message stringlengths 2 39.6k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 219 108k | cluster float64 11 11 | __index_level_0__ int64 438 217k |
|---|---|---|---|---|---|
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)
``` | instruction | 0 | 45,335 | 11 | 90,670 |
No | output | 1 | 45,335 | 11 | 90,671 |
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)
``` | instruction | 0 | 45,336 | 11 | 90,672 |
No | output | 1 | 45,336 | 11 | 90,673 |
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()
``` | instruction | 0 | 45,551 | 11 | 91,102 |
No | output | 1 | 45,551 | 11 | 91,103 |
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)
``` | instruction | 0 | 45,552 | 11 | 91,104 |
No | output | 1 | 45,552 | 11 | 91,105 |
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()
``` | instruction | 0 | 45,553 | 11 | 91,106 |
No | output | 1 | 45,553 | 11 | 91,107 |
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)
``` | instruction | 0 | 45,554 | 11 | 91,108 |
No | output | 1 | 45,554 | 11 | 91,109 |
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]))
``` | instruction | 0 | 45,660 | 11 | 91,320 |
Yes | output | 1 | 45,660 | 11 | 91,321 |
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=" ")
``` | instruction | 0 | 45,661 | 11 | 91,322 |
Yes | output | 1 | 45,661 | 11 | 91,323 |
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()
``` | instruction | 0 | 45,662 | 11 | 91,324 |
Yes | output | 1 | 45,662 | 11 | 91,325 |
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))
``` | instruction | 0 | 45,663 | 11 | 91,326 |
Yes | output | 1 | 45,663 | 11 | 91,327 |
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)
``` | instruction | 0 | 45,664 | 11 | 91,328 |
No | output | 1 | 45,664 | 11 | 91,329 |
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)
``` | instruction | 0 | 45,665 | 11 | 91,330 |
No | output | 1 | 45,665 | 11 | 91,331 |
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()
``` | instruction | 0 | 45,666 | 11 | 91,332 |
No | output | 1 | 45,666 | 11 | 91,333 |
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))
``` | instruction | 0 | 45,667 | 11 | 91,334 |
No | output | 1 | 45,667 | 11 | 91,335 |
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))
``` | instruction | 0 | 45,734 | 11 | 91,468 |
Yes | output | 1 | 45,734 | 11 | 91,469 |
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)]))
``` | instruction | 0 | 45,736 | 11 | 91,472 |
Yes | output | 1 | 45,736 | 11 | 91,473 |
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)
``` | instruction | 0 | 45,738 | 11 | 91,476 |
No | output | 1 | 45,738 | 11 | 91,477 |
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)
``` | instruction | 0 | 45,740 | 11 | 91,480 |
No | output | 1 | 45,740 | 11 | 91,481 |
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)))
``` | instruction | 0 | 46,016 | 11 | 92,032 |
Yes | output | 1 | 46,016 | 11 | 92,033 |
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)))
``` | instruction | 0 | 46,017 | 11 | 92,034 |
Yes | output | 1 | 46,017 | 11 | 92,035 |
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)))
``` | instruction | 0 | 46,018 | 11 | 92,036 |
Yes | output | 1 | 46,018 | 11 | 92,037 |
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))
``` | instruction | 0 | 46,019 | 11 | 92,038 |
Yes | output | 1 | 46,019 | 11 | 92,039 |
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
``` | instruction | 0 | 46,020 | 11 | 92,040 |
No | output | 1 | 46,020 | 11 | 92,041 |
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)))
``` | instruction | 0 | 46,021 | 11 | 92,042 |
No | output | 1 | 46,021 | 11 | 92,043 |
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)
``` | instruction | 0 | 46,022 | 11 | 92,044 |
No | output | 1 | 46,022 | 11 | 92,045 |
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)))
``` | instruction | 0 | 46,023 | 11 | 92,046 |
No | output | 1 | 46,023 | 11 | 92,047 |
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. | instruction | 0 | 46,398 | 11 | 92,796 |
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)
``` | output | 1 | 46,398 | 11 | 92,797 |
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. | instruction | 0 | 46,399 | 11 | 92,798 |
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)
``` | output | 1 | 46,399 | 11 | 92,799 |
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. | instruction | 0 | 46,400 | 11 | 92,800 |
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)
``` | output | 1 | 46,400 | 11 | 92,801 |
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. | instruction | 0 | 46,401 | 11 | 92,802 |
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
``` | output | 1 | 46,401 | 11 | 92,803 |
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. | instruction | 0 | 46,402 | 11 | 92,804 |
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)
``` | output | 1 | 46,402 | 11 | 92,805 |
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. | instruction | 0 | 46,403 | 11 | 92,806 |
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)
``` | output | 1 | 46,403 | 11 | 92,807 |
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. | instruction | 0 | 46,404 | 11 | 92,808 |
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)
``` | output | 1 | 46,404 | 11 | 92,809 |
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. | instruction | 0 | 46,405 | 11 | 92,810 |
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)
``` | output | 1 | 46,405 | 11 | 92,811 |
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)
``` | instruction | 0 | 46,406 | 11 | 92,812 |
Yes | output | 1 | 46,406 | 11 | 92,813 |
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)
``` | instruction | 0 | 46,407 | 11 | 92,814 |
Yes | output | 1 | 46,407 | 11 | 92,815 |
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)
``` | instruction | 0 | 46,408 | 11 | 92,816 |
Yes | output | 1 | 46,408 | 11 | 92,817 |
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()
``` | instruction | 0 | 46,409 | 11 | 92,818 |
Yes | output | 1 | 46,409 | 11 | 92,819 |
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)
``` | instruction | 0 | 46,410 | 11 | 92,820 |
No | output | 1 | 46,410 | 11 | 92,821 |
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])
``` | instruction | 0 | 46,411 | 11 | 92,822 |
No | output | 1 | 46,411 | 11 | 92,823 |
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])
``` | instruction | 0 | 46,412 | 11 | 92,824 |
No | output | 1 | 46,412 | 11 | 92,825 |
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)
``` | instruction | 0 | 46,413 | 11 | 92,826 |
No | output | 1 | 46,413 | 11 | 92,827 |
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'))
``` | instruction | 0 | 46,558 | 11 | 93,116 |
No | output | 1 | 46,558 | 11 | 93,117 |
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)
``` | instruction | 0 | 46,559 | 11 | 93,118 |
No | output | 1 | 46,559 | 11 | 93,119 |
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)])
``` | instruction | 0 | 46,561 | 11 | 93,122 |
No | output | 1 | 46,561 | 11 | 93,123 |
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?
?>>??
>>?
""" )
``` | instruction | 0 | 46,573 | 11 | 93,146 |
No | output | 1 | 46,573 | 11 | 93,147 |
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")
``` | instruction | 0 | 46,574 | 11 | 93,148 |
No | output | 1 | 46,574 | 11 | 93,149 |
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)+'?')
``` | instruction | 0 | 46,575 | 11 | 93,150 |
No | output | 1 | 46,575 | 11 | 93,151 |
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)))
``` | instruction | 0 | 46,737 | 11 | 93,474 |
No | output | 1 | 46,737 | 11 | 93,475 |
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);
``` | instruction | 0 | 46,738 | 11 | 93,476 |
No | output | 1 | 46,738 | 11 | 93,477 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.