res+=1
i+=1
while(res+1 in a):
res+=1
print(res)
t = inp()
for i in range(t):
solve()
```
Yes
| 46,409 | [
0.46875,
0.1429443359375,
-0.10833740234375,
0.336669921875,
-0.5849609375,
-0.482421875,
0.10321044921875,
0.46923828125,
0.1346435546875,
0.751953125,
0.6279296875,
-0.04864501953125,
0.3173828125,
-0.7578125,
-0.391845703125,
-0.20947265625,
-0.5302734375,
-0.77490234375,
-0.5... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dreamoon is a big fan of the Codeforces contests.
One day, he claimed that he will collect all the places from 1 to 54 after two more rated contests. It's amazing!
Based on this, you come up with the following problem:
There is a person who participated in n Codeforces rounds. His place in the first round is a_1, his place in the second round is a_2, ..., his place in the n-th round is a_n.
You are given a positive non-zero integer x.
Please, find the largest v such that this person can collect all the places from 1 to v after x more rated contests.
In other words, you need to find the largest v, such that it is possible, that after x more rated contests, for each 1 ≤ i ≤ v, there will exist a contest where this person took the i-th place.
For example, if n=6, x=2 and a=[3,1,1,5,7,10] then answer is v=5, because if on the next two contest he will take places 2 and 4, then he will collect all places from 1 to 5, so it is possible to get v=5.
Input
The first line contains an integer t (1 ≤ t ≤ 5) denoting the number of test cases in the input.
Each test case contains two lines. The first line contains two integers n, x (1 ≤ n, x ≤ 100). The second line contains n positive non-zero integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 100).
Output
For each test case print one line containing the largest v, such that it is possible that after x other contests, for each 1 ≤ i ≤ v, there will exist a contest where this person took the i-th place.
Example
Input
5
6 2
3 1 1 5 7 10
1 100
100
11 1
1 1 1 1 1 1 1 1 1 1 1
1 1
1
4 57
80 60 40 20
Output
5
101
2
2
60
Note
The first test case is described in the statement.
In the second test case, the person has one hundred future contests, so he can take place 1,2,…,99 and place 101 on them in some order, to collect places 1,2,…,101.
Submitted Solution:
```
from sys import stdin
a=int(stdin.readline())
for b in range(0,a):
c=stdin.readline().split()
d=int(c[0])
e=int(c[1])
g=stdin.readline().split()
A=set()
for f in range(0,d):
A.add(int(g[f]))
A=list(A)
A.append(0)
list.sort(A)
count=0
ans=0
while count<len(A)-1:
if A[count+1]-ans==1:
ans=A[count+1]
count+=1
continue
elif e-A[count+1]+A[count]+1>0:
ans=A[count+1]
e=e-A[count+1]+A[count]+1
count+=1
elif e-A[count+1]+A[count]+1<=0 and e>0:
ans=ans+e
e=0
count+=1
else:
count+=1
continue
print(ans+1)
```
No
| 46,410 | [
0.46875,
0.1429443359375,
-0.10833740234375,
0.336669921875,
-0.5849609375,
-0.482421875,
0.10321044921875,
0.46923828125,
0.1346435546875,
0.751953125,
0.6279296875,
-0.04864501953125,
0.3173828125,
-0.7578125,
-0.391845703125,
-0.20947265625,
-0.5302734375,
-0.77490234375,
-0.5... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dreamoon is a big fan of the Codeforces contests.
One day, he claimed that he will collect all the places from 1 to 54 after two more rated contests. It's amazing!
Based on this, you come up with the following problem:
There is a person who participated in n Codeforces rounds. His place in the first round is a_1, his place in the second round is a_2, ..., his place in the n-th round is a_n.
You are given a positive non-zero integer x.
Please, find the largest v such that this person can collect all the places from 1 to v after x more rated contests.
In other words, you need to find the largest v, such that it is possible, that after x more rated contests, for each 1 ≤ i ≤ v, there will exist a contest where this person took the i-th place.
For example, if n=6, x=2 and a=[3,1,1,5,7,10] then answer is v=5, because if on the next two contest he will take places 2 and 4, then he will collect all places from 1 to 5, so it is possible to get v=5.
Input
The first line contains an integer t (1 ≤ t ≤ 5) denoting the number of test cases in the input.
Each test case contains two lines. The first line contains two integers n, x (1 ≤ n, x ≤ 100). The second line contains n positive non-zero integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 100).
Output
For each test case print one line containing the largest v, such that it is possible that after x other contests, for each 1 ≤ i ≤ v, there will exist a contest where this person took the i-th place.
Example
Input
5
6 2
3 1 1 5 7 10
1 100
100
11 1
1 1 1 1 1 1 1 1 1 1 1
1 1
1
4 57
80 60 40 20
Output
5
101
2
2
60
Note
The first test case is described in the statement.
In the second test case, the person has one hundred future contests, so he can take place 1,2,…,99 and place 101 on them in some order, to collect places 1,2,…,101.
Submitted Solution:
```
for i in range(int(input())):
cnt=0
n,x=map(int,input().split())
a=sorted(list(map(int,input().split())))
for j in range(0,n):
if a[j]-a[j-1]<=x:
cnt=cnt+1
if x == 1 or n == 1: print(a[-1] + 1)
else:
print(a[cnt-2])
```
No
| 46,411 | [
0.46875,
0.1429443359375,
-0.10833740234375,
0.336669921875,
-0.5849609375,
-0.482421875,
0.10321044921875,
0.46923828125,
0.1346435546875,
0.751953125,
0.6279296875,
-0.04864501953125,
0.3173828125,
-0.7578125,
-0.391845703125,
-0.20947265625,
-0.5302734375,
-0.77490234375,
-0.5... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dreamoon is a big fan of the Codeforces contests.
One day, he claimed that he will collect all the places from 1 to 54 after two more rated contests. It's amazing!
Based on this, you come up with the following problem:
There is a person who participated in n Codeforces rounds. His place in the first round is a_1, his place in the second round is a_2, ..., his place in the n-th round is a_n.
You are given a positive non-zero integer x.
Please, find the largest v such that this person can collect all the places from 1 to v after x more rated contests.
In other words, you need to find the largest v, such that it is possible, that after x more rated contests, for each 1 ≤ i ≤ v, there will exist a contest where this person took the i-th place.
For example, if n=6, x=2 and a=[3,1,1,5,7,10] then answer is v=5, because if on the next two contest he will take places 2 and 4, then he will collect all places from 1 to 5, so it is possible to get v=5.
Input
The first line contains an integer t (1 ≤ t ≤ 5) denoting the number of test cases in the input.
Each test case contains two lines. The first line contains two integers n, x (1 ≤ n, x ≤ 100). The second line contains n positive non-zero integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 100).
Output
For each test case print one line containing the largest v, such that it is possible that after x other contests, for each 1 ≤ i ≤ v, there will exist a contest where this person took the i-th place.
Example
Input
5
6 2
3 1 1 5 7 10
1 100
100
11 1
1 1 1 1 1 1 1 1 1 1 1
1 1
1
4 57
80 60 40 20
Output
5
101
2
2
60
Note
The first test case is described in the statement.
In the second test case, the person has one hundred future contests, so he can take place 1,2,…,99 and place 101 on them in some order, to collect places 1,2,…,101.
Submitted Solution:
```
t = int(input())
for i in range(t):
n_m = list(map(int,input().split()))
n = n_m[0]
x = n_m[1]
a = list(map(int,input().split()))
a.sort()
flag=0
for j in range(n):
if(j==0):
diff = a[0]-1
else:
diff = a[j]-a[j-1]-1
if(diff>0 and x>=diff):
x-=diff
elif(x==0 and diff>0):
print(a[j-1])
flag=1
break
elif(diff>0 and x<diff):
if(j==0):
print(1)
else:
print(a[j-1]+x)
flag=1
break
if(x>0 and flag==0):
print(a[n-1]+x)
elif(x==0 and flag==0):
print(a[n-1])
```
No
| 46,412 | [
0.46875,
0.1429443359375,
-0.10833740234375,
0.336669921875,
-0.5849609375,
-0.482421875,
0.10321044921875,
0.46923828125,
0.1346435546875,
0.751953125,
0.6279296875,
-0.04864501953125,
0.3173828125,
-0.7578125,
-0.391845703125,
-0.20947265625,
-0.5302734375,
-0.77490234375,
-0.5... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dreamoon is a big fan of the Codeforces contests.
One day, he claimed that he will collect all the places from 1 to 54 after two more rated contests. It's amazing!
Based on this, you come up with the following problem:
There is a person who participated in n Codeforces rounds. His place in the first round is a_1, his place in the second round is a_2, ..., his place in the n-th round is a_n.
You are given a positive non-zero integer x.
Please, find the largest v such that this person can collect all the places from 1 to v after x more rated contests.
In other words, you need to find the largest v, such that it is possible, that after x more rated contests, for each 1 ≤ i ≤ v, there will exist a contest where this person took the i-th place.
For example, if n=6, x=2 and a=[3,1,1,5,7,10] then answer is v=5, because if on the next two contest he will take places 2 and 4, then he will collect all places from 1 to 5, so it is possible to get v=5.
Input
The first line contains an integer t (1 ≤ t ≤ 5) denoting the number of test cases in the input.
Each test case contains two lines. The first line contains two integers n, x (1 ≤ n, x ≤ 100). The second line contains n positive non-zero integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 100).
Output
For each test case print one line containing the largest v, such that it is possible that after x other contests, for each 1 ≤ i ≤ v, there will exist a contest where this person took the i-th place.
Example
Input
5
6 2
3 1 1 5 7 10
1 100
100
11 1
1 1 1 1 1 1 1 1 1 1 1
1 1
1
4 57
80 60 40 20
Output
5
101
2
2
60
Note
The first test case is described in the statement.
In the second test case, the person has one hundred future contests, so he can take place 1,2,…,99 and place 101 on them in some order, to collect places 1,2,…,101.
Submitted Solution:
```
t = int(input())
for _ in range(t):
n, x = map(int, input().split())
a = list(map(int, input().split()))
pref = [0 for _ in range(102)]
v = x
for i in range(n):
pref[a[i] - 1] = 1
for j in range(102):
if sum(pref[0 : j + 1]) + x >= j + 1:
v = max(j + 1, v)
print(v)
```
No
| 46,413 | [
0.46875,
0.1429443359375,
-0.10833740234375,
0.336669921875,
-0.5849609375,
-0.482421875,
0.10321044921875,
0.46923828125,
0.1346435546875,
0.751953125,
0.6279296875,
-0.04864501953125,
0.3173828125,
-0.7578125,
-0.391845703125,
-0.20947265625,
-0.5302734375,
-0.77490234375,
-0.5... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To learn as soon as possible the latest news about their favourite fundamentally new operating system, BolgenOS community from Nizhni Tagil decided to develop a scheme. According to this scheme a community member, who is the first to learn the news, calls some other member, the latter, in his turn, calls some third member, and so on; i.e. a person with index i got a person with index fi, to whom he has to call, if he learns the news. With time BolgenOS community members understood that their scheme doesn't work sometimes — there were cases when some members didn't learn the news at all. Now they want to supplement the scheme: they add into the scheme some instructions of type (xi, yi), which mean that person xi has to call person yi as well. What is the minimum amount of instructions that they need to add so, that at the end everyone learns the news, no matter who is the first to learn it?
Input
The first input line contains number n (2 ≤ n ≤ 105) — amount of BolgenOS community members. The second line contains n space-separated integer numbers fi (1 ≤ fi ≤ n, i ≠ fi) — index of a person, to whom calls a person with index i.
Output
In the first line output one number — the minimum amount of instructions to add. Then output one of the possible variants to add these instructions into the scheme, one instruction in each line. If the solution is not unique, output any.
Examples
Input
3
3 3 2
Output
1
3 1
Input
7
2 3 1 3 4 4 1
Output
3
2 5
2 6
3 7
Submitted Solution:
```
import sys
from array import array # noqa: F401
def input():
return sys.stdin.buffer.readline().decode('utf-8')
n = int(input())
a = [0] + list(map(int, input().split()))
rev = [[] for _ in range(n + 1)]
indeg = [0] * (n + 1)
for i in range(1, n + 1):
indeg[a[i]] += 1
rev[a[i]].append(i)
_indeg = indeg[:]
for i in range(1, n + 1):
v = i
while indeg[v] == 0:
indeg[a[v]] -= 1
v = a[v]
visited = [0] * (n + 1)
group = []
group_leaf = []
for i in range(1, n + 1):
if visited[i] or indeg[i] == 0:
continue
visited[i] = 1
leaves = []
stack = [i]
while stack:
v = stack.pop()
if _indeg[v] == 0:
leaves.append(v)
for dest in rev[v]:
if not visited[dest]:
visited[dest] = 1
stack.append(dest)
if not leaves:
leaves.append(i)
group.append(i)
group_leaf.append(leaves)
assert all(visited[1:])
ans = []
m = len(group)
if not (m == 1 and group_leaf[0][0] == group[0]):
for i in range(m):
k = group[(i + 1) % m]
for j in group_leaf[i]:
ans.append(f'{k} {j}')
ans_str = str(len(ans)) + '\n' + '\n'.join(ans)
sys.stdout.buffer.write(ans_str.encode('utf-8'))
```
No
| 46,558 | [
0.50146484375,
0.2276611328125,
-0.281982421875,
-0.02374267578125,
-0.7373046875,
-0.7216796875,
-0.1356201171875,
0.482666015625,
0.1678466796875,
0.91796875,
0.64453125,
-0.560546875,
0.301025390625,
-0.24072265625,
-0.802734375,
-0.10638427734375,
-0.6845703125,
-0.85546875,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To learn as soon as possible the latest news about their favourite fundamentally new operating system, BolgenOS community from Nizhni Tagil decided to develop a scheme. According to this scheme a community member, who is the first to learn the news, calls some other member, the latter, in his turn, calls some third member, and so on; i.e. a person with index i got a person with index fi, to whom he has to call, if he learns the news. With time BolgenOS community members understood that their scheme doesn't work sometimes — there were cases when some members didn't learn the news at all. Now they want to supplement the scheme: they add into the scheme some instructions of type (xi, yi), which mean that person xi has to call person yi as well. What is the minimum amount of instructions that they need to add so, that at the end everyone learns the news, no matter who is the first to learn it?
Input
The first input line contains number n (2 ≤ n ≤ 105) — amount of BolgenOS community members. The second line contains n space-separated integer numbers fi (1 ≤ fi ≤ n, i ≠ fi) — index of a person, to whom calls a person with index i.
Output
In the first line output one number — the minimum amount of instructions to add. Then output one of the possible variants to add these instructions into the scheme, one instruction in each line. If the solution is not unique, output any.
Examples
Input
3
3 3 2
Output
1
3 1
Input
7
2 3 1 3 4 4 1
Output
3
2 5
2 6
3 7
Submitted Solution:
```
def dfs_1(graph):
n=len(graph)
stack=[]
status=[[False,-1,0,0] for i in range(n)]
time=0
for i in range(n):
if not status[i][0]:
__dfs(graph,status,stack,i,time)
return stack
def __dfs(graph,status,stack,vert,time):
time+=1
status[vert][2]=time
status[vert][0]=True
for i in graph[vert]:
if not status[i][0]:
status[i][1]=vert
__dfs(graph,status,stack,i,time)
time+=1
status[vert][3]=time
stack.append(vert)
def reverse_graph(graph):
n=len(graph)
new_graph=[[]for i in range(n)]
pos =0
for i in graph:
for j in i:
new_graph[j].append(pos)
pos+=1
return new_graph
def dfs_2(graph,stack):
n=len(graph)
status=[ False for i in range(n)]
scc=0
dag=[-1 for i in range(n)]
while len(stack) > 0:
root=stack.pop()
if not status[root]:
__dfs_2(graph,status,scc,dag,root)
scc+=1
return (scc,dag)
def __dfs_2(graph,status,scc,dag,root):
status[root]=True
for i in graph[root]:
if not status[i]:
__dfs_2(graph,status,scc,dag,i)
dag[root]=scc
def SCC(graph):
stack=dfs_1(graph)
rvgraph=reverse_graph(graph)
return dfs_2(rvgraph,stack)
def get_dag_degrees(graph):
n_scc,GSCC= SCC(graph)
dag=[[]for i in range(n_scc)]
degres=[[0,0]for i in range(n_scc)]
count=0
for i in graph:
for j in i :
if GSCC[count]!=GSCC[j]:
dag[GSCC[count]].append(GSCC[j])
degres[GSCC[count]][1]+=1
degres[GSCC[j]][0]+=1
count+=1
return dag,degres,GSCC,n_scc
def part_degrees(degrees):
in_degree=[]
cout=0
for i in degrees:
if i[0]==0:
in_degree.append(cout)
cout+=1
return in_degree
def solve(graph):
dag,degrees,GSCC,n_scc=get_dag_degrees(graph)
if n_scc==1:
print(0)
return
mapping=[-1 for i in range(len(dag))]
in_degree=part_degrees(degrees)
for i in range(len(GSCC)):
if mapping[GSCC[i]]==-1:
mapping[GSCC[i]]=i
status=[False for i in range(len(dag))]
current_info=[0,-1]
upd_arcs=[]
last_vert=[-1]
for i in in_degree:
if not status[i]:
if last_vert[0] != -1:
current_info[0]+=1
upd_arcs.append((last_vert[0],i))
__dfs_solve(dag,status,degrees,upd_arcs,i,current_info,last_vert)
current_info[1]=-1
print(last_vert,current_info,upd_arcs)
upd_arcs.append((last_vert[0],in_degree[0]))
current_info[0]+=1
print(current_info[0])
for i in upd_arcs:
print(mapping[i[0]]+1,mapping[i[1]]+1)
def __dfs_solve(graph,status,degrees,upd_arcs,root,current_info,last_vert):
status[root]=True
for i in graph[root]:
if not status[i]:
__dfs_solve(graph,status,degrees,upd_arcs,i,current_info,last_vert)
if degrees[root][1]==0:
if current_info[1]!=-1:
current_info[0]+=1
upd_arcs.append((current_info[1],root))
current_info[1]=last_vert[0]=root
n=int(input())
graph=[[]for i in range(n)]
data_set=input().split()
for i in range(n):
graph[i].append(int(data_set[i])-1)
solve(graph)
```
No
| 46,559 | [
0.52978515625,
0.2081298828125,
-0.224365234375,
0.0428466796875,
-0.638671875,
-0.6279296875,
-0.173095703125,
0.494384765625,
0.19140625,
0.85595703125,
0.57861328125,
-0.6787109375,
0.44677734375,
-0.198486328125,
-0.8759765625,
-0.0227203369140625,
-0.8212890625,
-0.818359375,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To learn as soon as possible the latest news about their favourite fundamentally new operating system, BolgenOS community from Nizhni Tagil decided to develop a scheme. According to this scheme a community member, who is the first to learn the news, calls some other member, the latter, in his turn, calls some third member, and so on; i.e. a person with index i got a person with index fi, to whom he has to call, if he learns the news. With time BolgenOS community members understood that their scheme doesn't work sometimes — there were cases when some members didn't learn the news at all. Now they want to supplement the scheme: they add into the scheme some instructions of type (xi, yi), which mean that person xi has to call person yi as well. What is the minimum amount of instructions that they need to add so, that at the end everyone learns the news, no matter who is the first to learn it?
Input
The first input line contains number n (2 ≤ n ≤ 105) — amount of BolgenOS community members. The second line contains n space-separated integer numbers fi (1 ≤ fi ≤ n, i ≠ fi) — index of a person, to whom calls a person with index i.
Output
In the first line output one number — the minimum amount of instructions to add. Then output one of the possible variants to add these instructions into the scheme, one instruction in each line. If the solution is not unique, output any.
Examples
Input
3
3 3 2
Output
1
3 1
Input
7
2 3 1 3 4 4 1
Output
3
2 5
2 6
3 7
Submitted Solution:
```
#!/usr/bin/python3
import sys
from collections import Counter
input = lambda: sys.stdin.readline().strip()
n = int(input())
f = [0, *(int(x) for x in input().split())]
c = Counter(f)
component = [0 if i in c else i for i in range(n + 1)]
last = [0 for i in range(n + 1)]
roots = [i for i in range(1, n + 1) if i not in c]
q = [u for u in roots]
for u in q:
last[component[u]] = f[u]
if not component[f[u]]:
component[f[u]] = component[u]
q.append(f[u])
for u in range(1, n + 1):
if not component[u]:
component[u] = u
last[u] = u
roots.append(u)
if len(roots) == 1:
if last[roots[0]] == roots[0]:
print(0)
else:
print(1)
print(last[roots[0]], roots[0])
else:
print(len(roots))
for i in range(len(roots)):
print(last[roots[i]], roots[(i + 1) % len(roots)])
```
No
| 46,561 | [
0.5107421875,
0.251953125,
-0.337158203125,
-0.08233642578125,
-0.74169921875,
-0.63037109375,
-0.13330078125,
0.490234375,
0.1898193359375,
0.86181640625,
0.5830078125,
-0.61083984375,
0.263916015625,
-0.30859375,
-0.86376953125,
-0.09332275390625,
-0.75390625,
-0.8701171875,
-0... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yaroslav likes algorithms. We'll describe one of his favorite algorithms.
1. The algorithm receives a string as the input. We denote this input string as a.
2. The algorithm consists of some number of command. Сommand number i looks either as si >> wi, or as si <> wi, where si and wi are some possibly empty strings of length at most 7, consisting of digits and characters "?".
3. At each iteration, the algorithm looks for a command with the minimum index i, such that si occurs in a as a substring. If this command is not found the algorithm terminates.
4. Let's denote the number of the found command as k. In string a the first occurrence of the string sk is replaced by string wk. If the found command at that had form sk >> wk, then the algorithm continues its execution and proceeds to the next iteration. Otherwise, the algorithm terminates.
5. The value of string a after algorithm termination is considered to be the output of the algorithm.
Yaroslav has a set of n positive integers, he needs to come up with his favorite algorithm that will increase each of the given numbers by one. More formally, if we consider each number as a string representing the decimal representation of the number, then being run on each of these strings separately, the algorithm should receive the output string that is a recording of the corresponding number increased by one.
Help Yaroslav.
Input
The first line contains integer n (1 ≤ n ≤ 100) — the number of elements in the set. The next n lines contains one positive integer each. All the given numbers are less than 1025.
Output
Print the algorithm which can individually increase each number of the set. In the i-th line print the command number i without spaces.
Your algorithm will be launched for each of these numbers. The answer will be considered correct if:
* Each line will a correct algorithm command (see the description in the problem statement).
* The number of commands should not exceed 50.
* The algorithm will increase each of the given numbers by one.
* To get a respond, the algorithm will perform no more than 200 iterations for each number.
Examples
Input
2
10
79
Output
10<>11
79<>80
Submitted Solution:
```
print ( """0??<>1
1??<>2
2??<>3
3??<>4
4??<>5
5??<>6
6??<>7
7??<>8
8??<>9
9??<>??0
??<>1
?0>>0?
?1>>1?
?2>>2?
?3>>3?
?4>>4?
?5>>5?
?6>>6?
?7>>7?
?8>>8?
?9>>9?
?>>??
>>?
""" )
```
No
| 46,573 | [
0.39111328125,
-0.03857421875,
0.058563232421875,
0.48681640625,
-0.408203125,
-0.64111328125,
-0.2022705078125,
0.2452392578125,
0.031097412109375,
1.1806640625,
0.7001953125,
-0.1806640625,
0.08673095703125,
-0.89013671875,
-0.444580078125,
-0.2008056640625,
-0.447021484375,
-0.7... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yaroslav likes algorithms. We'll describe one of his favorite algorithms.
1. The algorithm receives a string as the input. We denote this input string as a.
2. The algorithm consists of some number of command. Сommand number i looks either as si >> wi, or as si <> wi, where si and wi are some possibly empty strings of length at most 7, consisting of digits and characters "?".
3. At each iteration, the algorithm looks for a command with the minimum index i, such that si occurs in a as a substring. If this command is not found the algorithm terminates.
4. Let's denote the number of the found command as k. In string a the first occurrence of the string sk is replaced by string wk. If the found command at that had form sk >> wk, then the algorithm continues its execution and proceeds to the next iteration. Otherwise, the algorithm terminates.
5. The value of string a after algorithm termination is considered to be the output of the algorithm.
Yaroslav has a set of n positive integers, he needs to come up with his favorite algorithm that will increase each of the given numbers by one. More formally, if we consider each number as a string representing the decimal representation of the number, then being run on each of these strings separately, the algorithm should receive the output string that is a recording of the corresponding number increased by one.
Help Yaroslav.
Input
The first line contains integer n (1 ≤ n ≤ 100) — the number of elements in the set. The next n lines contains one positive integer each. All the given numbers are less than 1025.
Output
Print the algorithm which can individually increase each number of the set. In the i-th line print the command number i without spaces.
Your algorithm will be launched for each of these numbers. The answer will be considered correct if:
* Each line will a correct algorithm command (see the description in the problem statement).
* The number of commands should not exceed 50.
* The algorithm will increase each of the given numbers by one.
* To get a respond, the algorithm will perform no more than 200 iterations for each number.
Examples
Input
2
10
79
Output
10<>11
79<>80
Submitted Solution:
```
print("0??<>1\n1??<>2\n2??<>3\n3??<>4\n4??<>5\n5??<>6\n6??<>7\n7??<>8\n8??<>9\n9??>>??0\n??<>1\n?0>>0?\n?1>>1?\n?2>>2?\n?3>>3?\n?4>>4?\n?5>>5?\n?6>>6?\n?7>>7?\n?8>>8?\n?9>>9?\n>>?\n")
```
No
| 46,574 | [
0.388427734375,
-0.040557861328125,
0.06390380859375,
0.485595703125,
-0.4072265625,
-0.6474609375,
-0.2015380859375,
0.2451171875,
0.027374267578125,
1.185546875,
0.69677734375,
-0.1812744140625,
0.087158203125,
-0.89404296875,
-0.44677734375,
-0.203125,
-0.44970703125,
-0.7216796... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yaroslav likes algorithms. We'll describe one of his favorite algorithms.
1. The algorithm receives a string as the input. We denote this input string as a.
2. The algorithm consists of some number of command. Сommand number i looks either as si >> wi, or as si <> wi, where si and wi are some possibly empty strings of length at most 7, consisting of digits and characters "?".
3. At each iteration, the algorithm looks for a command with the minimum index i, such that si occurs in a as a substring. If this command is not found the algorithm terminates.
4. Let's denote the number of the found command as k. In string a the first occurrence of the string sk is replaced by string wk. If the found command at that had form sk >> wk, then the algorithm continues its execution and proceeds to the next iteration. Otherwise, the algorithm terminates.
5. The value of string a after algorithm termination is considered to be the output of the algorithm.
Yaroslav has a set of n positive integers, he needs to come up with his favorite algorithm that will increase each of the given numbers by one. More formally, if we consider each number as a string representing the decimal representation of the number, then being run on each of these strings separately, the algorithm should receive the output string that is a recording of the corresponding number increased by one.
Help Yaroslav.
Input
The first line contains integer n (1 ≤ n ≤ 100) — the number of elements in the set. The next n lines contains one positive integer each. All the given numbers are less than 1025.
Output
Print the algorithm which can individually increase each number of the set. In the i-th line print the command number i without spaces.
Your algorithm will be launched for each of these numbers. The answer will be considered correct if:
* Each line will a correct algorithm command (see the description in the problem statement).
* The number of commands should not exceed 50.
* The algorithm will increase each of the given numbers by one.
* To get a respond, the algorithm will perform no more than 200 iterations for each number.
Examples
Input
2
10
79
Output
10<>11
79<>80
Submitted Solution:
```
print('9??>>??0')
for i in range(9):
print(str(i)+'??<>'+str(i+1)+'??')
print('??<>1')
for i in range(10):
print('?'+str(i)+'>>'+str(i)+'?')
print('?>>??')
for i in range(10):
print(str(i)+'>>'+str(i)+'?')
```
No
| 46,575 | [
0.385986328125,
-0.043212890625,
0.069580078125,
0.48388671875,
-0.403564453125,
-0.6484375,
-0.1961669921875,
0.2371826171875,
0.0283203125,
1.1845703125,
0.69189453125,
-0.18212890625,
0.084228515625,
-0.8984375,
-0.447509765625,
-0.21044921875,
-0.450927734375,
-0.72412109375,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. You should use flush operation after each printed line. For example, in C++ you should use fflush(stdout), in Java you should use System.out.flush(), and in Pascal — flush(output).
In this problem you should guess an array a which is unknown for you. The only information you have initially is the length n of the array a.
The only allowed action is to ask the sum of two elements by their indices. Formally, you can print two indices i and j (the indices should be distinct). Then your program should read the response: the single integer equals to ai + aj.
It is easy to prove that it is always possible to guess the array using at most n requests.
Write a program that will guess the array a by making at most n requests.
Interaction
In each test your program should guess a single array.
The input starts with a line containing integer n (3 ≤ n ≤ 5000) — the length of the array. Your program should read it at first.
After that your program should print to the standard output the requests about the sum of two elements or inform that the array is guessed.
* In case your program is making a request to ask the sum of two elements, it should print line in the format "? i j" (i and j are distinct integers between 1 and n), where i and j are indices in the array a.
* In case your program informs that the array is guessed, it should print line in the format "! a1 a2 ... an" (it is guaranteed that all ai are positive integers not exceeding 105), where ai is the i-th element of the array a.
The response on a request is a single integer equal to ai + aj, printed on a separate line.
Your program can do at most n requests. Note that the final line «! a1 a2 ... an» is not counted as a request.
Do not forget about flush operation after each printed line.
After you program prints the guessed array, it should terminate normally.
Example
Input
5
9
7
9
11
6
Output
? 1 5
? 2 3
? 4 1
? 5 2
? 3 4
! 4 6 1 5 5
Note
The format of a test to make a hack is:
* The first line contains an integer number n (3 ≤ n ≤ 5000) — the length of the array.
* The second line contains n numbers a1, a2, ..., an (1 ≤ ai ≤ 105) — the elements of the array to guess.
Submitted Solution:
```
# @author Nayara Souza
# UFCG - Universidade Federal de Campina Grande
# AA - Basico
import sys
f = sys.stdout.flush()
n = int(input())
print('? 1 2')
a = int(input())
f
print('? 2 3')
b = int(input())
f
print('? 1 3')
c = int(input())
f
x = ((a+b)-c)//2
y = a - x
z = b - x
r = [x,y,z]
for i in range(4,n+1):
print('? 1 ' + str(i -1) + str(i))
k = int(input())
r.append(k-r[i-2])
f
r = list(map(int, r))
print('! ' + ' '.join(map(str,r)))
```
No
| 46,737 | [
-0.03875732421875,
0.06329345703125,
-0.45263671875,
-0.02984619140625,
-0.65673828125,
-0.302001953125,
-0.12042236328125,
0.1314697265625,
0.29736328125,
1.064453125,
0.67822265625,
-0.04339599609375,
0.1204833984375,
-0.90869140625,
-0.8173828125,
-0.021331787109375,
-0.0995483398... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. You should use flush operation after each printed line. For example, in C++ you should use fflush(stdout), in Java you should use System.out.flush(), and in Pascal — flush(output).
In this problem you should guess an array a which is unknown for you. The only information you have initially is the length n of the array a.
The only allowed action is to ask the sum of two elements by their indices. Formally, you can print two indices i and j (the indices should be distinct). Then your program should read the response: the single integer equals to ai + aj.
It is easy to prove that it is always possible to guess the array using at most n requests.
Write a program that will guess the array a by making at most n requests.
Interaction
In each test your program should guess a single array.
The input starts with a line containing integer n (3 ≤ n ≤ 5000) — the length of the array. Your program should read it at first.
After that your program should print to the standard output the requests about the sum of two elements or inform that the array is guessed.
* In case your program is making a request to ask the sum of two elements, it should print line in the format "? i j" (i and j are distinct integers between 1 and n), where i and j are indices in the array a.
* In case your program informs that the array is guessed, it should print line in the format "! a1 a2 ... an" (it is guaranteed that all ai are positive integers not exceeding 105), where ai is the i-th element of the array a.
The response on a request is a single integer equal to ai + aj, printed on a separate line.
Your program can do at most n requests. Note that the final line «! a1 a2 ... an» is not counted as a request.
Do not forget about flush operation after each printed line.
After you program prints the guessed array, it should terminate normally.
Example
Input
5
9
7
9
11
6
Output
? 1 5
? 2 3
? 4 1
? 5 2
? 3 4
! 4 6 1 5 5
Note
The format of a test to make a hack is:
* The first line contains an integer number n (3 ≤ n ≤ 5000) — the length of the array.
* The second line contains n numbers a1, a2, ..., an (1 ≤ ai ≤ 105) — the elements of the array to guess.
Submitted Solution:
```
n=int(input())
a=[0]*n
if n%3==2:
b=[]
for i in range(5):
print('?', i+1, (i+1)%5+1, flush=True)
b+=[int(input())]
a[0]=(+b[0]-b[1]+b[2]-b[3]+b[4])>>1
a[1]=(+b[0]+b[1]-b[2]+b[3]-b[4])>>1
a[2]=(-b[0]+b[1]+b[2]-b[3]+b[4])>>1
a[3]=(+b[0]-b[1]+b[2]+b[3]-b[4])>>1
a[4]=(-b[0]+b[1]-b[2]+b[3]+b[4])>>1
l=5
elif n%3==1:
b=[]
for i in range(3):
print('?', 1, i+2, flush=True)
b+=[int(input())]
print('? 3 4', flush=True)
b+=[int(input())]
a[0]=(b[0]*0+b[1]+b[2]-b[3])>>1
a[1]=(b[0]*2-b[1]-b[2]+b[3])>>1
a[2]=(b[0]*0+b[1]-b[2]+b[3])>>1
a[3]=(b[0]*0-b[1]+b[2]+b[3])>>1
l=4
else:
l=0
while l<n:
b=[]
for i in range(3):
print('?', l+i+1, l+(i+1)%4+1, flush=True)
b+=[int(input())]
a[l+0]=(+b[0]-b[1]+b[2])>>1
a[l+1]=(+b[0]+b[1]-b[2])>>1
a[l+2]=(-b[0]+b[1]+b[2])>>1
l+=3
print('!', *a, flush=True);
```
No
| 46,738 | [
-0.03875732421875,
0.06329345703125,
-0.45263671875,
-0.02984619140625,
-0.65673828125,
-0.302001953125,
-0.12042236328125,
0.1314697265625,
0.29736328125,
1.064453125,
0.67822265625,
-0.04339599609375,
0.1204833984375,
-0.90869140625,
-0.8173828125,
-0.021331787109375,
-0.0995483398... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. You should use flush operation after each printed line. For example, in C++ you should use fflush(stdout), in Java you should use System.out.flush(), and in Pascal — flush(output).
In this problem you should guess an array a which is unknown for you. The only information you have initially is the length n of the array a.
The only allowed action is to ask the sum of two elements by their indices. Formally, you can print two indices i and j (the indices should be distinct). Then your program should read the response: the single integer equals to ai + aj.
It is easy to prove that it is always possible to guess the array using at most n requests.
Write a program that will guess the array a by making at most n requests.
Interaction
In each test your program should guess a single array.
The input starts with a line containing integer n (3 ≤ n ≤ 5000) — the length of the array. Your program should read it at first.
After that your program should print to the standard output the requests about the sum of two elements or inform that the array is guessed.
* In case your program is making a request to ask the sum of two elements, it should print line in the format "? i j" (i and j are distinct integers between 1 and n), where i and j are indices in the array a.
* In case your program informs that the array is guessed, it should print line in the format "! a1 a2 ... an" (it is guaranteed that all ai are positive integers not exceeding 105), where ai is the i-th element of the array a.
The response on a request is a single integer equal to ai + aj, printed on a separate line.
Your program can do at most n requests. Note that the final line «! a1 a2 ... an» is not counted as a request.
Do not forget about flush operation after each printed line.
After you program prints the guessed array, it should terminate normally.
Example
Input
5
9
7
9
11
6
Output
? 1 5
? 2 3
? 4 1
? 5 2
? 3 4
! 4 6 1 5 5
Note
The format of a test to make a hack is:
* The first line contains an integer number n (3 ≤ n ≤ 5000) — the length of the array.
* The second line contains n numbers a1, a2, ..., an (1 ≤ ai ≤ 105) — the elements of the array to guess.
Submitted Solution:
```
n = int(input())
arr = [0]
print("? 1 2")
firsteq = int(input())
print("? 1 3")
seceq = int(input())
print("? 2 3")
thirdeq = int(input())
first = ( firsteq + seceq - thirdeq ) // 2
arr.append(first)
arr.append(firsteq)
arr.append(seceq)
n -= 3
for i in range(n):
i = i + 3
strs = "? 1 " + str(i)
print(strs)
arr.append(int(input()))
ans = "! "
ans += str(first) + " "
for i in range(2, n + 3):
temp = arr[i] - first
ans += str(temp) + " "
print(ans)
```
No
| 46,739 | [
-0.03875732421875,
0.06329345703125,
-0.45263671875,
-0.02984619140625,
-0.65673828125,
-0.302001953125,
-0.12042236328125,
0.1314697265625,
0.29736328125,
1.064453125,
0.67822265625,
-0.04339599609375,
0.1204833984375,
-0.90869140625,
-0.8173828125,
-0.021331787109375,
-0.0995483398... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. You should use flush operation after each printed line. For example, in C++ you should use fflush(stdout), in Java you should use System.out.flush(), and in Pascal — flush(output).
In this problem you should guess an array a which is unknown for you. The only information you have initially is the length n of the array a.
The only allowed action is to ask the sum of two elements by their indices. Formally, you can print two indices i and j (the indices should be distinct). Then your program should read the response: the single integer equals to ai + aj.
It is easy to prove that it is always possible to guess the array using at most n requests.
Write a program that will guess the array a by making at most n requests.
Interaction
In each test your program should guess a single array.
The input starts with a line containing integer n (3 ≤ n ≤ 5000) — the length of the array. Your program should read it at first.
After that your program should print to the standard output the requests about the sum of two elements or inform that the array is guessed.
* In case your program is making a request to ask the sum of two elements, it should print line in the format "? i j" (i and j are distinct integers between 1 and n), where i and j are indices in the array a.
* In case your program informs that the array is guessed, it should print line in the format "! a1 a2 ... an" (it is guaranteed that all ai are positive integers not exceeding 105), where ai is the i-th element of the array a.
The response on a request is a single integer equal to ai + aj, printed on a separate line.
Your program can do at most n requests. Note that the final line «! a1 a2 ... an» is not counted as a request.
Do not forget about flush operation after each printed line.
After you program prints the guessed array, it should terminate normally.
Example
Input
5
9
7
9
11
6
Output
? 1 5
? 2 3
? 4 1
? 5 2
? 3 4
! 4 6 1 5 5
Note
The format of a test to make a hack is:
* The first line contains an integer number n (3 ≤ n ≤ 5000) — the length of the array.
* The second line contains n numbers a1, a2, ..., an (1 ≤ ai ≤ 105) — the elements of the array to guess.
Submitted Solution:
```
# @author Nayara Souza
# UFCG - Universidade Federal de Campina Grande
# AA - Basico
n = int(input())
print('? 1 2')
a = int(input())
print('? 2 3')
b = int(input())
print('? 1 3')
c = int(input())
x = (a+b-c)//2
y = a - x
z = b - x
r = [x,y,z]
for i in range(4,n+1):
print('? 1 ' + str(i))
k = int(input())
r.append(k-r[0])
print('! ' + ' '.join(map(str,r)))
```
No
| 46,740 | [
-0.03875732421875,
0.06329345703125,
-0.45263671875,
-0.02984619140625,
-0.65673828125,
-0.302001953125,
-0.12042236328125,
0.1314697265625,
0.29736328125,
1.064453125,
0.67822265625,
-0.04339599609375,
0.1204833984375,
-0.90869140625,
-0.8173828125,
-0.021331787109375,
-0.0995483398... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Example
Input
1+2*3+4
11
Output
M
Submitted Solution:
```
s = input()
t = int(input())
e1 = eval(s)
e2 = int(s[0])
i = 1
while i < len(s):
x = int(s[i+1])
if s[i] == '+':
e2 += x
else:
e2 *= x
i += 2
if e1 == t and e2 == t:
print("U")
elif e1 == t:
print("M")
elif e2 == t:
print("L")
else:
print("I")
```
Yes
| 46,999 | [
0.428466796875,
0.054901123046875,
-0.380615234375,
-0.137939453125,
-0.85693359375,
-0.50048828125,
-0.2178955078125,
0.2181396484375,
-0.01001739501953125,
0.79931640625,
0.3916015625,
0.336181640625,
0.06451416015625,
-0.69384765625,
-0.56005859375,
-0.337890625,
-0.354248046875,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Example
Input
1+2*3+4
11
Output
M
Submitted Solution:
```
# -*- coding: utf-8 -*-
S = input()
bob = int(input())
M = eval(S)
L = int(S[0])
for o, v in zip(S[1::2], S[2::2]):
if o == "*":
L *= int(v)
else:
L += int(v)
if M == L == bob:
print("U")
elif M == bob:
print("M")
elif L == bob:
print("L")
else:
print("I")
```
Yes
| 47,000 | [
0.4951171875,
-0.147216796875,
-0.326416015625,
-0.2037353515625,
-0.7431640625,
-0.463134765625,
0.307861328125,
0.21630859375,
0.00925445556640625,
0.828125,
0.454833984375,
0.2205810546875,
-0.11895751953125,
-0.62353515625,
-0.5556640625,
-0.32373046875,
-0.30029296875,
-0.8881... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Example
Input
1+2*3+4
11
Output
M
Submitted Solution:
```
import re
class c(int):
def __add__(self,n):
return c(int(self)+int(n))
def __mul__(self,n):
return c(int(self)*int(n))
s=input()
d=int(input())
a=eval(re.sub(r'(\d+)',r'c(\1)',s))
b=int(s[0])
for i in range(1,len(s),2):
if s[i]=='+':b+=int(s[i+1])
else:b*=int(s[i+1])
if(a==b==d):e='U'
elif(a==d):e='M'
elif(b==d):e='L'
else:e='I'
print(e)
```
Yes
| 47,001 | [
0.314453125,
-0.03460693359375,
-0.427734375,
-0.2392578125,
-0.7802734375,
-0.2393798828125,
-0.07525634765625,
0.12091064453125,
-0.1536865234375,
0.94189453125,
0.556640625,
0.1702880859375,
0.0390625,
-0.7724609375,
-0.60498046875,
-0.445556640625,
-0.47265625,
-0.87109375,
-... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Example
Input
1+2*3+4
11
Output
M
Submitted Solution:
```
def left2right(formula):
ans = int(formula[0])
for i in range(1,len(formula),2):
ans = eval(str(ans) + formula[i:i+2])
return ans
formula = input()
val = int(input())
if left2right(formula) == val:
print('U' if eval(formula) == val else 'L')
else:
print('M' if eval(formula) == val else 'I')
```
Yes
| 47,002 | [
0.2362060546875,
-0.150634765625,
-0.254638671875,
-0.158203125,
-0.80810546875,
-0.304443359375,
0.27197265625,
0.2164306640625,
-0.330810546875,
0.814453125,
0.447021484375,
0.307861328125,
-0.045654296875,
-0.404541015625,
-0.44189453125,
0.08050537109375,
-0.1729736328125,
-0.6... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nauuo is a girl who loves writing comments.
One day, she posted a comment on Codeforces, wondering whether she would get upvotes or downvotes.
It's known that there were x persons who would upvote, y persons who would downvote, and there were also another z persons who would vote, but you don't know whether they would upvote or downvote. Note that each of the x+y+z people would vote exactly one time.
There are three different results: if there are more people upvote than downvote, the result will be "+"; if there are more people downvote than upvote, the result will be "-"; otherwise the result will be "0".
Because of the z unknown persons, the result may be uncertain (i.e. there are more than one possible results). More formally, the result is uncertain if and only if there exist two different situations of how the z persons vote, that the results are different in the two situations.
Tell Nauuo the result or report that the result is uncertain.
Input
The only line contains three integers x, y, z (0≤ x,y,z≤100), corresponding to the number of persons who would upvote, downvote or unknown.
Output
If there is only one possible result, print the result : "+", "-" or "0".
Otherwise, print "?" to report that the result is uncertain.
Examples
Input
3 7 0
Output
-
Input
2 0 1
Output
+
Input
1 1 0
Output
0
Input
0 0 1
Output
?
Note
In the first example, Nauuo would definitely get three upvotes and seven downvotes, so the only possible result is "-".
In the second example, no matter the person unknown downvotes or upvotes, Nauuo would get more upvotes than downvotes. So the only possible result is "+".
In the third example, Nauuo would definitely get one upvote and one downvote, so the only possible result is "0".
In the fourth example, if the only one person upvoted, the result would be "+", otherwise, the result would be "-". There are two possible results, so the result is uncertain.
Tags: greedy
Correct Solution:
```
x,y,z=map(int,(input().split()))
if ((abs(x-y)-z)<=0 and z>0):
print('?')
elif(x-y)>0:
print('+')
elif(x-y)==0:
print('0')
elif(x-y)<0:
print('-')
```
| 47,083 | [
0.07684326171875,
0.093505859375,
-0.48388671875,
-0.229248046875,
-0.7109375,
-0.5859375,
0.41357421875,
0.381103515625,
0.2001953125,
0.82275390625,
0.68798828125,
-0.2059326171875,
0.32763671875,
-0.2763671875,
-0.6962890625,
0.07952880859375,
-0.5166015625,
-0.9189453125,
0.1... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nauuo is a girl who loves writing comments.
One day, she posted a comment on Codeforces, wondering whether she would get upvotes or downvotes.
It's known that there were x persons who would upvote, y persons who would downvote, and there were also another z persons who would vote, but you don't know whether they would upvote or downvote. Note that each of the x+y+z people would vote exactly one time.
There are three different results: if there are more people upvote than downvote, the result will be "+"; if there are more people downvote than upvote, the result will be "-"; otherwise the result will be "0".
Because of the z unknown persons, the result may be uncertain (i.e. there are more than one possible results). More formally, the result is uncertain if and only if there exist two different situations of how the z persons vote, that the results are different in the two situations.
Tell Nauuo the result or report that the result is uncertain.
Input
The only line contains three integers x, y, z (0≤ x,y,z≤100), corresponding to the number of persons who would upvote, downvote or unknown.
Output
If there is only one possible result, print the result : "+", "-" or "0".
Otherwise, print "?" to report that the result is uncertain.
Examples
Input
3 7 0
Output
-
Input
2 0 1
Output
+
Input
1 1 0
Output
0
Input
0 0 1
Output
?
Note
In the first example, Nauuo would definitely get three upvotes and seven downvotes, so the only possible result is "-".
In the second example, no matter the person unknown downvotes or upvotes, Nauuo would get more upvotes than downvotes. So the only possible result is "+".
In the third example, Nauuo would definitely get one upvote and one downvote, so the only possible result is "0".
In the fourth example, if the only one person upvoted, the result would be "+", otherwise, the result would be "-". There are two possible results, so the result is uncertain.
Tags: greedy
Correct Solution:
```
arr = list(map(int, input().split()))
diff = abs(arr[0] - arr[1])
if diff == 0:
if arr[2] == 0:
print("0")
else:
print("?")
else:
if arr[2] < diff:
print("+" if arr[0] > arr[1] else "-")
else:
print("?")
```
| 47,084 | [
0.07684326171875,
0.093505859375,
-0.48388671875,
-0.229248046875,
-0.7109375,
-0.5859375,
0.41357421875,
0.381103515625,
0.2001953125,
0.82275390625,
0.68798828125,
-0.2059326171875,
0.32763671875,
-0.2763671875,
-0.6962890625,
0.07952880859375,
-0.5166015625,
-0.9189453125,
0.1... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nauuo is a girl who loves writing comments.
One day, she posted a comment on Codeforces, wondering whether she would get upvotes or downvotes.
It's known that there were x persons who would upvote, y persons who would downvote, and there were also another z persons who would vote, but you don't know whether they would upvote or downvote. Note that each of the x+y+z people would vote exactly one time.
There are three different results: if there are more people upvote than downvote, the result will be "+"; if there are more people downvote than upvote, the result will be "-"; otherwise the result will be "0".
Because of the z unknown persons, the result may be uncertain (i.e. there are more than one possible results). More formally, the result is uncertain if and only if there exist two different situations of how the z persons vote, that the results are different in the two situations.
Tell Nauuo the result or report that the result is uncertain.
Input
The only line contains three integers x, y, z (0≤ x,y,z≤100), corresponding to the number of persons who would upvote, downvote or unknown.
Output
If there is only one possible result, print the result : "+", "-" or "0".
Otherwise, print "?" to report that the result is uncertain.
Examples
Input
3 7 0
Output
-
Input
2 0 1
Output
+
Input
1 1 0
Output
0
Input
0 0 1
Output
?
Note
In the first example, Nauuo would definitely get three upvotes and seven downvotes, so the only possible result is "-".
In the second example, no matter the person unknown downvotes or upvotes, Nauuo would get more upvotes than downvotes. So the only possible result is "+".
In the third example, Nauuo would definitely get one upvote and one downvote, so the only possible result is "0".
In the fourth example, if the only one person upvoted, the result would be "+", otherwise, the result would be "-". There are two possible results, so the result is uncertain.
Tags: greedy
Correct Solution:
```
x,y,z=map(int,input().split())
s=x-y
ma = s+z
mi = s-z
if ma > 0 and mi>0:
print("+")
elif ma < 0 and mi < 0 :
print("-")
elif mi==0 and ma==0:
print("0")
else:
print("?")
```
| 47,085 | [
0.07684326171875,
0.093505859375,
-0.48388671875,
-0.229248046875,
-0.7109375,
-0.5859375,
0.41357421875,
0.381103515625,
0.2001953125,
0.82275390625,
0.68798828125,
-0.2059326171875,
0.32763671875,
-0.2763671875,
-0.6962890625,
0.07952880859375,
-0.5166015625,
-0.9189453125,
0.1... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nauuo is a girl who loves writing comments.
One day, she posted a comment on Codeforces, wondering whether she would get upvotes or downvotes.
It's known that there were x persons who would upvote, y persons who would downvote, and there were also another z persons who would vote, but you don't know whether they would upvote or downvote. Note that each of the x+y+z people would vote exactly one time.
There are three different results: if there are more people upvote than downvote, the result will be "+"; if there are more people downvote than upvote, the result will be "-"; otherwise the result will be "0".
Because of the z unknown persons, the result may be uncertain (i.e. there are more than one possible results). More formally, the result is uncertain if and only if there exist two different situations of how the z persons vote, that the results are different in the two situations.
Tell Nauuo the result or report that the result is uncertain.
Input
The only line contains three integers x, y, z (0≤ x,y,z≤100), corresponding to the number of persons who would upvote, downvote or unknown.
Output
If there is only one possible result, print the result : "+", "-" or "0".
Otherwise, print "?" to report that the result is uncertain.
Examples
Input
3 7 0
Output
-
Input
2 0 1
Output
+
Input
1 1 0
Output
0
Input
0 0 1
Output
?
Note
In the first example, Nauuo would definitely get three upvotes and seven downvotes, so the only possible result is "-".
In the second example, no matter the person unknown downvotes or upvotes, Nauuo would get more upvotes than downvotes. So the only possible result is "+".
In the third example, Nauuo would definitely get one upvote and one downvote, so the only possible result is "0".
In the fourth example, if the only one person upvoted, the result would be "+", otherwise, the result would be "-". There are two possible results, so the result is uncertain.
Tags: greedy
Correct Solution:
```
def getN():
return int(input())
def getList():
return list(map(int, input().split()))
x,y,z = getList()
if z >= abs(x-y):
if z == 0:
print(0)
else:
print("?")
else:
if x > y:
print("+")
else:
print("-")
```
| 47,086 | [
0.07684326171875,
0.093505859375,
-0.48388671875,
-0.229248046875,
-0.7109375,
-0.5859375,
0.41357421875,
0.381103515625,
0.2001953125,
0.82275390625,
0.68798828125,
-0.2059326171875,
0.32763671875,
-0.2763671875,
-0.6962890625,
0.07952880859375,
-0.5166015625,
-0.9189453125,
0.1... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nauuo is a girl who loves writing comments.
One day, she posted a comment on Codeforces, wondering whether she would get upvotes or downvotes.
It's known that there were x persons who would upvote, y persons who would downvote, and there were also another z persons who would vote, but you don't know whether they would upvote or downvote. Note that each of the x+y+z people would vote exactly one time.
There are three different results: if there are more people upvote than downvote, the result will be "+"; if there are more people downvote than upvote, the result will be "-"; otherwise the result will be "0".
Because of the z unknown persons, the result may be uncertain (i.e. there are more than one possible results). More formally, the result is uncertain if and only if there exist two different situations of how the z persons vote, that the results are different in the two situations.
Tell Nauuo the result or report that the result is uncertain.
Input
The only line contains three integers x, y, z (0≤ x,y,z≤100), corresponding to the number of persons who would upvote, downvote or unknown.
Output
If there is only one possible result, print the result : "+", "-" or "0".
Otherwise, print "?" to report that the result is uncertain.
Examples
Input
3 7 0
Output
-
Input
2 0 1
Output
+
Input
1 1 0
Output
0
Input
0 0 1
Output
?
Note
In the first example, Nauuo would definitely get three upvotes and seven downvotes, so the only possible result is "-".
In the second example, no matter the person unknown downvotes or upvotes, Nauuo would get more upvotes than downvotes. So the only possible result is "+".
In the third example, Nauuo would definitely get one upvote and one downvote, so the only possible result is "0".
In the fourth example, if the only one person upvoted, the result would be "+", otherwise, the result would be "-". There are two possible results, so the result is uncertain.
Tags: greedy
Correct Solution:
```
X, Y, Z = map(int, input().split())
now = X - Y
if abs(now) <= Z:
if now == 0 and Z == 0: print('0')
else: print('?')
else:
if now > 0: print('+')
else: print('-')
```
| 47,087 | [
0.07684326171875,
0.093505859375,
-0.48388671875,
-0.229248046875,
-0.7109375,
-0.5859375,
0.41357421875,
0.381103515625,
0.2001953125,
0.82275390625,
0.68798828125,
-0.2059326171875,
0.32763671875,
-0.2763671875,
-0.6962890625,
0.07952880859375,
-0.5166015625,
-0.9189453125,
0.1... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nauuo is a girl who loves writing comments.
One day, she posted a comment on Codeforces, wondering whether she would get upvotes or downvotes.
It's known that there were x persons who would upvote, y persons who would downvote, and there were also another z persons who would vote, but you don't know whether they would upvote or downvote. Note that each of the x+y+z people would vote exactly one time.
There are three different results: if there are more people upvote than downvote, the result will be "+"; if there are more people downvote than upvote, the result will be "-"; otherwise the result will be "0".
Because of the z unknown persons, the result may be uncertain (i.e. there are more than one possible results). More formally, the result is uncertain if and only if there exist two different situations of how the z persons vote, that the results are different in the two situations.
Tell Nauuo the result or report that the result is uncertain.
Input
The only line contains three integers x, y, z (0≤ x,y,z≤100), corresponding to the number of persons who would upvote, downvote or unknown.
Output
If there is only one possible result, print the result : "+", "-" or "0".
Otherwise, print "?" to report that the result is uncertain.
Examples
Input
3 7 0
Output
-
Input
2 0 1
Output
+
Input
1 1 0
Output
0
Input
0 0 1
Output
?
Note
In the first example, Nauuo would definitely get three upvotes and seven downvotes, so the only possible result is "-".
In the second example, no matter the person unknown downvotes or upvotes, Nauuo would get more upvotes than downvotes. So the only possible result is "+".
In the third example, Nauuo would definitely get one upvote and one downvote, so the only possible result is "0".
In the fourth example, if the only one person upvoted, the result would be "+", otherwise, the result would be "-". There are two possible results, so the result is uncertain.
Tags: greedy
Correct Solution:
```
from sys import stdin
def main():
X, Y, Z = map(int, input().split())
n = sum([X, Y, Z]) // 2
if not Z and X == Y:
print(0)
elif n < X and n < X + Z:
print('+')
elif n < Y and n < Y + Z:
print('-')
else:
print('?')
input = lambda: stdin.readline()
main()
```
| 47,088 | [
0.07684326171875,
0.093505859375,
-0.48388671875,
-0.229248046875,
-0.7109375,
-0.5859375,
0.41357421875,
0.381103515625,
0.2001953125,
0.82275390625,
0.68798828125,
-0.2059326171875,
0.32763671875,
-0.2763671875,
-0.6962890625,
0.07952880859375,
-0.5166015625,
-0.9189453125,
0.1... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nauuo is a girl who loves writing comments.
One day, she posted a comment on Codeforces, wondering whether she would get upvotes or downvotes.
It's known that there were x persons who would upvote, y persons who would downvote, and there were also another z persons who would vote, but you don't know whether they would upvote or downvote. Note that each of the x+y+z people would vote exactly one time.
There are three different results: if there are more people upvote than downvote, the result will be "+"; if there are more people downvote than upvote, the result will be "-"; otherwise the result will be "0".
Because of the z unknown persons, the result may be uncertain (i.e. there are more than one possible results). More formally, the result is uncertain if and only if there exist two different situations of how the z persons vote, that the results are different in the two situations.
Tell Nauuo the result or report that the result is uncertain.
Input
The only line contains three integers x, y, z (0≤ x,y,z≤100), corresponding to the number of persons who would upvote, downvote or unknown.
Output
If there is only one possible result, print the result : "+", "-" or "0".
Otherwise, print "?" to report that the result is uncertain.
Examples
Input
3 7 0
Output
-
Input
2 0 1
Output
+
Input
1 1 0
Output
0
Input
0 0 1
Output
?
Note
In the first example, Nauuo would definitely get three upvotes and seven downvotes, so the only possible result is "-".
In the second example, no matter the person unknown downvotes or upvotes, Nauuo would get more upvotes than downvotes. So the only possible result is "+".
In the third example, Nauuo would definitely get one upvote and one downvote, so the only possible result is "0".
In the fourth example, if the only one person upvoted, the result would be "+", otherwise, the result would be "-". There are two possible results, so the result is uncertain.
Tags: greedy
Correct Solution:
```
x,y,z=map(int,input().split())
if(x>y and x>(y+z)):
res='+'
elif(y>x and y>(x+z)):
res='-'
elif(x==y and z==0):
res='0'
else:
res='?'
print(res)
```
| 47,089 | [
0.07684326171875,
0.093505859375,
-0.48388671875,
-0.229248046875,
-0.7109375,
-0.5859375,
0.41357421875,
0.381103515625,
0.2001953125,
0.82275390625,
0.68798828125,
-0.2059326171875,
0.32763671875,
-0.2763671875,
-0.6962890625,
0.07952880859375,
-0.5166015625,
-0.9189453125,
0.1... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nauuo is a girl who loves writing comments.
One day, she posted a comment on Codeforces, wondering whether she would get upvotes or downvotes.
It's known that there were x persons who would upvote, y persons who would downvote, and there were also another z persons who would vote, but you don't know whether they would upvote or downvote. Note that each of the x+y+z people would vote exactly one time.
There are three different results: if there are more people upvote than downvote, the result will be "+"; if there are more people downvote than upvote, the result will be "-"; otherwise the result will be "0".
Because of the z unknown persons, the result may be uncertain (i.e. there are more than one possible results). More formally, the result is uncertain if and only if there exist two different situations of how the z persons vote, that the results are different in the two situations.
Tell Nauuo the result or report that the result is uncertain.
Input
The only line contains three integers x, y, z (0≤ x,y,z≤100), corresponding to the number of persons who would upvote, downvote or unknown.
Output
If there is only one possible result, print the result : "+", "-" or "0".
Otherwise, print "?" to report that the result is uncertain.
Examples
Input
3 7 0
Output
-
Input
2 0 1
Output
+
Input
1 1 0
Output
0
Input
0 0 1
Output
?
Note
In the first example, Nauuo would definitely get three upvotes and seven downvotes, so the only possible result is "-".
In the second example, no matter the person unknown downvotes or upvotes, Nauuo would get more upvotes than downvotes. So the only possible result is "+".
In the third example, Nauuo would definitely get one upvote and one downvote, so the only possible result is "0".
In the fourth example, if the only one person upvoted, the result would be "+", otherwise, the result would be "-". There are two possible results, so the result is uncertain.
Tags: greedy
Correct Solution:
```
a,b,c=map(int,input().split())
if c==0:
if a>b:
print('+')
elif a<b:
print('-')
else:
print(0)
else:
if a-b>c:
print('+')
elif b-a>c:
print('-')
else:
print('?')
```
| 47,090 | [
0.07684326171875,
0.093505859375,
-0.48388671875,
-0.229248046875,
-0.7109375,
-0.5859375,
0.41357421875,
0.381103515625,
0.2001953125,
0.82275390625,
0.68798828125,
-0.2059326171875,
0.32763671875,
-0.2763671875,
-0.6962890625,
0.07952880859375,
-0.5166015625,
-0.9189453125,
0.1... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nauuo is a girl who loves writing comments.
One day, she posted a comment on Codeforces, wondering whether she would get upvotes or downvotes.
It's known that there were x persons who would upvote, y persons who would downvote, and there were also another z persons who would vote, but you don't know whether they would upvote or downvote. Note that each of the x+y+z people would vote exactly one time.
There are three different results: if there are more people upvote than downvote, the result will be "+"; if there are more people downvote than upvote, the result will be "-"; otherwise the result will be "0".
Because of the z unknown persons, the result may be uncertain (i.e. there are more than one possible results). More formally, the result is uncertain if and only if there exist two different situations of how the z persons vote, that the results are different in the two situations.
Tell Nauuo the result or report that the result is uncertain.
Input
The only line contains three integers x, y, z (0≤ x,y,z≤100), corresponding to the number of persons who would upvote, downvote or unknown.
Output
If there is only one possible result, print the result : "+", "-" or "0".
Otherwise, print "?" to report that the result is uncertain.
Examples
Input
3 7 0
Output
-
Input
2 0 1
Output
+
Input
1 1 0
Output
0
Input
0 0 1
Output
?
Note
In the first example, Nauuo would definitely get three upvotes and seven downvotes, so the only possible result is "-".
In the second example, no matter the person unknown downvotes or upvotes, Nauuo would get more upvotes than downvotes. So the only possible result is "+".
In the third example, Nauuo would definitely get one upvote and one downvote, so the only possible result is "0".
In the fourth example, if the only one person upvoted, the result would be "+", otherwise, the result would be "-". There are two possible results, so the result is uncertain.
Submitted Solution:
```
n, m, k = map(int, input().split())
if n > m and n > k + m:
print('+')
elif n < m and n + k < m:
print('-')
elif n == m and k == 0:
print('0')
else:
print('?')
```
Yes
| 47,091 | [
0.1199951171875,
0.1048583984375,
-0.58154296875,
-0.32568359375,
-0.75732421875,
-0.459716796875,
0.309814453125,
0.4921875,
0.1390380859375,
0.8486328125,
0.62060546875,
-0.07476806640625,
0.3388671875,
-0.306396484375,
-0.7763671875,
-0.038787841796875,
-0.5390625,
-0.7651367187... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nauuo is a girl who loves writing comments.
One day, she posted a comment on Codeforces, wondering whether she would get upvotes or downvotes.
It's known that there were x persons who would upvote, y persons who would downvote, and there were also another z persons who would vote, but you don't know whether they would upvote or downvote. Note that each of the x+y+z people would vote exactly one time.
There are three different results: if there are more people upvote than downvote, the result will be "+"; if there are more people downvote than upvote, the result will be "-"; otherwise the result will be "0".
Because of the z unknown persons, the result may be uncertain (i.e. there are more than one possible results). More formally, the result is uncertain if and only if there exist two different situations of how the z persons vote, that the results are different in the two situations.
Tell Nauuo the result or report that the result is uncertain.
Input
The only line contains three integers x, y, z (0≤ x,y,z≤100), corresponding to the number of persons who would upvote, downvote or unknown.
Output
If there is only one possible result, print the result : "+", "-" or "0".
Otherwise, print "?" to report that the result is uncertain.
Examples
Input
3 7 0
Output
-
Input
2 0 1
Output
+
Input
1 1 0
Output
0
Input
0 0 1
Output
?
Note
In the first example, Nauuo would definitely get three upvotes and seven downvotes, so the only possible result is "-".
In the second example, no matter the person unknown downvotes or upvotes, Nauuo would get more upvotes than downvotes. So the only possible result is "+".
In the third example, Nauuo would definitely get one upvote and one downvote, so the only possible result is "0".
In the fourth example, if the only one person upvoted, the result would be "+", otherwise, the result would be "-". There are two possible results, so the result is uncertain.
Submitted Solution:
```
from sys import stdin
line = stdin.readline().split(" ")
x = int(line[0])
y = int(line[1])
z = int(line[2])
if (y-z > x):
print("-")
elif (x-z > y):
print("+")
elif ((y-z == x) and (x-z == y)):
print("0")
else:
print("?")
```
Yes
| 47,092 | [
0.1199951171875,
0.1048583984375,
-0.58154296875,
-0.32568359375,
-0.75732421875,
-0.459716796875,
0.309814453125,
0.4921875,
0.1390380859375,
0.8486328125,
0.62060546875,
-0.07476806640625,
0.3388671875,
-0.306396484375,
-0.7763671875,
-0.038787841796875,
-0.5390625,
-0.7651367187... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nauuo is a girl who loves writing comments.
One day, she posted a comment on Codeforces, wondering whether she would get upvotes or downvotes.
It's known that there were x persons who would upvote, y persons who would downvote, and there were also another z persons who would vote, but you don't know whether they would upvote or downvote. Note that each of the x+y+z people would vote exactly one time.
There are three different results: if there are more people upvote than downvote, the result will be "+"; if there are more people downvote than upvote, the result will be "-"; otherwise the result will be "0".
Because of the z unknown persons, the result may be uncertain (i.e. there are more than one possible results). More formally, the result is uncertain if and only if there exist two different situations of how the z persons vote, that the results are different in the two situations.
Tell Nauuo the result or report that the result is uncertain.
Input
The only line contains three integers x, y, z (0≤ x,y,z≤100), corresponding to the number of persons who would upvote, downvote or unknown.
Output
If there is only one possible result, print the result : "+", "-" or "0".
Otherwise, print "?" to report that the result is uncertain.
Examples
Input
3 7 0
Output
-
Input
2 0 1
Output
+
Input
1 1 0
Output
0
Input
0 0 1
Output
?
Note
In the first example, Nauuo would definitely get three upvotes and seven downvotes, so the only possible result is "-".
In the second example, no matter the person unknown downvotes or upvotes, Nauuo would get more upvotes than downvotes. So the only possible result is "+".
In the third example, Nauuo would definitely get one upvote and one downvote, so the only possible result is "0".
In the fourth example, if the only one person upvoted, the result would be "+", otherwise, the result would be "-". There are two possible results, so the result is uncertain.
Submitted Solution:
```
a = [int(x) for x in input().split()]
if a[0] > a[1] and a[1] + a[2] < a[0]:
print("+")
elif a[1] > a[0] and a[0] + a[2] < a[1]:
print("-")
elif a[0] == a[1] and a[2] == 0:
print("0")
else:
print("?")
```
Yes
| 47,093 | [
0.1199951171875,
0.1048583984375,
-0.58154296875,
-0.32568359375,
-0.75732421875,
-0.459716796875,
0.309814453125,
0.4921875,
0.1390380859375,
0.8486328125,
0.62060546875,
-0.07476806640625,
0.3388671875,
-0.306396484375,
-0.7763671875,
-0.038787841796875,
-0.5390625,
-0.7651367187... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nauuo is a girl who loves writing comments.
One day, she posted a comment on Codeforces, wondering whether she would get upvotes or downvotes.
It's known that there were x persons who would upvote, y persons who would downvote, and there were also another z persons who would vote, but you don't know whether they would upvote or downvote. Note that each of the x+y+z people would vote exactly one time.
There are three different results: if there are more people upvote than downvote, the result will be "+"; if there are more people downvote than upvote, the result will be "-"; otherwise the result will be "0".
Because of the z unknown persons, the result may be uncertain (i.e. there are more than one possible results). More formally, the result is uncertain if and only if there exist two different situations of how the z persons vote, that the results are different in the two situations.
Tell Nauuo the result or report that the result is uncertain.
Input
The only line contains three integers x, y, z (0≤ x,y,z≤100), corresponding to the number of persons who would upvote, downvote or unknown.
Output
If there is only one possible result, print the result : "+", "-" or "0".
Otherwise, print "?" to report that the result is uncertain.
Examples
Input
3 7 0
Output
-
Input
2 0 1
Output
+
Input
1 1 0
Output
0
Input
0 0 1
Output
?
Note
In the first example, Nauuo would definitely get three upvotes and seven downvotes, so the only possible result is "-".
In the second example, no matter the person unknown downvotes or upvotes, Nauuo would get more upvotes than downvotes. So the only possible result is "+".
In the third example, Nauuo would definitely get one upvote and one downvote, so the only possible result is "0".
In the fourth example, if the only one person upvoted, the result would be "+", otherwise, the result would be "-". There are two possible results, so the result is uncertain.
Submitted Solution:
```
[x, y, z] = [int(x) for x in input().split()]
if x-y+z == 0 and x-y-z == 0: print('0')
elif x-y+z > 0 and x-y-z > 0: print('+')
elif x-y+z < 0 and x-y-z < 0: print('-')
else: print('?')
```
Yes
| 47,094 | [
0.1199951171875,
0.1048583984375,
-0.58154296875,
-0.32568359375,
-0.75732421875,
-0.459716796875,
0.309814453125,
0.4921875,
0.1390380859375,
0.8486328125,
0.62060546875,
-0.07476806640625,
0.3388671875,
-0.306396484375,
-0.7763671875,
-0.038787841796875,
-0.5390625,
-0.7651367187... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nauuo is a girl who loves writing comments.
One day, she posted a comment on Codeforces, wondering whether she would get upvotes or downvotes.
It's known that there were x persons who would upvote, y persons who would downvote, and there were also another z persons who would vote, but you don't know whether they would upvote or downvote. Note that each of the x+y+z people would vote exactly one time.
There are three different results: if there are more people upvote than downvote, the result will be "+"; if there are more people downvote than upvote, the result will be "-"; otherwise the result will be "0".
Because of the z unknown persons, the result may be uncertain (i.e. there are more than one possible results). More formally, the result is uncertain if and only if there exist two different situations of how the z persons vote, that the results are different in the two situations.
Tell Nauuo the result or report that the result is uncertain.
Input
The only line contains three integers x, y, z (0≤ x,y,z≤100), corresponding to the number of persons who would upvote, downvote or unknown.
Output
If there is only one possible result, print the result : "+", "-" or "0".
Otherwise, print "?" to report that the result is uncertain.
Examples
Input
3 7 0
Output
-
Input
2 0 1
Output
+
Input
1 1 0
Output
0
Input
0 0 1
Output
?
Note
In the first example, Nauuo would definitely get three upvotes and seven downvotes, so the only possible result is "-".
In the second example, no matter the person unknown downvotes or upvotes, Nauuo would get more upvotes than downvotes. So the only possible result is "+".
In the third example, Nauuo would definitely get one upvote and one downvote, so the only possible result is "0".
In the fourth example, if the only one person upvoted, the result would be "+", otherwise, the result would be "-". There are two possible results, so the result is uncertain.
Submitted Solution:
```
x,y,z=input().split()
x,y,z=int(x),int(y),int(z)
if x>y:
print("+")
if y>x:
print("-")
if x==y and x!=0 and y!=0:
print("0")
if z==1:
print("?")
```
No
| 47,095 | [
0.1199951171875,
0.1048583984375,
-0.58154296875,
-0.32568359375,
-0.75732421875,
-0.459716796875,
0.309814453125,
0.4921875,
0.1390380859375,
0.8486328125,
0.62060546875,
-0.07476806640625,
0.3388671875,
-0.306396484375,
-0.7763671875,
-0.038787841796875,
-0.5390625,
-0.7651367187... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nauuo is a girl who loves writing comments.
One day, she posted a comment on Codeforces, wondering whether she would get upvotes or downvotes.
It's known that there were x persons who would upvote, y persons who would downvote, and there were also another z persons who would vote, but you don't know whether they would upvote or downvote. Note that each of the x+y+z people would vote exactly one time.
There are three different results: if there are more people upvote than downvote, the result will be "+"; if there are more people downvote than upvote, the result will be "-"; otherwise the result will be "0".
Because of the z unknown persons, the result may be uncertain (i.e. there are more than one possible results). More formally, the result is uncertain if and only if there exist two different situations of how the z persons vote, that the results are different in the two situations.
Tell Nauuo the result or report that the result is uncertain.
Input
The only line contains three integers x, y, z (0≤ x,y,z≤100), corresponding to the number of persons who would upvote, downvote or unknown.
Output
If there is only one possible result, print the result : "+", "-" or "0".
Otherwise, print "?" to report that the result is uncertain.
Examples
Input
3 7 0
Output
-
Input
2 0 1
Output
+
Input
1 1 0
Output
0
Input
0 0 1
Output
?
Note
In the first example, Nauuo would definitely get three upvotes and seven downvotes, so the only possible result is "-".
In the second example, no matter the person unknown downvotes or upvotes, Nauuo would get more upvotes than downvotes. So the only possible result is "+".
In the third example, Nauuo would definitely get one upvote and one downvote, so the only possible result is "0".
In the fourth example, if the only one person upvoted, the result would be "+", otherwise, the result would be "-". There are two possible results, so the result is uncertain.
Submitted Solution:
```
a,b,c = input().split(' ')
a = int(a)
b = int(b)
c = int(c)
def solve(a,b,c):
if a == 0 and b == 0 and c > 0:
return '?'
if a > b + c:
return '+'
if a + c < b:
return '-'
return '?'
print(solve(a,b,c))
```
No
| 47,096 | [
0.1199951171875,
0.1048583984375,
-0.58154296875,
-0.32568359375,
-0.75732421875,
-0.459716796875,
0.309814453125,
0.4921875,
0.1390380859375,
0.8486328125,
0.62060546875,
-0.07476806640625,
0.3388671875,
-0.306396484375,
-0.7763671875,
-0.038787841796875,
-0.5390625,
-0.7651367187... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nauuo is a girl who loves writing comments.
One day, she posted a comment on Codeforces, wondering whether she would get upvotes or downvotes.
It's known that there were x persons who would upvote, y persons who would downvote, and there were also another z persons who would vote, but you don't know whether they would upvote or downvote. Note that each of the x+y+z people would vote exactly one time.
There are three different results: if there are more people upvote than downvote, the result will be "+"; if there are more people downvote than upvote, the result will be "-"; otherwise the result will be "0".
Because of the z unknown persons, the result may be uncertain (i.e. there are more than one possible results). More formally, the result is uncertain if and only if there exist two different situations of how the z persons vote, that the results are different in the two situations.
Tell Nauuo the result or report that the result is uncertain.
Input
The only line contains three integers x, y, z (0≤ x,y,z≤100), corresponding to the number of persons who would upvote, downvote or unknown.
Output
If there is only one possible result, print the result : "+", "-" or "0".
Otherwise, print "?" to report that the result is uncertain.
Examples
Input
3 7 0
Output
-
Input
2 0 1
Output
+
Input
1 1 0
Output
0
Input
0 0 1
Output
?
Note
In the first example, Nauuo would definitely get three upvotes and seven downvotes, so the only possible result is "-".
In the second example, no matter the person unknown downvotes or upvotes, Nauuo would get more upvotes than downvotes. So the only possible result is "+".
In the third example, Nauuo would definitely get one upvote and one downvote, so the only possible result is "0".
In the fourth example, if the only one person upvoted, the result would be "+", otherwise, the result would be "-". There are two possible results, so the result is uncertain.
Submitted Solution:
```
s = input().split()
for i in range(len(s)):
s[i] = int(s[i])
x = s[0]
y = s[1]
z = s[2]
if x > y:
if x > y+z:
print('+')
if x < y+z:
print('?')
if y > x:
if y > x+z:
print('-')
if y < x+z:
print('?')
if x == y:
if z > 0:
print('?')
if z == 0:
print('0')
```
No
| 47,097 | [
0.1199951171875,
0.1048583984375,
-0.58154296875,
-0.32568359375,
-0.75732421875,
-0.459716796875,
0.309814453125,
0.4921875,
0.1390380859375,
0.8486328125,
0.62060546875,
-0.07476806640625,
0.3388671875,
-0.306396484375,
-0.7763671875,
-0.038787841796875,
-0.5390625,
-0.7651367187... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nauuo is a girl who loves writing comments.
One day, she posted a comment on Codeforces, wondering whether she would get upvotes or downvotes.
It's known that there were x persons who would upvote, y persons who would downvote, and there were also another z persons who would vote, but you don't know whether they would upvote or downvote. Note that each of the x+y+z people would vote exactly one time.
There are three different results: if there are more people upvote than downvote, the result will be "+"; if there are more people downvote than upvote, the result will be "-"; otherwise the result will be "0".
Because of the z unknown persons, the result may be uncertain (i.e. there are more than one possible results). More formally, the result is uncertain if and only if there exist two different situations of how the z persons vote, that the results are different in the two situations.
Tell Nauuo the result or report that the result is uncertain.
Input
The only line contains three integers x, y, z (0≤ x,y,z≤100), corresponding to the number of persons who would upvote, downvote or unknown.
Output
If there is only one possible result, print the result : "+", "-" or "0".
Otherwise, print "?" to report that the result is uncertain.
Examples
Input
3 7 0
Output
-
Input
2 0 1
Output
+
Input
1 1 0
Output
0
Input
0 0 1
Output
?
Note
In the first example, Nauuo would definitely get three upvotes and seven downvotes, so the only possible result is "-".
In the second example, no matter the person unknown downvotes or upvotes, Nauuo would get more upvotes than downvotes. So the only possible result is "+".
In the third example, Nauuo would definitely get one upvote and one downvote, so the only possible result is "0".
In the fourth example, if the only one person upvoted, the result would be "+", otherwise, the result would be "-". There are two possible results, so the result is uncertain.
Submitted Solution:
```
x,y,z = map(int,input().split())
if x > y and z < x and z < y:
print("+")
elif x < y and z < x and z < y:
print("-")
elif x!=0 and x == y:
print(0)
elif z > x or z > y :
print('?')
```
No
| 47,098 | [
0.1199951171875,
0.1048583984375,
-0.58154296875,
-0.32568359375,
-0.75732421875,
-0.459716796875,
0.309814453125,
0.4921875,
0.1390380859375,
0.8486328125,
0.62060546875,
-0.07476806640625,
0.3388671875,
-0.306396484375,
-0.7763671875,
-0.038787841796875,
-0.5390625,
-0.7651367187... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you?
You are given a system of equations:
<image>
You should count, how many there are pairs of integers (a, b) (0 ≤ a, b) which satisfy the system.
Input
A single line contains two integers n, m (1 ≤ n, m ≤ 1000) — the parameters of the system. The numbers on the line are separated by a space.
Output
On a single line print the answer to the problem.
Examples
Input
9 3
Output
1
Input
14 28
Output
1
Input
4 20
Output
0
Note
In the first sample the suitable pair is integers (3, 0). In the second sample the suitable pair is integers (3, 5). In the third sample there is no suitable pair.
Submitted Solution:
```
import math
[n, m] = [int(x) for x in input().split()]
def check_a(a, b):
return (a*a)+b
def check_b(a, b):
return (b*b)+a
def __helper(n , m):
ans = 0
for i in range( int(math.sqrt(n)+1)):
for j in range(int(math.sqrt(m))+1):
# print(i, j)
if check_a(i, j) == n and check_b(i, j) == m:
ans += 1
return ans
ans = __helper(n, m)
print(ans)
```
Yes
| 47,325 | [
0.59716796875,
0.199951171875,
-0.33203125,
0.10699462890625,
-0.69189453125,
-0.44091796875,
-0.018524169921875,
0.28662109375,
0.077392578125,
0.96435546875,
0.6796875,
-0.0975341796875,
0.1309814453125,
-0.73193359375,
-0.5302734375,
0.0885009765625,
-0.572265625,
-0.546875,
-... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During a New Year special offer the "Sudislavl Bars" offered n promo codes. Each promo code consists of exactly six digits and gives right to one free cocktail at the bar "Mosquito Shelter". Of course, all the promocodes differ.
As the "Mosquito Shelter" opens only at 9, and partying in Sudislavl usually begins at as early as 6, many problems may arise as to how to type a promotional code without errors. It is necessary to calculate such maximum k, that the promotional code could be uniquely identified if it was typed with no more than k errors. At that, k = 0 means that the promotional codes must be entered exactly.
A mistake in this problem should be considered as entering the wrong numbers. For example, value "123465" contains two errors relative to promocode "123456". Regardless of the number of errors the entered value consists of exactly six digits.
Input
The first line of the output contains number n (1 ≤ n ≤ 1000) — the number of promocodes.
Each of the next n lines contains a single promocode, consisting of exactly 6 digits. It is guaranteed that all the promocodes are distinct. Promocodes can start from digit "0".
Output
Print the maximum k (naturally, not exceeding the length of the promocode), such that any promocode can be uniquely identified if it is typed with at most k mistakes.
Examples
Input
2
000000
999999
Output
2
Input
6
211111
212111
222111
111111
112111
121111
Output
0
Note
In the first sample k < 3, so if a bar customer types in value "090909", then it will be impossible to define which promocode exactly corresponds to it.
Submitted Solution:
```
#Here u can check fast your code!
def checcker(x, y):
p = 0
for mkT in range(6):
if x[mkT]==y[mkT]:
p=p+1
if p>=4:
break
return p
def viewer_finish(x, k):
if x!=1:
if k>=4: #2, 1, 0
print('0')
elif k==3 or k==2: #3 5
print('1')
elif k==1 or k==0:
print('2')
else:
print(6)
def checcker_main(kList, x):
k = 0
for fmN in range(x-1):
for KKT in range(fmN+1,x):
p = checcker(kList[fmN], kList[KKT])
if p>k:
k = p
if k>=4:
break
if k>=4:
break
return(k)
def main():
kList = []
x = int(input())
for i in range(x):
kList.append(input())
k = checcker_main(kList, x)
viewer_finish(x, k)
main()
```
Yes
| 47,510 | [
0.373291015625,
0.153564453125,
0.10504150390625,
0.254150390625,
-0.42822265625,
-0.4208984375,
-0.059967041015625,
0.380859375,
-0.46142578125,
0.6884765625,
0.54052734375,
-0.01093292236328125,
-0.0213623046875,
-0.4169921875,
-0.1964111328125,
0.09967041015625,
-0.767578125,
-0... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During a New Year special offer the "Sudislavl Bars" offered n promo codes. Each promo code consists of exactly six digits and gives right to one free cocktail at the bar "Mosquito Shelter". Of course, all the promocodes differ.
As the "Mosquito Shelter" opens only at 9, and partying in Sudislavl usually begins at as early as 6, many problems may arise as to how to type a promotional code without errors. It is necessary to calculate such maximum k, that the promotional code could be uniquely identified if it was typed with no more than k errors. At that, k = 0 means that the promotional codes must be entered exactly.
A mistake in this problem should be considered as entering the wrong numbers. For example, value "123465" contains two errors relative to promocode "123456". Regardless of the number of errors the entered value consists of exactly six digits.
Input
The first line of the output contains number n (1 ≤ n ≤ 1000) — the number of promocodes.
Each of the next n lines contains a single promocode, consisting of exactly 6 digits. It is guaranteed that all the promocodes are distinct. Promocodes can start from digit "0".
Output
Print the maximum k (naturally, not exceeding the length of the promocode), such that any promocode can be uniquely identified if it is typed with at most k mistakes.
Examples
Input
2
000000
999999
Output
2
Input
6
211111
212111
222111
111111
112111
121111
Output
0
Note
In the first sample k < 3, so if a bar customer types in value "090909", then it will be impossible to define which promocode exactly corresponds to it.
Submitted Solution:
```
n = int(input())
a = list(map(lambda i: input(), range(n)))
if n == 1:
print(6)
exit()
m = 7
for i in range(n):
for j in range(i + 1, n):
m = min(m, sum(a[i][k] != a[j][k] for k in range(6)))
print((m - 1) // 2)
```
Yes
| 47,511 | [
0.397216796875,
0.154541015625,
0.0703125,
0.21875,
-0.438720703125,
-0.40625,
-0.09686279296875,
0.34814453125,
-0.459228515625,
0.71875,
0.5185546875,
-0.00557708740234375,
-0.027862548828125,
-0.416015625,
-0.256591796875,
0.033416748046875,
-0.7802734375,
-0.467529296875,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During a New Year special offer the "Sudislavl Bars" offered n promo codes. Each promo code consists of exactly six digits and gives right to one free cocktail at the bar "Mosquito Shelter". Of course, all the promocodes differ.
As the "Mosquito Shelter" opens only at 9, and partying in Sudislavl usually begins at as early as 6, many problems may arise as to how to type a promotional code without errors. It is necessary to calculate such maximum k, that the promotional code could be uniquely identified if it was typed with no more than k errors. At that, k = 0 means that the promotional codes must be entered exactly.
A mistake in this problem should be considered as entering the wrong numbers. For example, value "123465" contains two errors relative to promocode "123456". Regardless of the number of errors the entered value consists of exactly six digits.
Input
The first line of the output contains number n (1 ≤ n ≤ 1000) — the number of promocodes.
Each of the next n lines contains a single promocode, consisting of exactly 6 digits. It is guaranteed that all the promocodes are distinct. Promocodes can start from digit "0".
Output
Print the maximum k (naturally, not exceeding the length of the promocode), such that any promocode can be uniquely identified if it is typed with at most k mistakes.
Examples
Input
2
000000
999999
Output
2
Input
6
211111
212111
222111
111111
112111
121111
Output
0
Note
In the first sample k < 3, so if a bar customer types in value "090909", then it will be impossible to define which promocode exactly corresponds to it.
Submitted Solution:
```
n = int(input())
m = 6
arr = []
for i in range(n):
arr.append(input())
for i in range(n - 1):
for j in range(i + 1, n):
d = 0
for z in range(6):
if arr[i][z] != arr[j][z]:
d += 1
if d == 6:
m = min(m, 2)
elif d == 5:
m = min(m, 2)
elif d == 4:
m = min(m, 1)
elif d == 3:
m = min(m, 1)
else:
m = 0
print(m)
```
Yes
| 47,512 | [
0.417236328125,
0.2171630859375,
0.07952880859375,
0.2373046875,
-0.4609375,
-0.43359375,
-0.060577392578125,
0.303955078125,
-0.45947265625,
0.7060546875,
0.6376953125,
-0.03704833984375,
-0.049072265625,
-0.454833984375,
-0.337890625,
0.0667724609375,
-0.7861328125,
-0.4514160156... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During a New Year special offer the "Sudislavl Bars" offered n promo codes. Each promo code consists of exactly six digits and gives right to one free cocktail at the bar "Mosquito Shelter". Of course, all the promocodes differ.
As the "Mosquito Shelter" opens only at 9, and partying in Sudislavl usually begins at as early as 6, many problems may arise as to how to type a promotional code without errors. It is necessary to calculate such maximum k, that the promotional code could be uniquely identified if it was typed with no more than k errors. At that, k = 0 means that the promotional codes must be entered exactly.
A mistake in this problem should be considered as entering the wrong numbers. For example, value "123465" contains two errors relative to promocode "123456". Regardless of the number of errors the entered value consists of exactly six digits.
Input
The first line of the output contains number n (1 ≤ n ≤ 1000) — the number of promocodes.
Each of the next n lines contains a single promocode, consisting of exactly 6 digits. It is guaranteed that all the promocodes are distinct. Promocodes can start from digit "0".
Output
Print the maximum k (naturally, not exceeding the length of the promocode), such that any promocode can be uniquely identified if it is typed with at most k mistakes.
Examples
Input
2
000000
999999
Output
2
Input
6
211111
212111
222111
111111
112111
121111
Output
0
Note
In the first sample k < 3, so if a bar customer types in value "090909", then it will be impossible to define which promocode exactly corresponds to it.
Submitted Solution:
```
ans = 12
l=[input() for i in range(int(input()))]
for i in range(len(l)):
for j in range(i+1,len(l)):
d=0
for w in range(6):
d+= l[i][w] != l[j][w]
ans=min(d-1,ans)
print(ans//2)
```
Yes
| 47,513 | [
0.40869140625,
0.1668701171875,
0.082275390625,
0.263916015625,
-0.4482421875,
-0.3994140625,
-0.040130615234375,
0.338134765625,
-0.471435546875,
0.6806640625,
0.51220703125,
-0.04644775390625,
-0.06072998046875,
-0.423095703125,
-0.260498046875,
0.06243896484375,
-0.7744140625,
-... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During a New Year special offer the "Sudislavl Bars" offered n promo codes. Each promo code consists of exactly six digits and gives right to one free cocktail at the bar "Mosquito Shelter". Of course, all the promocodes differ.
As the "Mosquito Shelter" opens only at 9, and partying in Sudislavl usually begins at as early as 6, many problems may arise as to how to type a promotional code without errors. It is necessary to calculate such maximum k, that the promotional code could be uniquely identified if it was typed with no more than k errors. At that, k = 0 means that the promotional codes must be entered exactly.
A mistake in this problem should be considered as entering the wrong numbers. For example, value "123465" contains two errors relative to promocode "123456". Regardless of the number of errors the entered value consists of exactly six digits.
Input
The first line of the output contains number n (1 ≤ n ≤ 1000) — the number of promocodes.
Each of the next n lines contains a single promocode, consisting of exactly 6 digits. It is guaranteed that all the promocodes are distinct. Promocodes can start from digit "0".
Output
Print the maximum k (naturally, not exceeding the length of the promocode), such that any promocode can be uniquely identified if it is typed with at most k mistakes.
Examples
Input
2
000000
999999
Output
2
Input
6
211111
212111
222111
111111
112111
121111
Output
0
Note
In the first sample k < 3, so if a bar customer types in value "090909", then it will be impossible to define which promocode exactly corresponds to it.
Submitted Solution:
```
def C():
n = int(input())
if n == 1:
return 6;
promocodes = [str(input()) for i in range(n)]
min_diff = 2
for i in promocodes[:-1]:
for j in promocodes[1:]:
count_diff = 0
for r, k in zip(i, j):
if r != k:
count_diff+=1
if count_diff < min_diff:
min_diff = count_diff
elif count_diff%2:
if ((count_diff-2) >= 0) and ((count_diff - 2) < min_diff):
min_diff = count_diff - 2
elif count_diff%2 == 0:
if ((count_diff-3) >= 0) and ((count_diff - 3) < min_diff):
min_diff = count_diff - 3
return min_diff
if __name__ == '__main__':
print(C())
```
No
| 47,514 | [
0.384521484375,
0.146728515625,
0.0855712890625,
0.24853515625,
-0.4638671875,
-0.399169921875,
-0.09375,
0.29541015625,
-0.449462890625,
0.728515625,
0.52392578125,
-0.00569915771484375,
-0.015899658203125,
-0.431640625,
-0.2685546875,
0.049835205078125,
-0.798828125,
-0.500488281... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During a New Year special offer the "Sudislavl Bars" offered n promo codes. Each promo code consists of exactly six digits and gives right to one free cocktail at the bar "Mosquito Shelter". Of course, all the promocodes differ.
As the "Mosquito Shelter" opens only at 9, and partying in Sudislavl usually begins at as early as 6, many problems may arise as to how to type a promotional code without errors. It is necessary to calculate such maximum k, that the promotional code could be uniquely identified if it was typed with no more than k errors. At that, k = 0 means that the promotional codes must be entered exactly.
A mistake in this problem should be considered as entering the wrong numbers. For example, value "123465" contains two errors relative to promocode "123456". Regardless of the number of errors the entered value consists of exactly six digits.
Input
The first line of the output contains number n (1 ≤ n ≤ 1000) — the number of promocodes.
Each of the next n lines contains a single promocode, consisting of exactly 6 digits. It is guaranteed that all the promocodes are distinct. Promocodes can start from digit "0".
Output
Print the maximum k (naturally, not exceeding the length of the promocode), such that any promocode can be uniquely identified if it is typed with at most k mistakes.
Examples
Input
2
000000
999999
Output
2
Input
6
211111
212111
222111
111111
112111
121111
Output
0
Note
In the first sample k < 3, so if a bar customer types in value "090909", then it will be impossible to define which promocode exactly corresponds to it.
Submitted Solution:
```
def func(a,b):
t = 0
for i in range(6):
if a[i] != b[i]:
print(1)
t+=1
return t
n = int(input())
A = [0] * n
for i in range(n):
A[i] = input()
per = 10
for i in range(0, n-1):
for j in range(i+1, n):
per = min(per, func(A[i], A[j]))
if n == 1:
print(6)
else:
if per in (5,6):
print(2)
elif per in (3,4):
print(1)
elif per in (0,1,2):
print(0)
```
No
| 47,515 | [
0.390625,
0.1419677734375,
0.050018310546875,
0.26318359375,
-0.466796875,
-0.431884765625,
-0.0699462890625,
0.348876953125,
-0.5009765625,
0.65625,
0.501953125,
-0.0194091796875,
-0.0703125,
-0.381591796875,
-0.286376953125,
0.098388671875,
-0.83544921875,
-0.4482421875,
-0.439... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During a New Year special offer the "Sudislavl Bars" offered n promo codes. Each promo code consists of exactly six digits and gives right to one free cocktail at the bar "Mosquito Shelter". Of course, all the promocodes differ.
As the "Mosquito Shelter" opens only at 9, and partying in Sudislavl usually begins at as early as 6, many problems may arise as to how to type a promotional code without errors. It is necessary to calculate such maximum k, that the promotional code could be uniquely identified if it was typed with no more than k errors. At that, k = 0 means that the promotional codes must be entered exactly.
A mistake in this problem should be considered as entering the wrong numbers. For example, value "123465" contains two errors relative to promocode "123456". Regardless of the number of errors the entered value consists of exactly six digits.
Input
The first line of the output contains number n (1 ≤ n ≤ 1000) — the number of promocodes.
Each of the next n lines contains a single promocode, consisting of exactly 6 digits. It is guaranteed that all the promocodes are distinct. Promocodes can start from digit "0".
Output
Print the maximum k (naturally, not exceeding the length of the promocode), such that any promocode can be uniquely identified if it is typed with at most k mistakes.
Examples
Input
2
000000
999999
Output
2
Input
6
211111
212111
222111
111111
112111
121111
Output
0
Note
In the first sample k < 3, so if a bar customer types in value "090909", then it will be impossible to define which promocode exactly corresponds to it.
Submitted Solution:
```
n = int(input())
codes = [input() for i in range(n)]
def dist(w1, w2):
assert len(w1) == len(w2)
d = 0
for i in range(len(w1)):
if w1[i] != w2[i]: d += 1
return max(0, d//2 - 1)
d = 6
for i in range(n):
for j in range(i+1, n):
d = min(d, dist(codes[i], codes[j]))
print(d)
```
No
| 47,516 | [
0.37353515625,
0.134521484375,
0.0789794921875,
0.226806640625,
-0.4697265625,
-0.396728515625,
-0.02899169921875,
0.316650390625,
-0.456787109375,
0.73046875,
0.495849609375,
-0.01259613037109375,
0.005706787109375,
-0.41455078125,
-0.24951171875,
0.06292724609375,
-0.84375,
-0.45... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During a New Year special offer the "Sudislavl Bars" offered n promo codes. Each promo code consists of exactly six digits and gives right to one free cocktail at the bar "Mosquito Shelter". Of course, all the promocodes differ.
As the "Mosquito Shelter" opens only at 9, and partying in Sudislavl usually begins at as early as 6, many problems may arise as to how to type a promotional code without errors. It is necessary to calculate such maximum k, that the promotional code could be uniquely identified if it was typed with no more than k errors. At that, k = 0 means that the promotional codes must be entered exactly.
A mistake in this problem should be considered as entering the wrong numbers. For example, value "123465" contains two errors relative to promocode "123456". Regardless of the number of errors the entered value consists of exactly six digits.
Input
The first line of the output contains number n (1 ≤ n ≤ 1000) — the number of promocodes.
Each of the next n lines contains a single promocode, consisting of exactly 6 digits. It is guaranteed that all the promocodes are distinct. Promocodes can start from digit "0".
Output
Print the maximum k (naturally, not exceeding the length of the promocode), such that any promocode can be uniquely identified if it is typed with at most k mistakes.
Examples
Input
2
000000
999999
Output
2
Input
6
211111
212111
222111
111111
112111
121111
Output
0
Note
In the first sample k < 3, so if a bar customer types in value "090909", then it will be impossible to define which promocode exactly corresponds to it.
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Sun Mar 13 19:40:03 2016
@author: Kostya S.
"""
from math import ceil
def hem(s,t):
r = 0
for u,v in zip(s,t):
r += u != v
return r
n = int(input())
code = set()
for _ in range(n):
c = (input())
code.add(c)
dist = set()
a = set()
k = 0
q = 0
for c1 in code:
for c2 in code:
if c1 != c2:
a.add(hem(c1,c2))
ans = 6 if n == 1 else max(0,(min(a)//2 - 1))
print(ans)
```
No
| 47,517 | [
0.373046875,
0.1683349609375,
0.06768798828125,
0.313232421875,
-0.46484375,
-0.43603515625,
-0.05413818359375,
0.333251953125,
-0.43701171875,
0.67578125,
0.537109375,
-0.07635498046875,
-0.055084228515625,
-0.42578125,
-0.19384765625,
0.1136474609375,
-0.73046875,
-0.454833984375... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya and Petya were tired of studying so they decided to play a game. Before the game begins Vasya looks at array a consisting of n integers. As soon as he remembers all elements of a the game begins. Vasya closes his eyes and Petya does q actions of one of two types:
1) Petya says 4 integers l1, r1, l2, r2 — boundaries of two non-intersecting segments. After that he swaps one random element from the [l1, r1] segment with another random element from the [l2, r2] segment.
2) Petya asks Vasya the sum of the elements of a in the [l, r] segment.
Vasya is a mathematician so he answers Petya the mathematical expectation of the sum of the elements in the segment.
Your task is to write a program which will answer the second type questions as Vasya would do it. In other words your program should print the mathematical expectation of the sum of the elements of a in the [l, r] segment for every second type query.
Input
The first line contains two integers n, q (2 ≤ n ≤ 105, 1 ≤ q ≤ 105) — the number of elements in the array and the number of queries you need to handle.
The second line contains n integers ai (1 ≤ ai ≤ 109) — elements of the array.
The next q lines contain Petya's actions of type 1 or 2.
If it is a type 1 action then the line contains 5 integers 1, l1, r1, l2, r2 (1 ≤ l1 ≤ r1 ≤ n, 1 ≤ l2 ≤ r2 ≤ n).
If it is a type 2 query then the line contains 3 integers 2, l, r (1 ≤ l ≤ r ≤ n).
It is guaranteed that there is at least one type 2 query and segments [l1, r1], [l2, r2] don't have common elements.
Output
For each type 2 query print one real number — the mathematical expectation of the sum of elements in the segment.
Your answer will be considered correct if its absolute or relative error doesn't exceed 10 - 4 — formally, the answer is correct if <image> where x is jury's answer and y is yours.
Examples
Input
4 4
1 1 2 2
1 2 2 3 3
2 1 2
1 1 2 3 4
2 1 2
Output
3.0000000
3.0000000
Input
10 5
1 1 1 1 1 2 2 2 2 2
1 1 5 6 10
2 1 5
1 1 5 6 10
1 1 5 6 10
2 6 10
Output
6.0000000
8.0400000
Input
10 10
1 2 3 4 5 6 7 8 9 10
1 1 5 6 10
1 1 5 6 10
2 1 5
1 1 3 6 9
2 1 3
1 5 7 8 10
1 1 1 10 10
2 1 5
2 7 10
2 1 10
Output
23.0000000
14.0000000
28.0133333
21.5733333
55.0000000
Submitted Solution:
```
n, k = input().split()
n, k = int(n), int(k)
array = input().split()
for i in range(len(array)):
array[i] = int(array[i])
for i in range(k):
v = input().split()
for a in range(len(v)):
v[a] = int(v[a])
if v[0] == 1:
array[(v[1]-1):v[2]], array[(v[3]-1):v[4]] = array[(v[3]-1):v[4]], array[(v[1]-1):v[2]]
elif v[0] == 2:
print(sum(array[(v[1]-1):v[2]]))
```
No
| 47,622 | [
0.14599609375,
0.267822265625,
-0.353515625,
0.2320556640625,
-0.65869140625,
-0.5087890625,
-0.058685302734375,
0.2125244140625,
-0.089599609375,
0.70849609375,
0.6494140625,
0.1461181640625,
0.10003662109375,
-0.66650390625,
-0.7197265625,
-0.09320068359375,
-0.5390625,
-1.127929... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya and Petya were tired of studying so they decided to play a game. Before the game begins Vasya looks at array a consisting of n integers. As soon as he remembers all elements of a the game begins. Vasya closes his eyes and Petya does q actions of one of two types:
1) Petya says 4 integers l1, r1, l2, r2 — boundaries of two non-intersecting segments. After that he swaps one random element from the [l1, r1] segment with another random element from the [l2, r2] segment.
2) Petya asks Vasya the sum of the elements of a in the [l, r] segment.
Vasya is a mathematician so he answers Petya the mathematical expectation of the sum of the elements in the segment.
Your task is to write a program which will answer the second type questions as Vasya would do it. In other words your program should print the mathematical expectation of the sum of the elements of a in the [l, r] segment for every second type query.
Input
The first line contains two integers n, q (2 ≤ n ≤ 105, 1 ≤ q ≤ 105) — the number of elements in the array and the number of queries you need to handle.
The second line contains n integers ai (1 ≤ ai ≤ 109) — elements of the array.
The next q lines contain Petya's actions of type 1 or 2.
If it is a type 1 action then the line contains 5 integers 1, l1, r1, l2, r2 (1 ≤ l1 ≤ r1 ≤ n, 1 ≤ l2 ≤ r2 ≤ n).
If it is a type 2 query then the line contains 3 integers 2, l, r (1 ≤ l ≤ r ≤ n).
It is guaranteed that there is at least one type 2 query and segments [l1, r1], [l2, r2] don't have common elements.
Output
For each type 2 query print one real number — the mathematical expectation of the sum of elements in the segment.
Your answer will be considered correct if its absolute or relative error doesn't exceed 10 - 4 — formally, the answer is correct if <image> where x is jury's answer and y is yours.
Examples
Input
4 4
1 1 2 2
1 2 2 3 3
2 1 2
1 1 2 3 4
2 1 2
Output
3.0000000
3.0000000
Input
10 5
1 1 1 1 1 2 2 2 2 2
1 1 5 6 10
2 1 5
1 1 5 6 10
1 1 5 6 10
2 6 10
Output
6.0000000
8.0400000
Input
10 10
1 2 3 4 5 6 7 8 9 10
1 1 5 6 10
1 1 5 6 10
2 1 5
1 1 3 6 9
2 1 3
1 5 7 8 10
1 1 1 10 10
2 1 5
2 7 10
2 1 10
Output
23.0000000
14.0000000
28.0133333
21.5733333
55.0000000
Submitted Solution:
```
def read():
return tuple(int(x) for x in input().split())
def main():
n, q = read()
nums = list(read())
for i in range(q):
op = read()
if op[0] == 1:
s1 = 0
for j in range(op[1] - 1, op[2]):
s1 += nums[j]
s1s = s1 / (op[2] + 1 - op[1])
s2 = 0
for j in range(op[3] - 1, op[4]):
s2 += nums[j]
s2s = s2 / (op[4] + 1 - op[3])
s1 += s2s - s1s
s2 += s1s - s2s
s1 /= (op[2] + 1 - op[1])
s2 /= (op[4] + 1 - op[3])
for j in range(op[1] - 1, op[2]):
nums[j] = s1
for j in range(op[3] - 1, op[4]):
nums[j] = s2
elif op[0] == 2:
s = 0
for j in range(op[1] - 1, op[2]):
s += nums[j]
print(s)
main()
```
No
| 47,623 | [
0.14599609375,
0.267822265625,
-0.353515625,
0.2320556640625,
-0.65869140625,
-0.5087890625,
-0.058685302734375,
0.2125244140625,
-0.089599609375,
0.70849609375,
0.6494140625,
0.1461181640625,
0.10003662109375,
-0.66650390625,
-0.7197265625,
-0.09320068359375,
-0.5390625,
-1.127929... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya and Petya were tired of studying so they decided to play a game. Before the game begins Vasya looks at array a consisting of n integers. As soon as he remembers all elements of a the game begins. Vasya closes his eyes and Petya does q actions of one of two types:
1) Petya says 4 integers l1, r1, l2, r2 — boundaries of two non-intersecting segments. After that he swaps one random element from the [l1, r1] segment with another random element from the [l2, r2] segment.
2) Petya asks Vasya the sum of the elements of a in the [l, r] segment.
Vasya is a mathematician so he answers Petya the mathematical expectation of the sum of the elements in the segment.
Your task is to write a program which will answer the second type questions as Vasya would do it. In other words your program should print the mathematical expectation of the sum of the elements of a in the [l, r] segment for every second type query.
Input
The first line contains two integers n, q (2 ≤ n ≤ 105, 1 ≤ q ≤ 105) — the number of elements in the array and the number of queries you need to handle.
The second line contains n integers ai (1 ≤ ai ≤ 109) — elements of the array.
The next q lines contain Petya's actions of type 1 or 2.
If it is a type 1 action then the line contains 5 integers 1, l1, r1, l2, r2 (1 ≤ l1 ≤ r1 ≤ n, 1 ≤ l2 ≤ r2 ≤ n).
If it is a type 2 query then the line contains 3 integers 2, l, r (1 ≤ l ≤ r ≤ n).
It is guaranteed that there is at least one type 2 query and segments [l1, r1], [l2, r2] don't have common elements.
Output
For each type 2 query print one real number — the mathematical expectation of the sum of elements in the segment.
Your answer will be considered correct if its absolute or relative error doesn't exceed 10 - 4 — formally, the answer is correct if <image> where x is jury's answer and y is yours.
Examples
Input
4 4
1 1 2 2
1 2 2 3 3
2 1 2
1 1 2 3 4
2 1 2
Output
3.0000000
3.0000000
Input
10 5
1 1 1 1 1 2 2 2 2 2
1 1 5 6 10
2 1 5
1 1 5 6 10
1 1 5 6 10
2 6 10
Output
6.0000000
8.0400000
Input
10 10
1 2 3 4 5 6 7 8 9 10
1 1 5 6 10
1 1 5 6 10
2 1 5
1 1 3 6 9
2 1 3
1 5 7 8 10
1 1 1 10 10
2 1 5
2 7 10
2 1 10
Output
23.0000000
14.0000000
28.0133333
21.5733333
55.0000000
Submitted Solution:
```
import random
n, q = map(int, input().split())
mas = list(map(int, input().split()))
for i in range(q):
s = list(map(int, input().split()))
if len(s) == 5:
one = random.choice(mas[s[1] - 1: s[2]])
two = random.choice(mas[s[3] - 1: s[4]])
mas[mas.index(one)], mas[mas.index(two)] = mas[mas.index(two)], mas[mas.index(one)]
print(mas)
else:
print(float(sum(mas[s[1] - 1: s[2]])))
```
No
| 47,624 | [
0.14599609375,
0.267822265625,
-0.353515625,
0.2320556640625,
-0.65869140625,
-0.5087890625,
-0.058685302734375,
0.2125244140625,
-0.089599609375,
0.70849609375,
0.6494140625,
0.1461181640625,
0.10003662109375,
-0.66650390625,
-0.7197265625,
-0.09320068359375,
-0.5390625,
-1.127929... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya and Petya were tired of studying so they decided to play a game. Before the game begins Vasya looks at array a consisting of n integers. As soon as he remembers all elements of a the game begins. Vasya closes his eyes and Petya does q actions of one of two types:
1) Petya says 4 integers l1, r1, l2, r2 — boundaries of two non-intersecting segments. After that he swaps one random element from the [l1, r1] segment with another random element from the [l2, r2] segment.
2) Petya asks Vasya the sum of the elements of a in the [l, r] segment.
Vasya is a mathematician so he answers Petya the mathematical expectation of the sum of the elements in the segment.
Your task is to write a program which will answer the second type questions as Vasya would do it. In other words your program should print the mathematical expectation of the sum of the elements of a in the [l, r] segment for every second type query.
Input
The first line contains two integers n, q (2 ≤ n ≤ 105, 1 ≤ q ≤ 105) — the number of elements in the array and the number of queries you need to handle.
The second line contains n integers ai (1 ≤ ai ≤ 109) — elements of the array.
The next q lines contain Petya's actions of type 1 or 2.
If it is a type 1 action then the line contains 5 integers 1, l1, r1, l2, r2 (1 ≤ l1 ≤ r1 ≤ n, 1 ≤ l2 ≤ r2 ≤ n).
If it is a type 2 query then the line contains 3 integers 2, l, r (1 ≤ l ≤ r ≤ n).
It is guaranteed that there is at least one type 2 query and segments [l1, r1], [l2, r2] don't have common elements.
Output
For each type 2 query print one real number — the mathematical expectation of the sum of elements in the segment.
Your answer will be considered correct if its absolute or relative error doesn't exceed 10 - 4 — formally, the answer is correct if <image> where x is jury's answer and y is yours.
Examples
Input
4 4
1 1 2 2
1 2 2 3 3
2 1 2
1 1 2 3 4
2 1 2
Output
3.0000000
3.0000000
Input
10 5
1 1 1 1 1 2 2 2 2 2
1 1 5 6 10
2 1 5
1 1 5 6 10
1 1 5 6 10
2 6 10
Output
6.0000000
8.0400000
Input
10 10
1 2 3 4 5 6 7 8 9 10
1 1 5 6 10
1 1 5 6 10
2 1 5
1 1 3 6 9
2 1 3
1 5 7 8 10
1 1 1 10 10
2 1 5
2 7 10
2 1 10
Output
23.0000000
14.0000000
28.0133333
21.5733333
55.0000000
Submitted Solution:
```
import copy
def swaps(l1,r1,l2,r2):
ss = []
for i in range(l1,r1+1):
for j in range(l2,r2+1):
ss.append([i,j])
return ss
def permute(arr,l1,r1,l2,r2):
arrays = []
ss = swaps(l1,r1,l2,r2)
for a,b in ss:
temp = copy.deepcopy(arr)
temp[a],temp[b] = temp[b],temp[a]
arrays.append(temp)
return arrays
def asum(arr,l,r):
temp = 0
for a in arr:
temp+= sum(a[l:r+1])
return temp/len(arr)
n,q = [int(i) for i in input().split()]
arr = [[int(i) for i in input().split()]]
while True:
newa = []
try:
stuff = input().split()
except:
break
typ,data = stuff[0],stuff[1:]
if typ == '1':
l1,r1,l2,r2 = [int(i)-1 for i in data]
for a in arr[:10000]:
newa += permute(a,l1,r1,l2,r2)
arr = newa
elif typ == '2':
l,r = [int(i)-1 for i in data]
print(asum(arr,l,r))
```
No
| 47,625 | [
0.14599609375,
0.267822265625,
-0.353515625,
0.2320556640625,
-0.65869140625,
-0.5087890625,
-0.058685302734375,
0.2125244140625,
-0.089599609375,
0.70849609375,
0.6494140625,
0.1461181640625,
0.10003662109375,
-0.66650390625,
-0.7197265625,
-0.09320068359375,
-0.5390625,
-1.127929... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a set of items, each having some integer weight not greater than 8. You denote that a subset of items is good if total weight of items in the subset does not exceed W.
You want to calculate the maximum possible weight of a good subset of items. Note that you have to consider the empty set and the original set when calculating the answer.
Input
The first line contains one integer W (0 ≤ W ≤ 10^{18}) — the maximum total weight of a good subset.
The second line denotes the set of items you have. It contains 8 integers cnt_1, cnt_2, ..., cnt_8 (0 ≤ cnt_i ≤ 10^{16}), where cnt_i is the number of items having weight i in the set.
Output
Print one integer — the maximum possible weight of a good subset of items.
Examples
Input
10
1 2 3 4 5 6 7 8
Output
10
Input
0
0 0 0 0 0 0 0 0
Output
0
Input
3
0 4 1 0 0 9 8 3
Output
3
Submitted Solution:
```
w = int(input())
l= list(map(int, input().split()))
maximum = 0
def fun(point, weight):
global maximum
if point == 8 and weight <= w:
maximum= max(weight, maximum)
else:
if weight < w:
fun(point + 1, weight + l[point])
fun(point + 1, weight)
fun(0, 0)
print(maximum)
```
No
| 47,986 | [
0.2294921875,
-0.002933502197265625,
-0.142578125,
0.310791015625,
-0.6865234375,
-0.47509765625,
0.1475830078125,
0.38916015625,
0.2301025390625,
0.77880859375,
0.6826171875,
0.01396942138671875,
0.1302490234375,
-0.86865234375,
-0.55859375,
0.1385498046875,
-0.94775390625,
-0.909... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a set of items, each having some integer weight not greater than 8. You denote that a subset of items is good if total weight of items in the subset does not exceed W.
You want to calculate the maximum possible weight of a good subset of items. Note that you have to consider the empty set and the original set when calculating the answer.
Input
The first line contains one integer W (0 ≤ W ≤ 10^{18}) — the maximum total weight of a good subset.
The second line denotes the set of items you have. It contains 8 integers cnt_1, cnt_2, ..., cnt_8 (0 ≤ cnt_i ≤ 10^{16}), where cnt_i is the number of items having weight i in the set.
Output
Print one integer — the maximum possible weight of a good subset of items.
Examples
Input
10
1 2 3 4 5 6 7 8
Output
10
Input
0
0 0 0 0 0 0 0 0
Output
0
Input
3
0 4 1 0 0 9 8 3
Output
3
Submitted Solution:
```
import time
import random
W = int(input())
M = [int(a) for a in input().split()]
A = [0] * 8
sTime = time.time()
s = 0
mi = 10**10
for i in range(8):
if s + M[i]*(i+1) <= W:
s += M[i]*(i+1)
A[i] = M[i]
else:
t = (W-s)//(i+1)
s += t*(i+1)
A[i] += t
if s <= W:
mi = min(mi, W-s)
while time.time() - sTime < 1.75:
i = random.randrange(8)
a = random.randrange(2)
if W-s >= 20 or (s-W < 10 and a == 0):
if A[i] < M[i]:
A[i] += 1
s += (i+1)
else:
if A[i] > 0:
A[i] -= 1
s -= (i+1)
if s <= W:
mi = min(mi, W-s)
print(W-mi)
```
No
| 47,989 | [
0.2200927734375,
0.11224365234375,
-0.20751953125,
0.185546875,
-0.65673828125,
-0.53955078125,
0.04022216796875,
0.32666015625,
0.33740234375,
0.87841796875,
0.609375,
-0.0972900390625,
0.1468505859375,
-0.93798828125,
-0.49462890625,
0.09814453125,
-1.0322265625,
-0.8955078125,
... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During the last Sereja's Codesecrof round the server crashed many times, so the round was decided to be made unrated for some participants.
Let's assume that n people took part in the contest. Let's assume that the participant who got the first place has rating a1, the second place participant has rating a2, ..., the n-th place participant has rating an. Then changing the rating on the Codesecrof site is calculated by the formula <image>.
After the round was over, the Codesecrof management published the participants' results table. They decided that if for a participant di < k, then the round can be considered unrated for him. But imagine the management's surprise when they found out that the participants' rating table is dynamic. In other words, when some participant is removed from the rating, he is removed from the results' table and the rating is recalculated according to the new table. And of course, all applications for exclusion from the rating are considered in view of the current table.
We know that among all the applications for exclusion from the rating the first application to consider is from the participant with the best rank (the rank with the minimum number), for who di < k. We also know that the applications for exclusion from rating were submitted by all participants.
Now Sereja wonders, what is the number of participants to be excluded from the contest rating, and the numbers of the participants in the original table in the order of their exclusion from the rating. Pay attention to the analysis of the first test case for a better understanding of the statement.
Input
The first line contains two integers n, k (1 ≤ n ≤ 2·105, - 109 ≤ k ≤ 0). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — ratings of the participants in the initial table.
Output
Print the numbers of participants in the order in which they were removed from the table. Print the initial numbers of the participants, that is, the numbers that the participants had in the initial table.
Examples
Input
5 0
5 3 4 1 2
Output
2
3
4
Input
10 -10
5 5 1 7 5 1 2 4 9 2
Output
2
4
5
7
8
9
Note
Consider the first test sample.
1. Initially the sequence of the contest participants' ratings equals [5, 3, 4, 1, 2]. You can use this sequence to calculate the sequence of rating changes: [0, -9, -13, 8, 14]. According to the problem statement, the application of the participant who won the second place will be considered first.
2. As soon as the second place winner is out from the ratings, the participants' rating sequence will equal [5, 4, 1, 2]. By this sequence you can count the new sequence of rating changes: [0, -8, 2, 6]. According to the problem statement, the application of the participant who won the second place will be considered. Initially this participant won third place.
3. The new rating sequence equals [5, 1, 2], the new sequence of rating changes equals [0, -1, 1]. The second place participant's application is taken into consideration, initially this participant won the fourth place.
4. The new rating sequence equals [5, 2], the new sequence of rating changes equals [0, 0]. No more applications will be considered.
Thus, you should print 2, 3, 4.
Tags: implementation
Correct Solution:
```
from sys import stdin,stdout
nmbr = lambda: int(stdin.readline())
lst = lambda: list(map(int,stdin.readline().split()))
def fn(a):
print('a',a)
n=len(a)
pos1=[0]
for i in range(1,n):
pos1+=[pos1[-1]+a[i]*i]
print('pos1',pos1)
neg=[]
for i in range(n):
neg+=[i*(n-i-1)*a[i]]
print('neg',neg)
d=[]
for i in range(1,n+1):
sm=0
for j in range(1,i):
sm+=a[j-1]*(j-1)-(n-i)*a[i-1]
d+=[sm]
print('d',d);
print('================================')
p=-1
for i in range(1,n):
if pos1[i-1]-neg[i]<k:
p=i
break
if p==-1:return
a.pop(p)
fn(a)
for _ in range(1):#nmbr()):
n,k=lst()
a=lst()
# fn(a)
# exit(0)
pos = [0]
for i in range(1, n):
pos += [pos[-1] + a[i] * i]
removed=0
positive_term=0
for i in range(1,n):
negative_term=(i-removed)*a[i]*(n-i-1)
# print(positive_term,positive_term-negative_term)
if (positive_term-negative_term)<k:
removed+=1
stdout.write(str(i+1)+'\n')
else:positive_term += (i - removed) * a[i]
```
| 48,242 | [
0.07452392578125,
-0.1990966796875,
-0.2069091796875,
0.4658203125,
-0.26904296875,
-0.6845703125,
-0.52001953125,
0.1756591796875,
-0.237548828125,
0.7900390625,
0.7080078125,
0.0361328125,
0.142333984375,
-0.98095703125,
-0.397216796875,
0.1455078125,
-0.78271484375,
-0.766601562... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During the last Sereja's Codesecrof round the server crashed many times, so the round was decided to be made unrated for some participants.
Let's assume that n people took part in the contest. Let's assume that the participant who got the first place has rating a1, the second place participant has rating a2, ..., the n-th place participant has rating an. Then changing the rating on the Codesecrof site is calculated by the formula <image>.
After the round was over, the Codesecrof management published the participants' results table. They decided that if for a participant di < k, then the round can be considered unrated for him. But imagine the management's surprise when they found out that the participants' rating table is dynamic. In other words, when some participant is removed from the rating, he is removed from the results' table and the rating is recalculated according to the new table. And of course, all applications for exclusion from the rating are considered in view of the current table.
We know that among all the applications for exclusion from the rating the first application to consider is from the participant with the best rank (the rank with the minimum number), for who di < k. We also know that the applications for exclusion from rating were submitted by all participants.
Now Sereja wonders, what is the number of participants to be excluded from the contest rating, and the numbers of the participants in the original table in the order of their exclusion from the rating. Pay attention to the analysis of the first test case for a better understanding of the statement.
Input
The first line contains two integers n, k (1 ≤ n ≤ 2·105, - 109 ≤ k ≤ 0). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — ratings of the participants in the initial table.
Output
Print the numbers of participants in the order in which they were removed from the table. Print the initial numbers of the participants, that is, the numbers that the participants had in the initial table.
Examples
Input
5 0
5 3 4 1 2
Output
2
3
4
Input
10 -10
5 5 1 7 5 1 2 4 9 2
Output
2
4
5
7
8
9
Note
Consider the first test sample.
1. Initially the sequence of the contest participants' ratings equals [5, 3, 4, 1, 2]. You can use this sequence to calculate the sequence of rating changes: [0, -9, -13, 8, 14]. According to the problem statement, the application of the participant who won the second place will be considered first.
2. As soon as the second place winner is out from the ratings, the participants' rating sequence will equal [5, 4, 1, 2]. By this sequence you can count the new sequence of rating changes: [0, -8, 2, 6]. According to the problem statement, the application of the participant who won the second place will be considered. Initially this participant won third place.
3. The new rating sequence equals [5, 1, 2], the new sequence of rating changes equals [0, -1, 1]. The second place participant's application is taken into consideration, initially this participant won the fourth place.
4. The new rating sequence equals [5, 2], the new sequence of rating changes equals [0, 0]. No more applications will be considered.
Thus, you should print 2, 3, 4.
Tags: implementation
Correct Solution:
```
n, k = map(int, input().split())
c, m, l, r = 0, 0, [], 0
for e in [int(i) for i in input().split()]:
d = m - c * (n - c - 1) * e
r+= 1
if d < k:
n -= 1
l += [r]
else:
m += c * e
c += 1
l.sort()
for e in l: print(e)
```
| 48,243 | [
0.07452392578125,
-0.1990966796875,
-0.2069091796875,
0.4658203125,
-0.26904296875,
-0.6845703125,
-0.52001953125,
0.1756591796875,
-0.237548828125,
0.7900390625,
0.7080078125,
0.0361328125,
0.142333984375,
-0.98095703125,
-0.397216796875,
0.1455078125,
-0.78271484375,
-0.766601562... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During the last Sereja's Codesecrof round the server crashed many times, so the round was decided to be made unrated for some participants.
Let's assume that n people took part in the contest. Let's assume that the participant who got the first place has rating a1, the second place participant has rating a2, ..., the n-th place participant has rating an. Then changing the rating on the Codesecrof site is calculated by the formula <image>.
After the round was over, the Codesecrof management published the participants' results table. They decided that if for a participant di < k, then the round can be considered unrated for him. But imagine the management's surprise when they found out that the participants' rating table is dynamic. In other words, when some participant is removed from the rating, he is removed from the results' table and the rating is recalculated according to the new table. And of course, all applications for exclusion from the rating are considered in view of the current table.
We know that among all the applications for exclusion from the rating the first application to consider is from the participant with the best rank (the rank with the minimum number), for who di < k. We also know that the applications for exclusion from rating were submitted by all participants.
Now Sereja wonders, what is the number of participants to be excluded from the contest rating, and the numbers of the participants in the original table in the order of their exclusion from the rating. Pay attention to the analysis of the first test case for a better understanding of the statement.
Input
The first line contains two integers n, k (1 ≤ n ≤ 2·105, - 109 ≤ k ≤ 0). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — ratings of the participants in the initial table.
Output
Print the numbers of participants in the order in which they were removed from the table. Print the initial numbers of the participants, that is, the numbers that the participants had in the initial table.
Examples
Input
5 0
5 3 4 1 2
Output
2
3
4
Input
10 -10
5 5 1 7 5 1 2 4 9 2
Output
2
4
5
7
8
9
Note
Consider the first test sample.
1. Initially the sequence of the contest participants' ratings equals [5, 3, 4, 1, 2]. You can use this sequence to calculate the sequence of rating changes: [0, -9, -13, 8, 14]. According to the problem statement, the application of the participant who won the second place will be considered first.
2. As soon as the second place winner is out from the ratings, the participants' rating sequence will equal [5, 4, 1, 2]. By this sequence you can count the new sequence of rating changes: [0, -8, 2, 6]. According to the problem statement, the application of the participant who won the second place will be considered. Initially this participant won third place.
3. The new rating sequence equals [5, 1, 2], the new sequence of rating changes equals [0, -1, 1]. The second place participant's application is taken into consideration, initially this participant won the fourth place.
4. The new rating sequence equals [5, 2], the new sequence of rating changes equals [0, 0]. No more applications will be considered.
Thus, you should print 2, 3, 4.
Tags: implementation
Correct Solution:
```
n, k = map(int, input().split())
arr = map(int, input().split())
s, j, all_res = 0, 0, []
for i, q in enumerate(arr, 1):
if s - j * (n - i) * q < k:
all_res.append(str(i))
else:
s += q * j
j += 1
print('\n'.join(all_res))
```
| 48,244 | [
0.07452392578125,
-0.1990966796875,
-0.2069091796875,
0.4658203125,
-0.26904296875,
-0.6845703125,
-0.52001953125,
0.1756591796875,
-0.237548828125,
0.7900390625,
0.7080078125,
0.0361328125,
0.142333984375,
-0.98095703125,
-0.397216796875,
0.1455078125,
-0.78271484375,
-0.766601562... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During the last Sereja's Codesecrof round the server crashed many times, so the round was decided to be made unrated for some participants.
Let's assume that n people took part in the contest. Let's assume that the participant who got the first place has rating a1, the second place participant has rating a2, ..., the n-th place participant has rating an. Then changing the rating on the Codesecrof site is calculated by the formula <image>.
After the round was over, the Codesecrof management published the participants' results table. They decided that if for a participant di < k, then the round can be considered unrated for him. But imagine the management's surprise when they found out that the participants' rating table is dynamic. In other words, when some participant is removed from the rating, he is removed from the results' table and the rating is recalculated according to the new table. And of course, all applications for exclusion from the rating are considered in view of the current table.
We know that among all the applications for exclusion from the rating the first application to consider is from the participant with the best rank (the rank with the minimum number), for who di < k. We also know that the applications for exclusion from rating were submitted by all participants.
Now Sereja wonders, what is the number of participants to be excluded from the contest rating, and the numbers of the participants in the original table in the order of their exclusion from the rating. Pay attention to the analysis of the first test case for a better understanding of the statement.
Input
The first line contains two integers n, k (1 ≤ n ≤ 2·105, - 109 ≤ k ≤ 0). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — ratings of the participants in the initial table.
Output
Print the numbers of participants in the order in which they were removed from the table. Print the initial numbers of the participants, that is, the numbers that the participants had in the initial table.
Examples
Input
5 0
5 3 4 1 2
Output
2
3
4
Input
10 -10
5 5 1 7 5 1 2 4 9 2
Output
2
4
5
7
8
9
Note
Consider the first test sample.
1. Initially the sequence of the contest participants' ratings equals [5, 3, 4, 1, 2]. You can use this sequence to calculate the sequence of rating changes: [0, -9, -13, 8, 14]. According to the problem statement, the application of the participant who won the second place will be considered first.
2. As soon as the second place winner is out from the ratings, the participants' rating sequence will equal [5, 4, 1, 2]. By this sequence you can count the new sequence of rating changes: [0, -8, 2, 6]. According to the problem statement, the application of the participant who won the second place will be considered. Initially this participant won third place.
3. The new rating sequence equals [5, 1, 2], the new sequence of rating changes equals [0, -1, 1]. The second place participant's application is taken into consideration, initially this participant won the fourth place.
4. The new rating sequence equals [5, 2], the new sequence of rating changes equals [0, 0]. No more applications will be considered.
Thus, you should print 2, 3, 4.
Tags: implementation
Correct Solution:
```
from sys import stdin,stdout
nmbr = lambda: int(stdin.readline())
lst = lambda: list(map(int,stdin.readline().split()))
for _ in range(1):#nmbr()):
n,k=lst()
a=lst()
removed=0
positive_term=0
for i in range(1,n):
negative_term=(i-removed)*a[i]*(n-i-1)
if (positive_term-negative_term)<k:
removed+=1
stdout.write(str(i+1)+'\n')
else:positive_term += (i - removed) * a[i]
```
| 48,245 | [
0.07452392578125,
-0.1990966796875,
-0.2069091796875,
0.4658203125,
-0.26904296875,
-0.6845703125,
-0.52001953125,
0.1756591796875,
-0.237548828125,
0.7900390625,
0.7080078125,
0.0361328125,
0.142333984375,
-0.98095703125,
-0.397216796875,
0.1455078125,
-0.78271484375,
-0.766601562... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During the last Sereja's Codesecrof round the server crashed many times, so the round was decided to be made unrated for some participants.
Let's assume that n people took part in the contest. Let's assume that the participant who got the first place has rating a1, the second place participant has rating a2, ..., the n-th place participant has rating an. Then changing the rating on the Codesecrof site is calculated by the formula <image>.
After the round was over, the Codesecrof management published the participants' results table. They decided that if for a participant di < k, then the round can be considered unrated for him. But imagine the management's surprise when they found out that the participants' rating table is dynamic. In other words, when some participant is removed from the rating, he is removed from the results' table and the rating is recalculated according to the new table. And of course, all applications for exclusion from the rating are considered in view of the current table.
We know that among all the applications for exclusion from the rating the first application to consider is from the participant with the best rank (the rank with the minimum number), for who di < k. We also know that the applications for exclusion from rating were submitted by all participants.
Now Sereja wonders, what is the number of participants to be excluded from the contest rating, and the numbers of the participants in the original table in the order of their exclusion from the rating. Pay attention to the analysis of the first test case for a better understanding of the statement.
Input
The first line contains two integers n, k (1 ≤ n ≤ 2·105, - 109 ≤ k ≤ 0). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — ratings of the participants in the initial table.
Output
Print the numbers of participants in the order in which they were removed from the table. Print the initial numbers of the participants, that is, the numbers that the participants had in the initial table.
Examples
Input
5 0
5 3 4 1 2
Output
2
3
4
Input
10 -10
5 5 1 7 5 1 2 4 9 2
Output
2
4
5
7
8
9
Note
Consider the first test sample.
1. Initially the sequence of the contest participants' ratings equals [5, 3, 4, 1, 2]. You can use this sequence to calculate the sequence of rating changes: [0, -9, -13, 8, 14]. According to the problem statement, the application of the participant who won the second place will be considered first.
2. As soon as the second place winner is out from the ratings, the participants' rating sequence will equal [5, 4, 1, 2]. By this sequence you can count the new sequence of rating changes: [0, -8, 2, 6]. According to the problem statement, the application of the participant who won the second place will be considered. Initially this participant won third place.
3. The new rating sequence equals [5, 1, 2], the new sequence of rating changes equals [0, -1, 1]. The second place participant's application is taken into consideration, initially this participant won the fourth place.
4. The new rating sequence equals [5, 2], the new sequence of rating changes equals [0, 0]. No more applications will be considered.
Thus, you should print 2, 3, 4.
Submitted Solution:
```
from sys import stdin,stdout
nmbr = lambda: int(stdin.readline())
lst = lambda: list(map(int,stdin.readline().split()))
for _ in range(1):#nmbr()):
n,k=lst()
a=lst()
pos=[0]
neg=[]
for i in range(1,n):
pos+=[pos[-1]+a[i]*i]
removed=0;sm=0
for i in range(1,n):
positive_term=pos[i-1]-sm
negative_term=(i-removed)*a[i]*(n-i-1)
if (positive_term-negative_term)<k:
sm+=i*a[i]
removed+=1
stdout.write(str(i+1)+'\n')
```
No
| 48,246 | [
0.1669921875,
-0.1617431640625,
-0.1416015625,
0.351806640625,
-0.383544921875,
-0.6142578125,
-0.55078125,
0.218994140625,
-0.25048828125,
0.73095703125,
0.62548828125,
0.10675048828125,
0.099609375,
-0.939453125,
-0.337890625,
0.1007080078125,
-0.7333984375,
-0.751953125,
-0.39... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given matrix with n rows and n columns filled with zeroes. You should put k ones in it in such a way that the resulting matrix is symmetrical with respect to the main diagonal (the diagonal that goes from the top left to the bottom right corner) and is lexicographically maximal.
One matrix is lexicographically greater than the other if the first different number in the first different row from the top in the first matrix is greater than the corresponding number in the second one.
If there exists no such matrix then output -1.
Input
The first line consists of two numbers n and k (1 ≤ n ≤ 100, 0 ≤ k ≤ 106).
Output
If the answer exists then output resulting matrix. Otherwise output -1.
Examples
Input
2 1
Output
1 0
0 0
Input
3 2
Output
1 0 0
0 1 0
0 0 0
Input
2 5
Output
-1
Submitted Solution:
```
import sys
import math
n,k = map(int, input().split())
ans = [[0 for i in range(n)] for i in range(n)]
row = 0;
col = 0;
scol = 0;
while(row < n):
col = scol
while(col < n):
if(col == row and k > 0):
ans[row][col] = 1
k -= 1
elif k > 0:
if k >= 2:
ans[row][col] = 1
ans[col][row] = 1
k -= 2
col += 1
row += 1
scol += 1
if k == 0:
for i in range(n):
for j in range(n):
print(ans[i][j], end = " ")
print()
else:
print(-1)
```
Yes
| 48,490 | [
0.415771484375,
0.146484375,
0.1346435546875,
0.031890869140625,
-0.52978515625,
-0.446044921875,
-0.09686279296875,
0.29443359375,
-0.2486572265625,
0.79736328125,
0.476318359375,
0.2919921875,
0.00008153915405273438,
-0.58984375,
-0.432373046875,
-0.050811767578125,
-0.403076171875... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given matrix with n rows and n columns filled with zeroes. You should put k ones in it in such a way that the resulting matrix is symmetrical with respect to the main diagonal (the diagonal that goes from the top left to the bottom right corner) and is lexicographically maximal.
One matrix is lexicographically greater than the other if the first different number in the first different row from the top in the first matrix is greater than the corresponding number in the second one.
If there exists no such matrix then output -1.
Input
The first line consists of two numbers n and k (1 ≤ n ≤ 100, 0 ≤ k ≤ 106).
Output
If the answer exists then output resulting matrix. Otherwise output -1.
Examples
Input
2 1
Output
1 0
0 0
Input
3 2
Output
1 0 0
0 1 0
0 0 0
Input
2 5
Output
-1
Submitted Solution:
```
import sys
def solve():
n, k = map(int, input().split())
if k > n**2:
print(-1)
else:
mat = [[0]*n for i in range(n)]
for i in range(n):
for j in range(i, n):
if k <= 0:
break
if i == j:
mat[i][i] = 1
k -= 1
else:
if k > 1:
mat[i][j] = mat[j][i] = 1
k -= 2
for mat_r in mat:
print(*mat_r)
if __name__ == '__main__':
solve()
```
Yes
| 48,491 | [
0.3681640625,
0.1644287109375,
0.1798095703125,
0.0614013671875,
-0.5048828125,
-0.3330078125,
-0.116943359375,
0.34228515625,
-0.1656494140625,
0.75390625,
0.419921875,
0.34033203125,
0.034332275390625,
-0.56201171875,
-0.466552734375,
-0.01324462890625,
-0.392333984375,
-0.727050... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given matrix with n rows and n columns filled with zeroes. You should put k ones in it in such a way that the resulting matrix is symmetrical with respect to the main diagonal (the diagonal that goes from the top left to the bottom right corner) and is lexicographically maximal.
One matrix is lexicographically greater than the other if the first different number in the first different row from the top in the first matrix is greater than the corresponding number in the second one.
If there exists no such matrix then output -1.
Input
The first line consists of two numbers n and k (1 ≤ n ≤ 100, 0 ≤ k ≤ 106).
Output
If the answer exists then output resulting matrix. Otherwise output -1.
Examples
Input
2 1
Output
1 0
0 0
Input
3 2
Output
1 0 0
0 1 0
0 0 0
Input
2 5
Output
-1
Submitted Solution:
```
import sys
def solve():
n, k = map(int, input().split())
mat = [[0]*n for i in range(n)]
for i in range(n):
for j in range(i, n):
if k == 0:
break
if i == j:
mat[i][j] = 1
k -= 1
elif k > 1:
mat[i][j] = mat[j][i] = 1
k -= 2
if k != 0:
print(-1)
return
for mat_r in mat:
print(*mat_r)
if __name__ == '__main__':
solve()
```
Yes
| 48,492 | [
0.39697265625,
0.1763916015625,
0.1695556640625,
0.0428466796875,
-0.492919921875,
-0.348876953125,
-0.10723876953125,
0.33740234375,
-0.1622314453125,
0.728515625,
0.40185546875,
0.324462890625,
0.037750244140625,
-0.5615234375,
-0.46923828125,
-0.0252685546875,
-0.40966796875,
-0... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given matrix with n rows and n columns filled with zeroes. You should put k ones in it in such a way that the resulting matrix is symmetrical with respect to the main diagonal (the diagonal that goes from the top left to the bottom right corner) and is lexicographically maximal.
One matrix is lexicographically greater than the other if the first different number in the first different row from the top in the first matrix is greater than the corresponding number in the second one.
If there exists no such matrix then output -1.
Input
The first line consists of two numbers n and k (1 ≤ n ≤ 100, 0 ≤ k ≤ 106).
Output
If the answer exists then output resulting matrix. Otherwise output -1.
Examples
Input
2 1
Output
1 0
0 0
Input
3 2
Output
1 0 0
0 1 0
0 0 0
Input
2 5
Output
-1
Submitted Solution:
```
#Bhargey Mehta (Sophomore)
#DA-IICT, Gandhinagar
import sys, math, queue
#sys.stdin = open("input.txt", "r")
MOD = 10**9+7
sys.setrecursionlimit(1000000)
n, k = map(int, input().split())
if k > n:
print(-1)
exit()
a = [[0 for i in range(n)] for i in range(n)]
if k == 1:
a[0][0] = 1
if k <= 1:
for ai in a:
print(*ai)
exit()
for i in range(n):
if k > 2:
a[i][i] = 1
k -= 1
elif k == 1:
a[i][i] = 1
break
for j in range(i+1, n):
if k < 2:
break
a[i][j] = 1
a[j][i] = 1
k -= 2
for ai in a:
print(*ai)
```
No
| 48,494 | [
0.31787109375,
0.22802734375,
0.24755859375,
0.10479736328125,
-0.47802734375,
-0.381591796875,
-0.1707763671875,
0.294189453125,
-0.31884765625,
0.771484375,
0.3681640625,
0.33544921875,
-0.11016845703125,
-0.66943359375,
-0.453369140625,
0.0297088623046875,
-0.4951171875,
-0.8046... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given matrix with n rows and n columns filled with zeroes. You should put k ones in it in such a way that the resulting matrix is symmetrical with respect to the main diagonal (the diagonal that goes from the top left to the bottom right corner) and is lexicographically maximal.
One matrix is lexicographically greater than the other if the first different number in the first different row from the top in the first matrix is greater than the corresponding number in the second one.
If there exists no such matrix then output -1.
Input
The first line consists of two numbers n and k (1 ≤ n ≤ 100, 0 ≤ k ≤ 106).
Output
If the answer exists then output resulting matrix. Otherwise output -1.
Examples
Input
2 1
Output
1 0
0 0
Input
3 2
Output
1 0 0
0 1 0
0 0 0
Input
2 5
Output
-1
Submitted Solution:
```
from collections import Counter;
import sys
n,k=map(int,input().split())
li=[[0]*n for _ in range(n)]
if k<=n:
i=0
while k>0:
li[i][i]=1
i+=1
k-=1
elif k>n and (n-k)==0:
while k>0:
for i in li:
for j in i:
if li[i][j]!=1:
li[i][j]=1
li[j][i]=1
k-=2
else:
li=None
if li==None:
print(-1)
else:
for i in li:
for j in i:
print(j,end=" ")
print()
```
No
| 48,495 | [
0.313720703125,
0.15283203125,
0.1490478515625,
0.08197021484375,
-0.481689453125,
-0.4072265625,
-0.126708984375,
0.3486328125,
-0.19970703125,
0.79150390625,
0.4140625,
0.311279296875,
0.0576171875,
-0.58837890625,
-0.440673828125,
-0.06549072265625,
-0.47119140625,
-0.7490234375... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given matrix with n rows and n columns filled with zeroes. You should put k ones in it in such a way that the resulting matrix is symmetrical with respect to the main diagonal (the diagonal that goes from the top left to the bottom right corner) and is lexicographically maximal.
One matrix is lexicographically greater than the other if the first different number in the first different row from the top in the first matrix is greater than the corresponding number in the second one.
If there exists no such matrix then output -1.
Input
The first line consists of two numbers n and k (1 ≤ n ≤ 100, 0 ≤ k ≤ 106).
Output
If the answer exists then output resulting matrix. Otherwise output -1.
Examples
Input
2 1
Output
1 0
0 0
Input
3 2
Output
1 0 0
0 1 0
0 0 0
Input
2 5
Output
-1
Submitted Solution:
```
def main():
n, k = map(int, input().split())
if n * n < k:
print(-1)
return
l = [['0'] * n for _ in range(n)]
for i in range(n):
q = (n - i) * 2 - 1
if q <= k:
for j in range(i, n):
l[i][j] = l[j][i] = '1'
k -= q
else:
for j in range(i, i + (k + 1) // 2):
l[i][j] = l[j][i] = '1'
if k & 0xfe:
l[i + 1][i + 1] = '1'
break
for r in l:
print(' '.join(r))
if __name__ == "__main__":
main()
```
No
| 48,497 | [
0.371337890625,
0.123779296875,
0.158447265625,
0.003025054931640625,
-0.463134765625,
-0.408935546875,
-0.11529541015625,
0.345703125,
-0.316162109375,
0.8017578125,
0.424072265625,
0.32177734375,
0.043792724609375,
-0.59765625,
-0.479736328125,
-0.06805419921875,
-0.517578125,
-0... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
Decades have passed since the beginning of AtCoder Beginner Contest.
The contests are labeled as `ABC001`, `ABC002`, ... from the first round, but after the 999-th round `ABC999`, a problem occurred: how the future rounds should be labeled?
In the end, the labels for the rounds from the 1000-th to the 1998-th are decided: `ABD001`, `ABD002`, ..., `ABD999`.
You are given an integer N between 1 and 1998 (inclusive). Print the first three characters of the label of the N-th round of AtCoder Beginner Contest.
Constraints
* 1 \leq N \leq 1998
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
Print the first three characters of the label of the N-th round of AtCoder Beginner Contest.
Examples
Input
999
Output
ABC
Input
1000
Output
ABD
Input
1481
Output
ABD
"Correct Solution:
```
n = int(input())
print('ABD' if n > 999 else 'ABC')
```
| 48,622 | [
0.4951171875,
-0.2401123046875,
-0.458984375,
0.09783935546875,
-0.417236328125,
-0.5712890625,
-0.0302581787109375,
0.1875,
0.12744140625,
0.367919921875,
0.67919921875,
-0.431640625,
-0.0916748046875,
-0.8515625,
-0.2274169921875,
0.29736328125,
-0.459716796875,
-0.763671875,
-... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
Decades have passed since the beginning of AtCoder Beginner Contest.
The contests are labeled as `ABC001`, `ABC002`, ... from the first round, but after the 999-th round `ABC999`, a problem occurred: how the future rounds should be labeled?
In the end, the labels for the rounds from the 1000-th to the 1998-th are decided: `ABD001`, `ABD002`, ..., `ABD999`.
You are given an integer N between 1 and 1998 (inclusive). Print the first three characters of the label of the N-th round of AtCoder Beginner Contest.
Constraints
* 1 \leq N \leq 1998
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
Print the first three characters of the label of the N-th round of AtCoder Beginner Contest.
Examples
Input
999
Output
ABC
Input
1000
Output
ABD
Input
1481
Output
ABD
"Correct Solution:
```
print(["ABC","ABD"][1000<=int(input())])
```
| 48,623 | [
0.52197265625,
-0.24951171875,
-0.42724609375,
0.073974609375,
-0.400146484375,
-0.541015625,
-0.045654296875,
0.16357421875,
0.13916015625,
0.34423828125,
0.689453125,
-0.466064453125,
-0.09332275390625,
-0.8291015625,
-0.20458984375,
0.317626953125,
-0.428466796875,
-0.755859375,... | 11 |
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