message stringlengths 2 65.1k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 0 108k | cluster float64 14 14 | __index_level_0__ int64 0 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
As a tradition, every year before IOI all the members of Natalia Fan Club are invited to Malek Dance Club to have a fun night together. Malek Dance Club has 2n members and coincidentally Natalia Fan Club also has 2n members. Each member of MDC is assigned a unique id i from 0 to 2n - 1. The same holds for each member of NFC.
One of the parts of this tradition is one by one dance, where each member of MDC dances with a member of NFC. A dance pair is a pair of numbers (a, b) such that member a from MDC dances with member b from NFC.
The complexity of a pairs' assignment is the number of pairs of dancing pairs (a, b) and (c, d) such that a < c and b > d.
You are given a binary number of length n named x. We know that member i from MDC dances with member <image> from NFC. Your task is to calculate the complexity of this assignment modulo 1000000007 (109 + 7).
Expression <image> denotes applying «XOR» to numbers x and y. This operation exists in all modern programming languages, for example, in C++ and Java it denotes as «^», in Pascal — «xor».
Input
The first line of input contains a binary number x of lenght n, (1 ≤ n ≤ 100).
This number may contain leading zeros.
Output
Print the complexity of the given dance assignent modulo 1000000007 (109 + 7).
Examples
Input
11
Output
6
Input
01
Output
2
Input
1
Output
1 | instruction | 0 | 88,720 | 14 | 177,440 |
Tags: combinatorics, math
Correct Solution:
```
s,m,ans=input()[::-1],int(1e9+7),0
for i in range(len(s)):
if s[i]=='1':ans=(ans+pow(2,i,m))%m
print(ans*pow(2,len(s)-1,m)%m)
``` | output | 1 | 88,720 | 14 | 177,441 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As a tradition, every year before IOI all the members of Natalia Fan Club are invited to Malek Dance Club to have a fun night together. Malek Dance Club has 2n members and coincidentally Natalia Fan Club also has 2n members. Each member of MDC is assigned a unique id i from 0 to 2n - 1. The same holds for each member of NFC.
One of the parts of this tradition is one by one dance, where each member of MDC dances with a member of NFC. A dance pair is a pair of numbers (a, b) such that member a from MDC dances with member b from NFC.
The complexity of a pairs' assignment is the number of pairs of dancing pairs (a, b) and (c, d) such that a < c and b > d.
You are given a binary number of length n named x. We know that member i from MDC dances with member <image> from NFC. Your task is to calculate the complexity of this assignment modulo 1000000007 (109 + 7).
Expression <image> denotes applying «XOR» to numbers x and y. This operation exists in all modern programming languages, for example, in C++ and Java it denotes as «^», in Pascal — «xor».
Input
The first line of input contains a binary number x of lenght n, (1 ≤ n ≤ 100).
This number may contain leading zeros.
Output
Print the complexity of the given dance assignent modulo 1000000007 (109 + 7).
Examples
Input
11
Output
6
Input
01
Output
2
Input
1
Output
1 | instruction | 0 | 88,721 | 14 | 177,442 |
Tags: combinatorics, math
Correct Solution:
```
n = input().strip()
s = len(n)
k = int(n,2)
start = 4 **( s-1)
zib = 2**(s-1)
step = 2**(s-1)
print((start+(k-zib)*step)%(10**9+7))
``` | output | 1 | 88,721 | 14 | 177,443 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As a tradition, every year before IOI all the members of Natalia Fan Club are invited to Malek Dance Club to have a fun night together. Malek Dance Club has 2n members and coincidentally Natalia Fan Club also has 2n members. Each member of MDC is assigned a unique id i from 0 to 2n - 1. The same holds for each member of NFC.
One of the parts of this tradition is one by one dance, where each member of MDC dances with a member of NFC. A dance pair is a pair of numbers (a, b) such that member a from MDC dances with member b from NFC.
The complexity of a pairs' assignment is the number of pairs of dancing pairs (a, b) and (c, d) such that a < c and b > d.
You are given a binary number of length n named x. We know that member i from MDC dances with member <image> from NFC. Your task is to calculate the complexity of this assignment modulo 1000000007 (109 + 7).
Expression <image> denotes applying «XOR» to numbers x and y. This operation exists in all modern programming languages, for example, in C++ and Java it denotes as «^», in Pascal — «xor».
Input
The first line of input contains a binary number x of lenght n, (1 ≤ n ≤ 100).
This number may contain leading zeros.
Output
Print the complexity of the given dance assignent modulo 1000000007 (109 + 7).
Examples
Input
11
Output
6
Input
01
Output
2
Input
1
Output
1 | instruction | 0 | 88,722 | 14 | 177,444 |
Tags: combinatorics, math
Correct Solution:
```
#!/usr/bin/env python3
p = 1000000007
s = input()
print(int(s,2) * pow(2, len(s) - 1) % p)
``` | output | 1 | 88,722 | 14 | 177,445 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As a tradition, every year before IOI all the members of Natalia Fan Club are invited to Malek Dance Club to have a fun night together. Malek Dance Club has 2n members and coincidentally Natalia Fan Club also has 2n members. Each member of MDC is assigned a unique id i from 0 to 2n - 1. The same holds for each member of NFC.
One of the parts of this tradition is one by one dance, where each member of MDC dances with a member of NFC. A dance pair is a pair of numbers (a, b) such that member a from MDC dances with member b from NFC.
The complexity of a pairs' assignment is the number of pairs of dancing pairs (a, b) and (c, d) such that a < c and b > d.
You are given a binary number of length n named x. We know that member i from MDC dances with member <image> from NFC. Your task is to calculate the complexity of this assignment modulo 1000000007 (109 + 7).
Expression <image> denotes applying «XOR» to numbers x and y. This operation exists in all modern programming languages, for example, in C++ and Java it denotes as «^», in Pascal — «xor».
Input
The first line of input contains a binary number x of lenght n, (1 ≤ n ≤ 100).
This number may contain leading zeros.
Output
Print the complexity of the given dance assignent modulo 1000000007 (109 + 7).
Examples
Input
11
Output
6
Input
01
Output
2
Input
1
Output
1 | instruction | 0 | 88,723 | 14 | 177,446 |
Tags: combinatorics, math
Correct Solution:
```
p = 1000000007
s = input()
print(int(s,2) * pow(2, len(s) - 1) % p)
``` | output | 1 | 88,723 | 14 | 177,447 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As a tradition, every year before IOI all the members of Natalia Fan Club are invited to Malek Dance Club to have a fun night together. Malek Dance Club has 2n members and coincidentally Natalia Fan Club also has 2n members. Each member of MDC is assigned a unique id i from 0 to 2n - 1. The same holds for each member of NFC.
One of the parts of this tradition is one by one dance, where each member of MDC dances with a member of NFC. A dance pair is a pair of numbers (a, b) such that member a from MDC dances with member b from NFC.
The complexity of a pairs' assignment is the number of pairs of dancing pairs (a, b) and (c, d) such that a < c and b > d.
You are given a binary number of length n named x. We know that member i from MDC dances with member <image> from NFC. Your task is to calculate the complexity of this assignment modulo 1000000007 (109 + 7).
Expression <image> denotes applying «XOR» to numbers x and y. This operation exists in all modern programming languages, for example, in C++ and Java it denotes as «^», in Pascal — «xor».
Input
The first line of input contains a binary number x of lenght n, (1 ≤ n ≤ 100).
This number may contain leading zeros.
Output
Print the complexity of the given dance assignent modulo 1000000007 (109 + 7).
Examples
Input
11
Output
6
Input
01
Output
2
Input
1
Output
1 | instruction | 0 | 88,724 | 14 | 177,448 |
Tags: combinatorics, math
Correct Solution:
```
#t=int(input())
mod = 10**9+7
for _ in range(1):
s=input()
c=1
ans = 0
for i in range(len(s)-1,-1,-1):
ans=(ans+c*int(s[i]))%mod
c=(c*2)%mod
d = pow(2,len(s),mod)
ans=(ans*d)%mod
ans=(ans*pow(2,mod-2,mod))%mod
print(ans)
``` | output | 1 | 88,724 | 14 | 177,449 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As a tradition, every year before IOI all the members of Natalia Fan Club are invited to Malek Dance Club to have a fun night together. Malek Dance Club has 2n members and coincidentally Natalia Fan Club also has 2n members. Each member of MDC is assigned a unique id i from 0 to 2n - 1. The same holds for each member of NFC.
One of the parts of this tradition is one by one dance, where each member of MDC dances with a member of NFC. A dance pair is a pair of numbers (a, b) such that member a from MDC dances with member b from NFC.
The complexity of a pairs' assignment is the number of pairs of dancing pairs (a, b) and (c, d) such that a < c and b > d.
You are given a binary number of length n named x. We know that member i from MDC dances with member <image> from NFC. Your task is to calculate the complexity of this assignment modulo 1000000007 (109 + 7).
Expression <image> denotes applying «XOR» to numbers x and y. This operation exists in all modern programming languages, for example, in C++ and Java it denotes as «^», in Pascal — «xor».
Input
The first line of input contains a binary number x of lenght n, (1 ≤ n ≤ 100).
This number may contain leading zeros.
Output
Print the complexity of the given dance assignent modulo 1000000007 (109 + 7).
Examples
Input
11
Output
6
Input
01
Output
2
Input
1
Output
1 | instruction | 0 | 88,725 | 14 | 177,450 |
Tags: combinatorics, math
Correct Solution:
```
s = input()
res = pow(2, len(s)-1)*(int(s, 2))
print (res%1000000007)
``` | output | 1 | 88,725 | 14 | 177,451 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As a tradition, every year before IOI all the members of Natalia Fan Club are invited to Malek Dance Club to have a fun night together. Malek Dance Club has 2n members and coincidentally Natalia Fan Club also has 2n members. Each member of MDC is assigned a unique id i from 0 to 2n - 1. The same holds for each member of NFC.
One of the parts of this tradition is one by one dance, where each member of MDC dances with a member of NFC. A dance pair is a pair of numbers (a, b) such that member a from MDC dances with member b from NFC.
The complexity of a pairs' assignment is the number of pairs of dancing pairs (a, b) and (c, d) such that a < c and b > d.
You are given a binary number of length n named x. We know that member i from MDC dances with member <image> from NFC. Your task is to calculate the complexity of this assignment modulo 1000000007 (109 + 7).
Expression <image> denotes applying «XOR» to numbers x and y. This operation exists in all modern programming languages, for example, in C++ and Java it denotes as «^», in Pascal — «xor».
Input
The first line of input contains a binary number x of lenght n, (1 ≤ n ≤ 100).
This number may contain leading zeros.
Output
Print the complexity of the given dance assignent modulo 1000000007 (109 + 7).
Examples
Input
11
Output
6
Input
01
Output
2
Input
1
Output
1
Submitted Solution:
```
MOD = 1000000007
s = input()
n = len(s)
p = [1]
for i in range(301):
p.append((2*p[-1]) %MOD)
res = 0
m = n
g = 0
for i in range(n):
if s[i] == '1':
res += (p[g] * (p[2*m-2]%MOD)) % MOD
res = res % MOD
g += 1
m -= 1
print(res)
``` | instruction | 0 | 88,726 | 14 | 177,452 |
Yes | output | 1 | 88,726 | 14 | 177,453 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As a tradition, every year before IOI all the members of Natalia Fan Club are invited to Malek Dance Club to have a fun night together. Malek Dance Club has 2n members and coincidentally Natalia Fan Club also has 2n members. Each member of MDC is assigned a unique id i from 0 to 2n - 1. The same holds for each member of NFC.
One of the parts of this tradition is one by one dance, where each member of MDC dances with a member of NFC. A dance pair is a pair of numbers (a, b) such that member a from MDC dances with member b from NFC.
The complexity of a pairs' assignment is the number of pairs of dancing pairs (a, b) and (c, d) such that a < c and b > d.
You are given a binary number of length n named x. We know that member i from MDC dances with member <image> from NFC. Your task is to calculate the complexity of this assignment modulo 1000000007 (109 + 7).
Expression <image> denotes applying «XOR» to numbers x and y. This operation exists in all modern programming languages, for example, in C++ and Java it denotes as «^», in Pascal — «xor».
Input
The first line of input contains a binary number x of lenght n, (1 ≤ n ≤ 100).
This number may contain leading zeros.
Output
Print the complexity of the given dance assignent modulo 1000000007 (109 + 7).
Examples
Input
11
Output
6
Input
01
Output
2
Input
1
Output
1
Submitted Solution:
```
mod = 10 ** 9 + 7
x = input()[::-1]
n = len(x)
ans = 0
for i, t in enumerate(x):
if t == '1':
ans = (ans + pow(2, n - 1 + i)) % mod
print(ans)
``` | instruction | 0 | 88,727 | 14 | 177,454 |
Yes | output | 1 | 88,727 | 14 | 177,455 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As a tradition, every year before IOI all the members of Natalia Fan Club are invited to Malek Dance Club to have a fun night together. Malek Dance Club has 2n members and coincidentally Natalia Fan Club also has 2n members. Each member of MDC is assigned a unique id i from 0 to 2n - 1. The same holds for each member of NFC.
One of the parts of this tradition is one by one dance, where each member of MDC dances with a member of NFC. A dance pair is a pair of numbers (a, b) such that member a from MDC dances with member b from NFC.
The complexity of a pairs' assignment is the number of pairs of dancing pairs (a, b) and (c, d) such that a < c and b > d.
You are given a binary number of length n named x. We know that member i from MDC dances with member <image> from NFC. Your task is to calculate the complexity of this assignment modulo 1000000007 (109 + 7).
Expression <image> denotes applying «XOR» to numbers x and y. This operation exists in all modern programming languages, for example, in C++ and Java it denotes as «^», in Pascal — «xor».
Input
The first line of input contains a binary number x of lenght n, (1 ≤ n ≤ 100).
This number may contain leading zeros.
Output
Print the complexity of the given dance assignent modulo 1000000007 (109 + 7).
Examples
Input
11
Output
6
Input
01
Output
2
Input
1
Output
1
Submitted Solution:
```
x = input()
n = len(x)
mod = 10 ** 9 + 7
ans = 0
for i in range(n):
if x[i] == '1':
pref = pow(2, i, mod)
suf = pow(2, n - i - 1, mod) ** 2
ans += (pref * suf) % mod
print(ans % mod)
``` | instruction | 0 | 88,728 | 14 | 177,456 |
Yes | output | 1 | 88,728 | 14 | 177,457 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As a tradition, every year before IOI all the members of Natalia Fan Club are invited to Malek Dance Club to have a fun night together. Malek Dance Club has 2n members and coincidentally Natalia Fan Club also has 2n members. Each member of MDC is assigned a unique id i from 0 to 2n - 1. The same holds for each member of NFC.
One of the parts of this tradition is one by one dance, where each member of MDC dances with a member of NFC. A dance pair is a pair of numbers (a, b) such that member a from MDC dances with member b from NFC.
The complexity of a pairs' assignment is the number of pairs of dancing pairs (a, b) and (c, d) such that a < c and b > d.
You are given a binary number of length n named x. We know that member i from MDC dances with member <image> from NFC. Your task is to calculate the complexity of this assignment modulo 1000000007 (109 + 7).
Expression <image> denotes applying «XOR» to numbers x and y. This operation exists in all modern programming languages, for example, in C++ and Java it denotes as «^», in Pascal — «xor».
Input
The first line of input contains a binary number x of lenght n, (1 ≤ n ≤ 100).
This number may contain leading zeros.
Output
Print the complexity of the given dance assignent modulo 1000000007 (109 + 7).
Examples
Input
11
Output
6
Input
01
Output
2
Input
1
Output
1
Submitted Solution:
```
M = 10 ** 9 + 7
def solve1(x):
n = len(x)
x = int(x, 2)
ans = 0
for a in range(2 ** n):
for c in range(2 ** n):
b = a ^ x
d = c ^ x
if a < c and b > d:
ans += 1
return ans % M
def solve2(x):
return int(x, 2) * pow(2, (len(x) - 1), M) % M
x = input()
# print(solve1(x))
print(solve2(x))
``` | instruction | 0 | 88,729 | 14 | 177,458 |
Yes | output | 1 | 88,729 | 14 | 177,459 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As a tradition, every year before IOI all the members of Natalia Fan Club are invited to Malek Dance Club to have a fun night together. Malek Dance Club has 2n members and coincidentally Natalia Fan Club also has 2n members. Each member of MDC is assigned a unique id i from 0 to 2n - 1. The same holds for each member of NFC.
One of the parts of this tradition is one by one dance, where each member of MDC dances with a member of NFC. A dance pair is a pair of numbers (a, b) such that member a from MDC dances with member b from NFC.
The complexity of a pairs' assignment is the number of pairs of dancing pairs (a, b) and (c, d) such that a < c and b > d.
You are given a binary number of length n named x. We know that member i from MDC dances with member <image> from NFC. Your task is to calculate the complexity of this assignment modulo 1000000007 (109 + 7).
Expression <image> denotes applying «XOR» to numbers x and y. This operation exists in all modern programming languages, for example, in C++ and Java it denotes as «^», in Pascal — «xor».
Input
The first line of input contains a binary number x of lenght n, (1 ≤ n ≤ 100).
This number may contain leading zeros.
Output
Print the complexity of the given dance assignent modulo 1000000007 (109 + 7).
Examples
Input
11
Output
6
Input
01
Output
2
Input
1
Output
1
Submitted Solution:
```
__inp = input()
len = len(__inp)
x = 0
for i in __inp:
x = x*2 + int(i)
print((len*x) % 1000000007)
``` | instruction | 0 | 88,730 | 14 | 177,460 |
No | output | 1 | 88,730 | 14 | 177,461 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As a tradition, every year before IOI all the members of Natalia Fan Club are invited to Malek Dance Club to have a fun night together. Malek Dance Club has 2n members and coincidentally Natalia Fan Club also has 2n members. Each member of MDC is assigned a unique id i from 0 to 2n - 1. The same holds for each member of NFC.
One of the parts of this tradition is one by one dance, where each member of MDC dances with a member of NFC. A dance pair is a pair of numbers (a, b) such that member a from MDC dances with member b from NFC.
The complexity of a pairs' assignment is the number of pairs of dancing pairs (a, b) and (c, d) such that a < c and b > d.
You are given a binary number of length n named x. We know that member i from MDC dances with member <image> from NFC. Your task is to calculate the complexity of this assignment modulo 1000000007 (109 + 7).
Expression <image> denotes applying «XOR» to numbers x and y. This operation exists in all modern programming languages, for example, in C++ and Java it denotes as «^», in Pascal — «xor».
Input
The first line of input contains a binary number x of lenght n, (1 ≤ n ≤ 100).
This number may contain leading zeros.
Output
Print the complexity of the given dance assignent modulo 1000000007 (109 + 7).
Examples
Input
11
Output
6
Input
01
Output
2
Input
1
Output
1
Submitted Solution:
```
x =input()
print(int(x,2) << (len(x) - 1))
``` | instruction | 0 | 88,731 | 14 | 177,462 |
No | output | 1 | 88,731 | 14 | 177,463 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As a tradition, every year before IOI all the members of Natalia Fan Club are invited to Malek Dance Club to have a fun night together. Malek Dance Club has 2n members and coincidentally Natalia Fan Club also has 2n members. Each member of MDC is assigned a unique id i from 0 to 2n - 1. The same holds for each member of NFC.
One of the parts of this tradition is one by one dance, where each member of MDC dances with a member of NFC. A dance pair is a pair of numbers (a, b) such that member a from MDC dances with member b from NFC.
The complexity of a pairs' assignment is the number of pairs of dancing pairs (a, b) and (c, d) such that a < c and b > d.
You are given a binary number of length n named x. We know that member i from MDC dances with member <image> from NFC. Your task is to calculate the complexity of this assignment modulo 1000000007 (109 + 7).
Expression <image> denotes applying «XOR» to numbers x and y. This operation exists in all modern programming languages, for example, in C++ and Java it denotes as «^», in Pascal — «xor».
Input
The first line of input contains a binary number x of lenght n, (1 ≤ n ≤ 100).
This number may contain leading zeros.
Output
Print the complexity of the given dance assignent modulo 1000000007 (109 + 7).
Examples
Input
11
Output
6
Input
01
Output
2
Input
1
Output
1
Submitted Solution:
```
MOD = int(10e9 + 7)
x = input()[::-1]
n = len(x)
res = 0
for i, t in enumerate(x):
if t == '1':
res = (res + (1 << (n - 1 + i))) % MOD
print(res)
``` | instruction | 0 | 88,732 | 14 | 177,464 |
No | output | 1 | 88,732 | 14 | 177,465 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As a tradition, every year before IOI all the members of Natalia Fan Club are invited to Malek Dance Club to have a fun night together. Malek Dance Club has 2n members and coincidentally Natalia Fan Club also has 2n members. Each member of MDC is assigned a unique id i from 0 to 2n - 1. The same holds for each member of NFC.
One of the parts of this tradition is one by one dance, where each member of MDC dances with a member of NFC. A dance pair is a pair of numbers (a, b) such that member a from MDC dances with member b from NFC.
The complexity of a pairs' assignment is the number of pairs of dancing pairs (a, b) and (c, d) such that a < c and b > d.
You are given a binary number of length n named x. We know that member i from MDC dances with member <image> from NFC. Your task is to calculate the complexity of this assignment modulo 1000000007 (109 + 7).
Expression <image> denotes applying «XOR» to numbers x and y. This operation exists in all modern programming languages, for example, in C++ and Java it denotes as «^», in Pascal — «xor».
Input
The first line of input contains a binary number x of lenght n, (1 ≤ n ≤ 100).
This number may contain leading zeros.
Output
Print the complexity of the given dance assignent modulo 1000000007 (109 + 7).
Examples
Input
11
Output
6
Input
01
Output
2
Input
1
Output
1
Submitted Solution:
```
a=input()
print((2**a.count('1')-1)*2**(len(a)-1))
``` | instruction | 0 | 88,733 | 14 | 177,466 |
No | output | 1 | 88,733 | 14 | 177,467 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Amr is a young coder who likes music a lot. He always wanted to learn how to play music but he was busy coding so he got an idea.
Amr has n instruments, it takes ai days to learn i-th instrument. Being busy, Amr dedicated k days to learn how to play the maximum possible number of instruments.
Amr asked for your help to distribute his free days between instruments so that he can achieve his goal.
Input
The first line contains two numbers n, k (1 ≤ n ≤ 100, 0 ≤ k ≤ 10 000), the number of instruments and number of days respectively.
The second line contains n integers ai (1 ≤ ai ≤ 100), representing number of days required to learn the i-th instrument.
Output
In the first line output one integer m representing the maximum number of instruments Amr can learn.
In the second line output m space-separated integers: the indices of instruments to be learnt. You may output indices in any order.
if there are multiple optimal solutions output any. It is not necessary to use all days for studying.
Examples
Input
4 10
4 3 1 2
Output
4
1 2 3 4
Input
5 6
4 3 1 1 2
Output
3
1 3 4
Input
1 3
4
Output
0
Note
In the first test Amr can learn all 4 instruments.
In the second test other possible solutions are: {2, 3, 5} or {3, 4, 5}.
In the third test Amr doesn't have enough time to learn the only presented instrument. | instruction | 0 | 88,817 | 14 | 177,634 |
Tags: greedy, implementation, sortings
Correct Solution:
```
n, k = map(int, input().split())
arr = list(map(int, input().split()))
arr = [(arr[i], i) for i in range(n)]
arr.sort()
t = 0
res = []
for i in arr:
res.append(i)
if(sum([i[0] for i in res])>k):
res = res[:-1]
break
if(len(res)==0):
print('0')
else:
print(len(res))
print(' '.join([str(i[1]+1) for i in res]))
``` | output | 1 | 88,817 | 14 | 177,635 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Amr is a young coder who likes music a lot. He always wanted to learn how to play music but he was busy coding so he got an idea.
Amr has n instruments, it takes ai days to learn i-th instrument. Being busy, Amr dedicated k days to learn how to play the maximum possible number of instruments.
Amr asked for your help to distribute his free days between instruments so that he can achieve his goal.
Input
The first line contains two numbers n, k (1 ≤ n ≤ 100, 0 ≤ k ≤ 10 000), the number of instruments and number of days respectively.
The second line contains n integers ai (1 ≤ ai ≤ 100), representing number of days required to learn the i-th instrument.
Output
In the first line output one integer m representing the maximum number of instruments Amr can learn.
In the second line output m space-separated integers: the indices of instruments to be learnt. You may output indices in any order.
if there are multiple optimal solutions output any. It is not necessary to use all days for studying.
Examples
Input
4 10
4 3 1 2
Output
4
1 2 3 4
Input
5 6
4 3 1 1 2
Output
3
1 3 4
Input
1 3
4
Output
0
Note
In the first test Amr can learn all 4 instruments.
In the second test other possible solutions are: {2, 3, 5} or {3, 4, 5}.
In the third test Amr doesn't have enough time to learn the only presented instrument. | instruction | 0 | 88,818 | 14 | 177,636 |
Tags: greedy, implementation, sortings
Correct Solution:
```
k, n = map(int,input().split())
temp = input().split()
data = []
i = 1
for item in temp:
data.append([int(item),i])
i += 1
data.sort()
cost = 0
answer = []
for item in data:
if (item[0]+cost)<=n:
cost += item[0]
answer.append(item[1])
else:
break
print(len(answer))
if(len(answer) != 0):
for i in range(len(answer)-1):
print(answer[i],end=' ')
print(answer[-1])
``` | output | 1 | 88,818 | 14 | 177,637 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Amr is a young coder who likes music a lot. He always wanted to learn how to play music but he was busy coding so he got an idea.
Amr has n instruments, it takes ai days to learn i-th instrument. Being busy, Amr dedicated k days to learn how to play the maximum possible number of instruments.
Amr asked for your help to distribute his free days between instruments so that he can achieve his goal.
Input
The first line contains two numbers n, k (1 ≤ n ≤ 100, 0 ≤ k ≤ 10 000), the number of instruments and number of days respectively.
The second line contains n integers ai (1 ≤ ai ≤ 100), representing number of days required to learn the i-th instrument.
Output
In the first line output one integer m representing the maximum number of instruments Amr can learn.
In the second line output m space-separated integers: the indices of instruments to be learnt. You may output indices in any order.
if there are multiple optimal solutions output any. It is not necessary to use all days for studying.
Examples
Input
4 10
4 3 1 2
Output
4
1 2 3 4
Input
5 6
4 3 1 1 2
Output
3
1 3 4
Input
1 3
4
Output
0
Note
In the first test Amr can learn all 4 instruments.
In the second test other possible solutions are: {2, 3, 5} or {3, 4, 5}.
In the third test Amr doesn't have enough time to learn the only presented instrument. | instruction | 0 | 88,819 | 14 | 177,638 |
Tags: greedy, implementation, sortings
Correct Solution:
```
n,k=map(int,input().split())
l=list(map(int,input().split()))
p=sorted(l)
s=0
a=[]
c=0
for i in range(n):
if(p[i]+s)<=k:
s=s+p[i]
c=c+1
z=l.index(p[i])
a.append(z+1)
l[z]=0
print(c)
print(*a)
``` | output | 1 | 88,819 | 14 | 177,639 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Amr is a young coder who likes music a lot. He always wanted to learn how to play music but he was busy coding so he got an idea.
Amr has n instruments, it takes ai days to learn i-th instrument. Being busy, Amr dedicated k days to learn how to play the maximum possible number of instruments.
Amr asked for your help to distribute his free days between instruments so that he can achieve his goal.
Input
The first line contains two numbers n, k (1 ≤ n ≤ 100, 0 ≤ k ≤ 10 000), the number of instruments and number of days respectively.
The second line contains n integers ai (1 ≤ ai ≤ 100), representing number of days required to learn the i-th instrument.
Output
In the first line output one integer m representing the maximum number of instruments Amr can learn.
In the second line output m space-separated integers: the indices of instruments to be learnt. You may output indices in any order.
if there are multiple optimal solutions output any. It is not necessary to use all days for studying.
Examples
Input
4 10
4 3 1 2
Output
4
1 2 3 4
Input
5 6
4 3 1 1 2
Output
3
1 3 4
Input
1 3
4
Output
0
Note
In the first test Amr can learn all 4 instruments.
In the second test other possible solutions are: {2, 3, 5} or {3, 4, 5}.
In the third test Amr doesn't have enough time to learn the only presented instrument. | instruction | 0 | 88,820 | 14 | 177,640 |
Tags: greedy, implementation, sortings
Correct Solution:
```
n, k = map(int, input().split())
a = [(i, e) for i, e in enumerate(map(int, input().split()))]
a.sort(key=lambda x: x[1])
s, d = 0, []
for i, e in a:
if s + e <= k:
s += e
d.append(i + 1)
else:
break
print(len(d))
for i in d:
print(i, end=' ')
``` | output | 1 | 88,820 | 14 | 177,641 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Amr is a young coder who likes music a lot. He always wanted to learn how to play music but he was busy coding so he got an idea.
Amr has n instruments, it takes ai days to learn i-th instrument. Being busy, Amr dedicated k days to learn how to play the maximum possible number of instruments.
Amr asked for your help to distribute his free days between instruments so that he can achieve his goal.
Input
The first line contains two numbers n, k (1 ≤ n ≤ 100, 0 ≤ k ≤ 10 000), the number of instruments and number of days respectively.
The second line contains n integers ai (1 ≤ ai ≤ 100), representing number of days required to learn the i-th instrument.
Output
In the first line output one integer m representing the maximum number of instruments Amr can learn.
In the second line output m space-separated integers: the indices of instruments to be learnt. You may output indices in any order.
if there are multiple optimal solutions output any. It is not necessary to use all days for studying.
Examples
Input
4 10
4 3 1 2
Output
4
1 2 3 4
Input
5 6
4 3 1 1 2
Output
3
1 3 4
Input
1 3
4
Output
0
Note
In the first test Amr can learn all 4 instruments.
In the second test other possible solutions are: {2, 3, 5} or {3, 4, 5}.
In the third test Amr doesn't have enough time to learn the only presented instrument. | instruction | 0 | 88,821 | 14 | 177,642 |
Tags: greedy, implementation, sortings
Correct Solution:
```
n,k=map(int,input().split())
l=sorted([(x,i) for i,x in enumerate(map(int,input().split()),1)])
s=[]
for i in range(n):
if l[i][0]>k: break
s+=[l[i][1]]
k-=l[i][0]
print(len(s))
print(' '.join(map(str,s)))
``` | output | 1 | 88,821 | 14 | 177,643 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Amr is a young coder who likes music a lot. He always wanted to learn how to play music but he was busy coding so he got an idea.
Amr has n instruments, it takes ai days to learn i-th instrument. Being busy, Amr dedicated k days to learn how to play the maximum possible number of instruments.
Amr asked for your help to distribute his free days between instruments so that he can achieve his goal.
Input
The first line contains two numbers n, k (1 ≤ n ≤ 100, 0 ≤ k ≤ 10 000), the number of instruments and number of days respectively.
The second line contains n integers ai (1 ≤ ai ≤ 100), representing number of days required to learn the i-th instrument.
Output
In the first line output one integer m representing the maximum number of instruments Amr can learn.
In the second line output m space-separated integers: the indices of instruments to be learnt. You may output indices in any order.
if there are multiple optimal solutions output any. It is not necessary to use all days for studying.
Examples
Input
4 10
4 3 1 2
Output
4
1 2 3 4
Input
5 6
4 3 1 1 2
Output
3
1 3 4
Input
1 3
4
Output
0
Note
In the first test Amr can learn all 4 instruments.
In the second test other possible solutions are: {2, 3, 5} or {3, 4, 5}.
In the third test Amr doesn't have enough time to learn the only presented instrument. | instruction | 0 | 88,822 | 14 | 177,644 |
Tags: greedy, implementation, sortings
Correct Solution:
```
n,k = map(int, input().split())
a = input().split()
def ke(n):
return n[1]
for i in range(n):
a[i] = [i,int(a[i])]
a.sort(key=ke)
s = 0
i = 0
answ = 0
answ_v=[]
while True:
if i == n:
break
if s+a[i][1]>k:
break
s += a[i][1]
answ+=1
answ_v.append(a[i][0]+1)
i+=1
print(answ)
for i in answ_v:
print(i,end=" ")
``` | output | 1 | 88,822 | 14 | 177,645 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Amr is a young coder who likes music a lot. He always wanted to learn how to play music but he was busy coding so he got an idea.
Amr has n instruments, it takes ai days to learn i-th instrument. Being busy, Amr dedicated k days to learn how to play the maximum possible number of instruments.
Amr asked for your help to distribute his free days between instruments so that he can achieve his goal.
Input
The first line contains two numbers n, k (1 ≤ n ≤ 100, 0 ≤ k ≤ 10 000), the number of instruments and number of days respectively.
The second line contains n integers ai (1 ≤ ai ≤ 100), representing number of days required to learn the i-th instrument.
Output
In the first line output one integer m representing the maximum number of instruments Amr can learn.
In the second line output m space-separated integers: the indices of instruments to be learnt. You may output indices in any order.
if there are multiple optimal solutions output any. It is not necessary to use all days for studying.
Examples
Input
4 10
4 3 1 2
Output
4
1 2 3 4
Input
5 6
4 3 1 1 2
Output
3
1 3 4
Input
1 3
4
Output
0
Note
In the first test Amr can learn all 4 instruments.
In the second test other possible solutions are: {2, 3, 5} or {3, 4, 5}.
In the third test Amr doesn't have enough time to learn the only presented instrument. | instruction | 0 | 88,823 | 14 | 177,646 |
Tags: greedy, implementation, sortings
Correct Solution:
```
n,k=map(int,input().split())
a=list(map(int,input().split()))
a=[(a[i],i+1) for i in range(len(a))]
#print(a)
l=0
a.sort()
for i in a:
k-=i[0]
if(k<0):
break
l+=1
if(l==0):
print(l)
else:
print(l)
for i in a[:l]:
print(i[1],end=" ")
``` | output | 1 | 88,823 | 14 | 177,647 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Amr is a young coder who likes music a lot. He always wanted to learn how to play music but he was busy coding so he got an idea.
Amr has n instruments, it takes ai days to learn i-th instrument. Being busy, Amr dedicated k days to learn how to play the maximum possible number of instruments.
Amr asked for your help to distribute his free days between instruments so that he can achieve his goal.
Input
The first line contains two numbers n, k (1 ≤ n ≤ 100, 0 ≤ k ≤ 10 000), the number of instruments and number of days respectively.
The second line contains n integers ai (1 ≤ ai ≤ 100), representing number of days required to learn the i-th instrument.
Output
In the first line output one integer m representing the maximum number of instruments Amr can learn.
In the second line output m space-separated integers: the indices of instruments to be learnt. You may output indices in any order.
if there are multiple optimal solutions output any. It is not necessary to use all days for studying.
Examples
Input
4 10
4 3 1 2
Output
4
1 2 3 4
Input
5 6
4 3 1 1 2
Output
3
1 3 4
Input
1 3
4
Output
0
Note
In the first test Amr can learn all 4 instruments.
In the second test other possible solutions are: {2, 3, 5} or {3, 4, 5}.
In the third test Amr doesn't have enough time to learn the only presented instrument. | instruction | 0 | 88,824 | 14 | 177,648 |
Tags: greedy, implementation, sortings
Correct Solution:
```
a,k = map(int,input().split())
ls = list(map(int,input().split()))
nls = []
for i in ls:
nls.append(i)
ls.sort()
ind = []
ins = 0
for i in range(a):
if ls[i] <= k and ins <= k:
if ins + ls[i]>k:
break
else:
ins += ls[i]
ind.append(nls.index(ls[i])+1)
nls[nls.index(ls[i])] = "l"
print(len(ind))
for i in sorted(ind):
print(i,end = " ")
``` | output | 1 | 88,824 | 14 | 177,649 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N people, conveniently numbered 1 through N. They were standing in a row yesterday, but now they are unsure of the order in which they were standing. However, each person remembered the following fact: the absolute difference of the number of the people who were standing to the left of that person, and the number of the people who were standing to the right of that person. According to their reports, the difference above for person i is A_i.
Based on these reports, find the number of the possible orders in which they were standing. Since it can be extremely large, print the answer modulo 10^9+7. Note that the reports may be incorrect and thus there may be no consistent order. In such a case, print 0.
Constraints
* 1≦N≦10^5
* 0≦A_i≦N-1
Input
The input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the number of the possible orders in which they were standing, modulo 10^9+7.
Examples
Input
5
2 4 4 0 2
Output
4
Input
7
6 4 0 2 4 0 2
Output
0
Input
8
7 5 1 1 7 3 5 3
Output
16
Submitted Solution:
```
import sys, os, math, bisect, itertools, collections, heapq, queue
# from scipy.sparse.csgraph import csgraph_from_dense, floyd_warshall
from decimal import Decimal
from collections import defaultdict, deque
sys.setrecursionlimit(10000000)
ii = lambda: int(sys.stdin.buffer.readline().rstrip())
il = lambda: list(map(int, sys.stdin.buffer.readline().split()))
fl = lambda: list(map(float, sys.stdin.buffer.readline().split()))
iln = lambda n: [int(sys.stdin.buffer.readline().rstrip()) for _ in range(n)]
iss = lambda: sys.stdin.buffer.readline().decode().rstrip()
sl = lambda: list(map(str, sys.stdin.buffer.readline().decode().split()))
isn = lambda n: [sys.stdin.buffer.readline().decode().rstrip() for _ in range(n)]
lcm = lambda x, y: (x * y) // math.gcd(x, y)
MOD = 10 ** 9 + 7
MAX = float('inf')
def main():
if os.getenv("LOCAL"):
sys.stdin = open("input.txt", "r")
N = ii()
A = il()
cnt = collections.Counter(A)
start = 1 if N % 2 == 0 else 0
m = 0
for n in range(start, N, 2):
if n == 0 and cnt[n] == 1:
continue
elif cnt[n] == 2:
m += 1
continue
else:
print(0)
exit()
print((m**2)%MOD)
if __name__ == '__main__':
main()
``` | instruction | 0 | 89,122 | 14 | 178,244 |
No | output | 1 | 89,122 | 14 | 178,245 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Athenaeus has just finished creating his latest musical composition and will present it tomorrow to the people of Athens. Unfortunately, the melody is rather dull and highly likely won't be met with a warm reception.
His song consists of n notes, which we will treat as positive integers. The diversity of a song is the number of different notes it contains. As a patron of music, Euterpe watches over composers and guides them throughout the process of creating new melodies. She decided to help Athenaeus by changing his song to make it more diverse.
Being a minor goddess, she cannot arbitrarily change the song. Instead, for each of the n notes in the song, she can either leave it as it is or increase it by 1.
Given the song as a sequence of integers describing the notes, find out the maximal, achievable diversity.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 10 000) — the number of test cases. Then t test cases follow, each one is described in two lines.
In the first line of each test case there is a single integer n (1 ≤ n ≤ 10^5) denoting the length of the song. The next line contains a sequence of n integers x_1, x_2, …, x_n (1 ≤ x_1 ≤ x_2 ≤ … ≤ x_n ≤ 2 ⋅ n), describing the song.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case, you should output a single line containing precisely one integer, the maximal diversity of the song, i.e. the maximal possible number of different elements in the final sequence.
Example
Input
5
6
1 2 2 2 5 6
2
4 4
6
1 1 3 4 4 5
1
1
6
1 1 1 2 2 2
Output
5
2
6
1
3
Note
In the first test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{3}, 2, 2, \underline{6}, \underline{7}, which has 5 different elements (increased elements are underlined).
In the second test case, Euterpe can increase the first element to obtain the sequence \underline{5}, 4, which has 2 different elements.
In the third test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{2}, 3, 4, \underline{5}, \underline{6}, which has 6 different elements. | instruction | 0 | 89,512 | 14 | 179,024 |
Tags: dp, greedy
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
a=list(map(int,input().split()))
for i in range(n-1, -1, -1):
if i == n-1:
a[i] += 1
else:
if a[i] < a[i+1]-1:
a[i] += 1
print(len(set(a)))
``` | output | 1 | 89,512 | 14 | 179,025 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Athenaeus has just finished creating his latest musical composition and will present it tomorrow to the people of Athens. Unfortunately, the melody is rather dull and highly likely won't be met with a warm reception.
His song consists of n notes, which we will treat as positive integers. The diversity of a song is the number of different notes it contains. As a patron of music, Euterpe watches over composers and guides them throughout the process of creating new melodies. She decided to help Athenaeus by changing his song to make it more diverse.
Being a minor goddess, she cannot arbitrarily change the song. Instead, for each of the n notes in the song, she can either leave it as it is or increase it by 1.
Given the song as a sequence of integers describing the notes, find out the maximal, achievable diversity.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 10 000) — the number of test cases. Then t test cases follow, each one is described in two lines.
In the first line of each test case there is a single integer n (1 ≤ n ≤ 10^5) denoting the length of the song. The next line contains a sequence of n integers x_1, x_2, …, x_n (1 ≤ x_1 ≤ x_2 ≤ … ≤ x_n ≤ 2 ⋅ n), describing the song.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case, you should output a single line containing precisely one integer, the maximal diversity of the song, i.e. the maximal possible number of different elements in the final sequence.
Example
Input
5
6
1 2 2 2 5 6
2
4 4
6
1 1 3 4 4 5
1
1
6
1 1 1 2 2 2
Output
5
2
6
1
3
Note
In the first test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{3}, 2, 2, \underline{6}, \underline{7}, which has 5 different elements (increased elements are underlined).
In the second test case, Euterpe can increase the first element to obtain the sequence \underline{5}, 4, which has 2 different elements.
In the third test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{2}, 3, 4, \underline{5}, \underline{6}, which has 6 different elements. | instruction | 0 | 89,513 | 14 | 179,026 |
Tags: dp, greedy
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
l=list(map(int,input().split()))
a=[0]*(2*n+1)
for i in l:
a[i]+=1
c=0
for i in range(1,2*n):
#print(a)
if a[i]>1 and a[i+1]==0:
a[i+1]=1
elif a[i]>1:
a[i+1]+=1
for i in a:
if i:
c+=1
if a[-1]>1:
c+=1
print(c)
``` | output | 1 | 89,513 | 14 | 179,027 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Athenaeus has just finished creating his latest musical composition and will present it tomorrow to the people of Athens. Unfortunately, the melody is rather dull and highly likely won't be met with a warm reception.
His song consists of n notes, which we will treat as positive integers. The diversity of a song is the number of different notes it contains. As a patron of music, Euterpe watches over composers and guides them throughout the process of creating new melodies. She decided to help Athenaeus by changing his song to make it more diverse.
Being a minor goddess, she cannot arbitrarily change the song. Instead, for each of the n notes in the song, she can either leave it as it is or increase it by 1.
Given the song as a sequence of integers describing the notes, find out the maximal, achievable diversity.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 10 000) — the number of test cases. Then t test cases follow, each one is described in two lines.
In the first line of each test case there is a single integer n (1 ≤ n ≤ 10^5) denoting the length of the song. The next line contains a sequence of n integers x_1, x_2, …, x_n (1 ≤ x_1 ≤ x_2 ≤ … ≤ x_n ≤ 2 ⋅ n), describing the song.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case, you should output a single line containing precisely one integer, the maximal diversity of the song, i.e. the maximal possible number of different elements in the final sequence.
Example
Input
5
6
1 2 2 2 5 6
2
4 4
6
1 1 3 4 4 5
1
1
6
1 1 1 2 2 2
Output
5
2
6
1
3
Note
In the first test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{3}, 2, 2, \underline{6}, \underline{7}, which has 5 different elements (increased elements are underlined).
In the second test case, Euterpe can increase the first element to obtain the sequence \underline{5}, 4, which has 2 different elements.
In the third test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{2}, 3, 4, \underline{5}, \underline{6}, which has 6 different elements. | instruction | 0 | 89,514 | 14 | 179,028 |
Tags: dp, greedy
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
note = list(map(int, input().rstrip().split()))
if n == 1:
print('1')
else:
if note[1] == note[0]:
note[1] += 1
for i in range(2,n):
if note[i] == note[i-1] or note[i] == note[i-2]:
note[i] += 1
print(len(set(note)))
``` | output | 1 | 89,514 | 14 | 179,029 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Athenaeus has just finished creating his latest musical composition and will present it tomorrow to the people of Athens. Unfortunately, the melody is rather dull and highly likely won't be met with a warm reception.
His song consists of n notes, which we will treat as positive integers. The diversity of a song is the number of different notes it contains. As a patron of music, Euterpe watches over composers and guides them throughout the process of creating new melodies. She decided to help Athenaeus by changing his song to make it more diverse.
Being a minor goddess, she cannot arbitrarily change the song. Instead, for each of the n notes in the song, she can either leave it as it is or increase it by 1.
Given the song as a sequence of integers describing the notes, find out the maximal, achievable diversity.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 10 000) — the number of test cases. Then t test cases follow, each one is described in two lines.
In the first line of each test case there is a single integer n (1 ≤ n ≤ 10^5) denoting the length of the song. The next line contains a sequence of n integers x_1, x_2, …, x_n (1 ≤ x_1 ≤ x_2 ≤ … ≤ x_n ≤ 2 ⋅ n), describing the song.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case, you should output a single line containing precisely one integer, the maximal diversity of the song, i.e. the maximal possible number of different elements in the final sequence.
Example
Input
5
6
1 2 2 2 5 6
2
4 4
6
1 1 3 4 4 5
1
1
6
1 1 1 2 2 2
Output
5
2
6
1
3
Note
In the first test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{3}, 2, 2, \underline{6}, \underline{7}, which has 5 different elements (increased elements are underlined).
In the second test case, Euterpe can increase the first element to obtain the sequence \underline{5}, 4, which has 2 different elements.
In the third test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{2}, 3, 4, \underline{5}, \underline{6}, which has 6 different elements. | instruction | 0 | 89,515 | 14 | 179,030 |
Tags: dp, greedy
Correct Solution:
```
# This is a sample Python script.
# Press ⌃R to execute it or replace it with your code.
# Press Double ⇧ to search everywhere for classes, files, tool windows, actions, and settings.
# Press the green button in the gutter to run the script.
import sys
if __name__ == '__main__':
# sys.stdin = open('/input.txt', 'r')
tc = int(input())
for _ in range(tc):
e = int(input())
rn = ans = 0
for n in list(map(int, input().split())):
if rn <= n:
rn = (rn + 1 if rn == n else n)
ans = ans + 1
print(ans)
``` | output | 1 | 89,515 | 14 | 179,031 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Athenaeus has just finished creating his latest musical composition and will present it tomorrow to the people of Athens. Unfortunately, the melody is rather dull and highly likely won't be met with a warm reception.
His song consists of n notes, which we will treat as positive integers. The diversity of a song is the number of different notes it contains. As a patron of music, Euterpe watches over composers and guides them throughout the process of creating new melodies. She decided to help Athenaeus by changing his song to make it more diverse.
Being a minor goddess, she cannot arbitrarily change the song. Instead, for each of the n notes in the song, she can either leave it as it is or increase it by 1.
Given the song as a sequence of integers describing the notes, find out the maximal, achievable diversity.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 10 000) — the number of test cases. Then t test cases follow, each one is described in two lines.
In the first line of each test case there is a single integer n (1 ≤ n ≤ 10^5) denoting the length of the song. The next line contains a sequence of n integers x_1, x_2, …, x_n (1 ≤ x_1 ≤ x_2 ≤ … ≤ x_n ≤ 2 ⋅ n), describing the song.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case, you should output a single line containing precisely one integer, the maximal diversity of the song, i.e. the maximal possible number of different elements in the final sequence.
Example
Input
5
6
1 2 2 2 5 6
2
4 4
6
1 1 3 4 4 5
1
1
6
1 1 1 2 2 2
Output
5
2
6
1
3
Note
In the first test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{3}, 2, 2, \underline{6}, \underline{7}, which has 5 different elements (increased elements are underlined).
In the second test case, Euterpe can increase the first element to obtain the sequence \underline{5}, 4, which has 2 different elements.
In the third test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{2}, 3, 4, \underline{5}, \underline{6}, which has 6 different elements. | instruction | 0 | 89,516 | 14 | 179,032 |
Tags: dp, greedy
Correct Solution:
```
from collections import Counter
for _ in range(int(input())):
n=int(input())
l=list(map(int,input().split()))
d=Counter(l)
c=0
for i in range(2*n+1,0,-1):
if(d[i]>0):
c+=1
for j in range(2*n+1,0,-1):
if(d[j]==1 and d[j+1]==0):
d[j+1]=1
d[j]-=1
elif(d[j]>1 and d[j+1]==0):
c+=1
d[j+1]=1
d[j]-=1
print(c)
``` | output | 1 | 89,516 | 14 | 179,033 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Athenaeus has just finished creating his latest musical composition and will present it tomorrow to the people of Athens. Unfortunately, the melody is rather dull and highly likely won't be met with a warm reception.
His song consists of n notes, which we will treat as positive integers. The diversity of a song is the number of different notes it contains. As a patron of music, Euterpe watches over composers and guides them throughout the process of creating new melodies. She decided to help Athenaeus by changing his song to make it more diverse.
Being a minor goddess, she cannot arbitrarily change the song. Instead, for each of the n notes in the song, she can either leave it as it is or increase it by 1.
Given the song as a sequence of integers describing the notes, find out the maximal, achievable diversity.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 10 000) — the number of test cases. Then t test cases follow, each one is described in two lines.
In the first line of each test case there is a single integer n (1 ≤ n ≤ 10^5) denoting the length of the song. The next line contains a sequence of n integers x_1, x_2, …, x_n (1 ≤ x_1 ≤ x_2 ≤ … ≤ x_n ≤ 2 ⋅ n), describing the song.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case, you should output a single line containing precisely one integer, the maximal diversity of the song, i.e. the maximal possible number of different elements in the final sequence.
Example
Input
5
6
1 2 2 2 5 6
2
4 4
6
1 1 3 4 4 5
1
1
6
1 1 1 2 2 2
Output
5
2
6
1
3
Note
In the first test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{3}, 2, 2, \underline{6}, \underline{7}, which has 5 different elements (increased elements are underlined).
In the second test case, Euterpe can increase the first element to obtain the sequence \underline{5}, 4, which has 2 different elements.
In the third test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{2}, 3, 4, \underline{5}, \underline{6}, which has 6 different elements. | instruction | 0 | 89,517 | 14 | 179,034 |
Tags: dp, greedy
Correct Solution:
```
n = int(input())
for _ in range(n):
m = int(input())
arr = list(map(int, input().split()))
dct = {}
so = 0
tong = 0
for i in arr:
if i == so:
tong += 1
so += 1
elif i > so:
tong += 1
so = i
print(tong)
``` | output | 1 | 89,517 | 14 | 179,035 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Athenaeus has just finished creating his latest musical composition and will present it tomorrow to the people of Athens. Unfortunately, the melody is rather dull and highly likely won't be met with a warm reception.
His song consists of n notes, which we will treat as positive integers. The diversity of a song is the number of different notes it contains. As a patron of music, Euterpe watches over composers and guides them throughout the process of creating new melodies. She decided to help Athenaeus by changing his song to make it more diverse.
Being a minor goddess, she cannot arbitrarily change the song. Instead, for each of the n notes in the song, she can either leave it as it is or increase it by 1.
Given the song as a sequence of integers describing the notes, find out the maximal, achievable diversity.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 10 000) — the number of test cases. Then t test cases follow, each one is described in two lines.
In the first line of each test case there is a single integer n (1 ≤ n ≤ 10^5) denoting the length of the song. The next line contains a sequence of n integers x_1, x_2, …, x_n (1 ≤ x_1 ≤ x_2 ≤ … ≤ x_n ≤ 2 ⋅ n), describing the song.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case, you should output a single line containing precisely one integer, the maximal diversity of the song, i.e. the maximal possible number of different elements in the final sequence.
Example
Input
5
6
1 2 2 2 5 6
2
4 4
6
1 1 3 4 4 5
1
1
6
1 1 1 2 2 2
Output
5
2
6
1
3
Note
In the first test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{3}, 2, 2, \underline{6}, \underline{7}, which has 5 different elements (increased elements are underlined).
In the second test case, Euterpe can increase the first element to obtain the sequence \underline{5}, 4, which has 2 different elements.
In the third test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{2}, 3, 4, \underline{5}, \underline{6}, which has 6 different elements. | instruction | 0 | 89,518 | 14 | 179,036 |
Tags: dp, greedy
Correct Solution:
```
from sys import stdin
def function(list):
list.sort()
for i in range(len(list)-1):
if list[i]==list[i-1] and list[i]!=list[i+1]:
list[i]+=1
if len(list)>1:
if list[-1]==list[-2]:
list[-1]+=1
return len(set(list))
n=int(input())
for i in range(n):
stdin.readline()
print(function([int(x) for x in stdin.readline().split()]))
``` | output | 1 | 89,518 | 14 | 179,037 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Athenaeus has just finished creating his latest musical composition and will present it tomorrow to the people of Athens. Unfortunately, the melody is rather dull and highly likely won't be met with a warm reception.
His song consists of n notes, which we will treat as positive integers. The diversity of a song is the number of different notes it contains. As a patron of music, Euterpe watches over composers and guides them throughout the process of creating new melodies. She decided to help Athenaeus by changing his song to make it more diverse.
Being a minor goddess, she cannot arbitrarily change the song. Instead, for each of the n notes in the song, she can either leave it as it is or increase it by 1.
Given the song as a sequence of integers describing the notes, find out the maximal, achievable diversity.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 10 000) — the number of test cases. Then t test cases follow, each one is described in two lines.
In the first line of each test case there is a single integer n (1 ≤ n ≤ 10^5) denoting the length of the song. The next line contains a sequence of n integers x_1, x_2, …, x_n (1 ≤ x_1 ≤ x_2 ≤ … ≤ x_n ≤ 2 ⋅ n), describing the song.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case, you should output a single line containing precisely one integer, the maximal diversity of the song, i.e. the maximal possible number of different elements in the final sequence.
Example
Input
5
6
1 2 2 2 5 6
2
4 4
6
1 1 3 4 4 5
1
1
6
1 1 1 2 2 2
Output
5
2
6
1
3
Note
In the first test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{3}, 2, 2, \underline{6}, \underline{7}, which has 5 different elements (increased elements are underlined).
In the second test case, Euterpe can increase the first element to obtain the sequence \underline{5}, 4, which has 2 different elements.
In the third test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{2}, 3, 4, \underline{5}, \underline{6}, which has 6 different elements. | instruction | 0 | 89,519 | 14 | 179,038 |
Tags: dp, greedy
Correct Solution:
```
t=int(input())
for tcs in range(t):
n=int(input())
l=list(map(int,input().split()))
c=0
if(n<3):
print(n)
else:
for i in range(1,n-1):
if(l[i]>l[i-1] and l[i]<l[i+1]):
c+=1
else:
if(l[i]<l[i+1]):
l[i]+=1
if(l[n-2]==l[n-1]):
l[n-1]+=1
print(len(set(l)))
``` | output | 1 | 89,519 | 14 | 179,039 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Athenaeus has just finished creating his latest musical composition and will present it tomorrow to the people of Athens. Unfortunately, the melody is rather dull and highly likely won't be met with a warm reception.
His song consists of n notes, which we will treat as positive integers. The diversity of a song is the number of different notes it contains. As a patron of music, Euterpe watches over composers and guides them throughout the process of creating new melodies. She decided to help Athenaeus by changing his song to make it more diverse.
Being a minor goddess, she cannot arbitrarily change the song. Instead, for each of the n notes in the song, she can either leave it as it is or increase it by 1.
Given the song as a sequence of integers describing the notes, find out the maximal, achievable diversity.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 10 000) — the number of test cases. Then t test cases follow, each one is described in two lines.
In the first line of each test case there is a single integer n (1 ≤ n ≤ 10^5) denoting the length of the song. The next line contains a sequence of n integers x_1, x_2, …, x_n (1 ≤ x_1 ≤ x_2 ≤ … ≤ x_n ≤ 2 ⋅ n), describing the song.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case, you should output a single line containing precisely one integer, the maximal diversity of the song, i.e. the maximal possible number of different elements in the final sequence.
Example
Input
5
6
1 2 2 2 5 6
2
4 4
6
1 1 3 4 4 5
1
1
6
1 1 1 2 2 2
Output
5
2
6
1
3
Note
In the first test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{3}, 2, 2, \underline{6}, \underline{7}, which has 5 different elements (increased elements are underlined).
In the second test case, Euterpe can increase the first element to obtain the sequence \underline{5}, 4, which has 2 different elements.
In the third test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{2}, 3, 4, \underline{5}, \underline{6}, which has 6 different elements.
Submitted Solution:
```
import collections
t=int(input())
while t:
t-=1
n=int(input())
l=list(map(int,input().split()))
s=collections.Counter(l)
c=0
k=list(s.keys())
for i in k:
if s[i]>1:
if i+1 in s:
s[i+1]+=1
else:
c+=1
print(len(s.keys())+c)
``` | instruction | 0 | 89,520 | 14 | 179,040 |
Yes | output | 1 | 89,520 | 14 | 179,041 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Athenaeus has just finished creating his latest musical composition and will present it tomorrow to the people of Athens. Unfortunately, the melody is rather dull and highly likely won't be met with a warm reception.
His song consists of n notes, which we will treat as positive integers. The diversity of a song is the number of different notes it contains. As a patron of music, Euterpe watches over composers and guides them throughout the process of creating new melodies. She decided to help Athenaeus by changing his song to make it more diverse.
Being a minor goddess, she cannot arbitrarily change the song. Instead, for each of the n notes in the song, she can either leave it as it is or increase it by 1.
Given the song as a sequence of integers describing the notes, find out the maximal, achievable diversity.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 10 000) — the number of test cases. Then t test cases follow, each one is described in two lines.
In the first line of each test case there is a single integer n (1 ≤ n ≤ 10^5) denoting the length of the song. The next line contains a sequence of n integers x_1, x_2, …, x_n (1 ≤ x_1 ≤ x_2 ≤ … ≤ x_n ≤ 2 ⋅ n), describing the song.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case, you should output a single line containing precisely one integer, the maximal diversity of the song, i.e. the maximal possible number of different elements in the final sequence.
Example
Input
5
6
1 2 2 2 5 6
2
4 4
6
1 1 3 4 4 5
1
1
6
1 1 1 2 2 2
Output
5
2
6
1
3
Note
In the first test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{3}, 2, 2, \underline{6}, \underline{7}, which has 5 different elements (increased elements are underlined).
In the second test case, Euterpe can increase the first element to obtain the sequence \underline{5}, 4, which has 2 different elements.
In the third test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{2}, 3, 4, \underline{5}, \underline{6}, which has 6 different elements.
Submitted Solution:
```
import sys
#First solution
"""
for _ in range(int(input())):
n = int(input())
s = set()
x = list(map(int, input().split()))
for i in x:
if i in s and i + 1 not in s:
s.add(i + 1)
s.add(i)
print(len(s))
"""
#Second solution
"""
for _ in range(int(input())):
n = int(input())
x = list(map(int, input().split()))
x.sort()
changed = 0
for i in range(n):
if x[i] == x[i - 1]:
x[i] += 1
changed = x[i]
else:
if changed == x[i]:
x[i] += 1
print(len(set(x)))
"""
#Third solution
for _ in range(int(input())):
n = int(input())
x = list(map(int, input().split()))
for i in range(1, n):
if x[i] <= x[i - 1]:
x[i] += 1
print(len(set(x)))
#Fourth solution
"""
for _ in range(int(input())):
n = int(input())
s = set()
for i in map(int, input().split()):
if i in s:
i += 1
s.add(i)
print(len(s))
"""
``` | instruction | 0 | 89,521 | 14 | 179,042 |
Yes | output | 1 | 89,521 | 14 | 179,043 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Athenaeus has just finished creating his latest musical composition and will present it tomorrow to the people of Athens. Unfortunately, the melody is rather dull and highly likely won't be met with a warm reception.
His song consists of n notes, which we will treat as positive integers. The diversity of a song is the number of different notes it contains. As a patron of music, Euterpe watches over composers and guides them throughout the process of creating new melodies. She decided to help Athenaeus by changing his song to make it more diverse.
Being a minor goddess, she cannot arbitrarily change the song. Instead, for each of the n notes in the song, she can either leave it as it is or increase it by 1.
Given the song as a sequence of integers describing the notes, find out the maximal, achievable diversity.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 10 000) — the number of test cases. Then t test cases follow, each one is described in two lines.
In the first line of each test case there is a single integer n (1 ≤ n ≤ 10^5) denoting the length of the song. The next line contains a sequence of n integers x_1, x_2, …, x_n (1 ≤ x_1 ≤ x_2 ≤ … ≤ x_n ≤ 2 ⋅ n), describing the song.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case, you should output a single line containing precisely one integer, the maximal diversity of the song, i.e. the maximal possible number of different elements in the final sequence.
Example
Input
5
6
1 2 2 2 5 6
2
4 4
6
1 1 3 4 4 5
1
1
6
1 1 1 2 2 2
Output
5
2
6
1
3
Note
In the first test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{3}, 2, 2, \underline{6}, \underline{7}, which has 5 different elements (increased elements are underlined).
In the second test case, Euterpe can increase the first element to obtain the sequence \underline{5}, 4, which has 2 different elements.
In the third test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{2}, 3, 4, \underline{5}, \underline{6}, which has 6 different elements.
Submitted Solution:
```
from collections import Counter
from collections import defaultdict
t = int(input())
for _ in range(t):
n = int(input())
arr = [int(i) for i in input().split()]
count = Counter(arr)
keys = list(count.keys())
Dict = defaultdict(int)
for i in keys:
if count[i] > 1:
Dict[i+1] += 1
Dict[i] += 1
else:
if Dict[i] > 0:
Dict[i+1] += 1
else:
Dict[i] += 1
ans = len(list(Dict.keys()))
print(ans)
``` | instruction | 0 | 89,522 | 14 | 179,044 |
Yes | output | 1 | 89,522 | 14 | 179,045 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Athenaeus has just finished creating his latest musical composition and will present it tomorrow to the people of Athens. Unfortunately, the melody is rather dull and highly likely won't be met with a warm reception.
His song consists of n notes, which we will treat as positive integers. The diversity of a song is the number of different notes it contains. As a patron of music, Euterpe watches over composers and guides them throughout the process of creating new melodies. She decided to help Athenaeus by changing his song to make it more diverse.
Being a minor goddess, she cannot arbitrarily change the song. Instead, for each of the n notes in the song, she can either leave it as it is or increase it by 1.
Given the song as a sequence of integers describing the notes, find out the maximal, achievable diversity.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 10 000) — the number of test cases. Then t test cases follow, each one is described in two lines.
In the first line of each test case there is a single integer n (1 ≤ n ≤ 10^5) denoting the length of the song. The next line contains a sequence of n integers x_1, x_2, …, x_n (1 ≤ x_1 ≤ x_2 ≤ … ≤ x_n ≤ 2 ⋅ n), describing the song.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case, you should output a single line containing precisely one integer, the maximal diversity of the song, i.e. the maximal possible number of different elements in the final sequence.
Example
Input
5
6
1 2 2 2 5 6
2
4 4
6
1 1 3 4 4 5
1
1
6
1 1 1 2 2 2
Output
5
2
6
1
3
Note
In the first test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{3}, 2, 2, \underline{6}, \underline{7}, which has 5 different elements (increased elements are underlined).
In the second test case, Euterpe can increase the first element to obtain the sequence \underline{5}, 4, which has 2 different elements.
In the third test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{2}, 3, 4, \underline{5}, \underline{6}, which has 6 different elements.
Submitted Solution:
```
n = int(input())
for _ in range(n):
x = int(input())
dic = {}
lis = list(map(int, input().split()))
for i in lis:
if i in dic:
dic[i] += 1
else:
dic[i] = 1
temp = list(dic.keys())
for d in temp:
if(dic[d]>1):
if d+1 not in dic:
dic[d+1] = 1
else:
dic[d] -= 1
dic[d+1] += 1
print(len(dic))
``` | instruction | 0 | 89,523 | 14 | 179,046 |
Yes | output | 1 | 89,523 | 14 | 179,047 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Athenaeus has just finished creating his latest musical composition and will present it tomorrow to the people of Athens. Unfortunately, the melody is rather dull and highly likely won't be met with a warm reception.
His song consists of n notes, which we will treat as positive integers. The diversity of a song is the number of different notes it contains. As a patron of music, Euterpe watches over composers and guides them throughout the process of creating new melodies. She decided to help Athenaeus by changing his song to make it more diverse.
Being a minor goddess, she cannot arbitrarily change the song. Instead, for each of the n notes in the song, she can either leave it as it is or increase it by 1.
Given the song as a sequence of integers describing the notes, find out the maximal, achievable diversity.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 10 000) — the number of test cases. Then t test cases follow, each one is described in two lines.
In the first line of each test case there is a single integer n (1 ≤ n ≤ 10^5) denoting the length of the song. The next line contains a sequence of n integers x_1, x_2, …, x_n (1 ≤ x_1 ≤ x_2 ≤ … ≤ x_n ≤ 2 ⋅ n), describing the song.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case, you should output a single line containing precisely one integer, the maximal diversity of the song, i.e. the maximal possible number of different elements in the final sequence.
Example
Input
5
6
1 2 2 2 5 6
2
4 4
6
1 1 3 4 4 5
1
1
6
1 1 1 2 2 2
Output
5
2
6
1
3
Note
In the first test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{3}, 2, 2, \underline{6}, \underline{7}, which has 5 different elements (increased elements are underlined).
In the second test case, Euterpe can increase the first element to obtain the sequence \underline{5}, 4, which has 2 different elements.
In the third test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{2}, 3, 4, \underline{5}, \underline{6}, which has 6 different elements.
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
a.sort()
a.reverse()
k=max(a)
for i in range(len(a)):
if a[i]==k:
a[i]=a[i]+1
if(a[i]+1 not in a):
a[i]=a[i]+1
b=list(set(a))
print(len(b))
``` | instruction | 0 | 89,524 | 14 | 179,048 |
No | output | 1 | 89,524 | 14 | 179,049 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Athenaeus has just finished creating his latest musical composition and will present it tomorrow to the people of Athens. Unfortunately, the melody is rather dull and highly likely won't be met with a warm reception.
His song consists of n notes, which we will treat as positive integers. The diversity of a song is the number of different notes it contains. As a patron of music, Euterpe watches over composers and guides them throughout the process of creating new melodies. She decided to help Athenaeus by changing his song to make it more diverse.
Being a minor goddess, she cannot arbitrarily change the song. Instead, for each of the n notes in the song, she can either leave it as it is or increase it by 1.
Given the song as a sequence of integers describing the notes, find out the maximal, achievable diversity.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 10 000) — the number of test cases. Then t test cases follow, each one is described in two lines.
In the first line of each test case there is a single integer n (1 ≤ n ≤ 10^5) denoting the length of the song. The next line contains a sequence of n integers x_1, x_2, …, x_n (1 ≤ x_1 ≤ x_2 ≤ … ≤ x_n ≤ 2 ⋅ n), describing the song.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case, you should output a single line containing precisely one integer, the maximal diversity of the song, i.e. the maximal possible number of different elements in the final sequence.
Example
Input
5
6
1 2 2 2 5 6
2
4 4
6
1 1 3 4 4 5
1
1
6
1 1 1 2 2 2
Output
5
2
6
1
3
Note
In the first test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{3}, 2, 2, \underline{6}, \underline{7}, which has 5 different elements (increased elements are underlined).
In the second test case, Euterpe can increase the first element to obtain the sequence \underline{5}, 4, which has 2 different elements.
In the third test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{2}, 3, 4, \underline{5}, \underline{6}, which has 6 different elements.
Submitted Solution:
```
def solve(n, arr):
for i in range(n-1):
if arr[i] == arr[i+1]:
arr[i+1] += 1
print(len(set(arr)))
t = int(input())
for _ in range(t):
n = int(input())
arr = list(map(int, input().split()))
solve(n, arr)
``` | instruction | 0 | 89,525 | 14 | 179,050 |
No | output | 1 | 89,525 | 14 | 179,051 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Athenaeus has just finished creating his latest musical composition and will present it tomorrow to the people of Athens. Unfortunately, the melody is rather dull and highly likely won't be met with a warm reception.
His song consists of n notes, which we will treat as positive integers. The diversity of a song is the number of different notes it contains. As a patron of music, Euterpe watches over composers and guides them throughout the process of creating new melodies. She decided to help Athenaeus by changing his song to make it more diverse.
Being a minor goddess, she cannot arbitrarily change the song. Instead, for each of the n notes in the song, she can either leave it as it is or increase it by 1.
Given the song as a sequence of integers describing the notes, find out the maximal, achievable diversity.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 10 000) — the number of test cases. Then t test cases follow, each one is described in two lines.
In the first line of each test case there is a single integer n (1 ≤ n ≤ 10^5) denoting the length of the song. The next line contains a sequence of n integers x_1, x_2, …, x_n (1 ≤ x_1 ≤ x_2 ≤ … ≤ x_n ≤ 2 ⋅ n), describing the song.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case, you should output a single line containing precisely one integer, the maximal diversity of the song, i.e. the maximal possible number of different elements in the final sequence.
Example
Input
5
6
1 2 2 2 5 6
2
4 4
6
1 1 3 4 4 5
1
1
6
1 1 1 2 2 2
Output
5
2
6
1
3
Note
In the first test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{3}, 2, 2, \underline{6}, \underline{7}, which has 5 different elements (increased elements are underlined).
In the second test case, Euterpe can increase the first element to obtain the sequence \underline{5}, 4, which has 2 different elements.
In the third test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{2}, 3, 4, \underline{5}, \underline{6}, which has 6 different elements.
Submitted Solution:
```
from itertools import permutations
#from fractions import Fraction
from collections import defaultdict
from math import*
import os
import sys
from io import BytesIO, IOBase
from heapq import nlargest
from bisect import*
import copy
import itertools
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
def input(): return sys.stdin.readline().rstrip("\r\n")
#-------------above part copied-----------------------
t=int(input())
while t:
t-=1
n=int(input())
arr=list(map(int,input().split()))
d=defaultdict(int)
for i in range(n):
if d[arr[i]]<2:
d[arr[i]]+=1
arr = list(set(arr))
arr.sort()
ans=0
si = d[arr[0]]
p=arr[0]
# print(arr)
# print(d)
for i in range(1,len(arr)):
if arr[i]-arr[i-1]==1:
si+=d[arr[i]]
# print(ans)
else:
ans+=min(si,arr[i]-p+2)
# print(ans,si,si-d[p]+1)
p = arr[i]
si = d[p]
#print(si,d[p])
ans+=min(si,arr[-1]-p+2)
print(ans)
``` | instruction | 0 | 89,526 | 14 | 179,052 |
No | output | 1 | 89,526 | 14 | 179,053 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Athenaeus has just finished creating his latest musical composition and will present it tomorrow to the people of Athens. Unfortunately, the melody is rather dull and highly likely won't be met with a warm reception.
His song consists of n notes, which we will treat as positive integers. The diversity of a song is the number of different notes it contains. As a patron of music, Euterpe watches over composers and guides them throughout the process of creating new melodies. She decided to help Athenaeus by changing his song to make it more diverse.
Being a minor goddess, she cannot arbitrarily change the song. Instead, for each of the n notes in the song, she can either leave it as it is or increase it by 1.
Given the song as a sequence of integers describing the notes, find out the maximal, achievable diversity.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 10 000) — the number of test cases. Then t test cases follow, each one is described in two lines.
In the first line of each test case there is a single integer n (1 ≤ n ≤ 10^5) denoting the length of the song. The next line contains a sequence of n integers x_1, x_2, …, x_n (1 ≤ x_1 ≤ x_2 ≤ … ≤ x_n ≤ 2 ⋅ n), describing the song.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case, you should output a single line containing precisely one integer, the maximal diversity of the song, i.e. the maximal possible number of different elements in the final sequence.
Example
Input
5
6
1 2 2 2 5 6
2
4 4
6
1 1 3 4 4 5
1
1
6
1 1 1 2 2 2
Output
5
2
6
1
3
Note
In the first test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{3}, 2, 2, \underline{6}, \underline{7}, which has 5 different elements (increased elements are underlined).
In the second test case, Euterpe can increase the first element to obtain the sequence \underline{5}, 4, which has 2 different elements.
In the third test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{2}, 3, 4, \underline{5}, \underline{6}, which has 6 different elements.
Submitted Solution:
```
from collections import Counter
for _ in range(int(input())):
n=int(input())
arr=list(map(int,input().split()))
y=Counter(arr)
s=set(arr)
counter=0
for item in s:
if y[item]==1:
counter+=1
for item in s:
if y[item]>1:
if(item+1 in s):
if(y[item+1]>1):
counter+=1
else:
counter+=2
else:
counter+=2
print(counter)
``` | instruction | 0 | 89,527 | 14 | 179,054 |
No | output | 1 | 89,527 | 14 | 179,055 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ilya got tired of sports programming, left university and got a job in the subway. He was given the task to determine the escalator load factor.
Let's assume that n people stand in the queue for the escalator. At each second one of the two following possibilities takes place: either the first person in the queue enters the escalator with probability p, or the first person in the queue doesn't move with probability (1 - p), paralyzed by his fear of escalators and making the whole queue wait behind him.
Formally speaking, the i-th person in the queue cannot enter the escalator until people with indices from 1 to i - 1 inclusive enter it. In one second only one person can enter the escalator. The escalator is infinite, so if a person enters it, he never leaves it, that is he will be standing on the escalator at any following second. Ilya needs to count the expected value of the number of people standing on the escalator after t seconds.
Your task is to help him solve this complicated task.
Input
The first line of the input contains three numbers n, p, t (1 ≤ n, t ≤ 2000, 0 ≤ p ≤ 1). Numbers n and t are integers, number p is real, given with exactly two digits after the decimal point.
Output
Print a single real number — the expected number of people who will be standing on the escalator after t seconds. The absolute or relative error mustn't exceed 10 - 6.
Examples
Input
1 0.50 1
Output
0.5
Input
1 0.50 4
Output
0.9375
Input
4 0.20 2
Output
0.4 | instruction | 0 | 89,674 | 14 | 179,348 |
Tags: combinatorics, dp, math, probabilities
Correct Solution:
```
# #1 .5 1 (n>=t)
# # 1 => .5
# #1 .5 4 (n<t)
# # 1 => (.5) + (.5^2) + (.5^3) + (.5^4) (n < t)
# #2 .4 4
# # 1 => 4*(.4)(.6)^3
# # 2 => (.4)(.4) + (.4)(.6)(.4) + (.4)(.6.)(.6)(.4) + (.6)(.4)(.4) + (.6)(.4)(.6)(.4) + (.6)(.6)(.4)(.4)
# # =(.4)^2 [1 + 2(.6) + 3(.6)^2 ]
# # if we had 3
# # everything before is basically t * p **
# # 3 => (.4)(.4)(.4) + (.4)(.4)(.6)(.4) + (.4)(.6)(.4)(.4) + (.6)(.4)(.4)(.4)
# # =(.4)^3 (1+3(.6))
# #2 p 5
# #2 => pp + pqp + pqqp + pqqqp + qpp + qpqp + qpqqp + qqpp + qqpqp + qqqpp
# # = p^2 (1 + 2q + 3qq + 4qqq)
# #4 .2 2 (n>t)
# #1 => (.2*(.8) + (.8*.2))
# #2 => .2^2*2
# #4 .2 4 (n >= t)
# #1 => (.2(.8)^3)*4
# #2 => (.2^2 (.8)^2) ()
# #(n >= t)
# #k * (p^k) * ((1-p)^(t-k))
# #(n < t)
# from math import factorial as fact
# def bin(n, k):
# return fact(n)/(fact(k)*fact(n-k))
# n,p,t = map(float, input().split())
# #closed form of Sum{p^k} = p/(-1+p)*(-1+p**n)
# # t-n number of spots
# # 1 + (n)(1-p) + (n+1)(1-p)^2oup = 0.0
# oup = 0.0
# if(n>=t):
# for i in range(1, int(t)+1):
# oup += bin(t, i)*i*(p**i)*((1-p)**(t-i))
# else:
# for i in range(1, int(n)):
# oup += bin(t, i)*i*(p**i)*((1-p)**(t-i))
# n_spots = t-n
# quant = 1
# for i in range(int(n_spots)):
# quant += (n+i)*(1-p)**(i+1)
# oup += p**(n)*quant*n
# print("%.10f"%(oup))
n, p, t = map(float, input().split())
mx = int(max(n,t)+1)
dp = [[0 for i in range(mx+10)]for j in range(mx+10)]
dp[0][0]=1
#dp[0][0] = 1
#dp[0][1] = dp[0][0]*(1-p)
#dp[0][2] = dp[0][1]*(1-p)
#dp[0][3] ...
#dp[1][1] = dp[0][0]*p + dp[1][0]*(1-p)
#dp[1][2] = dp[0][1]*p + dp[1][1]*(1-p)
#dp[n][m] = dp[n-1][m-1]*p + dp[n][m-1]*(1-p)
for i in range(mx):
for j in range(1,mx):
if(i-1>=0):
dp[i][j]+=dp[i-1][j-1]*p
if(i>=int(n)):
dp[i][j]+=dp[i][j-1]
else:
dp[i][j]+=dp[i][j-1]*(1-p)
oup = 0
prev = 0
for i in range(int(n)+1):
oup += (i)*dp[i][int(t)]
print(oup)
``` | output | 1 | 89,674 | 14 | 179,349 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ilya got tired of sports programming, left university and got a job in the subway. He was given the task to determine the escalator load factor.
Let's assume that n people stand in the queue for the escalator. At each second one of the two following possibilities takes place: either the first person in the queue enters the escalator with probability p, or the first person in the queue doesn't move with probability (1 - p), paralyzed by his fear of escalators and making the whole queue wait behind him.
Formally speaking, the i-th person in the queue cannot enter the escalator until people with indices from 1 to i - 1 inclusive enter it. In one second only one person can enter the escalator. The escalator is infinite, so if a person enters it, he never leaves it, that is he will be standing on the escalator at any following second. Ilya needs to count the expected value of the number of people standing on the escalator after t seconds.
Your task is to help him solve this complicated task.
Input
The first line of the input contains three numbers n, p, t (1 ≤ n, t ≤ 2000, 0 ≤ p ≤ 1). Numbers n and t are integers, number p is real, given with exactly two digits after the decimal point.
Output
Print a single real number — the expected number of people who will be standing on the escalator after t seconds. The absolute or relative error mustn't exceed 10 - 6.
Examples
Input
1 0.50 1
Output
0.5
Input
1 0.50 4
Output
0.9375
Input
4 0.20 2
Output
0.4 | instruction | 0 | 89,675 | 14 | 179,350 |
Tags: combinatorics, dp, math, probabilities
Correct Solution:
```
import sys
n, p, t = map(str, sys.stdin.readline().split())
n = int(n)
p = float(p)
t = int(t)
def CC(nn,k):
tmp = n
t = max(nn - k, k)
for i in range(1, min(nn - k, k) + 1):
tmp = tmp * (t + i) * (1 - p) / i
if k > nn - k:
tmp = tmp * pow(1-p,k + k - nn)
return tmp
def C(n, k):
tmp = 1
if n - k > k:
tmp = tmp * pow(1 - p, n - k - k)
else:
tmp = tmp * pow(p, k + k - n)
t = max(n - k, k)
for i in range(1, min(n - k, k) + 1):
tmp = tmp * (t + i) * p * (1 - p) / i
return tmp
if n >= t:
print(t * p)
elif p != 1 and p != 0:
a = 0
b = 0
for i in range(n):
q = C(t, i)
a = a + q * i
b = b + q
a = a + (1 - b) * n
print(a)
b = n
for i in range(t - n):
b = b + CC(i + 1,n + i)
b = b * pow(p,n)
#print(a + b)
else:
if p == 1:
print(n)
else:
print(0)
``` | output | 1 | 89,675 | 14 | 179,351 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ilya got tired of sports programming, left university and got a job in the subway. He was given the task to determine the escalator load factor.
Let's assume that n people stand in the queue for the escalator. At each second one of the two following possibilities takes place: either the first person in the queue enters the escalator with probability p, or the first person in the queue doesn't move with probability (1 - p), paralyzed by his fear of escalators and making the whole queue wait behind him.
Formally speaking, the i-th person in the queue cannot enter the escalator until people with indices from 1 to i - 1 inclusive enter it. In one second only one person can enter the escalator. The escalator is infinite, so if a person enters it, he never leaves it, that is he will be standing on the escalator at any following second. Ilya needs to count the expected value of the number of people standing on the escalator after t seconds.
Your task is to help him solve this complicated task.
Input
The first line of the input contains three numbers n, p, t (1 ≤ n, t ≤ 2000, 0 ≤ p ≤ 1). Numbers n and t are integers, number p is real, given with exactly two digits after the decimal point.
Output
Print a single real number — the expected number of people who will be standing on the escalator after t seconds. The absolute or relative error mustn't exceed 10 - 6.
Examples
Input
1 0.50 1
Output
0.5
Input
1 0.50 4
Output
0.9375
Input
4 0.20 2
Output
0.4 | instruction | 0 | 89,676 | 14 | 179,352 |
Tags: combinatorics, dp, math, probabilities
Correct Solution:
```
n,p,t = input().split()
n = int(n)
p = float(p)
t = int(t)
# dp[n][p][t]
dp = [[0 for _ in range(t+1)] for _ in range(n+1)]
dp[0][0] = 1
for n_p in range(n+1):
for t_p in range(t):
dp[n_p][t_p+1] = dp[n_p][t_p]*(1-p) + dp[n_p][t_p+1]
if n_p != n:
dp[n_p+1][t_p+1] = dp[n_p][t_p]*p + dp[n_p+1][t_p+1]
else:
dp[n_p][t_p+1] = dp[n_p][t_p]*p + dp[n_p][t_p+1]
ans = 0
for n_p in range(n+1):
ans += dp[n_p][-1] * n_p
# print(dp)
print(ans)
``` | output | 1 | 89,676 | 14 | 179,353 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ilya got tired of sports programming, left university and got a job in the subway. He was given the task to determine the escalator load factor.
Let's assume that n people stand in the queue for the escalator. At each second one of the two following possibilities takes place: either the first person in the queue enters the escalator with probability p, or the first person in the queue doesn't move with probability (1 - p), paralyzed by his fear of escalators and making the whole queue wait behind him.
Formally speaking, the i-th person in the queue cannot enter the escalator until people with indices from 1 to i - 1 inclusive enter it. In one second only one person can enter the escalator. The escalator is infinite, so if a person enters it, he never leaves it, that is he will be standing on the escalator at any following second. Ilya needs to count the expected value of the number of people standing on the escalator after t seconds.
Your task is to help him solve this complicated task.
Input
The first line of the input contains three numbers n, p, t (1 ≤ n, t ≤ 2000, 0 ≤ p ≤ 1). Numbers n and t are integers, number p is real, given with exactly two digits after the decimal point.
Output
Print a single real number — the expected number of people who will be standing on the escalator after t seconds. The absolute or relative error mustn't exceed 10 - 6.
Examples
Input
1 0.50 1
Output
0.5
Input
1 0.50 4
Output
0.9375
Input
4 0.20 2
Output
0.4 | instruction | 0 | 89,677 | 14 | 179,354 |
Tags: combinatorics, dp, math, probabilities
Correct Solution:
```
n, p, t = map(float, input().split())
n, t = int(n), int(t)
dp = [[0 for i in range(t + 1)] for j in range(n + 1)]
for i in range(1, n + 1):
for j in range(1, t + 1):
dp[i][j] = p * (dp[i - 1][j - 1] + 1) + (1 - p) * dp[i][j - 1]
print(dp[n][t])
``` | output | 1 | 89,677 | 14 | 179,355 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ilya got tired of sports programming, left university and got a job in the subway. He was given the task to determine the escalator load factor.
Let's assume that n people stand in the queue for the escalator. At each second one of the two following possibilities takes place: either the first person in the queue enters the escalator with probability p, or the first person in the queue doesn't move with probability (1 - p), paralyzed by his fear of escalators and making the whole queue wait behind him.
Formally speaking, the i-th person in the queue cannot enter the escalator until people with indices from 1 to i - 1 inclusive enter it. In one second only one person can enter the escalator. The escalator is infinite, so if a person enters it, he never leaves it, that is he will be standing on the escalator at any following second. Ilya needs to count the expected value of the number of people standing on the escalator after t seconds.
Your task is to help him solve this complicated task.
Input
The first line of the input contains three numbers n, p, t (1 ≤ n, t ≤ 2000, 0 ≤ p ≤ 1). Numbers n and t are integers, number p is real, given with exactly two digits after the decimal point.
Output
Print a single real number — the expected number of people who will be standing on the escalator after t seconds. The absolute or relative error mustn't exceed 10 - 6.
Examples
Input
1 0.50 1
Output
0.5
Input
1 0.50 4
Output
0.9375
Input
4 0.20 2
Output
0.4 | instruction | 0 | 89,678 | 14 | 179,356 |
Tags: combinatorics, dp, math, probabilities
Correct Solution:
```
def main():
tmp = input().split()
n, p, t = int(tmp[0]), float(tmp[1]), int(tmp[2])
q = 1. - p
res = [0.] * (n + 1)
res[0] = 1.
for _ in range(t):
tmp = [x * q for x in res]
tmp[-1] = res[-1]
for i in range(n):
tmp[i + 1] += res[i] * p
res = tmp
print(sum(x * i for i, x in enumerate(res)))
if __name__ == '__main__':
main()
``` | output | 1 | 89,678 | 14 | 179,357 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ilya got tired of sports programming, left university and got a job in the subway. He was given the task to determine the escalator load factor.
Let's assume that n people stand in the queue for the escalator. At each second one of the two following possibilities takes place: either the first person in the queue enters the escalator with probability p, or the first person in the queue doesn't move with probability (1 - p), paralyzed by his fear of escalators and making the whole queue wait behind him.
Formally speaking, the i-th person in the queue cannot enter the escalator until people with indices from 1 to i - 1 inclusive enter it. In one second only one person can enter the escalator. The escalator is infinite, so if a person enters it, he never leaves it, that is he will be standing on the escalator at any following second. Ilya needs to count the expected value of the number of people standing on the escalator after t seconds.
Your task is to help him solve this complicated task.
Input
The first line of the input contains three numbers n, p, t (1 ≤ n, t ≤ 2000, 0 ≤ p ≤ 1). Numbers n and t are integers, number p is real, given with exactly two digits after the decimal point.
Output
Print a single real number — the expected number of people who will be standing on the escalator after t seconds. The absolute or relative error mustn't exceed 10 - 6.
Examples
Input
1 0.50 1
Output
0.5
Input
1 0.50 4
Output
0.9375
Input
4 0.20 2
Output
0.4 | instruction | 0 | 89,679 | 14 | 179,358 |
Tags: combinatorics, dp, math, probabilities
Correct Solution:
```
n, p, t = input().split()
n = int(n)
t = int(t)
p = float(p)
ans = 0.0
s = t
if abs(1.0 - p) > 1e-6:
a1 = 1.0
for i in range(1, t + 1):
a1 = a1 * (t - i + 1) / i if i <= t else 0
a1 = a1 * p
while (a1 > 1e5) and (s > i):
a1 *= (1 - p)
s -= 1
while (s < i):
a1 /= (1 - p)
s += 1
a2 = (1 - p) ** (s - i)
#print(a1, a2, s, i)
ans += a1 * a2 * min(i, n)
else:
ans = min(n, t)
print(ans)
``` | output | 1 | 89,679 | 14 | 179,359 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ilya got tired of sports programming, left university and got a job in the subway. He was given the task to determine the escalator load factor.
Let's assume that n people stand in the queue for the escalator. At each second one of the two following possibilities takes place: either the first person in the queue enters the escalator with probability p, or the first person in the queue doesn't move with probability (1 - p), paralyzed by his fear of escalators and making the whole queue wait behind him.
Formally speaking, the i-th person in the queue cannot enter the escalator until people with indices from 1 to i - 1 inclusive enter it. In one second only one person can enter the escalator. The escalator is infinite, so if a person enters it, he never leaves it, that is he will be standing on the escalator at any following second. Ilya needs to count the expected value of the number of people standing on the escalator after t seconds.
Your task is to help him solve this complicated task.
Input
The first line of the input contains three numbers n, p, t (1 ≤ n, t ≤ 2000, 0 ≤ p ≤ 1). Numbers n and t are integers, number p is real, given with exactly two digits after the decimal point.
Output
Print a single real number — the expected number of people who will be standing on the escalator after t seconds. The absolute or relative error mustn't exceed 10 - 6.
Examples
Input
1 0.50 1
Output
0.5
Input
1 0.50 4
Output
0.9375
Input
4 0.20 2
Output
0.4 | instruction | 0 | 89,680 | 14 | 179,360 |
Tags: combinatorics, dp, math, probabilities
Correct Solution:
```
MAXN = 2005
dp = [[0 for col in range(MAXN)] for row in range(MAXN)]
n, q, t = input().split()
n = int(n)
t = int(t)
q = float(q)
if t <= n:
print(t*q)
else:
dp[0][0] = 1
for i in range(t+1):
for j in range(min(t, n)+1):
if j == n:
dp[i+1][j] += dp[i][j]
else:
dp[i+1][j] += (1-q) * dp[i][j]
dp[i+1][j+1] += q * dp[i][j]
ans = 0
for i in range(n+1):
ans += i * dp[t][i]
print(ans)
``` | output | 1 | 89,680 | 14 | 179,361 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ilya got tired of sports programming, left university and got a job in the subway. He was given the task to determine the escalator load factor.
Let's assume that n people stand in the queue for the escalator. At each second one of the two following possibilities takes place: either the first person in the queue enters the escalator with probability p, or the first person in the queue doesn't move with probability (1 - p), paralyzed by his fear of escalators and making the whole queue wait behind him.
Formally speaking, the i-th person in the queue cannot enter the escalator until people with indices from 1 to i - 1 inclusive enter it. In one second only one person can enter the escalator. The escalator is infinite, so if a person enters it, he never leaves it, that is he will be standing on the escalator at any following second. Ilya needs to count the expected value of the number of people standing on the escalator after t seconds.
Your task is to help him solve this complicated task.
Input
The first line of the input contains three numbers n, p, t (1 ≤ n, t ≤ 2000, 0 ≤ p ≤ 1). Numbers n and t are integers, number p is real, given with exactly two digits after the decimal point.
Output
Print a single real number — the expected number of people who will be standing on the escalator after t seconds. The absolute or relative error mustn't exceed 10 - 6.
Examples
Input
1 0.50 1
Output
0.5
Input
1 0.50 4
Output
0.9375
Input
4 0.20 2
Output
0.4 | instruction | 0 | 89,681 | 14 | 179,362 |
Tags: combinatorics, dp, math, probabilities
Correct Solution:
```
inputs = [x for x in input().split()]
n,p,t = int(inputs[0]), float(inputs[1]), int(inputs[2])
saved = [[0 for _ in range(n)] for _ in range(t)]
saved[0][0] = p
for i in range(1, t):
saved[i][0] = saved[i-1][0] + ((1 - saved[i-1][0]) * p)
for i in range(1, t):
for j in range(1, n):
saved[i][j] = saved[i - 1][j - 1] * p + (saved[i - 1][j]) * (1 -p)
#print(saved)
print(sum(saved[-1]))
``` | output | 1 | 89,681 | 14 | 179,363 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ilya got tired of sports programming, left university and got a job in the subway. He was given the task to determine the escalator load factor.
Let's assume that n people stand in the queue for the escalator. At each second one of the two following possibilities takes place: either the first person in the queue enters the escalator with probability p, or the first person in the queue doesn't move with probability (1 - p), paralyzed by his fear of escalators and making the whole queue wait behind him.
Formally speaking, the i-th person in the queue cannot enter the escalator until people with indices from 1 to i - 1 inclusive enter it. In one second only one person can enter the escalator. The escalator is infinite, so if a person enters it, he never leaves it, that is he will be standing on the escalator at any following second. Ilya needs to count the expected value of the number of people standing on the escalator after t seconds.
Your task is to help him solve this complicated task.
Input
The first line of the input contains three numbers n, p, t (1 ≤ n, t ≤ 2000, 0 ≤ p ≤ 1). Numbers n and t are integers, number p is real, given with exactly two digits after the decimal point.
Output
Print a single real number — the expected number of people who will be standing on the escalator after t seconds. The absolute or relative error mustn't exceed 10 - 6.
Examples
Input
1 0.50 1
Output
0.5
Input
1 0.50 4
Output
0.9375
Input
4 0.20 2
Output
0.4
Submitted Solution:
```
def find_expected_people(people_count, probability, seconds):
# the probability after i seconds to have k people on the escalator
current_row = [0.] * (seconds + 1)
current_row[0] = 1.
for sec in range(1, seconds + 1):
next_row = [x * (1. - probability) for x in current_row]
next_row[-1] = current_row[-1]
for person in range(people_count):
next_row[person + 1] += current_row[person] * probability
current_row = next_row
return sum(index * value for index, value in enumerate(current_row))
def main():
raw_input = input().split()
people_count = int(raw_input[0])
probability = float(raw_input[1])
seconds = int(raw_input[2])
print(find_expected_people(people_count, probability, seconds))
if __name__ == '__main__':
main()
``` | instruction | 0 | 89,686 | 14 | 179,372 |
No | output | 1 | 89,686 | 14 | 179,373 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Well, the series which Stepan watched for a very long time, ended. In total, the series had n episodes. For each of them, Stepan remembers either that he definitely has watched it, or that he definitely hasn't watched it, or he is unsure, has he watched this episode or not.
Stepan's dissatisfaction is the maximum number of consecutive series that Stepan did not watch.
Your task is to determine according to Stepan's memories if his dissatisfaction could be exactly equal to k.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100, 0 ≤ k ≤ n) — the number of episodes in the series and the dissatisfaction which should be checked.
The second line contains the sequence which consists of n symbols "Y", "N" and "?". If the i-th symbol equals "Y", Stepan remembers that he has watched the episode number i. If the i-th symbol equals "N", Stepan remembers that he hasn't watched the epizode number i. If the i-th symbol equals "?", Stepan doesn't exactly remember if he has watched the episode number i or not.
Output
If Stepan's dissatisfaction can be exactly equal to k, then print "YES" (without qoutes). Otherwise print "NO" (without qoutes).
Examples
Input
5 2
NYNNY
Output
YES
Input
6 1
????NN
Output
NO
Note
In the first test Stepan remembers about all the episodes whether he has watched them or not. His dissatisfaction is 2, because he hasn't watch two episodes in a row — the episode number 3 and the episode number 4. The answer is "YES", because k = 2.
In the second test k = 1, Stepan's dissatisfaction is greater than or equal to 2 (because he remembers that he hasn't watch at least two episodes in a row — number 5 and number 6), even if he has watched the episodes from the first to the fourth, inclusive. | instruction | 0 | 89,775 | 14 | 179,550 |
Tags: *special, dp
Correct Solution:
```
from sys import stdin
from functools import lru_cache
from collections import defaultdict
infile = stdin
N, K = map(int, infile.readline().split())
watched = infile.readline().strip()
mem = set()
mem.add((0,0,0))
for i, letter in enumerate(watched):
for d in range(K+1):
for c in range(K+1):
if (i,d,c) in mem:
if letter in ('N', '?'):
mem.add((i+1, max(d, c+1), c+1))
if letter in ('Y', '?'):
mem.add((i+1, d, 0))
def good():
for i,d,c in mem:
if i == N and d == K:
return 'YES'
return 'NO'
print(good())
``` | output | 1 | 89,775 | 14 | 179,551 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Well, the series which Stepan watched for a very long time, ended. In total, the series had n episodes. For each of them, Stepan remembers either that he definitely has watched it, or that he definitely hasn't watched it, or he is unsure, has he watched this episode or not.
Stepan's dissatisfaction is the maximum number of consecutive series that Stepan did not watch.
Your task is to determine according to Stepan's memories if his dissatisfaction could be exactly equal to k.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100, 0 ≤ k ≤ n) — the number of episodes in the series and the dissatisfaction which should be checked.
The second line contains the sequence which consists of n symbols "Y", "N" and "?". If the i-th symbol equals "Y", Stepan remembers that he has watched the episode number i. If the i-th symbol equals "N", Stepan remembers that he hasn't watched the epizode number i. If the i-th symbol equals "?", Stepan doesn't exactly remember if he has watched the episode number i or not.
Output
If Stepan's dissatisfaction can be exactly equal to k, then print "YES" (without qoutes). Otherwise print "NO" (without qoutes).
Examples
Input
5 2
NYNNY
Output
YES
Input
6 1
????NN
Output
NO
Note
In the first test Stepan remembers about all the episodes whether he has watched them or not. His dissatisfaction is 2, because he hasn't watch two episodes in a row — the episode number 3 and the episode number 4. The answer is "YES", because k = 2.
In the second test k = 1, Stepan's dissatisfaction is greater than or equal to 2 (because he remembers that he hasn't watch at least two episodes in a row — number 5 and number 6), even if he has watched the episodes from the first to the fourth, inclusive. | instruction | 0 | 89,776 | 14 | 179,552 |
Tags: *special, dp
Correct Solution:
```
def process(S, k):
n = len(S)
if k > n:
return 'NO'
current = 0
for i in range(n):
if S[i]=='N':
current+=1
if current > k:
return 'NO'
else:
current = 0
start_work = True
for i in range(k):
if S[i]=='Y':
start_work = False
if start_work and (k==n or k < n and S[k] != 'N'):
return 'YES'
if n==1:
if k==1 and S[0] in ['?N']:
return 'YES'
if k==0 and S[0] in ['?Y']:
return 'YES'
return 'NO'
if k==n:
if 'Y' in S:
return 'NO'
return 'YES'
Y_count = 0
p1 = 0
for i in range(1, k+1):
if S[i]=='Y':
Y_count+=1
p2 = i
for i in range(k+1, n):
# print(p1, Y_count, p2)
if Y_count==0 and S[p1] != 'N' and (p2==n-1 or S[p2+1] != 'N'):
return 'YES'
p1+=1
p2+=1
if p2 < n and S[p2]=='Y':
Y_count+=1
if p1 < n and S[p1]=='Y':
Y_count-=1
# print(p1, Y_count, p2)
if Y_count==0 and p1 < n and S[p1] != 'N':
return 'YES'
return 'NO'
n, k = [int(x) for x in input().split()]
S = input()
print(process(S, k))
``` | output | 1 | 89,776 | 14 | 179,553 |
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