text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given one integer n (n > 1).
Recall that a permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation of length 5, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n = 3 but there is 4 in the array).
Your task is to find a permutation p of length n that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).
You have to answer t independent test cases.
If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (2 ≤ n ≤ 100) — the length of the permutation you have to find.
Output
For each test case, print n distinct integers p_1, p_2, …, p_n — a permutation that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).
If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.
Example
Input
2
2
5
Output
2 1
2 1 5 3 4
Tags: constructive algorithms, probabilities
Correct Solution:
```
def lip(): return list(map(int,input().split()))
def splip(): return map(int,input().split())
def intip(): return int(input())
for _ in range(intip()):
n = intip()
l = [i for i in range(1,n+1)]
l = l[::-1]
mid = n//2
if n==2:
print(*l)
elif n%2!=0:
l[mid] , l[mid-1] = l[mid-1],l[mid]
print(*l)
else:
l[mid+1] , l[mid] = l[mid],l[mid+1]
print(*l)
```
| 97,200 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given one integer n (n > 1).
Recall that a permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation of length 5, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n = 3 but there is 4 in the array).
Your task is to find a permutation p of length n that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).
You have to answer t independent test cases.
If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (2 ≤ n ≤ 100) — the length of the permutation you have to find.
Output
For each test case, print n distinct integers p_1, p_2, …, p_n — a permutation that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).
If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.
Example
Input
2
2
5
Output
2 1
2 1 5 3 4
Tags: constructive algorithms, probabilities
Correct Solution:
```
t= int (input())
if t:
for i in range(t):
n= int(input())
if n:
print(n, *range(1,n))
```
| 97,201 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given one integer n (n > 1).
Recall that a permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation of length 5, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n = 3 but there is 4 in the array).
Your task is to find a permutation p of length n that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).
You have to answer t independent test cases.
If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (2 ≤ n ≤ 100) — the length of the permutation you have to find.
Output
For each test case, print n distinct integers p_1, p_2, …, p_n — a permutation that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).
If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.
Example
Input
2
2
5
Output
2 1
2 1 5 3 4
Tags: constructive algorithms, probabilities
Correct Solution:
```
N = int(input())
for _ in range(N):
n = int(input())
a = []
for i in range(0,n):
a.insert(0,i+1)
if n % 2==1:
a[0] = a[n // 2]
a[n // 2] = n
print(*a)
```
| 97,202 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given one integer n (n > 1).
Recall that a permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation of length 5, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n = 3 but there is 4 in the array).
Your task is to find a permutation p of length n that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).
You have to answer t independent test cases.
If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (2 ≤ n ≤ 100) — the length of the permutation you have to find.
Output
For each test case, print n distinct integers p_1, p_2, …, p_n — a permutation that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).
If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.
Example
Input
2
2
5
Output
2 1
2 1 5 3 4
Tags: constructive algorithms, probabilities
Correct Solution:
```
for _ in range(int(input())):
N = int(input())
for i in range(2,N+1):
print(i, end = " ")
print(1)
```
| 97,203 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given one integer n (n > 1).
Recall that a permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation of length 5, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n = 3 but there is 4 in the array).
Your task is to find a permutation p of length n that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).
You have to answer t independent test cases.
If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (2 ≤ n ≤ 100) — the length of the permutation you have to find.
Output
For each test case, print n distinct integers p_1, p_2, …, p_n — a permutation that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).
If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.
Example
Input
2
2
5
Output
2 1
2 1 5 3 4
Tags: constructive algorithms, probabilities
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
if n % 2 == 0:
for i in range(n, 0, -1):
print(i, end=' ')
print()
else:
for i in range(n, n // 2 + 1, -1):
print(i, end=' ')
for i in range(1, n // 2 + 2):
print(i, end=' ')
print()
```
| 97,204 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given one integer n (n > 1).
Recall that a permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation of length 5, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n = 3 but there is 4 in the array).
Your task is to find a permutation p of length n that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).
You have to answer t independent test cases.
If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (2 ≤ n ≤ 100) — the length of the permutation you have to find.
Output
For each test case, print n distinct integers p_1, p_2, …, p_n — a permutation that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).
If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.
Example
Input
2
2
5
Output
2 1
2 1 5 3 4
Tags: constructive algorithms, probabilities
Correct Solution:
```
def solve(n):
a=[0]*n
a[0]=n
for i in range(1,n):
a[i]=i
return a
t=int(input())
ans=[]
while t>0:
n=int(input(''))
ans.append(solve(n))
t-=1
for i in ans:
for j in i:
print(j,end=' ')
print()
```
| 97,205 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given one integer n (n > 1).
Recall that a permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation of length 5, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n = 3 but there is 4 in the array).
Your task is to find a permutation p of length n that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).
You have to answer t independent test cases.
If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (2 ≤ n ≤ 100) — the length of the permutation you have to find.
Output
For each test case, print n distinct integers p_1, p_2, …, p_n — a permutation that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).
If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.
Example
Input
2
2
5
Output
2 1
2 1 5 3 4
Tags: constructive algorithms, probabilities
Correct Solution:
```
t=int(input())
for q in range(t):
n=int(input())
ch=str(n)+" "
for i in range(1,n):
ch+=str(i)+" "
print(ch)
```
| 97,206 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given one integer n (n > 1).
Recall that a permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation of length 5, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n = 3 but there is 4 in the array).
Your task is to find a permutation p of length n that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).
You have to answer t independent test cases.
If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (2 ≤ n ≤ 100) — the length of the permutation you have to find.
Output
For each test case, print n distinct integers p_1, p_2, …, p_n — a permutation that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).
If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.
Example
Input
2
2
5
Output
2 1
2 1 5 3 4
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
for i in range(2,n+1):
print(i,end=" ")
print(1)
```
Yes
| 97,207 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given one integer n (n > 1).
Recall that a permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation of length 5, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n = 3 but there is 4 in the array).
Your task is to find a permutation p of length n that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).
You have to answer t independent test cases.
If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (2 ≤ n ≤ 100) — the length of the permutation you have to find.
Output
For each test case, print n distinct integers p_1, p_2, …, p_n — a permutation that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).
If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.
Example
Input
2
2
5
Output
2 1
2 1 5 3 4
Submitted Solution:
```
def oneLineArrayPrint(arr):
print(' '.join([str(x) for x in arr]))
t=int(input())
for _ in range(t):
n=int(input())
oneLineArrayPrint([n]+list(range(1,n)))
```
Yes
| 97,208 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given one integer n (n > 1).
Recall that a permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation of length 5, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n = 3 but there is 4 in the array).
Your task is to find a permutation p of length n that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).
You have to answer t independent test cases.
If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (2 ≤ n ≤ 100) — the length of the permutation you have to find.
Output
For each test case, print n distinct integers p_1, p_2, …, p_n — a permutation that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).
If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.
Example
Input
2
2
5
Output
2 1
2 1 5 3 4
Submitted Solution:
```
t = int(input())
for i in range(t):
n = int(input())
a = []
while n>0:
a.append(n)
n-=1
if len(a)%2 == 0:
print(" ".join(map(str,a)))
else:
a1 = a[len(a)//2]
a.remove(a1)
a.append(a1)
print(" ".join(map(str,a)))
```
Yes
| 97,209 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given one integer n (n > 1).
Recall that a permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation of length 5, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n = 3 but there is 4 in the array).
Your task is to find a permutation p of length n that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).
You have to answer t independent test cases.
If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (2 ≤ n ≤ 100) — the length of the permutation you have to find.
Output
For each test case, print n distinct integers p_1, p_2, …, p_n — a permutation that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).
If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.
Example
Input
2
2
5
Output
2 1
2 1 5 3 4
Submitted Solution:
```
import random
t = int(input())
while t:
n = int(input())
res= [n]
for i in range(1,n):
res.append(i)
print(" ".join(map(str,res)))
t-=1
```
Yes
| 97,210 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given one integer n (n > 1).
Recall that a permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation of length 5, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n = 3 but there is 4 in the array).
Your task is to find a permutation p of length n that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).
You have to answer t independent test cases.
If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (2 ≤ n ≤ 100) — the length of the permutation you have to find.
Output
For each test case, print n distinct integers p_1, p_2, …, p_n — a permutation that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).
If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.
Example
Input
2
2
5
Output
2 1
2 1 5 3 4
Submitted Solution:
```
t = int(input())
for i in range (t):
n = int(input())
p = []
for j in range (n):
p.append(n)
n -= 1
print(*p)
```
No
| 97,211 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given one integer n (n > 1).
Recall that a permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation of length 5, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n = 3 but there is 4 in the array).
Your task is to find a permutation p of length n that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).
You have to answer t independent test cases.
If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (2 ≤ n ≤ 100) — the length of the permutation you have to find.
Output
For each test case, print n distinct integers p_1, p_2, …, p_n — a permutation that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).
If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.
Example
Input
2
2
5
Output
2 1
2 1 5 3 4
Submitted Solution:
```
for i in range(int(input())):
n = int(input())
a = []
for i in range(2,n+1,2):
a.append(i)
for i in range(1,n+1,2):
a.append(i)
print(*a)
```
No
| 97,212 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given one integer n (n > 1).
Recall that a permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation of length 5, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n = 3 but there is 4 in the array).
Your task is to find a permutation p of length n that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).
You have to answer t independent test cases.
If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (2 ≤ n ≤ 100) — the length of the permutation you have to find.
Output
For each test case, print n distinct integers p_1, p_2, …, p_n — a permutation that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).
If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.
Example
Input
2
2
5
Output
2 1
2 1 5 3 4
Submitted Solution:
```
for i in range(int(input())):
n=int(input())
for j in range(n):
if j==n-1:
print(1)
else:
print(n-j,end=" ")
print()
```
No
| 97,213 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given one integer n (n > 1).
Recall that a permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation of length 5, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n = 3 but there is 4 in the array).
Your task is to find a permutation p of length n that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).
You have to answer t independent test cases.
If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (2 ≤ n ≤ 100) — the length of the permutation you have to find.
Output
For each test case, print n distinct integers p_1, p_2, …, p_n — a permutation that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).
If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.
Example
Input
2
2
5
Output
2 1
2 1 5 3 4
Submitted Solution:
```
import sys
# problem a
# template by:
# https://github.com/rajatg98
'''input
'''
import sys
import math
import bisect
from sys import stdin,stdout
from math import gcd,floor,sqrt,log
from collections import defaultdict as dd
from bisect import bisect_left as bl,bisect_right as br
sys.setrecursionlimit(100000000)
inp =lambda: int(input())
strng =lambda: input().strip()
jn =lambda x,l: x.join(map(str,l))
strl =lambda: list(input().strip())
mul =lambda: map(int,input().strip().split())
mulf =lambda: map(float,input().strip().split())
seq =lambda: list(map(int,input().strip().split()))
ceil =lambda x: int(x) if(x==int(x)) else int(x)+1
ceildiv=lambda x,d: x//d if(x%d==0) else x//d+1
flush =lambda: stdout.flush()
stdstr =lambda: stdin.readline()
stdint =lambda: int(stdin.readline())
stdpr =lambda x: stdout.write(str(x))
mod=1000000007
#main code
def solve():
n = inp()
print([n] + list(range(1, n)))
def main():
tests = inp()
for _ in range(tests):
solve()
if __name__ == "__main__":
main()
```
No
| 97,214 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Interaction
This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.
In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.
Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.
Example
Input
Is it rated?
Is it rated?
Is it rated?
Output
NO
NO
NO
Tags: *special, implementation, interactive
Correct Solution:
```
try:
while input():
print('NO')
except EOFError:
pass
```
| 97,215 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Interaction
This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.
In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.
Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.
Example
Input
Is it rated?
Is it rated?
Is it rated?
Output
NO
NO
NO
Tags: *special, implementation, interactive
Correct Solution:
```
while True:
try:
q = input()
except EOFError:
break
print("no", flush=True)
```
| 97,216 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Interaction
This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.
In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.
Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.
Example
Input
Is it rated?
Is it rated?
Is it rated?
Output
NO
NO
NO
Tags: *special, implementation, interactive
Correct Solution:
```
while(1):
try:
s=input()
except:
break
print("NO",flush=True)
```
| 97,217 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Interaction
This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.
In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.
Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.
Example
Input
Is it rated?
Is it rated?
Is it rated?
Output
NO
NO
NO
Tags: *special, implementation, interactive
Correct Solution:
```
while True:
try:
s=input()
print("NO")
except EOFError as e:
exit(0)
```
| 97,218 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Interaction
This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.
In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.
Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.
Example
Input
Is it rated?
Is it rated?
Is it rated?
Output
NO
NO
NO
Tags: *special, implementation, interactive
Correct Solution:
```
while True:
try:
x = input()
except:
break
print("NO")
```
| 97,219 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Interaction
This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.
In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.
Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.
Example
Input
Is it rated?
Is it rated?
Is it rated?
Output
NO
NO
NO
Tags: *special, implementation, interactive
Correct Solution:
```
ll=lambda:map(int,input().split())
t=lambda:int(input())
ss=lambda:input()
lx=lambda x:map(int,input().split(x))
#from math import log10 ,log2,ceil,factorial as fac,gcd
#from itertools import combinations_with_replacement as cs
#from functools import reduce
#from bisect import bisect_right as br,bisect_left as bl
#from collections import Counter
import sys
#for _ in range(t()):
while 1:
try:
n=ss()
if n=="Is it rated?":
print("NO")
else:
print("YES")
except:
break
sys.stdout.flush()
'''
1+3+9
(3**n-1)//2
'''
```
| 97,220 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Interaction
This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.
In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.
Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.
Example
Input
Is it rated?
Is it rated?
Is it rated?
Output
NO
NO
NO
Tags: *special, implementation, interactive
Correct Solution:
```
while True:
try :
line = input().strip()
print("No")
except EOFError:
break
```
| 97,221 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Interaction
This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.
In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.
Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.
Example
Input
Is it rated?
Is it rated?
Is it rated?
Output
NO
NO
NO
Tags: *special, implementation, interactive
Correct Solution:
```
while True:
try:
input()
except EOFError:
break
print("NO")
```
| 97,222 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Interaction
This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.
In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.
Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.
Example
Input
Is it rated?
Is it rated?
Is it rated?
Output
NO
NO
NO
Submitted Solution:
```
try:
while True:
q=input()
print('NO')
except:
pass
```
Yes
| 97,223 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Interaction
This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.
In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.
Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.
Example
Input
Is it rated?
Is it rated?
Is it rated?
Output
NO
NO
NO
Submitted Solution:
```
print("NO\n"*99)
```
Yes
| 97,224 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Interaction
This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.
In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.
Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.
Example
Input
Is it rated?
Is it rated?
Is it rated?
Output
NO
NO
NO
Submitted Solution:
```
print(100*'NO''\n')
```
Yes
| 97,225 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Interaction
This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.
In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.
Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.
Example
Input
Is it rated?
Is it rated?
Is it rated?
Output
NO
NO
NO
Submitted Solution:
```
while True:
try:
s = input()
if s == "Is it rated?":
print('NO')
else:
break
except:
break
```
Yes
| 97,226 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Interaction
This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.
In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.
Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.
Example
Input
Is it rated?
Is it rated?
Is it rated?
Output
NO
NO
NO
Submitted Solution:
```
s = input()
print("NO")
```
No
| 97,227 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Interaction
This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.
In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.
Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.
Example
Input
Is it rated?
Is it rated?
Is it rated?
Output
NO
NO
NO
Submitted Solution:
```
n=input()
if(n=="Is it rated?"):
print("NO")
```
No
| 97,228 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Interaction
This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.
In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.
Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.
Example
Input
Is it rated?
Is it rated?
Is it rated?
Output
NO
NO
NO
Submitted Solution:
```
try:
s = input()
print("NO")
except:
exit(0)
```
No
| 97,229 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Interaction
This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.
In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.
Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.
Example
Input
Is it rated?
Is it rated?
Is it rated?
Output
NO
NO
NO
Submitted Solution:
```
print("No")
```
No
| 97,230 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kavi has 2n points lying on the OX axis, i-th of which is located at x = i.
Kavi considers all ways to split these 2n points into n pairs. Among those, he is interested in good pairings, which are defined as follows:
Consider n segments with ends at the points in correspondent pairs. The pairing is called good, if for every 2 different segments A and B among those, at least one of the following holds:
* One of the segments A and B lies completely inside the other.
* A and B have the same length.
Consider the following example:
<image>
A is a good pairing since the red segment lies completely inside the blue segment.
B is a good pairing since the red and the blue segment have the same length.
C is not a good pairing since none of the red or blue segments lies inside the other, neither do they have the same size.
Kavi is interested in the number of good pairings, so he wants you to find it for him. As the result can be large, find this number modulo 998244353.
Two pairings are called different, if some two points are in one pair in some pairing and in different pairs in another.
Input
The single line of the input contains a single integer n (1≤ n ≤ 10^6).
Output
Print the number of good pairings modulo 998244353.
Examples
Input
1
Output
1
Input
2
Output
3
Input
3
Output
6
Input
100
Output
688750769
Note
The good pairings for the second example are:
<image>
In the third example, the good pairings are:
<image>
Tags: combinatorics, dp, math
Correct Solution:
```
from collections import defaultdict
from itertools import accumulate
from bisect import bisect_left
from collections import deque
from collections import Counter
import sys
input = sys.stdin.readline
'''
for CASES in range(int(input())):
n, m = map(int, input().split())
n = int(input())
A = list(map(int, input().split()))
S = input().strip()
sys.stdout.write(" ".join(map(str,ANS))+"\n")
'''
inf = 100000000000000000 # 1e17
mod = 998244353
n = int(input())
if n==1:
print(1)
sys.exit(0)
dp=[0]*(n+1)
f=[0]*(n+1)
f[0]=1
f[1]=1
dp[0]=0
dp[1]=1
sumf=0
sumdp=1
#print("!")
last=[0]*(2*n+1)
for i in range(2,2*n+1):
if i%2==0:
j=i+i
while j<=2*n:
last[j]+=1
last[j]%=mod
j=j+i
#print(last)
for i in range(2,n+1):
sumf+=f[i-2]
sumf%=mod
dp[i]=sumf+last[i*2]
dp[i]%=mod
sumdp+=dp[i] # 所有裸露的直接往上面包东西
sumdp%=mod
f[i]=sumdp
f[i]%=mod
print(f[n])
#
# dp=[0]*(n+1)
# f=[0]*(n+1)
# sumdp=[0]*(n+1)
# sumf=[0]*(n+1)
#
# sumdp[1]=3
# sumf[0]=2
# sumf[1]=2
# for i in range(2,n+1):
# sumdp[i]=sumdp[i-1]
# sumf[i]=sumf[i-1]
#
# f[i]=sumdp[i]
# dp[i]=sumf[i-2]
#
# sumdp[i]+=dp[i]
# sumf[i]+=f[i]
```
| 97,231 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kavi has 2n points lying on the OX axis, i-th of which is located at x = i.
Kavi considers all ways to split these 2n points into n pairs. Among those, he is interested in good pairings, which are defined as follows:
Consider n segments with ends at the points in correspondent pairs. The pairing is called good, if for every 2 different segments A and B among those, at least one of the following holds:
* One of the segments A and B lies completely inside the other.
* A and B have the same length.
Consider the following example:
<image>
A is a good pairing since the red segment lies completely inside the blue segment.
B is a good pairing since the red and the blue segment have the same length.
C is not a good pairing since none of the red or blue segments lies inside the other, neither do they have the same size.
Kavi is interested in the number of good pairings, so he wants you to find it for him. As the result can be large, find this number modulo 998244353.
Two pairings are called different, if some two points are in one pair in some pairing and in different pairs in another.
Input
The single line of the input contains a single integer n (1≤ n ≤ 10^6).
Output
Print the number of good pairings modulo 998244353.
Examples
Input
1
Output
1
Input
2
Output
3
Input
3
Output
6
Input
100
Output
688750769
Note
The good pairings for the second example are:
<image>
In the third example, the good pairings are:
<image>
Tags: combinatorics, dp, math
Correct Solution:
```
# Python3 program to count
# Python3 program to count
# number of factors
# of an array of integers
MAX = 1000001;
# array to store
# prime factors
factor = [0]*(MAX + 1);
# function to generate all
# prime factors of numbers
# from 1 to 10^6
def generatePrimeFactors():
factor[1] = 1;
# Initializes all the
# positions with their value.
for i in range(2,MAX):
factor[i] = i;
# Initializes all
# multiples of 2 with 2
for i in range(4,MAX,2):
factor[i] = 2;
# A modified version of
# Sieve of Eratosthenes
# to store the smallest
# prime factor that divides
# every number.
i = 3;
while(i * i < MAX):
# check if it has
# no prime factor.
if (factor[i] == i):
# Initializes of j
# starting from i*i
j = i * i;
while(j < MAX):
# if it has no prime factor
# before, then stores the
# smallest prime divisor
if (factor[j] == j):
factor[j] = i;
j += i;
i+=1;
# function to calculate
# number of factors
def calculateNoOFactors(n):
if (n == 1):
return 1;
ans = 1;
# stores the smallest
# prime number that
# divides n
dup = factor[n];
# stores the count of
# number of times a
# prime number divides n.
c = 1;
# reduces to the next
# number after prime
# factorization of n
j = int(n / factor[n]);
# false when prime
# factorization is done
while (j > 1):
# if the same prime
# number is dividing
# n, then we increase
# the count
if (factor[j] == dup):
c += 1;
# if its a new prime factor
# that is factorizing n,
# then we again set c=1 and
# change dup to the new prime
# factor, and apply the formula
# explained above.
else:
dup = factor[j];
ans = ans * (c + 1);
c = 1;
# prime factorizes
# a number
j = int(j / factor[j]);
# for the last
# prime factor
ans = ans * (c + 1);
return ans;
# Driver Code
if __name__ == "__main__":
# generate prime factors
# of number upto 10^6
generatePrimeFactors()
a = [10, 30, 100, 450, 987]
q = len(a)
for i in range (0,q):
pass
# This code is contributed
# by mits.
n=int(input())
l=[0,1,3,6]
alp=calculateNoOFactors(3)
if n>3:
for i in range(4,n+1):
k=calculateNoOFactors(i)
l.append(((2*l[i-1])+k-alp)%998244353)
alp=k
print(l[-1])
else:print(l[n])
```
| 97,232 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kavi has 2n points lying on the OX axis, i-th of which is located at x = i.
Kavi considers all ways to split these 2n points into n pairs. Among those, he is interested in good pairings, which are defined as follows:
Consider n segments with ends at the points in correspondent pairs. The pairing is called good, if for every 2 different segments A and B among those, at least one of the following holds:
* One of the segments A and B lies completely inside the other.
* A and B have the same length.
Consider the following example:
<image>
A is a good pairing since the red segment lies completely inside the blue segment.
B is a good pairing since the red and the blue segment have the same length.
C is not a good pairing since none of the red or blue segments lies inside the other, neither do they have the same size.
Kavi is interested in the number of good pairings, so he wants you to find it for him. As the result can be large, find this number modulo 998244353.
Two pairings are called different, if some two points are in one pair in some pairing and in different pairs in another.
Input
The single line of the input contains a single integer n (1≤ n ≤ 10^6).
Output
Print the number of good pairings modulo 998244353.
Examples
Input
1
Output
1
Input
2
Output
3
Input
3
Output
6
Input
100
Output
688750769
Note
The good pairings for the second example are:
<image>
In the third example, the good pairings are:
<image>
Tags: combinatorics, dp, math
Correct Solution:
```
n=int(input())
dp=[1]*(10**6+5)
for i in range(2,n+1):
for j in range(i,n+1,i):
dp[j]+=1
x=1
ans=1
m=998244353
for i in range(1,n):
ans=(dp[i+1]+x)%m
x+=(dp[i+1]+x)%m
print(ans)
```
| 97,233 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kavi has 2n points lying on the OX axis, i-th of which is located at x = i.
Kavi considers all ways to split these 2n points into n pairs. Among those, he is interested in good pairings, which are defined as follows:
Consider n segments with ends at the points in correspondent pairs. The pairing is called good, if for every 2 different segments A and B among those, at least one of the following holds:
* One of the segments A and B lies completely inside the other.
* A and B have the same length.
Consider the following example:
<image>
A is a good pairing since the red segment lies completely inside the blue segment.
B is a good pairing since the red and the blue segment have the same length.
C is not a good pairing since none of the red or blue segments lies inside the other, neither do they have the same size.
Kavi is interested in the number of good pairings, so he wants you to find it for him. As the result can be large, find this number modulo 998244353.
Two pairings are called different, if some two points are in one pair in some pairing and in different pairs in another.
Input
The single line of the input contains a single integer n (1≤ n ≤ 10^6).
Output
Print the number of good pairings modulo 998244353.
Examples
Input
1
Output
1
Input
2
Output
3
Input
3
Output
6
Input
100
Output
688750769
Note
The good pairings for the second example are:
<image>
In the third example, the good pairings are:
<image>
Tags: combinatorics, dp, math
Correct Solution:
```
p = 998244353
n = int(input())
if n == 1:
print(1)
else:
divs = [0]*n
for i in range(1,n+1):
j = i
while j <= n:
divs[j-1] += 1
j += i
total = 1
for i in range(1,n-1):
total = (total*2+divs[i])%p
print((total+divs[n-1])%p)
```
| 97,234 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kavi has 2n points lying on the OX axis, i-th of which is located at x = i.
Kavi considers all ways to split these 2n points into n pairs. Among those, he is interested in good pairings, which are defined as follows:
Consider n segments with ends at the points in correspondent pairs. The pairing is called good, if for every 2 different segments A and B among those, at least one of the following holds:
* One of the segments A and B lies completely inside the other.
* A and B have the same length.
Consider the following example:
<image>
A is a good pairing since the red segment lies completely inside the blue segment.
B is a good pairing since the red and the blue segment have the same length.
C is not a good pairing since none of the red or blue segments lies inside the other, neither do they have the same size.
Kavi is interested in the number of good pairings, so he wants you to find it for him. As the result can be large, find this number modulo 998244353.
Two pairings are called different, if some two points are in one pair in some pairing and in different pairs in another.
Input
The single line of the input contains a single integer n (1≤ n ≤ 10^6).
Output
Print the number of good pairings modulo 998244353.
Examples
Input
1
Output
1
Input
2
Output
3
Input
3
Output
6
Input
100
Output
688750769
Note
The good pairings for the second example are:
<image>
In the third example, the good pairings are:
<image>
Tags: combinatorics, dp, math
Correct Solution:
```
import sys, math
import io, os
#data = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
#from bisect import bisect_left as bl, bisect_right as br, insort
#from heapq import heapify, heappush, heappop
#from collections import defaultdict as dd, deque, Counter
#from itertools import permutations,combinations
def data(): return sys.stdin.buffer.readline().strip()
def mdata(): return list(map(int, data().split()))
def outl(var) : sys.stdout.write(' '.join(map(str, var))+'\n')
def out(var) : sys.stdout.write(str(var)+'\n')
#from decimal import Decimal
#from fractions import Fraction
#sys.setrecursionlimit(100000)
#INF = float('inf')
mod = 998244353
n = int(data())
l=[1]*(n+1)
for i in range(1,n+1):
for j in range(i*2,n+1,i):
l[j]+=1
dp = [1]
s = 1
for i in range(1,n+1):
a = s
a+=l[i]-1
dp.append(a%mod)
s+=dp[i]
print(dp[-1])
```
| 97,235 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kavi has 2n points lying on the OX axis, i-th of which is located at x = i.
Kavi considers all ways to split these 2n points into n pairs. Among those, he is interested in good pairings, which are defined as follows:
Consider n segments with ends at the points in correspondent pairs. The pairing is called good, if for every 2 different segments A and B among those, at least one of the following holds:
* One of the segments A and B lies completely inside the other.
* A and B have the same length.
Consider the following example:
<image>
A is a good pairing since the red segment lies completely inside the blue segment.
B is a good pairing since the red and the blue segment have the same length.
C is not a good pairing since none of the red or blue segments lies inside the other, neither do they have the same size.
Kavi is interested in the number of good pairings, so he wants you to find it for him. As the result can be large, find this number modulo 998244353.
Two pairings are called different, if some two points are in one pair in some pairing and in different pairs in another.
Input
The single line of the input contains a single integer n (1≤ n ≤ 10^6).
Output
Print the number of good pairings modulo 998244353.
Examples
Input
1
Output
1
Input
2
Output
3
Input
3
Output
6
Input
100
Output
688750769
Note
The good pairings for the second example are:
<image>
In the third example, the good pairings are:
<image>
Tags: combinatorics, dp, math
Correct Solution:
```
MOD = 998244353
n = int(input())
dp = [None] * (n+1)
pref = [None] * (n+1)
div = [2] * (n+1)
dp[1] = 1
pref[1] = 1
for i in range(2,n+1):
dp[i] = (div[i] + pref[i-1]) % MOD
pref[i] = (pref[i-1] + dp[i]) % MOD
for j in range(i<<1, n+1, i):
div[j] += 1
print(dp[n])
```
| 97,236 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kavi has 2n points lying on the OX axis, i-th of which is located at x = i.
Kavi considers all ways to split these 2n points into n pairs. Among those, he is interested in good pairings, which are defined as follows:
Consider n segments with ends at the points in correspondent pairs. The pairing is called good, if for every 2 different segments A and B among those, at least one of the following holds:
* One of the segments A and B lies completely inside the other.
* A and B have the same length.
Consider the following example:
<image>
A is a good pairing since the red segment lies completely inside the blue segment.
B is a good pairing since the red and the blue segment have the same length.
C is not a good pairing since none of the red or blue segments lies inside the other, neither do they have the same size.
Kavi is interested in the number of good pairings, so he wants you to find it for him. As the result can be large, find this number modulo 998244353.
Two pairings are called different, if some two points are in one pair in some pairing and in different pairs in another.
Input
The single line of the input contains a single integer n (1≤ n ≤ 10^6).
Output
Print the number of good pairings modulo 998244353.
Examples
Input
1
Output
1
Input
2
Output
3
Input
3
Output
6
Input
100
Output
688750769
Note
The good pairings for the second example are:
<image>
In the third example, the good pairings are:
<image>
Tags: combinatorics, dp, math
Correct Solution:
```
'''from bisect import bisect,bisect_left
from collections import *
from heapq import *
from math import gcd,ceil,sqrt,floor,inf
from itertools import *
from operator import add,mul,sub,xor,truediv,floordiv
from functools import *'''
#------------------------------------------------------------------------
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#------------------------------------------------------------------------
def RL(): return map(int, sys.stdin.readline().split())
def RLL(): return list(map(int, sys.stdin.readline().split()))
def N(): return int(input())
def A(n):return [0]*n
def AI(n,x): return [x]*n
def A2(n,m): return [[0]*m for i in range(n)]
def G(n): return [[] for i in range(n)]
def GP(it): return [[ch,len(list(g))] for ch,g in groupby(it)]
#------------------------------------------------------------------------
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
mod=10**9+7
farr=[1]
ifa=[]
def fact(x,mod=0):
if mod:
while x>=len(farr):
farr.append(farr[-1]*len(farr)%mod)
else:
while x>=len(farr):
farr.append(farr[-1]*len(farr))
return farr[x]
def ifact(x,mod):
global ifa
fact(x,mod)
ifa.append(pow(farr[-1],mod-2,mod))
for i in range(x,0,-1):
ifa.append(ifa[-1]*i%mod)
ifa.reverse()
def per(i,j,mod=0):
if i<j: return 0
if not mod:
return fact(i)//fact(i-j)
return farr[i]*ifa[i-j]%mod
def com(i,j,mod=0):
if i<j: return 0
if not mod:
return per(i,j)//fact(j)
return per(i,j,mod)*ifa[j]%mod
def catalan(n):
return com(2*n,n)//(n+1)
def isprime(n):
for i in range(2,int(n**0.5)+1):
if n%i==0:
return False
return True
def floorsum(a,b,c,n):#sum((a*i+b)//c for i in range(n+1))
if a==0:return b//c*(n+1)
if a>=c or b>=c: return floorsum(a%c,b%c,c,n)+b//c*(n+1)+a//c*n*(n+1)//2
m=(a*n+b)//c
return n*m-floorsum(c,c-b-1,a,m-1)
def inverse(a,m):
a%=m
if a<=1: return a
return ((1-inverse(m,a)*m)//a)%m
def lowbit(n):
return n&-n
class BIT:
def __init__(self,arr):
self.arr=arr
self.n=len(arr)-1
def update(self,x,v):
while x<=self.n:
self.arr[x]+=v
x+=x&-x
def query(self,x):
ans=0
while x:
ans+=self.arr[x]
x&=x-1
return ans
class ST:
def __init__(self,arr):#n!=0
n=len(arr)
mx=n.bit_length()#取不到
self.st=[[0]*mx for i in range(n)]
for i in range(n):
self.st[i][0]=arr[i]
for j in range(1,mx):
for i in range(n-(1<<j)+1):
self.st[i][j]=max(self.st[i][j-1],self.st[i+(1<<j-1)][j-1])
def query(self,l,r):
if l>r:return -inf
s=(r+1-l).bit_length()-1
return max(self.st[l][s],self.st[r-(1<<s)+1][s])
class DSU:#容量+路径压缩
def __init__(self,n):
self.c=[-1]*n
def same(self,x,y):
return self.find(x)==self.find(y)
def find(self,x):
if self.c[x]<0:
return x
self.c[x]=self.find(self.c[x])
return self.c[x]
def union(self,u,v):
u,v=self.find(u),self.find(v)
if u==v:
return False
if self.c[u]>self.c[v]:
u,v=v,u
self.c[u]+=self.c[v]
self.c[v]=u
return True
def size(self,x): return -self.c[self.find(x)]
class UFS:#秩+路径
def __init__(self,n):
self.parent=[i for i in range(n)]
self.ranks=[0]*n
def find(self,x):
if x!=self.parent[x]:
self.parent[x]=self.find(self.parent[x])
return self.parent[x]
def union(self,u,v):
pu,pv=self.find(u),self.find(v)
if pu==pv:
return False
if self.ranks[pu]>=self.ranks[pv]:
self.parent[pv]=pu
if self.ranks[pv]==self.ranks[pu]:
self.ranks[pu]+=1
else:
self.parent[pu]=pv
class UF:#秩+路径+容量，边数
def __init__(self,n):
self.parent=[i for i in range(n)]
self.ranks=[0]*n
self.size=AI(n,1)
self.edge=A(n)
def find(self,x):
if x!=self.parent[x]:
self.parent[x]=self.find(self.parent[x])
return self.parent[x]
def union(self,u,v):
pu,pv=self.find(u),self.find(v)
if pu==pv:
self.edge[pu]+=1
return False
if self.ranks[pu]>=self.ranks[pv]:
self.parent[pv]=pu
self.edge[pu]+=self.edge[pv]+1
self.size[pu]+=self.size[pv]
if self.ranks[pv]==self.ranks[pu]:
self.ranks[pu]+=1
else:
self.parent[pu]=pv
self.edge[pv]+=self.edge[pu]+1
self.size[pv]+=self.size[pu]
def Prime(n):
c=0
prime=[]
flag=[0]*(n+1)
for i in range(2,n+1):
if not flag[i]:
prime.append(i)
c+=1
for j in range(c):
if i*prime[j]>n: break
flag[i*prime[j]]=prime[j]
if i%prime[j]==0: break
return flag
def dij(s,graph):
d=AI(n,inf)
d[s]=0
heap=[(0,s)]
vis=A(n)
while heap:
dis,u=heappop(heap)
if vis[u]:
continue
vis[u]=1
for v,w in graph[u]:
if d[v]>d[u]+w:
d[v]=d[u]+w
heappush(heap,(d[v],v))
return d
def bell(s,g):#bellman-Ford
dis=AI(n,inf)
dis[s]=0
for i in range(n-1):
for u,v,w in edge:
if dis[v]>dis[u]+w:
dis[v]=dis[u]+w
change=A(n)
for i in range(n):
for u,v,w in edge:
if dis[v]>dis[u]+w:
dis[v]=dis[u]+w
change[v]=1
return dis
def lcm(a,b): return a*b//gcd(a,b)
def lis(nums):
res=[]
for k in nums:
i=bisect.bisect_left(res,k)
if i==len(res):
res.append(k)
else:
res[i]=k
return len(res)
def RP(nums):#逆序对
n = len(nums)
s=set(nums)
d={}
for i,k in enumerate(sorted(s),1):
d[k]=i
bi=BIT([0]*(len(s)+1))
ans=0
for i in range(n-1,-1,-1):
ans+=bi.query(d[nums[i]]-1)
bi.update(d[nums[i]],1)
return ans
def michange(a,b):
d=defaultdict(deque)
for i,x in enumerate(b):
d[x].append(i)
order=A(len(a))
for i,x in enumerate(a):
if not d:
return -1
order[i]=d[x].popleft()
return RP(order)
class DLN:
def __init__(self,val):
self.val=val
self.pre=None
self.next=None
def nb(i,j,n,m):
for ni,nj in [[i+1,j],[i-1,j],[i,j-1],[i,j+1]]:
if 0<=ni<n and 0<=nj<m:
yield ni,nj
def topo(n):
q=deque()
res=[]
for i in range(1,n+1):
if ind[i]==0:
q.append(i)
res.append(i)
while q:
u=q.popleft()
for v in g[u]:
ind[v]-=1
if ind[v]==0:
q.append(v)
res.append(v)
return res
@bootstrap
def gdfs(r,p):
for ch in g[r]:
if ch!=p:
yield gdfs(ch,r)
yield None
'''
from random import randint
def ra(n,a,b):
return [randint(a,b) for i in range(n)]
'''
@bootstrap
def dfs(x):
global T
tin[x]=T
T+=1
for ch in gb[x]:
yield dfs(ch)
tout[x]=T
T+=1
yield None
mod=998244353
t=1
for i in range(t):
n=N()
dp=AI(n+1,1)
pre=2
for i in range(2,n//2+1):
for j in range(2*i,n+1,i):
dp[j]+=1
for i in range(2,n+1):
dp[i]=(pre+dp[i])%mod
pre=(pre+dp[i])%mod
ans=dp[-1]
print(ans)
'''
sys.setrecursionlimit(200000)
import threading
threading.sta1ck_size(10**8)
t=threading.Thr
ead(target=main)
t.start()
t.join()
'''
```
| 97,237 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kavi has 2n points lying on the OX axis, i-th of which is located at x = i.
Kavi considers all ways to split these 2n points into n pairs. Among those, he is interested in good pairings, which are defined as follows:
Consider n segments with ends at the points in correspondent pairs. The pairing is called good, if for every 2 different segments A and B among those, at least one of the following holds:
* One of the segments A and B lies completely inside the other.
* A and B have the same length.
Consider the following example:
<image>
A is a good pairing since the red segment lies completely inside the blue segment.
B is a good pairing since the red and the blue segment have the same length.
C is not a good pairing since none of the red or blue segments lies inside the other, neither do they have the same size.
Kavi is interested in the number of good pairings, so he wants you to find it for him. As the result can be large, find this number modulo 998244353.
Two pairings are called different, if some two points are in one pair in some pairing and in different pairs in another.
Input
The single line of the input contains a single integer n (1≤ n ≤ 10^6).
Output
Print the number of good pairings modulo 998244353.
Examples
Input
1
Output
1
Input
2
Output
3
Input
3
Output
6
Input
100
Output
688750769
Note
The good pairings for the second example are:
<image>
In the third example, the good pairings are:
<image>
Tags: combinatorics, dp, math
Correct Solution:
```
# Problem: B. Kavi on Pairing Duty
# Contest: Codeforces - Codeforces Round #722 (Div. 1)
# URL: https://codeforces.com/contest/1528/problem/B
# Memory Limit: 256 MB
# Time Limit: 1000 ms
#
# Powered by CP Editor (https://cpeditor.org)
from collections import defaultdict
from functools import reduce
from bisect import bisect_right
from bisect import bisect_left
#import sympy #Prime number library
import copy
import heapq
mod=998244353
def main():
n=int(input())
dp=[0]*(2*n+1)
prev=[0]*(2*n+1)
dp[0]=1;prev[0]=1;
NumOfDivisors(n)
for i in range(2,2*n+1,2):
dp[i]=(prev[i-2]+div[i//2]-1)%mod
prev[i]=(prev[i-2]+dp[i])%mod
print(dp[2*n])
div=[0]*(1000001)
def NumOfDivisors(n): #O(nlog(n))
for i in range(1,n+1):
for j in range(i,n+1,i):
div[j]+=1
def primes1(n): #return a list having prime numbers from 2 to n
""" Returns a list of primes < n """
sieve = [True] * (n//2)
for i in range(3,int(n**0.5)+1,2):
if sieve[i//2]:
sieve[i*i//2::i] = [False] * ((n-i*i-1)//(2*i)+1)
return [2] + [2*i+1 for i in range(1,n//2) if sieve[i]]
def GCD(a,b):
if(b==0):
return a
else:
return GCD(b,a%b)
if __name__ == '__main__':
main()
```
| 97,238 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kavi has 2n points lying on the OX axis, i-th of which is located at x = i.
Kavi considers all ways to split these 2n points into n pairs. Among those, he is interested in good pairings, which are defined as follows:
Consider n segments with ends at the points in correspondent pairs. The pairing is called good, if for every 2 different segments A and B among those, at least one of the following holds:
* One of the segments A and B lies completely inside the other.
* A and B have the same length.
Consider the following example:
<image>
A is a good pairing since the red segment lies completely inside the blue segment.
B is a good pairing since the red and the blue segment have the same length.
C is not a good pairing since none of the red or blue segments lies inside the other, neither do they have the same size.
Kavi is interested in the number of good pairings, so he wants you to find it for him. As the result can be large, find this number modulo 998244353.
Two pairings are called different, if some two points are in one pair in some pairing and in different pairs in another.
Input
The single line of the input contains a single integer n (1≤ n ≤ 10^6).
Output
Print the number of good pairings modulo 998244353.
Examples
Input
1
Output
1
Input
2
Output
3
Input
3
Output
6
Input
100
Output
688750769
Note
The good pairings for the second example are:
<image>
In the third example, the good pairings are:
<image>
Submitted Solution:
```
# cook your dish here
n=int(input())
if n==1:
print(1)
else:
summ=1
mod=998244353
divisor=[0]*n
for i in range(1,n+1):
for j in range(i,n+1,i):
divisor[j-1]+=1
for i in range(2,n+1):
divisor[i-1]+=summ
if divisor[i-1]>mod:
divisor[i-1]-=mod
summ+=divisor[i-1]
if summ>mod:
summ-=mod
print(divisor[n-1])
```
Yes
| 97,239 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kavi has 2n points lying on the OX axis, i-th of which is located at x = i.
Kavi considers all ways to split these 2n points into n pairs. Among those, he is interested in good pairings, which are defined as follows:
Consider n segments with ends at the points in correspondent pairs. The pairing is called good, if for every 2 different segments A and B among those, at least one of the following holds:
* One of the segments A and B lies completely inside the other.
* A and B have the same length.
Consider the following example:
<image>
A is a good pairing since the red segment lies completely inside the blue segment.
B is a good pairing since the red and the blue segment have the same length.
C is not a good pairing since none of the red or blue segments lies inside the other, neither do they have the same size.
Kavi is interested in the number of good pairings, so he wants you to find it for him. As the result can be large, find this number modulo 998244353.
Two pairings are called different, if some two points are in one pair in some pairing and in different pairs in another.
Input
The single line of the input contains a single integer n (1≤ n ≤ 10^6).
Output
Print the number of good pairings modulo 998244353.
Examples
Input
1
Output
1
Input
2
Output
3
Input
3
Output
6
Input
100
Output
688750769
Note
The good pairings for the second example are:
<image>
In the third example, the good pairings are:
<image>
Submitted Solution:
```
# by the authority of GOD author: manhar singh sachdev #
def some_random_function():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function5():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
import os,sys
from io import BytesIO,IOBase
from array import array
# 2D list [[0]*large_index for _ in range(small_index)]
# switch from integers to floats if all integers are ≤ 2^52 and > 32 bit int
# fast mod https://github.com/cheran-senthil/PyRival/blob/master/pyrival/misc/mod.py
def main():
mod = 998244353
n = int(input())
dp = array('i',[1]+[0]*n)
fac = [0 for _ in range(n+1)]
for i in range(1,n//2+1):
for j in range(i+i,n+1,i):
fac[j] += 1
su = 1
for i in range(1,n+1):
dp[i] = (su+fac[i])%mod
su = (su+dp[i])%mod
print(dp[n])
#Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def some_random_function1():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function2():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function3():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function4():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function6():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function7():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function8():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
if __name__ == '__main__':
main()
```
Yes
| 97,240 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kavi has 2n points lying on the OX axis, i-th of which is located at x = i.
Kavi considers all ways to split these 2n points into n pairs. Among those, he is interested in good pairings, which are defined as follows:
Consider n segments with ends at the points in correspondent pairs. The pairing is called good, if for every 2 different segments A and B among those, at least one of the following holds:
* One of the segments A and B lies completely inside the other.
* A and B have the same length.
Consider the following example:
<image>
A is a good pairing since the red segment lies completely inside the blue segment.
B is a good pairing since the red and the blue segment have the same length.
C is not a good pairing since none of the red or blue segments lies inside the other, neither do they have the same size.
Kavi is interested in the number of good pairings, so he wants you to find it for him. As the result can be large, find this number modulo 998244353.
Two pairings are called different, if some two points are in one pair in some pairing and in different pairs in another.
Input
The single line of the input contains a single integer n (1≤ n ≤ 10^6).
Output
Print the number of good pairings modulo 998244353.
Examples
Input
1
Output
1
Input
2
Output
3
Input
3
Output
6
Input
100
Output
688750769
Note
The good pairings for the second example are:
<image>
In the third example, the good pairings are:
<image>
Submitted Solution:
```
N=int(input())
mod=998244353
D=[0]*(N+1)
for i in range(1,N+1):
for j in range(i,N+1,i):
D[j]+=1
X=[]
for i in range(N):
X.append(D[i+1]-D[i])
x=1
DP=[0]*(N+1)
for i in range(N):
DP[i+1]=(DP[i]*2+X[i])%mod
print(DP[N])
```
Yes
| 97,241 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kavi has 2n points lying on the OX axis, i-th of which is located at x = i.
Kavi considers all ways to split these 2n points into n pairs. Among those, he is interested in good pairings, which are defined as follows:
Consider n segments with ends at the points in correspondent pairs. The pairing is called good, if for every 2 different segments A and B among those, at least one of the following holds:
* One of the segments A and B lies completely inside the other.
* A and B have the same length.
Consider the following example:
<image>
A is a good pairing since the red segment lies completely inside the blue segment.
B is a good pairing since the red and the blue segment have the same length.
C is not a good pairing since none of the red or blue segments lies inside the other, neither do they have the same size.
Kavi is interested in the number of good pairings, so he wants you to find it for him. As the result can be large, find this number modulo 998244353.
Two pairings are called different, if some two points are in one pair in some pairing and in different pairs in another.
Input
The single line of the input contains a single integer n (1≤ n ≤ 10^6).
Output
Print the number of good pairings modulo 998244353.
Examples
Input
1
Output
1
Input
2
Output
3
Input
3
Output
6
Input
100
Output
688750769
Note
The good pairings for the second example are:
<image>
In the third example, the good pairings are:
<image>
Submitted Solution:
```
n,a,b,i=int(input())+1,0,0,1;d=[0]*n
while i<n:
for j in range(i,n,i):d[j]+=1
a=b%998244353+d[i];b+=a;i+=1
print(a)
```
Yes
| 97,242 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kavi has 2n points lying on the OX axis, i-th of which is located at x = i.
Kavi considers all ways to split these 2n points into n pairs. Among those, he is interested in good pairings, which are defined as follows:
Consider n segments with ends at the points in correspondent pairs. The pairing is called good, if for every 2 different segments A and B among those, at least one of the following holds:
* One of the segments A and B lies completely inside the other.
* A and B have the same length.
Consider the following example:
<image>
A is a good pairing since the red segment lies completely inside the blue segment.
B is a good pairing since the red and the blue segment have the same length.
C is not a good pairing since none of the red or blue segments lies inside the other, neither do they have the same size.
Kavi is interested in the number of good pairings, so he wants you to find it for him. As the result can be large, find this number modulo 998244353.
Two pairings are called different, if some two points are in one pair in some pairing and in different pairs in another.
Input
The single line of the input contains a single integer n (1≤ n ≤ 10^6).
Output
Print the number of good pairings modulo 998244353.
Examples
Input
1
Output
1
Input
2
Output
3
Input
3
Output
6
Input
100
Output
688750769
Note
The good pairings for the second example are:
<image>
In the third example, the good pairings are:
<image>
Submitted Solution:
```
p = 998244353
n = int(input())
divs = [0]*n
for i in range(1,n+1):
j = i
while j <= n:
divs[j-1] += 1
j += i
total = 1
for i in range(1,n-1):
total = (total*2+divs[i])%p
print((total+divs[n-1])%p)
```
No
| 97,243 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kavi has 2n points lying on the OX axis, i-th of which is located at x = i.
Kavi considers all ways to split these 2n points into n pairs. Among those, he is interested in good pairings, which are defined as follows:
Consider n segments with ends at the points in correspondent pairs. The pairing is called good, if for every 2 different segments A and B among those, at least one of the following holds:
* One of the segments A and B lies completely inside the other.
* A and B have the same length.
Consider the following example:
<image>
A is a good pairing since the red segment lies completely inside the blue segment.
B is a good pairing since the red and the blue segment have the same length.
C is not a good pairing since none of the red or blue segments lies inside the other, neither do they have the same size.
Kavi is interested in the number of good pairings, so he wants you to find it for him. As the result can be large, find this number modulo 998244353.
Two pairings are called different, if some two points are in one pair in some pairing and in different pairs in another.
Input
The single line of the input contains a single integer n (1≤ n ≤ 10^6).
Output
Print the number of good pairings modulo 998244353.
Examples
Input
1
Output
1
Input
2
Output
3
Input
3
Output
6
Input
100
Output
688750769
Note
The good pairings for the second example are:
<image>
In the third example, the good pairings are:
<image>
Submitted Solution:
```
import sys
input = sys.stdin.readline
n=int(input())
mod=998244353
DP=[0,1,3,6]
SUM=[0,1,4,10]
PLUS=[1]*(10**6+5)
for i in range(2,10**6):
for j in range(i,10**6+5,i):
PLUS[j]+=1
#print(PLUS[:10])
for i in range(10**6):
DP.append((SUM[-1]+PLUS[i+4])%mod)
SUM.append((SUM[-1]+DP[-1])%mod)
print(DP[n])
```
No
| 97,244 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kavi has 2n points lying on the OX axis, i-th of which is located at x = i.
Kavi considers all ways to split these 2n points into n pairs. Among those, he is interested in good pairings, which are defined as follows:
Consider n segments with ends at the points in correspondent pairs. The pairing is called good, if for every 2 different segments A and B among those, at least one of the following holds:
* One of the segments A and B lies completely inside the other.
* A and B have the same length.
Consider the following example:
<image>
A is a good pairing since the red segment lies completely inside the blue segment.
B is a good pairing since the red and the blue segment have the same length.
C is not a good pairing since none of the red or blue segments lies inside the other, neither do they have the same size.
Kavi is interested in the number of good pairings, so he wants you to find it for him. As the result can be large, find this number modulo 998244353.
Two pairings are called different, if some two points are in one pair in some pairing and in different pairs in another.
Input
The single line of the input contains a single integer n (1≤ n ≤ 10^6).
Output
Print the number of good pairings modulo 998244353.
Examples
Input
1
Output
1
Input
2
Output
3
Input
3
Output
6
Input
100
Output
688750769
Note
The good pairings for the second example are:
<image>
In the third example, the good pairings are:
<image>
Submitted Solution:
```
#!/usr/bin/env python3
import sys
import getpass # not available on codechef
import math, random
import functools, itertools, collections, heapq, bisect
from collections import Counter, defaultdict, deque
input = sys.stdin.readline # to read input quickly
# available on Google, AtCoder Python3, not available on Codeforces
# import numpy as np
# import scipy
M9 = 998244353
yes, no = "YES", "NO"
# d4 = [(1,0),(0,1),(-1,0),(0,-1)]
# d8 = [(1,0),(1,1),(0,1),(-1,1),(-1,0),(-1,-1),(0,-1),(1,-1)]
# d6 = [(2,0),(1,1),(-1,1),(-2,0),(-1,-1),(1,-1)] # hexagonal layout
MAXINT = sys.maxsize
# if testing locally, print to terminal with a different color
OFFLINE_TEST = getpass.getuser() == "hkmac"
# OFFLINE_TEST = False # codechef does not allow getpass
def log(*args):
if OFFLINE_TEST:
print('\033[36m', *args, '\033[0m', file=sys.stderr)
def solve(*args):
# screen input
if OFFLINE_TEST:
log("----- solving ------")
log(*args)
log("----- ------- ------")
return solve_(*args)
def read_matrix(rows):
return [list(map(int,input().split())) for _ in range(rows)]
def read_strings(rows):
return [input().strip() for _ in range(rows)]
def minus_one(arr):
return [x-1 for x in arr]
def minus_one_matrix(mrr):
return [[x-1 for x in row] for row in mrr]
# ---------------------------- template ends here ----------------------------
def solve_(k):
# your solution here
if k == 1:
return 1
if not OFFLINE_TEST:
if k == 100:
return 688750769
res = 1 # series
for x in range(1,k-1):
log(x)
res += pow(2,x-1,M9)
res += pow(2,k-1,M9)
return res%M9
for case_num in [0]: # no loop over test case
# for case_num in range(100): # if the number of test cases is specified
# for case_num in range(int(input())):
# read line as an integer
k = int(input())
# read line as a string
# srr = input().strip()
# read one line and parse each word as a string
# lst = input().split()
# read one line and parse each word as an integer
# a,b,c = list(map(int,input().split()))
# lst = list(map(int,input().split()))
# lst = minus_one(lst)
# read multiple rows
# arr = read_strings(k) # and return as a list of str
# mrr = read_matrix(k) # and return as a list of list of int
# mrr = minus_one_matrix(mrr)
res = solve(k) # include input here
# print length if applicable
# print(len(res))
# parse result
# res = " ".join(str(x) for x in res)
# res = "\n".join(str(x) for x in res)
# res = "\n".join(" ".join(str(x) for x in row) for row in res)
# print result
# print("Case #{}: {}".format(case_num+1, res)) # Google and Facebook - case number required
print(res)
```
No
| 97,245 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kavi has 2n points lying on the OX axis, i-th of which is located at x = i.
Kavi considers all ways to split these 2n points into n pairs. Among those, he is interested in good pairings, which are defined as follows:
Consider n segments with ends at the points in correspondent pairs. The pairing is called good, if for every 2 different segments A and B among those, at least one of the following holds:
* One of the segments A and B lies completely inside the other.
* A and B have the same length.
Consider the following example:
<image>
A is a good pairing since the red segment lies completely inside the blue segment.
B is a good pairing since the red and the blue segment have the same length.
C is not a good pairing since none of the red or blue segments lies inside the other, neither do they have the same size.
Kavi is interested in the number of good pairings, so he wants you to find it for him. As the result can be large, find this number modulo 998244353.
Two pairings are called different, if some two points are in one pair in some pairing and in different pairs in another.
Input
The single line of the input contains a single integer n (1≤ n ≤ 10^6).
Output
Print the number of good pairings modulo 998244353.
Examples
Input
1
Output
1
Input
2
Output
3
Input
3
Output
6
Input
100
Output
688750769
Note
The good pairings for the second example are:
<image>
In the third example, the good pairings are:
<image>
Submitted Solution:
```
import math
import sys
sys.setrecursionlimit(10**6+100)
def div(n) :
cnt = 0
for i in range(1, (int)(math.sqrt(n)) + 1) :
if (n % i == 0) :
if (n / i == i) :
cnt = cnt + 1
else : # Otherwise count both
cnt = cnt + 2
return cnt
def ans(n,pre,pr_ans):
if(n==1):
return 1
elif(n==2):
return 3
elif(n==3):
return 6
else:
return (2*pr_ans+div(n)-div(n-1))
#return(2*ans(n-1)+div(n)-div(n-1))
n=int(input())
if n==1:
print(1)
elif(n==2):
print(3)
elif(n==3):
print(6)
elif(n==333625):
print(668510586)
elif(n==1000000):
print(123)
elif(n>500000):
res=7993472373202089015346647259458696082532062862039876127112380610002294466097669420037495436014591213822038512726371749787635107773503232815107353201490561733068556198634847487052336375482771716440411860106504919923562698197474940717127996272801574351759103670244013302239415293457942065023154673574287881689064254399390214425177532534724504542132860236020385100915569282371191306329419371012535306844448580425305073331002510251604376156745018741535036236158474033088461882232816490362398455282999089900777611178701070578263349930751866527191899521750327259931635264949627609097326867245406307548062359611360456200585578685198529107240759809674346888474129081507302179820852158871688492052143428447146408884629001561715613126946492846050043700708012396025694566940049025702264541133543552570009198512135914634616270793670649086755437746294562338681811659055571053773936719489490866376590116022752817096127301926735453981640946449738923700816159879300620418544068515572461086348779252484673906370928303878981364152931180836741601754303560146135040878557646910891702601595496323831093367385164741130605712453031642619262741374572551937392865276572092910751970467030924869960564754562214745267467123766122647523188367482572659366827119395979133050779070475415999478381726825752723458727062962175679544429793309486904955151123773276088959584186136260914906675356533754416423735003677657919597753135215479694394753387739976719844624811251825155613699140719712918587009184939256583869807607323431930999369356297145440636860748846648445060967392387477396002937177384754299464397599037639346915864896263655987235856260830567654433576858661698515077511243395990156628569371514509954328735162481834133799531420136265978578446272807576640213659342565699272273526388328432811597525002655935303554530003297243494487097505292057245356072184267495073133337945179019479106174699236530290672500893567368740860311832438547491457093840620682536965693517918553951227851384080598619289647140632641704666936961373383992940988910964492195841938067636392239511609837490266391908917135845124496992739570560966972180993782994341156239912474384407570826593983299825469050423480513103660584465632895472212985958560872992900927611751999880044377825418309021208714668982676874567216877970432132683652204490869259219957866144482537278405557967358291453406204237366272824769915868958720954954613861229920937958377646247376027568161666055122108149714620455086237489051472689473393357194163439393384902337791096896632353937721797853193110903255812607246444735216588758824285124880490687753611915080221903797559718568987976261983815913349660959513865810142828803990013587967731454871947234705943992890490310113363176235353986986819033078048800956715012138425304808901519312676376473152826451140689306132299722068223053751675945032000436609761224533196350568047555606572142133442653106328920949002605391877706507944401227965945453630524937569321994974550298793156905811442968542251997563714174297026752463498238554414158954951361339256233871076078757493224859058252728342588828483616060213933655669468598307162549310377996342385506669668328880256808848725172682082877978103234928014111254999230341982482073170398593920749422254753345504268868482753286911209138894278309548379295578525315452486681009657910894349084859073919932580150393192398207823754893574629283233583762524934937851288111180809434169689666348952088771821484195118173093695409984889851627958800810557679276183773444910752006119236451394764143246375869216540645327643716115268499829917375105154435295823409751918845629831962531454068626527275301203287987546766408465750211115476874361476540936981205892211201332286353497373812655145473919586734592394709477616541469942984247244178178906270780992188986174724024053594972686081576130571449368167769501520360079004105755826222094909089023772215710031604299643123238191622864858519497870148714500586415613777608563570183227699982156757947181846691484140781885134063654177794797201549831440409383444758130150483821442159283518933200459590316058989345047420738318598972395290804055040909992880914795311459451914738805010306325112508381353530051185304657336911321100584439143461596870138347178592045784220956941750070870787416669599780102365066484846934808938454976184160984969759655925522581007488644414005290962609183698163469877509420621331023369365280750530022367930413229229437636542719743554387933460582966677012990860579453223805003291924381659535482203431424036026857225969378872246755851402629466052534105682953709824361008049830316902843262497096560518865638768144130452689321861636499748393608095738140051509735006832318894160596880164932799408761843201084957098717173329568250490499782667386974362133887079926521459010748493214312180965012141618378372257551571657408500710372091275321454698216478597012523643374876963391801558695383666361159000252175715156339688890083550690606608639075469373892633649885780659352268562011426208638470673720848872467661750641544494815320794338527814671929431902306885392313648522371690119248444373757306217652708451057571208298538480006227266817301780959689616480820255528933418307373932304808638536667448855704230521702061183879759256800911455806707045481165120836981663084806159718564536209198480334365828238050410934498634076655679158063207252187940105841917097752049338637875735694917471533430265919043137827986213925121547970402026374302911839293287510592213835990343536193998259122523801881987137572273247163443185582329937081174190486688749215484448234947416078991960235611190323878578275996888233385463800925425339434386324686521438726587103573345212033360479687695317279052926872466755285783717665943853405655744646768944515984612168218159771827681881884374519570794728927372782821391349048064725288011809103419759716123710189933533646964674675239646387505894266175128680509944177187747686555801597446844990946287078286479212959915474448636398804445196881992927916241371700402997983231753273835842126222481437822740325304040293285096049678976503458372626771144815882542074982717455411937902122251671682579404234898720717822787425285266974279202914015180456245719128583410335657235158029495481550932876363597831728587709081487727913762829850007807642560273977216874668720069808523057572777583985825130829642724321004886693737154871122578280309613683824107657919428404979031873939428723871364655771175794812037013832997509459077323333105725926015305559662301154374343501054069320944872931439825919342540975026803804161775864079017007654011285838915099478794667858617932572362941180713315214215792723977451642619487480432969685764140491085403339538604311079889348246958447970318179814845507751095461558928103239886604428409429147508579884645110072411760405667152264605977471917734128352440544304295716510620987099162919728023582427113786066815921638449380093132691747719429724048414793949286017833548033901324969519776787933718730264712783309750787389491329049096282782178863966985880861095008455163257371875772965076428837811697664413763058250408181636396294387478128726008827351313712389100049488965432602707791269047370837102575331730582551172731234001161875357354396443408074485270811798380280897645295048067942238046405964875614295299080153434448977410859999689093592501378460932467335733327793211124199357157554177408075193131702585825512006194935034923075875650597030572306995420300577771567098047421335294852013845005565267112849372626081897819042791969276296931882025610152454662994495108145172968081550975664927627313573554424361782469066089380839892447087215184824351924213982225071633071573902019790614460799299354834271625046478126639231549225754275595882463915278563646072373758836118183925108641730253701343664763608306425748394134595580779205123735693671718317514717178117087982239395417244619570164263616675475461594122074841074047388282527090151981226146377554773706982563848701766888632660045636849604070705754298355719638807997524170815868571925018365555898301089364372751767605349745881582957935661722518570218500091681582865095619795457060300195560122306456005388606304574173497771532742531686311661628808934478144521274373515464848402118179224837767024145128790657734052605652977935776394751452628734841281411790733255285069105408171509337873817243350830197373571142535126415148107268580693728656826344595451102807242000914505348212696643558452280710872006463136389692441202415821110316776771224170359398041368644186743641066212794135227309948003682870295505634458591470598291215556354877869384056069470841495138204162393219256954463207498829913770437765678322293662239593654606194092570217177731021155049298054899675375818899125558938676030359915958065148199968681862608919294371185276075284267843161489117007657678543517344127566356803997272958188331720064069752153893598660574117854992485764176214222487380195272405684330522659431930257015182189731469305425940851447477776396238763495199637729321100841406165627500875654873319206451783083341519146837460512334850341682326700600576857633757561751143934881260359841737175391094415912827491135073151644843893569741902605597359064209627765187582482781707156857559412143366699513225492254331682830253785426910382601220662574549673255805243278245653806046114333135685217049452879360072543610512743608990551266011696813741204364957166952649738684848494446135738311240486930911625923310517484571222885964508989408343036626585868891376434036474845416228850353325876799744439415055812432656547612929109666833977803709581801829262625715290391134636453949421173283532428057409173588534402148878442324317032465009786827469058231957061364260719593889430323184834060125080056487863271944917733710447163279666172662793046919768778224371218026412177115555958377175778025665282567006452028152490794793661576162397228682459891825131679148709529555880730137233394601604817145660067710782886511369145348732921183314626022058652804198240864706734173199026353258718878032030934167297123014545632702546448853418300830119018260932630179438830965642028804536255615061825732052284192374553661616584201302020334257677394672684333941794955831793759234229363083090084179895205007149146230831848046760805976400745541220880070481934458222169516716027599256657325985286728975715091189987428115112994120702722258419661476010283495322420194948533043000250578622433718214164915
for m in range(500001,n+1):
res=ans(m,m-1,res)
print(res%998244353)
elif(n>400000):
res=8001457083120437458753816741146583024273796972656886861406341277697147332169727739984583243708670181166646895440620312588421517440527754710898942201543441778588894546483519857276111568042433258766354072799055355906783803080039425650358300937309119867754933317619566395584469220318740953828336650192442211292492672453268682426698901306251410732966459023475853800910392301413829588925093057170521779822703271586680117299048890731754835211168807805288440130319490011732933559838092096900345305721102795182447254579982163137307957410924240383494256181715502594403239898134521985732805251295359790377012837697766867351725964335656116945931644288523731788358633971489643588074619443437427547024422066874463967868146275469462192312292396211807526797172788880611796054300969411841752781245887340183107403001601080072145514795664306927659835851800913303433008938725275822810793128791392527558749912615317013135880860326457378072781883610906352110212053107453668632920581514699476359998495250020871760096211283225465018560570984439643743420417950983350887259917686690493105104304625396773170431263159743400990298742404683768509564610734868655559726902015503303218515160802429401424015649139391562639926499644343959274804666635968259632412561835977958883508612822790678806043579701283803878588173138420188188732732547570461964647958434120477814316042694779966549239503538198878783697311983934787878751821604111743953858500068519965798602107692950137752508623270100232952221725829929079402604400938995983086178082779390052290115363534088822724997625314702946144329592223005303955336791369599448060639837357241062481121737344750658933845450930742658169624692129926689433258193231467532361293462848264320108555391647946299439365879490170496902473082786117558115744411484363047759470263087127288900530840436243790049937912606955570323572577885572093759260448805621285972002612113000440754154648959199864480848317149949733563960269519016118355435657946148496937224825155536436303817557410663885078257661385520497162530740171407523189060859147736308117128883474576283259847821719868959283163536010994435057224558115953738514867850822956825622903242878757475275987474747645123687185979072813433465216340431989634818781389456345939475739709257075242915016693799618100462936155981685536085509493612124096859496516338767642448120416957335079112545111756647712400192335777334468809225215450649553036910107451816306444474770579007222973718941429886147162465670582994383460628094260809271356597769821256910661256870231981524968582735020037132604712848587452194977538501005001259015812285537411441459573486602513642379704366928166629241854644982117585005314239270259052031992130310137659782368402892661797087968966163256044854180750618691017852565802965528559577686751323434286623451953407223107267085365292452109819535288015892105808195960168771537860545740972567845246566158332910373294349375195335763185441228146967471077921554208870531507784641551782650110578574277146330467283015610071187126096924157673233187304774233713039476241886985255757796724229998901873773585136398374299400651367076963225624792039913550946668685109382100844589250022368124333935637763582322095382016420145412782242286924131580285148125427595931815056000484012940535287867496154800877211695949610887753821158740076291873356780119156190234925452115145408119009583025779103838994915199690882899933438456640696923089559891925250427346286186555637281294067616490770102949784741629947546423511792527088030299686434541916256529458060582120820263054951436084289157221807911309067480941213581049302717966211705255053046290281846059653768642315593625398209436046382813088455414214693272946686942067380192952184984914645947574330050979040001447187639837314752197558099595708555077226890954846218125788227577036222491582779397116700310329090355382036451222213020656407881100889748154829439540356787962432883682440919619634042418838992495777653291964264422189908802817908293157827841249158936574410588859670995581777118407112844234484779424207197643609137278009629293179226760542536721430685529208825692532083179441183460277737326062510916873055175893651151895266495347448410950425610561107342780071290994559360998135656338626864469027483155982558046978992949863352965646275757949294951851832842919298987579758095959656439985880380522841485428828059698878644550214875001378335833572959593280577469606590365825494611767482085589060739225018207390002510472970919411184749965434611203618786483274629773833702042545866160110427505547092147873506794823675196310007728408164422741909244972417125774062145508429917723526510877819739294194418322785299667849485257916557378764777264003269647826529808710432360711543336635614558841953214619366504633153749725456291557657211047828697581678481487904909485439659737722827170811470076399359809736028384034565366663282683447889467673128884021931567447469146832472560539883547900836418097872287745480598119435186443562043171980475457186105345253863175502717722466423759795149009085362122190652144103238784156917045245075039494657867197357099510123081690193128934855439720912373141732431010671843884488333965763173192982590747459073452943889509973493457533539386340883262267699691989044602755407589941548713919529936520353701902791716450443215000649383629318334044958999191906330270666941999085161383038369926616369327396448278436875598969309294186646965637403463424758116949589085764804523333860540089289501211433476142017169555521672762136030977096547322706162766686700809802770998555399547823044080207998529128242646085974354941777668056086810852194293699531986038767463861627292818254475181752527879053255991030707778981639731470791248537975967081594782579105478650321459598181344978227299686099403039084449461518607112866168643836758957295494774414522942433874396091635036516389744305994117215326890168659475493766439725799013950849442475495215633992168820415930085650665394461942038449757870256568121273748485770617085139512799483534222825967778025902049122890581114625650900942444183985546042653471953892949498604995897641723970684719198088531160234471868397455550313992152257934417297312886155673536088845542551406522184644443146564315647398967824559975059754032568980740871306216351016911697758542968106261476904391804398617842498605776848534879882966495853404973319564051517606267906659747856001411557798757446598948862434465056768005767719611558522973167196310731104808320233271295010867784185564475088492679804407421631856387110553975433331344182030776212042989770004051081657359099687981715124487402917543037484145699879601284282091556889280217223754107481620001268897467985090635221634052067678189421803756453332554825832970299881045197289386992627100969559979288772658715044615814120415916714633427148100121907404447067172166287611003177926330873538652981495560784810494577704004440326531053204834791070742348676095331892282607182319736736122660437535068721991066123129247837406276805682549602209725430462589762516386659503457675753609378476841531712822881690690955408278079234450321895611286667792662373062950254097233951768820849775160194439257738421440661027178712080911233800460590093903893645755898830049209423936156386827253705922411492576597315993348768029865902313917613125138572747725887278878858473856966960146776878170796621247124320079537945257647639048144391932617776494615922679061111396801223452623230790845535293822371213846454479015663313639220424171905806605772305901461592502622731311159411982526526951542559660802671373181159413313446453264598629008245039543927053590535474865101541855686088018543149026158893423740196289190055237132339409273623137445040460034544548235173122948884477594843989414145971736991194889183037299522260953354553029929250393585939549036377675026017828875972450250693211978688741696656459412094568819033756018102994930695517661494973027841345120956748296873114008488327787162232871019530445562144423370785663658793737129567214305880284590211756920717829725730638354891356996827019326215758897702407722049295720662061011506834218108504609600867638403171258222068923242499343759341641941672831798826095247071564642234050575085557288839717181159113154880686337258873648807706545909361490882646056554538355402280403020227397909067709154134138284110977366337675312461024989335910262944311720021825726335342905261894665434866636227199510553702177193154970445819778085564064964374137777415108242424883704628544617470391851699244693208282247948761396791001852782435500388121485745537693302119301593124554442743338599332387622728005123251692529215258665220202604531254216493155240698924443978880256921565076004297892412385544317707751695333545860054402083309868054239982651637930876471479274828109003764042684883194249056018653056407933143729247052512487950818878380884675579648709402071927240114624582618813674646266396261348354420965668052659729681711057985259417250267160635299963715576691398449869181893271919405093736979341127800712852087829789921508329545844734367452930938800446055932438844622312526665249995052951669440875157471925887034135193951752681550261452363529690030963758485191449947122209248889972788569039860294229372600234733992617900554616418529416101279868116833791057303723037307025512321342152725679177347437469331810346871301508344487501662865765528556172238094541458423944747452216296613602464680959498935800243282739121266586578263085338468304397190045165422693779924289534592876688086082276512921190506407945740716768147141485867663867491092838145130386093015318343833761129652509104497487260145995546050580513145772242279424118851255612876573837960014069771514311781009484109101575033304027684457206295023972014350260534755415045757344812430780778561578238922217090718443805597698415400638107441937044128817966821841654448692635986319766072847749635164360891925794905436686970157675548127180285036379222585545083112460586285703191018304888711924294561323055745947325531337962193605285692843439043825183754056975702341163477087688763158980360627222962407621737885711357357810768386900495113701973856923604404619801561223602320223817305671284230949231412950944375884586766439697567867223981881699216048306479761534957038672413287041434726295010804758339379959887481340504254910177148750175076592000832000983877550640707747607101088318274044835379293563134588614232081520237236920593747967
for m in range(400001,n+1):
res=ans(m,m-1,res)
print(res%998244353)
elif(n>300000):
res=8009449768995854811411655849707156308854078268005173975012032013846509806603736441663361906458873267429544610288897915707850299207449140672451059356628278326072583844915060937394167486088087284632792951873526975509261947761133873711500928594520565654392638068701460868976992791175135208774946725041104924509715175757133841863441452210769126061168119938454561427431364095143085915604705538470950444723513074701496601410677742435184271246873396766493630909186924930010495107220949828332892452528090752604447304470885659739018925696066297325159035503830822784532506951495131197806353629203709475173235006516630574997675821558416070083084521467175276654083027879510483051034851485683184078383567723072792108928462853519189970140603132937336199496828158069971341607170448361594544809389678088969937074720085589700784318737955370613385761740883970451696552328711248155866564164917214402968343220174655409056845113127363557324916300890039721622802596560737437884382659658513016649026633655160920956145889183236946069576447533300545324102210404933138099052421741464355084633037800169274181600612736940484097214947231511671974319592759264171846996301689708891657674114343490635548875299137844191019088977863377777709763645019799826602335030497643443256054459328628688815433769086696572328352849925318106577544133789448230189966360416828693501415354454632868365135770286162764989978539019384262453255437586310709202990093114323500997660436894784996639486821321816872877437656208577266111083648251145544150940843420015843998429748968919222719803793960265106858209270868977774707178709033782835859973098230435673628705333724109508330540936980496462274269353171965565584936274595633345453294042307055180360177721104090088004267521478354672929229108625322223441700154125544106233534733257906629158225269659788648667851231917425360138795613167704538674753149689229034203584744348817311673965581667390409948537499630207287146569690107415261163936402682695458346967129483264490533277008469672763958652036882796803809695385039712410082200712894001461205810753456611370503552231881461782546727087993513710129201110156393314548425353265410401731414615793770835486967416799089915209279646977250526604971847025990596026894537021863337824196234506783903522223709126407792207710233205007565027373871546216256853349081173090586718754290060924696519849090640456643040687463798311022849880166593739977170187370397347834273925621188474060260437925390953545404147931798391088116822997964279198229074984883972875641547043825342150375108280947659639470298831217564485631260047034041880544924187926641687084749909908498799497679065678518574978920726114375320580766913183294580727713622276576376780323250631415408828547047294606207343038394414868789783917053379964887888732542163791555904314607582902036336859427005066231107110894929741267205038036485279595957685849512534327565301879096441341133044227817437873809323296295041045166529668326206553320579573922041681398762456425198601897470193746794348314787180680654480215090917987767715235816785198618355810057563879724409213047749481959154187428238287920614219754275641096085934252368676372385794900677437152534669625694697393622241228185985650889980324375384702947843795226832895796402156310005252445834828506387139074289493155533571790231696139117569590064182100974272559767362666334832973738183953037285194253946058527985816605116536625045219735495720577015185434079403547968369386629501104708969613510235517035952733369737589856836092723270902308803084005481705408768228319047781724160121894832036047305077199210360385784976644928283027443259579011262193066489263671518381336404077679185915918279706531858549870971491681972036296391178343611035122701937674282712481396812931214798793632027360857225153047999845680049428353754214758552849785641025482164983376642407145261851890642501121327430841461643237367656528137153167583981795647068490772331920205375706639763152261001075692714210882919108935484842403116706508177684308289972969599416291339842563059564271512875839333722188785441287248446117045821404137605619621806849363858546612125823371272995566905100904351540446239018141615592096185869645596362577521789183893613677601007577422829649774998677115896430748465573220225120443914068768098983766142058697597953800580677662162924304756290488081558797351130101863915928615002181385804561107591426916152868926234161809576553779562881942495263729612959057577210682959015521441937526009463190069065847037878683654626392146641671208138749169432458109178809436688238333907131785711486677751698504535436101115054516022549413919256771910988934789421510707087280231030223888438586069533930820134633719401954924773858072079420875954861114304127544669910477133555471221991849112104826958701223808688008142836079602035164918196217294506314353698617043608232694354280236022054684426043582022561194315343833891464489053253861244374206299764732759355834796604210295015378758903656394921039707207997431114618117348767861650627428507375736230309683239538491964235089931658364970554601674178738934102458390551133949633602330186402184177393111377079087281652505969922954307819835280212324526018701473088126778199605996300952343980849530498876402981125040826432724524307444528975451233785274093242150161695608402331250527037493674924002354045655906035805885035324117620106384282927877040288227149072438689355449380581473044417297375995081479054133386615442628637011314724288548580219164897234959461426405089700209038341280627342197717659286215116692135149008581412386087126223637914809521252370367640037905643834808694083135462767589032678193728894111253291510231802114920009042852139297600537929185634475315383935365248539073470288234842312478041970214390624865935754459618668453491570612427752116458501857361555234943797338657757125397826216249310927970249903627427669361254178882508523994231134594735313963389064408845332123181421157671941404597938977749876936866532519436997009448658376407547772533241242322699385262924738307106828358677862857414467852086194401278013025085725798804045651148207232993878804897742829655111447133262824032472990319421335991416121876627283034412493759035418580266970504821683384120448953776824130726604450537049539733050306235644870077621760850332103528109709575255940234084353247867223980770002017031090646732725359431473589384370772917759107677878358904111337488943965072623324969260984069572547312190178130464210562138022592130871434855852344689972342820321066504737121293804797535468461984818427632981837115289069042146975289439638890661509438846354638897967877497699099678187802219048732391729803095075060072891647765902932805267396366837680994415281650759417673213479659856240437516283654795954111555106793146811207167290702371276699405778690034101522588472724988248011331884701213019605286064911030159905848605902252993921818488090049411925630669081806884437793976501136772369299116568351429956979798445999485415962230262981577144525090815387273476298619534955795444704221507106245250256771211124114273042081139770480825633981371318794924954289809922400980807194050313728410629368972697584120446942022021336483608585393153543750031803135090917861181195368521114168542927530368580834778603089395395175630340009421981391237295616573745876350839796899841699997651605771046173825487484279594583587798515247988067089485723289148246254854812551252434918560003383464172309596964149669495034426872267910586945874450159106594409345932525700507710900568734772760664873200563785853691049831332252385781364632197978194163434490675973336984271713403048019134178776473424842493174328844329091696956372904460795208153486318644086438898241943642191192173657618752941869386081751661311864680508367738285448909891154288270001499444815033098489591254888340532173405317263056837147194142227802227270559767163192158407124207785260845424557080297523836778022860075267767393356976783497469455080778939882789800039977639774911644382332352122803400052369557310689484087573096000564122051464587657142552367772804125657640951633865000733917242638427969228515153218549439155329040330261316163519718166219264447023717624071135031682068185389429584994316543897771919965654075104065345135358834199510627259973946403677514508658002642995207119922180370046907584056995382036748785529734542656641589466381134897773314185247501543493070833450940059961591772076280976530313959876630742919601757967835400413003832068083137100848300682248035035592767980255101737601837037834214412170853683020257378335976178292667429213438041779646552205816663111454752021798837829021580725995953430818880193018910037948746585678393038367984189238486439222159167260905649373996940657672361119750462455166503945436756701551872987465429907026925878809473853352066925720260911290023090123430265077676233807728098607000307866906435331246300732417157249588740908645709866761213632640766388061872915761886333091495391937494483931140542567263349413761537247328823920369883701405524832120403458467699784585420810400092814984731958758583790874789508875736089040108423942695902066809848089982444912497965903765202520923188170406434176053612077599140886117304937437410965435011915198481704156122016579588020365205311979241439462931561285121431994899759976183266911427157198326676701309093399168566923778711045564403729780791707139774827752880264923913705167704697849165465067230872728594403236994134477986450891203238667071691726461227327788944284015999417099200068962885296074874497104269883497091603275671000777698985052203071547701346845130137150286335215360025004116218027488507257071800498742509696205699155349786566482427353096318371724084140610639909642864631556758520062184151434143972876705140433021343436253486951678310437327740361677501180130953267485252102755543740241498424336952149427820261726751902720804988779470291469582269057693833175114501347214421868577757642398161163803373280837923189112824607070550398981996839581667714578209162655966410293400662995030551339968370887080270967186767914234356366837701363568223289827213527288550027142414364858064229579275095515463985517081508696467182011969548577456907224103695134196437680649075185472067090216931018214055037659097605054523593972858696210194897657538932443418242067264696193174404329877869967792996161438353727000619872893532864377343110528578807939883678523264950179076588121107815506
for m in range(300001,n+1):
res=ans(m,m-1,res)
print(res%998244353)
elif(n>200000):
res=8017450438795554887675207521553510501188498678892269166400761455404971455643009722039082368787972975718346114711857804924096304157284280181103201392717477822850648800553442718703698648893965360073768596726570674857214778409088461558346729643111377019454123694625142197008371688554148810174889625458234724090633831231375746915164145396752372878736507493518350222085660344164320985707811847309764837453317790946513837583885717135608928772367893279345558464995625964301024554170064366345606221348870108744136424702745874458347620699099163717947931910681962956277162535967337682829551028667591493197931880173534721891383661470137491732441498350672149611360321490353164372129895633575034880967066756015810467831054608631085316588583225492852234722642443325245819070484277176946554817396448841700335998514147214807365457907176887150660493519485490489941591252583549669480591926886849738382553552273827147927696518879811414744222476816936050732128272545170430901557867948639434947286662601241340653610997619579247784967402446587975558693165377431217545184045238472756107415164045735808665770633779588319844372068948710752136892500488088265605509689311040565986942262205435070486001142495480804085487080333319013418983320841204295160353038010185445460006839038621797033055750424672818552735751345252328028760793162157538264779505927649636257111050560551202786395322957418616200731544071589617450454938573316150492148364623000477786037382254885760901240008827983995641794687986200777437852839042549198429379853521078964299889132664054823807699510820857941344414462454838337332212689924333718645948666758559629510250447266874104806366211243667070219145883486282935134548642909934883981449994973941151287932853575371702347305026322355127819791425692480282918890419547172323280996077072826020214642673839189676357590653751781806717703758893450320510712229304233654693177025430957343871850458836411231433018046271683847596939732283009625599118607766279704332827277611974477879514397633729923817878784618573078859938774466961456739254926937866591685581175114439649314390874908488806062637404346335659557959175869786363975692443374128663355871506457711203215762066074918058471285902268588783753414209033503712733279447291379975377217806408097036253444816911221103900512363392538692482921627645487843457413369160642698445562195397695215597778630952686623486988107142830285357615357988189802496913421753715952443741726991573366438253092047848637904296986442151454711223706410202282601618964013397462899476975330962193581949919980618666512208266281854818302961481114767743957391321887162000259657012369027637341086033959086448870772459811533870722698422517707107168020113440902628279388926880476381443253711961327161657608861004892617747707554707448683165134331364249357668779758063574722513185024141233349289959082974593618830451079291819046593314526374078332990007402895762600874126511741430087084681822130613352072436576572936585061340667562408238276325302179855972173325781548982134708128858036752954669233102384927086933372819510609121785415668156478850892524044325221867355546164576073161780122848961384670419454147216721514942749105857795766330904006767161385504995459613897773770049367326937624267878703289314893197283740924728479594761907079756291406651604797630717658918629489333227655743346987210620266774669229840160276433831894203635764811341296363205149353252180494977442869578285638635620076582880533619200277612350638374951776287931881226464062783294163175038311658524978352115383756794998640793777783311646469077975626442250606165073471329799677856060695647528606991189515518547523134126482539957183377392277721555775310379157215786139759155088762464800693219590086473576427041925324866603653533090751254153006945793497728646900103546798271023879404826904223841705606857906318314105454708304136667749562875129753525851370417932332443838153634518308909406167773569505823225493962858251224392113866672598417783244509423138232689571525803308864430269141202487684313979924404368575950057862959269819329603494853407308613831775362965444674274227474920864725514172047818376656547965795727147133016281658134991318667661013749745725226665569367236393106313661763717798125422469055322031245207927471349611049669585500080209131144519623060867150104358254170344201844692053719420875739359286236821891348886283830073006552355275276927792810024611781063530114307042561535191336026120768919159129029579879982100190662239464131810925466762134159985916843773906883531461855743655178706492825346016493855682503779251104522915147326683585076626900200726606674047788234796443664423738146541345691257772235864488536106868837528756243751673635010302139836424861885057738046988196462757471913944010044240758976974064265926641322899907597006631290291336685962116230168402535420505658094640581720445802457600422111806218842806921412637509066532309584636542268959254554501942590034595128881303893635625137705739345405791763228361605133970215006106158774671407561295800908160521231627103228094770390614963832459677111126649213263498057405687816034852229166395729238723118163546495651923573737885919315393252602514719162615452140884932992015795458201681702815170905677736077047686244295783303476878181026049029764287172716074552777356425510167887525875804322094481714506696657268921644097020026438737372919585186587230816629101967145095533794817253917043482376072129391295846951070670023105469528650817784665130002390255904820672282297946140422909833748414580380867338899342832982162446626240333555095703047977919659859421797276938056621299579040426787169317052479210420457170464562555215764075817117731616261161298436419984097872860920706685306763055593507856287715702489984630662283345609176910967297046798405759953065041670595961586640625598611115528404850208680480812057442973171019491780134289087517888509165184877037659580676528378260661083266978931353759329032700000685707347803895796832359354487643434957429697351005327618776365330337885136739356515130811417520162332509106112571535898578702926497576473982517147201539948965949100951015021624425788122934551047047529256925631012407339977535457927169436439250824605547135950708071081781592306136262199869170945105025370474639041679950134965069716494096949712432457768919121409364841520588482741002306188982242126104387920111119651487236342741843897261664476638233910275668584493381516819409878992041686266293856967151838063209633417004685953527863414647118566580754007372072338689846375019554434989627553799787569732074108406467650708983466165835056193300463293282904876506049598994048234942184852891009449948847935348981144712596259188315660318145690050150408618809011871849996799365520827374627985379556737012327298543599078075147101638791964979040876712128162278142906893425353965050633296747578706176747290697783535251850988233392627032317528884165685115888409815319042372811521851603602562949439072726901846308522307256136253623750270731721352728710154377342357939849715500319474389341913162001990104534304939191457825227399910822786852345317153289062182605159172635375263087964714474649435605948425786291038106609684089491511968046223407427917922702941795809422188255065444358556338366765003480506910582119069422147244165081524568969654952751517142640964972881840167548432239441689209262069250233670706046200330005943442889786401205328109605598010616721355191305652734003200132615980484895975896926354396049456558891631150298915458734487727606392931120546206171208212874937454295681119267387367711584770665318997691245470364685000707858302364956697651029374824145887038030665671087154785131590386475730554009564001538539019397734263457404104756247559207111242508938712135751455194837864676072267138654906486125377697350846770786222875467211084915851852636308970452165846876708645086927074816765378957396843807744418251564729466868749172123544858757405478238749472806015746984822100540738228121093941842860790213958666320813255942339583708139389616537043582527205247279821171765470596311437786907037923722732896686199188054266185080172985137091859255640535398955379827287864996065712747168271539490461557244868162898367519234115725676447720732521799780284251214571246885925126621766943716912893380206907504900086904340266662932425812186334124087212874142687760609691465428875546769271153832658597008505425783195261176421182774267935589422373170886634085458836178857593843402139516996101408716159950558851304776395445954843183445810034096496811576788490421735291587472943826855890198152371193826986683066333205796627038801102828366948462976948926921084622875636026369556612811332734981832410159134115045369934489622939641859453113457241793643622148200866879433232906428324823181953074964274549713362108949676981983417827554461771783211846271186143479320653218613679860786915548984629767414749928546995234546023844846878860320859075459522064521396840888290851810116331365423233506995875904401246920916298648305386639644628427356170888941431347944040257881162410830002959942363361455734466639961871346111809589050992206960872881064676858154445014977090747081023408216537501402772649527349583568524360384733257335585413478567425819492251685262863116496428105802254799267949513160265269688780012038344029210028763717905471595895192395031519723290892955322988381863528549742602835374506529154324071893916149762297138409010938882563079791590514743412905723437809998559843688865285705828022700340219851684519112272152880783720603828478005301157924171601135349965750423523234480470674551833779604101368041826188433757412418705287092576861133006078996366897915826389989075034850532725205801508582279188498482592350170639828287807841483132950168280667487251273989540757859112452150362428084154768142807864617822219523736709175527791696190639979946822997907126047106173772441200437942437095437049646296816296230157311037532927651097835386444108208571148899515534889174379871197503821380684586035304408558716352355165119292421263304370036288731025355447600631424112610520870732738258604009934937930638404144740181011437162408180467499131980896670050096686797703814794085803470588858164047792308119552879695993739954534914328655045618338762274553467980157889733570150371035180107586462628484379278952290787520937306151643463631363609970541651298012896126198877992583785667985261315470401676452234854844045697733763164995
for m in range(200001,n+1):
res=ans(m,m-1,res)
print(res%998244353)
else:
res=3
for m in range(3,n+1):
res=ans(m,m-1,res)
print(res%998244353)
```
No
| 97,246 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A new cottage village called «Flatville» is being built in Flatland. By now they have already built in «Flatville» n square houses with the centres on the Оx-axis. The houses' sides are parallel to the coordinate axes. It's known that no two houses overlap, but they can touch each other.
The architect bureau, where Peter works, was commissioned to build a new house in «Flatville». The customer wants his future house to be on the Оx-axis, to be square in shape, have a side t, and touch at least one of the already built houses. For sure, its sides should be parallel to the coordinate axes, its centre should be on the Ox-axis and it shouldn't overlap any of the houses in the village.
Peter was given a list of all the houses in «Flatville». Would you help him find the amount of possible positions of the new house?
Input
The first line of the input data contains numbers n and t (1 ≤ n, t ≤ 1000). Then there follow n lines, each of them contains two space-separated integer numbers: xi ai, where xi — x-coordinate of the centre of the i-th house, and ai — length of its side ( - 1000 ≤ xi ≤ 1000, 1 ≤ ai ≤ 1000).
Output
Output the amount of possible positions of the new house.
Examples
Input
2 2
0 4
6 2
Output
4
Input
2 2
0 4
5 2
Output
3
Input
2 3
0 4
5 2
Output
2
Note
It is possible for the x-coordinate of the new house to have non-integer value.
Tags: implementation, sortings
Correct Solution:
```
n, t = [int(item) for item in input().split(' ')]
cont = []
for i in range(n):
center, house_len = [int(item) for item in input().split(' ')]
cont.append([center-house_len / 2, center + house_len / 2])
cont.sort(key=lambda item: item[0])
ans = 2
for i in range(n - 1):
gap = cont[i + 1][0] - cont[i][1]
if gap > t:
ans += 2
elif gap == t:
ans += 1
print(ans)
```
| 97,247 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A new cottage village called «Flatville» is being built in Flatland. By now they have already built in «Flatville» n square houses with the centres on the Оx-axis. The houses' sides are parallel to the coordinate axes. It's known that no two houses overlap, but they can touch each other.
The architect bureau, where Peter works, was commissioned to build a new house in «Flatville». The customer wants his future house to be on the Оx-axis, to be square in shape, have a side t, and touch at least one of the already built houses. For sure, its sides should be parallel to the coordinate axes, its centre should be on the Ox-axis and it shouldn't overlap any of the houses in the village.
Peter was given a list of all the houses in «Flatville». Would you help him find the amount of possible positions of the new house?
Input
The first line of the input data contains numbers n and t (1 ≤ n, t ≤ 1000). Then there follow n lines, each of them contains two space-separated integer numbers: xi ai, where xi — x-coordinate of the centre of the i-th house, and ai — length of its side ( - 1000 ≤ xi ≤ 1000, 1 ≤ ai ≤ 1000).
Output
Output the amount of possible positions of the new house.
Examples
Input
2 2
0 4
6 2
Output
4
Input
2 2
0 4
5 2
Output
3
Input
2 3
0 4
5 2
Output
2
Note
It is possible for the x-coordinate of the new house to have non-integer value.
Tags: implementation, sortings
Correct Solution:
```
n,t = map(int, input().split())
lista_punktow = list()
lista_odleglosci = list()
suma= 0
for i in range(n):
x,a = map(int, input().split())
lista_punktow.append(float(x-a/2))
lista_punktow.append(float(x+a/2))
lista_punktow.sort()
#print (lista_punktow)
for i in range(0,len(lista_punktow)-2,2):
w = lista_punktow[i+2] - lista_punktow[i+1]
lista_odleglosci.append(w)
#print(lista_odleglosci)
for i in range(len(lista_odleglosci)):
if (lista_odleglosci[i] - t) > 0:
d = (lista_odleglosci[i] - t)
suma += 2
elif (lista_odleglosci[i] - t) == 0:
suma +=1
print (suma+2)
```
| 97,248 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A new cottage village called «Flatville» is being built in Flatland. By now they have already built in «Flatville» n square houses with the centres on the Оx-axis. The houses' sides are parallel to the coordinate axes. It's known that no two houses overlap, but they can touch each other.
The architect bureau, where Peter works, was commissioned to build a new house in «Flatville». The customer wants his future house to be on the Оx-axis, to be square in shape, have a side t, and touch at least one of the already built houses. For sure, its sides should be parallel to the coordinate axes, its centre should be on the Ox-axis and it shouldn't overlap any of the houses in the village.
Peter was given a list of all the houses in «Flatville». Would you help him find the amount of possible positions of the new house?
Input
The first line of the input data contains numbers n and t (1 ≤ n, t ≤ 1000). Then there follow n lines, each of them contains two space-separated integer numbers: xi ai, where xi — x-coordinate of the centre of the i-th house, and ai — length of its side ( - 1000 ≤ xi ≤ 1000, 1 ≤ ai ≤ 1000).
Output
Output the amount of possible positions of the new house.
Examples
Input
2 2
0 4
6 2
Output
4
Input
2 2
0 4
5 2
Output
3
Input
2 3
0 4
5 2
Output
2
Note
It is possible for the x-coordinate of the new house to have non-integer value.
Tags: implementation, sortings
Correct Solution:
```
temp = input().split()
n, t, ans = int(temp[0]), int(temp[1]), 2
cont = []
for i in range(n):
# string_arr = input().split()
# temp = input().split(' ')
# center, house_len = float(temp[0]), float(temp[1]),
# temp = [float(item) for item in input().split(' ')]
# house_center, house_len = temp[0], temp[1]
temp = list(map(float, input().split()))
house_center, house_len = temp[0], temp[1]
cont.append([house_center - house_len / 2, house_center + house_len / 2])
cont.sort(key=lambda item: item[0])
for i in range(n - 1):
helper = cont[i + 1][0] - cont[i][1]
if helper > t:
ans += 2
elif helper == t:
ans += 1
print(ans)
# print(cont)
```
| 97,249 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A new cottage village called «Flatville» is being built in Flatland. By now they have already built in «Flatville» n square houses with the centres on the Оx-axis. The houses' sides are parallel to the coordinate axes. It's known that no two houses overlap, but they can touch each other.
The architect bureau, where Peter works, was commissioned to build a new house in «Flatville». The customer wants his future house to be on the Оx-axis, to be square in shape, have a side t, and touch at least one of the already built houses. For sure, its sides should be parallel to the coordinate axes, its centre should be on the Ox-axis and it shouldn't overlap any of the houses in the village.
Peter was given a list of all the houses in «Flatville». Would you help him find the amount of possible positions of the new house?
Input
The first line of the input data contains numbers n and t (1 ≤ n, t ≤ 1000). Then there follow n lines, each of them contains two space-separated integer numbers: xi ai, where xi — x-coordinate of the centre of the i-th house, and ai — length of its side ( - 1000 ≤ xi ≤ 1000, 1 ≤ ai ≤ 1000).
Output
Output the amount of possible positions of the new house.
Examples
Input
2 2
0 4
6 2
Output
4
Input
2 2
0 4
5 2
Output
3
Input
2 3
0 4
5 2
Output
2
Note
It is possible for the x-coordinate of the new house to have non-integer value.
Tags: implementation, sortings
Correct Solution:
```
n,t = list(map(int,input().split()))
cottages = []
for i in range(n):
c,d = list(map(int,input().split()))
cottages.append([c-d/2,c+d/2])
cottages.sort()
possibilities = 0
for i in range(1,len(cottages)):
interval = cottages[i][0]-cottages[i-1][1]
if interval > t:
possibilities += 2
elif interval == t:
possibilities += 1
else:
pass
possibilities += 2
print(possibilities)
```
| 97,250 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A new cottage village called «Flatville» is being built in Flatland. By now they have already built in «Flatville» n square houses with the centres on the Оx-axis. The houses' sides are parallel to the coordinate axes. It's known that no two houses overlap, but they can touch each other.
The architect bureau, where Peter works, was commissioned to build a new house in «Flatville». The customer wants his future house to be on the Оx-axis, to be square in shape, have a side t, and touch at least one of the already built houses. For sure, its sides should be parallel to the coordinate axes, its centre should be on the Ox-axis and it shouldn't overlap any of the houses in the village.
Peter was given a list of all the houses in «Flatville». Would you help him find the amount of possible positions of the new house?
Input
The first line of the input data contains numbers n and t (1 ≤ n, t ≤ 1000). Then there follow n lines, each of them contains two space-separated integer numbers: xi ai, where xi — x-coordinate of the centre of the i-th house, and ai — length of its side ( - 1000 ≤ xi ≤ 1000, 1 ≤ ai ≤ 1000).
Output
Output the amount of possible positions of the new house.
Examples
Input
2 2
0 4
6 2
Output
4
Input
2 2
0 4
5 2
Output
3
Input
2 3
0 4
5 2
Output
2
Note
It is possible for the x-coordinate of the new house to have non-integer value.
Tags: implementation, sortings
Correct Solution:
```
n, t = map(int, input().split())
x, var = sorted(list(map(int, input().split())) for i in range(n)), 2
for i in range(n - 1):
distance = 2 * x[i + 1][0] - x[i + 1][1] - 2 * x[i][0] - x[i][1] #calculate diatance btw existing houses
if distance > 2 * t: #case when we have enough distance for 2 positions
var += 2
elif distance == 2 * t: #case when we have only one position available
var += 1
print(var) #print answer
```
| 97,251 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A new cottage village called «Flatville» is being built in Flatland. By now they have already built in «Flatville» n square houses with the centres on the Оx-axis. The houses' sides are parallel to the coordinate axes. It's known that no two houses overlap, but they can touch each other.
The architect bureau, where Peter works, was commissioned to build a new house in «Flatville». The customer wants his future house to be on the Оx-axis, to be square in shape, have a side t, and touch at least one of the already built houses. For sure, its sides should be parallel to the coordinate axes, its centre should be on the Ox-axis and it shouldn't overlap any of the houses in the village.
Peter was given a list of all the houses in «Flatville». Would you help him find the amount of possible positions of the new house?
Input
The first line of the input data contains numbers n and t (1 ≤ n, t ≤ 1000). Then there follow n lines, each of them contains two space-separated integer numbers: xi ai, where xi — x-coordinate of the centre of the i-th house, and ai — length of its side ( - 1000 ≤ xi ≤ 1000, 1 ≤ ai ≤ 1000).
Output
Output the amount of possible positions of the new house.
Examples
Input
2 2
0 4
6 2
Output
4
Input
2 2
0 4
5 2
Output
3
Input
2 3
0 4
5 2
Output
2
Note
It is possible for the x-coordinate of the new house to have non-integer value.
Tags: implementation, sortings
Correct Solution:
```
L = input()
L = L.split()
n = int(L[0])
t = int(L[1])
A = []
for k in range (n):
h = input()
h = h.split()
center = int(h[0])
side = int(h[1])
A.append(center-(side/2))
A.append(center+(side/2))
moves = 2
A.sort()
for k in range (1,len(A)-1,2):
if A[k+1]-A[k]>t:
moves += 2
elif A[k+1]-A[k]==t:
moves +=1
print(moves)
```
| 97,252 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A new cottage village called «Flatville» is being built in Flatland. By now they have already built in «Flatville» n square houses with the centres on the Оx-axis. The houses' sides are parallel to the coordinate axes. It's known that no two houses overlap, but they can touch each other.
The architect bureau, where Peter works, was commissioned to build a new house in «Flatville». The customer wants his future house to be on the Оx-axis, to be square in shape, have a side t, and touch at least one of the already built houses. For sure, its sides should be parallel to the coordinate axes, its centre should be on the Ox-axis and it shouldn't overlap any of the houses in the village.
Peter was given a list of all the houses in «Flatville». Would you help him find the amount of possible positions of the new house?
Input
The first line of the input data contains numbers n and t (1 ≤ n, t ≤ 1000). Then there follow n lines, each of them contains two space-separated integer numbers: xi ai, where xi — x-coordinate of the centre of the i-th house, and ai — length of its side ( - 1000 ≤ xi ≤ 1000, 1 ≤ ai ≤ 1000).
Output
Output the amount of possible positions of the new house.
Examples
Input
2 2
0 4
6 2
Output
4
Input
2 2
0 4
5 2
Output
3
Input
2 3
0 4
5 2
Output
2
Note
It is possible for the x-coordinate of the new house to have non-integer value.
Tags: implementation, sortings
Correct Solution:
```
def overlap_check(arr,c,t):
#print("ASd")
for i in arr:
c1 = i[0]
d1 = i[1]
dc = abs(c1-c)
#print(dc,t,d1)
if(t+d1>dc):
return False
#print(c,t)
return True
n,t = map(int,input().split())
arr = []
for i in range(n):
c,d = map(int,input().split())
d = d/2
arr.append((c,d))
ans = set()
t/=2
for i in arr:
#left
c = i[0]-i[1] - t
if overlap_check(arr,c,t):
ans.add(c)
#right
c = i[0]+i[1]+t
if overlap_check(arr,c,t):
ans.add(c)
print(len(ans))
```
| 97,253 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A new cottage village called «Flatville» is being built in Flatland. By now they have already built in «Flatville» n square houses with the centres on the Оx-axis. The houses' sides are parallel to the coordinate axes. It's known that no two houses overlap, but they can touch each other.
The architect bureau, where Peter works, was commissioned to build a new house in «Flatville». The customer wants his future house to be on the Оx-axis, to be square in shape, have a side t, and touch at least one of the already built houses. For sure, its sides should be parallel to the coordinate axes, its centre should be on the Ox-axis and it shouldn't overlap any of the houses in the village.
Peter was given a list of all the houses in «Flatville». Would you help him find the amount of possible positions of the new house?
Input
The first line of the input data contains numbers n and t (1 ≤ n, t ≤ 1000). Then there follow n lines, each of them contains two space-separated integer numbers: xi ai, where xi — x-coordinate of the centre of the i-th house, and ai — length of its side ( - 1000 ≤ xi ≤ 1000, 1 ≤ ai ≤ 1000).
Output
Output the amount of possible positions of the new house.
Examples
Input
2 2
0 4
6 2
Output
4
Input
2 2
0 4
5 2
Output
3
Input
2 3
0 4
5 2
Output
2
Note
It is possible for the x-coordinate of the new house to have non-integer value.
Tags: implementation, sortings
Correct Solution:
```
import sys
import math
import bisect
def solve(A, m):
n = len(A)
A.sort()
#ke print('A: %s' % (str(A)))
ans = 2
for i in range(1, n):
delta = A[i][0] - A[i-1][1]
if delta == m:
ans += 1
elif delta > m:
ans += 2
return ans
def main():
n, m = map(int, input().split())
m <<= 1
A = []
for i in range(n):
a, b = map(int, input().split())
A.append((2 * a - b, 2 * a + b))
ans = solve(A, m)
print(ans)
if __name__ == "__main__":
main()
```
| 97,254 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A new cottage village called «Flatville» is being built in Flatland. By now they have already built in «Flatville» n square houses with the centres on the Оx-axis. The houses' sides are parallel to the coordinate axes. It's known that no two houses overlap, but they can touch each other.
The architect bureau, where Peter works, was commissioned to build a new house in «Flatville». The customer wants his future house to be on the Оx-axis, to be square in shape, have a side t, and touch at least one of the already built houses. For sure, its sides should be parallel to the coordinate axes, its centre should be on the Ox-axis and it shouldn't overlap any of the houses in the village.
Peter was given a list of all the houses in «Flatville». Would you help him find the amount of possible positions of the new house?
Input
The first line of the input data contains numbers n and t (1 ≤ n, t ≤ 1000). Then there follow n lines, each of them contains two space-separated integer numbers: xi ai, where xi — x-coordinate of the centre of the i-th house, and ai — length of its side ( - 1000 ≤ xi ≤ 1000, 1 ≤ ai ≤ 1000).
Output
Output the amount of possible positions of the new house.
Examples
Input
2 2
0 4
6 2
Output
4
Input
2 2
0 4
5 2
Output
3
Input
2 3
0 4
5 2
Output
2
Note
It is possible for the x-coordinate of the new house to have non-integer value.
Submitted Solution:
```
n,t = map(int,input().split())
xa = []
for i in range(n):
xi,ai = map(int,input().split())
xa.append([xi,ai])
xa.sort()
for i in range(n):
xa[i] = [xa[i][0]-xa[i][1]/2,xa[i][0]+xa[i][1]/2]
count = 2
for i in range(n-1):
k = xa[i+1][0] - xa[i][1]
if k < t:
continue
if k == t:
count += 1
else:
count += 2
print(count)
```
Yes
| 97,255 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A new cottage village called «Flatville» is being built in Flatland. By now they have already built in «Flatville» n square houses with the centres on the Оx-axis. The houses' sides are parallel to the coordinate axes. It's known that no two houses overlap, but they can touch each other.
The architect bureau, where Peter works, was commissioned to build a new house in «Flatville». The customer wants his future house to be on the Оx-axis, to be square in shape, have a side t, and touch at least one of the already built houses. For sure, its sides should be parallel to the coordinate axes, its centre should be on the Ox-axis and it shouldn't overlap any of the houses in the village.
Peter was given a list of all the houses in «Flatville». Would you help him find the amount of possible positions of the new house?
Input
The first line of the input data contains numbers n and t (1 ≤ n, t ≤ 1000). Then there follow n lines, each of them contains two space-separated integer numbers: xi ai, where xi — x-coordinate of the centre of the i-th house, and ai — length of its side ( - 1000 ≤ xi ≤ 1000, 1 ≤ ai ≤ 1000).
Output
Output the amount of possible positions of the new house.
Examples
Input
2 2
0 4
6 2
Output
4
Input
2 2
0 4
5 2
Output
3
Input
2 3
0 4
5 2
Output
2
Note
It is possible for the x-coordinate of the new house to have non-integer value.
Submitted Solution:
```
from math import floor
n, t = [int(x) for x in input().split()]
centers = []
houses = {}
answer = 2
place = 0
for i in range(n):
x, a = [float(x) for x in input().split()]
centers.append(x)
houses[str(x)] = (a / 2)
centers.sort()
for i in range(1, n):
place = (centers[i] - centers[i - 1] - houses[str(centers[i])] - houses[str(centers[i - 1])])
if place == t:
answer += 1
if place > t:
answer += 2
#print(centers)
#print(houses)
print(answer)
```
Yes
| 97,256 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A new cottage village called «Flatville» is being built in Flatland. By now they have already built in «Flatville» n square houses with the centres on the Оx-axis. The houses' sides are parallel to the coordinate axes. It's known that no two houses overlap, but they can touch each other.
The architect bureau, where Peter works, was commissioned to build a new house in «Flatville». The customer wants his future house to be on the Оx-axis, to be square in shape, have a side t, and touch at least one of the already built houses. For sure, its sides should be parallel to the coordinate axes, its centre should be on the Ox-axis and it shouldn't overlap any of the houses in the village.
Peter was given a list of all the houses in «Flatville». Would you help him find the amount of possible positions of the new house?
Input
The first line of the input data contains numbers n and t (1 ≤ n, t ≤ 1000). Then there follow n lines, each of them contains two space-separated integer numbers: xi ai, where xi — x-coordinate of the centre of the i-th house, and ai — length of its side ( - 1000 ≤ xi ≤ 1000, 1 ≤ ai ≤ 1000).
Output
Output the amount of possible positions of the new house.
Examples
Input
2 2
0 4
6 2
Output
4
Input
2 2
0 4
5 2
Output
3
Input
2 3
0 4
5 2
Output
2
Note
It is possible for the x-coordinate of the new house to have non-integer value.
Submitted Solution:
```
N, T = map(int, input().split())
def read_houses():
for _ in range(N):
yield tuple(map(int, input().split()))
houses = list(read_houses())
houses.sort()
count = 2 # borders left and right
for (a, x), (b, y) in zip(houses, houses[1:]):
if b-a - (x/2+y/2) > T:
count += 2
if b-a - (x/2+y/2) == T:
count += 1
print(count)
```
Yes
| 97,257 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A new cottage village called «Flatville» is being built in Flatland. By now they have already built in «Flatville» n square houses with the centres on the Оx-axis. The houses' sides are parallel to the coordinate axes. It's known that no two houses overlap, but they can touch each other.
The architect bureau, where Peter works, was commissioned to build a new house in «Flatville». The customer wants his future house to be on the Оx-axis, to be square in shape, have a side t, and touch at least one of the already built houses. For sure, its sides should be parallel to the coordinate axes, its centre should be on the Ox-axis and it shouldn't overlap any of the houses in the village.
Peter was given a list of all the houses in «Flatville». Would you help him find the amount of possible positions of the new house?
Input
The first line of the input data contains numbers n and t (1 ≤ n, t ≤ 1000). Then there follow n lines, each of them contains two space-separated integer numbers: xi ai, where xi — x-coordinate of the centre of the i-th house, and ai — length of its side ( - 1000 ≤ xi ≤ 1000, 1 ≤ ai ≤ 1000).
Output
Output the amount of possible positions of the new house.
Examples
Input
2 2
0 4
6 2
Output
4
Input
2 2
0 4
5 2
Output
3
Input
2 3
0 4
5 2
Output
2
Note
It is possible for the x-coordinate of the new house to have non-integer value.
Submitted Solution:
```
n,t = list(map(int,input().split()))
cottages = []
for i in range(n):
c,d = list(map(int,input().split()))
cottages.append([c-d/2,c+d/2])
cottages.sort()
psbLocs = 0
for i in range(1,len(cottages)):
interval = cottages[i][0]-cottages[i-1][1]
if interval > t:
psbLocs += 2
elif interval == t:
psbLocs += 1
else:
pass
psbLocs += 2
print(psbLocs)
```
Yes
| 97,258 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A new cottage village called «Flatville» is being built in Flatland. By now they have already built in «Flatville» n square houses with the centres on the Оx-axis. The houses' sides are parallel to the coordinate axes. It's known that no two houses overlap, but they can touch each other.
The architect bureau, where Peter works, was commissioned to build a new house in «Flatville». The customer wants his future house to be on the Оx-axis, to be square in shape, have a side t, and touch at least one of the already built houses. For sure, its sides should be parallel to the coordinate axes, its centre should be on the Ox-axis and it shouldn't overlap any of the houses in the village.
Peter was given a list of all the houses in «Flatville». Would you help him find the amount of possible positions of the new house?
Input
The first line of the input data contains numbers n and t (1 ≤ n, t ≤ 1000). Then there follow n lines, each of them contains two space-separated integer numbers: xi ai, where xi — x-coordinate of the centre of the i-th house, and ai — length of its side ( - 1000 ≤ xi ≤ 1000, 1 ≤ ai ≤ 1000).
Output
Output the amount of possible positions of the new house.
Examples
Input
2 2
0 4
6 2
Output
4
Input
2 2
0 4
5 2
Output
3
Input
2 3
0 4
5 2
Output
2
Note
It is possible for the x-coordinate of the new house to have non-integer value.
Submitted Solution:
```
n,t=list(map(int,input().split()))
amount=2
for i in range(n):
c,l=list(map(int,input().split()))
l=l//2
start,end=c-l,c+l
if i:
if start-prev_end>t:
amount+=2
elif start-prev_end==t:
amount+=1
prev_end=end
print(amount)
```
No
| 97,259 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A new cottage village called «Flatville» is being built in Flatland. By now they have already built in «Flatville» n square houses with the centres on the Оx-axis. The houses' sides are parallel to the coordinate axes. It's known that no two houses overlap, but they can touch each other.
The architect bureau, where Peter works, was commissioned to build a new house in «Flatville». The customer wants his future house to be on the Оx-axis, to be square in shape, have a side t, and touch at least one of the already built houses. For sure, its sides should be parallel to the coordinate axes, its centre should be on the Ox-axis and it shouldn't overlap any of the houses in the village.
Peter was given a list of all the houses in «Flatville». Would you help him find the amount of possible positions of the new house?
Input
The first line of the input data contains numbers n and t (1 ≤ n, t ≤ 1000). Then there follow n lines, each of them contains two space-separated integer numbers: xi ai, where xi — x-coordinate of the centre of the i-th house, and ai — length of its side ( - 1000 ≤ xi ≤ 1000, 1 ≤ ai ≤ 1000).
Output
Output the amount of possible positions of the new house.
Examples
Input
2 2
0 4
6 2
Output
4
Input
2 2
0 4
5 2
Output
3
Input
2 3
0 4
5 2
Output
2
Note
It is possible for the x-coordinate of the new house to have non-integer value.
Submitted Solution:
```
n, t = map(int, input().split())
houses = [list(map(int, input().split())) for i in range(n)]
ans = 2
for i in range(n - 1):
x = houses[i][0] + houses[i][1] / 2
y = houses[i + 1][0] - houses[i + 1][1] / 2
if y - x == t:
ans += 1
elif y - x > t:
ans += 2
print(ans)
```
No
| 97,260 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A new cottage village called «Flatville» is being built in Flatland. By now they have already built in «Flatville» n square houses with the centres on the Оx-axis. The houses' sides are parallel to the coordinate axes. It's known that no two houses overlap, but they can touch each other.
The architect bureau, where Peter works, was commissioned to build a new house in «Flatville». The customer wants his future house to be on the Оx-axis, to be square in shape, have a side t, and touch at least one of the already built houses. For sure, its sides should be parallel to the coordinate axes, its centre should be on the Ox-axis and it shouldn't overlap any of the houses in the village.
Peter was given a list of all the houses in «Flatville». Would you help him find the amount of possible positions of the new house?
Input
The first line of the input data contains numbers n and t (1 ≤ n, t ≤ 1000). Then there follow n lines, each of them contains two space-separated integer numbers: xi ai, where xi — x-coordinate of the centre of the i-th house, and ai — length of its side ( - 1000 ≤ xi ≤ 1000, 1 ≤ ai ≤ 1000).
Output
Output the amount of possible positions of the new house.
Examples
Input
2 2
0 4
6 2
Output
4
Input
2 2
0 4
5 2
Output
3
Input
2 3
0 4
5 2
Output
2
Note
It is possible for the x-coordinate of the new house to have non-integer value.
Submitted Solution:
```
#15A - Cottage Village
# n - how many square houses we have on the x-axis
# t - the side of the main house
n, t = map(int, input().split())
#Variable that sorts square houses by x-axis coordinates in ascending order
#Input: house's center on the x-axis and the house's side length
houses = sorted([list(map(int, input().split())) for i in range(n)], key=lambda x: x[0])
print(houses)
#Because there's at least 1 other house in the village, we have 2 possibilities
#by default, cause the main house can touch one of the 2 exposed sides of the other house
ans = 2
#The next loop computes the number of houses that fit between 2 adjacent houses
for i in range(n - 1):
x = houses[i][0] + houses[i][1] / 2
y = houses[i + 1][0] - houses[i + 1][1] / 2
#If the space between the 2 houses is equal to the main house's side,
#the only way we can place our house is to touch both houses at the same time
if y - x == t:
ans += 1
#If the space between the houses is bigger than the main house's side,
#then we consider the 2 houses as 2 separate cases, each with 2 possibilities
#of their own
elif y - x > t:
ans += 2
print(ans)
```
No
| 97,261 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A new cottage village called «Flatville» is being built in Flatland. By now they have already built in «Flatville» n square houses with the centres on the Оx-axis. The houses' sides are parallel to the coordinate axes. It's known that no two houses overlap, but they can touch each other.
The architect bureau, where Peter works, was commissioned to build a new house in «Flatville». The customer wants his future house to be on the Оx-axis, to be square in shape, have a side t, and touch at least one of the already built houses. For sure, its sides should be parallel to the coordinate axes, its centre should be on the Ox-axis and it shouldn't overlap any of the houses in the village.
Peter was given a list of all the houses in «Flatville». Would you help him find the amount of possible positions of the new house?
Input
The first line of the input data contains numbers n and t (1 ≤ n, t ≤ 1000). Then there follow n lines, each of them contains two space-separated integer numbers: xi ai, where xi — x-coordinate of the centre of the i-th house, and ai — length of its side ( - 1000 ≤ xi ≤ 1000, 1 ≤ ai ≤ 1000).
Output
Output the amount of possible positions of the new house.
Examples
Input
2 2
0 4
6 2
Output
4
Input
2 2
0 4
5 2
Output
3
Input
2 3
0 4
5 2
Output
2
Note
It is possible for the x-coordinate of the new house to have non-integer value.
Submitted Solution:
```
n, t = map(int, input().split())
li = []
for _ in range(n):
tx = tuple(map(int, input().split()))
li.append(tx)
p = 2
li.sort(key = lambda x: x[0])
for i in range(len(li)-1):
k = (li[i][1]+li[i+1][1])
if k%2 != 0:
d = abs(li[i][0]-li[i+1][0])-int(k/2+.5)
else:
d = abs(li[i][0]-li[i+1][0])-(k/2)
if d > t:
p += 2
elif d == t:
p += 1
print(p)
```
No
| 97,262 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A widely known among some people Belarusian sport programmer Lesha decided to make some money to buy a one square meter larger flat. To do this, he wants to make and carry out a Super Rated Match (SRM) on the site Torcoder.com. But there's a problem — a severe torcoder coordinator Ivan does not accept any Lesha's problem, calling each of them an offensive word "duped" (that is, duplicated). And one day they nearely quarrelled over yet another problem Ivan wouldn't accept.
You are invited to act as a fair judge and determine whether the problem is indeed brand new, or Ivan is right and the problem bears some resemblance to those used in the previous SRMs.
You are given the descriptions of Lesha's problem and each of Torcoder.com archive problems. The description of each problem is a sequence of words. Besides, it is guaranteed that Lesha's problem has no repeated words, while the description of an archive problem may contain any number of repeated words.
The "similarity" between Lesha's problem and some archive problem can be found as follows. Among all permutations of words in Lesha's problem we choose the one that occurs in the archive problem as a subsequence. If there are multiple such permutations, we choose the one with the smallest number of inversions. Then the "similarity" of a problem can be written as <image>, where n is the number of words in Lesha's problem and x is the number of inversions in the chosen permutation. Note that the "similarity" p is always a positive integer.
The problem is called brand new if there is not a single problem in Ivan's archive which contains a permutation of words from Lesha's problem as a subsequence.
Help the boys and determine whether the proposed problem is new, or specify the problem from the archive which resembles Lesha's problem the most, otherwise.
Input
The first line contains a single integer n (1 ≤ n ≤ 15) — the number of words in Lesha's problem. The second line contains n space-separated words — the short description of the problem.
The third line contains a single integer m (1 ≤ m ≤ 10) — the number of problems in the Torcoder.com archive. Next m lines contain the descriptions of the problems as "k s1 s2 ... sk", where k (1 ≤ k ≤ 500000) is the number of words in the problem and si is a word of the problem description.
All words from all problem descriptions contain no more than 10 lowercase English letters. It is guaranteed that the total length of words in all problem descriptions does not exceed 500015.
Output
If Lesha's problem is brand new, print string "Brand new problem!" (without quotes).
Otherwise, on the first line print the index of the archive problem which resembles Lesha's problem most. If there are multiple such problems, print the one with the smallest index. On the second line print a string consisting of characters [:, character | repeated p times, and characters :], where p is the "similarity" between this problem and Lesha's one. The archive problems are numbered starting from one in the order in which they are given in the input.
Examples
Input
4
find the next palindrome
1
10 find the previous palindrome or print better luck next time
Output
1
[:||||||:]
Input
3
add two numbers
3
1 add
2 two two
3 numbers numbers numbers
Output
Brand new problem!
Input
4
these papers are formulas
3
6 what are these formulas and papers
5 papers are driving me crazy
4 crazy into the night
Output
1
[:||||:]
Input
3
add two decimals
5
4 please two decimals add
5 decimals want to be added
4 two add decimals add
4 add one two three
7 one plus two plus three equals six
Output
3
[:|||:]
Note
Let us remind you that the number of inversions is the number of pairs of words that follow in the permutation not in their original order. Thus, for example, if the original problem is "add two numbers", then permutation "numbers add two" contains two inversions — pairs of words "numbers" and "add", "numbers" and "two".
Sequence b1, b2, ..., bk is a subsequence of sequence a1, a2, ..., an if there exists such a set of indices 1 ≤ i1 < i2 < ... < ik ≤ n that aij = bj (in other words, if sequence b can be obtained from a by deleting some of its elements).
In the first test case the first problem contains the "find the palindrome next" permutation as a subsequence, in which the number of inversions equals 1 (words "palindrome" and "next").
In the second test case there is no problem that contains a permutation of words from Lesha's problem as a subsequence.
Tags: bitmasks, brute force, dp
Correct Solution:
```
import itertools
def count_inversions(enumerate_seq):
tmp = list(enumerate_seq[:])
result = 0
for i in range(len(tmp)):
for j in range(len(tmp) - 1):
if tmp[j][0] > tmp[j + 1][0]:
result += 1
tmp[j], tmp[j + 1] = tmp[j + 1], tmp[j]
return result
def sub_seq(a, b):
i, j = 0, 0
while i < len(a) and j < len(b):
if a[i] == b[j]:
i += 1
j += 1
return i == len(a)
n = int(input())
A = input().split()
B = []
m = int(input())
for _ in range(m):
B += [input().split()[1:]]
brand_new = True
best_perms = 0
best_seq = []
best_ind = -1
for i, b in enumerate(B):
for tmp in itertools.permutations(enumerate(A)):
if sub_seq([x for y, x in tmp], b):
brand_new = False
inversions = count_inversions(tmp)
similarity = n * (n - 1) // 2 - inversions + 1
if best_perms < similarity:
best_perms = similarity
best_seq = [x for y, x in tmp]
best_ind = i
if not brand_new:
print(best_ind + 1)
print('[:' + '|' * best_perms + ':]')
else:
print("Brand new problem!")
```
| 97,263 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A widely known among some people Belarusian sport programmer Lesha decided to make some money to buy a one square meter larger flat. To do this, he wants to make and carry out a Super Rated Match (SRM) on the site Torcoder.com. But there's a problem — a severe torcoder coordinator Ivan does not accept any Lesha's problem, calling each of them an offensive word "duped" (that is, duplicated). And one day they nearely quarrelled over yet another problem Ivan wouldn't accept.
You are invited to act as a fair judge and determine whether the problem is indeed brand new, or Ivan is right and the problem bears some resemblance to those used in the previous SRMs.
You are given the descriptions of Lesha's problem and each of Torcoder.com archive problems. The description of each problem is a sequence of words. Besides, it is guaranteed that Lesha's problem has no repeated words, while the description of an archive problem may contain any number of repeated words.
The "similarity" between Lesha's problem and some archive problem can be found as follows. Among all permutations of words in Lesha's problem we choose the one that occurs in the archive problem as a subsequence. If there are multiple such permutations, we choose the one with the smallest number of inversions. Then the "similarity" of a problem can be written as <image>, where n is the number of words in Lesha's problem and x is the number of inversions in the chosen permutation. Note that the "similarity" p is always a positive integer.
The problem is called brand new if there is not a single problem in Ivan's archive which contains a permutation of words from Lesha's problem as a subsequence.
Help the boys and determine whether the proposed problem is new, or specify the problem from the archive which resembles Lesha's problem the most, otherwise.
Input
The first line contains a single integer n (1 ≤ n ≤ 15) — the number of words in Lesha's problem. The second line contains n space-separated words — the short description of the problem.
The third line contains a single integer m (1 ≤ m ≤ 10) — the number of problems in the Torcoder.com archive. Next m lines contain the descriptions of the problems as "k s1 s2 ... sk", where k (1 ≤ k ≤ 500000) is the number of words in the problem and si is a word of the problem description.
All words from all problem descriptions contain no more than 10 lowercase English letters. It is guaranteed that the total length of words in all problem descriptions does not exceed 500015.
Output
If Lesha's problem is brand new, print string "Brand new problem!" (without quotes).
Otherwise, on the first line print the index of the archive problem which resembles Lesha's problem most. If there are multiple such problems, print the one with the smallest index. On the second line print a string consisting of characters [:, character | repeated p times, and characters :], where p is the "similarity" between this problem and Lesha's one. The archive problems are numbered starting from one in the order in which they are given in the input.
Examples
Input
4
find the next palindrome
1
10 find the previous palindrome or print better luck next time
Output
1
[:||||||:]
Input
3
add two numbers
3
1 add
2 two two
3 numbers numbers numbers
Output
Brand new problem!
Input
4
these papers are formulas
3
6 what are these formulas and papers
5 papers are driving me crazy
4 crazy into the night
Output
1
[:||||:]
Input
3
add two decimals
5
4 please two decimals add
5 decimals want to be added
4 two add decimals add
4 add one two three
7 one plus two plus three equals six
Output
3
[:|||:]
Note
Let us remind you that the number of inversions is the number of pairs of words that follow in the permutation not in their original order. Thus, for example, if the original problem is "add two numbers", then permutation "numbers add two" contains two inversions — pairs of words "numbers" and "add", "numbers" and "two".
Sequence b1, b2, ..., bk is a subsequence of sequence a1, a2, ..., an if there exists such a set of indices 1 ≤ i1 < i2 < ... < ik ≤ n that aij = bj (in other words, if sequence b can be obtained from a by deleting some of its elements).
In the first test case the first problem contains the "find the palindrome next" permutation as a subsequence, in which the number of inversions equals 1 (words "palindrome" and "next").
In the second test case there is no problem that contains a permutation of words from Lesha's problem as a subsequence.
Tags: bitmasks, brute force, dp
Correct Solution:
```
import itertools
def check(curr_words, line):
if curr_words == []:
return True
for i in range(len(line)):
if line[i] == curr_words[0]:
return check(curr_words[1:], line[i+1:])
return False
n = int(input())
words = input().split()
m = int(input())
res, idx = 0, 0
for i in range(m):
line = input().split()[1:]
for p in itertools.permutations(range(n)):
curr_words = [words[j] for j in p]
cnt = 0
for j in range(n):
cnt += len([k for k in range(j+1, n) if p[k] < p[j]])
v = n * (n-1) // 2 - cnt + 1
if check(curr_words, line[:]) and v > res:
res, idx = v, i+1
if res > 0:
print(idx)
print('[:'+str('|'*res)+':]')
else:
print('Brand new problem!')
```
| 97,264 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A widely known among some people Belarusian sport programmer Lesha decided to make some money to buy a one square meter larger flat. To do this, he wants to make and carry out a Super Rated Match (SRM) on the site Torcoder.com. But there's a problem — a severe torcoder coordinator Ivan does not accept any Lesha's problem, calling each of them an offensive word "duped" (that is, duplicated). And one day they nearely quarrelled over yet another problem Ivan wouldn't accept.
You are invited to act as a fair judge and determine whether the problem is indeed brand new, or Ivan is right and the problem bears some resemblance to those used in the previous SRMs.
You are given the descriptions of Lesha's problem and each of Torcoder.com archive problems. The description of each problem is a sequence of words. Besides, it is guaranteed that Lesha's problem has no repeated words, while the description of an archive problem may contain any number of repeated words.
The "similarity" between Lesha's problem and some archive problem can be found as follows. Among all permutations of words in Lesha's problem we choose the one that occurs in the archive problem as a subsequence. If there are multiple such permutations, we choose the one with the smallest number of inversions. Then the "similarity" of a problem can be written as <image>, where n is the number of words in Lesha's problem and x is the number of inversions in the chosen permutation. Note that the "similarity" p is always a positive integer.
The problem is called brand new if there is not a single problem in Ivan's archive which contains a permutation of words from Lesha's problem as a subsequence.
Help the boys and determine whether the proposed problem is new, or specify the problem from the archive which resembles Lesha's problem the most, otherwise.
Input
The first line contains a single integer n (1 ≤ n ≤ 15) — the number of words in Lesha's problem. The second line contains n space-separated words — the short description of the problem.
The third line contains a single integer m (1 ≤ m ≤ 10) — the number of problems in the Torcoder.com archive. Next m lines contain the descriptions of the problems as "k s1 s2 ... sk", where k (1 ≤ k ≤ 500000) is the number of words in the problem and si is a word of the problem description.
All words from all problem descriptions contain no more than 10 lowercase English letters. It is guaranteed that the total length of words in all problem descriptions does not exceed 500015.
Output
If Lesha's problem is brand new, print string "Brand new problem!" (without quotes).
Otherwise, on the first line print the index of the archive problem which resembles Lesha's problem most. If there are multiple such problems, print the one with the smallest index. On the second line print a string consisting of characters [:, character | repeated p times, and characters :], where p is the "similarity" between this problem and Lesha's one. The archive problems are numbered starting from one in the order in which they are given in the input.
Examples
Input
4
find the next palindrome
1
10 find the previous palindrome or print better luck next time
Output
1
[:||||||:]
Input
3
add two numbers
3
1 add
2 two two
3 numbers numbers numbers
Output
Brand new problem!
Input
4
these papers are formulas
3
6 what are these formulas and papers
5 papers are driving me crazy
4 crazy into the night
Output
1
[:||||:]
Input
3
add two decimals
5
4 please two decimals add
5 decimals want to be added
4 two add decimals add
4 add one two three
7 one plus two plus three equals six
Output
3
[:|||:]
Note
Let us remind you that the number of inversions is the number of pairs of words that follow in the permutation not in their original order. Thus, for example, if the original problem is "add two numbers", then permutation "numbers add two" contains two inversions — pairs of words "numbers" and "add", "numbers" and "two".
Sequence b1, b2, ..., bk is a subsequence of sequence a1, a2, ..., an if there exists such a set of indices 1 ≤ i1 < i2 < ... < ik ≤ n that aij = bj (in other words, if sequence b can be obtained from a by deleting some of its elements).
In the first test case the first problem contains the "find the palindrome next" permutation as a subsequence, in which the number of inversions equals 1 (words "palindrome" and "next").
In the second test case there is no problem that contains a permutation of words from Lesha's problem as a subsequence.
Submitted Solution:
```
data = []
i = int(input())
words = list(map(str, input().split(' ')))
j = int(input())
for t in range(j):
ok = list(map(str, input().split(' ')))
data.append([int(ok[0]), ok[1:]])
if i == 1:
inv = []
for x in range(len(data)):
if words[0] in data[x][1]:
print(x)
print('[:|:]')
quit()
print("Brand new problem!")
elif i == 2:
inv = []
a = words[0]
b = words[1]
for x in range(len(data)):
invx = 0
for i in range(data[x][0]):
for j in range(data[x][0]):
if i != j:
if data[x][1][i] == a and data[x][1][j] == b:
if i > j:
invx+=1
inv.append([x, invx])
inv.sort(key = lambda x : x[1])
if inv == []:
print("Brand new problem!")
else:
print(inv[0][0]+1)
print('[:' + (1+1-inv[0][1]) * '|' + ':]')
elif i == 3:
inv = []
a = words[0]
b = words[1]
c = words[2]
for x in range(len(data)):
invx = 0
for i in range(data[x][0]):
for j in range(data[x][0]):
for k in range(data[x][0]):
if i != j and j != k and i != k:
if data[x][1][i] == a and data[x][1][j] == b and data[x][1][k] == c:
if i > j:
invx+=1
if i > k:
invx += 1
if j > k:
invx += 1
inv.append([x, invx])
inv.sort(key = lambda x : x[1])
if inv == []:
print("Brand new problem!")
else:
print(inv[0][0]+1)
print('[:' + (3+1-inv[0][1]) * '|' + ':]')
elif i == 4:
inv = []
a = words[0]
b = words[1]
c = words[2]
d = words[3]
for x in range(len(data)):
invx = 0
for i in range(data[x][0]):
for j in range(data[x][0]):
for k in range(data[x][0]):
for l in range(data[x][0]):
if i != j and j != k and i != k and i != l and j != l and k != l:
if data[x][1][i] == a and data[x][1][j] == b and data[x][1][k] == c and data[x][1][l] == d:
if i > j:
invx+=1
if i > k:
invx += 1
if j > k:
invx += 1
if j > l:
invx += 1
if i > l:
invx += 1
if k > l:
invx += 1
inv.append([x, invx])
inv.sort(key = lambda x : x[1])
if inv == []:
print("Brand new problem!")
else:
print(inv[0][0]+1)
print('[:' + (6+1-inv[0][1]) * '|' + ':]')
```
No
| 97,265 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A widely known among some people Belarusian sport programmer Lesha decided to make some money to buy a one square meter larger flat. To do this, he wants to make and carry out a Super Rated Match (SRM) on the site Torcoder.com. But there's a problem — a severe torcoder coordinator Ivan does not accept any Lesha's problem, calling each of them an offensive word "duped" (that is, duplicated). And one day they nearely quarrelled over yet another problem Ivan wouldn't accept.
You are invited to act as a fair judge and determine whether the problem is indeed brand new, or Ivan is right and the problem bears some resemblance to those used in the previous SRMs.
You are given the descriptions of Lesha's problem and each of Torcoder.com archive problems. The description of each problem is a sequence of words. Besides, it is guaranteed that Lesha's problem has no repeated words, while the description of an archive problem may contain any number of repeated words.
The "similarity" between Lesha's problem and some archive problem can be found as follows. Among all permutations of words in Lesha's problem we choose the one that occurs in the archive problem as a subsequence. If there are multiple such permutations, we choose the one with the smallest number of inversions. Then the "similarity" of a problem can be written as <image>, where n is the number of words in Lesha's problem and x is the number of inversions in the chosen permutation. Note that the "similarity" p is always a positive integer.
The problem is called brand new if there is not a single problem in Ivan's archive which contains a permutation of words from Lesha's problem as a subsequence.
Help the boys and determine whether the proposed problem is new, or specify the problem from the archive which resembles Lesha's problem the most, otherwise.
Input
The first line contains a single integer n (1 ≤ n ≤ 15) — the number of words in Lesha's problem. The second line contains n space-separated words — the short description of the problem.
The third line contains a single integer m (1 ≤ m ≤ 10) — the number of problems in the Torcoder.com archive. Next m lines contain the descriptions of the problems as "k s1 s2 ... sk", where k (1 ≤ k ≤ 500000) is the number of words in the problem and si is a word of the problem description.
All words from all problem descriptions contain no more than 10 lowercase English letters. It is guaranteed that the total length of words in all problem descriptions does not exceed 500015.
Output
If Lesha's problem is brand new, print string "Brand new problem!" (without quotes).
Otherwise, on the first line print the index of the archive problem which resembles Lesha's problem most. If there are multiple such problems, print the one with the smallest index. On the second line print a string consisting of characters [:, character | repeated p times, and characters :], where p is the "similarity" between this problem and Lesha's one. The archive problems are numbered starting from one in the order in which they are given in the input.
Examples
Input
4
find the next palindrome
1
10 find the previous palindrome or print better luck next time
Output
1
[:||||||:]
Input
3
add two numbers
3
1 add
2 two two
3 numbers numbers numbers
Output
Brand new problem!
Input
4
these papers are formulas
3
6 what are these formulas and papers
5 papers are driving me crazy
4 crazy into the night
Output
1
[:||||:]
Input
3
add two decimals
5
4 please two decimals add
5 decimals want to be added
4 two add decimals add
4 add one two three
7 one plus two plus three equals six
Output
3
[:|||:]
Note
Let us remind you that the number of inversions is the number of pairs of words that follow in the permutation not in their original order. Thus, for example, if the original problem is "add two numbers", then permutation "numbers add two" contains two inversions — pairs of words "numbers" and "add", "numbers" and "two".
Sequence b1, b2, ..., bk is a subsequence of sequence a1, a2, ..., an if there exists such a set of indices 1 ≤ i1 < i2 < ... < ik ≤ n that aij = bj (in other words, if sequence b can be obtained from a by deleting some of its elements).
In the first test case the first problem contains the "find the palindrome next" permutation as a subsequence, in which the number of inversions equals 1 (words "palindrome" and "next").
In the second test case there is no problem that contains a permutation of words from Lesha's problem as a subsequence.
Submitted Solution:
```
import itertools
def check(curr_words, line):
if curr_words == []:
return True
for i in range(len(line)):
if line[i] == curr_words[0]:
return check(curr_words[1:], line[i+1:])
return False
n = int(input())
words = input().split()
m = int(input())
res, idx = 0, 0
for i in range(m):
line = input().split()[1:]
for p in itertools.permutations(range(n)):
curr_words = [words[j] for j in p]
cnt = 0
for j in range(n):
cnt += len([k for k in range(j+1, n) if p[k] < p[j]])
v = n * (n-1) // 2 - cnt + 1
if check(curr_words, line[:]) and v > res:
res, idx = v, i+1
if res > 0:
print(idx)
print('[:'+str('|'*res)+':]')
else:
print('Brand new wordslem!')
```
No
| 97,266 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus has been working in the analytic department of the "F.R.A.U.D." company for as much as n days. Right now his task is to make a series of reports about the company's performance for the last n days. We know that the main information in a day report is value ai, the company's profit on the i-th day. If ai is negative, then the company suffered losses on the i-th day.
Polycarpus should sort the daily reports into folders. Each folder should include data on the company's performance for several consecutive days. Of course, the information on each of the n days should be exactly in one folder. Thus, Polycarpus puts information on the first few days in the first folder. The information on the several following days goes to the second folder, and so on.
It is known that the boss reads one daily report folder per day. If one folder has three or more reports for the days in which the company suffered losses (ai < 0), he loses his temper and his wrath is terrible.
Therefore, Polycarpus wants to prepare the folders so that none of them contains information on three or more days with the loss, and the number of folders is minimal.
Write a program that, given sequence ai, will print the minimum number of folders.
Input
The first line contains integer n (1 ≤ n ≤ 100), n is the number of days. The second line contains a sequence of integers a1, a2, ..., an (|ai| ≤ 100), where ai means the company profit on the i-th day. It is possible that the company has no days with the negative ai.
Output
Print an integer k — the required minimum number of folders. In the second line print a sequence of integers b1, b2, ..., bk, where bj is the number of day reports in the j-th folder.
If there are multiple ways to sort the reports into k days, print any of them.
Examples
Input
11
1 2 3 -4 -5 -6 5 -5 -6 -7 6
Output
3
5 3 3
Input
5
0 -1 100 -1 0
Output
1
5
Note
Here goes a way to sort the reports from the first sample into three folders:
1 2 3 -4 -5 | -6 5 -5 | -6 -7 6
In the second sample you can put all five reports in one folder.
Tags: greedy
Correct Solution:
```
n=int(input())
l=list(map(int,input().split()))
c,x=0,0
d=[]
for i in range(n):
if(l[i]<0):
c=c+1
x=x+1
if(x==2):
d.append(c)
x=0
c=0
else:
c=c+1
if(d==[]):
d.append(c)
c=0
if(c>0 and x==0):
d[-1]+=c
elif(c>0 and x!=0):
d.append(c)
print(len(d))
print(*d)
```
| 97,267 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus has been working in the analytic department of the "F.R.A.U.D." company for as much as n days. Right now his task is to make a series of reports about the company's performance for the last n days. We know that the main information in a day report is value ai, the company's profit on the i-th day. If ai is negative, then the company suffered losses on the i-th day.
Polycarpus should sort the daily reports into folders. Each folder should include data on the company's performance for several consecutive days. Of course, the information on each of the n days should be exactly in one folder. Thus, Polycarpus puts information on the first few days in the first folder. The information on the several following days goes to the second folder, and so on.
It is known that the boss reads one daily report folder per day. If one folder has three or more reports for the days in which the company suffered losses (ai < 0), he loses his temper and his wrath is terrible.
Therefore, Polycarpus wants to prepare the folders so that none of them contains information on three or more days with the loss, and the number of folders is minimal.
Write a program that, given sequence ai, will print the minimum number of folders.
Input
The first line contains integer n (1 ≤ n ≤ 100), n is the number of days. The second line contains a sequence of integers a1, a2, ..., an (|ai| ≤ 100), where ai means the company profit on the i-th day. It is possible that the company has no days with the negative ai.
Output
Print an integer k — the required minimum number of folders. In the second line print a sequence of integers b1, b2, ..., bk, where bj is the number of day reports in the j-th folder.
If there are multiple ways to sort the reports into k days, print any of them.
Examples
Input
11
1 2 3 -4 -5 -6 5 -5 -6 -7 6
Output
3
5 3 3
Input
5
0 -1 100 -1 0
Output
1
5
Note
Here goes a way to sort the reports from the first sample into three folders:
1 2 3 -4 -5 | -6 5 -5 | -6 -7 6
In the second sample you can put all five reports in one folder.
Tags: greedy
Correct Solution:
```
n = input()
s = input().split(" ")
rugi, jumlah, berkas = 0, 0, []
for j, i in enumerate(s):
if int(i) < 0:
rugi += 1
if rugi == 3:
rugi = 1
berkas.append(str(jumlah))
jumlah = 1
else:
jumlah += 1
else:
jumlah += 1
if j == len(s)-1:
berkas.append(str(jumlah))
print(len(berkas), " ".join(berkas), sep='\n')
```
| 97,268 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus has been working in the analytic department of the "F.R.A.U.D." company for as much as n days. Right now his task is to make a series of reports about the company's performance for the last n days. We know that the main information in a day report is value ai, the company's profit on the i-th day. If ai is negative, then the company suffered losses on the i-th day.
Polycarpus should sort the daily reports into folders. Each folder should include data on the company's performance for several consecutive days. Of course, the information on each of the n days should be exactly in one folder. Thus, Polycarpus puts information on the first few days in the first folder. The information on the several following days goes to the second folder, and so on.
It is known that the boss reads one daily report folder per day. If one folder has three or more reports for the days in which the company suffered losses (ai < 0), he loses his temper and his wrath is terrible.
Therefore, Polycarpus wants to prepare the folders so that none of them contains information on three or more days with the loss, and the number of folders is minimal.
Write a program that, given sequence ai, will print the minimum number of folders.
Input
The first line contains integer n (1 ≤ n ≤ 100), n is the number of days. The second line contains a sequence of integers a1, a2, ..., an (|ai| ≤ 100), where ai means the company profit on the i-th day. It is possible that the company has no days with the negative ai.
Output
Print an integer k — the required minimum number of folders. In the second line print a sequence of integers b1, b2, ..., bk, where bj is the number of day reports in the j-th folder.
If there are multiple ways to sort the reports into k days, print any of them.
Examples
Input
11
1 2 3 -4 -5 -6 5 -5 -6 -7 6
Output
3
5 3 3
Input
5
0 -1 100 -1 0
Output
1
5
Note
Here goes a way to sort the reports from the first sample into three folders:
1 2 3 -4 -5 | -6 5 -5 | -6 -7 6
In the second sample you can put all five reports in one folder.
Tags: greedy
Correct Solution:
```
n = int(input())
dig = list( map(int, input().split() ))
flag = 0
index = 0
folder = []
for i in range(n):
folder.append( 0 )
for i in dig:
if i<0:
flag+=1
if flag == 3:
flag = 1
index +=1
folder[index]+=1
print( index+1 )
for i in range(index+1):
print(folder[i], end = " ")
```
| 97,269 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus has been working in the analytic department of the "F.R.A.U.D." company for as much as n days. Right now his task is to make a series of reports about the company's performance for the last n days. We know that the main information in a day report is value ai, the company's profit on the i-th day. If ai is negative, then the company suffered losses on the i-th day.
Polycarpus should sort the daily reports into folders. Each folder should include data on the company's performance for several consecutive days. Of course, the information on each of the n days should be exactly in one folder. Thus, Polycarpus puts information on the first few days in the first folder. The information on the several following days goes to the second folder, and so on.
It is known that the boss reads one daily report folder per day. If one folder has three or more reports for the days in which the company suffered losses (ai < 0), he loses his temper and his wrath is terrible.
Therefore, Polycarpus wants to prepare the folders so that none of them contains information on three or more days with the loss, and the number of folders is minimal.
Write a program that, given sequence ai, will print the minimum number of folders.
Input
The first line contains integer n (1 ≤ n ≤ 100), n is the number of days. The second line contains a sequence of integers a1, a2, ..., an (|ai| ≤ 100), where ai means the company profit on the i-th day. It is possible that the company has no days with the negative ai.
Output
Print an integer k — the required minimum number of folders. In the second line print a sequence of integers b1, b2, ..., bk, where bj is the number of day reports in the j-th folder.
If there are multiple ways to sort the reports into k days, print any of them.
Examples
Input
11
1 2 3 -4 -5 -6 5 -5 -6 -7 6
Output
3
5 3 3
Input
5
0 -1 100 -1 0
Output
1
5
Note
Here goes a way to sort the reports from the first sample into three folders:
1 2 3 -4 -5 | -6 5 -5 | -6 -7 6
In the second sample you can put all five reports in one folder.
Tags: greedy
Correct Solution:
```
n=eval(input())
import math
l=list(map(int,input().split()))
i=0
c=0
while(i<n):
if(l[i]<0):
c+=1
i+=1
d=math.ceil(c/2)
if(c==0 or c==1):
print(1)
else:
print(d)
c=0
i=0
s=0
while(i<n):
if(l[i]<0):
c+=1
else:
s+=1
if(c==3):
print(s+c-1,end=" ")
c=1
s=0
i+=1
if(s+c!=0):
print(s+c,end=" ")
```
| 97,270 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus has been working in the analytic department of the "F.R.A.U.D." company for as much as n days. Right now his task is to make a series of reports about the company's performance for the last n days. We know that the main information in a day report is value ai, the company's profit on the i-th day. If ai is negative, then the company suffered losses on the i-th day.
Polycarpus should sort the daily reports into folders. Each folder should include data on the company's performance for several consecutive days. Of course, the information on each of the n days should be exactly in one folder. Thus, Polycarpus puts information on the first few days in the first folder. The information on the several following days goes to the second folder, and so on.
It is known that the boss reads one daily report folder per day. If one folder has three or more reports for the days in which the company suffered losses (ai < 0), he loses his temper and his wrath is terrible.
Therefore, Polycarpus wants to prepare the folders so that none of them contains information on three or more days with the loss, and the number of folders is minimal.
Write a program that, given sequence ai, will print the minimum number of folders.
Input
The first line contains integer n (1 ≤ n ≤ 100), n is the number of days. The second line contains a sequence of integers a1, a2, ..., an (|ai| ≤ 100), where ai means the company profit on the i-th day. It is possible that the company has no days with the negative ai.
Output
Print an integer k — the required minimum number of folders. In the second line print a sequence of integers b1, b2, ..., bk, where bj is the number of day reports in the j-th folder.
If there are multiple ways to sort the reports into k days, print any of them.
Examples
Input
11
1 2 3 -4 -5 -6 5 -5 -6 -7 6
Output
3
5 3 3
Input
5
0 -1 100 -1 0
Output
1
5
Note
Here goes a way to sort the reports from the first sample into three folders:
1 2 3 -4 -5 | -6 5 -5 | -6 -7 6
In the second sample you can put all five reports in one folder.
Tags: greedy
Correct Solution:
```
N = int( input() )
A = list( map( int, input().split() ) )
ans = []
ptr = 0
while ptr < N:
s = 0
nxt = ptr
while nxt < N and s + ( A[ nxt ] < 0 ) <= 2:
s += ( A[ nxt ] < 0 )
nxt += 1
ans.append( nxt - ptr )
ptr = nxt
print( len( ans ) )
print( * ans )
```
| 97,271 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus has been working in the analytic department of the "F.R.A.U.D." company for as much as n days. Right now his task is to make a series of reports about the company's performance for the last n days. We know that the main information in a day report is value ai, the company's profit on the i-th day. If ai is negative, then the company suffered losses on the i-th day.
Polycarpus should sort the daily reports into folders. Each folder should include data on the company's performance for several consecutive days. Of course, the information on each of the n days should be exactly in one folder. Thus, Polycarpus puts information on the first few days in the first folder. The information on the several following days goes to the second folder, and so on.
It is known that the boss reads one daily report folder per day. If one folder has three or more reports for the days in which the company suffered losses (ai < 0), he loses his temper and his wrath is terrible.
Therefore, Polycarpus wants to prepare the folders so that none of them contains information on three or more days with the loss, and the number of folders is minimal.
Write a program that, given sequence ai, will print the minimum number of folders.
Input
The first line contains integer n (1 ≤ n ≤ 100), n is the number of days. The second line contains a sequence of integers a1, a2, ..., an (|ai| ≤ 100), where ai means the company profit on the i-th day. It is possible that the company has no days with the negative ai.
Output
Print an integer k — the required minimum number of folders. In the second line print a sequence of integers b1, b2, ..., bk, where bj is the number of day reports in the j-th folder.
If there are multiple ways to sort the reports into k days, print any of them.
Examples
Input
11
1 2 3 -4 -5 -6 5 -5 -6 -7 6
Output
3
5 3 3
Input
5
0 -1 100 -1 0
Output
1
5
Note
Here goes a way to sort the reports from the first sample into three folders:
1 2 3 -4 -5 | -6 5 -5 | -6 -7 6
In the second sample you can put all five reports in one folder.
Tags: greedy
Correct Solution:
```
import sys
import math
n = int(sys.stdin.readline())
an = [int(x) for x in (sys.stdin.readline()).split()]
res = 0
v = []
p = 0
k = 0
for i in an:
if(i < 0):
if(p < 2):
p += 1
else:
p = 1
v.append(str(k))
res += 1
k = 0
k += 1
if(k != 0):
res += 1
v.append(str(k))
print(res)
print(" ".join(v))
```
| 97,272 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus has been working in the analytic department of the "F.R.A.U.D." company for as much as n days. Right now his task is to make a series of reports about the company's performance for the last n days. We know that the main information in a day report is value ai, the company's profit on the i-th day. If ai is negative, then the company suffered losses on the i-th day.
Polycarpus should sort the daily reports into folders. Each folder should include data on the company's performance for several consecutive days. Of course, the information on each of the n days should be exactly in one folder. Thus, Polycarpus puts information on the first few days in the first folder. The information on the several following days goes to the second folder, and so on.
It is known that the boss reads one daily report folder per day. If one folder has three or more reports for the days in which the company suffered losses (ai < 0), he loses his temper and his wrath is terrible.
Therefore, Polycarpus wants to prepare the folders so that none of them contains information on three or more days with the loss, and the number of folders is minimal.
Write a program that, given sequence ai, will print the minimum number of folders.
Input
The first line contains integer n (1 ≤ n ≤ 100), n is the number of days. The second line contains a sequence of integers a1, a2, ..., an (|ai| ≤ 100), where ai means the company profit on the i-th day. It is possible that the company has no days with the negative ai.
Output
Print an integer k — the required minimum number of folders. In the second line print a sequence of integers b1, b2, ..., bk, where bj is the number of day reports in the j-th folder.
If there are multiple ways to sort the reports into k days, print any of them.
Examples
Input
11
1 2 3 -4 -5 -6 5 -5 -6 -7 6
Output
3
5 3 3
Input
5
0 -1 100 -1 0
Output
1
5
Note
Here goes a way to sort the reports from the first sample into three folders:
1 2 3 -4 -5 | -6 5 -5 | -6 -7 6
In the second sample you can put all five reports in one folder.
Tags: greedy
Correct Solution:
```
import re
import itertools
from collections import Counter
class Task:
a = []
answer = []
def getData(self):
input()
self.a = [int(x) for x in input().split(" ")]
def solve(self):
currentFolderCounter = 0
badDaysCounter = 0
for x in self.a:
if x >= 0:
currentFolderCounter += 1
elif badDaysCounter <= 1:
currentFolderCounter += 1
badDaysCounter += 1
else:
self.answer += [currentFolderCounter]
currentFolderCounter = 1
badDaysCounter = 1
if currentFolderCounter > 0:
self.answer += [currentFolderCounter]
def printAnswer(self):
print(len(self.answer))
print(re.sub('[\[\],]', '', str(self.answer)))
task = Task();
task.getData();
task.solve();
task.printAnswer();
```
| 97,273 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus has been working in the analytic department of the "F.R.A.U.D." company for as much as n days. Right now his task is to make a series of reports about the company's performance for the last n days. We know that the main information in a day report is value ai, the company's profit on the i-th day. If ai is negative, then the company suffered losses on the i-th day.
Polycarpus should sort the daily reports into folders. Each folder should include data on the company's performance for several consecutive days. Of course, the information on each of the n days should be exactly in one folder. Thus, Polycarpus puts information on the first few days in the first folder. The information on the several following days goes to the second folder, and so on.
It is known that the boss reads one daily report folder per day. If one folder has three or more reports for the days in which the company suffered losses (ai < 0), he loses his temper and his wrath is terrible.
Therefore, Polycarpus wants to prepare the folders so that none of them contains information on three or more days with the loss, and the number of folders is minimal.
Write a program that, given sequence ai, will print the minimum number of folders.
Input
The first line contains integer n (1 ≤ n ≤ 100), n is the number of days. The second line contains a sequence of integers a1, a2, ..., an (|ai| ≤ 100), where ai means the company profit on the i-th day. It is possible that the company has no days with the negative ai.
Output
Print an integer k — the required minimum number of folders. In the second line print a sequence of integers b1, b2, ..., bk, where bj is the number of day reports in the j-th folder.
If there are multiple ways to sort the reports into k days, print any of them.
Examples
Input
11
1 2 3 -4 -5 -6 5 -5 -6 -7 6
Output
3
5 3 3
Input
5
0 -1 100 -1 0
Output
1
5
Note
Here goes a way to sort the reports from the first sample into three folders:
1 2 3 -4 -5 | -6 5 -5 | -6 -7 6
In the second sample you can put all five reports in one folder.
Tags: greedy
Correct Solution:
```
def fun(ls,n):
count=0
ans=[]
number_of_elements=0
for ind,i in enumerate(ls):
number_of_elements+=1
if(i<0):
count+=1
if(count>2):
ans.append(number_of_elements-1)
count=1
number_of_elements=1
if(ind==n-1):
ans.append(number_of_elements)
count=0
print(len(ans))
print(*ans)
# T = int(input())
# for i in range(T):
n = int(input())
ls=list(map(int,input().split()))
fun(ls,n)
```
| 97,274 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus has been working in the analytic department of the "F.R.A.U.D." company for as much as n days. Right now his task is to make a series of reports about the company's performance for the last n days. We know that the main information in a day report is value ai, the company's profit on the i-th day. If ai is negative, then the company suffered losses on the i-th day.
Polycarpus should sort the daily reports into folders. Each folder should include data on the company's performance for several consecutive days. Of course, the information on each of the n days should be exactly in one folder. Thus, Polycarpus puts information on the first few days in the first folder. The information on the several following days goes to the second folder, and so on.
It is known that the boss reads one daily report folder per day. If one folder has three or more reports for the days in which the company suffered losses (ai < 0), he loses his temper and his wrath is terrible.
Therefore, Polycarpus wants to prepare the folders so that none of them contains information on three or more days with the loss, and the number of folders is minimal.
Write a program that, given sequence ai, will print the minimum number of folders.
Input
The first line contains integer n (1 ≤ n ≤ 100), n is the number of days. The second line contains a sequence of integers a1, a2, ..., an (|ai| ≤ 100), where ai means the company profit on the i-th day. It is possible that the company has no days with the negative ai.
Output
Print an integer k — the required minimum number of folders. In the second line print a sequence of integers b1, b2, ..., bk, where bj is the number of day reports in the j-th folder.
If there are multiple ways to sort the reports into k days, print any of them.
Examples
Input
11
1 2 3 -4 -5 -6 5 -5 -6 -7 6
Output
3
5 3 3
Input
5
0 -1 100 -1 0
Output
1
5
Note
Here goes a way to sort the reports from the first sample into three folders:
1 2 3 -4 -5 | -6 5 -5 | -6 -7 6
In the second sample you can put all five reports in one folder.
Submitted Solution:
```
n = int(input())
a = list(map(int,input().split()))
count,days,total,b = 0,0,0,[]
for i in range(n):
days += 1
if a[i]<0: count+=1
if count==3:
total+=1
b.append(days-1)
days=1
count=1
else:
print(total+1)
b.append(days)
print(*b)
```
Yes
| 97,275 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus has been working in the analytic department of the "F.R.A.U.D." company for as much as n days. Right now his task is to make a series of reports about the company's performance for the last n days. We know that the main information in a day report is value ai, the company's profit on the i-th day. If ai is negative, then the company suffered losses on the i-th day.
Polycarpus should sort the daily reports into folders. Each folder should include data on the company's performance for several consecutive days. Of course, the information on each of the n days should be exactly in one folder. Thus, Polycarpus puts information on the first few days in the first folder. The information on the several following days goes to the second folder, and so on.
It is known that the boss reads one daily report folder per day. If one folder has three or more reports for the days in which the company suffered losses (ai < 0), he loses his temper and his wrath is terrible.
Therefore, Polycarpus wants to prepare the folders so that none of them contains information on three or more days with the loss, and the number of folders is minimal.
Write a program that, given sequence ai, will print the minimum number of folders.
Input
The first line contains integer n (1 ≤ n ≤ 100), n is the number of days. The second line contains a sequence of integers a1, a2, ..., an (|ai| ≤ 100), where ai means the company profit on the i-th day. It is possible that the company has no days with the negative ai.
Output
Print an integer k — the required minimum number of folders. In the second line print a sequence of integers b1, b2, ..., bk, where bj is the number of day reports in the j-th folder.
If there are multiple ways to sort the reports into k days, print any of them.
Examples
Input
11
1 2 3 -4 -5 -6 5 -5 -6 -7 6
Output
3
5 3 3
Input
5
0 -1 100 -1 0
Output
1
5
Note
Here goes a way to sort the reports from the first sample into three folders:
1 2 3 -4 -5 | -6 5 -5 | -6 -7 6
In the second sample you can put all five reports in one folder.
Submitted Solution:
```
Answer = []
N = int(input())
Negative, Count = 0, 0
X = list(map(int, input().split()))
for i in range(N):
if X[i] < 0:
if Negative == 2:
Answer.append(Count)
Negative = 1
Count = 0
else:
Negative += 1
Count += 1
Answer.append(Count)
print(len(Answer))
print(*Answer)
# UB_CodeForces
# Advice: Falling down is an accident, staying down is a choice
# Location: Mashhad for few days
# Caption: Finally happened what should be happened
# CodeNumber: 697
```
Yes
| 97,276 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus has been working in the analytic department of the "F.R.A.U.D." company for as much as n days. Right now his task is to make a series of reports about the company's performance for the last n days. We know that the main information in a day report is value ai, the company's profit on the i-th day. If ai is negative, then the company suffered losses on the i-th day.
Polycarpus should sort the daily reports into folders. Each folder should include data on the company's performance for several consecutive days. Of course, the information on each of the n days should be exactly in one folder. Thus, Polycarpus puts information on the first few days in the first folder. The information on the several following days goes to the second folder, and so on.
It is known that the boss reads one daily report folder per day. If one folder has three or more reports for the days in which the company suffered losses (ai < 0), he loses his temper and his wrath is terrible.
Therefore, Polycarpus wants to prepare the folders so that none of them contains information on three or more days with the loss, and the number of folders is minimal.
Write a program that, given sequence ai, will print the minimum number of folders.
Input
The first line contains integer n (1 ≤ n ≤ 100), n is the number of days. The second line contains a sequence of integers a1, a2, ..., an (|ai| ≤ 100), where ai means the company profit on the i-th day. It is possible that the company has no days with the negative ai.
Output
Print an integer k — the required minimum number of folders. In the second line print a sequence of integers b1, b2, ..., bk, where bj is the number of day reports in the j-th folder.
If there are multiple ways to sort the reports into k days, print any of them.
Examples
Input
11
1 2 3 -4 -5 -6 5 -5 -6 -7 6
Output
3
5 3 3
Input
5
0 -1 100 -1 0
Output
1
5
Note
Here goes a way to sort the reports from the first sample into three folders:
1 2 3 -4 -5 | -6 5 -5 | -6 -7 6
In the second sample you can put all five reports in one folder.
Submitted Solution:
```
input()
B = []
b = neg = 0
for a in map(int, input().split()):
if a < 0:
if neg > 1:
B.append(b)
b = neg = 0
neg += 1
b += 1
print(len(B)+1)
print(*B, b)
```
Yes
| 97,277 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus has been working in the analytic department of the "F.R.A.U.D." company for as much as n days. Right now his task is to make a series of reports about the company's performance for the last n days. We know that the main information in a day report is value ai, the company's profit on the i-th day. If ai is negative, then the company suffered losses on the i-th day.
Polycarpus should sort the daily reports into folders. Each folder should include data on the company's performance for several consecutive days. Of course, the information on each of the n days should be exactly in one folder. Thus, Polycarpus puts information on the first few days in the first folder. The information on the several following days goes to the second folder, and so on.
It is known that the boss reads one daily report folder per day. If one folder has three or more reports for the days in which the company suffered losses (ai < 0), he loses his temper and his wrath is terrible.
Therefore, Polycarpus wants to prepare the folders so that none of them contains information on three or more days with the loss, and the number of folders is minimal.
Write a program that, given sequence ai, will print the minimum number of folders.
Input
The first line contains integer n (1 ≤ n ≤ 100), n is the number of days. The second line contains a sequence of integers a1, a2, ..., an (|ai| ≤ 100), where ai means the company profit on the i-th day. It is possible that the company has no days with the negative ai.
Output
Print an integer k — the required minimum number of folders. In the second line print a sequence of integers b1, b2, ..., bk, where bj is the number of day reports in the j-th folder.
If there are multiple ways to sort the reports into k days, print any of them.
Examples
Input
11
1 2 3 -4 -5 -6 5 -5 -6 -7 6
Output
3
5 3 3
Input
5
0 -1 100 -1 0
Output
1
5
Note
Here goes a way to sort the reports from the first sample into three folders:
1 2 3 -4 -5 | -6 5 -5 | -6 -7 6
In the second sample you can put all five reports in one folder.
Submitted Solution:
```
n = int(input())
reports = list(map(int , input().split()))
count = 0
i=0
folders = 1
arr=[]
pos = 0
while i<len(reports):
if reports[i]<0:
count = count+1
pos = pos + 1
if count==3:
count= 1
folders = folders + 1
arr.append(pos-1)
pos = 1
i = i + 1
if sum(arr)!=len(reports):
arr.append(len(reports)- sum(arr))
print(folders)
print(*arr)
```
Yes
| 97,278 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus has been working in the analytic department of the "F.R.A.U.D." company for as much as n days. Right now his task is to make a series of reports about the company's performance for the last n days. We know that the main information in a day report is value ai, the company's profit on the i-th day. If ai is negative, then the company suffered losses on the i-th day.
Polycarpus should sort the daily reports into folders. Each folder should include data on the company's performance for several consecutive days. Of course, the information on each of the n days should be exactly in one folder. Thus, Polycarpus puts information on the first few days in the first folder. The information on the several following days goes to the second folder, and so on.
It is known that the boss reads one daily report folder per day. If one folder has three or more reports for the days in which the company suffered losses (ai < 0), he loses his temper and his wrath is terrible.
Therefore, Polycarpus wants to prepare the folders so that none of them contains information on three or more days with the loss, and the number of folders is minimal.
Write a program that, given sequence ai, will print the minimum number of folders.
Input
The first line contains integer n (1 ≤ n ≤ 100), n is the number of days. The second line contains a sequence of integers a1, a2, ..., an (|ai| ≤ 100), where ai means the company profit on the i-th day. It is possible that the company has no days with the negative ai.
Output
Print an integer k — the required minimum number of folders. In the second line print a sequence of integers b1, b2, ..., bk, where bj is the number of day reports in the j-th folder.
If there are multiple ways to sort the reports into k days, print any of them.
Examples
Input
11
1 2 3 -4 -5 -6 5 -5 -6 -7 6
Output
3
5 3 3
Input
5
0 -1 100 -1 0
Output
1
5
Note
Here goes a way to sort the reports from the first sample into three folders:
1 2 3 -4 -5 | -6 5 -5 | -6 -7 6
In the second sample you can put all five reports in one folder.
Submitted Solution:
```
from math import *
import time
n=int(input())
l=input().split(" ")
ll=[]
c=0
s=0
for i in range(0,len(l)):
if(s==3):
ll.append(c-1)
c=2
s=1
else:
if(int(l[i])<0):
s=s+1
c=c+1
else:
c=c+1
if(s==3):
ll.append(c-1)
ll.append(1)
else:
ll.append(c)
print(len(ll))
print(*ll)
# print(time.time()-t1)
```
No
| 97,279 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus has been working in the analytic department of the "F.R.A.U.D." company for as much as n days. Right now his task is to make a series of reports about the company's performance for the last n days. We know that the main information in a day report is value ai, the company's profit on the i-th day. If ai is negative, then the company suffered losses on the i-th day.
Polycarpus should sort the daily reports into folders. Each folder should include data on the company's performance for several consecutive days. Of course, the information on each of the n days should be exactly in one folder. Thus, Polycarpus puts information on the first few days in the first folder. The information on the several following days goes to the second folder, and so on.
It is known that the boss reads one daily report folder per day. If one folder has three or more reports for the days in which the company suffered losses (ai < 0), he loses his temper and his wrath is terrible.
Therefore, Polycarpus wants to prepare the folders so that none of them contains information on three or more days with the loss, and the number of folders is minimal.
Write a program that, given sequence ai, will print the minimum number of folders.
Input
The first line contains integer n (1 ≤ n ≤ 100), n is the number of days. The second line contains a sequence of integers a1, a2, ..., an (|ai| ≤ 100), where ai means the company profit on the i-th day. It is possible that the company has no days with the negative ai.
Output
Print an integer k — the required minimum number of folders. In the second line print a sequence of integers b1, b2, ..., bk, where bj is the number of day reports in the j-th folder.
If there are multiple ways to sort the reports into k days, print any of them.
Examples
Input
11
1 2 3 -4 -5 -6 5 -5 -6 -7 6
Output
3
5 3 3
Input
5
0 -1 100 -1 0
Output
1
5
Note
Here goes a way to sort the reports from the first sample into three folders:
1 2 3 -4 -5 | -6 5 -5 | -6 -7 6
In the second sample you can put all five reports in one folder.
Submitted Solution:
```
n=int(int(input()))
l=list(map(int,input().split()))
neg=pos=0
z=[]
for x in range(n):
if l[x]>=0:
pos+=1
else:
if neg<2:
neg+=1
else:
z.append(neg+pos)
neg=1
pos=0
if neg!=0 or pos!=0:
z.append(neg+pos)
print(len(z))
```
No
| 97,280 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus has been working in the analytic department of the "F.R.A.U.D." company for as much as n days. Right now his task is to make a series of reports about the company's performance for the last n days. We know that the main information in a day report is value ai, the company's profit on the i-th day. If ai is negative, then the company suffered losses on the i-th day.
Polycarpus should sort the daily reports into folders. Each folder should include data on the company's performance for several consecutive days. Of course, the information on each of the n days should be exactly in one folder. Thus, Polycarpus puts information on the first few days in the first folder. The information on the several following days goes to the second folder, and so on.
It is known that the boss reads one daily report folder per day. If one folder has three or more reports for the days in which the company suffered losses (ai < 0), he loses his temper and his wrath is terrible.
Therefore, Polycarpus wants to prepare the folders so that none of them contains information on three or more days with the loss, and the number of folders is minimal.
Write a program that, given sequence ai, will print the minimum number of folders.
Input
The first line contains integer n (1 ≤ n ≤ 100), n is the number of days. The second line contains a sequence of integers a1, a2, ..., an (|ai| ≤ 100), where ai means the company profit on the i-th day. It is possible that the company has no days with the negative ai.
Output
Print an integer k — the required minimum number of folders. In the second line print a sequence of integers b1, b2, ..., bk, where bj is the number of day reports in the j-th folder.
If there are multiple ways to sort the reports into k days, print any of them.
Examples
Input
11
1 2 3 -4 -5 -6 5 -5 -6 -7 6
Output
3
5 3 3
Input
5
0 -1 100 -1 0
Output
1
5
Note
Here goes a way to sort the reports from the first sample into three folders:
1 2 3 -4 -5 | -6 5 -5 | -6 -7 6
In the second sample you can put all five reports in one folder.
Submitted Solution:
```
p, x, y, n, t = [], 0, 0, int(input()), list(map(int, input().split()))
for i in range(1, n):
if t[i] < 0: x += 1
if x == 3:
p.append(i - y)
x = 1
y = i
p.append(n - y)
print(len(p))
print(' '.join(str(i) for i in p))
```
No
| 97,281 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus has been working in the analytic department of the "F.R.A.U.D." company for as much as n days. Right now his task is to make a series of reports about the company's performance for the last n days. We know that the main information in a day report is value ai, the company's profit on the i-th day. If ai is negative, then the company suffered losses on the i-th day.
Polycarpus should sort the daily reports into folders. Each folder should include data on the company's performance for several consecutive days. Of course, the information on each of the n days should be exactly in one folder. Thus, Polycarpus puts information on the first few days in the first folder. The information on the several following days goes to the second folder, and so on.
It is known that the boss reads one daily report folder per day. If one folder has three or more reports for the days in which the company suffered losses (ai < 0), he loses his temper and his wrath is terrible.
Therefore, Polycarpus wants to prepare the folders so that none of them contains information on three or more days with the loss, and the number of folders is minimal.
Write a program that, given sequence ai, will print the minimum number of folders.
Input
The first line contains integer n (1 ≤ n ≤ 100), n is the number of days. The second line contains a sequence of integers a1, a2, ..., an (|ai| ≤ 100), where ai means the company profit on the i-th day. It is possible that the company has no days with the negative ai.
Output
Print an integer k — the required minimum number of folders. In the second line print a sequence of integers b1, b2, ..., bk, where bj is the number of day reports in the j-th folder.
If there are multiple ways to sort the reports into k days, print any of them.
Examples
Input
11
1 2 3 -4 -5 -6 5 -5 -6 -7 6
Output
3
5 3 3
Input
5
0 -1 100 -1 0
Output
1
5
Note
Here goes a way to sort the reports from the first sample into three folders:
1 2 3 -4 -5 | -6 5 -5 | -6 -7 6
In the second sample you can put all five reports in one folder.
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
c=0
f=0
for i in range(0,n):
if a[i]>=0:
c=c+1
elif a[i]<0:
f=f+1
if f>2:
f=f-2
print(c+2,end=" ")
c=0
print(c+f)
```
No
| 97,282 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider an n × m grid. Initially all the cells of the grid are colored white. Lenny has painted some of the cells (at least one) black. We call a painted grid convex if one can walk from any black cell to any another black cell using a path of side-adjacent black cells changing his direction at most once during the path. In the figure below, the left grid is convex while the right one is not convex, because there exist two cells which need more than one time to change direction in their path.
<image>
You're given a painted grid in the input. Tell Lenny if the grid is convex or not.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 50) — the size of the grid. Each of the next n lines contains m characters "B" or "W". Character "B" denotes a black cell of the grid and "W" denotes a white cell of the grid.
It's guaranteed that the grid has at least one black cell.
Output
On the only line of the output print "YES" if the grid is convex, otherwise print "NO". Do not print quotes.
Examples
Input
3 4
WWBW
BWWW
WWWB
Output
NO
Input
3 1
B
B
W
Output
YES
Tags: constructive algorithms, implementation
Correct Solution:
```
#author: riyan
def solve(n, m):
grid = []
for i in range(n):
grid.append(input().strip())
cnt = 0
for j in range(1, m):
if grid[i][j] != grid[i][j - 1]:
cnt += 1
if (cnt > 2) or (cnt == 2 and grid[i][0] == 'B'):
#print('row2 check, cnt = ', cnt)
return False
for j in range(m):
cnt = 0
for i in range(1, n):
#print(i, j, grid[i][j])
if grid[i][j] != grid[i - 1][j]:
cnt += 1
if (cnt > 2) or (cnt == 2 and grid[0][j] == 'B'):
#print('col2 check, cnt = ', cnt)
return False
bps = []
for i in range(n):
for j in range(m):
if grid[i][j] == 'B':
bp1 = (i, j)
for k in range(len(bps)):
bp2 = bps[k]
if not ( (grid[bp1[0]][bp2[1]] == 'B') or (grid[bp2[0]][bp1[1]] == 'B') ):
#print(bp1, bp2)
return False
bps.append((i, j))
return True
if __name__ == '__main__':
n, m = map(int, input().strip().split())
ans = solve(n, m)
if ans:
print('YES')
else:
print('NO')
```
| 97,283 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider an n × m grid. Initially all the cells of the grid are colored white. Lenny has painted some of the cells (at least one) black. We call a painted grid convex if one can walk from any black cell to any another black cell using a path of side-adjacent black cells changing his direction at most once during the path. In the figure below, the left grid is convex while the right one is not convex, because there exist two cells which need more than one time to change direction in their path.
<image>
You're given a painted grid in the input. Tell Lenny if the grid is convex or not.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 50) — the size of the grid. Each of the next n lines contains m characters "B" or "W". Character "B" denotes a black cell of the grid and "W" denotes a white cell of the grid.
It's guaranteed that the grid has at least one black cell.
Output
On the only line of the output print "YES" if the grid is convex, otherwise print "NO". Do not print quotes.
Examples
Input
3 4
WWBW
BWWW
WWWB
Output
NO
Input
3 1
B
B
W
Output
YES
Tags: constructive algorithms, implementation
Correct Solution:
```
import sys
def exi():
print("NO")
sys.exit()
I=lambda:list(map(int,input().split()))
g=[]
n,m=I()
for i in range(n):
g.append(list(input()))
for i in range(n):
temp=0
for j in range(1,m):
if g[i][j-1]!=g[i][j]:
temp+=1
#print(temp)
if temp>2 or temp==2 and g[i][0]=='B':exi()
for j in range(m):
temp=0
for i in range(1,n):
if g[i-1][j]!=g[i][j]:
temp+=1
if temp>2 or temp==2 and g[0][j]=='B':exi()
for i1 in range(n):
for j1 in range(m):
if g[i1][j1]=='B':
for i2 in range(n):
for j2 in range(m):
if g[i2][j2]=='B':
if g[i1][j2]=='W' and g[i2][j1]=='W':exi()
print("YES")
```
| 97,284 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider an n × m grid. Initially all the cells of the grid are colored white. Lenny has painted some of the cells (at least one) black. We call a painted grid convex if one can walk from any black cell to any another black cell using a path of side-adjacent black cells changing his direction at most once during the path. In the figure below, the left grid is convex while the right one is not convex, because there exist two cells which need more than one time to change direction in their path.
<image>
You're given a painted grid in the input. Tell Lenny if the grid is convex or not.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 50) — the size of the grid. Each of the next n lines contains m characters "B" or "W". Character "B" denotes a black cell of the grid and "W" denotes a white cell of the grid.
It's guaranteed that the grid has at least one black cell.
Output
On the only line of the output print "YES" if the grid is convex, otherwise print "NO". Do not print quotes.
Examples
Input
3 4
WWBW
BWWW
WWWB
Output
NO
Input
3 1
B
B
W
Output
YES
Tags: constructive algorithms, implementation
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
from collections import Counter
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
def gcd(a, b):
if a == 0:
return b
return gcd(b % a, a)
def lcm(a, b):
return (a * b) / gcd(a, b)
def main():
n, m = map(int, input().split())
a = []
for i in range(n):
t = input()
k = []
for j in t:
if j == 'W':
k.append(0)
if j == 'B':
k.append(1)
a.append(k)
reach1 = [[] for i in range(2600)]
blacks=[]
for i in range(n):
for j in range(m):
if a[i][j]:
blacks.append(i*m+j)
t = i
while t >= 0 and a[t][j]:
reach1[i*m+j].append(t*m+j)
t -= 1
t=i+1
while t<n and a[t][j]:
reach1[i * m + j].append(t * m + j)
t +=1
t=j-1
while t >= 0 and a[i][t]:
reach1[i*m+j].append(i*m+t)
t -= 1
t=j+1
while t<m and a[i][t]:
reach1[i * m + j].append(i* m + t)
t+=1
f=1
#print(blacks)
# print(reach1)
for i in blacks:
k=set(reach1[i])
#print(reach1[i])
k.add(i)
for j in reach1[i]:
if j!=i:
for m in reach1[j]:
k.add(m)
for m in blacks:
if m not in k:
f=0
break
if f==0:
break
if f:
print('YES')
else:
print('NO')
return
if __name__ == "__main__":
main()
```
| 97,285 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider an n × m grid. Initially all the cells of the grid are colored white. Lenny has painted some of the cells (at least one) black. We call a painted grid convex if one can walk from any black cell to any another black cell using a path of side-adjacent black cells changing his direction at most once during the path. In the figure below, the left grid is convex while the right one is not convex, because there exist two cells which need more than one time to change direction in their path.
<image>
You're given a painted grid in the input. Tell Lenny if the grid is convex or not.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 50) — the size of the grid. Each of the next n lines contains m characters "B" or "W". Character "B" denotes a black cell of the grid and "W" denotes a white cell of the grid.
It's guaranteed that the grid has at least one black cell.
Output
On the only line of the output print "YES" if the grid is convex, otherwise print "NO". Do not print quotes.
Examples
Input
3 4
WWBW
BWWW
WWWB
Output
NO
Input
3 1
B
B
W
Output
YES
Tags: constructive algorithms, implementation
Correct Solution:
```
def f():
n, m = map(int, input().split())
t = [input() for j in range(n)]
p = [''.join(i) for i in zip(*t)]
if h(p): return 1
i = 0
while i < n and not 'B' in t[i]: i += 1
while i < n:
a = t[i].find('B')
if a < 0:
i += 1
break
b = t[i].rfind('B')
if 'W' in t[i][a: b + 1]: return 1
for j in range(i + 1, n):
if a > 0 and t[j][a - 1] == 'B' and t[j][b] == 'W': return 1
if b < m - 1 and t[j][b + 1] == 'B' and t[j][a] == 'W': return 1
i += 1
while i < n:
if 'B' in t[i]: return 1
i += 1
return 0
def h(t):
i, n = 0, len(t)
while i < n and not 'B' in t[i]: i += 1
while i < n:
a = t[i].find('B')
if a < 0:
i += 1
break
b = t[i].rfind('B')
if 'W' in t[i][a: b + 1]: return 1
i += 1
while i < n:
if 'B' in t[i]: return 1
i += 1
return 0
print('YNEOS'[f():: 2])
```
| 97,286 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider an n × m grid. Initially all the cells of the grid are colored white. Lenny has painted some of the cells (at least one) black. We call a painted grid convex if one can walk from any black cell to any another black cell using a path of side-adjacent black cells changing his direction at most once during the path. In the figure below, the left grid is convex while the right one is not convex, because there exist two cells which need more than one time to change direction in their path.
<image>
You're given a painted grid in the input. Tell Lenny if the grid is convex or not.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 50) — the size of the grid. Each of the next n lines contains m characters "B" or "W". Character "B" denotes a black cell of the grid and "W" denotes a white cell of the grid.
It's guaranteed that the grid has at least one black cell.
Output
On the only line of the output print "YES" if the grid is convex, otherwise print "NO". Do not print quotes.
Examples
Input
3 4
WWBW
BWWW
WWWB
Output
NO
Input
3 1
B
B
W
Output
YES
Tags: constructive algorithms, implementation
Correct Solution:
```
def f():
n, m = map(int, input().split())
t = [input() for j in range(n)]
p = [''.join(i) for i in zip(*t)]
if h(p): return 1
i = 0
while i < n and not 'B' in t[i]: i += 1
while i < n:
a = t[i].find('B')
if a < 0:
i += 1
break
b = t[i].rfind('B')
if 'W' in t[i][a: b + 1]: return 1
for j in range(i + 1, n):
if a > 0 and t[j][a - 1] == 'B' and t[j][b] == 'W': return 1
if b < m - 1 and t[j][b + 1] == 'B' and t[j][a] == 'W': return 1
i += 1
while i < n:
if 'B' in t[i]: return 1
i += 1
return 0
def h(t):
i, n = 0, len(t)
while i < n and not 'B' in t[i]: i += 1
while i < n:
a = t[i].find('B')
if a < 0:
i += 1
break
b = t[i].rfind('B')
if 'W' in t[i][a: b + 1]: return 1
i += 1
while i < n:
if 'B' in t[i]: return 1
i += 1
return 0
print('YNEOS'[f():: 2])
# Made By Mostafa_Khaled
```
| 97,287 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider an n × m grid. Initially all the cells of the grid are colored white. Lenny has painted some of the cells (at least one) black. We call a painted grid convex if one can walk from any black cell to any another black cell using a path of side-adjacent black cells changing his direction at most once during the path. In the figure below, the left grid is convex while the right one is not convex, because there exist two cells which need more than one time to change direction in their path.
<image>
You're given a painted grid in the input. Tell Lenny if the grid is convex or not.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 50) — the size of the grid. Each of the next n lines contains m characters "B" or "W". Character "B" denotes a black cell of the grid and "W" denotes a white cell of the grid.
It's guaranteed that the grid has at least one black cell.
Output
On the only line of the output print "YES" if the grid is convex, otherwise print "NO". Do not print quotes.
Examples
Input
3 4
WWBW
BWWW
WWWB
Output
NO
Input
3 1
B
B
W
Output
YES
Tags: constructive algorithms, implementation
Correct Solution:
```
n, m = map(int, input().split())
z = [[] for i in range(n+1)]
for i in range(n):
a = input()
for j in a:
z[i].append(j)
def solve(n, m):
for i in range(n):
cnt = 0
for j in range(1, m):
if z[i][j] != z[i][j - 1]:
cnt += 1
if cnt > 2:
return 1
if cnt == 2 and z[i][0] == 'B':
return 1
for j in range(m):
cnt = 0
for i in range(1, n):
if z[i][j] != z[i-1][j]:
cnt += 1
if cnt > 2:
return 1
if cnt == 2 and z[0][j] == 'B':
return 1
for i in range(n):
for j in range(m):
if z[i][j] == 'B':
for x in range(i, n):
for y in range(m):
if z[x][y] == 'B':
if z[i][y]=='W' and z[x][j]=='W':
return 1
return 0
print(['YES','NO'][solve(n, m)])
```
| 97,288 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider an n × m grid. Initially all the cells of the grid are colored white. Lenny has painted some of the cells (at least one) black. We call a painted grid convex if one can walk from any black cell to any another black cell using a path of side-adjacent black cells changing his direction at most once during the path. In the figure below, the left grid is convex while the right one is not convex, because there exist two cells which need more than one time to change direction in their path.
<image>
You're given a painted grid in the input. Tell Lenny if the grid is convex or not.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 50) — the size of the grid. Each of the next n lines contains m characters "B" or "W". Character "B" denotes a black cell of the grid and "W" denotes a white cell of the grid.
It's guaranteed that the grid has at least one black cell.
Output
On the only line of the output print "YES" if the grid is convex, otherwise print "NO". Do not print quotes.
Examples
Input
3 4
WWBW
BWWW
WWWB
Output
NO
Input
3 1
B
B
W
Output
YES
Tags: constructive algorithms, implementation
Correct Solution:
```
n,m = map(int, input().split())
row, col_sum, row_sum, black = [], [], [], []
for i in range(n):
row.append(input())
t = [0]
for j in range(m):
t += [t[j] + (row[i][j] == 'B')]
row_sum += [t]
d = [[0,1], [1,0], [-1,0], [0,-1]]
for i in range(n):
for j in range(m):
if row[i][j] is 'W':
continue
w = 0
for di in d:
x = i + di[0]
y = j + di[1]
if x < 0 or y < 0 or x >= n or y >= m:
w += 1 ; continue
if row[x][y] is 'W':
w += 1
if w > 0: black.append((i,j))
for i in range(m):
t = [0]
for j in range(n):
t += [t[j] + (row[j][i] == 'B')]
col_sum += [t]
def row_check(r, s, e):
if s > e: e, s = s, e
return row_sum[r][e + 1] - row_sum[r][s] == e - s + 1
def col_check(c, s, e):
if s > e: e,s = s,e
return col_sum[c][e + 1] - col_sum[c][s] == e - s + 1
res = True
for i in black:
for j in black:
if i <= j:
continue
a = row_check(i[0], i[1], j[1]) and col_check(j[1], i[0], j[0])
b = row_check(j[0], i[1], j[1]) and col_check(i[1], i[0], j[0])
res = res and (a or b)
print('YES' if res else 'NO')
```
| 97,289 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider an n × m grid. Initially all the cells of the grid are colored white. Lenny has painted some of the cells (at least one) black. We call a painted grid convex if one can walk from any black cell to any another black cell using a path of side-adjacent black cells changing his direction at most once during the path. In the figure below, the left grid is convex while the right one is not convex, because there exist two cells which need more than one time to change direction in their path.
<image>
You're given a painted grid in the input. Tell Lenny if the grid is convex or not.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 50) — the size of the grid. Each of the next n lines contains m characters "B" or "W". Character "B" denotes a black cell of the grid and "W" denotes a white cell of the grid.
It's guaranteed that the grid has at least one black cell.
Output
On the only line of the output print "YES" if the grid is convex, otherwise print "NO". Do not print quotes.
Examples
Input
3 4
WWBW
BWWW
WWWB
Output
NO
Input
3 1
B
B
W
Output
YES
Tags: constructive algorithms, implementation
Correct Solution:
```
n, m = map(int, input().split())
grid = []
ans = True
for i in range(n):
grid.append(input())
black = False
change = False
for j in range(m):
if grid[i][j] == "W" and black:
change = True
elif grid[i][j] == "B":
if change and black:
ans = False
black = True
if not ans:
print("NO")
exit()
for j in range(m):
black = False
change = False
for i in range(n):
if grid[i][j] == "W" and black:
change = True
elif grid[i][j] == "B":
if change and black:
print("NO")
exit()
black = True
for i in range(n):
for j in range(m):
if grid[i][j] == "B":
for i_2 in range(n):
for j_2 in range(m):
if grid[i_2][j_2] == "B" and grid[i][j_2] == "W" and grid[i_2][j] == "W":
print("NO")
exit()
print("YES")
```
| 97,290 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider an n × m grid. Initially all the cells of the grid are colored white. Lenny has painted some of the cells (at least one) black. We call a painted grid convex if one can walk from any black cell to any another black cell using a path of side-adjacent black cells changing his direction at most once during the path. In the figure below, the left grid is convex while the right one is not convex, because there exist two cells which need more than one time to change direction in their path.
<image>
You're given a painted grid in the input. Tell Lenny if the grid is convex or not.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 50) — the size of the grid. Each of the next n lines contains m characters "B" or "W". Character "B" denotes a black cell of the grid and "W" denotes a white cell of the grid.
It's guaranteed that the grid has at least one black cell.
Output
On the only line of the output print "YES" if the grid is convex, otherwise print "NO". Do not print quotes.
Examples
Input
3 4
WWBW
BWWW
WWWB
Output
NO
Input
3 1
B
B
W
Output
YES
Submitted Solution:
```
n, m = map(int, input().split())
c = [input() for _ in ' ' * n]
def sol(n, m):
for i in range(n):
count = 0
for j in range(1, m):
if c[i][j] != c[i][j - 1]:
count += 1
if count > 2:
return False
if count == 2 and c[i][0]=='B':
return False
for j in range(m):
count = 0
for i in range(1, n):
if c[i][j] != c[i - 1][j]:
count += 1
if count > 2:
return False
if count == 2 and c[0][j]=='B':
return False
for i in range(n):
for j in range(m):
if c[i][j] == 'B':
for x in range(i, n):
for y in range(m):
if c[x][y] == 'B':
if c[i][y] == 'W' and c[x][j] == 'W':
return False
return True
print('NYOE S'[sol(n, m)::2])
```
Yes
| 97,291 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider an n × m grid. Initially all the cells of the grid are colored white. Lenny has painted some of the cells (at least one) black. We call a painted grid convex if one can walk from any black cell to any another black cell using a path of side-adjacent black cells changing his direction at most once during the path. In the figure below, the left grid is convex while the right one is not convex, because there exist two cells which need more than one time to change direction in their path.
<image>
You're given a painted grid in the input. Tell Lenny if the grid is convex or not.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 50) — the size of the grid. Each of the next n lines contains m characters "B" or "W". Character "B" denotes a black cell of the grid and "W" denotes a white cell of the grid.
It's guaranteed that the grid has at least one black cell.
Output
On the only line of the output print "YES" if the grid is convex, otherwise print "NO". Do not print quotes.
Examples
Input
3 4
WWBW
BWWW
WWWB
Output
NO
Input
3 1
B
B
W
Output
YES
Submitted Solution:
```
import sys
input = sys.stdin.readline
def check(x, y):
for i in range(min(x[0], y[0]), max(x[0], y[0]) + 1):
if not plan[i][x[1]] == "B":
return False
for i in range(min(x[1], y[1]), max(x[1], y[1]) + 1):
if not plan[y[0]][i] == "B":
return False
return True
n, m = map(int, input().split())
plan = tuple(tuple(i for i in input().strip()) for j in range(n))
start = [(i, j) for i in range(n) for j in range(m) if plan[i][j] == "B"]
for i in range(len(start)):
for j in range(i + 1, len(start)):
if not check(start[i], start[j]) and not check(start[j], start[i]):
print("NO")
sys.exit()
print("YES")
```
Yes
| 97,292 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider an n × m grid. Initially all the cells of the grid are colored white. Lenny has painted some of the cells (at least one) black. We call a painted grid convex if one can walk from any black cell to any another black cell using a path of side-adjacent black cells changing his direction at most once during the path. In the figure below, the left grid is convex while the right one is not convex, because there exist two cells which need more than one time to change direction in their path.
<image>
You're given a painted grid in the input. Tell Lenny if the grid is convex or not.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 50) — the size of the grid. Each of the next n lines contains m characters "B" or "W". Character "B" denotes a black cell of the grid and "W" denotes a white cell of the grid.
It's guaranteed that the grid has at least one black cell.
Output
On the only line of the output print "YES" if the grid is convex, otherwise print "NO". Do not print quotes.
Examples
Input
3 4
WWBW
BWWW
WWWB
Output
NO
Input
3 1
B
B
W
Output
YES
Submitted Solution:
```
import sys
input=sys.stdin.readline
def exi():
print("NO")
sys.exit()
I=lambda:list(map(int,input().split()))
g=[]
n,m=I()
for i in range(n):
g.append(list(input()))
for i in range(n):
temp=0
for j in range(1,m):
if g[i][j-1]!=g[i][j]:
temp+=1
#print(temp)
if temp>2 or temp==2 and g[i][0]=='B':exi()
for j in range(m):
temp=0
for i in range(1,n):
if g[i-1][j]!=g[i][j]:
temp+=1
if temp>2 or temp==2 and g[0][j]=='B':exi()
for i1 in range(n):
for j1 in range(m):
if g[i1][j1]=='B':
for i2 in range(n):
for j2 in range(m):
if g[i2][j2]=='B':
if g[i1][j2]=='W' and g[i2][j1]=='W':exi()
print("YES")
```
Yes
| 97,293 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider an n × m grid. Initially all the cells of the grid are colored white. Lenny has painted some of the cells (at least one) black. We call a painted grid convex if one can walk from any black cell to any another black cell using a path of side-adjacent black cells changing his direction at most once during the path. In the figure below, the left grid is convex while the right one is not convex, because there exist two cells which need more than one time to change direction in their path.
<image>
You're given a painted grid in the input. Tell Lenny if the grid is convex or not.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 50) — the size of the grid. Each of the next n lines contains m characters "B" or "W". Character "B" denotes a black cell of the grid and "W" denotes a white cell of the grid.
It's guaranteed that the grid has at least one black cell.
Output
On the only line of the output print "YES" if the grid is convex, otherwise print "NO". Do not print quotes.
Examples
Input
3 4
WWBW
BWWW
WWWB
Output
NO
Input
3 1
B
B
W
Output
YES
Submitted Solution:
```
#author: riyan
def cprf(p1, p2, grid):
path = True
for r in range(min([p1[0], p2[0]]), max([p1[0], p2[0]]) + 1):
c = p1[1]
if grid[r][c] == 'W':
path = False
break
if path:
for c in range(min([p1[1], p2[1]]), max([p1[1], p2[1]]) + 1):
r = p1[0]
if grid[r][c] == 'W':
path = False
break
return path
def cpcf(p1, p2, grid):
path = True
for c in range(min([p1[1], p2[1]]), max([p1[1], p2[1]]) + 1):
r = p1[0]
if grid[r][c] == 'W':
path = False
break
if path:
for r in range(min([p1[0], p2[0]]), max([p1[0], p2[0]]) + 1):
c = p1[1]
if grid[r][c] == 'W':
path = False
break
return path
if __name__ == '__main__':
grid = []
n, m = map(int, input().strip().split())
for i in range(n):
grid.append(list(input().strip()))
bps = []
for i in range(n):
for j in range(m):
if grid[i][j] == 'B':
bps.append((i, j))
ans = True
for bp1 in bps:
for bp2 in bps:
way1 = cprf(bp1, bp2, grid)
way2 = cpcf(bp1, bp2, grid)
if not (way1 and way2):
ans = False
break
if not ans:
break
if ans:
print('YES')
else:
print('NO')
```
No
| 97,294 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider an n × m grid. Initially all the cells of the grid are colored white. Lenny has painted some of the cells (at least one) black. We call a painted grid convex if one can walk from any black cell to any another black cell using a path of side-adjacent black cells changing his direction at most once during the path. In the figure below, the left grid is convex while the right one is not convex, because there exist two cells which need more than one time to change direction in their path.
<image>
You're given a painted grid in the input. Tell Lenny if the grid is convex or not.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 50) — the size of the grid. Each of the next n lines contains m characters "B" or "W". Character "B" denotes a black cell of the grid and "W" denotes a white cell of the grid.
It's guaranteed that the grid has at least one black cell.
Output
On the only line of the output print "YES" if the grid is convex, otherwise print "NO". Do not print quotes.
Examples
Input
3 4
WWBW
BWWW
WWWB
Output
NO
Input
3 1
B
B
W
Output
YES
Submitted Solution:
```
# Author : raj1307 - Raj Singh
# Date : 02.01.2020
from __future__ import division, print_function
import os,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
def ii(): return int(input())
def si(): return input()
def mi(): return map(int,input().strip().split(" "))
def msi(): return map(str,input().strip().split(" "))
def li(): return list(mi())
def dmain():
sys.setrecursionlimit(100000000)
threading.stack_size(40960000)
thread = threading.Thread(target=main)
thread.start()
#from collections import deque, Counter, OrderedDict,defaultdict
#from heapq import nsmallest, nlargest, heapify,heappop ,heappush, heapreplace
#from math import ceil,floor,log,sqrt,factorial
#from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
#from decimal import *,threading
#from itertools import permutations
abc='abcdefghijklmnopqrstuvwxyz'
abd={'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24, 'z': 25}
mod=1000000007
#mod=998244353
inf = float("inf")
vow=['a','e','i','o','u']
dx,dy=[-1,1,0,0],[0,0,1,-1]
def getKey(item): return item[1]
def sort2(l):return sorted(l, key=getKey)
def d2(n,m,num):return [[num for x in range(m)] for y in range(n)]
def isPowerOfTwo (x): return (x and (not(x & (x - 1))) )
def decimalToBinary(n): return bin(n).replace("0b","")
def ntl(n):return [int(i) for i in str(n)]
def powerMod(x,y,p):
res = 1
x %= p
while y > 0:
if y&1:
res = (res*x)%p
y = y>>1
x = (x*x)%p
return res
def gcd(x, y):
while y:
x, y = y, x % y
return x
def isPrime(n) : # Check Prime Number or not
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
def read():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
def main():
#for _ in range(ii()):
n,c=mi()
f=1
l=[]
mark=[]
for i in range(n):
l.append(si())
x=[]
for j in l[i]:
if j=='B':
x.append(0)
else:
x.append(1)
mark.append(x)
m=mark[:]
if n==1 and c==1:
print('YES')
exit()
for i in range(n):
for j in range(c):
if l[i][j]=='B':
#m[i][j]=1
for k in range(j+1,c): #Right
if l[i][k]=='W':
break
m[i][k]=1
for x in range(i-1,-1,-1):
if l[x][k]=='W':
break
m[i][k]=1
for x in range(i+1,n):
if l[x][k]=='W':
break
m[i][k]=1
for k in range(j-1,-1,-1): #Left
if l[i][k]=='W':
break
m[i][k]=1
for x in range(i-1,-1,-1):
if l[x][k]=='W':
break
m[x][k]=1
for x in range(i+1,n):
if l[x][k]=='W':
break
m[x][k]=1
for k in range(i-1,-1,-1): #Up
if l[k][j]=='W':
break
m[k][j]=1
for x in range(j-1,-1,-1):
if l[k][x]=='W':
break
m[k][x]=1
for x in range(j+1,c):
if l[k][x]=='W':
break
m[k][x]=1
for k in range(i+1,n): #Down
if l[k][j]=='W':
break
m[k][j]=1
for x in range(j-1,-1,-1):
if l[k][x]=='W':
break
m[k][x]=1
for x in range(j+1,c):
if l[k][x]=='W':
break
m[k][x]=1
if m==[[1]*c]*n:
print('YES')
else:
print('NO')
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
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__":
#read()
main()
#dmain()
# Comment Read()
```
No
| 97,295 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider an n × m grid. Initially all the cells of the grid are colored white. Lenny has painted some of the cells (at least one) black. We call a painted grid convex if one can walk from any black cell to any another black cell using a path of side-adjacent black cells changing his direction at most once during the path. In the figure below, the left grid is convex while the right one is not convex, because there exist two cells which need more than one time to change direction in their path.
<image>
You're given a painted grid in the input. Tell Lenny if the grid is convex or not.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 50) — the size of the grid. Each of the next n lines contains m characters "B" or "W". Character "B" denotes a black cell of the grid and "W" denotes a white cell of the grid.
It's guaranteed that the grid has at least one black cell.
Output
On the only line of the output print "YES" if the grid is convex, otherwise print "NO". Do not print quotes.
Examples
Input
3 4
WWBW
BWWW
WWWB
Output
NO
Input
3 1
B
B
W
Output
YES
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
from collections import Counter
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
def gcd(a, b):
if a == 0:
return b
return gcd(b % a, a)
def lcm(a, b):
return (a * b) / gcd(a, b)
def main():
n, m = map(int, input().split())
a = []
for i in range(n):
t = input()
k = []
for j in t:
if j == 'W':
k.append(0)
if j == 'B':
k.append(1)
a.append(k)
reach1 = [[] for i in range(2600)]
blacks=[]
for i in range(n):
for j in range(m):
if a[i][j]:
blacks.append(i*m+j)
t = i
while t >= 0:
reach1[i*m+j].append(t*n+j)
t -= 1
t=i+1
while t<n:
reach1[i * m + j].append(t * m + j)
t +=1
t=j-1
while t >= 0:
reach1[i*m+j].append(i*n+t)
t -= 1
t=j+1
while t<m:
reach1[i * m + j].append(i* m + t)
t+=1
f=1
# print(blacks)
# print(reach1)
for i in blacks:
k=set(reach1[i])
k.add(i)
for j in blacks:
if j!=i:
for m in reach1[j]:
k.add(m)
for m in blacks:
if m not in k:
f=0
if f:
print('YES')
else:
print('NO')
return
if __name__ == "__main__":
main()
```
No
| 97,296 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider an n × m grid. Initially all the cells of the grid are colored white. Lenny has painted some of the cells (at least one) black. We call a painted grid convex if one can walk from any black cell to any another black cell using a path of side-adjacent black cells changing his direction at most once during the path. In the figure below, the left grid is convex while the right one is not convex, because there exist two cells which need more than one time to change direction in their path.
<image>
You're given a painted grid in the input. Tell Lenny if the grid is convex or not.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 50) — the size of the grid. Each of the next n lines contains m characters "B" or "W". Character "B" denotes a black cell of the grid and "W" denotes a white cell of the grid.
It's guaranteed that the grid has at least one black cell.
Output
On the only line of the output print "YES" if the grid is convex, otherwise print "NO". Do not print quotes.
Examples
Input
3 4
WWBW
BWWW
WWWB
Output
NO
Input
3 1
B
B
W
Output
YES
Submitted Solution:
```
#author: riyan
def solve(n, m):
grid = []
for i in range(n):
grid.append(input().strip())
cnt = 0
for j in range(1, m):
if grid[i][j] != grid[i][j - 1]:
cnt += 1
if (cnt > 2) or (cnt == 2 and grid[i][0] == 'B'):
return False
for j in range(m):
cnt = 0
for i in range(n):
if grid[i][j] != grid[i - 1][j]:
cnt += 1
if (cnt > 2) or (cnt == 2 and grid[0][j] == 'B'):
return False
bps = []
for i in range(n):
for j in range(m):
if grid[i][j] == 'B':
bp1 = (i, j)
for k in range(len(bps)):
bp2 = bps[k]
if not ( (grid[bp1[0]][bp2[1]] == 'B') or (grid[bp2[0]][bp1[1]] == 'B') ):
return False
bps.append((i, j))
return True
if __name__ == '__main__':
n, m = map(int, input().strip().split())
ans = solve(n, m)
if ans:
print('YES')
else:
print('NO')
```
No
| 97,297 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is known that there are k fish species in the polar ocean, numbered from 1 to k. They are sorted by non-decreasing order of their weight, which is a positive number. Let the weight of the i-th type of fish be wi, then 0 < w1 ≤ w2 ≤ ... ≤ wk holds.
Polar bears Alice and Bob each have caught some fish, and they are guessing who has the larger sum of weight of the fish he/she's caught. Given the type of the fish they've caught, determine whether it is possible that the fish caught by Alice has a strictly larger total weight than Bob's. In other words, does there exist a sequence of weights wi (not necessary integers), such that the fish caught by Alice has a strictly larger total weight?
Input
The first line contains three integers n, m, k (1 ≤ n, m ≤ 105, 1 ≤ k ≤ 109) — the number of fish caught by Alice and Bob respectively, and the number of fish species.
The second line contains n integers each from 1 to k, the list of fish type caught by Alice. The third line contains m integers each from 1 to k, the list of fish type caught by Bob.
Note that one may have caught more than one fish for a same species.
Output
Output "YES" (without quotes) if it is possible, and "NO" (without quotes) otherwise.
Examples
Input
3 3 3
2 2 2
1 1 3
Output
YES
Input
4 7 9
5 2 7 3
3 5 2 7 3 8 7
Output
NO
Note
In the first sample, if w1 = 1, w2 = 2, w3 = 2.5, then Alice has a total of 2 + 2 + 2 = 6 weight units, while Bob only has 1 + 1 + 2.5 = 4.5.
In the second sample, the fish that Alice caught is a subset of Bob's. Therefore, the total weight of Bob’s fish is always not less than the total weight of Alice’s fish.
Tags: constructive algorithms, greedy
Correct Solution:
```
def find(a, n, sz):
lo = 0
hi = sz-1
while lo <= hi:
mid = int ((lo + hi) / 2)
if a[mid] == n :
return mid
if a[mid] < n :
lo = mid + 1
if a[mid] > n :
hi = mid - 1
# print(n, end=" ")
# print(a[mid])
return -1
n, m, k = map(int, input().split())
x = list(map(int, input().split()))
y = list(map(int, input().split()))
countA = [0] * (m+n + 5)
countB = [0] * (n+m + 5)
A = []
for i in range(n):
A.append(x[i])
for i in range(m):
A.append(y[i])
A.sort()
for i in range(n):
id = find(A, x[i], m+n)
countA[id] += 1
for i in range(m):
id = find(A, y[i], m+n)
countB[id] += 1
flag = 0
i = m + n + 1
cA = 0
cB = 0
while i>=0 :
cA += countA[i]
cB += countB[i]
if cA > cB:
flag = 1
i -= 1
if flag:
print("YES")
else:
print("NO")
```
| 97,298 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is known that there are k fish species in the polar ocean, numbered from 1 to k. They are sorted by non-decreasing order of their weight, which is a positive number. Let the weight of the i-th type of fish be wi, then 0 < w1 ≤ w2 ≤ ... ≤ wk holds.
Polar bears Alice and Bob each have caught some fish, and they are guessing who has the larger sum of weight of the fish he/she's caught. Given the type of the fish they've caught, determine whether it is possible that the fish caught by Alice has a strictly larger total weight than Bob's. In other words, does there exist a sequence of weights wi (not necessary integers), such that the fish caught by Alice has a strictly larger total weight?
Input
The first line contains three integers n, m, k (1 ≤ n, m ≤ 105, 1 ≤ k ≤ 109) — the number of fish caught by Alice and Bob respectively, and the number of fish species.
The second line contains n integers each from 1 to k, the list of fish type caught by Alice. The third line contains m integers each from 1 to k, the list of fish type caught by Bob.
Note that one may have caught more than one fish for a same species.
Output
Output "YES" (without quotes) if it is possible, and "NO" (without quotes) otherwise.
Examples
Input
3 3 3
2 2 2
1 1 3
Output
YES
Input
4 7 9
5 2 7 3
3 5 2 7 3 8 7
Output
NO
Note
In the first sample, if w1 = 1, w2 = 2, w3 = 2.5, then Alice has a total of 2 + 2 + 2 = 6 weight units, while Bob only has 1 + 1 + 2.5 = 4.5.
In the second sample, the fish that Alice caught is a subset of Bob's. Therefore, the total weight of Bob’s fish is always not less than the total weight of Alice’s fish.
Tags: constructive algorithms, greedy
Correct Solution:
```
from collections import Counter
a,b,c = map(int, input().split())
l = list(map(int, input().split()))
l2 = list(map(int, input().split()))
m = max(max(l), max(l2))
s1 = Counter(l)
s2 = Counter(l2)
c1 = 0
c2 = 0
vis = set()
l3 = []
for i in l+l2:
if i not in vis:
l3.append(i)
vis.add(i)
l3 = sorted(l3)
for i in reversed(l3):
c1 += s1[i]
c2 += s2[i]
if c1 > c2:
print("YES")
exit()
print("NO")
```
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