message stringlengths 2 22.7k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 145 109k | cluster float64 9 9 | __index_level_0__ int64 290 217k |
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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions).
A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen.
Write a program that, when S is given, prints the price of the corresponding bowl of ramen.
Constraints
* S is a string of length 3.
* Each character in S is `o` or `x`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the price of the bowl of ramen corresponding to S.
Examples
Input
oxo
Output
900
Input
ooo
Output
1000
Input
xxx
Output
700
Submitted Solution:
```
s=input()
print(700+100*int(s.coount("o")))
``` | instruction | 0 | 4,059 | 9 | 8,118 |
No | output | 1 | 4,059 | 9 | 8,119 |
Provide a correct Python 3 solution for this coding contest problem.
Problem statement
2D, who is good at cooking, is trying to make lunch. Cooking requires all N ingredients a_ {0}, a_ {1},…, a_ {N−1}.
Now, 2D's refrigerator doesn't contain any ingredients, so I have to go to the supermarket to buy it. At the supermarket, you can buy the material a_ {i} for the price x_ {i} yen.
2D is also a wizard and can use M types of magic. The i-th magic can be changed to the material t_ {i} by applying it to the material s_ {i}, and conversely to s_ {i} by applying it to the material t_ {i}. In addition, you can repeatedly use multiple spells on a single material. For example, you can get r from p using the magic of changing from p to q and the magic of changing from q to r.
2D decided to use the power of magic to prepare the materials as cheaply as possible. Find the minimum sum of the prices of the ingredients that 2D needs to buy to complete the dish.
input
The input is given in the following format.
N
a_ {0} x_ {0}
...
a_ {N−1} x_ {N−1}
M
s_ {0} t_ {0}
...
s_ {M−1} t_ {M−1}
Constraint
* All numbers are integers
* All material names consist of at least 1 and no more than 10 lowercase letters.
* If i ≠ j, then a_ {i} ≠ a_ {j}
* 1 \ ≤ x_ {i} \ ≤ 1,000
* 1 \ ≤ N \ ≤ 5,000
* 0 \ ≤ M \ ≤ {\ rm min} (N (N−1) / 2, 1000)
* s_ {i} ≠ t_ {i}
* There is no duplication in the pair of s_ {i}, t_ {i}
* s_ {i}, t_ {i} are included in a_ {0},…, a_ {N−1}
output
Print the answer in one line.
sample
Sample input 1
2
tako 2
yaki 1
1
tako yaki
You can magically turn a cheap yaki into a tako, so buy two yaki.
Sample output 1
2
Sample input 2
Five
a 1
b 2
c 2
d 4
e 3
Five
b a
a c
c d
e b
c b
Sample output 2
Five
As shown below, all materials can be changed from a.
* a: a as it is
* b: a-> c-> b
* c: a-> c
* d: a-> c-> d
* e: a-> b-> e
<image>
Example
Input
2
tako 2
yaki 1
1
tako yaki
Output
2 | instruction | 0 | 4,128 | 9 | 8,256 |
"Correct Solution:
```
n = int(input())
dic = {}
price = []
for i in range(n):
a, x = input().split()
dic[a] = i
price.append(int(x))
parent = [i for i in range(n)]
def find(x):
if parent[x] == x:return x
parent[x] = find(parent[x])
return parent[x]
m = int(input())
for _ in range(m):
s, t = input().split()
si, ti = dic[s], dic[t]
ps = find(si)
pt = find(ti)
parent[ps] = parent[pt]
groups = {}
for i in range(n):
if find(i) in groups:
min_price, cnt = groups[find(i)]
groups[find(i)] = (min(min_price, price[i]), cnt + 1)
else:
groups[find(i)] = (price[i], 1)
print(sum([v[0] * v[1] for _, v in groups.items()]))
``` | output | 1 | 4,128 | 9 | 8,257 |
Provide a correct Python 3 solution for this coding contest problem.
Problem statement
2D, who is good at cooking, is trying to make lunch. Cooking requires all N ingredients a_ {0}, a_ {1},…, a_ {N−1}.
Now, 2D's refrigerator doesn't contain any ingredients, so I have to go to the supermarket to buy it. At the supermarket, you can buy the material a_ {i} for the price x_ {i} yen.
2D is also a wizard and can use M types of magic. The i-th magic can be changed to the material t_ {i} by applying it to the material s_ {i}, and conversely to s_ {i} by applying it to the material t_ {i}. In addition, you can repeatedly use multiple spells on a single material. For example, you can get r from p using the magic of changing from p to q and the magic of changing from q to r.
2D decided to use the power of magic to prepare the materials as cheaply as possible. Find the minimum sum of the prices of the ingredients that 2D needs to buy to complete the dish.
input
The input is given in the following format.
N
a_ {0} x_ {0}
...
a_ {N−1} x_ {N−1}
M
s_ {0} t_ {0}
...
s_ {M−1} t_ {M−1}
Constraint
* All numbers are integers
* All material names consist of at least 1 and no more than 10 lowercase letters.
* If i ≠ j, then a_ {i} ≠ a_ {j}
* 1 \ ≤ x_ {i} \ ≤ 1,000
* 1 \ ≤ N \ ≤ 5,000
* 0 \ ≤ M \ ≤ {\ rm min} (N (N−1) / 2, 1000)
* s_ {i} ≠ t_ {i}
* There is no duplication in the pair of s_ {i}, t_ {i}
* s_ {i}, t_ {i} are included in a_ {0},…, a_ {N−1}
output
Print the answer in one line.
sample
Sample input 1
2
tako 2
yaki 1
1
tako yaki
You can magically turn a cheap yaki into a tako, so buy two yaki.
Sample output 1
2
Sample input 2
Five
a 1
b 2
c 2
d 4
e 3
Five
b a
a c
c d
e b
c b
Sample output 2
Five
As shown below, all materials can be changed from a.
* a: a as it is
* b: a-> c-> b
* c: a-> c
* d: a-> c-> d
* e: a-> b-> e
<image>
Example
Input
2
tako 2
yaki 1
1
tako yaki
Output
2 | instruction | 0 | 4,129 | 9 | 8,258 |
"Correct Solution:
```
import sys
from collections import defaultdict,deque
def root(x):
if par[x] == x:
return x
par[x] = root(par[x])
return par[x]
def unite(x,y):
x = root(x)
y = root(y)
if rank[x] < rank[y]:
par[x] = y
else:
par[y] = x
if rank[x] == rank[y]:
rank[x] += 1
f = defaultdict(int)
par = defaultdict(int)
rank = defaultdict(int)
n = int(sys.stdin.readline())
for i in range(n):
a,x = sys.stdin.readline().split()
f[a] = int(x)
par[a] = a
rank[a] = 0
m = int(sys.stdin.readline())
for i in range(m):
x,y = sys.stdin.readline().split()
if root(x) != root(y):
unite(x,y)
ans = 0
for i in f.keys():
root(i)
f[par[i]] = min(f[par[i]],f[i])
for i in f.keys():
ans += f[root(i)]
print(ans)
``` | output | 1 | 4,129 | 9 | 8,259 |
Provide tags and a correct Python 3 solution for this coding contest problem.
JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | instruction | 0 | 5,045 | 9 | 10,090 |
Tags: greedy, implementation, math
Correct Solution:
```
n,q = list(map(int, input().split()))
a = input()
Q = []
for _ in range(q):
Q.append(list(map(int, input().split())))
d = [0]
ab = 0
for i in a:
if i == '1':
ab += 1
d.append(ab)
mod = int(1e9 + 7)
p = [1]
i = 1
for _ in range(n):
i = (i*2)%mod
p.append(i)
for l,r in Q:
y = r-l + 1
x = d[r] - d[l-1]
y -= x
print(((p[x]-1)*p[y])%mod)
``` | output | 1 | 5,045 | 9 | 10,091 |
Provide tags and a correct Python 3 solution for this coding contest problem.
JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | instruction | 0 | 5,046 | 9 | 10,092 |
Tags: greedy, implementation, math
Correct Solution:
```
import sys
r = sys.stdin.readlines()
M = 10 ** 9 + 7
n, q = map(int, r[0].strip().split())
s = list(map(int, r[1].strip()))
p = [0]
for i in range(n):
p.append(p[i] + int(s[i]))
ans = []
for k in range(q):
a, b = map(int, r[k + 2].strip().split())
l = b - a + 1
one = p[b] - p[a - 1]
zero = l - one
ans.append(str(((pow(2, one, M) - 1) * pow(2, zero, M)) % M))
sys.stdout.write("\n".join(ans))
``` | output | 1 | 5,046 | 9 | 10,093 |
Provide tags and a correct Python 3 solution for this coding contest problem.
JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | instruction | 0 | 5,047 | 9 | 10,094 |
Tags: greedy, implementation, math
Correct Solution:
```
#Wolve
from sys import *
m = 1000000007
n, q = map(int, stdin.readline().split())
a = stdin.readline()
ans = []
t = []
count = 0
for i in a:
if i == '1':
count+=1
t.append(count)
for _ in range(q):
x,y=map(int,input().split())
if(x==1):
p=t[y-1]
else:
p=t[y-1]-t[x-2]
q=(y-x+1-p)
s=pow(2,p+q,m)%m
s=(((s%m)-(pow(2,q,m)%m))%m)
ans.append(s)
stdout.write('\n'.join(map(str, ans)))
``` | output | 1 | 5,047 | 9 | 10,095 |
Provide tags and a correct Python 3 solution for this coding contest problem.
JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | instruction | 0 | 5,048 | 9 | 10,096 |
Tags: greedy, implementation, math
Correct Solution:
```
n,q = map(int,input().split())
a = input()
sums = []
degrees = [1,2]
res = []
d = 2
if (a[0] == "1"):
s = 1
else:
s = 0
sums.append(s)
for i in range(1,n):
if (a[i] == "1"):
s += 1
d = (d*2)%1000000007
sums.append(s)
degrees.append(d)
for i in range(0,q):
l,r = map(int,input().split())
k0 = (r - sums[r-1]) - (l - sums[l-1])
k1 = sums[r-1] - sums[l-1]
if (a[l-1] == "0"):
k0 += 1
else:
k1 += 1
R = (degrees[k0]*(degrees[k1]-1))%1000000007
res.append(R)
for i in range(0,q):
print(res[i])
``` | output | 1 | 5,048 | 9 | 10,097 |
Provide tags and a correct Python 3 solution for this coding contest problem.
JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | instruction | 0 | 5,049 | 9 | 10,098 |
Tags: greedy, implementation, math
Correct Solution:
```
from sys import stdin, stdout
n,q = tuple(map(int,stdin.readline().split()))
arr = list(str(stdin.readline()))[:-1]
arr[0] = int(arr[0])
for i in range(1,n):
arr[i] = int(arr[i]) + arr[i-1]
arr.insert(0,0)
mod = int(1e9+7)
inputs = [ tuple(map(int,line.split())) for line in stdin ]
ans = []
for l,r in inputs:
o = (arr[r] - arr[l-1])
z = (r - l + 1 - o)
e_o = (pow(2,o,mod) - 1)%mod
e_z = e_o * (pow(2,z,mod)-1)%mod
ans.append(str(int((e_o + e_z)%mod)))
stdout.write("\n".join(ans))
``` | output | 1 | 5,049 | 9 | 10,099 |
Provide tags and a correct Python 3 solution for this coding contest problem.
JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | instruction | 0 | 5,050 | 9 | 10,100 |
Tags: greedy, implementation, math
Correct Solution:
```
import sys
MOD = 10 ** 9 + 7
r = sys.stdin.readlines()
n, q = r[0].split(' ')
n = int(n)
q = int(q)
s = r[1]
c = [0] * (n + 1)
for i in range(n):
c[i + 1] = c[i] + (s[i] == '1')
p2 = [1] * (2 * n + 1)
for i in range(1, 2 * n + 1):
p2[i] = p2[i - 1] * 2 % MOD
out = []
for qq in range(q):
a, b = r[qq + 2].split(' ')
a = int(a)
b = int(b)
o = c[b] - c[a - 1]
z = (b - a + 1) - o
ans = (p2[o + z] - p2[z]) % MOD
out.append(str(ans))
sys.stdout.write("\n".join(out))
``` | output | 1 | 5,050 | 9 | 10,101 |
Provide tags and a correct Python 3 solution for this coding contest problem.
JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | instruction | 0 | 5,051 | 9 | 10,102 |
Tags: greedy, implementation, math
Correct Solution:
```
import sys
import math
input=sys.stdin.readline
def power(x, y, p) :
res = 1 # Initialize result
# Update x if it is more
# than or equal to p
x = x % p
while (y > 0) :
# If y is odd, multiply
# x with result
if ((y & 1) == 1) :
res = (res * x) % p
# y must be even now
y = y >> 1 # y = y/2
x = (x * x) % p
return res
modi=1000000007
ones=[0]*(100001)
for i in range(1,100001):
ones[i]=(2*ones[i-1]+1)%modi
zeroes=[1]*(100001)
for i in range(1,100001):
zeroes[i]=(2*zeroes[i-1])%modi
n,q=map(int,input().split())
s=input()
cones=[0]*(n+1)
for i in range(1,n+1):
if(s[i-1]=='1'):
cones[i]=cones[i-1]+1
else:
cones[i]=cones[i-1]
for i in range(q):
l,r=map(int,input().split())
curr=cones[r]-cones[l-1]
ans=(ones[curr])%modi
ans=((ans%modi)*(zeroes[r-l+1-curr]%modi))%modi
print(ans)
``` | output | 1 | 5,051 | 9 | 10,103 |
Provide tags and a correct Python 3 solution for this coding contest problem.
JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | instruction | 0 | 5,052 | 9 | 10,104 |
Tags: greedy, implementation, math
Correct Solution:
```
from sys import *
m = 1000000007
n, q = map(int, stdin.readline().split())
a = stdin.readline()
ans = []
t = []
count = 0
for i in a:
if i == '1':
count+=1
t.append(count)
for _ in range(q):
x,y=map(int,input().split())
if(x==1):
p=t[y-1]
else:
p=t[y-1]-t[x-2]
q=(y-x+1-p)
s=pow(2,p+q,m) - (pow(2,q,m))
ans.append(s%m)
stdout.write('\n'.join(map(str, ans)))
``` | output | 1 | 5,052 | 9 | 10,105 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer.
Submitted Solution:
```
from sys import stdin,stdout
n,q=map(int,input().split())
mod=1000000007
o=[]
s=[]
r=m=0
a=input()
for i in a:
if i=='0':
r+=1
else:
m+=1
o.append(r)
s.append(m)
z=[1]
#print(o)
for i in range(100000):
z.append((z[-1]*2)%mod)
for j in range(q):
l,r=(int(j) for j in stdin.readline().split())
m=r-l+1
zs=o[r-1]-o[l-1]+(a[l-1]=='0')
os=m-zs
#print(zs,os)
if zs!=0:
print((((z[os]-1)%mod)*((z[zs])%mod))%mod)
else:
print(((z[os]-1)%mod))
``` | instruction | 0 | 5,053 | 9 | 10,106 |
Yes | output | 1 | 5,053 | 9 | 10,107 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer.
Submitted Solution:
```
import sys
input = sys.stdin.readline
n, q = map(int, input().split())
s = input()
pref = [0 for i in range(n + 1)]
for i in range(1, n + 1):
pref[i] = pref[i - 1] + (s[i - 1] == '1')
mod = 1000000007
ans = []
for i in range(q):
a, b = map(int, input().split())
k = pref[b] - pref[a - 1];
N = b - a + 1
z = N - k
ans.append((pow(2, k, mod) - 1) * pow(2, z, mod) % mod)
print(*ans)
``` | instruction | 0 | 5,054 | 9 | 10,108 |
Yes | output | 1 | 5,054 | 9 | 10,109 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer.
Submitted Solution:
```
import sys
r = sys.stdin.readlines()
n, q = r[0].split(' ')
n = int(n)
q = int(q)
M = 10 ** 9 + 7
twopow = [1] * (2 * n + 1)
for j in range(1, (2 * n + 1)):
twopow[j] = twopow[j - 1] * 2 % M
s = r[1]
p = [0] * (n + 1)
for v in range(n):
p[v + 1] = p[v] + (s[v] == '1')
ans = [0] * q
for k in range(q):
a, b = r[k + 2].split(' ')
a = int(a)
b = int(b)
l = b - a + 1
one = p[b] - p[a - 1]
v = (twopow[l] - twopow[l - one] + M) % M
ans[k] = str(v)
sys.stdout.write("\n".join(ans))
``` | instruction | 0 | 5,055 | 9 | 10,110 |
Yes | output | 1 | 5,055 | 9 | 10,111 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer.
Submitted Solution:
```
import sys
r = sys.stdin.readlines()
M = 10 ** 9 + 7
n, q = r[0].strip().split()
n = int(n)
q = int(q)
s = r[1]
p = [0]
k = 0
for v in range(n):
d = int(s[v])
k += d
p.append(k)
ans = []
for k in range(q):
a, b = r[k + 2].strip().split()
a = int(a)
b = int(b)
l = b - a + 1
one = p[b] - p[a - 1]
ans.append((pow(2, l, M) - pow(2, l - one, M) + M) % M)
sys.stdout.write("\n".join(map(str, ans)))
``` | instruction | 0 | 5,056 | 9 | 10,112 |
Yes | output | 1 | 5,056 | 9 | 10,113 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer.
Submitted Solution:
```
from collections import defaultdict, deque, Counter
from sys import stdin, stdout
from heapq import heappush, heappop
import math
import io
import os
import math
import bisect
#?############################################################
def gcd(a,b):
if a == 0:
return b
return gcd(b % a, a)
def lcm(a,b):
return (a // gcd(a,b))* b
def isPrime(x):
for i in range(2, x):
if i*i > x:
break
if (x % i == 0):
return False
return True
#?############################################################
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den, p - 2, p)) % p
#?############################################################
def primeFactors(n):
l = []
while n % 2 == 0:
l.append(2)
n = n / 2
for i in range(3, int(math.sqrt(n))+1, 2):
while n % i == 0:
l.append(int(i))
n = n / i
if n > 2:
l.append(n)
return list(set(l))
#?############################################################
def power(x, y, p):
res = 1
x = x % p
if (x == 0):
return 0
while (y > 0):
if ((y & 1) == 1):
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
#?############################################################
def sieve(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
#?############################################################
def digits(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
#?############################################################
def ceil(n, x):
if (n % x == 0):
return n//x
return n//x+1
#?############################################################
def mapin():
return map(int, input().split())
#?############################################################
# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
# python3 15.py<i:n>op
n, q = map(int, input().split())
s = input()
nn = 1000000009
N = 1000003
l = [0]*N
ll = [0]*(len(s)+1)
tt = 0
for i in range(len(s)):
if(s[i] == '0'):
tt+=1
ll[i+1] = tt
temp = 1
for i in range(N):
l[i] = temp%nn
temp = temp+temp
temp = temp%nn
# print(l[0:116])
# print(ll)/
for i in range(q):
a, b = map(int, input().split())
d = ll[b] - ll[a-1]
c = b-a+1-d
e = l[c]-1
f = l[d]
print((e*f)%nn)
``` | instruction | 0 | 5,057 | 9 | 10,114 |
No | output | 1 | 5,057 | 9 | 10,115 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer.
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Sat Jun 8 11:24:26 2019
@author: plosi
"""
import math
def compute(l,r,B):
if l==0:
n=B[r]
else:
n=B[r]-B[l-1]
j=r-(l-1)-n
return (math.pow(2,n)-1)*math.pow(2,j)
def main():
n,q=input().split(" ")
n=int(n)
q=int(q)
arr=input()
narr=[0 for i in range(n)]
for i in range(n):
narr[i]=int(arr[i])
B=[0 for i in range(n)]
B[0]=narr[0]
for i in range(1,len(arr)):
B[i]=B[i-1]+narr[i]
for i in range(q):
l,r=input().split(" ")
l=int(l)-1
r=int(r)-1
print(compute(l,r,B))
main()
``` | instruction | 0 | 5,058 | 9 | 10,116 |
No | output | 1 | 5,058 | 9 | 10,117 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer.
Submitted Solution:
```
n,q = map(int,input().split())
a = input()
sums = []
degrees = [1,2]
res = []
d = 2
if (a[0] == "1"):
s = 1
else:
s = 0
sums.append(s)
for i in range(1,n):
if (a[i] == "1"):
s += 1
d = d%1000000007*2
sums.append(s)
degrees.append(d)
for i in range(0,q):
l,r = map(int,input().split())
k0 = (r - sums[r-1]) - (l - sums[l-1])
k1 = sums[r-1] - sums[l-1]
if (a[l-1] == "0"):
k0 += 1
else:
k1 += 1
R = degrees[k0]*(degrees[k1]-1)
res.append(R)
for i in range(0,q):
print(res[i])
``` | instruction | 0 | 5,059 | 9 | 10,118 |
No | output | 1 | 5,059 | 9 | 10,119 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer.
Submitted Solution:
```
mod=1000000007
n,m=map(int,input().split())
s=list(input())
l=[0]*(n+1)
for i in range(1,n+1):
l[i]=l[i-1]+int(s[i-1])
print(l)
for i in range(m):
l1,r=map(int,input().split())
s=r-l1
no=l[r]-l[l1-1]
ans=pow(2,s+1,mod)-pow(2,s-no+1,mod)
print(ans)
``` | instruction | 0 | 5,060 | 9 | 10,120 |
No | output | 1 | 5,060 | 9 | 10,121 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr. Apple, a gourmet, works as editor-in-chief of a gastronomic periodical. He travels around the world, tasting new delights of famous chefs from the most fashionable restaurants. Mr. Apple has his own signature method of review — in each restaurant Mr. Apple orders two sets of dishes on two different days. All the dishes are different, because Mr. Apple doesn't like to eat the same food. For each pair of dishes from different days he remembers exactly which was better, or that they were of the same quality. After this the gourmet evaluates each dish with a positive integer.
Once, during a revision of a restaurant of Celtic medieval cuisine named «Poisson», that serves chestnut soup with fir, warm soda bread, spicy lemon pie and other folk food, Mr. Apple was very pleasantly surprised the gourmet with its variety of menu, and hence ordered too much. Now he's confused about evaluating dishes.
The gourmet tasted a set of n dishes on the first day and a set of m dishes on the second day. He made a table a of size n × m, in which he described his impressions. If, according to the expert, dish i from the first set was better than dish j from the second set, then a_{ij} is equal to ">", in the opposite case a_{ij} is equal to "<". Dishes also may be equally good, in this case a_{ij} is "=".
Now Mr. Apple wants you to help him to evaluate every dish. Since Mr. Apple is very strict, he will evaluate the dishes so that the maximal number used is as small as possible. But Mr. Apple also is very fair, so he never evaluates the dishes so that it goes against his feelings. In other words, if a_{ij} is "<", then the number assigned to dish i from the first set should be less than the number of dish j from the second set, if a_{ij} is ">", then it should be greater, and finally if a_{ij} is "=", then the numbers should be the same.
Help Mr. Apple to evaluate each dish from both sets so that it is consistent with his feelings, or determine that this is impossible.
Input
The first line contains integers n and m (1 ≤ n, m ≤ 1000) — the number of dishes in both days.
Each of the next n lines contains a string of m symbols. The j-th symbol on i-th line is a_{ij}. All strings consist only of "<", ">" and "=".
Output
The first line of output should contain "Yes", if it's possible to do a correct evaluation for all the dishes, or "No" otherwise.
If case an answer exist, on the second line print n integers — evaluations of dishes from the first set, and on the third line print m integers — evaluations of dishes from the second set.
Examples
Input
3 4
>>>>
>>>>
>>>>
Output
Yes
2 2 2
1 1 1 1
Input
3 3
>>>
<<<
>>>
Output
Yes
3 1 3
2 2 2
Input
3 2
==
=<
==
Output
No
Note
In the first sample, all dishes of the first day are better than dishes of the second day. So, the highest score will be 2, for all dishes of the first day.
In the third sample, the table is contradictory — there is no possible evaluation of the dishes that satisfies it. | instruction | 0 | 5,080 | 9 | 10,160 |
Tags: dfs and similar, dp, dsu, graphs, greedy
Correct Solution:
```
import sys
import collections
def main():
input = iter(sys.stdin)
N, M = map(int, next(input).split())
dsu = DSU(N + M)
edges = []
for i in range(N):
line = next(input).strip()
for j, sign in enumerate(line):
j = j + N
if sign == '=':
dsu.merge(i, j)
else:
edges.append((i, j) if sign == '<' else (j, i))
incoming = collections.defaultdict(set)
outgoing = collections.defaultdict(set)
nodes = sorted(set([dsu.root(i) for i in range(N + M)]))
for i, j in edges:
incoming[dsu.root(i)].add(dsu.root(j))
outgoing[dsu.root(j)].add(dsu.root(i))
no_incoming = list(set(i for i in nodes if i not in incoming))
ordering = []
while no_incoming:
node = no_incoming.pop(-1)
ordering.append(node)
for n in outgoing[node]:
incoming[n].discard(node)
if not incoming[n]:
no_incoming.append(n)
if len(ordering) != len(nodes):
return 'NO'
root_to_index = {}
for n in reversed(ordering):
if not outgoing[n]:
root_to_index[n] = 0
else:
root_to_index[n] = max(root_to_index[x] for x in outgoing[n]) + 1
s = ['YES']
s.append(' '.join(str(root_to_index[dsu.root(i)] + 1) for i in range(N)))
s.append(' '.join(str(root_to_index[dsu.root(i + N)] + 1) for i in range(M)))
return '\n'.join(s)
class DSU():
def __init__(self, n):
self.parents = [i for i in range(n)]
def root(self, i):
try:
if self.parents[i] != i:
self.parents[i] = self.root(self.parents[i])
except:
print(i, len(self.parents))
raise
return self.parents[i]
def merge(self, i, j):
r1 = self.root(i)
r2 = self.root(j)
if r1 != r2:
self.parents[r1] = r2
if __name__ == '__main__':
print(main())
``` | output | 1 | 5,080 | 9 | 10,161 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr. Apple, a gourmet, works as editor-in-chief of a gastronomic periodical. He travels around the world, tasting new delights of famous chefs from the most fashionable restaurants. Mr. Apple has his own signature method of review — in each restaurant Mr. Apple orders two sets of dishes on two different days. All the dishes are different, because Mr. Apple doesn't like to eat the same food. For each pair of dishes from different days he remembers exactly which was better, or that they were of the same quality. After this the gourmet evaluates each dish with a positive integer.
Once, during a revision of a restaurant of Celtic medieval cuisine named «Poisson», that serves chestnut soup with fir, warm soda bread, spicy lemon pie and other folk food, Mr. Apple was very pleasantly surprised the gourmet with its variety of menu, and hence ordered too much. Now he's confused about evaluating dishes.
The gourmet tasted a set of n dishes on the first day and a set of m dishes on the second day. He made a table a of size n × m, in which he described his impressions. If, according to the expert, dish i from the first set was better than dish j from the second set, then a_{ij} is equal to ">", in the opposite case a_{ij} is equal to "<". Dishes also may be equally good, in this case a_{ij} is "=".
Now Mr. Apple wants you to help him to evaluate every dish. Since Mr. Apple is very strict, he will evaluate the dishes so that the maximal number used is as small as possible. But Mr. Apple also is very fair, so he never evaluates the dishes so that it goes against his feelings. In other words, if a_{ij} is "<", then the number assigned to dish i from the first set should be less than the number of dish j from the second set, if a_{ij} is ">", then it should be greater, and finally if a_{ij} is "=", then the numbers should be the same.
Help Mr. Apple to evaluate each dish from both sets so that it is consistent with his feelings, or determine that this is impossible.
Input
The first line contains integers n and m (1 ≤ n, m ≤ 1000) — the number of dishes in both days.
Each of the next n lines contains a string of m symbols. The j-th symbol on i-th line is a_{ij}. All strings consist only of "<", ">" and "=".
Output
The first line of output should contain "Yes", if it's possible to do a correct evaluation for all the dishes, or "No" otherwise.
If case an answer exist, on the second line print n integers — evaluations of dishes from the first set, and on the third line print m integers — evaluations of dishes from the second set.
Examples
Input
3 4
>>>>
>>>>
>>>>
Output
Yes
2 2 2
1 1 1 1
Input
3 3
>>>
<<<
>>>
Output
Yes
3 1 3
2 2 2
Input
3 2
==
=<
==
Output
No
Note
In the first sample, all dishes of the first day are better than dishes of the second day. So, the highest score will be 2, for all dishes of the first day.
In the third sample, the table is contradictory — there is no possible evaluation of the dishes that satisfies it. | instruction | 0 | 5,081 | 9 | 10,162 |
Tags: dfs and similar, dp, dsu, graphs, greedy
Correct Solution:
```
class UnionFind:
def __init__(self, n):
self.follow = [-1]*(n+1)
self.num_follower = [1]*(n+1)
def root_index_of(self, a):
r = a
while self.follow[r] > -1:
r = self.follow[r]
return r
def connected(self, a, b):
return self.root_index_of(a) == self.root_index_of(b)
def connect(self, a, b):
ra = self.root_index_of(a)
rb = self.root_index_of(b)
if ra == rb:
return
if self.num_follower[ra] < self.num_follower[rb]:
self.follow[ra] = rb
self.follow[a] = rb
self.num_follower[rb] += self.num_follower[ra]
else:
self.follow[rb] = ra
self.follow[b] = ra
self.num_follower[ra] += self.num_follower[rb]
import sys
from collections import deque
input = sys.stdin.readline
n, m = map(int, input().split())
s = [input() for _ in range(n)]
uf = UnionFind(n+m)
v = [[] for _ in range(n+m)]
c = [0]*(n+m)
for i in range(n):
for j in range(m):
if s[i][j] == "=":
uf.connect(i, j+n)
for i in range(n):
for j in range(m):
if s[i][j] == "<":
v[uf.root_index_of(i)].append(uf.root_index_of(n+j))
c[uf.root_index_of(n+j)] += 1
elif s[i][j] == ">":
v[uf.root_index_of(n+j)].append(uf.root_index_of(i))
c[uf.root_index_of(i)] += 1
ans = [0]*(n+m)
used = [0]*(n+m)
que = deque([])
for i in range(n+m):
if c[i] == 0:
ans[i] = 1
used[i] = 1
que.append(i)
while que:
x = que.popleft()
for i in v[x]:
if used[i]:
continue
ans[i] = max(ans[x]+1, ans[i])
c[i] -= 1
if c[i] == 0:
used[i] = 1
que.append(i)
for i in range(n+m):
if used[i] == 0:
print("No")
exit()
for i in range(n):
for j in range(m):
if s[i][j] == "=":
if ans[uf.root_index_of(i)] != ans[uf.root_index_of(n+j)]:
print("No")
exit()
for i in range(n):
for j in range(m):
if s[i][j] == "<":
if ans[uf.root_index_of(i)] >= ans[uf.root_index_of(n+j)]:
print("No")
exit()
for i in range(n):
for j in range(m):
if s[i][j] == ">":
if ans[uf.root_index_of(i)] <= ans[uf.root_index_of(n+j)]:
print("No")
exit()
print("Yes")
for i in range(n):
print(ans[uf.root_index_of(i)], end=" ")
print()
for j in range(m):
print(ans[uf.root_index_of(j+n)], end=" ")
print()
``` | output | 1 | 5,081 | 9 | 10,163 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr. Apple, a gourmet, works as editor-in-chief of a gastronomic periodical. He travels around the world, tasting new delights of famous chefs from the most fashionable restaurants. Mr. Apple has his own signature method of review — in each restaurant Mr. Apple orders two sets of dishes on two different days. All the dishes are different, because Mr. Apple doesn't like to eat the same food. For each pair of dishes from different days he remembers exactly which was better, or that they were of the same quality. After this the gourmet evaluates each dish with a positive integer.
Once, during a revision of a restaurant of Celtic medieval cuisine named «Poisson», that serves chestnut soup with fir, warm soda bread, spicy lemon pie and other folk food, Mr. Apple was very pleasantly surprised the gourmet with its variety of menu, and hence ordered too much. Now he's confused about evaluating dishes.
The gourmet tasted a set of n dishes on the first day and a set of m dishes on the second day. He made a table a of size n × m, in which he described his impressions. If, according to the expert, dish i from the first set was better than dish j from the second set, then a_{ij} is equal to ">", in the opposite case a_{ij} is equal to "<". Dishes also may be equally good, in this case a_{ij} is "=".
Now Mr. Apple wants you to help him to evaluate every dish. Since Mr. Apple is very strict, he will evaluate the dishes so that the maximal number used is as small as possible. But Mr. Apple also is very fair, so he never evaluates the dishes so that it goes against his feelings. In other words, if a_{ij} is "<", then the number assigned to dish i from the first set should be less than the number of dish j from the second set, if a_{ij} is ">", then it should be greater, and finally if a_{ij} is "=", then the numbers should be the same.
Help Mr. Apple to evaluate each dish from both sets so that it is consistent with his feelings, or determine that this is impossible.
Input
The first line contains integers n and m (1 ≤ n, m ≤ 1000) — the number of dishes in both days.
Each of the next n lines contains a string of m symbols. The j-th symbol on i-th line is a_{ij}. All strings consist only of "<", ">" and "=".
Output
The first line of output should contain "Yes", if it's possible to do a correct evaluation for all the dishes, or "No" otherwise.
If case an answer exist, on the second line print n integers — evaluations of dishes from the first set, and on the third line print m integers — evaluations of dishes from the second set.
Examples
Input
3 4
>>>>
>>>>
>>>>
Output
Yes
2 2 2
1 1 1 1
Input
3 3
>>>
<<<
>>>
Output
Yes
3 1 3
2 2 2
Input
3 2
==
=<
==
Output
No
Note
In the first sample, all dishes of the first day are better than dishes of the second day. So, the highest score will be 2, for all dishes of the first day.
In the third sample, the table is contradictory — there is no possible evaluation of the dishes that satisfies it. | instruction | 0 | 5,082 | 9 | 10,164 |
Tags: dfs and similar, dp, dsu, graphs, greedy
Correct Solution:
```
import time
debug = False
n1, m2 = map(int, input().split())
tests = []
for i in range(n1):
tests.append(list(input()))
if debug:
print (tests)
begin = time.time()
if debug:
print("---")
marks1 = []
result1 = []
for i in range(n1):
marks1.append([i,0.0])
result1.append(0)
marks2 = []
result2 = []
for j in range(m2):
marks2.append([j,0.0])
result2.append(0)
for i in range(n1):
for j in range(m2):
test = tests[i][j]
if test == ">":
marks1[i][1] += 1.0
elif test == "<":
marks2[j][1] += 1.0
else:
marks1[i][1] += 0.0001
marks2[j][1] += 0.0001
marks1.sort(key=lambda val: val[1])
marks2.sort(key=lambda val: val[1])
if debug:
print(marks1)
print(marks2)
i = 0
j = 0
value = 0
lastmark = -1
lastItem = [0,0]
while i < n1 or j < m2:
LetAdd = 0
if i < n1 and j < m2:
test = tests[marks1[i][0]][marks2[j][0]]
if test == ">":
LetAdd = 2
else:
LetAdd = 1
elif i < n1:
LetAdd = 1
else:
LetAdd = 2
if LetAdd == 1:
if marks1[i][1] != lastmark and lastItem[0] != 2 or lastItem[0] == 2 and tests[marks1[i][0]][lastItem[1]] != '=':
if debug:
if lastItem[0] == 2:
print(1, lastmark, lastItem, marks1[i][0], tests[marks1[i][0]][lastItem[1]])
else:
print(1, lastmark, lastItem, marks1[i][0])
value += 1
lastmark = marks1[i][1]
result1[marks1[i][0]] = value
lastItem = [1,marks1[i][0]]
i += 1
else:
if marks2[j][1] != lastmark and lastItem[0] != 1 or lastItem[0] == 1 and tests[lastItem[1]][marks2[j][0]] != '=':
if debug:
if lastItem[0] == 1:
print(2, lastmark, lastItem, marks2[j][0], tests[lastItem[1]][marks2[j][0]])
else:
print(2, lastmark, lastItem, marks2[j][0])
value += 1
lastmark = marks2[j][1]
result2[marks2[j][0]] = value
lastItem = [2,marks2[j][0]]
j += 1
if debug:
print("Set ", lastItem, " to ", value)
CheckCorrect = True
for i in range(n1):
for j in range(m2):
test = tests[i][j]
if test == ">":
if result1[i] <= result2[j]:
CheckCorrect = False
elif test == "<":
if result1[i] >= result2[j]:
CheckCorrect = False
else:
if result1[i] != result2[j]:
CheckCorrect = False
if debug:
print("---")
if debug:
print("Time: ", time.time() - begin)
if CheckCorrect:
print ("Yes")
else:
print ("No")
if CheckCorrect or debug:
print (*result1)
print (*result2)
``` | output | 1 | 5,082 | 9 | 10,165 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr. Apple, a gourmet, works as editor-in-chief of a gastronomic periodical. He travels around the world, tasting new delights of famous chefs from the most fashionable restaurants. Mr. Apple has his own signature method of review — in each restaurant Mr. Apple orders two sets of dishes on two different days. All the dishes are different, because Mr. Apple doesn't like to eat the same food. For each pair of dishes from different days he remembers exactly which was better, or that they were of the same quality. After this the gourmet evaluates each dish with a positive integer.
Once, during a revision of a restaurant of Celtic medieval cuisine named «Poisson», that serves chestnut soup with fir, warm soda bread, spicy lemon pie and other folk food, Mr. Apple was very pleasantly surprised the gourmet with its variety of menu, and hence ordered too much. Now he's confused about evaluating dishes.
The gourmet tasted a set of n dishes on the first day and a set of m dishes on the second day. He made a table a of size n × m, in which he described his impressions. If, according to the expert, dish i from the first set was better than dish j from the second set, then a_{ij} is equal to ">", in the opposite case a_{ij} is equal to "<". Dishes also may be equally good, in this case a_{ij} is "=".
Now Mr. Apple wants you to help him to evaluate every dish. Since Mr. Apple is very strict, he will evaluate the dishes so that the maximal number used is as small as possible. But Mr. Apple also is very fair, so he never evaluates the dishes so that it goes against his feelings. In other words, if a_{ij} is "<", then the number assigned to dish i from the first set should be less than the number of dish j from the second set, if a_{ij} is ">", then it should be greater, and finally if a_{ij} is "=", then the numbers should be the same.
Help Mr. Apple to evaluate each dish from both sets so that it is consistent with his feelings, or determine that this is impossible.
Input
The first line contains integers n and m (1 ≤ n, m ≤ 1000) — the number of dishes in both days.
Each of the next n lines contains a string of m symbols. The j-th symbol on i-th line is a_{ij}. All strings consist only of "<", ">" and "=".
Output
The first line of output should contain "Yes", if it's possible to do a correct evaluation for all the dishes, or "No" otherwise.
If case an answer exist, on the second line print n integers — evaluations of dishes from the first set, and on the third line print m integers — evaluations of dishes from the second set.
Examples
Input
3 4
>>>>
>>>>
>>>>
Output
Yes
2 2 2
1 1 1 1
Input
3 3
>>>
<<<
>>>
Output
Yes
3 1 3
2 2 2
Input
3 2
==
=<
==
Output
No
Note
In the first sample, all dishes of the first day are better than dishes of the second day. So, the highest score will be 2, for all dishes of the first day.
In the third sample, the table is contradictory — there is no possible evaluation of the dishes that satisfies it. | instruction | 0 | 5,083 | 9 | 10,166 |
Tags: dfs and similar, dp, dsu, graphs, greedy
Correct Solution:
```
def prov(mass, now):
check = True
for i in range(n):
for k in range(m):
if now[i][k] == '>' and mass[i] <= mass[n + k]:
check = False
break
elif now[i][k] == '<' and mass[i] >= mass[n + k]:
check = False
break
elif now[i][k] == '=' and mass[i] != mass[n + k]:
check = False
break
if not check:
break
return check
def prog(mass, n, m):
prov = True
for i in range(1, m):
for k in range(n):
if mass[i][k] < mass[i - 1][k]:
prov = False
break
if not prov:
break
if not prov:
return False
else:
mass_new = []
for i in range(1, m):
mass_n = []
for k in range(n):
mass_n.append(mass[i][k] - mass[i - 1][k])
mass_new.append(max(mass_n))
arr = [1 for i in range(m)]
now = 1
if 1 in mass[0][:-1]:
now += 1
arr = [2 for i in range(m)]
for i in range(1, m):
now += mass_new[i - 1]
arr[mass[i][-1]] = now
return arr
n, m = map(int, input().split())
if n + m <= 6:
now = []
for i in range(n):
now.append(input())
ppp = True
for i1 in range(n + m):
for i2 in range(n + m):
for i3 in range(n + m):
for i4 in range(n + m):
for i5 in range(n + m):
for i6 in range(n + m):
mass = [i1 + 1, i2 + 1, i3 + 1, i4 + 1, i5 + 1, i6 + 1][:n + m]
if prov(mass, now) and ppp:
print('Yes')
print(*mass[:n])
print(*mass[n:])
ppp = False
if ppp:
print('No')
else:
mass = [[] for i in range(m)]
mass1 = [[] for i in range(n)]
for i in range(n):
now = input()
for k in range(m):
if now[k] == '<':
mass[k].append(1)
mass1[i].append(-1)
elif now[k] == '=':
mass[k].append(0)
mass1[i].append(0)
else:
mass[k].append(-1)
mass1[i].append(1)
for i in range(m):
mass[i].append(i)
for i in range(n):
mass1[i].append(i)
mass.sort()
mass1.sort()
arr = prog(mass, n, m)
arr1 = prog(mass1, m, n)
if arr == False or arr1 == False:
print('No')
else:
print('Yes')
print(*arr1)
print(*arr)
``` | output | 1 | 5,083 | 9 | 10,167 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr. Apple, a gourmet, works as editor-in-chief of a gastronomic periodical. He travels around the world, tasting new delights of famous chefs from the most fashionable restaurants. Mr. Apple has his own signature method of review — in each restaurant Mr. Apple orders two sets of dishes on two different days. All the dishes are different, because Mr. Apple doesn't like to eat the same food. For each pair of dishes from different days he remembers exactly which was better, or that they were of the same quality. After this the gourmet evaluates each dish with a positive integer.
Once, during a revision of a restaurant of Celtic medieval cuisine named «Poisson», that serves chestnut soup with fir, warm soda bread, spicy lemon pie and other folk food, Mr. Apple was very pleasantly surprised the gourmet with its variety of menu, and hence ordered too much. Now he's confused about evaluating dishes.
The gourmet tasted a set of n dishes on the first day and a set of m dishes on the second day. He made a table a of size n × m, in which he described his impressions. If, according to the expert, dish i from the first set was better than dish j from the second set, then a_{ij} is equal to ">", in the opposite case a_{ij} is equal to "<". Dishes also may be equally good, in this case a_{ij} is "=".
Now Mr. Apple wants you to help him to evaluate every dish. Since Mr. Apple is very strict, he will evaluate the dishes so that the maximal number used is as small as possible. But Mr. Apple also is very fair, so he never evaluates the dishes so that it goes against his feelings. In other words, if a_{ij} is "<", then the number assigned to dish i from the first set should be less than the number of dish j from the second set, if a_{ij} is ">", then it should be greater, and finally if a_{ij} is "=", then the numbers should be the same.
Help Mr. Apple to evaluate each dish from both sets so that it is consistent with his feelings, or determine that this is impossible.
Input
The first line contains integers n and m (1 ≤ n, m ≤ 1000) — the number of dishes in both days.
Each of the next n lines contains a string of m symbols. The j-th symbol on i-th line is a_{ij}. All strings consist only of "<", ">" and "=".
Output
The first line of output should contain "Yes", if it's possible to do a correct evaluation for all the dishes, or "No" otherwise.
If case an answer exist, on the second line print n integers — evaluations of dishes from the first set, and on the third line print m integers — evaluations of dishes from the second set.
Examples
Input
3 4
>>>>
>>>>
>>>>
Output
Yes
2 2 2
1 1 1 1
Input
3 3
>>>
<<<
>>>
Output
Yes
3 1 3
2 2 2
Input
3 2
==
=<
==
Output
No
Note
In the first sample, all dishes of the first day are better than dishes of the second day. So, the highest score will be 2, for all dishes of the first day.
In the third sample, the table is contradictory — there is no possible evaluation of the dishes that satisfies it. | instruction | 0 | 5,084 | 9 | 10,168 |
Tags: dfs and similar, dp, dsu, graphs, greedy
Correct Solution:
```
n, m = (int(t) for t in input().split(' '))
def direction(c):
if c == '=': return 0
if c == '>': return 1
return -1
mx = [[direction(c) for c in input()] for _ in range(n)]
index = 0
class DSet(object):
def __init__(self, value):
global index
self.values = set([value])
self.index = index
index += 1
self.edges_to = set()
self.edges_from = set()
def __len__(self):
return len(self.values)
def update(self, other_dset):
self.values.update(other_dset.values)
def add_edge_to(self, i):
self.edges_to.add(i)
def add_edge_from(self, i):
self.edges_from.add(i)
def remove_edge_from(self, i):
self.edges_from.remove(i)
dsu = [DSet(i) for i in range(n+m)]
def union(v1, v2):
if len(dsu[v1]) > len(dsu[v2]):
d = dsu[v1]
d.update(dsu[v2])
else:
d = dsu[v2]
d.update(dsu[v1])
dsu[v1] = dsu[v2] = d
for i in range(n):
for j in range(m):
if not mx[i][j]:
union(i, n + j)
for i in range(n):
for j in range(m):
if mx[i][j] > 0:
dsu[i].add_edge_from(dsu[n + j].index)
dsu[n+j].add_edge_to(dsu[i].index)
elif mx[i][j] < 0:
dsu[n + j].add_edge_from(dsu[i].index)
dsu[i].add_edge_to(dsu[n+j].index)
weights = [1] * (n+m)
dsu_by_index = { d.index: d for d in dsu }
while True:
try:
v = next(d for d in range(n+m) if not len(dsu[d].edges_from) and len(dsu[d].edges_to))
except:
break
dsuv = dsu[v]
for edge in dsu[v].edges_to:
dsue = dsu_by_index[edge]
dsue.remove_edge_from(dsuv.index)
for value in dsue.values:
weights[value] = weights[v] + 1
dsu[v].edges_to.clear()
ok = all(not len(d.edges_from) and not len(d.edges_to) for d in dsu)
if ok:
print('Yes')
print(*weights[0:n])
print(*weights[n:])
else:
print('No')
``` | output | 1 | 5,084 | 9 | 10,169 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr. Apple, a gourmet, works as editor-in-chief of a gastronomic periodical. He travels around the world, tasting new delights of famous chefs from the most fashionable restaurants. Mr. Apple has his own signature method of review — in each restaurant Mr. Apple orders two sets of dishes on two different days. All the dishes are different, because Mr. Apple doesn't like to eat the same food. For each pair of dishes from different days he remembers exactly which was better, or that they were of the same quality. After this the gourmet evaluates each dish with a positive integer.
Once, during a revision of a restaurant of Celtic medieval cuisine named «Poisson», that serves chestnut soup with fir, warm soda bread, spicy lemon pie and other folk food, Mr. Apple was very pleasantly surprised the gourmet with its variety of menu, and hence ordered too much. Now he's confused about evaluating dishes.
The gourmet tasted a set of n dishes on the first day and a set of m dishes on the second day. He made a table a of size n × m, in which he described his impressions. If, according to the expert, dish i from the first set was better than dish j from the second set, then a_{ij} is equal to ">", in the opposite case a_{ij} is equal to "<". Dishes also may be equally good, in this case a_{ij} is "=".
Now Mr. Apple wants you to help him to evaluate every dish. Since Mr. Apple is very strict, he will evaluate the dishes so that the maximal number used is as small as possible. But Mr. Apple also is very fair, so he never evaluates the dishes so that it goes against his feelings. In other words, if a_{ij} is "<", then the number assigned to dish i from the first set should be less than the number of dish j from the second set, if a_{ij} is ">", then it should be greater, and finally if a_{ij} is "=", then the numbers should be the same.
Help Mr. Apple to evaluate each dish from both sets so that it is consistent with his feelings, or determine that this is impossible.
Input
The first line contains integers n and m (1 ≤ n, m ≤ 1000) — the number of dishes in both days.
Each of the next n lines contains a string of m symbols. The j-th symbol on i-th line is a_{ij}. All strings consist only of "<", ">" and "=".
Output
The first line of output should contain "Yes", if it's possible to do a correct evaluation for all the dishes, or "No" otherwise.
If case an answer exist, on the second line print n integers — evaluations of dishes from the first set, and on the third line print m integers — evaluations of dishes from the second set.
Examples
Input
3 4
>>>>
>>>>
>>>>
Output
Yes
2 2 2
1 1 1 1
Input
3 3
>>>
<<<
>>>
Output
Yes
3 1 3
2 2 2
Input
3 2
==
=<
==
Output
No
Note
In the first sample, all dishes of the first day are better than dishes of the second day. So, the highest score will be 2, for all dishes of the first day.
In the third sample, the table is contradictory — there is no possible evaluation of the dishes that satisfies it. | instruction | 0 | 5,085 | 9 | 10,170 |
Tags: dfs and similar, dp, dsu, graphs, greedy
Correct Solution:
```
#!/usr/bin/env pypy
import collections
#import random
import heapq
import bisect
import math
import time
class Solution2:
def solve(self, A1, A2):
pass
def gcd(a, b):
if not b: return a
return gcd(b, a%b)
def lcm(a, b):
return b*a//gcd(b,a)
class Solution:
def solve(self, grid):
def union(i,j):
leader_i, leader_j = find(i), find(j)
sets[leader_j] = sets[i] = sets[j] = leader_i
def find(i):
while i != sets[i]:
i = sets[i]
return i
N = len(grid) + len(grid[0])
sets = list(range(N))
for i in range(len(grid)):
for j in range(len(grid[0])):
if grid[i][j] == '=': union(i,j+len(grid))
graph = collections.defaultdict(set)
inc = collections.defaultdict(set)
for i in range(len(grid)):
for j in range(len(grid[0])):
leader_i, leader_j = find(i), find(j+len(grid))
if grid[i][j] == '>':
if leader_i == leader_j:
print("No")
return
graph[leader_j].add(leader_i)
inc[leader_i].add(leader_j)
elif grid[i][j] == '<':
if leader_i == leader_j:
print("No")
return
graph[leader_i].add(leader_j)
inc[leader_j].add(leader_i)
self.levels = [0]*N
def dfs(node, level):
self.levels[node] = max(self.levels[node],level)
if not inc[node]:
seen.add(node)
for next_node in graph[node]:
inc[next_node].discard(node)
dfs(next_node,self.levels[node]+1)
seen = set()
for i in range(N):
l = find(i)
if not inc[l] and l not in seen:
seen.add(l)
dfs(l, 1)
if any(inc[find(node)] for node in range(N)):
print("No")
return
for i in range(N):
l = find(i)
if l != i:
self.levels[i] = self.levels[l]
print("Yes")
print(' '.join(str(o) for o in self.levels[:len(grid)]))
print(' '.join(str(o) for o in self.levels[len(grid):]))
sol = Solution()
sol2 = Solution2()
#TT = int(input())
for test_case in range(1):
N, M = input().split()
a = []
for _ in range(int(N)):
a.append(input())
#b = [int(c) for c in input().split()]
out = sol.solve(a)
#print(' '.join([str(o) for o in out]))
#print(str(out))
# out2 = sol2.solve(s)
# for _ in range(100000):
# rand = [random.randrange(60) for _ in range(10)]
# out1 = sol.solve(rand)
# out2 = sol2.solve(rand)
# if out1 != out2:
# print(rand, out1, out2)
# break
``` | output | 1 | 5,085 | 9 | 10,171 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr. Apple, a gourmet, works as editor-in-chief of a gastronomic periodical. He travels around the world, tasting new delights of famous chefs from the most fashionable restaurants. Mr. Apple has his own signature method of review — in each restaurant Mr. Apple orders two sets of dishes on two different days. All the dishes are different, because Mr. Apple doesn't like to eat the same food. For each pair of dishes from different days he remembers exactly which was better, or that they were of the same quality. After this the gourmet evaluates each dish with a positive integer.
Once, during a revision of a restaurant of Celtic medieval cuisine named «Poisson», that serves chestnut soup with fir, warm soda bread, spicy lemon pie and other folk food, Mr. Apple was very pleasantly surprised the gourmet with its variety of menu, and hence ordered too much. Now he's confused about evaluating dishes.
The gourmet tasted a set of n dishes on the first day and a set of m dishes on the second day. He made a table a of size n × m, in which he described his impressions. If, according to the expert, dish i from the first set was better than dish j from the second set, then a_{ij} is equal to ">", in the opposite case a_{ij} is equal to "<". Dishes also may be equally good, in this case a_{ij} is "=".
Now Mr. Apple wants you to help him to evaluate every dish. Since Mr. Apple is very strict, he will evaluate the dishes so that the maximal number used is as small as possible. But Mr. Apple also is very fair, so he never evaluates the dishes so that it goes against his feelings. In other words, if a_{ij} is "<", then the number assigned to dish i from the first set should be less than the number of dish j from the second set, if a_{ij} is ">", then it should be greater, and finally if a_{ij} is "=", then the numbers should be the same.
Help Mr. Apple to evaluate each dish from both sets so that it is consistent with his feelings, or determine that this is impossible.
Input
The first line contains integers n and m (1 ≤ n, m ≤ 1000) — the number of dishes in both days.
Each of the next n lines contains a string of m symbols. The j-th symbol on i-th line is a_{ij}. All strings consist only of "<", ">" and "=".
Output
The first line of output should contain "Yes", if it's possible to do a correct evaluation for all the dishes, or "No" otherwise.
If case an answer exist, on the second line print n integers — evaluations of dishes from the first set, and on the third line print m integers — evaluations of dishes from the second set.
Examples
Input
3 4
>>>>
>>>>
>>>>
Output
Yes
2 2 2
1 1 1 1
Input
3 3
>>>
<<<
>>>
Output
Yes
3 1 3
2 2 2
Input
3 2
==
=<
==
Output
No
Note
In the first sample, all dishes of the first day are better than dishes of the second day. So, the highest score will be 2, for all dishes of the first day.
In the third sample, the table is contradictory — there is no possible evaluation of the dishes that satisfies it. | instruction | 0 | 5,086 | 9 | 10,172 |
Tags: dfs and similar, dp, dsu, graphs, greedy
Correct Solution:
```
n, m = map(int, input().split())
dishes = [0 for _ in range(n + m)]
father = [-1 for _ in range(n + m)]
e_out = dict()
v_in = [0 for _ in range(n + m)]
def get_father(n):
if father[n] == -1:
return n
else:
father[n] = get_father(father[n])
return father[n]
compare_matrix = []
for i in range(n):
compare_matrix.append(input())
for i in range(n):
for j in range(m):
if compare_matrix[i][j] == "=":
fi = get_father(i)
fj = get_father(j + n)
if fi != fj:
father[fj] = fi
children = dict()
for i in range(n + m):
fi = get_father(i)
if fi != i:
if fi not in children:
children[fi] = [i]
else:
children[fi].append(i)
for i in range(n):
for j in range(m):
if compare_matrix[i][j] == "=":
continue
fi = get_father(i)
fj = get_father(j + n)
if fi == fj:
print("NO")
exit(0)
if compare_matrix[i][j] == ">":
v_in[fi] += 1
if fj in e_out:
e_out[fj].append(fi)
else:
e_out[fj] = [fi]
if compare_matrix[i][j] == "<":
v_in[fj] += 1
if fi in e_out:
e_out[fi].append(fj)
else:
e_out[fi] = [fj]
# print(v_in)
# print(e_out)
score = 1
visited = [False for _ in range(n + m)]
v_total = 0
q = [v for v in range(n + m) if v_in[v] == 0]
while q:
t = []
for i in q:
dishes[i] = score
v_total += 1
if i in children:
for j in children[i]:
dishes[j] = score
v_total += 1
if i in e_out:
for j in e_out[i]:
v_in[j] -= 1
if v_in[j] == 0:
t.append(j)
q = t
score += 1
if v_total < n + m:
print("NO")
exit(0)
print("YES")
for i in dishes[:n]:
print(i, end=" ")
print()
for i in dishes[n:n + m]:
print(i, end=" ")
``` | output | 1 | 5,086 | 9 | 10,173 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr. Apple, a gourmet, works as editor-in-chief of a gastronomic periodical. He travels around the world, tasting new delights of famous chefs from the most fashionable restaurants. Mr. Apple has his own signature method of review — in each restaurant Mr. Apple orders two sets of dishes on two different days. All the dishes are different, because Mr. Apple doesn't like to eat the same food. For each pair of dishes from different days he remembers exactly which was better, or that they were of the same quality. After this the gourmet evaluates each dish with a positive integer.
Once, during a revision of a restaurant of Celtic medieval cuisine named «Poisson», that serves chestnut soup with fir, warm soda bread, spicy lemon pie and other folk food, Mr. Apple was very pleasantly surprised the gourmet with its variety of menu, and hence ordered too much. Now he's confused about evaluating dishes.
The gourmet tasted a set of n dishes on the first day and a set of m dishes on the second day. He made a table a of size n × m, in which he described his impressions. If, according to the expert, dish i from the first set was better than dish j from the second set, then a_{ij} is equal to ">", in the opposite case a_{ij} is equal to "<". Dishes also may be equally good, in this case a_{ij} is "=".
Now Mr. Apple wants you to help him to evaluate every dish. Since Mr. Apple is very strict, he will evaluate the dishes so that the maximal number used is as small as possible. But Mr. Apple also is very fair, so he never evaluates the dishes so that it goes against his feelings. In other words, if a_{ij} is "<", then the number assigned to dish i from the first set should be less than the number of dish j from the second set, if a_{ij} is ">", then it should be greater, and finally if a_{ij} is "=", then the numbers should be the same.
Help Mr. Apple to evaluate each dish from both sets so that it is consistent with his feelings, or determine that this is impossible.
Input
The first line contains integers n and m (1 ≤ n, m ≤ 1000) — the number of dishes in both days.
Each of the next n lines contains a string of m symbols. The j-th symbol on i-th line is a_{ij}. All strings consist only of "<", ">" and "=".
Output
The first line of output should contain "Yes", if it's possible to do a correct evaluation for all the dishes, or "No" otherwise.
If case an answer exist, on the second line print n integers — evaluations of dishes from the first set, and on the third line print m integers — evaluations of dishes from the second set.
Examples
Input
3 4
>>>>
>>>>
>>>>
Output
Yes
2 2 2
1 1 1 1
Input
3 3
>>>
<<<
>>>
Output
Yes
3 1 3
2 2 2
Input
3 2
==
=<
==
Output
No
Note
In the first sample, all dishes of the first day are better than dishes of the second day. So, the highest score will be 2, for all dishes of the first day.
In the third sample, the table is contradictory — there is no possible evaluation of the dishes that satisfies it. | instruction | 0 | 5,087 | 9 | 10,174 |
Tags: dfs and similar, dp, dsu, graphs, greedy
Correct Solution:
```
"""Codeforces 1131D
"""
from sys import *
from typing import *
from collections import *
r, c = [int(i) for i in stdin.readline().strip().split(' ')]
a = [stdin.readline().strip() for i in range(r)]
def L(i: int) -> int:
return i
def R(i: int) -> int:
return r + i
parent = [0] * (r + c)
for i in range(r):
parent[L(i)] = L(i)
for i in range(c):
parent[R(i)] = R(i)
def find(i: int) -> int:
parent[i] = i if parent[i] == i else find(parent[i])
return parent[i]
def union(i: int, j: int):
parent[find(i)] = find(j)
for i in range(r):
for j in range(c):
if a[i][j] == '=':
union(L(i), R(j))
less_than = [[] for i in range(r + c)]
for i in range(r):
for j in range(c):
if a[i][j] == '<':
less_than[find(R(j))].append(find(L(i)))
elif a[i][j] == '>':
less_than[find(L(i))].append(find(R(j)))
score = {}
color = {}
no_solution = set()
def f(i: int) -> int:
if i not in color:
color[i] = 'gray'
elif color[i] == 'gray':
no_solution.add(True)
else:
return score[i]
if no_solution:
return 0
score[i] = 1 + max([f(j) for j in less_than[i]], default=0)
color[i] = 'black'
return score[i]
ans1 = [f(find(L(i))) for i in range(r)]
ans2 = [f(find(R(i))) for i in range(c)]
if no_solution:
print('No')
else:
print('Yes')
print(*ans1)
print(*ans2)
``` | output | 1 | 5,087 | 9 | 10,175 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Apple, a gourmet, works as editor-in-chief of a gastronomic periodical. He travels around the world, tasting new delights of famous chefs from the most fashionable restaurants. Mr. Apple has his own signature method of review — in each restaurant Mr. Apple orders two sets of dishes on two different days. All the dishes are different, because Mr. Apple doesn't like to eat the same food. For each pair of dishes from different days he remembers exactly which was better, or that they were of the same quality. After this the gourmet evaluates each dish with a positive integer.
Once, during a revision of a restaurant of Celtic medieval cuisine named «Poisson», that serves chestnut soup with fir, warm soda bread, spicy lemon pie and other folk food, Mr. Apple was very pleasantly surprised the gourmet with its variety of menu, and hence ordered too much. Now he's confused about evaluating dishes.
The gourmet tasted a set of n dishes on the first day and a set of m dishes on the second day. He made a table a of size n × m, in which he described his impressions. If, according to the expert, dish i from the first set was better than dish j from the second set, then a_{ij} is equal to ">", in the opposite case a_{ij} is equal to "<". Dishes also may be equally good, in this case a_{ij} is "=".
Now Mr. Apple wants you to help him to evaluate every dish. Since Mr. Apple is very strict, he will evaluate the dishes so that the maximal number used is as small as possible. But Mr. Apple also is very fair, so he never evaluates the dishes so that it goes against his feelings. In other words, if a_{ij} is "<", then the number assigned to dish i from the first set should be less than the number of dish j from the second set, if a_{ij} is ">", then it should be greater, and finally if a_{ij} is "=", then the numbers should be the same.
Help Mr. Apple to evaluate each dish from both sets so that it is consistent with his feelings, or determine that this is impossible.
Input
The first line contains integers n and m (1 ≤ n, m ≤ 1000) — the number of dishes in both days.
Each of the next n lines contains a string of m symbols. The j-th symbol on i-th line is a_{ij}. All strings consist only of "<", ">" and "=".
Output
The first line of output should contain "Yes", if it's possible to do a correct evaluation for all the dishes, or "No" otherwise.
If case an answer exist, on the second line print n integers — evaluations of dishes from the first set, and on the third line print m integers — evaluations of dishes from the second set.
Examples
Input
3 4
>>>>
>>>>
>>>>
Output
Yes
2 2 2
1 1 1 1
Input
3 3
>>>
<<<
>>>
Output
Yes
3 1 3
2 2 2
Input
3 2
==
=<
==
Output
No
Note
In the first sample, all dishes of the first day are better than dishes of the second day. So, the highest score will be 2, for all dishes of the first day.
In the third sample, the table is contradictory — there is no possible evaluation of the dishes that satisfies it.
Submitted Solution:
```
import os
from io import BytesIO
#input = BytesIO(os.read(0, os.fstat(0).st_size)).readline
from itertools import product
class CYCLE(Exception):
...
debug_print = lambda *args: None
def solve(n, m, gs):
def union(u, v):
p, q = find(u), find(v)
if p == q: return
if rank[p] < rank[q]: p, q = q, p
if rank[p] == rank[q]: rank[p] += 1
parent[q] = p
def find(u):
if parent[u] == u: return u
else:
p = find(parent[u])
parent[u] = p
return p
rank = [0] * (n + m)
parent = list(range(n + m))
for i, j in product(range(n), range(m)):
if gs[i][j] == "=":
union(i, n + j)
vertices = set(parent)
v_sz = len(vertices)
g = [set() for _ in range(n + m)]
for i, j in product(range(n), range(m)):
c = gs[i][j]
i, j = parent[i], parent[n + j]
if c == "<":
g[i].add(j)
elif c == ">":
g[j].add(i)
debug_print(vertices, g)
NIL, VISITED, FINAL = 0, 1, 2
state = [NIL] * (n + m)
toposort_stack = []
def visit(v):
debug_print(v)
if state[v] == VISITED:
raise CYCLE
state[v] = VISITED
for u in g[v]:
if state[u] != FINAL:
visit(u)
state[v] = FINAL
toposort_stack.append(v)
try:
for v in vertices:
if state[v] == FINAL:
continue
visit(v)
except CYCLE:
print('No')
return
debug_print(toposort_stack)
layer = [1] * (n + m)
while toposort_stack:
v = toposort_stack.pop()
for u in g[v]:
layer[u] = max(layer[u], layer[v]+1)
print('Yes')
out = []
for i in range(n):
out.append(str(layer[parent[i]]))
print(' '.join(out))
out = []
for j in range(m):
out.append(str(layer[parent[n + j]]))
print(' '.join(out))
def solve_from_stdin():
n, m = map(int, input().split())
gs = []
for _ in range(n):
gs.append(input())
solve(n, m, gs)
solve_from_stdin()
``` | instruction | 0 | 5,088 | 9 | 10,176 |
Yes | output | 1 | 5,088 | 9 | 10,177 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Apple, a gourmet, works as editor-in-chief of a gastronomic periodical. He travels around the world, tasting new delights of famous chefs from the most fashionable restaurants. Mr. Apple has his own signature method of review — in each restaurant Mr. Apple orders two sets of dishes on two different days. All the dishes are different, because Mr. Apple doesn't like to eat the same food. For each pair of dishes from different days he remembers exactly which was better, or that they were of the same quality. After this the gourmet evaluates each dish with a positive integer.
Once, during a revision of a restaurant of Celtic medieval cuisine named «Poisson», that serves chestnut soup with fir, warm soda bread, spicy lemon pie and other folk food, Mr. Apple was very pleasantly surprised the gourmet with its variety of menu, and hence ordered too much. Now he's confused about evaluating dishes.
The gourmet tasted a set of n dishes on the first day and a set of m dishes on the second day. He made a table a of size n × m, in which he described his impressions. If, according to the expert, dish i from the first set was better than dish j from the second set, then a_{ij} is equal to ">", in the opposite case a_{ij} is equal to "<". Dishes also may be equally good, in this case a_{ij} is "=".
Now Mr. Apple wants you to help him to evaluate every dish. Since Mr. Apple is very strict, he will evaluate the dishes so that the maximal number used is as small as possible. But Mr. Apple also is very fair, so he never evaluates the dishes so that it goes against his feelings. In other words, if a_{ij} is "<", then the number assigned to dish i from the first set should be less than the number of dish j from the second set, if a_{ij} is ">", then it should be greater, and finally if a_{ij} is "=", then the numbers should be the same.
Help Mr. Apple to evaluate each dish from both sets so that it is consistent with his feelings, or determine that this is impossible.
Input
The first line contains integers n and m (1 ≤ n, m ≤ 1000) — the number of dishes in both days.
Each of the next n lines contains a string of m symbols. The j-th symbol on i-th line is a_{ij}. All strings consist only of "<", ">" and "=".
Output
The first line of output should contain "Yes", if it's possible to do a correct evaluation for all the dishes, or "No" otherwise.
If case an answer exist, on the second line print n integers — evaluations of dishes from the first set, and on the third line print m integers — evaluations of dishes from the second set.
Examples
Input
3 4
>>>>
>>>>
>>>>
Output
Yes
2 2 2
1 1 1 1
Input
3 3
>>>
<<<
>>>
Output
Yes
3 1 3
2 2 2
Input
3 2
==
=<
==
Output
No
Note
In the first sample, all dishes of the first day are better than dishes of the second day. So, the highest score will be 2, for all dishes of the first day.
In the third sample, the table is contradictory — there is no possible evaluation of the dishes that satisfies it.
Submitted Solution:
```
class mergefind:
def __init__(self,n):
self.parent = list(range(n))
self.size = [1]*n
self.num_sets = n
def find(self,a):
to_update = []
while a != self.parent[a]:
to_update.append(a)
a = self.parent[a]
for b in to_update:
self.parent[b] = a
return self.parent[a]
def merge(self,a,b):
a = self.find(a)
b = self.find(b)
if a==b:
return
if self.size[a]<self.size[b]:
a,b = b,a
self.num_sets -= 1
self.parent[b] = a
self.size[a] += self.size[b]
def set_size(self, a):
return self.size[self.find(a)]
def __len__(self):
return self.num_sets
def toposort(C, n):
indeg = [0]*n
for i,neighs in enumerate(C):
for neigh in neighs:
indeg[neigh] += 1
S = [i for i in range(n) if indeg[i] == 0]
nparent = indeg[:]
topo = []
while S:
cur = S.pop()
topo.append(cur)
for neigh in C[cur]:
nparent[neigh] -= 1
if nparent[neigh] == 0:
S.append(neigh)
nparent[cur] = -1
return topo
n,m = map(int,input().split())
A = [input() for _ in range(n)]
mf = mergefind(n+m)
# merge equal elements
for i in range(n):
for j in range(m):
if A[i][j] == '=':
mf.merge(i,n+j)
# Connections: smaller -> larger
C = [set() for _ in range(n+m)]
for i in range(n):
for j in range(m):
if A[i][j] == '<':
C[mf.find(i)].add(mf.find(n+j))
elif A[i][j] == '>':
C[mf.find(n+j)].add(mf.find(i))
# Walk through graph in toposort order
# What I'm pointing to must be at least
# my value + 1
D = [1]*(n+m)
for cur in toposort(C, n+m):
for neigh in C[cur]:
D[neigh] = max(D[neigh], D[cur]+1)
# Propagate values within equal clusters
D = [D[mf.find(i)] for i in range(n+m)]
# Validate answer
ok = True
for i in range(n):
for j in range(m):
if A[i][j] == '<':
if D[i] >= D[n+j]:
ok = False
elif A[i][j] == '>':
if D[i] <= D[n+j]:
ok = False
else:
if D[i] != D[n+j]:
ok = False
if ok:
print('Yes')
print(*D[:n])
print(*D[n:])
else:
print('No')
``` | instruction | 0 | 5,089 | 9 | 10,178 |
Yes | output | 1 | 5,089 | 9 | 10,179 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Apple, a gourmet, works as editor-in-chief of a gastronomic periodical. He travels around the world, tasting new delights of famous chefs from the most fashionable restaurants. Mr. Apple has his own signature method of review — in each restaurant Mr. Apple orders two sets of dishes on two different days. All the dishes are different, because Mr. Apple doesn't like to eat the same food. For each pair of dishes from different days he remembers exactly which was better, or that they were of the same quality. After this the gourmet evaluates each dish with a positive integer.
Once, during a revision of a restaurant of Celtic medieval cuisine named «Poisson», that serves chestnut soup with fir, warm soda bread, spicy lemon pie and other folk food, Mr. Apple was very pleasantly surprised the gourmet with its variety of menu, and hence ordered too much. Now he's confused about evaluating dishes.
The gourmet tasted a set of n dishes on the first day and a set of m dishes on the second day. He made a table a of size n × m, in which he described his impressions. If, according to the expert, dish i from the first set was better than dish j from the second set, then a_{ij} is equal to ">", in the opposite case a_{ij} is equal to "<". Dishes also may be equally good, in this case a_{ij} is "=".
Now Mr. Apple wants you to help him to evaluate every dish. Since Mr. Apple is very strict, he will evaluate the dishes so that the maximal number used is as small as possible. But Mr. Apple also is very fair, so he never evaluates the dishes so that it goes against his feelings. In other words, if a_{ij} is "<", then the number assigned to dish i from the first set should be less than the number of dish j from the second set, if a_{ij} is ">", then it should be greater, and finally if a_{ij} is "=", then the numbers should be the same.
Help Mr. Apple to evaluate each dish from both sets so that it is consistent with his feelings, or determine that this is impossible.
Input
The first line contains integers n and m (1 ≤ n, m ≤ 1000) — the number of dishes in both days.
Each of the next n lines contains a string of m symbols. The j-th symbol on i-th line is a_{ij}. All strings consist only of "<", ">" and "=".
Output
The first line of output should contain "Yes", if it's possible to do a correct evaluation for all the dishes, or "No" otherwise.
If case an answer exist, on the second line print n integers — evaluations of dishes from the first set, and on the third line print m integers — evaluations of dishes from the second set.
Examples
Input
3 4
>>>>
>>>>
>>>>
Output
Yes
2 2 2
1 1 1 1
Input
3 3
>>>
<<<
>>>
Output
Yes
3 1 3
2 2 2
Input
3 2
==
=<
==
Output
No
Note
In the first sample, all dishes of the first day are better than dishes of the second day. So, the highest score will be 2, for all dishes of the first day.
In the third sample, the table is contradictory — there is no possible evaluation of the dishes that satisfies it.
Submitted Solution:
```
import sys
sys.setrecursionlimit(2000)
def dfs1(v, mintime):
localtime = mintime
vis1[v] = 1
for v2 in range(m):
if a[v][v2] == ">":
if not vis2[v2]:
dfs2(v2, 1)
localtime = max(localtime, time2[v2] + 1)
for v2 in range(m):
if a[v][v2] == "=":
if not vis2[v2]:
dfs2(v2, localtime)
localtime = max(localtime, time2[v2])
time1[v] = localtime
def dfs2(v, mintime):
localtime = mintime
vis2[v] = 1
for v2 in range(n):
if a[v2][v] == "<":
if not vis1[v2]:
dfs1(v2, 1)
localtime = max(localtime, time1[v2] + 1)
for v2 in range(n):
if a[v2][v] == "=":
if not vis1[v2]:
dfs1(v2, localtime)
localtime = max(localtime, time1[v2])
time2[v] = localtime
n, m = map(int, input().split())
a = [input() for i in range(n)]
time1 = [0] * n
time2 = [0] * m
vis1 = [0] * n
vis2 = [0] * m
time = 0
try:
for v in range(n):
if not time1[v]:
dfs1(v, 1)
for v in range(m):
if not time2[v]:
dfs2(v, 1)
correct = True
for v1 in range(n):
for v2 in range(m):
if a[v1][v2] == "=" and time1[v1] != time2[v2]:
correct = False
if a[v1][v2] == ">" and time1[v1] <= time2[v2]:
correct = False
if a[v1][v2] == "<" and time1[v1] >= time2[v2]:
correct = False
if correct:
print("Yes")
print(*time1)
print(*time2)
else:
print("No")
except RecursionError:
print("No")
``` | instruction | 0 | 5,090 | 9 | 10,180 |
Yes | output | 1 | 5,090 | 9 | 10,181 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Apple, a gourmet, works as editor-in-chief of a gastronomic periodical. He travels around the world, tasting new delights of famous chefs from the most fashionable restaurants. Mr. Apple has his own signature method of review — in each restaurant Mr. Apple orders two sets of dishes on two different days. All the dishes are different, because Mr. Apple doesn't like to eat the same food. For each pair of dishes from different days he remembers exactly which was better, or that they were of the same quality. After this the gourmet evaluates each dish with a positive integer.
Once, during a revision of a restaurant of Celtic medieval cuisine named «Poisson», that serves chestnut soup with fir, warm soda bread, spicy lemon pie and other folk food, Mr. Apple was very pleasantly surprised the gourmet with its variety of menu, and hence ordered too much. Now he's confused about evaluating dishes.
The gourmet tasted a set of n dishes on the first day and a set of m dishes on the second day. He made a table a of size n × m, in which he described his impressions. If, according to the expert, dish i from the first set was better than dish j from the second set, then a_{ij} is equal to ">", in the opposite case a_{ij} is equal to "<". Dishes also may be equally good, in this case a_{ij} is "=".
Now Mr. Apple wants you to help him to evaluate every dish. Since Mr. Apple is very strict, he will evaluate the dishes so that the maximal number used is as small as possible. But Mr. Apple also is very fair, so he never evaluates the dishes so that it goes against his feelings. In other words, if a_{ij} is "<", then the number assigned to dish i from the first set should be less than the number of dish j from the second set, if a_{ij} is ">", then it should be greater, and finally if a_{ij} is "=", then the numbers should be the same.
Help Mr. Apple to evaluate each dish from both sets so that it is consistent with his feelings, or determine that this is impossible.
Input
The first line contains integers n and m (1 ≤ n, m ≤ 1000) — the number of dishes in both days.
Each of the next n lines contains a string of m symbols. The j-th symbol on i-th line is a_{ij}. All strings consist only of "<", ">" and "=".
Output
The first line of output should contain "Yes", if it's possible to do a correct evaluation for all the dishes, or "No" otherwise.
If case an answer exist, on the second line print n integers — evaluations of dishes from the first set, and on the third line print m integers — evaluations of dishes from the second set.
Examples
Input
3 4
>>>>
>>>>
>>>>
Output
Yes
2 2 2
1 1 1 1
Input
3 3
>>>
<<<
>>>
Output
Yes
3 1 3
2 2 2
Input
3 2
==
=<
==
Output
No
Note
In the first sample, all dishes of the first day are better than dishes of the second day. So, the highest score will be 2, for all dishes of the first day.
In the third sample, the table is contradictory — there is no possible evaluation of the dishes that satisfies it.
Submitted Solution:
```
class mergefind:
def __init__(self,n):
self.parent = list(range(n))
self.size = [1]*n
self.num_sets = n
def find(self,a):
to_update = []
while a != self.parent[a]:
to_update.append(a)
a = self.parent[a]
for b in to_update:
self.parent[b] = a
return self.parent[a]
def merge(self,a,b):
a = self.find(a)
b = self.find(b)
if a==b:
return
if self.size[a]<self.size[b]:
a,b = b,a
self.num_sets -= 1
self.parent[b] = a
self.size[a] += self.size[b]
def set_size(self, a):
return self.size[self.find(a)]
def __len__(self):
return self.num_sets
def toposort(C, n):
indeg = [0]*n
for i,neighs in enumerate(C):
for neigh in neighs:
indeg[neigh] += 1
S = [i for i in range(n) if indeg[i] == 0]
nparent = indeg[:]
topo = []
while S:
cur = S.pop()
topo.append(cur)
for neigh in C[cur]:
nparent[neigh] -= 1
if nparent[neigh] == 0:
S.append(neigh)
# nparent[cur] = -1
return topo
n,m = map(int,input().split())
A = [input() for _ in range(n)]
mf = mergefind(n+m)
# merge equal elements
for i in range(n):
for j in range(m):
if A[i][j] == '=':
mf.merge(i,n+j)
# Connections: smaller -> larger
C = [set() for _ in range(n+m)]
for i in range(n):
for j in range(m):
if A[i][j] == '<':
C[mf.find(i)].add(mf.find(n+j))
elif A[i][j] == '>':
C[mf.find(n+j)].add(mf.find(i))
# Walk through graph in toposort order
# What I'm pointing to must be at least
# my value + 1
D = [1]*(n+m)
for cur in toposort(C, n+m):
for neigh in C[cur]:
D[neigh] = max(D[neigh], D[cur]+1)
# Propagate values within equal clusters
D = [D[mf.find(i)] for i in range(n+m)]
# Validate answer
ok = True
for i in range(n):
for j in range(m):
if A[i][j] == '<':
if D[i] >= D[n+j]:
ok = False
elif A[i][j] == '>':
if D[i] <= D[n+j]:
ok = False
else:
if D[i] != D[n+j]:
ok = False
if ok:
print('Yes')
print(*D[:n])
print(*D[n:])
else:
print('No')
``` | instruction | 0 | 5,091 | 9 | 10,182 |
Yes | output | 1 | 5,091 | 9 | 10,183 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Apple, a gourmet, works as editor-in-chief of a gastronomic periodical. He travels around the world, tasting new delights of famous chefs from the most fashionable restaurants. Mr. Apple has his own signature method of review — in each restaurant Mr. Apple orders two sets of dishes on two different days. All the dishes are different, because Mr. Apple doesn't like to eat the same food. For each pair of dishes from different days he remembers exactly which was better, or that they were of the same quality. After this the gourmet evaluates each dish with a positive integer.
Once, during a revision of a restaurant of Celtic medieval cuisine named «Poisson», that serves chestnut soup with fir, warm soda bread, spicy lemon pie and other folk food, Mr. Apple was very pleasantly surprised the gourmet with its variety of menu, and hence ordered too much. Now he's confused about evaluating dishes.
The gourmet tasted a set of n dishes on the first day and a set of m dishes on the second day. He made a table a of size n × m, in which he described his impressions. If, according to the expert, dish i from the first set was better than dish j from the second set, then a_{ij} is equal to ">", in the opposite case a_{ij} is equal to "<". Dishes also may be equally good, in this case a_{ij} is "=".
Now Mr. Apple wants you to help him to evaluate every dish. Since Mr. Apple is very strict, he will evaluate the dishes so that the maximal number used is as small as possible. But Mr. Apple also is very fair, so he never evaluates the dishes so that it goes against his feelings. In other words, if a_{ij} is "<", then the number assigned to dish i from the first set should be less than the number of dish j from the second set, if a_{ij} is ">", then it should be greater, and finally if a_{ij} is "=", then the numbers should be the same.
Help Mr. Apple to evaluate each dish from both sets so that it is consistent with his feelings, or determine that this is impossible.
Input
The first line contains integers n and m (1 ≤ n, m ≤ 1000) — the number of dishes in both days.
Each of the next n lines contains a string of m symbols. The j-th symbol on i-th line is a_{ij}. All strings consist only of "<", ">" and "=".
Output
The first line of output should contain "Yes", if it's possible to do a correct evaluation for all the dishes, or "No" otherwise.
If case an answer exist, on the second line print n integers — evaluations of dishes from the first set, and on the third line print m integers — evaluations of dishes from the second set.
Examples
Input
3 4
>>>>
>>>>
>>>>
Output
Yes
2 2 2
1 1 1 1
Input
3 3
>>>
<<<
>>>
Output
Yes
3 1 3
2 2 2
Input
3 2
==
=<
==
Output
No
Note
In the first sample, all dishes of the first day are better than dishes of the second day. So, the highest score will be 2, for all dishes of the first day.
In the third sample, the table is contradictory — there is no possible evaluation of the dishes that satisfies it.
Submitted Solution:
```
n, m = map(int, input().split())
g = [[] for i in range(n + m)]
table = []
for i in range(n):
table.append(input().strip())
for i in range(n):
for j in range(m):
if table[i][j] == "<":
g[i].append((n + j, False))
elif table[i][j] == ">":
g[n + j].append((i, False))
else:
g[n + j].append((i, True))
g[i].append((n + j, True))
sorted_ver = []
visited = [False] * (n + m)
def dfs(v):
visited[v] = True
for i in g[v]:
if not visited[i[0]]:
dfs(i[0])
sorted_ver.append(v)
for i in range(n + m):
if not visited[i]:
dfs(i)
sorted_ver.reverse()
scores = [-1] * (m + n)
visited = [False] * (n + m)
res = []
def dfs(v, score):
scores[v] = score
for i in g[v]:
if i[1]:
if scores[i[0]] != score and scores[i[0]] != -1:
print("No")
exit()
elif scores[i[0]] == -1:
scores[i[0]] = score
dfs(i[0], score)
else:
if scores[i[0]] == -1:
dfs(i[0], score + 1)
for i in sorted_ver:
if scores[i] == -1:
dfs(i, 1)
for i in range(n):
print(scores[i], end=" ")
print()
for i in range(n, n + m):
print(scores[i], end=" ")
``` | instruction | 0 | 5,092 | 9 | 10,184 |
No | output | 1 | 5,092 | 9 | 10,185 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Apple, a gourmet, works as editor-in-chief of a gastronomic periodical. He travels around the world, tasting new delights of famous chefs from the most fashionable restaurants. Mr. Apple has his own signature method of review — in each restaurant Mr. Apple orders two sets of dishes on two different days. All the dishes are different, because Mr. Apple doesn't like to eat the same food. For each pair of dishes from different days he remembers exactly which was better, or that they were of the same quality. After this the gourmet evaluates each dish with a positive integer.
Once, during a revision of a restaurant of Celtic medieval cuisine named «Poisson», that serves chestnut soup with fir, warm soda bread, spicy lemon pie and other folk food, Mr. Apple was very pleasantly surprised the gourmet with its variety of menu, and hence ordered too much. Now he's confused about evaluating dishes.
The gourmet tasted a set of n dishes on the first day and a set of m dishes on the second day. He made a table a of size n × m, in which he described his impressions. If, according to the expert, dish i from the first set was better than dish j from the second set, then a_{ij} is equal to ">", in the opposite case a_{ij} is equal to "<". Dishes also may be equally good, in this case a_{ij} is "=".
Now Mr. Apple wants you to help him to evaluate every dish. Since Mr. Apple is very strict, he will evaluate the dishes so that the maximal number used is as small as possible. But Mr. Apple also is very fair, so he never evaluates the dishes so that it goes against his feelings. In other words, if a_{ij} is "<", then the number assigned to dish i from the first set should be less than the number of dish j from the second set, if a_{ij} is ">", then it should be greater, and finally if a_{ij} is "=", then the numbers should be the same.
Help Mr. Apple to evaluate each dish from both sets so that it is consistent with his feelings, or determine that this is impossible.
Input
The first line contains integers n and m (1 ≤ n, m ≤ 1000) — the number of dishes in both days.
Each of the next n lines contains a string of m symbols. The j-th symbol on i-th line is a_{ij}. All strings consist only of "<", ">" and "=".
Output
The first line of output should contain "Yes", if it's possible to do a correct evaluation for all the dishes, or "No" otherwise.
If case an answer exist, on the second line print n integers — evaluations of dishes from the first set, and on the third line print m integers — evaluations of dishes from the second set.
Examples
Input
3 4
>>>>
>>>>
>>>>
Output
Yes
2 2 2
1 1 1 1
Input
3 3
>>>
<<<
>>>
Output
Yes
3 1 3
2 2 2
Input
3 2
==
=<
==
Output
No
Note
In the first sample, all dishes of the first day are better than dishes of the second day. So, the highest score will be 2, for all dishes of the first day.
In the third sample, the table is contradictory — there is no possible evaluation of the dishes that satisfies it.
Submitted Solution:
```
r,c = map(int, input().split())
R = [None for i in range(0,r)]
C = [None for i in range(0,c)]
ans = True
for i in range(0,r):
s = input().split()[0]
s = list(s)
for j in range(0,c):
if s[j] == '=':
if R[i] is not None and C[j] is not None and R[i] != C[j]:
ans = False
elif R[i] is not None:
C[j] = R[i]
elif C[j] is not None:
R[i] = C[j]
else:
R[i] = 1
C[j] = 1
elif s[j] == '>':
if R[i] is not None and C[j] is not None and R[i] <= C[j]:
ans = False
elif R[i] is not None:
C[j] = R[i] - 1
elif C[j] is not None:
R[i] = C[j] + 1
else:
R[i] = 2
C[j] = 1
elif s[j] == '<':
if R[i] is not None and C[j] is not None and R[i] >= C[j]:
ans = False
elif R[i] is not None:
C[j] = R[i] + 1
elif C[j] is not None:
R[i] = C[j] - 1
else:
R[i] = 1
C[j] = 2
mini = 1
for i in R:
mini = min(mini, i)
for i in C:
mini = min(mini, i)
add = 1-mini
for i in range(0,len(R)):
R[i]+=add
for j in range(0,len(C)):
C[j]+=add
if ans:
print("Yes")
print(str(R).replace(",","").replace("[",'').replace("]",''))
print(str(C).replace(",","").replace("[",'').replace("]",''))
else:
print("No")
``` | instruction | 0 | 5,093 | 9 | 10,186 |
No | output | 1 | 5,093 | 9 | 10,187 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Apple, a gourmet, works as editor-in-chief of a gastronomic periodical. He travels around the world, tasting new delights of famous chefs from the most fashionable restaurants. Mr. Apple has his own signature method of review — in each restaurant Mr. Apple orders two sets of dishes on two different days. All the dishes are different, because Mr. Apple doesn't like to eat the same food. For each pair of dishes from different days he remembers exactly which was better, or that they were of the same quality. After this the gourmet evaluates each dish with a positive integer.
Once, during a revision of a restaurant of Celtic medieval cuisine named «Poisson», that serves chestnut soup with fir, warm soda bread, spicy lemon pie and other folk food, Mr. Apple was very pleasantly surprised the gourmet with its variety of menu, and hence ordered too much. Now he's confused about evaluating dishes.
The gourmet tasted a set of n dishes on the first day and a set of m dishes on the second day. He made a table a of size n × m, in which he described his impressions. If, according to the expert, dish i from the first set was better than dish j from the second set, then a_{ij} is equal to ">", in the opposite case a_{ij} is equal to "<". Dishes also may be equally good, in this case a_{ij} is "=".
Now Mr. Apple wants you to help him to evaluate every dish. Since Mr. Apple is very strict, he will evaluate the dishes so that the maximal number used is as small as possible. But Mr. Apple also is very fair, so he never evaluates the dishes so that it goes against his feelings. In other words, if a_{ij} is "<", then the number assigned to dish i from the first set should be less than the number of dish j from the second set, if a_{ij} is ">", then it should be greater, and finally if a_{ij} is "=", then the numbers should be the same.
Help Mr. Apple to evaluate each dish from both sets so that it is consistent with his feelings, or determine that this is impossible.
Input
The first line contains integers n and m (1 ≤ n, m ≤ 1000) — the number of dishes in both days.
Each of the next n lines contains a string of m symbols. The j-th symbol on i-th line is a_{ij}. All strings consist only of "<", ">" and "=".
Output
The first line of output should contain "Yes", if it's possible to do a correct evaluation for all the dishes, or "No" otherwise.
If case an answer exist, on the second line print n integers — evaluations of dishes from the first set, and on the third line print m integers — evaluations of dishes from the second set.
Examples
Input
3 4
>>>>
>>>>
>>>>
Output
Yes
2 2 2
1 1 1 1
Input
3 3
>>>
<<<
>>>
Output
Yes
3 1 3
2 2 2
Input
3 2
==
=<
==
Output
No
Note
In the first sample, all dishes of the first day are better than dishes of the second day. So, the highest score will be 2, for all dishes of the first day.
In the third sample, the table is contradictory — there is no possible evaluation of the dishes that satisfies it.
Submitted Solution:
```
import time
n1, m2 = map(int, input().split())
tests = []
for i in range(n1):
tests.append(list(input()))
#print (tests)
begin = time.time()
#print("---")
marks1 = []
result1 = []
for i in range(n1):
marks1.append([i,0.0])
result1.append(0)
marks2 = []
result2 = []
for j in range(m2):
marks2.append([j,0.0])
result2.append(0)
for i in range(n1):
for j in range(m2):
test = tests[i][j]
if test == ">":
marks1[i][1] += 1.0
elif test == "<":
marks2[j][1] += 1.0
else:
marks1[i][1] += 0.0001
marks2[j][1] += 0.0001
marks1.sort(key=lambda val: val[1])
marks2.sort(key=lambda val: val[1])
#print(marks1)
#print(marks2)
i = 0
j = 0
value = 0
lastmark = -1
lastItem = [0,0]
while i < n1 or j < m2:
LetAdd = 0
if i < n1 and j < m2:
if marks1[i][1] <= marks2[j][1]:
LetAdd = 1
else:
LetAdd = 2
elif i < n1:
LetAdd = 1
else:
LetAdd = 2
if LetAdd == 1:
if marks1[i][1] != lastmark and lastItem[0] != 2 or lastItem[0] == 2 and tests[marks1[i][0]][lastItem[1]] != '=':
lastmark = marks1[i][1]
value += 1
result1[marks1[i][0]] = value
lastItem = [1,i]
i += 1
else:
if marks2[j][1] != lastmark and lastItem[0] != 1 or lastItem[0] == 1 and tests[lastItem[1]][marks2[j][0]] != '=':
# print(lastItem[1], j, tests[lastItem[1]][marks2[j][0]])
lastmark = marks2[j][1]
value += 1
result2[marks2[j][0]] = value
lastItem = [2,j]
j += 1
CheckCorrect = True
for i in range(n1):
for j in range(m2):
test = tests[i][j]
if test == ">":
if result1[i] <= result2[j]:
CheckCorrect = False
elif test == "<":
if result1[i] >= result2[j]:
CheckCorrect = False
else:
if result1[i] != result2[j]:
CheckCorrect = False
#print("---")
#print("Time: ", time.time() - begin)
if CheckCorrect:
print ("Yes")
print (*result1)
print (*result2)
else:
print ("No")
``` | instruction | 0 | 5,094 | 9 | 10,188 |
No | output | 1 | 5,094 | 9 | 10,189 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Apple, a gourmet, works as editor-in-chief of a gastronomic periodical. He travels around the world, tasting new delights of famous chefs from the most fashionable restaurants. Mr. Apple has his own signature method of review — in each restaurant Mr. Apple orders two sets of dishes on two different days. All the dishes are different, because Mr. Apple doesn't like to eat the same food. For each pair of dishes from different days he remembers exactly which was better, or that they were of the same quality. After this the gourmet evaluates each dish with a positive integer.
Once, during a revision of a restaurant of Celtic medieval cuisine named «Poisson», that serves chestnut soup with fir, warm soda bread, spicy lemon pie and other folk food, Mr. Apple was very pleasantly surprised the gourmet with its variety of menu, and hence ordered too much. Now he's confused about evaluating dishes.
The gourmet tasted a set of n dishes on the first day and a set of m dishes on the second day. He made a table a of size n × m, in which he described his impressions. If, according to the expert, dish i from the first set was better than dish j from the second set, then a_{ij} is equal to ">", in the opposite case a_{ij} is equal to "<". Dishes also may be equally good, in this case a_{ij} is "=".
Now Mr. Apple wants you to help him to evaluate every dish. Since Mr. Apple is very strict, he will evaluate the dishes so that the maximal number used is as small as possible. But Mr. Apple also is very fair, so he never evaluates the dishes so that it goes against his feelings. In other words, if a_{ij} is "<", then the number assigned to dish i from the first set should be less than the number of dish j from the second set, if a_{ij} is ">", then it should be greater, and finally if a_{ij} is "=", then the numbers should be the same.
Help Mr. Apple to evaluate each dish from both sets so that it is consistent with his feelings, or determine that this is impossible.
Input
The first line contains integers n and m (1 ≤ n, m ≤ 1000) — the number of dishes in both days.
Each of the next n lines contains a string of m symbols. The j-th symbol on i-th line is a_{ij}. All strings consist only of "<", ">" and "=".
Output
The first line of output should contain "Yes", if it's possible to do a correct evaluation for all the dishes, or "No" otherwise.
If case an answer exist, on the second line print n integers — evaluations of dishes from the first set, and on the third line print m integers — evaluations of dishes from the second set.
Examples
Input
3 4
>>>>
>>>>
>>>>
Output
Yes
2 2 2
1 1 1 1
Input
3 3
>>>
<<<
>>>
Output
Yes
3 1 3
2 2 2
Input
3 2
==
=<
==
Output
No
Note
In the first sample, all dishes of the first day are better than dishes of the second day. So, the highest score will be 2, for all dishes of the first day.
In the third sample, the table is contradictory — there is no possible evaluation of the dishes that satisfies it.
Submitted Solution:
```
import sys
input=sys.stdin.readline
n,m=map(int,input().split())
grid=[]
for i in range(n):
s=input()
grid.append(s)
a=[]
b=[]
for i in range(n):
tmp=0
tmp2=0
for j in range(m):
if grid[i][j]=='<':
tmp+=1
if grid[i][j]=='>':
tmp2-=1
a.append([i,tmp,tmp2])
for i in range(m):
tmp=0
tmp2=0
for j in range(n):
if grid[j][i]=='<':
tmp+=1
if grid[j][i]=='>':
tmp2-=1
b.append([i,tmp,tmp2])
a.sort(key=lambda thing: thing[1])
b.sort(key=lambda thing: thing[1])
a.sort(key=lambda thing: thing[2])
b.sort(key=lambda thing: thing[2])
a.reverse()
b.reverse()
pointer1=0
pointer2=0
ans1=[0]*n
ans2=[0]*m
curr=1
last1=-1
last2=-1
flag=0
while pointer1<n and pointer2<m:
k=a[pointer1][0]
l=b[pointer2][0]
t=grid[k][l]
r=grid[a[pointer1-1][0]][l]
s=grid[k][b[pointer2-1][0]]
if t=='<':
if pointer1!=0 and r!='>':
curr-=1
ans1[k]=curr
last=r
curr+=1
pointer1+=1
elif t=='>':
if pointer2!=0 and s!='<':
curr-=1
ans2[l]=curr
curr+=1
pointer2+=1
else:
if (pointer1!=0 and r=='=') or (pointer2!=0 and s=='='):
curr-=1
ans1[k]=curr
ans2[l]=curr
curr+=1
pointer1+=1
pointer2+=1
if pointer1<n:
if pointer1!=0 and grid[a[pointer1-1][0]][b[m-1][0]]!='>':
curr-=1
tl=0
if not tl and grid[a[pointer1][0]][b[m-1][0]]=='>':
tl=1
while pointer1<n:
if not tl and grid[a[pointer1][0]][b[m-1][0]]=='>':
curr+=1
tl=1
k=a[pointer1][0]
ans1[k]=curr
pointer1+=1
while pointer2<m:
if pointer2!=0 and grid[a[n-1][0]][b[pointer2-1][0]]!='<':
curr-=1
tl=0
if not tl and grid[a[n-1][0]][b[pointer2-1][0]]=='<':
tl=1
while pointer2<m:
if not tl and grid[a[n-1][0]][b[pointer2-1][0]]=='<':
curr+=1
tl=1
k=b[pointer2][0]
ans2[k]=curr
pointer+=1
for i in range(n):
if ans1[i]==0:
ans1[i]=1
for i in range(m):
if ans2[i]==0:
ans2[i]=1
for i in range(n):
for j in range(m):
p=grid[i][j]
if p=='<':
if ans1[i]>=ans2[j]:
flag=1
break
elif p=='>':
if ans1[i]<=ans2[j]:
flag=1
break
else:
if ans1[i]!=ans2[j]:
flag=1
break
if flag:
break
if flag:
print('No')
else:
print('Yes')
print(' '.join(map(str,ans1)))
print(' '.join(map(str,ans2)))
``` | instruction | 0 | 5,095 | 9 | 10,190 |
No | output | 1 | 5,095 | 9 | 10,191 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Piegirl is buying stickers for a project. Stickers come on sheets, and each sheet of stickers contains exactly n stickers. Each sticker has exactly one character printed on it, so a sheet of stickers can be described by a string of length n. Piegirl wants to create a string s using stickers. She may buy as many sheets of stickers as she wants, and may specify any string of length n for the sheets, but all the sheets must be identical, so the string is the same for all sheets. Once she attains the sheets of stickers, she will take some of the stickers from the sheets and arrange (in any order) them to form s. Determine the minimum number of sheets she has to buy, and provide a string describing a possible sheet of stickers she should buy.
Input
The first line contains string s (1 ≤ |s| ≤ 1000), consisting of lowercase English characters only. The second line contains an integer n (1 ≤ n ≤ 1000).
Output
On the first line, print the minimum number of sheets Piegirl has to buy. On the second line, print a string consisting of n lower case English characters. This string should describe a sheet of stickers that Piegirl can buy in order to minimize the number of sheets. If Piegirl cannot possibly form the string s, print instead a single line with the number -1.
Examples
Input
banana
4
Output
2
baan
Input
banana
3
Output
3
nab
Input
banana
2
Output
-1
Note
In the second example, Piegirl can order 3 sheets of stickers with the characters "nab". She can take characters "nab" from the first sheet, "na" from the second, and "a" from the third, and arrange them to from "banana". | instruction | 0 | 5,340 | 9 | 10,680 |
Tags: binary search, constructive algorithms, greedy
Correct Solution:
```
from math import ceil
p = {i: 0 for i in 'abcdefghijklmnopqrstuvwxyz'}
t = input()
for i in t: p[i] += 1
p = {i: p[i] for i in p if p[i] > 0}
n = int(input())
if len(p) > n: print(-1)
elif len(t) > n:
r = [[p[i], p[i], 1, i] for i in p]
for i in range(n - len(p)):
j = max(r)
j[2] += 1
j[0] = j[1] / j[2]
print(ceil(max(r)[0]))
print(''.join(j[3] * j[2] for j in r))
else: print('1\n' + t * (n // len(t)) + t[: n % len(t)])
``` | output | 1 | 5,340 | 9 | 10,681 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Piegirl is buying stickers for a project. Stickers come on sheets, and each sheet of stickers contains exactly n stickers. Each sticker has exactly one character printed on it, so a sheet of stickers can be described by a string of length n. Piegirl wants to create a string s using stickers. She may buy as many sheets of stickers as she wants, and may specify any string of length n for the sheets, but all the sheets must be identical, so the string is the same for all sheets. Once she attains the sheets of stickers, she will take some of the stickers from the sheets and arrange (in any order) them to form s. Determine the minimum number of sheets she has to buy, and provide a string describing a possible sheet of stickers she should buy.
Input
The first line contains string s (1 ≤ |s| ≤ 1000), consisting of lowercase English characters only. The second line contains an integer n (1 ≤ n ≤ 1000).
Output
On the first line, print the minimum number of sheets Piegirl has to buy. On the second line, print a string consisting of n lower case English characters. This string should describe a sheet of stickers that Piegirl can buy in order to minimize the number of sheets. If Piegirl cannot possibly form the string s, print instead a single line with the number -1.
Examples
Input
banana
4
Output
2
baan
Input
banana
3
Output
3
nab
Input
banana
2
Output
-1
Note
In the second example, Piegirl can order 3 sheets of stickers with the characters "nab". She can take characters "nab" from the first sheet, "na" from the second, and "a" from the third, and arrange them to from "banana". | instruction | 0 | 5,346 | 9 | 10,692 |
Tags: binary search, constructive algorithms, greedy
Correct Solution:
```
s = input()
n = int(input())
freq = [0 for i in range(0, 300)]
raport = [0 for i in range(0, 300)]
differentLetters = 0
tickets = 0
sol = ''
for c in s: freq[ord(c)] += 1
for i in freq:
if i > 0: differentLetters += 1
if differentLetters > n:
print('-1')
exit()
for i in 'abcdefghijklmnopqrstuvwxyz':
if freq[ord(i)] == 0: continue
sol += i
freq[ord(i)] -= 1
raport[ord(i)] = freq[ord(i)]
for i in range(differentLetters, n):
#pun litera cu cea mai mare frecventa
maxRaport = raport[ord('z')]
chosenLetter = 'z'
for j in 'abcdefghijklmnopqrstuvwxyz':
if raport[ord(j)] > maxRaport:
maxRaport = raport[ord(j)]
chosenLetter = j
sol += chosenLetter
raport[ord(chosenLetter)] = freq[ord(chosenLetter)] / sol.count(chosenLetter)
for i in sol:
a = s.count(i)
b = sol.count(i)
if a%b == 0: tickets = max(tickets, int(a//b))
else: tickets = max(tickets, int(a//b) + 1)
print(tickets)
print(sol)
``` | output | 1 | 5,346 | 9 | 10,693 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Piegirl is buying stickers for a project. Stickers come on sheets, and each sheet of stickers contains exactly n stickers. Each sticker has exactly one character printed on it, so a sheet of stickers can be described by a string of length n. Piegirl wants to create a string s using stickers. She may buy as many sheets of stickers as she wants, and may specify any string of length n for the sheets, but all the sheets must be identical, so the string is the same for all sheets. Once she attains the sheets of stickers, she will take some of the stickers from the sheets and arrange (in any order) them to form s. Determine the minimum number of sheets she has to buy, and provide a string describing a possible sheet of stickers she should buy.
Input
The first line contains string s (1 ≤ |s| ≤ 1000), consisting of lowercase English characters only. The second line contains an integer n (1 ≤ n ≤ 1000).
Output
On the first line, print the minimum number of sheets Piegirl has to buy. On the second line, print a string consisting of n lower case English characters. This string should describe a sheet of stickers that Piegirl can buy in order to minimize the number of sheets. If Piegirl cannot possibly form the string s, print instead a single line with the number -1.
Examples
Input
banana
4
Output
2
baan
Input
banana
3
Output
3
nab
Input
banana
2
Output
-1
Note
In the second example, Piegirl can order 3 sheets of stickers with the characters "nab". She can take characters "nab" from the first sheet, "na" from the second, and "a" from the third, and arrange them to from "banana".
Submitted Solution:
```
from collections import Counter
from heapq import heappushpop
def main():
cnt, n = Counter(input()), int(input())
if n < len(cnt):
print(-1)
return
h = list((1 / v, 1, c) for c, v in cnt.most_common())
res = list(cnt.keys())
_, v, c = h.pop(0)
for _ in range(n - len(cnt)):
res.append(c)
v += 1
_, v, c = heappushpop(h, (v / cnt[c], v, c))
print((cnt[c] + v - 1) // v)
print(''.join(res))
if __name__ == '__main__':
main()
``` | instruction | 0 | 5,347 | 9 | 10,694 |
Yes | output | 1 | 5,347 | 9 | 10,695 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Piegirl is buying stickers for a project. Stickers come on sheets, and each sheet of stickers contains exactly n stickers. Each sticker has exactly one character printed on it, so a sheet of stickers can be described by a string of length n. Piegirl wants to create a string s using stickers. She may buy as many sheets of stickers as she wants, and may specify any string of length n for the sheets, but all the sheets must be identical, so the string is the same for all sheets. Once she attains the sheets of stickers, she will take some of the stickers from the sheets and arrange (in any order) them to form s. Determine the minimum number of sheets she has to buy, and provide a string describing a possible sheet of stickers she should buy.
Input
The first line contains string s (1 ≤ |s| ≤ 1000), consisting of lowercase English characters only. The second line contains an integer n (1 ≤ n ≤ 1000).
Output
On the first line, print the minimum number of sheets Piegirl has to buy. On the second line, print a string consisting of n lower case English characters. This string should describe a sheet of stickers that Piegirl can buy in order to minimize the number of sheets. If Piegirl cannot possibly form the string s, print instead a single line with the number -1.
Examples
Input
banana
4
Output
2
baan
Input
banana
3
Output
3
nab
Input
banana
2
Output
-1
Note
In the second example, Piegirl can order 3 sheets of stickers with the characters "nab". She can take characters "nab" from the first sheet, "na" from the second, and "a" from the third, and arrange them to from "banana".
Submitted Solution:
```
#!/usr/bin/env python
from __future__ import division, print_function
import os
import sys
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
from collections import Counter
import heapq as pq
from math import ceil
def solve():
s = insr()
n = inp()
count_map = Counter(s)
if n < len(count_map):
print(-1)
return
copy_map = {el: 1 for el in count_map.keys()}
used = len(count_map)
q = [(-el, key) for key, el in count_map.items()]
pq.heapify(q)
while used < n and q:
copy, c = pq.heappop(q)
copy_map[c] += 1
copies = ceil(count_map[c] / copy_map[c])
if copies > 1:
pq.heappush(q, (-copies, c))
used += 1
if not q:
print(1)
else:
print(pq.heappop(q)[0] * -1)
res = []
for el, copies in copy_map.items():
res.extend([el] * copies)
if len(res) < n:
res.extend(['a'] * (n - len(res)))
print("".join(res))
def main():
t = 1
for i in range(t):
solve()
BUFSIZE = 8192
def inp():
return (int(input()))
def inlt():
return (list(map(int, input().split())))
def insr():
return (input().strip())
def invr():
return (map(int, input().split()))
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
``` | instruction | 0 | 5,348 | 9 | 10,696 |
Yes | output | 1 | 5,348 | 9 | 10,697 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Piegirl is buying stickers for a project. Stickers come on sheets, and each sheet of stickers contains exactly n stickers. Each sticker has exactly one character printed on it, so a sheet of stickers can be described by a string of length n. Piegirl wants to create a string s using stickers. She may buy as many sheets of stickers as she wants, and may specify any string of length n for the sheets, but all the sheets must be identical, so the string is the same for all sheets. Once she attains the sheets of stickers, she will take some of the stickers from the sheets and arrange (in any order) them to form s. Determine the minimum number of sheets she has to buy, and provide a string describing a possible sheet of stickers she should buy.
Input
The first line contains string s (1 ≤ |s| ≤ 1000), consisting of lowercase English characters only. The second line contains an integer n (1 ≤ n ≤ 1000).
Output
On the first line, print the minimum number of sheets Piegirl has to buy. On the second line, print a string consisting of n lower case English characters. This string should describe a sheet of stickers that Piegirl can buy in order to minimize the number of sheets. If Piegirl cannot possibly form the string s, print instead a single line with the number -1.
Examples
Input
banana
4
Output
2
baan
Input
banana
3
Output
3
nab
Input
banana
2
Output
-1
Note
In the second example, Piegirl can order 3 sheets of stickers with the characters "nab". She can take characters "nab" from the first sheet, "na" from the second, and "a" from the third, and arrange them to from "banana".
Submitted Solution:
```
s = input()
n = int(input())
cnt = {}
for c in s:
if cnt.get(c) == None:
cnt[c] = 1
else:
cnt[c] += 1
if (n < len(cnt)):
print(-1)
else:
ansNum = 0;
while(True):
ansNum+=1
l = 0;
char = []
for c, v in cnt.items():
need = (v + ansNum -1)//ansNum
l+= need
char.append((c, need))
if (l > n):
continue
print(ansNum)
ans = "".join([str(c[0])*c[1] for c in char])
ans = ans + 'a'*(n - len(ans))
print(ans)
break
``` | instruction | 0 | 5,349 | 9 | 10,698 |
Yes | output | 1 | 5,349 | 9 | 10,699 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Piegirl is buying stickers for a project. Stickers come on sheets, and each sheet of stickers contains exactly n stickers. Each sticker has exactly one character printed on it, so a sheet of stickers can be described by a string of length n. Piegirl wants to create a string s using stickers. She may buy as many sheets of stickers as she wants, and may specify any string of length n for the sheets, but all the sheets must be identical, so the string is the same for all sheets. Once she attains the sheets of stickers, she will take some of the stickers from the sheets and arrange (in any order) them to form s. Determine the minimum number of sheets she has to buy, and provide a string describing a possible sheet of stickers she should buy.
Input
The first line contains string s (1 ≤ |s| ≤ 1000), consisting of lowercase English characters only. The second line contains an integer n (1 ≤ n ≤ 1000).
Output
On the first line, print the minimum number of sheets Piegirl has to buy. On the second line, print a string consisting of n lower case English characters. This string should describe a sheet of stickers that Piegirl can buy in order to minimize the number of sheets. If Piegirl cannot possibly form the string s, print instead a single line with the number -1.
Examples
Input
banana
4
Output
2
baan
Input
banana
3
Output
3
nab
Input
banana
2
Output
-1
Note
In the second example, Piegirl can order 3 sheets of stickers with the characters "nab". She can take characters "nab" from the first sheet, "na" from the second, and "a" from the third, and arrange them to from "banana".
Submitted Solution:
```
s = input()
n = int(input())
from collections import Counter
c = Counter(s)
out = Counter()
contrib = Counter()
for letter in c:
out[letter] = 1
contrib[letter] = c[letter]
sum_vals = sum(out.values())
from math import ceil
from fractions import Fraction
if sum_vals > n:
print(-1)
else:
while sum_vals < n:
el, _ = contrib.most_common(1)[0]
out[el] += 1
sum_vals += 1
contrib[el] = ceil(Fraction(c[el], out[el]))
print(max(contrib.values()))
print(''.join(out.elements()))
``` | instruction | 0 | 5,350 | 9 | 10,700 |
Yes | output | 1 | 5,350 | 9 | 10,701 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Piegirl is buying stickers for a project. Stickers come on sheets, and each sheet of stickers contains exactly n stickers. Each sticker has exactly one character printed on it, so a sheet of stickers can be described by a string of length n. Piegirl wants to create a string s using stickers. She may buy as many sheets of stickers as she wants, and may specify any string of length n for the sheets, but all the sheets must be identical, so the string is the same for all sheets. Once she attains the sheets of stickers, she will take some of the stickers from the sheets and arrange (in any order) them to form s. Determine the minimum number of sheets she has to buy, and provide a string describing a possible sheet of stickers she should buy.
Input
The first line contains string s (1 ≤ |s| ≤ 1000), consisting of lowercase English characters only. The second line contains an integer n (1 ≤ n ≤ 1000).
Output
On the first line, print the minimum number of sheets Piegirl has to buy. On the second line, print a string consisting of n lower case English characters. This string should describe a sheet of stickers that Piegirl can buy in order to minimize the number of sheets. If Piegirl cannot possibly form the string s, print instead a single line with the number -1.
Examples
Input
banana
4
Output
2
baan
Input
banana
3
Output
3
nab
Input
banana
2
Output
-1
Note
In the second example, Piegirl can order 3 sheets of stickers with the characters "nab". She can take characters "nab" from the first sheet, "na" from the second, and "a" from the third, and arrange them to from "banana".
Submitted Solution:
```
from operator import itemgetter
class CodeforcesTask335ASolution:
def __init__(self):
self.result = ''
self.s = ''
self.n = 0
def read_input(self):
self.s = input()
self.n = int(input())
def process_task(self):
letters = 'abcdefghijklmnopqrstuvwxyz'
occs = [[c, self.s.count(c)] for c in letters if self.s.count(c)]
occs.sort(key=itemgetter(1), reverse=True)
if len(occs) > self.n:
self.result = "-1"
else:
resulting = "".join([x[0] for x in occs])
while len(resulting) < self.n:
priorities = [[c, resulting.count(c)] for c in letters if resulting.count(c)]
priorities.sort(key=itemgetter(1), reverse=True)
lel = []
for x in range(len(priorities)):
lel.append([occs[x][0], occs[x][1] / priorities[x][1]])
#print(resulting, priorities, occs, lel)
#print(lel)
lel.sort(key=itemgetter(1), reverse=True)
resulting += lel[0][0]
priorities = [[c, resulting.count(c)] for c in letters if resulting.count(c)]
#print(priorities)
priorities.sort(key=itemgetter(1), reverse=True)
lel = []
for x in range(len(priorities)):
lel.append([occs[x][0], occs[x][1] / priorities[x][1]])
#print(lel)
self.result = "{0}\n{1}".format(int(max([x[1] for x in lel])), resulting)
def get_result(self):
return self.result
if __name__ == "__main__":
Solution = CodeforcesTask335ASolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())
``` | instruction | 0 | 5,351 | 9 | 10,702 |
No | output | 1 | 5,351 | 9 | 10,703 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Piegirl is buying stickers for a project. Stickers come on sheets, and each sheet of stickers contains exactly n stickers. Each sticker has exactly one character printed on it, so a sheet of stickers can be described by a string of length n. Piegirl wants to create a string s using stickers. She may buy as many sheets of stickers as she wants, and may specify any string of length n for the sheets, but all the sheets must be identical, so the string is the same for all sheets. Once she attains the sheets of stickers, she will take some of the stickers from the sheets and arrange (in any order) them to form s. Determine the minimum number of sheets she has to buy, and provide a string describing a possible sheet of stickers she should buy.
Input
The first line contains string s (1 ≤ |s| ≤ 1000), consisting of lowercase English characters only. The second line contains an integer n (1 ≤ n ≤ 1000).
Output
On the first line, print the minimum number of sheets Piegirl has to buy. On the second line, print a string consisting of n lower case English characters. This string should describe a sheet of stickers that Piegirl can buy in order to minimize the number of sheets. If Piegirl cannot possibly form the string s, print instead a single line with the number -1.
Examples
Input
banana
4
Output
2
baan
Input
banana
3
Output
3
nab
Input
banana
2
Output
-1
Note
In the second example, Piegirl can order 3 sheets of stickers with the characters "nab". She can take characters "nab" from the first sheet, "na" from the second, and "a" from the third, and arrange them to from "banana".
Submitted Solution:
```
from collections import Counter
import math
s = input()
n = int(input())
c = Counter(s)
if len(c) > n:
print(-1)
exit()
ans = ""
d = Counter()
for x in c:
ans += x
d[x] += 1
c.most_common()
# print(c)
for x in c:
if len(ans) < n:
t = min(n - len(ans), c[x] - 1)
ans += x * t
d[x] += t
else:
break
maxm = 1
for x in d:
maxm = max(c[x] / d[x], maxm)
print(int(math.ceil(maxm)))
print(ans)
``` | instruction | 0 | 5,352 | 9 | 10,704 |
No | output | 1 | 5,352 | 9 | 10,705 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Piegirl is buying stickers for a project. Stickers come on sheets, and each sheet of stickers contains exactly n stickers. Each sticker has exactly one character printed on it, so a sheet of stickers can be described by a string of length n. Piegirl wants to create a string s using stickers. She may buy as many sheets of stickers as she wants, and may specify any string of length n for the sheets, but all the sheets must be identical, so the string is the same for all sheets. Once she attains the sheets of stickers, she will take some of the stickers from the sheets and arrange (in any order) them to form s. Determine the minimum number of sheets she has to buy, and provide a string describing a possible sheet of stickers she should buy.
Input
The first line contains string s (1 ≤ |s| ≤ 1000), consisting of lowercase English characters only. The second line contains an integer n (1 ≤ n ≤ 1000).
Output
On the first line, print the minimum number of sheets Piegirl has to buy. On the second line, print a string consisting of n lower case English characters. This string should describe a sheet of stickers that Piegirl can buy in order to minimize the number of sheets. If Piegirl cannot possibly form the string s, print instead a single line with the number -1.
Examples
Input
banana
4
Output
2
baan
Input
banana
3
Output
3
nab
Input
banana
2
Output
-1
Note
In the second example, Piegirl can order 3 sheets of stickers with the characters "nab". She can take characters "nab" from the first sheet, "na" from the second, and "a" from the third, and arrange them to from "banana".
Submitted Solution:
```
s = input()
n = int(input())
d = {}
r = 0
for a in s:
d.setdefault(a, 0)
d[a] += 1
if(d[a] > r):
r=d[a]
if (len(d) > n):
print(-1)
else:
l = 0
while r - l > 1:
k = (l + r) // 2
cur = 0
for x in d.values():
cur += (x+k-1) // k
if cur > n:
l = k
else:
r = k
print(r)
s = ''
for a in d.keys():
s += a * ((d[a]+r-1) // r)
print(s)
``` | instruction | 0 | 5,353 | 9 | 10,706 |
No | output | 1 | 5,353 | 9 | 10,707 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Piegirl is buying stickers for a project. Stickers come on sheets, and each sheet of stickers contains exactly n stickers. Each sticker has exactly one character printed on it, so a sheet of stickers can be described by a string of length n. Piegirl wants to create a string s using stickers. She may buy as many sheets of stickers as she wants, and may specify any string of length n for the sheets, but all the sheets must be identical, so the string is the same for all sheets. Once she attains the sheets of stickers, she will take some of the stickers from the sheets and arrange (in any order) them to form s. Determine the minimum number of sheets she has to buy, and provide a string describing a possible sheet of stickers she should buy.
Input
The first line contains string s (1 ≤ |s| ≤ 1000), consisting of lowercase English characters only. The second line contains an integer n (1 ≤ n ≤ 1000).
Output
On the first line, print the minimum number of sheets Piegirl has to buy. On the second line, print a string consisting of n lower case English characters. This string should describe a sheet of stickers that Piegirl can buy in order to minimize the number of sheets. If Piegirl cannot possibly form the string s, print instead a single line with the number -1.
Examples
Input
banana
4
Output
2
baan
Input
banana
3
Output
3
nab
Input
banana
2
Output
-1
Note
In the second example, Piegirl can order 3 sheets of stickers with the characters "nab". She can take characters "nab" from the first sheet, "na" from the second, and "a" from the third, and arrange them to from "banana".
Submitted Solution:
```
import math
S=input()
N=int(input())
Srep={}
ansrep={}
for item in "abcdefghijklmnopqrstuvwxyz":
Srep[item]=0
ansrep[item]=0
for item in S:
Srep[item]+=1
ansrep[item]+=1
Q=list(set(S))
if(len(Q)>N):
print(-1)
else:
n=len(Q)
ans=list(S)
num=1
req=1
n=len(ans)
while(len(ans)>N):
n=len(ans)
for item in ans:
if(ansrep[item]==1):
continue
if(math.ceil(Srep[item]/(ansrep[item]-1))>req):
continue
else:
ansrep[item]-=1
ans.remove(item)
break
if(n==len(ans)):
for item in ans:
if(ansrep[item]==1):
continue
ansrep[item]-=1
ans.remove(item)
req+=1
break
g=""
if(len(ans)<N):
g=S[0]*(N-len(ans))
print(req)
for item in ans:
print(item,end="")
print(g)
``` | instruction | 0 | 5,354 | 9 | 10,708 |
No | output | 1 | 5,354 | 9 | 10,709 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole n ordered piles of worms such that i-th pile contains ai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to a1, worms in second pile are labeled with numbers a1 + 1 to a1 + a2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
Input
The first line contains a single integer n (1 ≤ n ≤ 105), the number of piles.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 103, a1 + a2 + ... + an ≤ 106), where ai is the number of worms in the i-th pile.
The third line contains single integer m (1 ≤ m ≤ 105), the number of juicy worms said by Marmot.
The fourth line contains m integers q1, q2, ..., qm (1 ≤ qi ≤ a1 + a2 + ... + an), the labels of the juicy worms.
Output
Print m lines to the standard output. The i-th line should contain an integer, representing the number of the pile where the worm labeled with the number qi is.
Examples
Input
5
2 7 3 4 9
3
1 25 11
Output
1
5
3
Note
For the sample input:
* The worms with labels from [1, 2] are in the first pile.
* The worms with labels from [3, 9] are in the second pile.
* The worms with labels from [10, 12] are in the third pile.
* The worms with labels from [13, 16] are in the fourth pile.
* The worms with labels from [17, 25] are in the fifth pile.
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
k=0
l=[]
c=1
for i in range(len(a)):
x=a[i]
for j in range(k,k+x):
l.append(c)
c+=1
k=k+x
t=int(input())
b=list(map(int,input().split()))
for i in b:
print(l[i-1])
``` | instruction | 0 | 5,399 | 9 | 10,798 |
Yes | output | 1 | 5,399 | 9 | 10,799 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole n ordered piles of worms such that i-th pile contains ai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to a1, worms in second pile are labeled with numbers a1 + 1 to a1 + a2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
Input
The first line contains a single integer n (1 ≤ n ≤ 105), the number of piles.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 103, a1 + a2 + ... + an ≤ 106), where ai is the number of worms in the i-th pile.
The third line contains single integer m (1 ≤ m ≤ 105), the number of juicy worms said by Marmot.
The fourth line contains m integers q1, q2, ..., qm (1 ≤ qi ≤ a1 + a2 + ... + an), the labels of the juicy worms.
Output
Print m lines to the standard output. The i-th line should contain an integer, representing the number of the pile where the worm labeled with the number qi is.
Examples
Input
5
2 7 3 4 9
3
1 25 11
Output
1
5
3
Note
For the sample input:
* The worms with labels from [1, 2] are in the first pile.
* The worms with labels from [3, 9] are in the second pile.
* The worms with labels from [10, 12] are in the third pile.
* The worms with labels from [13, 16] are in the fourth pile.
* The worms with labels from [17, 25] are in the fifth pile.
Submitted Solution:
```
# Worms
def fst(tuple):
return tuple[0]
def snd(tuple):
return tuple[1]
n = input()
piles = list(map(int, input().split()))
w = input()
worms = list(map(int, input().split()))
piles[0] = (1, piles[0])
ans = [0 for i in range(10**6 + 2)]
for i in range(len(piles)):
if i > 0:
size = piles[i]
piles[i] = (snd(piles[i-1]) + 1, snd(piles[i-1]) + size)
for j in range(fst(piles[i]), snd(piles[i]) + 1):
ans[j] = i + 1
for w in worms:
print(ans[w])
``` | instruction | 0 | 5,400 | 9 | 10,800 |
Yes | output | 1 | 5,400 | 9 | 10,801 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole n ordered piles of worms such that i-th pile contains ai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to a1, worms in second pile are labeled with numbers a1 + 1 to a1 + a2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
Input
The first line contains a single integer n (1 ≤ n ≤ 105), the number of piles.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 103, a1 + a2 + ... + an ≤ 106), where ai is the number of worms in the i-th pile.
The third line contains single integer m (1 ≤ m ≤ 105), the number of juicy worms said by Marmot.
The fourth line contains m integers q1, q2, ..., qm (1 ≤ qi ≤ a1 + a2 + ... + an), the labels of the juicy worms.
Output
Print m lines to the standard output. The i-th line should contain an integer, representing the number of the pile where the worm labeled with the number qi is.
Examples
Input
5
2 7 3 4 9
3
1 25 11
Output
1
5
3
Note
For the sample input:
* The worms with labels from [1, 2] are in the first pile.
* The worms with labels from [3, 9] are in the second pile.
* The worms with labels from [10, 12] are in the third pile.
* The worms with labels from [13, 16] are in the fourth pile.
* The worms with labels from [17, 25] are in the fifth pile.
Submitted Solution:
```
from bisect import bisect_left
from itertools import accumulate
n = int(input())
a = list(map(int, input().split()))
m = int(input())
q = list(map(int, input().split()))
sums = list(accumulate(a))
ans = list()
for e in q:
print(bisect_left(sums,e)+1)
``` | instruction | 0 | 5,401 | 9 | 10,802 |
Yes | output | 1 | 5,401 | 9 | 10,803 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole n ordered piles of worms such that i-th pile contains ai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to a1, worms in second pile are labeled with numbers a1 + 1 to a1 + a2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
Input
The first line contains a single integer n (1 ≤ n ≤ 105), the number of piles.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 103, a1 + a2 + ... + an ≤ 106), where ai is the number of worms in the i-th pile.
The third line contains single integer m (1 ≤ m ≤ 105), the number of juicy worms said by Marmot.
The fourth line contains m integers q1, q2, ..., qm (1 ≤ qi ≤ a1 + a2 + ... + an), the labels of the juicy worms.
Output
Print m lines to the standard output. The i-th line should contain an integer, representing the number of the pile where the worm labeled with the number qi is.
Examples
Input
5
2 7 3 4 9
3
1 25 11
Output
1
5
3
Note
For the sample input:
* The worms with labels from [1, 2] are in the first pile.
* The worms with labels from [3, 9] are in the second pile.
* The worms with labels from [10, 12] are in the third pile.
* The worms with labels from [13, 16] are in the fourth pile.
* The worms with labels from [17, 25] are in the fifth pile.
Submitted Solution:
```
n = int(input())
piles = [int(a) for a in input().split()]
q = int(input())
queries = [int(b) for b in input().split()]
worms = {}
i = 1
for p in range(n):
for k in range(piles[p]):
worms[i] = p+1
i+=1
for j in queries:
print(worms[j])
``` | instruction | 0 | 5,402 | 9 | 10,804 |
Yes | output | 1 | 5,402 | 9 | 10,805 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole n ordered piles of worms such that i-th pile contains ai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to a1, worms in second pile are labeled with numbers a1 + 1 to a1 + a2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
Input
The first line contains a single integer n (1 ≤ n ≤ 105), the number of piles.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 103, a1 + a2 + ... + an ≤ 106), where ai is the number of worms in the i-th pile.
The third line contains single integer m (1 ≤ m ≤ 105), the number of juicy worms said by Marmot.
The fourth line contains m integers q1, q2, ..., qm (1 ≤ qi ≤ a1 + a2 + ... + an), the labels of the juicy worms.
Output
Print m lines to the standard output. The i-th line should contain an integer, representing the number of the pile where the worm labeled with the number qi is.
Examples
Input
5
2 7 3 4 9
3
1 25 11
Output
1
5
3
Note
For the sample input:
* The worms with labels from [1, 2] are in the first pile.
* The worms with labels from [3, 9] are in the second pile.
* The worms with labels from [10, 12] are in the third pile.
* The worms with labels from [13, 16] are in the fourth pile.
* The worms with labels from [17, 25] are in the fifth pile.
Submitted Solution:
```
def find(arr, value):
low = 0
high = len(arr) - 1
while high - low != 0:
med = int((low + high) / 2)
if value > arr[med]:
low = med
else:
high = med
if high - low == 1:
if value < arr[low]:
return low + 1
else:
return high + 1
n = int(input())
L = list(map(int, input().split(' ')))
R = []
R.append(L[0])
for i in range(1,n):
R.append(R[i-1] + L[i])
m = int(input())
L = list(map(int, input().split(' ')))
for x in L:
print(find(R, x))
``` | instruction | 0 | 5,403 | 9 | 10,806 |
No | output | 1 | 5,403 | 9 | 10,807 |
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