text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
As you probably know, Anton goes to school. One of the school subjects that Anton studies is Bracketology. On the Bracketology lessons students usually learn different sequences that consist of round brackets (characters "(" and ")" (without quotes)).
On the last lesson Anton learned about the regular simple bracket sequences (RSBS). A bracket sequence s of length n is an RSBS if the following conditions are met:
* It is not empty (that is n β 0).
* The length of the sequence is even.
* First <image> charactes of the sequence are equal to "(".
* Last <image> charactes of the sequence are equal to ")".
For example, the sequence "((()))" is an RSBS but the sequences "((())" and "(()())" are not RSBS.
Elena Ivanovna, Anton's teacher, gave him the following task as a homework. Given a bracket sequence s. Find the number of its distinct subsequences such that they are RSBS. Note that a subsequence of s is a string that can be obtained from s by deleting some of its elements. Two subsequences are considered distinct if distinct sets of positions are deleted.
Because the answer can be very big and Anton's teacher doesn't like big numbers, she asks Anton to find the answer modulo 109 + 7.
Anton thought of this task for a very long time, but he still doesn't know how to solve it. Help Anton to solve this task and write a program that finds the answer for it!
Input
The only line of the input contains a string s β the bracket sequence given in Anton's homework. The string consists only of characters "(" and ")" (without quotes). It's guaranteed that the string is not empty and its length doesn't exceed 200 000.
Output
Output one number β the answer for the task modulo 109 + 7.
Examples
Input
)(()()
Output
6
Input
()()()
Output
7
Input
)))
Output
0
Note
In the first sample the following subsequences are possible:
* If we delete characters at the positions 1 and 5 (numbering starts with one), we will get the subsequence "(())".
* If we delete characters at the positions 1, 2, 3 and 4, we will get the subsequence "()".
* If we delete characters at the positions 1, 2, 4 and 5, we will get the subsequence "()".
* If we delete characters at the positions 1, 2, 5 and 6, we will get the subsequence "()".
* If we delete characters at the positions 1, 3, 4 and 5, we will get the subsequence "()".
* If we delete characters at the positions 1, 3, 5 and 6, we will get the subsequence "()".
The rest of the subsequnces are not RSBS. So we got 6 distinct subsequences that are RSBS, so the answer is 6.
Tags: combinatorics, dp, math, number theory
Correct Solution:
```
mod = 10 ** 9 + 7
s = input()
n = len(s)
f = [1] * (n + 1)
for i in range(1, n + 1): f[i] = i * f[i - 1] % mod
finv = [pow(x, mod - 2, mod) for x in f]
op = 0
cl = s.count(')')
ans = 0
if cl > 0:
for c in s:
if c == '(':
op += 1
ans += f[op + cl - 1] * finv[cl - 1] * finv[op]
elif cl <= 1:
break
else:
cl -= 1
print(ans % mod)
#bruno
```
| 90,600 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you probably know, Anton goes to school. One of the school subjects that Anton studies is Bracketology. On the Bracketology lessons students usually learn different sequences that consist of round brackets (characters "(" and ")" (without quotes)).
On the last lesson Anton learned about the regular simple bracket sequences (RSBS). A bracket sequence s of length n is an RSBS if the following conditions are met:
* It is not empty (that is n β 0).
* The length of the sequence is even.
* First <image> charactes of the sequence are equal to "(".
* Last <image> charactes of the sequence are equal to ")".
For example, the sequence "((()))" is an RSBS but the sequences "((())" and "(()())" are not RSBS.
Elena Ivanovna, Anton's teacher, gave him the following task as a homework. Given a bracket sequence s. Find the number of its distinct subsequences such that they are RSBS. Note that a subsequence of s is a string that can be obtained from s by deleting some of its elements. Two subsequences are considered distinct if distinct sets of positions are deleted.
Because the answer can be very big and Anton's teacher doesn't like big numbers, she asks Anton to find the answer modulo 109 + 7.
Anton thought of this task for a very long time, but he still doesn't know how to solve it. Help Anton to solve this task and write a program that finds the answer for it!
Input
The only line of the input contains a string s β the bracket sequence given in Anton's homework. The string consists only of characters "(" and ")" (without quotes). It's guaranteed that the string is not empty and its length doesn't exceed 200 000.
Output
Output one number β the answer for the task modulo 109 + 7.
Examples
Input
)(()()
Output
6
Input
()()()
Output
7
Input
)))
Output
0
Note
In the first sample the following subsequences are possible:
* If we delete characters at the positions 1 and 5 (numbering starts with one), we will get the subsequence "(())".
* If we delete characters at the positions 1, 2, 3 and 4, we will get the subsequence "()".
* If we delete characters at the positions 1, 2, 4 and 5, we will get the subsequence "()".
* If we delete characters at the positions 1, 2, 5 and 6, we will get the subsequence "()".
* If we delete characters at the positions 1, 3, 4 and 5, we will get the subsequence "()".
* If we delete characters at the positions 1, 3, 5 and 6, we will get the subsequence "()".
The rest of the subsequnces are not RSBS. So we got 6 distinct subsequences that are RSBS, so the answer is 6.
Tags: combinatorics, dp, math, number theory
Correct Solution:
```
p = 10**9+7
def power(x, y, p):
b = bin(y)[2:]
answer = 1
start = x % p
for i in range(len(b)):
if b[len(b)-1-i]=='1':
answer = (answer*start) % p
start = (start*start) % p
return answer
def process(S):
answer = 0
n = len(S)
left_count = 0
right_count = 0
p_lc = 1
p_rc = 1
p_num = 1
num = 0
for c in S:
if c==')':
right_count+=1
p_rc = (p_rc*(right_count+1)) % p
num+=1
p_num = (p_num*num) % p
for i in range(n):
if S[i]=='(':
left_count+=1
if left_count > 1:
p_lc = (p_lc*(left_count-1)) % p
num+=1
p_num = (p_num*num) % p
else:
p_rc = (p_rc*power(right_count+1, p-2, p)) % p
p_num = (p_num*power(num, p-2, p)) % p
right_count-=1
num-=1
#(left_count choose n, right_count choose n-1)
#(left_count choose n, right_count choose rc-n+1))
#(lc+rc choose rc+1)
if left_count > 0:
entry = (p_num*power(p_lc, p-2, p)*power(p_rc, p-2, p)) % p
answer = (answer+entry) % p
return answer
S = input()
print(process(S))
```
| 90,601 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you probably know, Anton goes to school. One of the school subjects that Anton studies is Bracketology. On the Bracketology lessons students usually learn different sequences that consist of round brackets (characters "(" and ")" (without quotes)).
On the last lesson Anton learned about the regular simple bracket sequences (RSBS). A bracket sequence s of length n is an RSBS if the following conditions are met:
* It is not empty (that is n β 0).
* The length of the sequence is even.
* First <image> charactes of the sequence are equal to "(".
* Last <image> charactes of the sequence are equal to ")".
For example, the sequence "((()))" is an RSBS but the sequences "((())" and "(()())" are not RSBS.
Elena Ivanovna, Anton's teacher, gave him the following task as a homework. Given a bracket sequence s. Find the number of its distinct subsequences such that they are RSBS. Note that a subsequence of s is a string that can be obtained from s by deleting some of its elements. Two subsequences are considered distinct if distinct sets of positions are deleted.
Because the answer can be very big and Anton's teacher doesn't like big numbers, she asks Anton to find the answer modulo 109 + 7.
Anton thought of this task for a very long time, but he still doesn't know how to solve it. Help Anton to solve this task and write a program that finds the answer for it!
Input
The only line of the input contains a string s β the bracket sequence given in Anton's homework. The string consists only of characters "(" and ")" (without quotes). It's guaranteed that the string is not empty and its length doesn't exceed 200 000.
Output
Output one number β the answer for the task modulo 109 + 7.
Examples
Input
)(()()
Output
6
Input
()()()
Output
7
Input
)))
Output
0
Note
In the first sample the following subsequences are possible:
* If we delete characters at the positions 1 and 5 (numbering starts with one), we will get the subsequence "(())".
* If we delete characters at the positions 1, 2, 3 and 4, we will get the subsequence "()".
* If we delete characters at the positions 1, 2, 4 and 5, we will get the subsequence "()".
* If we delete characters at the positions 1, 2, 5 and 6, we will get the subsequence "()".
* If we delete characters at the positions 1, 3, 4 and 5, we will get the subsequence "()".
* If we delete characters at the positions 1, 3, 5 and 6, we will get the subsequence "()".
The rest of the subsequnces are not RSBS. So we got 6 distinct subsequences that are RSBS, so the answer is 6.
Tags: combinatorics, dp, math, number theory
Correct Solution:
```
t = input()
n, m = len(t) + 1, 1000000007
x, y = 0, t.count(')') - 1
fact = [1] * n
for i in range(2, n): fact[i] = i * fact[i - 1] % m
invMult = [pow(x, m - 2, m) for x in fact]
s = 0
for b in t:
if y < 0: break
if b == '(':
x += 1
s += fact[x + y] * invMult[x] * invMult[y] % m
else: y -= 1
print(s % m)
```
| 90,602 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you probably know, Anton goes to school. One of the school subjects that Anton studies is Bracketology. On the Bracketology lessons students usually learn different sequences that consist of round brackets (characters "(" and ")" (without quotes)).
On the last lesson Anton learned about the regular simple bracket sequences (RSBS). A bracket sequence s of length n is an RSBS if the following conditions are met:
* It is not empty (that is n β 0).
* The length of the sequence is even.
* First <image> charactes of the sequence are equal to "(".
* Last <image> charactes of the sequence are equal to ")".
For example, the sequence "((()))" is an RSBS but the sequences "((())" and "(()())" are not RSBS.
Elena Ivanovna, Anton's teacher, gave him the following task as a homework. Given a bracket sequence s. Find the number of its distinct subsequences such that they are RSBS. Note that a subsequence of s is a string that can be obtained from s by deleting some of its elements. Two subsequences are considered distinct if distinct sets of positions are deleted.
Because the answer can be very big and Anton's teacher doesn't like big numbers, she asks Anton to find the answer modulo 109 + 7.
Anton thought of this task for a very long time, but he still doesn't know how to solve it. Help Anton to solve this task and write a program that finds the answer for it!
Input
The only line of the input contains a string s β the bracket sequence given in Anton's homework. The string consists only of characters "(" and ")" (without quotes). It's guaranteed that the string is not empty and its length doesn't exceed 200 000.
Output
Output one number β the answer for the task modulo 109 + 7.
Examples
Input
)(()()
Output
6
Input
()()()
Output
7
Input
)))
Output
0
Note
In the first sample the following subsequences are possible:
* If we delete characters at the positions 1 and 5 (numbering starts with one), we will get the subsequence "(())".
* If we delete characters at the positions 1, 2, 3 and 4, we will get the subsequence "()".
* If we delete characters at the positions 1, 2, 4 and 5, we will get the subsequence "()".
* If we delete characters at the positions 1, 2, 5 and 6, we will get the subsequence "()".
* If we delete characters at the positions 1, 3, 4 and 5, we will get the subsequence "()".
* If we delete characters at the positions 1, 3, 5 and 6, we will get the subsequence "()".
The rest of the subsequnces are not RSBS. So we got 6 distinct subsequences that are RSBS, so the answer is 6.
Tags: combinatorics, dp, math, number theory
Correct Solution:
```
s = input()
MOD = 10 ** 9 + 7
num = 0
N = len(s)
R = [0] * (N + 1)
L = [0] * (N + 1)
d = []
for i in range(N):
if s[i] == ')':
R[i + 1] = 1
else:
L[i + 1] = 1
d.append(i)
for i in range(1, N + 1):
R[i] += R[i - 1]
L[i] += L[i - 1]
M = 200005
fact = [0] * M
fact[0] = 1
for i in range(1, M):
fact[i] = fact[i - 1] * i
fact[i] %= MOD
rfact = [0] * M
rfact[M - 1] = pow(fact[M - 1], MOD - 2, MOD)
for i in range(M - 2, -1, -1):
rfact[i] = rfact[i + 1] * (i + 1)
rfact[i] %= MOD
def comb(n, k):
if k < 0 or k > n:
return 0
return fact[n] * rfact[n - k] * rfact[k] % MOD
for i in d:
if s[i] == '(':
l = L[i + 1] - 1
r = R[N] - R[i + 1]
num += comb(l + r, l + 1)
num %= MOD
print(num)
```
| 90,603 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you probably know, Anton goes to school. One of the school subjects that Anton studies is Bracketology. On the Bracketology lessons students usually learn different sequences that consist of round brackets (characters "(" and ")" (without quotes)).
On the last lesson Anton learned about the regular simple bracket sequences (RSBS). A bracket sequence s of length n is an RSBS if the following conditions are met:
* It is not empty (that is n β 0).
* The length of the sequence is even.
* First <image> charactes of the sequence are equal to "(".
* Last <image> charactes of the sequence are equal to ")".
For example, the sequence "((()))" is an RSBS but the sequences "((())" and "(()())" are not RSBS.
Elena Ivanovna, Anton's teacher, gave him the following task as a homework. Given a bracket sequence s. Find the number of its distinct subsequences such that they are RSBS. Note that a subsequence of s is a string that can be obtained from s by deleting some of its elements. Two subsequences are considered distinct if distinct sets of positions are deleted.
Because the answer can be very big and Anton's teacher doesn't like big numbers, she asks Anton to find the answer modulo 109 + 7.
Anton thought of this task for a very long time, but he still doesn't know how to solve it. Help Anton to solve this task and write a program that finds the answer for it!
Input
The only line of the input contains a string s β the bracket sequence given in Anton's homework. The string consists only of characters "(" and ")" (without quotes). It's guaranteed that the string is not empty and its length doesn't exceed 200 000.
Output
Output one number β the answer for the task modulo 109 + 7.
Examples
Input
)(()()
Output
6
Input
()()()
Output
7
Input
)))
Output
0
Note
In the first sample the following subsequences are possible:
* If we delete characters at the positions 1 and 5 (numbering starts with one), we will get the subsequence "(())".
* If we delete characters at the positions 1, 2, 3 and 4, we will get the subsequence "()".
* If we delete characters at the positions 1, 2, 4 and 5, we will get the subsequence "()".
* If we delete characters at the positions 1, 2, 5 and 6, we will get the subsequence "()".
* If we delete characters at the positions 1, 3, 4 and 5, we will get the subsequence "()".
* If we delete characters at the positions 1, 3, 5 and 6, we will get the subsequence "()".
The rest of the subsequnces are not RSBS. So we got 6 distinct subsequences that are RSBS, so the answer is 6.
Submitted Solution:
```
mod = 10 ** 9 + 7
F, inv, iF = [1,1], [0, 1], [1, 1]
for i in range(2, 200005):
F.append(F[-1] * i % mod)
inv.append(inv[mod%i] * (mod - mod // i) % mod)
iF.append(iF[-1] * inv[-1] % mod)
def C(n, k):
if k < 0 or k > n:
return 0
return F[n] * iF[k] * iF[n - k]
s = input()
open, close = 0, s.count(')')
ans = 0
for c in s:
if c == '(':
open += 1
ans += C(close + open - 1, open)
else:
close -= 1
print(ans % mod)
```
Yes
| 90,604 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you probably know, Anton goes to school. One of the school subjects that Anton studies is Bracketology. On the Bracketology lessons students usually learn different sequences that consist of round brackets (characters "(" and ")" (without quotes)).
On the last lesson Anton learned about the regular simple bracket sequences (RSBS). A bracket sequence s of length n is an RSBS if the following conditions are met:
* It is not empty (that is n β 0).
* The length of the sequence is even.
* First <image> charactes of the sequence are equal to "(".
* Last <image> charactes of the sequence are equal to ")".
For example, the sequence "((()))" is an RSBS but the sequences "((())" and "(()())" are not RSBS.
Elena Ivanovna, Anton's teacher, gave him the following task as a homework. Given a bracket sequence s. Find the number of its distinct subsequences such that they are RSBS. Note that a subsequence of s is a string that can be obtained from s by deleting some of its elements. Two subsequences are considered distinct if distinct sets of positions are deleted.
Because the answer can be very big and Anton's teacher doesn't like big numbers, she asks Anton to find the answer modulo 109 + 7.
Anton thought of this task for a very long time, but he still doesn't know how to solve it. Help Anton to solve this task and write a program that finds the answer for it!
Input
The only line of the input contains a string s β the bracket sequence given in Anton's homework. The string consists only of characters "(" and ")" (without quotes). It's guaranteed that the string is not empty and its length doesn't exceed 200 000.
Output
Output one number β the answer for the task modulo 109 + 7.
Examples
Input
)(()()
Output
6
Input
()()()
Output
7
Input
)))
Output
0
Note
In the first sample the following subsequences are possible:
* If we delete characters at the positions 1 and 5 (numbering starts with one), we will get the subsequence "(())".
* If we delete characters at the positions 1, 2, 3 and 4, we will get the subsequence "()".
* If we delete characters at the positions 1, 2, 4 and 5, we will get the subsequence "()".
* If we delete characters at the positions 1, 2, 5 and 6, we will get the subsequence "()".
* If we delete characters at the positions 1, 3, 4 and 5, we will get the subsequence "()".
* If we delete characters at the positions 1, 3, 5 and 6, we will get the subsequence "()".
The rest of the subsequnces are not RSBS. So we got 6 distinct subsequences that are RSBS, so the answer is 6.
Submitted Solution:
```
mod = 10 ** 9 + 7
maxn = 200001
def C(n, k):
if k < 0 or k > n:
return 0
return fact[n] * invfact[k] * invfact[n - k] % mod
fact = [1, 1]
inv = [0, 1]
invfact = [1, 1]
for i in range(2, maxn):
fact.append(fact[-1] * i % mod)
inv.append(inv[mod % i] * (mod - mod // i) % mod)
invfact.append(invfact[-1] * inv[-1] % mod)
s = input()
op = 0
cl = s.count(')')
ans = 0
for x in s:
if x == '(':
op += 1
cur = C(cl + op - 1, op) % mod
ans += cur % mod
else:
cl -= 1
print(ans % mod)
```
Yes
| 90,605 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you probably know, Anton goes to school. One of the school subjects that Anton studies is Bracketology. On the Bracketology lessons students usually learn different sequences that consist of round brackets (characters "(" and ")" (without quotes)).
On the last lesson Anton learned about the regular simple bracket sequences (RSBS). A bracket sequence s of length n is an RSBS if the following conditions are met:
* It is not empty (that is n β 0).
* The length of the sequence is even.
* First <image> charactes of the sequence are equal to "(".
* Last <image> charactes of the sequence are equal to ")".
For example, the sequence "((()))" is an RSBS but the sequences "((())" and "(()())" are not RSBS.
Elena Ivanovna, Anton's teacher, gave him the following task as a homework. Given a bracket sequence s. Find the number of its distinct subsequences such that they are RSBS. Note that a subsequence of s is a string that can be obtained from s by deleting some of its elements. Two subsequences are considered distinct if distinct sets of positions are deleted.
Because the answer can be very big and Anton's teacher doesn't like big numbers, she asks Anton to find the answer modulo 109 + 7.
Anton thought of this task for a very long time, but he still doesn't know how to solve it. Help Anton to solve this task and write a program that finds the answer for it!
Input
The only line of the input contains a string s β the bracket sequence given in Anton's homework. The string consists only of characters "(" and ")" (without quotes). It's guaranteed that the string is not empty and its length doesn't exceed 200 000.
Output
Output one number β the answer for the task modulo 109 + 7.
Examples
Input
)(()()
Output
6
Input
()()()
Output
7
Input
)))
Output
0
Note
In the first sample the following subsequences are possible:
* If we delete characters at the positions 1 and 5 (numbering starts with one), we will get the subsequence "(())".
* If we delete characters at the positions 1, 2, 3 and 4, we will get the subsequence "()".
* If we delete characters at the positions 1, 2, 4 and 5, we will get the subsequence "()".
* If we delete characters at the positions 1, 2, 5 and 6, we will get the subsequence "()".
* If we delete characters at the positions 1, 3, 4 and 5, we will get the subsequence "()".
* If we delete characters at the positions 1, 3, 5 and 6, we will get the subsequence "()".
The rest of the subsequnces are not RSBS. So we got 6 distinct subsequences that are RSBS, so the answer is 6.
Submitted Solution:
```
from datetime import datetime
cache = {}
def pascal(n, k):
if k == 0 or k == n:
return 1
elif (n,k) in cache:
return cache[(n,k)]
else:
value = pascal(n-1, k-1) + pascal(n-1, k)
cache[(n,k)] = value
return value
sentence=input()
startTime = datetime.now()
s=0
x=0
for i in range(len(sentence)):
if sentence[i]=="(":
x+=1
cut=sentence[i+1:]
y=cut.count(")")
if y>=1:
m=min(x,y)
for a in range(m+1):
if a!=0:
add=int(pascal(y,a))*int(pascal(x-1,a-1))
s+=add
print(s)
print(datetime.now() - startTime)
```
No
| 90,606 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you probably know, Anton goes to school. One of the school subjects that Anton studies is Bracketology. On the Bracketology lessons students usually learn different sequences that consist of round brackets (characters "(" and ")" (without quotes)).
On the last lesson Anton learned about the regular simple bracket sequences (RSBS). A bracket sequence s of length n is an RSBS if the following conditions are met:
* It is not empty (that is n β 0).
* The length of the sequence is even.
* First <image> charactes of the sequence are equal to "(".
* Last <image> charactes of the sequence are equal to ")".
For example, the sequence "((()))" is an RSBS but the sequences "((())" and "(()())" are not RSBS.
Elena Ivanovna, Anton's teacher, gave him the following task as a homework. Given a bracket sequence s. Find the number of its distinct subsequences such that they are RSBS. Note that a subsequence of s is a string that can be obtained from s by deleting some of its elements. Two subsequences are considered distinct if distinct sets of positions are deleted.
Because the answer can be very big and Anton's teacher doesn't like big numbers, she asks Anton to find the answer modulo 109 + 7.
Anton thought of this task for a very long time, but he still doesn't know how to solve it. Help Anton to solve this task and write a program that finds the answer for it!
Input
The only line of the input contains a string s β the bracket sequence given in Anton's homework. The string consists only of characters "(" and ")" (without quotes). It's guaranteed that the string is not empty and its length doesn't exceed 200 000.
Output
Output one number β the answer for the task modulo 109 + 7.
Examples
Input
)(()()
Output
6
Input
()()()
Output
7
Input
)))
Output
0
Note
In the first sample the following subsequences are possible:
* If we delete characters at the positions 1 and 5 (numbering starts with one), we will get the subsequence "(())".
* If we delete characters at the positions 1, 2, 3 and 4, we will get the subsequence "()".
* If we delete characters at the positions 1, 2, 4 and 5, we will get the subsequence "()".
* If we delete characters at the positions 1, 2, 5 and 6, we will get the subsequence "()".
* If we delete characters at the positions 1, 3, 4 and 5, we will get the subsequence "()".
* If we delete characters at the positions 1, 3, 5 and 6, we will get the subsequence "()".
The rest of the subsequnces are not RSBS. So we got 6 distinct subsequences that are RSBS, so the answer is 6.
Submitted Solution:
```
p = 10**9+7
def power(x, y, p):
b = bin(y)[2:]
answer = 1
start = x % p
for i in range(len(b)):
if b[len(b)-1-i]=='1':
answer = (answer*start) % p
start = (start*start) % p
return answer
def process(S):
answer = 0
n = len(S)
left_count = 0
right_count = 0
p_lc = 1
p_rc = 1
p_num = 1
num = 0
for c in S:
if c==')':
right_count+=1
p_rc = (p_rc*(right_count+1)) % p
num+=1
p_num = (p_num*num) % p
for i in range(n):
if S[i]=='(':
left_count+=1
if left_count > 1:
p_lc = (p_lc*(left_count-1)) % p
num+=1
p_num = (p_num*num) % p
else:
p_rc = (p_rc*power(right_count+1, p-2, p)) % p
p_num = (p_num*power(num, p-2, p)) % p
right_count-=1
num-=1
#(left_count choose n, right_count choose n-1)
#(left_count choose n, right_count choose rc-n+1))
#(lc+rc choose rc+1)
if left_count > 0:
entry = (p_num*power(p_lc, p-2, p)*power(p_rc, p-2, p)) % p
answer = (answer+entry) % p
return answer
```
No
| 90,607 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you probably know, Anton goes to school. One of the school subjects that Anton studies is Bracketology. On the Bracketology lessons students usually learn different sequences that consist of round brackets (characters "(" and ")" (without quotes)).
On the last lesson Anton learned about the regular simple bracket sequences (RSBS). A bracket sequence s of length n is an RSBS if the following conditions are met:
* It is not empty (that is n β 0).
* The length of the sequence is even.
* First <image> charactes of the sequence are equal to "(".
* Last <image> charactes of the sequence are equal to ")".
For example, the sequence "((()))" is an RSBS but the sequences "((())" and "(()())" are not RSBS.
Elena Ivanovna, Anton's teacher, gave him the following task as a homework. Given a bracket sequence s. Find the number of its distinct subsequences such that they are RSBS. Note that a subsequence of s is a string that can be obtained from s by deleting some of its elements. Two subsequences are considered distinct if distinct sets of positions are deleted.
Because the answer can be very big and Anton's teacher doesn't like big numbers, she asks Anton to find the answer modulo 109 + 7.
Anton thought of this task for a very long time, but he still doesn't know how to solve it. Help Anton to solve this task and write a program that finds the answer for it!
Input
The only line of the input contains a string s β the bracket sequence given in Anton's homework. The string consists only of characters "(" and ")" (without quotes). It's guaranteed that the string is not empty and its length doesn't exceed 200 000.
Output
Output one number β the answer for the task modulo 109 + 7.
Examples
Input
)(()()
Output
6
Input
()()()
Output
7
Input
)))
Output
0
Note
In the first sample the following subsequences are possible:
* If we delete characters at the positions 1 and 5 (numbering starts with one), we will get the subsequence "(())".
* If we delete characters at the positions 1, 2, 3 and 4, we will get the subsequence "()".
* If we delete characters at the positions 1, 2, 4 and 5, we will get the subsequence "()".
* If we delete characters at the positions 1, 2, 5 and 6, we will get the subsequence "()".
* If we delete characters at the positions 1, 3, 4 and 5, we will get the subsequence "()".
* If we delete characters at the positions 1, 3, 5 and 6, we will get the subsequence "()".
The rest of the subsequnces are not RSBS. So we got 6 distinct subsequences that are RSBS, so the answer is 6.
Submitted Solution:
```
s = input()
MAXN = 200005
MOD = 1000000007
Fac = [ 1 for i in range(0, MAXN)]
for i in range(1, MAXN): Fac[i] = (i*Fac[i-1])%MOD
f = lambda n : Fac[n] if n >= 0 else 0
suf = [ 0 for i in range(0, len(s)+1)]
for i in range(len(s)-1, -1, -1):
suf[i] = suf[i+1] + 1 if s[i] is ')' else suf[i+1]
def ncr(n, k):
return 0 if (f(n-k)*f(k)) == 0 else (f(n)* pow(f(n-k)*f(k), MOD-2, MOD) ) % MOD
ans = 0
pre = 0
for i in range(0, len(s)):
if s[i] is not '(':
continue
pre += 1
ans += ncr(suf[i+1]+pre-1, pre)
print(ans % MOD)
```
No
| 90,608 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you probably know, Anton goes to school. One of the school subjects that Anton studies is Bracketology. On the Bracketology lessons students usually learn different sequences that consist of round brackets (characters "(" and ")" (without quotes)).
On the last lesson Anton learned about the regular simple bracket sequences (RSBS). A bracket sequence s of length n is an RSBS if the following conditions are met:
* It is not empty (that is n β 0).
* The length of the sequence is even.
* First <image> charactes of the sequence are equal to "(".
* Last <image> charactes of the sequence are equal to ")".
For example, the sequence "((()))" is an RSBS but the sequences "((())" and "(()())" are not RSBS.
Elena Ivanovna, Anton's teacher, gave him the following task as a homework. Given a bracket sequence s. Find the number of its distinct subsequences such that they are RSBS. Note that a subsequence of s is a string that can be obtained from s by deleting some of its elements. Two subsequences are considered distinct if distinct sets of positions are deleted.
Because the answer can be very big and Anton's teacher doesn't like big numbers, she asks Anton to find the answer modulo 109 + 7.
Anton thought of this task for a very long time, but he still doesn't know how to solve it. Help Anton to solve this task and write a program that finds the answer for it!
Input
The only line of the input contains a string s β the bracket sequence given in Anton's homework. The string consists only of characters "(" and ")" (without quotes). It's guaranteed that the string is not empty and its length doesn't exceed 200 000.
Output
Output one number β the answer for the task modulo 109 + 7.
Examples
Input
)(()()
Output
6
Input
()()()
Output
7
Input
)))
Output
0
Note
In the first sample the following subsequences are possible:
* If we delete characters at the positions 1 and 5 (numbering starts with one), we will get the subsequence "(())".
* If we delete characters at the positions 1, 2, 3 and 4, we will get the subsequence "()".
* If we delete characters at the positions 1, 2, 4 and 5, we will get the subsequence "()".
* If we delete characters at the positions 1, 2, 5 and 6, we will get the subsequence "()".
* If we delete characters at the positions 1, 3, 4 and 5, we will get the subsequence "()".
* If we delete characters at the positions 1, 3, 5 and 6, we will get the subsequence "()".
The rest of the subsequnces are not RSBS. So we got 6 distinct subsequences that are RSBS, so the answer is 6.
Submitted Solution:
```
#!/usr/bin/env python3
def ri():
return map(int, input().split())
def prefac(n):
fac[0] = 1
for i in range(1,n):
fac[i] = fac[i-1]*i%(10**9+7)
s = input()
n = len(s)
o = [0 for i in range(len(s))]
c = [0 for i in range(len(s))]
fac = [0 for i in range(n)]
prefac(n)
if s[0] == '(':
o[0] = 1
for i in range(1,n):
if s[i] == '(':
o[i] = o[i-1] + 1
else:
o[i] = o[i-1]
if s[n-1] == ')':
c[n-1] = 1
for i in range(n-2, -1, -1):
if s[i] == ')':
c[i] = c[i+1] + 1
else:
c[i] = c[i+1]
ans = 0
for i in range(n):
if s[i] == '(':
a = o[i]
b = c[i]
if a != 0 and b != 0:
ans += fac[a+b-1]//fac[a]//fac[b-1]
ans %= 10**9+7
print(ans)
```
No
| 90,609 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya and Petya take part in a Codeforces round. The round lasts for two hours and contains five problems.
For this round the dynamic problem scoring is used. If you were lucky not to participate in any Codeforces round with dynamic problem scoring, here is what it means. The maximum point value of the problem depends on the ratio of the number of participants who solved the problem to the total number of round participants. Everyone who made at least one submission is considered to be participating in the round.
<image>
Pay attention to the range bounds. For example, if 40 people are taking part in the round, and 10 of them solve a particular problem, then the solvers fraction is equal to 1 / 4, and the problem's maximum point value is equal to 1500.
If the problem's maximum point value is equal to x, then for each whole minute passed from the beginning of the contest to the moment of the participant's correct submission, the participant loses x / 250 points. For example, if the problem's maximum point value is 2000, and the participant submits a correct solution to it 40 minutes into the round, this participant will be awarded with 2000Β·(1 - 40 / 250) = 1680 points for this problem.
There are n participants in the round, including Vasya and Petya. For each participant and each problem, the number of minutes which passed between the beginning of the contest and the submission of this participant to this problem is known. It's also possible that this participant made no submissions to this problem.
With two seconds until the end of the round, all participants' submissions have passed pretests, and not a single hack attempt has been made. Vasya believes that no more submissions or hack attempts will be made in the remaining two seconds, and every submission will pass the system testing.
Unfortunately, Vasya is a cheater. He has registered 109 + 7 new accounts for the round. Now Vasya can submit any of his solutions from these new accounts in order to change the maximum point values of the problems. Vasya can also submit any wrong solutions to any problems. Note that Vasya can not submit correct solutions to the problems he hasn't solved.
Vasya seeks to score strictly more points than Petya in the current round. Vasya has already prepared the scripts which allow to obfuscate his solutions and submit them into the system from any of the new accounts in just fractions of seconds. However, Vasya doesn't want to make his cheating too obvious, so he wants to achieve his goal while making submissions from the smallest possible number of new accounts.
Find the smallest number of new accounts Vasya needs in order to beat Petya (provided that Vasya's assumptions are correct), or report that Vasya can't achieve his goal.
Input
The first line contains a single integer n (2 β€ n β€ 120) β the number of round participants, including Vasya and Petya.
Each of the next n lines contains five integers ai, 1, ai, 2..., ai, 5 ( - 1 β€ ai, j β€ 119) β the number of minutes passed between the beginning of the round and the submission of problem j by participant i, or -1 if participant i hasn't solved problem j.
It is guaranteed that each participant has made at least one successful submission.
Vasya is listed as participant number 1, Petya is listed as participant number 2, all the other participants are listed in no particular order.
Output
Output a single integer β the number of new accounts Vasya needs to beat Petya, or -1 if Vasya can't achieve his goal.
Examples
Input
2
5 15 40 70 115
50 45 40 30 15
Output
2
Input
3
55 80 10 -1 -1
15 -1 79 60 -1
42 -1 13 -1 -1
Output
3
Input
5
119 119 119 119 119
0 0 0 0 -1
20 65 12 73 77
78 112 22 23 11
1 78 60 111 62
Output
27
Input
4
-1 20 40 77 119
30 10 73 50 107
21 29 -1 64 98
117 65 -1 -1 -1
Output
-1
Note
In the first example, Vasya's optimal strategy is to submit the solutions to the last three problems from two new accounts. In this case the first two problems will have the maximum point value of 1000, while the last three problems will have the maximum point value of 500. Vasya's score will be equal to 980 + 940 + 420 + 360 + 270 = 2970 points, while Petya will score just 800 + 820 + 420 + 440 + 470 = 2950 points.
In the second example, Vasya has to make a single unsuccessful submission to any problem from two new accounts, and a single successful submission to the first problem from the third new account. In this case, the maximum point values of the problems will be equal to 500, 1500, 1000, 1500, 3000. Vasya will score 2370 points, while Petya will score just 2294 points.
In the third example, Vasya can achieve his goal by submitting the solutions to the first four problems from 27 new accounts. The maximum point values of the problems will be equal to 500, 500, 500, 500, 2000. Thanks to the high cost of the fifth problem, Vasya will manage to beat Petya who solved the first four problems very quickly, but couldn't solve the fifth one.
Tags: brute force, greedy
Correct Solution:
```
n = int(input())
t = [0] * n
for i in range(n):
t[i] = list(map(int, input().split()))
value = [0] * 5
solve = [0] * 5
for i in range(5):
solved = 0
for j in range(n):
if t[j][i] != -1:
solved += 1
else:
t[j][i] = 250
if (solved * 2 > n):
value[i] = 500
elif (solved * 4 > n):
value[i] = 1000
elif (solved * 8 > n):
value[i] = 1500
elif (solved * 16 > n):
value[i] = 2000
elif (solved * 32 > n):
value[i] = 2500
else:
value[i] = 3000
solve[i] = solved
vasya = 0
for i in range(5):
vasya += value[i] - value[i] // 250 * t[0][i]
petya = 0
for i in range(5):
petya += value[i] - value[i] // 250 * t[1][i]
pot = [[0] * 20000 for i in range(5)]
for problem in range(5):
if t[0][problem] < t[1][problem]:
cur_ac = 0
while 1:
if solve[problem] * 32 <= n + cur_ac:
win = 3000 - 3000 // 250 * t[0][problem] - (3000 - 3000 // 250 * t[1][problem]) - (value[problem] - value[problem] // 250 * t[0][problem] -(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 16 <= n + cur_ac:
win = 2500 - 2500 // 250 * t[0][problem] - (2500 - 2500 // 250 * t[1][problem]) - (value[problem] - value[problem] // 250 * t[0][problem] -(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 8 <= n + cur_ac:
win = 2000 - 2000 // 250 * t[0][problem] - (2000 - 2000 // 250 * t[1][problem]) - (value[problem] - value[problem] // 250 * t[0][problem] -(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 4 <= n + cur_ac:
win = 1500 - 1500 // 250 * t[0][problem] - (1500 - 1500 // 250 * t[1][problem]) - (value[problem] - value[problem] // 250 * t[0][problem] -(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 2 <= n + cur_ac:
win = 1000 - 1000 // 250 * t[0][problem] - (1000 - 1000 // 250 * t[1][problem]) - (value[problem] - value[problem] // 250 * t[0][problem] -(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
else:
win = 500 - 500 // 250 * t[0][problem] - (500 - 500 // 250 * t[1][problem]) - (value[problem] - value[problem] // 250 * t[0][problem] -(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
cur_ac += 1
if cur_ac == 20000:
break
elif t[0][problem] != 250:
cur_ac = 0
while 1:
if solve[problem] * 32 + cur_ac <= n + cur_ac:
win = 3000 - 3000 // 250 * t[0][problem] - (3000 - 3000 // 250 * t[1][problem]) - (
value[problem] - value[problem] // 250 * t[0][problem] - (
value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 16 + cur_ac <= n + cur_ac:
win = 2500 - 2500 // 250 * t[0][problem] - (2500 - 2500 // 250 * t[1][problem]) - (
value[problem] - value[problem] // 250 * t[0][problem] - (
value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 8 + cur_ac <= n + cur_ac:
win = 2000 - 2000 // 250 * t[0][problem] - (2000 - 2000 // 250 * t[1][problem]) - (
value[problem] - value[problem] // 250 * t[0][problem] - (
value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 4 + cur_ac <= n + cur_ac:
win = 1500 - 1500 // 250 * t[0][problem] - (1500 - 1500 // 250 * t[1][problem]) - (
value[problem] - value[problem] // 250 * t[0][problem] - (
value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 2 + cur_ac <= n + cur_ac:
win = 1000 - 1000 // 250 * t[0][problem] - (1000 - 1000 // 250 * t[1][problem]) - (
value[problem] - value[problem] // 250 * t[0][problem] - (
value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
else:
win = 500 - 500 // 250 * t[0][problem] - (500 - 500 // 250 * t[1][problem]) - (
value[problem] - value[problem] // 250 * t[0][problem] - (
value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
cur_ac += 1
if cur_ac == 20000:
break
else:
cur_ac = 0
while 1:
if solve[problem] * 32 <= n + cur_ac:
win = -(3000 - 3000 // 250 * t[1][problem]) -(
-(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 16 <= n + cur_ac:
win = -(2500 - 2500 // 250 * t[1][problem]) - (
-(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 8 <= n + cur_ac:
win = -(2000 - 2000 // 250 * t[1][problem]) - (
-(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 4 <= n + cur_ac:
win = -(1500 - 1500 // 250 * t[1][problem]) - (
-(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 2 <= n + cur_ac:
win = -(1000 - 1000 // 250 * t[1][problem]) - (
-(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
else:
win = -(500 - 500 // 250 * t[1][problem]) - (
-(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
cur_ac += 1
if cur_ac == 20000:
break
res = -1
maxi = 0
for i in range(5):
maxi = max(maxi, len(pot[i]))
for i in range(maxi):
tmp = 0
for pr in range(5):
if len(pot[pr]) <= i:
tmp += pot[pr][-1]
else:
tmp += pot[pr][i]
if tmp > petya - vasya:
res = i
break
print(res)
```
| 90,610 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya and Petya take part in a Codeforces round. The round lasts for two hours and contains five problems.
For this round the dynamic problem scoring is used. If you were lucky not to participate in any Codeforces round with dynamic problem scoring, here is what it means. The maximum point value of the problem depends on the ratio of the number of participants who solved the problem to the total number of round participants. Everyone who made at least one submission is considered to be participating in the round.
<image>
Pay attention to the range bounds. For example, if 40 people are taking part in the round, and 10 of them solve a particular problem, then the solvers fraction is equal to 1 / 4, and the problem's maximum point value is equal to 1500.
If the problem's maximum point value is equal to x, then for each whole minute passed from the beginning of the contest to the moment of the participant's correct submission, the participant loses x / 250 points. For example, if the problem's maximum point value is 2000, and the participant submits a correct solution to it 40 minutes into the round, this participant will be awarded with 2000Β·(1 - 40 / 250) = 1680 points for this problem.
There are n participants in the round, including Vasya and Petya. For each participant and each problem, the number of minutes which passed between the beginning of the contest and the submission of this participant to this problem is known. It's also possible that this participant made no submissions to this problem.
With two seconds until the end of the round, all participants' submissions have passed pretests, and not a single hack attempt has been made. Vasya believes that no more submissions or hack attempts will be made in the remaining two seconds, and every submission will pass the system testing.
Unfortunately, Vasya is a cheater. He has registered 109 + 7 new accounts for the round. Now Vasya can submit any of his solutions from these new accounts in order to change the maximum point values of the problems. Vasya can also submit any wrong solutions to any problems. Note that Vasya can not submit correct solutions to the problems he hasn't solved.
Vasya seeks to score strictly more points than Petya in the current round. Vasya has already prepared the scripts which allow to obfuscate his solutions and submit them into the system from any of the new accounts in just fractions of seconds. However, Vasya doesn't want to make his cheating too obvious, so he wants to achieve his goal while making submissions from the smallest possible number of new accounts.
Find the smallest number of new accounts Vasya needs in order to beat Petya (provided that Vasya's assumptions are correct), or report that Vasya can't achieve his goal.
Input
The first line contains a single integer n (2 β€ n β€ 120) β the number of round participants, including Vasya and Petya.
Each of the next n lines contains five integers ai, 1, ai, 2..., ai, 5 ( - 1 β€ ai, j β€ 119) β the number of minutes passed between the beginning of the round and the submission of problem j by participant i, or -1 if participant i hasn't solved problem j.
It is guaranteed that each participant has made at least one successful submission.
Vasya is listed as participant number 1, Petya is listed as participant number 2, all the other participants are listed in no particular order.
Output
Output a single integer β the number of new accounts Vasya needs to beat Petya, or -1 if Vasya can't achieve his goal.
Examples
Input
2
5 15 40 70 115
50 45 40 30 15
Output
2
Input
3
55 80 10 -1 -1
15 -1 79 60 -1
42 -1 13 -1 -1
Output
3
Input
5
119 119 119 119 119
0 0 0 0 -1
20 65 12 73 77
78 112 22 23 11
1 78 60 111 62
Output
27
Input
4
-1 20 40 77 119
30 10 73 50 107
21 29 -1 64 98
117 65 -1 -1 -1
Output
-1
Note
In the first example, Vasya's optimal strategy is to submit the solutions to the last three problems from two new accounts. In this case the first two problems will have the maximum point value of 1000, while the last three problems will have the maximum point value of 500. Vasya's score will be equal to 980 + 940 + 420 + 360 + 270 = 2970 points, while Petya will score just 800 + 820 + 420 + 440 + 470 = 2950 points.
In the second example, Vasya has to make a single unsuccessful submission to any problem from two new accounts, and a single successful submission to the first problem from the third new account. In this case, the maximum point values of the problems will be equal to 500, 1500, 1000, 1500, 3000. Vasya will score 2370 points, while Petya will score just 2294 points.
In the third example, Vasya can achieve his goal by submitting the solutions to the first four problems from 27 new accounts. The maximum point values of the problems will be equal to 500, 500, 500, 500, 2000. Thanks to the high cost of the fifth problem, Vasya will manage to beat Petya who solved the first four problems very quickly, but couldn't solve the fifth one.
Tags: brute force, greedy
Correct Solution:
```
from sys import stdin, stdout
fractionBoundaries = [1/2,1/4,1/8,1/16,1/32,0]
scoreList = [500,1000,1500,2000,2500,3000]
def assessScores(n,Vsol,Psol,solvers):
vScore=0
pScore=0
for i in range(5):
for j in range(6):
if solvers[i]/n>fractionBoundaries[j]:
qnScore=scoreList[j]
break
if Vsol[i]>=0:
vScore+=qnScore-(qnScore*Vsol[i]//250)
if Psol[i]>=0:
pScore+=qnScore-(qnScore*Psol[i]//250)
return vScore>pScore
def main():
n = int(stdin.readline().rstrip())
Vsol = [int(x) for x in stdin.readline().rstrip().split()]
Psol = [int(x) for x in stdin.readline().rstrip().split()]
solvers = [0]*5
for _ in range(n-2):
a = [int(x) for x in stdin.readline().rstrip().split()]
for i in range(5):
if a[i]>=0:
solvers[i]+=1
for i in range(5):
if Vsol[i]>=0:
solvers[i]+=1
if Psol[i]>=0:
solvers[i]+=1
newSolvers=0
vWins=[]
pWins=[]
for i in range(5):
if (Vsol[i]<Psol[i] and Vsol[i]>=0) or (Vsol[i]>=0 and Psol[i]<0):
vWins.append(i)
elif (Psol[i]<Vsol[i] and Psol[i]>=0 and Vsol[i]>=0):
pWins.append(i)
if len(vWins)==0:
print(-1)
else:
while not assessScores(n+newSolvers,Vsol,Psol,solvers) and newSolvers<=100000007:
solversNeeded=9999999999999
for i in range(5):
if i in vWins:
currentRatio = solvers[i]/(newSolvers+n)
for j in range(6):
if solvers[i]/(newSolvers+n)>fractionBoundaries[j]:
nextBoundary = fractionBoundaries[j]
break
if nextBoundary!=0:
if solvers[i]%nextBoundary==0:
solversNeeded = min([solvers[i]//nextBoundary - (newSolvers+n),solversNeeded])
else:
solversNeeded = min([solvers[i]//nextBoundary - (newSolvers+n)+1,solversNeeded])
elif i in pWins and Vsol[i]>0:
currentRatio = solvers[i]/(newSolvers+n)
for j in range(6):
if solvers[i]/(newSolvers+n)>fractionBoundaries[j]:
if j>0:
nextBoundary = fractionBoundaries[j-1]
solversNeeded = min([(nextBoundary*(newSolvers+n)-solvers[i])//(1-nextBoundary)+1,solversNeeded])
break
newSolvers+=solversNeeded
for x in pWins:
solvers[x]+=solversNeeded
if newSolvers>1000000007:
print(-1)
else:
print(int(newSolvers))
main()
```
| 90,611 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya and Petya take part in a Codeforces round. The round lasts for two hours and contains five problems.
For this round the dynamic problem scoring is used. If you were lucky not to participate in any Codeforces round with dynamic problem scoring, here is what it means. The maximum point value of the problem depends on the ratio of the number of participants who solved the problem to the total number of round participants. Everyone who made at least one submission is considered to be participating in the round.
<image>
Pay attention to the range bounds. For example, if 40 people are taking part in the round, and 10 of them solve a particular problem, then the solvers fraction is equal to 1 / 4, and the problem's maximum point value is equal to 1500.
If the problem's maximum point value is equal to x, then for each whole minute passed from the beginning of the contest to the moment of the participant's correct submission, the participant loses x / 250 points. For example, if the problem's maximum point value is 2000, and the participant submits a correct solution to it 40 minutes into the round, this participant will be awarded with 2000Β·(1 - 40 / 250) = 1680 points for this problem.
There are n participants in the round, including Vasya and Petya. For each participant and each problem, the number of minutes which passed between the beginning of the contest and the submission of this participant to this problem is known. It's also possible that this participant made no submissions to this problem.
With two seconds until the end of the round, all participants' submissions have passed pretests, and not a single hack attempt has been made. Vasya believes that no more submissions or hack attempts will be made in the remaining two seconds, and every submission will pass the system testing.
Unfortunately, Vasya is a cheater. He has registered 109 + 7 new accounts for the round. Now Vasya can submit any of his solutions from these new accounts in order to change the maximum point values of the problems. Vasya can also submit any wrong solutions to any problems. Note that Vasya can not submit correct solutions to the problems he hasn't solved.
Vasya seeks to score strictly more points than Petya in the current round. Vasya has already prepared the scripts which allow to obfuscate his solutions and submit them into the system from any of the new accounts in just fractions of seconds. However, Vasya doesn't want to make his cheating too obvious, so he wants to achieve his goal while making submissions from the smallest possible number of new accounts.
Find the smallest number of new accounts Vasya needs in order to beat Petya (provided that Vasya's assumptions are correct), or report that Vasya can't achieve his goal.
Input
The first line contains a single integer n (2 β€ n β€ 120) β the number of round participants, including Vasya and Petya.
Each of the next n lines contains five integers ai, 1, ai, 2..., ai, 5 ( - 1 β€ ai, j β€ 119) β the number of minutes passed between the beginning of the round and the submission of problem j by participant i, or -1 if participant i hasn't solved problem j.
It is guaranteed that each participant has made at least one successful submission.
Vasya is listed as participant number 1, Petya is listed as participant number 2, all the other participants are listed in no particular order.
Output
Output a single integer β the number of new accounts Vasya needs to beat Petya, or -1 if Vasya can't achieve his goal.
Examples
Input
2
5 15 40 70 115
50 45 40 30 15
Output
2
Input
3
55 80 10 -1 -1
15 -1 79 60 -1
42 -1 13 -1 -1
Output
3
Input
5
119 119 119 119 119
0 0 0 0 -1
20 65 12 73 77
78 112 22 23 11
1 78 60 111 62
Output
27
Input
4
-1 20 40 77 119
30 10 73 50 107
21 29 -1 64 98
117 65 -1 -1 -1
Output
-1
Note
In the first example, Vasya's optimal strategy is to submit the solutions to the last three problems from two new accounts. In this case the first two problems will have the maximum point value of 1000, while the last three problems will have the maximum point value of 500. Vasya's score will be equal to 980 + 940 + 420 + 360 + 270 = 2970 points, while Petya will score just 800 + 820 + 420 + 440 + 470 = 2950 points.
In the second example, Vasya has to make a single unsuccessful submission to any problem from two new accounts, and a single successful submission to the first problem from the third new account. In this case, the maximum point values of the problems will be equal to 500, 1500, 1000, 1500, 3000. Vasya will score 2370 points, while Petya will score just 2294 points.
In the third example, Vasya can achieve his goal by submitting the solutions to the first four problems from 27 new accounts. The maximum point values of the problems will be equal to 500, 500, 500, 500, 2000. Thanks to the high cost of the fifth problem, Vasya will manage to beat Petya who solved the first four problems very quickly, but couldn't solve the fifth one.
Tags: brute force, greedy
Correct Solution:
```
import sys
inf = 10**9 + 7
def solve():
n = int(sys.stdin.readline())
v = [int(vi) for vi in sys.stdin.readline().split()] # Vesya
p = [int(pi) for pi in sys.stdin.readline().split()] # Petya
cnt = [0]*5
for i in range(5):
if v[i] != -1:
cnt[i] += 1
if p[i] != -1:
cnt[i] += 1
for i in range(n - 2):
a = [int(ai) for ai in sys.stdin.readline().split()]
for j in range(5):
if a[j] != -1:
cnt[j] += 1
for i in range(4000):
if check(n, v, p, cnt, i):
print(i)
return
print(-1)
def check(n, v, p, cnt, m):
tot = n + m
solved = cnt[:]
dif = 0
for i in range(5):
if p[i] != -1 and v[i] > p[i]:
solved[i] += m
for i in range(5):
if solved[i]*2 > tot:
max_score = 500
elif solved[i]*4 > tot:
max_score = 1000
elif solved[i]*8 > tot:
max_score = 1500
elif solved[i]*16 > tot:
max_score = 2000
elif solved[i]*32 > tot:
max_score = 2500
else:
max_score = 3000
if v[i] == p[i] == -1:
pass
elif v[i] == -1:
dif -= max_score * (250 - p[i]) // 250
elif p[i] == -1:
dif += max_score * (250 - v[i]) // 250
else:
dif += max_score * (p[i] - v[i]) // 250
return dif > 0
if __name__ == '__main__':
solve()
```
| 90,612 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya and Petya take part in a Codeforces round. The round lasts for two hours and contains five problems.
For this round the dynamic problem scoring is used. If you were lucky not to participate in any Codeforces round with dynamic problem scoring, here is what it means. The maximum point value of the problem depends on the ratio of the number of participants who solved the problem to the total number of round participants. Everyone who made at least one submission is considered to be participating in the round.
<image>
Pay attention to the range bounds. For example, if 40 people are taking part in the round, and 10 of them solve a particular problem, then the solvers fraction is equal to 1 / 4, and the problem's maximum point value is equal to 1500.
If the problem's maximum point value is equal to x, then for each whole minute passed from the beginning of the contest to the moment of the participant's correct submission, the participant loses x / 250 points. For example, if the problem's maximum point value is 2000, and the participant submits a correct solution to it 40 minutes into the round, this participant will be awarded with 2000Β·(1 - 40 / 250) = 1680 points for this problem.
There are n participants in the round, including Vasya and Petya. For each participant and each problem, the number of minutes which passed between the beginning of the contest and the submission of this participant to this problem is known. It's also possible that this participant made no submissions to this problem.
With two seconds until the end of the round, all participants' submissions have passed pretests, and not a single hack attempt has been made. Vasya believes that no more submissions or hack attempts will be made in the remaining two seconds, and every submission will pass the system testing.
Unfortunately, Vasya is a cheater. He has registered 109 + 7 new accounts for the round. Now Vasya can submit any of his solutions from these new accounts in order to change the maximum point values of the problems. Vasya can also submit any wrong solutions to any problems. Note that Vasya can not submit correct solutions to the problems he hasn't solved.
Vasya seeks to score strictly more points than Petya in the current round. Vasya has already prepared the scripts which allow to obfuscate his solutions and submit them into the system from any of the new accounts in just fractions of seconds. However, Vasya doesn't want to make his cheating too obvious, so he wants to achieve his goal while making submissions from the smallest possible number of new accounts.
Find the smallest number of new accounts Vasya needs in order to beat Petya (provided that Vasya's assumptions are correct), or report that Vasya can't achieve his goal.
Input
The first line contains a single integer n (2 β€ n β€ 120) β the number of round participants, including Vasya and Petya.
Each of the next n lines contains five integers ai, 1, ai, 2..., ai, 5 ( - 1 β€ ai, j β€ 119) β the number of minutes passed between the beginning of the round and the submission of problem j by participant i, or -1 if participant i hasn't solved problem j.
It is guaranteed that each participant has made at least one successful submission.
Vasya is listed as participant number 1, Petya is listed as participant number 2, all the other participants are listed in no particular order.
Output
Output a single integer β the number of new accounts Vasya needs to beat Petya, or -1 if Vasya can't achieve his goal.
Examples
Input
2
5 15 40 70 115
50 45 40 30 15
Output
2
Input
3
55 80 10 -1 -1
15 -1 79 60 -1
42 -1 13 -1 -1
Output
3
Input
5
119 119 119 119 119
0 0 0 0 -1
20 65 12 73 77
78 112 22 23 11
1 78 60 111 62
Output
27
Input
4
-1 20 40 77 119
30 10 73 50 107
21 29 -1 64 98
117 65 -1 -1 -1
Output
-1
Note
In the first example, Vasya's optimal strategy is to submit the solutions to the last three problems from two new accounts. In this case the first two problems will have the maximum point value of 1000, while the last three problems will have the maximum point value of 500. Vasya's score will be equal to 980 + 940 + 420 + 360 + 270 = 2970 points, while Petya will score just 800 + 820 + 420 + 440 + 470 = 2950 points.
In the second example, Vasya has to make a single unsuccessful submission to any problem from two new accounts, and a single successful submission to the first problem from the third new account. In this case, the maximum point values of the problems will be equal to 500, 1500, 1000, 1500, 3000. Vasya will score 2370 points, while Petya will score just 2294 points.
In the third example, Vasya can achieve his goal by submitting the solutions to the first four problems from 27 new accounts. The maximum point values of the problems will be equal to 500, 500, 500, 500, 2000. Thanks to the high cost of the fifth problem, Vasya will manage to beat Petya who solved the first four problems very quickly, but couldn't solve the fifth one.
Tags: brute force, greedy
Correct Solution:
```
def get_value(parters, solved):
if solved * 32 <= parters:
return 3000
if solved * 16 <= parters:
return 2500
if solved * 8 <= parters:
return 2000
if solved * 4 <= parters:
return 1500
if solved * 2 <= parters:
return 1000
return 500
def points(value, time):
return value - value // 500 * time
n = int(input())
t = [0] * n
for i in range(n):
t[i] = list(map(int, input().split()))
value = [0] * 5
solve = [0] * 5
for i in range(5):
solved = 0
for j in range(n):
if t[j][i] != -1:
solved += 1
else:
t[j][i] = 250
value[i] = get_value(n, solved)
solve[i] = solved
vasya = 0
for i in range(5):
vasya += points(value[i], t[0][i])
petya = 0
for i in range(5):
petya += points(value[i], t[1][i])
pot = [[0] * 20000 for i in range(5)]
for problem in range(5):
for cur_ac in range(20000):
new_value = 0
if t[0][problem] < t[1][problem]:
new_value = get_value(n + cur_ac, solve[problem])
elif t[0][problem] != 250:
new_value = get_value(n + cur_ac, solve[problem] + cur_ac)
else:
new_value = get_value(n + cur_ac, solve[problem])
win = points(new_value, t[0][problem]) - points(new_value, t[1][problem]) - points(value[problem], t[0][problem]) + points(value[problem], t[1][problem])
pot[problem][cur_ac] = win
res = -1
for i in range(20000):
tmp = 0
for pr in range(5):
if len(pot[pr]) <= i:
tmp += pot[pr][-1]
else:
tmp += pot[pr][i]
if tmp > petya - vasya:
res = i
break
print(res)
```
| 90,613 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya and Petya take part in a Codeforces round. The round lasts for two hours and contains five problems.
For this round the dynamic problem scoring is used. If you were lucky not to participate in any Codeforces round with dynamic problem scoring, here is what it means. The maximum point value of the problem depends on the ratio of the number of participants who solved the problem to the total number of round participants. Everyone who made at least one submission is considered to be participating in the round.
<image>
Pay attention to the range bounds. For example, if 40 people are taking part in the round, and 10 of them solve a particular problem, then the solvers fraction is equal to 1 / 4, and the problem's maximum point value is equal to 1500.
If the problem's maximum point value is equal to x, then for each whole minute passed from the beginning of the contest to the moment of the participant's correct submission, the participant loses x / 250 points. For example, if the problem's maximum point value is 2000, and the participant submits a correct solution to it 40 minutes into the round, this participant will be awarded with 2000Β·(1 - 40 / 250) = 1680 points for this problem.
There are n participants in the round, including Vasya and Petya. For each participant and each problem, the number of minutes which passed between the beginning of the contest and the submission of this participant to this problem is known. It's also possible that this participant made no submissions to this problem.
With two seconds until the end of the round, all participants' submissions have passed pretests, and not a single hack attempt has been made. Vasya believes that no more submissions or hack attempts will be made in the remaining two seconds, and every submission will pass the system testing.
Unfortunately, Vasya is a cheater. He has registered 109 + 7 new accounts for the round. Now Vasya can submit any of his solutions from these new accounts in order to change the maximum point values of the problems. Vasya can also submit any wrong solutions to any problems. Note that Vasya can not submit correct solutions to the problems he hasn't solved.
Vasya seeks to score strictly more points than Petya in the current round. Vasya has already prepared the scripts which allow to obfuscate his solutions and submit them into the system from any of the new accounts in just fractions of seconds. However, Vasya doesn't want to make his cheating too obvious, so he wants to achieve his goal while making submissions from the smallest possible number of new accounts.
Find the smallest number of new accounts Vasya needs in order to beat Petya (provided that Vasya's assumptions are correct), or report that Vasya can't achieve his goal.
Input
The first line contains a single integer n (2 β€ n β€ 120) β the number of round participants, including Vasya and Petya.
Each of the next n lines contains five integers ai, 1, ai, 2..., ai, 5 ( - 1 β€ ai, j β€ 119) β the number of minutes passed between the beginning of the round and the submission of problem j by participant i, or -1 if participant i hasn't solved problem j.
It is guaranteed that each participant has made at least one successful submission.
Vasya is listed as participant number 1, Petya is listed as participant number 2, all the other participants are listed in no particular order.
Output
Output a single integer β the number of new accounts Vasya needs to beat Petya, or -1 if Vasya can't achieve his goal.
Examples
Input
2
5 15 40 70 115
50 45 40 30 15
Output
2
Input
3
55 80 10 -1 -1
15 -1 79 60 -1
42 -1 13 -1 -1
Output
3
Input
5
119 119 119 119 119
0 0 0 0 -1
20 65 12 73 77
78 112 22 23 11
1 78 60 111 62
Output
27
Input
4
-1 20 40 77 119
30 10 73 50 107
21 29 -1 64 98
117 65 -1 -1 -1
Output
-1
Note
In the first example, Vasya's optimal strategy is to submit the solutions to the last three problems from two new accounts. In this case the first two problems will have the maximum point value of 1000, while the last three problems will have the maximum point value of 500. Vasya's score will be equal to 980 + 940 + 420 + 360 + 270 = 2970 points, while Petya will score just 800 + 820 + 420 + 440 + 470 = 2950 points.
In the second example, Vasya has to make a single unsuccessful submission to any problem from two new accounts, and a single successful submission to the first problem from the third new account. In this case, the maximum point values of the problems will be equal to 500, 1500, 1000, 1500, 3000. Vasya will score 2370 points, while Petya will score just 2294 points.
In the third example, Vasya can achieve his goal by submitting the solutions to the first four problems from 27 new accounts. The maximum point values of the problems will be equal to 500, 500, 500, 500, 2000. Thanks to the high cost of the fifth problem, Vasya will manage to beat Petya who solved the first four problems very quickly, but couldn't solve the fifth one.
Tags: brute force, greedy
Correct Solution:
```
import sys
inf = 10**9 + 7
def solve():
def check(mid):
tot = n + mid
dif = 0
solved = cnt[:]
for i in range(5):
if v[i] != -1 and p[i] != -1 and p[i] < v[i]:
solved[i] += mid
for i in range(5):
if solved[i]*2 > tot:
max_score = 500
elif solved[i]*4 > tot:
max_score = 1000
elif solved[i]*8 > tot:
max_score = 1500
elif solved[i]*16 > tot:
max_score = 2000
elif solved[i]*32 > tot:
max_score = 2500
else:
max_score = 3000
if v[i] == p[i]:
pass
elif v[i] == -1:
dif += max_score * (250 - p[i]) // 250
elif p[i] == -1:
dif += -max_score * (250 - v[i]) // 250
else:
dif += max_score * (-p[i] + v[i]) // 250
# print(mid, dif)
return dif < 0
n = int(sys.stdin.readline())
cnt = [0]*5
v = [int(i) for i in sys.stdin.readline().split()]
for i in range(5):
if v[i] != -1:
cnt[i] += 1
p = [int(i) for i in sys.stdin.readline().split()]
for i in range(5):
if p[i] != -1:
cnt[i] += 1
for i in range(n - 2):
a = [int(ai) for ai in sys.stdin.readline().split()]
for j in range(5):
if a[j] != -1:
cnt[j] += 1
for i in range(6000):
if check(i):
print(i)
return
print(-1)
if __name__ == '__main__':
solve()
```
| 90,614 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya and Petya take part in a Codeforces round. The round lasts for two hours and contains five problems.
For this round the dynamic problem scoring is used. If you were lucky not to participate in any Codeforces round with dynamic problem scoring, here is what it means. The maximum point value of the problem depends on the ratio of the number of participants who solved the problem to the total number of round participants. Everyone who made at least one submission is considered to be participating in the round.
<image>
Pay attention to the range bounds. For example, if 40 people are taking part in the round, and 10 of them solve a particular problem, then the solvers fraction is equal to 1 / 4, and the problem's maximum point value is equal to 1500.
If the problem's maximum point value is equal to x, then for each whole minute passed from the beginning of the contest to the moment of the participant's correct submission, the participant loses x / 250 points. For example, if the problem's maximum point value is 2000, and the participant submits a correct solution to it 40 minutes into the round, this participant will be awarded with 2000Β·(1 - 40 / 250) = 1680 points for this problem.
There are n participants in the round, including Vasya and Petya. For each participant and each problem, the number of minutes which passed between the beginning of the contest and the submission of this participant to this problem is known. It's also possible that this participant made no submissions to this problem.
With two seconds until the end of the round, all participants' submissions have passed pretests, and not a single hack attempt has been made. Vasya believes that no more submissions or hack attempts will be made in the remaining two seconds, and every submission will pass the system testing.
Unfortunately, Vasya is a cheater. He has registered 109 + 7 new accounts for the round. Now Vasya can submit any of his solutions from these new accounts in order to change the maximum point values of the problems. Vasya can also submit any wrong solutions to any problems. Note that Vasya can not submit correct solutions to the problems he hasn't solved.
Vasya seeks to score strictly more points than Petya in the current round. Vasya has already prepared the scripts which allow to obfuscate his solutions and submit them into the system from any of the new accounts in just fractions of seconds. However, Vasya doesn't want to make his cheating too obvious, so he wants to achieve his goal while making submissions from the smallest possible number of new accounts.
Find the smallest number of new accounts Vasya needs in order to beat Petya (provided that Vasya's assumptions are correct), or report that Vasya can't achieve his goal.
Input
The first line contains a single integer n (2 β€ n β€ 120) β the number of round participants, including Vasya and Petya.
Each of the next n lines contains five integers ai, 1, ai, 2..., ai, 5 ( - 1 β€ ai, j β€ 119) β the number of minutes passed between the beginning of the round and the submission of problem j by participant i, or -1 if participant i hasn't solved problem j.
It is guaranteed that each participant has made at least one successful submission.
Vasya is listed as participant number 1, Petya is listed as participant number 2, all the other participants are listed in no particular order.
Output
Output a single integer β the number of new accounts Vasya needs to beat Petya, or -1 if Vasya can't achieve his goal.
Examples
Input
2
5 15 40 70 115
50 45 40 30 15
Output
2
Input
3
55 80 10 -1 -1
15 -1 79 60 -1
42 -1 13 -1 -1
Output
3
Input
5
119 119 119 119 119
0 0 0 0 -1
20 65 12 73 77
78 112 22 23 11
1 78 60 111 62
Output
27
Input
4
-1 20 40 77 119
30 10 73 50 107
21 29 -1 64 98
117 65 -1 -1 -1
Output
-1
Note
In the first example, Vasya's optimal strategy is to submit the solutions to the last three problems from two new accounts. In this case the first two problems will have the maximum point value of 1000, while the last three problems will have the maximum point value of 500. Vasya's score will be equal to 980 + 940 + 420 + 360 + 270 = 2970 points, while Petya will score just 800 + 820 + 420 + 440 + 470 = 2950 points.
In the second example, Vasya has to make a single unsuccessful submission to any problem from two new accounts, and a single successful submission to the first problem from the third new account. In this case, the maximum point values of the problems will be equal to 500, 1500, 1000, 1500, 3000. Vasya will score 2370 points, while Petya will score just 2294 points.
In the third example, Vasya can achieve his goal by submitting the solutions to the first four problems from 27 new accounts. The maximum point values of the problems will be equal to 500, 500, 500, 500, 2000. Thanks to the high cost of the fifth problem, Vasya will manage to beat Petya who solved the first four problems very quickly, but couldn't solve the fifth one.
Tags: brute force, greedy
Correct Solution:
```
import sys
inf = 10**9 + 7
def solve():
n = int(sys.stdin.readline())
v = [int(vi) for vi in sys.stdin.readline().split()] # Vesya
p = [int(pi) for pi in sys.stdin.readline().split()] # Petya
cnt = [0]*5
for i in range(5):
if v[i] != -1:
cnt[i] += 1
if p[i] != -1:
cnt[i] += 1
for i in range(n - 2):
a = [int(ai) for ai in sys.stdin.readline().split()]
for j in range(5):
if a[j] != -1:
cnt[j] += 1
for i in range(10**5):
if check(n, v, p, cnt, i):
print(i)
return
print(-1)
def check(n, v, p, cnt, m):
tot = n + m
solved = cnt[:]
dif = 0
for i in range(5):
if p[i] != -1 and v[i] > p[i]:
solved[i] += m
for i in range(5):
if solved[i]*2 > tot:
max_score = 500
elif solved[i]*4 > tot:
max_score = 1000
elif solved[i]*8 > tot:
max_score = 1500
elif solved[i]*16 > tot:
max_score = 2000
elif solved[i]*32 > tot:
max_score = 2500
else:
max_score = 3000
if v[i] == p[i] == -1:
pass
elif v[i] == -1:
dif -= max_score * (250 - p[i]) // 250
elif p[i] == -1:
dif += max_score * (250 - v[i]) // 250
else:
dif += max_score * (p[i] - v[i]) // 250
return dif > 0
if __name__ == '__main__':
solve()
```
| 90,615 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya and Petya take part in a Codeforces round. The round lasts for two hours and contains five problems.
For this round the dynamic problem scoring is used. If you were lucky not to participate in any Codeforces round with dynamic problem scoring, here is what it means. The maximum point value of the problem depends on the ratio of the number of participants who solved the problem to the total number of round participants. Everyone who made at least one submission is considered to be participating in the round.
<image>
Pay attention to the range bounds. For example, if 40 people are taking part in the round, and 10 of them solve a particular problem, then the solvers fraction is equal to 1 / 4, and the problem's maximum point value is equal to 1500.
If the problem's maximum point value is equal to x, then for each whole minute passed from the beginning of the contest to the moment of the participant's correct submission, the participant loses x / 250 points. For example, if the problem's maximum point value is 2000, and the participant submits a correct solution to it 40 minutes into the round, this participant will be awarded with 2000Β·(1 - 40 / 250) = 1680 points for this problem.
There are n participants in the round, including Vasya and Petya. For each participant and each problem, the number of minutes which passed between the beginning of the contest and the submission of this participant to this problem is known. It's also possible that this participant made no submissions to this problem.
With two seconds until the end of the round, all participants' submissions have passed pretests, and not a single hack attempt has been made. Vasya believes that no more submissions or hack attempts will be made in the remaining two seconds, and every submission will pass the system testing.
Unfortunately, Vasya is a cheater. He has registered 109 + 7 new accounts for the round. Now Vasya can submit any of his solutions from these new accounts in order to change the maximum point values of the problems. Vasya can also submit any wrong solutions to any problems. Note that Vasya can not submit correct solutions to the problems he hasn't solved.
Vasya seeks to score strictly more points than Petya in the current round. Vasya has already prepared the scripts which allow to obfuscate his solutions and submit them into the system from any of the new accounts in just fractions of seconds. However, Vasya doesn't want to make his cheating too obvious, so he wants to achieve his goal while making submissions from the smallest possible number of new accounts.
Find the smallest number of new accounts Vasya needs in order to beat Petya (provided that Vasya's assumptions are correct), or report that Vasya can't achieve his goal.
Input
The first line contains a single integer n (2 β€ n β€ 120) β the number of round participants, including Vasya and Petya.
Each of the next n lines contains five integers ai, 1, ai, 2..., ai, 5 ( - 1 β€ ai, j β€ 119) β the number of minutes passed between the beginning of the round and the submission of problem j by participant i, or -1 if participant i hasn't solved problem j.
It is guaranteed that each participant has made at least one successful submission.
Vasya is listed as participant number 1, Petya is listed as participant number 2, all the other participants are listed in no particular order.
Output
Output a single integer β the number of new accounts Vasya needs to beat Petya, or -1 if Vasya can't achieve his goal.
Examples
Input
2
5 15 40 70 115
50 45 40 30 15
Output
2
Input
3
55 80 10 -1 -1
15 -1 79 60 -1
42 -1 13 -1 -1
Output
3
Input
5
119 119 119 119 119
0 0 0 0 -1
20 65 12 73 77
78 112 22 23 11
1 78 60 111 62
Output
27
Input
4
-1 20 40 77 119
30 10 73 50 107
21 29 -1 64 98
117 65 -1 -1 -1
Output
-1
Note
In the first example, Vasya's optimal strategy is to submit the solutions to the last three problems from two new accounts. In this case the first two problems will have the maximum point value of 1000, while the last three problems will have the maximum point value of 500. Vasya's score will be equal to 980 + 940 + 420 + 360 + 270 = 2970 points, while Petya will score just 800 + 820 + 420 + 440 + 470 = 2950 points.
In the second example, Vasya has to make a single unsuccessful submission to any problem from two new accounts, and a single successful submission to the first problem from the third new account. In this case, the maximum point values of the problems will be equal to 500, 1500, 1000, 1500, 3000. Vasya will score 2370 points, while Petya will score just 2294 points.
In the third example, Vasya can achieve his goal by submitting the solutions to the first four problems from 27 new accounts. The maximum point values of the problems will be equal to 500, 500, 500, 500, 2000. Thanks to the high cost of the fifth problem, Vasya will manage to beat Petya who solved the first four problems very quickly, but couldn't solve the fifth one.
Submitted Solution:
```
n = int(input())
t = [0] * n
for i in range(n):
t[i] = list(map(int, input().split()))
value = [0] * 5
solve = [0] * 5
for i in range(5):
solved = 0
for j in range(n):
if t[j][i] != -1:
solved += 1
else:
t[j][i] = 250
if (solved * 2 > n):
value[i] = 500
elif (solved * 4 > n):
value[i] = 1000
elif (solved * 8 > n):
value[i] = 1500
elif (solved * 16 > n):
value[i] = 2000
elif (solved * 32 > n):
value[i] = 2500
else:
value[i] = 3000
solve[i] = solved
vasya = 0
for i in range(5):
vasya += value[i] - value[i] // 250 * t[0][i]
petya = 0
for i in range(5):
petya += value[i] - value[i] // 250 * t[1][i]
pot = [[0] * 20000 for i in range(5)]
for problem in range(5):
if t[0][problem] < t[1][problem]:
cur_ac = 0
while 1:
if solve[problem] * 32 <= n + cur_ac:
win = 3000 - 3000 // 250 * t[0][problem] - (3000 - 3000 // 250 * t[1][problem]) - (value[problem] - value[problem] // 250 * t[0][problem] -(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 16 <= n + cur_ac:
win = 2500 - 2500 // 250 * t[0][problem] - (2500 - 2500 // 250 * t[1][problem]) - (value[problem] - value[problem] // 250 * t[0][problem] -(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 8 <= n + cur_ac:
win = 2000 - 2000 // 250 * t[0][problem] - (2000 - 2000 // 250 * t[1][problem]) - (value[problem] - value[problem] // 250 * t[0][problem] -(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 4 <= n + cur_ac:
win = 1500 - 1500 // 250 * t[0][problem] - (1500 - 1500 // 250 * t[1][problem]) - (value[problem] - value[problem] // 250 * t[0][problem] -(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 2 <= n + cur_ac:
win = 1000 - 1000 // 250 * t[0][problem] - (1000 - 1000 // 250 * t[1][problem]) - (value[problem] - value[problem] // 250 * t[0][problem] -(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
else:
win = 500 - 500 // 250 * t[0][problem] - (500 - 500 // 250 * t[1][problem]) - (value[problem] - value[problem] // 250 * t[0][problem] -(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
cur_ac += 1
if cur_ac == 20000:
break
elif t[0][problem] != 250:
cur_ac = 0
while 1:
if solve[problem] * 32 + cur_ac <= n + cur_ac:
win = 3000 - 3000 // 250 * t[0][problem] - (3000 - 3000 // 250 * t[1][problem]) - (
value[problem] - value[problem] // 250 * t[0][problem] - (
value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 16 + cur_ac <= n + cur_ac:
win = 2500 - 2500 // 250 * t[0][problem] - (2500 - 2500 // 250 * t[1][problem]) - (
value[problem] - value[problem] // 250 * t[0][problem] - (
value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 8 + cur_ac <= n + cur_ac:
win = 2000 - 2000 // 250 * t[0][problem] - (2000 - 2000 // 250 * t[1][problem]) - (
value[problem] - value[problem] // 250 * t[0][problem] - (
value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 4 + cur_ac <= n + cur_ac:
win = 1500 - 1500 // 250 * t[0][problem] - (1500 - 1500 // 250 * t[1][problem]) - (
value[problem] - value[problem] // 250 * t[0][problem] - (
value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 2 + cur_ac <= n + cur_ac:
win = 1000 - 1000 // 250 * t[0][problem] - (1000 - 1000 // 250 * t[1][problem]) - (
value[problem] - value[problem] // 250 * t[0][problem] - (
value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
else:
win = 500 - 500 // 250 * t[0][problem] - (500 - 500 // 250 * t[1][problem]) - (
value[problem] - value[problem] // 250 * t[0][problem] - (
value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
cur_ac += 1
if cur_ac == 20000:
break
else:
cur_ac = 0
while 1:
if solve[problem] * 32 <= n + cur_ac:
win = -(3000 - 3000 // 250 * t[1][problem]) -(
-(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
break
elif solve[problem] * 16 <= n + cur_ac:
win = -(2500 - 2500 // 250 * t[1][problem]) - (
-(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 8 <= n + cur_ac:
win = -(2000 - 2000 // 250 * t[1][problem]) - (
-(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 4 <= n + cur_ac:
win = -(1500 - 1500 // 250 * t[1][problem]) - (
-(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 2 <= n + cur_ac:
win = -(1000 - 1000 // 250 * t[1][problem]) - (
-(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
else:
win = -(500 - 500 // 250 * t[1][problem]) - (
-(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
cur_ac += 1
if cur_ac == 20000:
break
res = -1
maxi = 0
for i in range(5):
maxi = max(maxi, len(pot[i]))
for i in range(maxi):
tmp = 0
for pr in range(5):
if len(pot[pr]) <= i:
tmp += pot[pr][-1]
else:
tmp += pot[pr][i]
if tmp > petya - vasya:
res = i
break
print(res)
```
No
| 90,616 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya and Petya take part in a Codeforces round. The round lasts for two hours and contains five problems.
For this round the dynamic problem scoring is used. If you were lucky not to participate in any Codeforces round with dynamic problem scoring, here is what it means. The maximum point value of the problem depends on the ratio of the number of participants who solved the problem to the total number of round participants. Everyone who made at least one submission is considered to be participating in the round.
<image>
Pay attention to the range bounds. For example, if 40 people are taking part in the round, and 10 of them solve a particular problem, then the solvers fraction is equal to 1 / 4, and the problem's maximum point value is equal to 1500.
If the problem's maximum point value is equal to x, then for each whole minute passed from the beginning of the contest to the moment of the participant's correct submission, the participant loses x / 250 points. For example, if the problem's maximum point value is 2000, and the participant submits a correct solution to it 40 minutes into the round, this participant will be awarded with 2000Β·(1 - 40 / 250) = 1680 points for this problem.
There are n participants in the round, including Vasya and Petya. For each participant and each problem, the number of minutes which passed between the beginning of the contest and the submission of this participant to this problem is known. It's also possible that this participant made no submissions to this problem.
With two seconds until the end of the round, all participants' submissions have passed pretests, and not a single hack attempt has been made. Vasya believes that no more submissions or hack attempts will be made in the remaining two seconds, and every submission will pass the system testing.
Unfortunately, Vasya is a cheater. He has registered 109 + 7 new accounts for the round. Now Vasya can submit any of his solutions from these new accounts in order to change the maximum point values of the problems. Vasya can also submit any wrong solutions to any problems. Note that Vasya can not submit correct solutions to the problems he hasn't solved.
Vasya seeks to score strictly more points than Petya in the current round. Vasya has already prepared the scripts which allow to obfuscate his solutions and submit them into the system from any of the new accounts in just fractions of seconds. However, Vasya doesn't want to make his cheating too obvious, so he wants to achieve his goal while making submissions from the smallest possible number of new accounts.
Find the smallest number of new accounts Vasya needs in order to beat Petya (provided that Vasya's assumptions are correct), or report that Vasya can't achieve his goal.
Input
The first line contains a single integer n (2 β€ n β€ 120) β the number of round participants, including Vasya and Petya.
Each of the next n lines contains five integers ai, 1, ai, 2..., ai, 5 ( - 1 β€ ai, j β€ 119) β the number of minutes passed between the beginning of the round and the submission of problem j by participant i, or -1 if participant i hasn't solved problem j.
It is guaranteed that each participant has made at least one successful submission.
Vasya is listed as participant number 1, Petya is listed as participant number 2, all the other participants are listed in no particular order.
Output
Output a single integer β the number of new accounts Vasya needs to beat Petya, or -1 if Vasya can't achieve his goal.
Examples
Input
2
5 15 40 70 115
50 45 40 30 15
Output
2
Input
3
55 80 10 -1 -1
15 -1 79 60 -1
42 -1 13 -1 -1
Output
3
Input
5
119 119 119 119 119
0 0 0 0 -1
20 65 12 73 77
78 112 22 23 11
1 78 60 111 62
Output
27
Input
4
-1 20 40 77 119
30 10 73 50 107
21 29 -1 64 98
117 65 -1 -1 -1
Output
-1
Note
In the first example, Vasya's optimal strategy is to submit the solutions to the last three problems from two new accounts. In this case the first two problems will have the maximum point value of 1000, while the last three problems will have the maximum point value of 500. Vasya's score will be equal to 980 + 940 + 420 + 360 + 270 = 2970 points, while Petya will score just 800 + 820 + 420 + 440 + 470 = 2950 points.
In the second example, Vasya has to make a single unsuccessful submission to any problem from two new accounts, and a single successful submission to the first problem from the third new account. In this case, the maximum point values of the problems will be equal to 500, 1500, 1000, 1500, 3000. Vasya will score 2370 points, while Petya will score just 2294 points.
In the third example, Vasya can achieve his goal by submitting the solutions to the first four problems from 27 new accounts. The maximum point values of the problems will be equal to 500, 500, 500, 500, 2000. Thanks to the high cost of the fifth problem, Vasya will manage to beat Petya who solved the first four problems very quickly, but couldn't solve the fifth one.
Submitted Solution:
```
import sys
inf = 10**9 + 7
def solve():
def check(mid):
tot = n + mid
dif = 0
solved = cnt[:]
for i in range(5):
if p[i] != -1 and p[i] < v[i]:
solved[i] += mid
for i in range(5):
if solved[i]*2 > tot:
max_score = 500
elif solved[i]*4 > tot:
max_score = 1000
elif solved[i]*8 > tot:
max_score = 1500
elif solved[i]*16 > tot:
max_score = 2000
elif solved[i]*32 > tot:
max_score = 2500
else:
max_score = 3000
if v[i] == p[i]:
pass
elif v[i] == -1:
dif += max_score * (250 - p[i]) // 250
elif p[i] == -1:
dif += -max_score * (250 - v[i]) // 250
else:
dif += max_score * (-p[i] + v[i]) // 250
# print(mid, dif)
return dif < 0
n = int(sys.stdin.readline())
cnt = [0]*5
v = [int(i) for i in sys.stdin.readline().split()]
for i in range(5):
if v[i] != -1:
cnt[i] += 1
p = [int(i) for i in sys.stdin.readline().split()]
for i in range(5):
if p[i] != -1:
cnt[i] += 1
for i in range(n - 2):
a = [int(ai) for ai in sys.stdin.readline().split()]
for j in range(5):
if a[j] != -1:
cnt[j] += 1
btm = -1
top = inf + 2
while top - btm > 1:
mid = (top + btm) // 2
if check(mid):
top = mid
else:
btm = mid
ans = top if top < inf + 1 else -1
print(ans)
if __name__ == '__main__':
solve()
```
No
| 90,617 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya and Petya take part in a Codeforces round. The round lasts for two hours and contains five problems.
For this round the dynamic problem scoring is used. If you were lucky not to participate in any Codeforces round with dynamic problem scoring, here is what it means. The maximum point value of the problem depends on the ratio of the number of participants who solved the problem to the total number of round participants. Everyone who made at least one submission is considered to be participating in the round.
<image>
Pay attention to the range bounds. For example, if 40 people are taking part in the round, and 10 of them solve a particular problem, then the solvers fraction is equal to 1 / 4, and the problem's maximum point value is equal to 1500.
If the problem's maximum point value is equal to x, then for each whole minute passed from the beginning of the contest to the moment of the participant's correct submission, the participant loses x / 250 points. For example, if the problem's maximum point value is 2000, and the participant submits a correct solution to it 40 minutes into the round, this participant will be awarded with 2000Β·(1 - 40 / 250) = 1680 points for this problem.
There are n participants in the round, including Vasya and Petya. For each participant and each problem, the number of minutes which passed between the beginning of the contest and the submission of this participant to this problem is known. It's also possible that this participant made no submissions to this problem.
With two seconds until the end of the round, all participants' submissions have passed pretests, and not a single hack attempt has been made. Vasya believes that no more submissions or hack attempts will be made in the remaining two seconds, and every submission will pass the system testing.
Unfortunately, Vasya is a cheater. He has registered 109 + 7 new accounts for the round. Now Vasya can submit any of his solutions from these new accounts in order to change the maximum point values of the problems. Vasya can also submit any wrong solutions to any problems. Note that Vasya can not submit correct solutions to the problems he hasn't solved.
Vasya seeks to score strictly more points than Petya in the current round. Vasya has already prepared the scripts which allow to obfuscate his solutions and submit them into the system from any of the new accounts in just fractions of seconds. However, Vasya doesn't want to make his cheating too obvious, so he wants to achieve his goal while making submissions from the smallest possible number of new accounts.
Find the smallest number of new accounts Vasya needs in order to beat Petya (provided that Vasya's assumptions are correct), or report that Vasya can't achieve his goal.
Input
The first line contains a single integer n (2 β€ n β€ 120) β the number of round participants, including Vasya and Petya.
Each of the next n lines contains five integers ai, 1, ai, 2..., ai, 5 ( - 1 β€ ai, j β€ 119) β the number of minutes passed between the beginning of the round and the submission of problem j by participant i, or -1 if participant i hasn't solved problem j.
It is guaranteed that each participant has made at least one successful submission.
Vasya is listed as participant number 1, Petya is listed as participant number 2, all the other participants are listed in no particular order.
Output
Output a single integer β the number of new accounts Vasya needs to beat Petya, or -1 if Vasya can't achieve his goal.
Examples
Input
2
5 15 40 70 115
50 45 40 30 15
Output
2
Input
3
55 80 10 -1 -1
15 -1 79 60 -1
42 -1 13 -1 -1
Output
3
Input
5
119 119 119 119 119
0 0 0 0 -1
20 65 12 73 77
78 112 22 23 11
1 78 60 111 62
Output
27
Input
4
-1 20 40 77 119
30 10 73 50 107
21 29 -1 64 98
117 65 -1 -1 -1
Output
-1
Note
In the first example, Vasya's optimal strategy is to submit the solutions to the last three problems from two new accounts. In this case the first two problems will have the maximum point value of 1000, while the last three problems will have the maximum point value of 500. Vasya's score will be equal to 980 + 940 + 420 + 360 + 270 = 2970 points, while Petya will score just 800 + 820 + 420 + 440 + 470 = 2950 points.
In the second example, Vasya has to make a single unsuccessful submission to any problem from two new accounts, and a single successful submission to the first problem from the third new account. In this case, the maximum point values of the problems will be equal to 500, 1500, 1000, 1500, 3000. Vasya will score 2370 points, while Petya will score just 2294 points.
In the third example, Vasya can achieve his goal by submitting the solutions to the first four problems from 27 new accounts. The maximum point values of the problems will be equal to 500, 500, 500, 500, 2000. Thanks to the high cost of the fifth problem, Vasya will manage to beat Petya who solved the first four problems very quickly, but couldn't solve the fifth one.
Submitted Solution:
```
import sys
inf = 10**9 + 7
def solve():
def check(mid):
tot = n + mid
dif = 0
solved = cnt[:]
for i in range(5):
if v[i] != -1 and p[i] != -1 and p[i] < v[i]:
solved[i] += mid
for i in range(5):
if solved[i]*2 > tot:
max_score = 500
elif solved[i]*4 > tot:
max_score = 1000
elif solved[i]*8 > tot:
max_score = 1500
elif solved[i]*16 > tot:
max_score = 2000
elif solved[i]*32 > tot:
max_score = 2500
else:
max_score = 3000
if v[i] == p[i]:
pass
elif v[i] == -1:
dif += max_score * (250 - p[i]) // 250
elif p[i] == -1:
dif += -max_score * (250 - v[i]) // 250
else:
dif += max_score * (-p[i] + v[i]) // 250
# print(mid, dif)
return dif < 0
n = int(sys.stdin.readline())
cnt = [0]*5
v = [int(i) for i in sys.stdin.readline().split()]
for i in range(5):
if v[i] != -1:
cnt[i] += 1
p = [int(i) for i in sys.stdin.readline().split()]
for i in range(5):
if p[i] != -1:
cnt[i] += 1
for i in range(n - 2):
a = [int(ai) for ai in sys.stdin.readline().split()]
for j in range(5):
if a[j] != -1:
cnt[j] += 1
btm = -1
top = inf + 2
while top - btm > 1:
mid = (top + btm) // 2
if check(mid):
top = mid
else:
btm = mid
ans = top if top < inf + 1 else -1
print(ans)
if __name__ == '__main__':
solve()
```
No
| 90,618 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya and Petya take part in a Codeforces round. The round lasts for two hours and contains five problems.
For this round the dynamic problem scoring is used. If you were lucky not to participate in any Codeforces round with dynamic problem scoring, here is what it means. The maximum point value of the problem depends on the ratio of the number of participants who solved the problem to the total number of round participants. Everyone who made at least one submission is considered to be participating in the round.
<image>
Pay attention to the range bounds. For example, if 40 people are taking part in the round, and 10 of them solve a particular problem, then the solvers fraction is equal to 1 / 4, and the problem's maximum point value is equal to 1500.
If the problem's maximum point value is equal to x, then for each whole minute passed from the beginning of the contest to the moment of the participant's correct submission, the participant loses x / 250 points. For example, if the problem's maximum point value is 2000, and the participant submits a correct solution to it 40 minutes into the round, this participant will be awarded with 2000Β·(1 - 40 / 250) = 1680 points for this problem.
There are n participants in the round, including Vasya and Petya. For each participant and each problem, the number of minutes which passed between the beginning of the contest and the submission of this participant to this problem is known. It's also possible that this participant made no submissions to this problem.
With two seconds until the end of the round, all participants' submissions have passed pretests, and not a single hack attempt has been made. Vasya believes that no more submissions or hack attempts will be made in the remaining two seconds, and every submission will pass the system testing.
Unfortunately, Vasya is a cheater. He has registered 109 + 7 new accounts for the round. Now Vasya can submit any of his solutions from these new accounts in order to change the maximum point values of the problems. Vasya can also submit any wrong solutions to any problems. Note that Vasya can not submit correct solutions to the problems he hasn't solved.
Vasya seeks to score strictly more points than Petya in the current round. Vasya has already prepared the scripts which allow to obfuscate his solutions and submit them into the system from any of the new accounts in just fractions of seconds. However, Vasya doesn't want to make his cheating too obvious, so he wants to achieve his goal while making submissions from the smallest possible number of new accounts.
Find the smallest number of new accounts Vasya needs in order to beat Petya (provided that Vasya's assumptions are correct), or report that Vasya can't achieve his goal.
Input
The first line contains a single integer n (2 β€ n β€ 120) β the number of round participants, including Vasya and Petya.
Each of the next n lines contains five integers ai, 1, ai, 2..., ai, 5 ( - 1 β€ ai, j β€ 119) β the number of minutes passed between the beginning of the round and the submission of problem j by participant i, or -1 if participant i hasn't solved problem j.
It is guaranteed that each participant has made at least one successful submission.
Vasya is listed as participant number 1, Petya is listed as participant number 2, all the other participants are listed in no particular order.
Output
Output a single integer β the number of new accounts Vasya needs to beat Petya, or -1 if Vasya can't achieve his goal.
Examples
Input
2
5 15 40 70 115
50 45 40 30 15
Output
2
Input
3
55 80 10 -1 -1
15 -1 79 60 -1
42 -1 13 -1 -1
Output
3
Input
5
119 119 119 119 119
0 0 0 0 -1
20 65 12 73 77
78 112 22 23 11
1 78 60 111 62
Output
27
Input
4
-1 20 40 77 119
30 10 73 50 107
21 29 -1 64 98
117 65 -1 -1 -1
Output
-1
Note
In the first example, Vasya's optimal strategy is to submit the solutions to the last three problems from two new accounts. In this case the first two problems will have the maximum point value of 1000, while the last three problems will have the maximum point value of 500. Vasya's score will be equal to 980 + 940 + 420 + 360 + 270 = 2970 points, while Petya will score just 800 + 820 + 420 + 440 + 470 = 2950 points.
In the second example, Vasya has to make a single unsuccessful submission to any problem from two new accounts, and a single successful submission to the first problem from the third new account. In this case, the maximum point values of the problems will be equal to 500, 1500, 1000, 1500, 3000. Vasya will score 2370 points, while Petya will score just 2294 points.
In the third example, Vasya can achieve his goal by submitting the solutions to the first four problems from 27 new accounts. The maximum point values of the problems will be equal to 500, 500, 500, 500, 2000. Thanks to the high cost of the fifth problem, Vasya will manage to beat Petya who solved the first four problems very quickly, but couldn't solve the fifth one.
Submitted Solution:
```
n = int(input())
t = [0] * n
for i in range(n):
t[i] = list(map(int, input().split()))
value = [0] * 5
solve = [0] * 5
for i in range(5):
solved = 0
for j in range(n):
if t[j][i] != -1:
solved += 1
else:
t[j][i] = 250
if (solved * 2 > n):
value[i] = 500
elif (solved * 4 > n):
value[i] = 1000
elif (solved * 8 > n):
value[i] = 1500
elif (solved * 16 > n):
value[i] = 2000
elif (solved * 32 > n):
value[i] = 2500
else:
value[i] = 3000
solve[i] = solved
vasya = 0
for i in range(5):
vasya += value[i] - value[i] // 250 * t[0][i]
petya = 0
for i in range(5):
petya += value[i] - value[i] // 250 * t[1][i]
pot = [[0] * 200000 for i in range(5)]
for problem in range(5):
if t[0][problem] < t[1][problem]:
cur_ac = 0
while 1:
if solve[problem] * 32 <= n + cur_ac:
win = 3000 - 3000 // 250 * t[0][problem] - (3000 - 3000 // 250 * t[1][problem]) - (value[problem] - value[problem] // 250 * t[0][problem] -(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
break
elif solve[problem] * 16 <= n + cur_ac:
win = 2500 - 2500 // 250 * t[0][problem] - (2500 - 2500 // 250 * t[1][problem]) - (value[problem] - value[problem] // 250 * t[0][problem] -(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 8 <= n + cur_ac:
win = 2000 - 2000 // 250 * t[0][problem] - (2000 - 2000 // 250 * t[1][problem]) - (value[problem] - value[problem] // 250 * t[0][problem] -(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 4 <= n + cur_ac:
win = 1500 - 1500 // 250 * t[0][problem] - (1500 - 1500 // 250 * t[1][problem]) - (value[problem] - value[problem] // 250 * t[0][problem] -(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 2 <= n + cur_ac:
win = 1000 - 1000 // 250 * t[0][problem] - (1000 - 1000 // 250 * t[1][problem]) - (value[problem] - value[problem] // 250 * t[0][problem] -(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
else:
win = 500 - 500 // 250 * t[0][problem] - (500 - 500 // 250 * t[1][problem]) - (value[problem] - value[problem] // 250 * t[0][problem] -(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
cur_ac += 1
elif t[0][problem] != 250:
cur_ac = 0
while 1:
if solve[problem] * 32 + cur_ac <= n + cur_ac:
win = 3000 - 3000 // 250 * t[0][problem] - (3000 - 3000 // 250 * t[1][problem]) - (
value[problem] - value[problem] // 250 * t[0][problem] - (
value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 16 + cur_ac <= n + cur_ac:
win = 2500 - 2500 // 250 * t[0][problem] - (2500 - 2500 // 250 * t[1][problem]) - (
value[problem] - value[problem] // 250 * t[0][problem] - (
value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 8 + cur_ac <= n + cur_ac:
win = 2000 - 2000 // 250 * t[0][problem] - (2000 - 2000 // 250 * t[1][problem]) - (
value[problem] - value[problem] // 250 * t[0][problem] - (
value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 4 + cur_ac <= n + cur_ac:
win = 1500 - 1500 // 250 * t[0][problem] - (1500 - 1500 // 250 * t[1][problem]) - (
value[problem] - value[problem] // 250 * t[0][problem] - (
value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 2 + cur_ac <= n + cur_ac:
win = 1000 - 1000 // 250 * t[0][problem] - (1000 - 1000 // 250 * t[1][problem]) - (
value[problem] - value[problem] // 250 * t[0][problem] - (
value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
else:
win = 500 - 500 // 250 * t[0][problem] - (500 - 500 // 250 * t[1][problem]) - (
value[problem] - value[problem] // 250 * t[0][problem] - (
value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
break
cur_ac += 1
else:
cur_ac = 0
while 1:
if solve[problem] * 32 <= n + cur_ac:
win = -(3000 - 3000 // 250 * t[1][problem]) -(
-(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
break
elif solve[problem] * 16 <= n + cur_ac:
win = -(2500 - 2500 // 250 * t[1][problem]) - (
-(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 8 <= n + cur_ac:
win = -(2000 - 2000 // 250 * t[1][problem]) - (
-(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 4 <= n + cur_ac:
win = -(1500 - 1500 // 250 * t[1][problem]) - (
-(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
elif solve[problem] * 2 <= n + cur_ac:
win = -(1000 - 1000 // 250 * t[1][problem]) - (
-(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
else:
win = -(500 - 500 // 250 * t[1][problem]) - (
-(value[problem] - value[problem] // 250 * t[1][problem]))
pot[problem][cur_ac] = win
cur_ac += 1
res = -1
maxi = 0
for i in range(5):
maxi = max(maxi, len(pot[i]))
for i in range(maxi):
tmp = 0
for pr in range(5):
if len(pot[pr]) <= i:
tmp += pot[pr][-1]
else:
tmp += pot[pr][i]
if tmp > petya - vasya:
res = i
break
print(res)
```
No
| 90,619 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasily has a deck of cards consisting of n cards. There is an integer on each of the cards, this integer is between 1 and 100 000, inclusive. It is possible that some cards have the same integers on them.
Vasily decided to sort the cards. To do this, he repeatedly takes the top card from the deck, and if the number on it equals the minimum number written on the cards in the deck, then he places the card away. Otherwise, he puts it under the deck and takes the next card from the top, and so on. The process ends as soon as there are no cards in the deck. You can assume that Vasily always knows the minimum number written on some card in the remaining deck, but doesn't know where this card (or these cards) is.
You are to determine the total number of times Vasily takes the top card from the deck.
Input
The first line contains single integer n (1 β€ n β€ 100 000) β the number of cards in the deck.
The second line contains a sequence of n integers a1, a2, ..., an (1 β€ ai β€ 100 000), where ai is the number written on the i-th from top card in the deck.
Output
Print the total number of times Vasily takes the top card from the deck.
Examples
Input
4
6 3 1 2
Output
7
Input
1
1000
Output
1
Input
7
3 3 3 3 3 3 3
Output
7
Note
In the first example Vasily at first looks at the card with number 6 on it, puts it under the deck, then on the card with number 3, puts it under the deck, and then on the card with number 1. He places away the card with 1, because the number written on it is the minimum among the remaining cards. After that the cards from top to bottom are [2, 6, 3]. Then Vasily looks at the top card with number 2 and puts it away. After that the cards from top to bottom are [6, 3]. Then Vasily looks at card 6, puts it under the deck, then at card 3 and puts it away. Then there is only one card with number 6 on it, and Vasily looks at it and puts it away. Thus, in total Vasily looks at 7 cards.
Tags: data structures, implementation, sortings
Correct Solution:
```
# -*- coding: utf-8 -*-
import sys
# fout = open("output.txt", "w")
fin = sys.stdin
# fin = open("input.txt", "r")
fout = sys.stdout
n = int(fin.readline())
a = list(map(int, fin.readline().split()))
def solution(n, a):
sorted_arr = [(i, elem) for i, elem in enumerate(a)]
sorted_arr.sort(key=lambda x: (x[1], x[0]))
sorted_indexes = [x[0] for x in sorted_arr]
cnt = 0
current_n = n
prev_index = sorted_indexes[0]
prev_elem = sorted_arr[0][1]
cur_len = 1
i = 1
while i < n:
cur_index = sorted_indexes[i]
cur_elem = sorted_arr[i][1]
if prev_index < cur_index:
cur_len += 1
prev_index = sorted_indexes[i]
prev_elem = sorted_arr[i][1]
elif i+1<n and cur_elem == sorted_arr[i+1][1]: # Π·Π΄Π΅ΡΡ ΠΊΠΎΡΡΠΊ
penalty = 1
last_penalty_ind = sorted_indexes[i]
while i+1 < n and sorted_arr[i+1][1] == cur_elem:
if sorted_arr[i+1][0] >= prev_index:
cur_len += 1
else:
penalty += 1
last_penalty_ind = sorted_indexes[i+1]
i += 1
cnt += current_n
current_n -= cur_len
cur_len = penalty
prev_elem = cur_elem
prev_index = last_penalty_ind
else:
cnt += current_n
current_n -= cur_len
cur_len = 1
prev_index = sorted_indexes[i]
prev_elem = sorted_arr[i][1]
i += 1
cnt += current_n
return cnt
cnt = solution(n, a)
fout.write(str(cnt))
fout.close()
```
| 90,620 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasily has a deck of cards consisting of n cards. There is an integer on each of the cards, this integer is between 1 and 100 000, inclusive. It is possible that some cards have the same integers on them.
Vasily decided to sort the cards. To do this, he repeatedly takes the top card from the deck, and if the number on it equals the minimum number written on the cards in the deck, then he places the card away. Otherwise, he puts it under the deck and takes the next card from the top, and so on. The process ends as soon as there are no cards in the deck. You can assume that Vasily always knows the minimum number written on some card in the remaining deck, but doesn't know where this card (or these cards) is.
You are to determine the total number of times Vasily takes the top card from the deck.
Input
The first line contains single integer n (1 β€ n β€ 100 000) β the number of cards in the deck.
The second line contains a sequence of n integers a1, a2, ..., an (1 β€ ai β€ 100 000), where ai is the number written on the i-th from top card in the deck.
Output
Print the total number of times Vasily takes the top card from the deck.
Examples
Input
4
6 3 1 2
Output
7
Input
1
1000
Output
1
Input
7
3 3 3 3 3 3 3
Output
7
Note
In the first example Vasily at first looks at the card with number 6 on it, puts it under the deck, then on the card with number 3, puts it under the deck, and then on the card with number 1. He places away the card with 1, because the number written on it is the minimum among the remaining cards. After that the cards from top to bottom are [2, 6, 3]. Then Vasily looks at the top card with number 2 and puts it away. After that the cards from top to bottom are [6, 3]. Then Vasily looks at card 6, puts it under the deck, then at card 3 and puts it away. Then there is only one card with number 6 on it, and Vasily looks at it and puts it away. Thus, in total Vasily looks at 7 cards.
Submitted Solution:
```
# -*- coding: utf-8 -*-
import sys
# fout = open("output.txt", "w")
fin = sys.stdin
# fin = open("input.txt", "r")
fout = sys.stdout
n = int(fin.readline())
a = list(map(int, fin.readline().split()))
def solution(n, a):
sorted_arr = [(i, elem) for i, elem in enumerate(a)]
sorted_arr.sort(key=lambda x: (x[1], x[0]))
sorted_indexes = [x[0] for x in sorted_arr]
cnt = 0
current_n = n
prev = sorted_indexes[0]
cur_len = 1
for i in range(1, n):
if prev <= sorted_indexes[i]:
cur_len += 1
prev = sorted_indexes[i]
else:
cnt += current_n
current_n -= cur_len
cur_len = 1
prev = sorted_indexes[i]
cnt += current_n
return cnt
cnt = solution(n, a)
fout.write(str(cnt))
fout.close()
```
No
| 90,621 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider an array A with N elements, all being the same integer a.
Define the product transformation as a simultaneous update Ai = AiΒ·Ai + 1, that is multiplying each element to the element right to it for <image>, with the last number AN remaining the same. For example, if we start with an array A with a = 2 and N = 4, then after one product transformation A = [4, 4, 4, 2], and after two product transformations A = [16, 16, 8, 2].
Your simple task is to calculate the array A after M product transformations. Since the numbers can get quite big you should output them modulo Q.
Input
The first and only line of input contains four integers N, M, a, Q (7 β€ Q β€ 109 + 123, 2 β€ a β€ 106 + 123, <image>, <image> is prime), where <image> is the multiplicative order of the integer a modulo Q, see notes for definition.
Output
You should output the array A from left to right.
Example
Input
2 2 2 7
Output
1 2
Note
The multiplicative order of a number a modulo Q <image>, is the smallest natural number x such that ax mod Q = 1. For example, <image>.
Submitted Solution:
```
# #include <cstdio>
# using namespace std;
# #define N 2000050
# int n,m,a,q,i,pw[N],ni[N],c[N];
# int main()
# {
# scanf("%d%d%d%d",&n,&m,&a,&q);
# int y=1;pw[0]=1;pw[1]=a;
# while (pw[y]!=1) y++,pw[y]=1LL*pw[y-1]*a%q;
# ni[1]=1;
# for (i=2;i<y;i++) ni[i]=-1LL*(y/i)*ni[y%i]%y;
# int x=1;c[0]=1;
# for (i=1;i<n;i++) x=1LL*x*(m-i+1)%y*ni[i]%y,c[i]=(c[i-1]+x)%y;
# for (i=1;i<=n;i++) printf("%d ",pw[(c[n-i]+y)%y]);puts("");
# return 0;
# }
N = 2000050
n,m,a,q = map(int,input().split())
pw = [0]*N
ni = [0]*N
c = [0]*N
y = 1
pw[0] = 1
pw[1] = a
while(pw[y]!=1):
y+=1
pw[y] = pw[y-1]*a % q
ni[1] = 1
for i in range(2,y):
ni[i] = -(y/i)*ni[y%i]%y
x = 1
c[0] = 1
for i in range(1,n):
x *= (m-i+1) % y * ni[i] % y
for i in range(1,n+1):
print(pw[(c[n-i]+y)%y])
```
No
| 90,622 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider an array A with N elements, all being the same integer a.
Define the product transformation as a simultaneous update Ai = AiΒ·Ai + 1, that is multiplying each element to the element right to it for <image>, with the last number AN remaining the same. For example, if we start with an array A with a = 2 and N = 4, then after one product transformation A = [4, 4, 4, 2], and after two product transformations A = [16, 16, 8, 2].
Your simple task is to calculate the array A after M product transformations. Since the numbers can get quite big you should output them modulo Q.
Input
The first and only line of input contains four integers N, M, a, Q (7 β€ Q β€ 109 + 123, 2 β€ a β€ 106 + 123, <image>, <image> is prime), where <image> is the multiplicative order of the integer a modulo Q, see notes for definition.
Output
You should output the array A from left to right.
Example
Input
2 2 2 7
Output
1 2
Note
The multiplicative order of a number a modulo Q <image>, is the smallest natural number x such that ax mod Q = 1. For example, <image>.
Submitted Solution:
```
print("1 "+"2")
```
No
| 90,623 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider an array A with N elements, all being the same integer a.
Define the product transformation as a simultaneous update Ai = AiΒ·Ai + 1, that is multiplying each element to the element right to it for <image>, with the last number AN remaining the same. For example, if we start with an array A with a = 2 and N = 4, then after one product transformation A = [4, 4, 4, 2], and after two product transformations A = [16, 16, 8, 2].
Your simple task is to calculate the array A after M product transformations. Since the numbers can get quite big you should output them modulo Q.
Input
The first and only line of input contains four integers N, M, a, Q (7 β€ Q β€ 109 + 123, 2 β€ a β€ 106 + 123, <image>, <image> is prime), where <image> is the multiplicative order of the integer a modulo Q, see notes for definition.
Output
You should output the array A from left to right.
Example
Input
2 2 2 7
Output
1 2
Note
The multiplicative order of a number a modulo Q <image>, is the smallest natural number x such that ax mod Q = 1. For example, <image>.
Submitted Solution:
```
C = 2000050
fact = [0]*C
nInverse = [0]*C
def main():
n,m,a,Q = map(int,input().split())
mulOrder = 1
for i in range(1,C): #finding multiplicative order of a mod Q
if(pow(a,i,Q) == 1):
mulOrder = i
break
fact[0] = 1
nInverse[0] = 1
for i in range(1,C): #calculating factorial and n-inverses
fact[i] = (fact[i-1]*i) % mulOrder
nInverse[i] = pow(fact[i],mulOrder - 2,mulOrder) #given that mulOrder is a prime number and using Fermat's Little Theorem
D = [0]*(n+1)
for i in range(1,n+1):
D[i] = (cn(m,i-1,mulOrder) + D[i-1]) % mulOrder #calculating exponent for each term
for i in reversed(range(n,0)):
print(pow(a,D[i],Q))
def cn(n,k,Q):
if(k>n or k<0):
return 0
il,a,b,s = 1
while (n>0 or k>0):
a = n%Q
b = k%Q
s = sn(a,b,Q)
il = (il * s)%Q
k/=Q
n/=Q
return il
def sn(n,k,Q):
if(k>n or k<0):
return 0
return (((fact[n] * nInverse[k]) % Q) * nInverse[n-k]) % Q
```
No
| 90,624 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider an array A with N elements, all being the same integer a.
Define the product transformation as a simultaneous update Ai = AiΒ·Ai + 1, that is multiplying each element to the element right to it for <image>, with the last number AN remaining the same. For example, if we start with an array A with a = 2 and N = 4, then after one product transformation A = [4, 4, 4, 2], and after two product transformations A = [16, 16, 8, 2].
Your simple task is to calculate the array A after M product transformations. Since the numbers can get quite big you should output them modulo Q.
Input
The first and only line of input contains four integers N, M, a, Q (7 β€ Q β€ 109 + 123, 2 β€ a β€ 106 + 123, <image>, <image> is prime), where <image> is the multiplicative order of the integer a modulo Q, see notes for definition.
Output
You should output the array A from left to right.
Example
Input
2 2 2 7
Output
1 2
Note
The multiplicative order of a number a modulo Q <image>, is the smallest natural number x such that ax mod Q = 1. For example, <image>.
Submitted Solution:
```
print("1 2")
```
No
| 90,625 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day Alex was creating a contest about his friends, but accidentally deleted it. Fortunately, all the problems were saved, but now he needs to find them among other problems.
But there are too many problems, to do it manually. Alex asks you to write a program, which will determine if a problem is from this contest by its name.
It is known, that problem is from this contest if and only if its name contains one of Alex's friends' name exactly once. His friends' names are "Danil", "Olya", "Slava", "Ann" and "Nikita".
Names are case sensitive.
Input
The only line contains string from lowercase and uppercase letters and "_" symbols of length, not more than 100 β the name of the problem.
Output
Print "YES", if problem is from this contest, and "NO" otherwise.
Examples
Input
Alex_and_broken_contest
Output
NO
Input
NikitaAndString
Output
YES
Input
Danil_and_Olya
Output
NO
Tags: implementation, strings
Correct Solution:
```
s = input()
# print(s)
l = ["Danil", "Olya", "Slava", "Ann", "Nikita"]
cnt = 0
for n in l:
cnt = cnt + s.count(n)
if cnt == 1:
print("YES")
else:
print("NO")
```
| 90,626 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day Alex was creating a contest about his friends, but accidentally deleted it. Fortunately, all the problems were saved, but now he needs to find them among other problems.
But there are too many problems, to do it manually. Alex asks you to write a program, which will determine if a problem is from this contest by its name.
It is known, that problem is from this contest if and only if its name contains one of Alex's friends' name exactly once. His friends' names are "Danil", "Olya", "Slava", "Ann" and "Nikita".
Names are case sensitive.
Input
The only line contains string from lowercase and uppercase letters and "_" symbols of length, not more than 100 β the name of the problem.
Output
Print "YES", if problem is from this contest, and "NO" otherwise.
Examples
Input
Alex_and_broken_contest
Output
NO
Input
NikitaAndString
Output
YES
Input
Danil_and_Olya
Output
NO
Tags: implementation, strings
Correct Solution:
```
inp = input()
names = ["Danil", "Olya", "Slava", "Ann", "Nikita"]
rs = 0
for name in names:
rs += inp.count(name)
print("YES" if rs == 1 else "NO")
```
| 90,627 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day Alex was creating a contest about his friends, but accidentally deleted it. Fortunately, all the problems were saved, but now he needs to find them among other problems.
But there are too many problems, to do it manually. Alex asks you to write a program, which will determine if a problem is from this contest by its name.
It is known, that problem is from this contest if and only if its name contains one of Alex's friends' name exactly once. His friends' names are "Danil", "Olya", "Slava", "Ann" and "Nikita".
Names are case sensitive.
Input
The only line contains string from lowercase and uppercase letters and "_" symbols of length, not more than 100 β the name of the problem.
Output
Print "YES", if problem is from this contest, and "NO" otherwise.
Examples
Input
Alex_and_broken_contest
Output
NO
Input
NikitaAndString
Output
YES
Input
Danil_and_Olya
Output
NO
Tags: implementation, strings
Correct Solution:
```
str = input()
FRIENDS = ["Danil", "Olya", "Slava", "Ann", "Nikita"]
sum = 0
for name in FRIENDS:
sum += str.count(name)
if sum == 1:
print('YES')
else:
print('NO')
```
| 90,628 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day Alex was creating a contest about his friends, but accidentally deleted it. Fortunately, all the problems were saved, but now he needs to find them among other problems.
But there are too many problems, to do it manually. Alex asks you to write a program, which will determine if a problem is from this contest by its name.
It is known, that problem is from this contest if and only if its name contains one of Alex's friends' name exactly once. His friends' names are "Danil", "Olya", "Slava", "Ann" and "Nikita".
Names are case sensitive.
Input
The only line contains string from lowercase and uppercase letters and "_" symbols of length, not more than 100 β the name of the problem.
Output
Print "YES", if problem is from this contest, and "NO" otherwise.
Examples
Input
Alex_and_broken_contest
Output
NO
Input
NikitaAndString
Output
YES
Input
Danil_and_Olya
Output
NO
Tags: implementation, strings
Correct Solution:
```
ls=["Danil","Olya","Slava","Ann","Nikita"]
s=input()
ctr=0
for a in ls:
ctr+=s.count(a)
if ctr==1:
print("YES")
else:
print("NO")
```
| 90,629 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day Alex was creating a contest about his friends, but accidentally deleted it. Fortunately, all the problems were saved, but now he needs to find them among other problems.
But there are too many problems, to do it manually. Alex asks you to write a program, which will determine if a problem is from this contest by its name.
It is known, that problem is from this contest if and only if its name contains one of Alex's friends' name exactly once. His friends' names are "Danil", "Olya", "Slava", "Ann" and "Nikita".
Names are case sensitive.
Input
The only line contains string from lowercase and uppercase letters and "_" symbols of length, not more than 100 β the name of the problem.
Output
Print "YES", if problem is from this contest, and "NO" otherwise.
Examples
Input
Alex_and_broken_contest
Output
NO
Input
NikitaAndString
Output
YES
Input
Danil_and_Olya
Output
NO
Tags: implementation, strings
Correct Solution:
```
from sys import stdin, stdout
s = stdin.readline().strip()
challengers = ['Danil', 'Olya', 'Slava', 'Ann', 'Nikita']
n = len(s)
cnt = 0
s += '#' * 50
for i in range(n):
for f in challengers:
if s[i: i + len(f)] == f:
cnt += 1
if cnt == 1:
stdout.write('YES')
else:
stdout.write('NO')
```
| 90,630 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day Alex was creating a contest about his friends, but accidentally deleted it. Fortunately, all the problems were saved, but now he needs to find them among other problems.
But there are too many problems, to do it manually. Alex asks you to write a program, which will determine if a problem is from this contest by its name.
It is known, that problem is from this contest if and only if its name contains one of Alex's friends' name exactly once. His friends' names are "Danil", "Olya", "Slava", "Ann" and "Nikita".
Names are case sensitive.
Input
The only line contains string from lowercase and uppercase letters and "_" symbols of length, not more than 100 β the name of the problem.
Output
Print "YES", if problem is from this contest, and "NO" otherwise.
Examples
Input
Alex_and_broken_contest
Output
NO
Input
NikitaAndString
Output
YES
Input
Danil_and_Olya
Output
NO
Tags: implementation, strings
Correct Solution:
```
s = input()
print('YES' if sum(s.count(n) for n in ["Danil", "Olya", "Slava", "Ann", "Nikita"]) == 1 else 'NO')
```
| 90,631 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day Alex was creating a contest about his friends, but accidentally deleted it. Fortunately, all the problems were saved, but now he needs to find them among other problems.
But there are too many problems, to do it manually. Alex asks you to write a program, which will determine if a problem is from this contest by its name.
It is known, that problem is from this contest if and only if its name contains one of Alex's friends' name exactly once. His friends' names are "Danil", "Olya", "Slava", "Ann" and "Nikita".
Names are case sensitive.
Input
The only line contains string from lowercase and uppercase letters and "_" symbols of length, not more than 100 β the name of the problem.
Output
Print "YES", if problem is from this contest, and "NO" otherwise.
Examples
Input
Alex_and_broken_contest
Output
NO
Input
NikitaAndString
Output
YES
Input
Danil_and_Olya
Output
NO
Tags: implementation, strings
Correct Solution:
```
s = input()
lis = ["Danil", "Olya", "Slava", "Ann" , "Nikita"]
r = sum([s.count(i) for i in lis])
print('NO YES'.split()[r==1])
```
| 90,632 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day Alex was creating a contest about his friends, but accidentally deleted it. Fortunately, all the problems were saved, but now he needs to find them among other problems.
But there are too many problems, to do it manually. Alex asks you to write a program, which will determine if a problem is from this contest by its name.
It is known, that problem is from this contest if and only if its name contains one of Alex's friends' name exactly once. His friends' names are "Danil", "Olya", "Slava", "Ann" and "Nikita".
Names are case sensitive.
Input
The only line contains string from lowercase and uppercase letters and "_" symbols of length, not more than 100 β the name of the problem.
Output
Print "YES", if problem is from this contest, and "NO" otherwise.
Examples
Input
Alex_and_broken_contest
Output
NO
Input
NikitaAndString
Output
YES
Input
Danil_and_Olya
Output
NO
Tags: implementation, strings
Correct Solution:
```
def get_ans(s, t):
return s.count(t)
s = input()
a = get_ans(s, "Ann") + get_ans(s, "Danil") + get_ans(s, "Olya") + get_ans(s, "Slava") + get_ans(s, "Nikita")
if a == 1:
print("YES")
else:
print("NO")
```
| 90,633 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day Alex was creating a contest about his friends, but accidentally deleted it. Fortunately, all the problems were saved, but now he needs to find them among other problems.
But there are too many problems, to do it manually. Alex asks you to write a program, which will determine if a problem is from this contest by its name.
It is known, that problem is from this contest if and only if its name contains one of Alex's friends' name exactly once. His friends' names are "Danil", "Olya", "Slava", "Ann" and "Nikita".
Names are case sensitive.
Input
The only line contains string from lowercase and uppercase letters and "_" symbols of length, not more than 100 β the name of the problem.
Output
Print "YES", if problem is from this contest, and "NO" otherwise.
Examples
Input
Alex_and_broken_contest
Output
NO
Input
NikitaAndString
Output
YES
Input
Danil_and_Olya
Output
NO
Submitted Solution:
```
s = input()
a = ['Olya', 'Danil', 'Slava', 'Nikita', 'Ann']
res = 0
for i in range(5):
if a[i] in s:
res += s.count(a[i])
print('YES' if res == 1 else 'NO')
```
Yes
| 90,634 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day Alex was creating a contest about his friends, but accidentally deleted it. Fortunately, all the problems were saved, but now he needs to find them among other problems.
But there are too many problems, to do it manually. Alex asks you to write a program, which will determine if a problem is from this contest by its name.
It is known, that problem is from this contest if and only if its name contains one of Alex's friends' name exactly once. His friends' names are "Danil", "Olya", "Slava", "Ann" and "Nikita".
Names are case sensitive.
Input
The only line contains string from lowercase and uppercase letters and "_" symbols of length, not more than 100 β the name of the problem.
Output
Print "YES", if problem is from this contest, and "NO" otherwise.
Examples
Input
Alex_and_broken_contest
Output
NO
Input
NikitaAndString
Output
YES
Input
Danil_and_Olya
Output
NO
Submitted Solution:
```
# http://codeforces.com/problemset/problem/877/A
def count_in(smstr):
friends = ['Danil', 'Olya', 'Slava', 'Ann', 'Nikita']
countm = 0
for x in friends:
countm += smstr.count(x)
return countm
def main():
inp = input()
f1 = count_in(inp)
if f1 == 1:
return "YES"
return "NO"
if __name__ == "__main__":
print(main())
# input()
```
Yes
| 90,635 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day Alex was creating a contest about his friends, but accidentally deleted it. Fortunately, all the problems were saved, but now he needs to find them among other problems.
But there are too many problems, to do it manually. Alex asks you to write a program, which will determine if a problem is from this contest by its name.
It is known, that problem is from this contest if and only if its name contains one of Alex's friends' name exactly once. His friends' names are "Danil", "Olya", "Slava", "Ann" and "Nikita".
Names are case sensitive.
Input
The only line contains string from lowercase and uppercase letters and "_" symbols of length, not more than 100 β the name of the problem.
Output
Print "YES", if problem is from this contest, and "NO" otherwise.
Examples
Input
Alex_and_broken_contest
Output
NO
Input
NikitaAndString
Output
YES
Input
Danil_and_Olya
Output
NO
Submitted Solution:
```
str=input()
a=str.count("Danil")
a+=str.count("Olya")
a+=str.count("Slava")
a+=str.count("Ann")
a+=str.count("Nikita")
if(a==1):
print("YES")
else:
print("NO")
```
Yes
| 90,636 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day Alex was creating a contest about his friends, but accidentally deleted it. Fortunately, all the problems were saved, but now he needs to find them among other problems.
But there are too many problems, to do it manually. Alex asks you to write a program, which will determine if a problem is from this contest by its name.
It is known, that problem is from this contest if and only if its name contains one of Alex's friends' name exactly once. His friends' names are "Danil", "Olya", "Slava", "Ann" and "Nikita".
Names are case sensitive.
Input
The only line contains string from lowercase and uppercase letters and "_" symbols of length, not more than 100 β the name of the problem.
Output
Print "YES", if problem is from this contest, and "NO" otherwise.
Examples
Input
Alex_and_broken_contest
Output
NO
Input
NikitaAndString
Output
YES
Input
Danil_and_Olya
Output
NO
Submitted Solution:
```
ns = ["Danil", "Olya", "Slava", "Ann", "Nikita"]
s = input()
c = 0
for n in ns:
c += s.count(n)
print("YNEOS"[c != 1::2])
```
Yes
| 90,637 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day Alex was creating a contest about his friends, but accidentally deleted it. Fortunately, all the problems were saved, but now he needs to find them among other problems.
But there are too many problems, to do it manually. Alex asks you to write a program, which will determine if a problem is from this contest by its name.
It is known, that problem is from this contest if and only if its name contains one of Alex's friends' name exactly once. His friends' names are "Danil", "Olya", "Slava", "Ann" and "Nikita".
Names are case sensitive.
Input
The only line contains string from lowercase and uppercase letters and "_" symbols of length, not more than 100 β the name of the problem.
Output
Print "YES", if problem is from this contest, and "NO" otherwise.
Examples
Input
Alex_and_broken_contest
Output
NO
Input
NikitaAndString
Output
YES
Input
Danil_and_Olya
Output
NO
Submitted Solution:
```
name=input()
num=name.count('Danil')
num=name.count('Olya')
num=name.count('Slava')
num=name.count('Ann')
num=name.count('Nikita')
if num==1:
print('YES')
else:
print('NO')
```
No
| 90,638 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day Alex was creating a contest about his friends, but accidentally deleted it. Fortunately, all the problems were saved, but now he needs to find them among other problems.
But there are too many problems, to do it manually. Alex asks you to write a program, which will determine if a problem is from this contest by its name.
It is known, that problem is from this contest if and only if its name contains one of Alex's friends' name exactly once. His friends' names are "Danil", "Olya", "Slava", "Ann" and "Nikita".
Names are case sensitive.
Input
The only line contains string from lowercase and uppercase letters and "_" symbols of length, not more than 100 β the name of the problem.
Output
Print "YES", if problem is from this contest, and "NO" otherwise.
Examples
Input
Alex_and_broken_contest
Output
NO
Input
NikitaAndString
Output
YES
Input
Danil_and_Olya
Output
NO
Submitted Solution:
```
def main():
name = ["DANIL","OLYA","SLAVA","ANN","NIKITA"]
t = 0
x = input()
for i in name:
if i in x.upper():
t += 1
if t == 1:
print("YES")
else:
print("NO")
main()
```
No
| 90,639 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day Alex was creating a contest about his friends, but accidentally deleted it. Fortunately, all the problems were saved, but now he needs to find them among other problems.
But there are too many problems, to do it manually. Alex asks you to write a program, which will determine if a problem is from this contest by its name.
It is known, that problem is from this contest if and only if its name contains one of Alex's friends' name exactly once. His friends' names are "Danil", "Olya", "Slava", "Ann" and "Nikita".
Names are case sensitive.
Input
The only line contains string from lowercase and uppercase letters and "_" symbols of length, not more than 100 β the name of the problem.
Output
Print "YES", if problem is from this contest, and "NO" otherwise.
Examples
Input
Alex_and_broken_contest
Output
NO
Input
NikitaAndString
Output
YES
Input
Danil_and_Olya
Output
NO
Submitted Solution:
```
s = input()
a = ['Danil', 'Olya', 'Slava', 'Ann', 'Nikita']
an = 0
for i in range(len(a)):
if a[i] in s:
an += 1
if an == 0:
print('YES')
else:
print('NO')
```
No
| 90,640 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day Alex was creating a contest about his friends, but accidentally deleted it. Fortunately, all the problems were saved, but now he needs to find them among other problems.
But there are too many problems, to do it manually. Alex asks you to write a program, which will determine if a problem is from this contest by its name.
It is known, that problem is from this contest if and only if its name contains one of Alex's friends' name exactly once. His friends' names are "Danil", "Olya", "Slava", "Ann" and "Nikita".
Names are case sensitive.
Input
The only line contains string from lowercase and uppercase letters and "_" symbols of length, not more than 100 β the name of the problem.
Output
Print "YES", if problem is from this contest, and "NO" otherwise.
Examples
Input
Alex_and_broken_contest
Output
NO
Input
NikitaAndString
Output
YES
Input
Danil_and_Olya
Output
NO
Submitted Solution:
```
import sys
def input(): return sys.stdin.readline().strip()
def iinput(): return int(input())
def minput(): return map(int, input().split())
def listinput(): return list(map(int, input().split()))
s=input()
if s.count("Danil")+s.count("Olya")+s.count("Slava")+s.count("Ann")+s.count("Nikita"):
print("YES")
else:
print("NO")
```
No
| 90,641 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers a, b, c (1 β€ a < b β€ 105, 0 β€ c β€ 9).
Output
Print position of the first occurrence of digit c into the fraction. Positions are numbered from 1 after decimal point. It there is no such position, print -1.
Examples
Input
1 2 0
Output
2
Input
2 3 7
Output
-1
Note
The fraction in the first example has the following decimal notation: <image>. The first zero stands on second position.
The fraction in the second example has the following decimal notation: <image>. There is no digit 7 in decimal notation of the fraction.
Tags: math, number theory
Correct Solution:
```
a, b, c = map(int, input().split())
pos = -1
for i in range(1, 309):
if int((a * pow(10, i) // b)) % 10 == c:
pos = i
break
print(pos)
#128 7 0
```
| 90,642 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers a, b, c (1 β€ a < b β€ 105, 0 β€ c β€ 9).
Output
Print position of the first occurrence of digit c into the fraction. Positions are numbered from 1 after decimal point. It there is no such position, print -1.
Examples
Input
1 2 0
Output
2
Input
2 3 7
Output
-1
Note
The fraction in the first example has the following decimal notation: <image>. The first zero stands on second position.
The fraction in the second example has the following decimal notation: <image>. There is no digit 7 in decimal notation of the fraction.
Tags: math, number theory
Correct Solution:
```
import decimal
a, b, c = map(int, input().split())
decimal.getcontext().prec = 3 * b + 100
s = str(decimal.Decimal(a) / decimal.Decimal(b)).ljust(3 * b + 100, '0')
r = s.find(str(c), 2, 2 * b + 8) - 1
if r == -2:
print(-1)
else:
print(r)
```
| 90,643 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers a, b, c (1 β€ a < b β€ 105, 0 β€ c β€ 9).
Output
Print position of the first occurrence of digit c into the fraction. Positions are numbered from 1 after decimal point. It there is no such position, print -1.
Examples
Input
1 2 0
Output
2
Input
2 3 7
Output
-1
Note
The fraction in the first example has the following decimal notation: <image>. The first zero stands on second position.
The fraction in the second example has the following decimal notation: <image>. There is no digit 7 in decimal notation of the fraction.
Tags: math, number theory
Correct Solution:
```
from decimal import *
getcontext().prec = 300
split_strings = input().split()
[a, b, c] = map(lambda x: int(x), split_strings)
a = Decimal(a)
b = Decimal(b)
decimal = (a / b) * 10
for i in range(1, 300):
ones = int(decimal)
fractional = decimal % 1
if ones == c:
print(i)
exit()
decimal = fractional * 10
print(-1)
```
| 90,644 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers a, b, c (1 β€ a < b β€ 105, 0 β€ c β€ 9).
Output
Print position of the first occurrence of digit c into the fraction. Positions are numbered from 1 after decimal point. It there is no such position, print -1.
Examples
Input
1 2 0
Output
2
Input
2 3 7
Output
-1
Note
The fraction in the first example has the following decimal notation: <image>. The first zero stands on second position.
The fraction in the second example has the following decimal notation: <image>. There is no digit 7 in decimal notation of the fraction.
Tags: math, number theory
Correct Solution:
```
a,b,c = map(int,input().split(" "))
a = a%b
i = 1
while i <= b :
if a==0:
if c==0 :
print(i)
exit()
else :
print(-1)
exit()
if a<b : a *= 10
while(a<b) :
if c==0 :
print(i)
exit()
a *= 10
i += 1
d = a//b
if d==c:
print(i)
exit()
a -= d*b
i += 1
print(-1)
```
| 90,645 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers a, b, c (1 β€ a < b β€ 105, 0 β€ c β€ 9).
Output
Print position of the first occurrence of digit c into the fraction. Positions are numbered from 1 after decimal point. It there is no such position, print -1.
Examples
Input
1 2 0
Output
2
Input
2 3 7
Output
-1
Note
The fraction in the first example has the following decimal notation: <image>. The first zero stands on second position.
The fraction in the second example has the following decimal notation: <image>. There is no digit 7 in decimal notation of the fraction.
Tags: math, number theory
Correct Solution:
```
a, b, c = input().split()
a = int(a)
b = int(b)
c = int(c)
num = []
cnt = 0
while True:
cnt += 1
num.append( a )
n = a*10//b
if n == c:
break
a = a*10 - n*b
if a in num:
cnt = -1
break
print( cnt )
```
| 90,646 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers a, b, c (1 β€ a < b β€ 105, 0 β€ c β€ 9).
Output
Print position of the first occurrence of digit c into the fraction. Positions are numbered from 1 after decimal point. It there is no such position, print -1.
Examples
Input
1 2 0
Output
2
Input
2 3 7
Output
-1
Note
The fraction in the first example has the following decimal notation: <image>. The first zero stands on second position.
The fraction in the second example has the following decimal notation: <image>. There is no digit 7 in decimal notation of the fraction.
Tags: math, number theory
Correct Solution:
```
from decimal import *
getcontext().prec = 10000
a, b, c = map(int, input().split())
d = Decimal(a)/Decimal(b)
# print(d)
s = str(d)[2:]
if len(s) > 900:
s = s[:-1]
else:
s += '0'
# print(s)
try:
ans = s.index(str(c))
except ValueError:
print(-1)
else:
print(ans+1)
```
| 90,647 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers a, b, c (1 β€ a < b β€ 105, 0 β€ c β€ 9).
Output
Print position of the first occurrence of digit c into the fraction. Positions are numbered from 1 after decimal point. It there is no such position, print -1.
Examples
Input
1 2 0
Output
2
Input
2 3 7
Output
-1
Note
The fraction in the first example has the following decimal notation: <image>. The first zero stands on second position.
The fraction in the second example has the following decimal notation: <image>. There is no digit 7 in decimal notation of the fraction.
Tags: math, number theory
Correct Solution:
```
a,b,c=map(int,input().split())
r=a%b
t=b
cnt=1
f=0
while(t!=0):
r*=10
k=r//b
r=r%b
if(k==c):
f=1
break
cnt+=1
t-=1
if(f):
print(cnt)
else:
print(-1)
```
| 90,648 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers a, b, c (1 β€ a < b β€ 105, 0 β€ c β€ 9).
Output
Print position of the first occurrence of digit c into the fraction. Positions are numbered from 1 after decimal point. It there is no such position, print -1.
Examples
Input
1 2 0
Output
2
Input
2 3 7
Output
-1
Note
The fraction in the first example has the following decimal notation: <image>. The first zero stands on second position.
The fraction in the second example has the following decimal notation: <image>. There is no digit 7 in decimal notation of the fraction.
Tags: math, number theory
Correct Solution:
```
from decimal import *
getcontext().prec = 2048
getcontext().rounding = ROUND_DOWN
num, denom, target = [int(x) for x in input().split()]
a = Decimal(num)/Decimal(denom)
a = str(a)
a = a[a.find('.') : ]
if len(a) < 2047:
a = a + '0'
print(a.find(str(target)))
```
| 90,649 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers a, b, c (1 β€ a < b β€ 105, 0 β€ c β€ 9).
Output
Print position of the first occurrence of digit c into the fraction. Positions are numbered from 1 after decimal point. It there is no such position, print -1.
Examples
Input
1 2 0
Output
2
Input
2 3 7
Output
-1
Note
The fraction in the first example has the following decimal notation: <image>. The first zero stands on second position.
The fraction in the second example has the following decimal notation: <image>. There is no digit 7 in decimal notation of the fraction.
Submitted Solution:
```
from fractions import Fraction
(a, b, c) = tuple(map(int, input().split()))
f = Fraction(a, b)
for i in range(900):
for j in range(10):
if f < Fraction(j+1, 10):
if j == c:
print(i+1)
exit(0)
f -= Fraction(j, 10)
f *= 10
break
print(-1)
```
Yes
| 90,650 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers a, b, c (1 β€ a < b β€ 105, 0 β€ c β€ 9).
Output
Print position of the first occurrence of digit c into the fraction. Positions are numbered from 1 after decimal point. It there is no such position, print -1.
Examples
Input
1 2 0
Output
2
Input
2 3 7
Output
-1
Note
The fraction in the first example has the following decimal notation: <image>. The first zero stands on second position.
The fraction in the second example has the following decimal notation: <image>. There is no digit 7 in decimal notation of the fraction.
Submitted Solution:
```
from decimal import *
def main():
getcontext().prec = 500
a, b, c = map(int, input().split())
d = str(Decimal(a) / Decimal(b))
d += "0" * (502 - (len(d)))
cnt = [-2] * 10
for idx, item in enumerate(d[2:-1]):
if cnt[int(item)] == -2:
cnt[int(item)] = idx
print(cnt[c] + 1)
if __name__ == "__main__":
main()
```
Yes
| 90,651 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers a, b, c (1 β€ a < b β€ 105, 0 β€ c β€ 9).
Output
Print position of the first occurrence of digit c into the fraction. Positions are numbered from 1 after decimal point. It there is no such position, print -1.
Examples
Input
1 2 0
Output
2
Input
2 3 7
Output
-1
Note
The fraction in the first example has the following decimal notation: <image>. The first zero stands on second position.
The fraction in the second example has the following decimal notation: <image>. There is no digit 7 in decimal notation of the fraction.
Submitted Solution:
```
arr = list(map(int, input().split()))
a=arr[0]
b=arr[1]
c=arr[2]
for i in range(1,b+1):
a*=10
if a//b==c:
print(i)
break
a%=b
else:
print(-1)
```
Yes
| 90,652 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers a, b, c (1 β€ a < b β€ 105, 0 β€ c β€ 9).
Output
Print position of the first occurrence of digit c into the fraction. Positions are numbered from 1 after decimal point. It there is no such position, print -1.
Examples
Input
1 2 0
Output
2
Input
2 3 7
Output
-1
Note
The fraction in the first example has the following decimal notation: <image>. The first zero stands on second position.
The fraction in the second example has the following decimal notation: <image>. There is no digit 7 in decimal notation of the fraction.
Submitted Solution:
```
a, b, c = input().split()
a = int(a)
b = int(b)
c = int(c)
d = (10**10000) * a // b
x = 10 ** (10000 - 1)
for i in range(1, 1001, 1):
if d // x == c:
print(i)
exit(0)
d %= x
x //= 10
print(-1)
```
Yes
| 90,653 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers a, b, c (1 β€ a < b β€ 105, 0 β€ c β€ 9).
Output
Print position of the first occurrence of digit c into the fraction. Positions are numbered from 1 after decimal point. It there is no such position, print -1.
Examples
Input
1 2 0
Output
2
Input
2 3 7
Output
-1
Note
The fraction in the first example has the following decimal notation: <image>. The first zero stands on second position.
The fraction in the second example has the following decimal notation: <image>. There is no digit 7 in decimal notation of the fraction.
Submitted Solution:
```
a,b,c=map(int,input().split())
a=(a%b)
count=0
l=0
m=0
n=0
k=0
while True:
n=k
a*=10
k=a//b
a=(a%b)
count+=1
if k==c:
l+=1
print(count)
break
if a==0 and c!=0 :
m+=1
break
if n==k and c==k:
print(count-1)
break
if n==k:
m+=1
break
if m>l:
print('-1')
```
No
| 90,654 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers a, b, c (1 β€ a < b β€ 105, 0 β€ c β€ 9).
Output
Print position of the first occurrence of digit c into the fraction. Positions are numbered from 1 after decimal point. It there is no such position, print -1.
Examples
Input
1 2 0
Output
2
Input
2 3 7
Output
-1
Note
The fraction in the first example has the following decimal notation: <image>. The first zero stands on second position.
The fraction in the second example has the following decimal notation: <image>. There is no digit 7 in decimal notation of the fraction.
Submitted Solution:
```
a, b,c = map(int, input().split())
ans = float(a/b)
ans = format(ans, '.20f')
ans = str(ans)
ans = ans[2:len(ans)]
if(ans.count(str(c)) > 0):
print(str(ans.index(str(c))+1))
else:
print("-1")
```
No
| 90,655 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers a, b, c (1 β€ a < b β€ 105, 0 β€ c β€ 9).
Output
Print position of the first occurrence of digit c into the fraction. Positions are numbered from 1 after decimal point. It there is no such position, print -1.
Examples
Input
1 2 0
Output
2
Input
2 3 7
Output
-1
Note
The fraction in the first example has the following decimal notation: <image>. The first zero stands on second position.
The fraction in the second example has the following decimal notation: <image>. There is no digit 7 in decimal notation of the fraction.
Submitted Solution:
```
a, b, c = map(int, input().split())
for i in range (b):
z = a*10
if z // b == c :
print (i+1)
break
a %= b
else: print(-1)
```
No
| 90,656 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers a, b, c (1 β€ a < b β€ 105, 0 β€ c β€ 9).
Output
Print position of the first occurrence of digit c into the fraction. Positions are numbered from 1 after decimal point. It there is no such position, print -1.
Examples
Input
1 2 0
Output
2
Input
2 3 7
Output
-1
Note
The fraction in the first example has the following decimal notation: <image>. The first zero stands on second position.
The fraction in the second example has the following decimal notation: <image>. There is no digit 7 in decimal notation of the fraction.
Submitted Solution:
```
a,b,c=input().split()
a=int(a)
b=int(b)
c=int(c)
l=[]
rem=a
j=0
while 1:
ans=rem*10//b
l.append(ans)
j+=1
rem=rem*10%b
if j==1:
continue
if ans==l[0] or rem==0:
if rem==0:
l.append(0)
break
j=1
find=False
for i in l:
if c==i:
print(j)
find=True
break
j+=1
if not find:
print(-1)
```
No
| 90,657 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Imp is really pleased that you helped him. But it you solve the last problem, his gladness would raise even more.
<image> Let's define <image> for some set of integers <image> as the number of pairs a, b in <image>, such that:
* a is strictly less than b;
* a divides b without a remainder.
You are to find such a set <image>, which is a subset of {1, 2, ..., n} (the set that contains all positive integers not greater than n), that <image>.
Input
The only line contains two integers n and k <image>.
Output
If there is no answer, print "No".
Otherwise, in the first line print "Yes", in the second β an integer m that denotes the size of the set <image> you have found, in the second line print m integers β the elements of the set <image>, in any order.
If there are multiple answers, print any of them.
Examples
Input
3 3
Output
No
Input
6 6
Output
Yes
5
1 2 4 5 6
Input
8 3
Output
Yes
4
2 4 5 8
Note
In the second sample, the valid pairs in the output set are (1, 2), (1, 4), (1, 5), (1, 6), (2, 4), (2, 6). Thus, <image>.
In the third example, the valid pairs in the output set are (2, 4), (4, 8), (2, 8). Thus, <image>.
Submitted Solution:
```
import math
def div(n,lim) :
ans=0
for i in range(1, int(math.sqrt(n) + 1)) :
if (n % i == 0) :
# Check if divisors are equal
if (n / i == i) :
if i>=lim:
ans+=1
else :
# Otherwise print both
if i>=lim:
ans+=1
if (int(n / i))>=lim:
ans+=1
return ans
global x
x=[]
n,k=list(map(int,input().split()))
if n==300000:
print("No")
else:
s=[0 for i in range(n//2)]
for i in range(n//2):
s[i]=(n//(i+1))-1
a=0
lim=0
for i in range(1,len(s)+1):
a+=s[-i]
if a>=k:
lim=len(s)-i+1
break
dif=a-k
r=n//2
r+=1
no=[]
f=[]
for i in range((n//2)+1,n+1):
x=div(i,lim)-1
if x<=dif:
dif-=x
no.append(i)
if dif==0:
break
if dif!=0 or n==300000 :
print("No")
else:
print("Yes")
ans=[i for i in range(lim,n+1)]
for i in no:
ans.remove(i)
print(len(ans))
for i in ans:
print(i,end=" ")
```
No
| 90,658 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Imp is really pleased that you helped him. But it you solve the last problem, his gladness would raise even more.
<image> Let's define <image> for some set of integers <image> as the number of pairs a, b in <image>, such that:
* a is strictly less than b;
* a divides b without a remainder.
You are to find such a set <image>, which is a subset of {1, 2, ..., n} (the set that contains all positive integers not greater than n), that <image>.
Input
The only line contains two integers n and k <image>.
Output
If there is no answer, print "No".
Otherwise, in the first line print "Yes", in the second β an integer m that denotes the size of the set <image> you have found, in the second line print m integers β the elements of the set <image>, in any order.
If there are multiple answers, print any of them.
Examples
Input
3 3
Output
No
Input
6 6
Output
Yes
5
1 2 4 5 6
Input
8 3
Output
Yes
4
2 4 5 8
Note
In the second sample, the valid pairs in the output set are (1, 2), (1, 4), (1, 5), (1, 6), (2, 4), (2, 6). Thus, <image>.
In the third example, the valid pairs in the output set are (2, 4), (4, 8), (2, 8). Thus, <image>.
Submitted Solution:
```
global x
x=[]
def criba(n):
global x,prime
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * 2, n+1, p):
prime[i] = False
p += 1
for i in range(2,n+1):
if prime[i]:
x.append(i)
n,k=list(map(int,input().split()))
criba(n)
s=[0 for i in range(n)]
for i in range(n//2):
s[i]=(n//(i+1))-1
i=n//2
i=i-1
y=0
while i>=0 and y<k:
y+=s[i]
i-=1
dif=y-k
lim=i+2
no=[]
ans=[]
if dif<0:
print("No")
else:
print("Yes")
for j in range(len(x)):
if i==-1:
if s[x[j]-1]+1<=dif:
dif-=s[x[j]-1]+1
no.append(x[j])
else:
if s[x[j]-1]<=dif:
dif-=s[x[j]-1]
no.append(x[j])
if dif==0: break
for j in range(lim,n+1):
if len(no)==0 or j!=no[0]:
ans.append(j)
else:
no.remove(no[0])
if dif!=0:
t=lim**2
t=t+lim
t1=lim**3
while dif!=0:
if t<t1:
dif-=1
no.append(t)
else:
t=t1+lim
t1=t1*lim
t+=lim
ans1=[]
for j in ans:
if len(no)==0 or j!=no[0]:
ans1.append(j)
else:
no.remove(no[0])
print(len(ans1))
for i in ans1:
print(i,end=" ")
else:
for i in ans:
print(i,end=" ")
```
No
| 90,659 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Imp is really pleased that you helped him. But it you solve the last problem, his gladness would raise even more.
<image> Let's define <image> for some set of integers <image> as the number of pairs a, b in <image>, such that:
* a is strictly less than b;
* a divides b without a remainder.
You are to find such a set <image>, which is a subset of {1, 2, ..., n} (the set that contains all positive integers not greater than n), that <image>.
Input
The only line contains two integers n and k <image>.
Output
If there is no answer, print "No".
Otherwise, in the first line print "Yes", in the second β an integer m that denotes the size of the set <image> you have found, in the second line print m integers β the elements of the set <image>, in any order.
If there are multiple answers, print any of them.
Examples
Input
3 3
Output
No
Input
6 6
Output
Yes
5
1 2 4 5 6
Input
8 3
Output
Yes
4
2 4 5 8
Note
In the second sample, the valid pairs in the output set are (1, 2), (1, 4), (1, 5), (1, 6), (2, 4), (2, 6). Thus, <image>.
In the third example, the valid pairs in the output set are (2, 4), (4, 8), (2, 8). Thus, <image>.
Submitted Solution:
```
n,k=list(map(int,input().split()))
if ( n==300000 and k==1000000000)or ( n==300000 and k==3529833):
if ( n==300000 and k==1000000000):
print("No")
else:
print(n)
for i in range(1,n+1):
print(i, end=" ")
else:
s=[1 for i in range(n)]
s[0]=0
for i in range(2,n//2+1):
x=i+i
while x<=n:
s[x-1]+=1
x+=i
k1=0; lim=0
for i in range(n):
k1+=s[i]
if k1 >=k:
lim =i
break
dif=k1-k
if dif<0 :
print("No")
else:
print("Yes")
for i in range(lim//2+1):
s[i]+=(lim+1)//(i+1)-1
t=True
print(lim)
for i in range(lim+1):
if t:
if s[i]!=dif:
print(i+1,end=" ")
else:
t=False
else:
print(i+1,end=" ")
```
No
| 90,660 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Imp is really pleased that you helped him. But it you solve the last problem, his gladness would raise even more.
<image> Let's define <image> for some set of integers <image> as the number of pairs a, b in <image>, such that:
* a is strictly less than b;
* a divides b without a remainder.
You are to find such a set <image>, which is a subset of {1, 2, ..., n} (the set that contains all positive integers not greater than n), that <image>.
Input
The only line contains two integers n and k <image>.
Output
If there is no answer, print "No".
Otherwise, in the first line print "Yes", in the second β an integer m that denotes the size of the set <image> you have found, in the second line print m integers β the elements of the set <image>, in any order.
If there are multiple answers, print any of them.
Examples
Input
3 3
Output
No
Input
6 6
Output
Yes
5
1 2 4 5 6
Input
8 3
Output
Yes
4
2 4 5 8
Note
In the second sample, the valid pairs in the output set are (1, 2), (1, 4), (1, 5), (1, 6), (2, 4), (2, 6). Thus, <image>.
In the third example, the valid pairs in the output set are (2, 4), (4, 8), (2, 8). Thus, <image>.
Submitted Solution:
```
global x
x=[]
def criba(n):
global x,prime
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * 2, n+1, p):
prime[i] = False
p += 1
for i in range((n//2)+1,n+1):
if prime[i]:
x.append(i)
n,k=list(map(int,input().split()))
s=[0 for i in range(n)]
for i in range(n):
s[i]=(n//(i+1))-1
a=0
lim=0
for i in range(n//2):
a+=s[i]
if a>=k:
lim=i+1
break
dif=a-k
criba(n)
no=[]
print(dif)
for i in x:
if dif>0 and i>lim:
dif-=1
no.append(i)
elif dif==0:
break
if dif!=0:
print("No")
else:
print("Yes")
ans=[i for i in range(1,lim+1)]
ans1=[i for i in range((n//2)+1,n+1)]
ans=ans+ans1
for i in no:
ans.remove(i)
print(len(ans))
for i in ans:
print(i,end=" ")
```
No
| 90,661 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Hacker Zhorik wants to decipher two secret messages he intercepted yesterday. Yeah message is a sequence of encrypted blocks, each of them consists of several bytes of information.
Zhorik knows that each of the messages is an archive containing one or more files. Zhorik knows how each of these archives was transferred through the network: if an archive consists of k files of sizes l1, l2, ..., lk bytes, then the i-th file is split to one or more blocks bi, 1, bi, 2, ..., bi, mi (here the total length of the blocks bi, 1 + bi, 2 + ... + bi, mi is equal to the length of the file li), and after that all blocks are transferred through the network, maintaining the order of files in the archive.
Zhorik thinks that the two messages contain the same archive, because their total lengths are equal. However, each file can be split in blocks in different ways in the two messages.
You are given the lengths of blocks in each of the two messages. Help Zhorik to determine what is the maximum number of files could be in the archive, if the Zhorik's assumption is correct.
Input
The first line contains two integers n, m (1 β€ n, m β€ 105) β the number of blocks in the first and in the second messages.
The second line contains n integers x1, x2, ..., xn (1 β€ xi β€ 106) β the length of the blocks that form the first message.
The third line contains m integers y1, y2, ..., ym (1 β€ yi β€ 106) β the length of the blocks that form the second message.
It is guaranteed that x1 + ... + xn = y1 + ... + ym. Also, it is guaranteed that x1 + ... + xn β€ 106.
Output
Print the maximum number of files the intercepted array could consist of.
Examples
Input
7 6
2 5 3 1 11 4 4
7 8 2 4 1 8
Output
3
Input
3 3
1 10 100
1 100 10
Output
2
Input
1 4
4
1 1 1 1
Output
1
Note
In the first example the maximum number of files in the archive is 3. For example, it is possible that in the archive are three files of sizes 2 + 5 = 7, 15 = 3 + 1 + 11 = 8 + 2 + 4 + 1 and 4 + 4 = 8.
In the second example it is possible that the archive contains two files of sizes 1 and 110 = 10 + 100 = 100 + 10. Note that the order of files is kept while transferring archives through the network, so we can't say that there are three files of sizes 1, 10 and 100.
In the third example the only possibility is that the archive contains a single file of size 4.
Tags: greedy, implementation
Correct Solution:
```
n, m = map(int, input().split())
xs = list(map(int, input().split()))
ys = list(map(int, input().split()))
sm1 = sm2 = 0
i = j = 0
cnt = 0
moveX = moveY = True
while i < len(xs):
if moveX:
sm1 += xs[i]
if moveY:
sm2 += ys[j]
if sm1 < sm2:
i += 1
moveX = True
moveY = False
elif sm1 > sm2:
j += 1
moveX = False
moveY = True
else:
sm1 = sm2 = 0
cnt += 1
i += 1
j += 1
moveX = True
moveY = True
print(cnt)
```
| 90,662 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Hacker Zhorik wants to decipher two secret messages he intercepted yesterday. Yeah message is a sequence of encrypted blocks, each of them consists of several bytes of information.
Zhorik knows that each of the messages is an archive containing one or more files. Zhorik knows how each of these archives was transferred through the network: if an archive consists of k files of sizes l1, l2, ..., lk bytes, then the i-th file is split to one or more blocks bi, 1, bi, 2, ..., bi, mi (here the total length of the blocks bi, 1 + bi, 2 + ... + bi, mi is equal to the length of the file li), and after that all blocks are transferred through the network, maintaining the order of files in the archive.
Zhorik thinks that the two messages contain the same archive, because their total lengths are equal. However, each file can be split in blocks in different ways in the two messages.
You are given the lengths of blocks in each of the two messages. Help Zhorik to determine what is the maximum number of files could be in the archive, if the Zhorik's assumption is correct.
Input
The first line contains two integers n, m (1 β€ n, m β€ 105) β the number of blocks in the first and in the second messages.
The second line contains n integers x1, x2, ..., xn (1 β€ xi β€ 106) β the length of the blocks that form the first message.
The third line contains m integers y1, y2, ..., ym (1 β€ yi β€ 106) β the length of the blocks that form the second message.
It is guaranteed that x1 + ... + xn = y1 + ... + ym. Also, it is guaranteed that x1 + ... + xn β€ 106.
Output
Print the maximum number of files the intercepted array could consist of.
Examples
Input
7 6
2 5 3 1 11 4 4
7 8 2 4 1 8
Output
3
Input
3 3
1 10 100
1 100 10
Output
2
Input
1 4
4
1 1 1 1
Output
1
Note
In the first example the maximum number of files in the archive is 3. For example, it is possible that in the archive are three files of sizes 2 + 5 = 7, 15 = 3 + 1 + 11 = 8 + 2 + 4 + 1 and 4 + 4 = 8.
In the second example it is possible that the archive contains two files of sizes 1 and 110 = 10 + 100 = 100 + 10. Note that the order of files is kept while transferring archives through the network, so we can't say that there are three files of sizes 1, 10 and 100.
In the third example the only possibility is that the archive contains a single file of size 4.
Tags: greedy, implementation
Correct Solution:
```
n,m = map(int, input().split())
x = list(map(int, input().split()))
y = list(map(int, input().split()))
i1 = 0
i2 = 0
sum1 = x[0]
sum2 = y[0]
count = 0
while i1 != n and i2 != m:
while sum1 < sum2:
i1 += 1
if i1 == n:
break
sum1 += x[i1]
if i1 == n:
break
if (sum1 == sum2):
count += 1
i1 += 1
if i1 == n:
break
sum1 += x[i1]
while sum2 < sum1:
i2 += 1
if i2 == m:
break
sum2 += y[i2]
if i1 == n:
break
if (sum1 == sum2):
count += 1
i2 += 1
if i2 == m:
break
sum2 += y[i2]
print(count)
```
| 90,663 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Hacker Zhorik wants to decipher two secret messages he intercepted yesterday. Yeah message is a sequence of encrypted blocks, each of them consists of several bytes of information.
Zhorik knows that each of the messages is an archive containing one or more files. Zhorik knows how each of these archives was transferred through the network: if an archive consists of k files of sizes l1, l2, ..., lk bytes, then the i-th file is split to one or more blocks bi, 1, bi, 2, ..., bi, mi (here the total length of the blocks bi, 1 + bi, 2 + ... + bi, mi is equal to the length of the file li), and after that all blocks are transferred through the network, maintaining the order of files in the archive.
Zhorik thinks that the two messages contain the same archive, because their total lengths are equal. However, each file can be split in blocks in different ways in the two messages.
You are given the lengths of blocks in each of the two messages. Help Zhorik to determine what is the maximum number of files could be in the archive, if the Zhorik's assumption is correct.
Input
The first line contains two integers n, m (1 β€ n, m β€ 105) β the number of blocks in the first and in the second messages.
The second line contains n integers x1, x2, ..., xn (1 β€ xi β€ 106) β the length of the blocks that form the first message.
The third line contains m integers y1, y2, ..., ym (1 β€ yi β€ 106) β the length of the blocks that form the second message.
It is guaranteed that x1 + ... + xn = y1 + ... + ym. Also, it is guaranteed that x1 + ... + xn β€ 106.
Output
Print the maximum number of files the intercepted array could consist of.
Examples
Input
7 6
2 5 3 1 11 4 4
7 8 2 4 1 8
Output
3
Input
3 3
1 10 100
1 100 10
Output
2
Input
1 4
4
1 1 1 1
Output
1
Note
In the first example the maximum number of files in the archive is 3. For example, it is possible that in the archive are three files of sizes 2 + 5 = 7, 15 = 3 + 1 + 11 = 8 + 2 + 4 + 1 and 4 + 4 = 8.
In the second example it is possible that the archive contains two files of sizes 1 and 110 = 10 + 100 = 100 + 10. Note that the order of files is kept while transferring archives through the network, so we can't say that there are three files of sizes 1, 10 and 100.
In the third example the only possibility is that the archive contains a single file of size 4.
Tags: greedy, implementation
Correct Solution:
```
n, m = map(int, input().split())
x = list(map(int, input().split()))
y = list(map(int, input().split()))
ans = 0
i = 0
j = 0
sumx = 0
sumy = 0
while i < n and j < m:
if sumx <= sumy:
sumx += x[i]
i += 1
if sumy < sumx:
sumy += y[j]
j += 1
if sumx == sumy:
ans += 1
sumx = 0
sumy = 0
if i < n or j < m:
ans += 1
print(ans)
```
| 90,664 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Hacker Zhorik wants to decipher two secret messages he intercepted yesterday. Yeah message is a sequence of encrypted blocks, each of them consists of several bytes of information.
Zhorik knows that each of the messages is an archive containing one or more files. Zhorik knows how each of these archives was transferred through the network: if an archive consists of k files of sizes l1, l2, ..., lk bytes, then the i-th file is split to one or more blocks bi, 1, bi, 2, ..., bi, mi (here the total length of the blocks bi, 1 + bi, 2 + ... + bi, mi is equal to the length of the file li), and after that all blocks are transferred through the network, maintaining the order of files in the archive.
Zhorik thinks that the two messages contain the same archive, because their total lengths are equal. However, each file can be split in blocks in different ways in the two messages.
You are given the lengths of blocks in each of the two messages. Help Zhorik to determine what is the maximum number of files could be in the archive, if the Zhorik's assumption is correct.
Input
The first line contains two integers n, m (1 β€ n, m β€ 105) β the number of blocks in the first and in the second messages.
The second line contains n integers x1, x2, ..., xn (1 β€ xi β€ 106) β the length of the blocks that form the first message.
The third line contains m integers y1, y2, ..., ym (1 β€ yi β€ 106) β the length of the blocks that form the second message.
It is guaranteed that x1 + ... + xn = y1 + ... + ym. Also, it is guaranteed that x1 + ... + xn β€ 106.
Output
Print the maximum number of files the intercepted array could consist of.
Examples
Input
7 6
2 5 3 1 11 4 4
7 8 2 4 1 8
Output
3
Input
3 3
1 10 100
1 100 10
Output
2
Input
1 4
4
1 1 1 1
Output
1
Note
In the first example the maximum number of files in the archive is 3. For example, it is possible that in the archive are three files of sizes 2 + 5 = 7, 15 = 3 + 1 + 11 = 8 + 2 + 4 + 1 and 4 + 4 = 8.
In the second example it is possible that the archive contains two files of sizes 1 and 110 = 10 + 100 = 100 + 10. Note that the order of files is kept while transferring archives through the network, so we can't say that there are three files of sizes 1, 10 and 100.
In the third example the only possibility is that the archive contains a single file of size 4.
Tags: greedy, implementation
Correct Solution:
```
n, m = list(map(int, input().split()))
x = list(map(int, input().split()))
y = list(map(int, input().split()))
filesCnt = 0
xs = 0
ys = 0
xcnt = 0
ycnt = 0
while n >= xcnt or m >= ycnt:
# print(xs, ys)
if xs == ys and xs != 0:
filesCnt += 1
xs = 0
ys = 0
if n > xcnt or m > ycnt:
continue
else:
break
else:
if xs > ys:
ys += y[ycnt]
ycnt += 1
else:
xs += x[xcnt]
xcnt += 1
print(filesCnt)
```
| 90,665 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Hacker Zhorik wants to decipher two secret messages he intercepted yesterday. Yeah message is a sequence of encrypted blocks, each of them consists of several bytes of information.
Zhorik knows that each of the messages is an archive containing one or more files. Zhorik knows how each of these archives was transferred through the network: if an archive consists of k files of sizes l1, l2, ..., lk bytes, then the i-th file is split to one or more blocks bi, 1, bi, 2, ..., bi, mi (here the total length of the blocks bi, 1 + bi, 2 + ... + bi, mi is equal to the length of the file li), and after that all blocks are transferred through the network, maintaining the order of files in the archive.
Zhorik thinks that the two messages contain the same archive, because their total lengths are equal. However, each file can be split in blocks in different ways in the two messages.
You are given the lengths of blocks in each of the two messages. Help Zhorik to determine what is the maximum number of files could be in the archive, if the Zhorik's assumption is correct.
Input
The first line contains two integers n, m (1 β€ n, m β€ 105) β the number of blocks in the first and in the second messages.
The second line contains n integers x1, x2, ..., xn (1 β€ xi β€ 106) β the length of the blocks that form the first message.
The third line contains m integers y1, y2, ..., ym (1 β€ yi β€ 106) β the length of the blocks that form the second message.
It is guaranteed that x1 + ... + xn = y1 + ... + ym. Also, it is guaranteed that x1 + ... + xn β€ 106.
Output
Print the maximum number of files the intercepted array could consist of.
Examples
Input
7 6
2 5 3 1 11 4 4
7 8 2 4 1 8
Output
3
Input
3 3
1 10 100
1 100 10
Output
2
Input
1 4
4
1 1 1 1
Output
1
Note
In the first example the maximum number of files in the archive is 3. For example, it is possible that in the archive are three files of sizes 2 + 5 = 7, 15 = 3 + 1 + 11 = 8 + 2 + 4 + 1 and 4 + 4 = 8.
In the second example it is possible that the archive contains two files of sizes 1 and 110 = 10 + 100 = 100 + 10. Note that the order of files is kept while transferring archives through the network, so we can't say that there are three files of sizes 1, 10 and 100.
In the third example the only possibility is that the archive contains a single file of size 4.
Tags: greedy, implementation
Correct Solution:
```
n, m = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
i = 0
j = 0
s = a[0]
t = b[0]
k = 0
while i != n and j != m:
while s < t:
i += 1
s += a[i]
while t < s:
j += 1
t += b[j]
if s == t:
i += 1
j += 1
k += 1
if i != n:
s = a[i]
t = b[j]
print(k)
```
| 90,666 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Hacker Zhorik wants to decipher two secret messages he intercepted yesterday. Yeah message is a sequence of encrypted blocks, each of them consists of several bytes of information.
Zhorik knows that each of the messages is an archive containing one or more files. Zhorik knows how each of these archives was transferred through the network: if an archive consists of k files of sizes l1, l2, ..., lk bytes, then the i-th file is split to one or more blocks bi, 1, bi, 2, ..., bi, mi (here the total length of the blocks bi, 1 + bi, 2 + ... + bi, mi is equal to the length of the file li), and after that all blocks are transferred through the network, maintaining the order of files in the archive.
Zhorik thinks that the two messages contain the same archive, because their total lengths are equal. However, each file can be split in blocks in different ways in the two messages.
You are given the lengths of blocks in each of the two messages. Help Zhorik to determine what is the maximum number of files could be in the archive, if the Zhorik's assumption is correct.
Input
The first line contains two integers n, m (1 β€ n, m β€ 105) β the number of blocks in the first and in the second messages.
The second line contains n integers x1, x2, ..., xn (1 β€ xi β€ 106) β the length of the blocks that form the first message.
The third line contains m integers y1, y2, ..., ym (1 β€ yi β€ 106) β the length of the blocks that form the second message.
It is guaranteed that x1 + ... + xn = y1 + ... + ym. Also, it is guaranteed that x1 + ... + xn β€ 106.
Output
Print the maximum number of files the intercepted array could consist of.
Examples
Input
7 6
2 5 3 1 11 4 4
7 8 2 4 1 8
Output
3
Input
3 3
1 10 100
1 100 10
Output
2
Input
1 4
4
1 1 1 1
Output
1
Note
In the first example the maximum number of files in the archive is 3. For example, it is possible that in the archive are three files of sizes 2 + 5 = 7, 15 = 3 + 1 + 11 = 8 + 2 + 4 + 1 and 4 + 4 = 8.
In the second example it is possible that the archive contains two files of sizes 1 and 110 = 10 + 100 = 100 + 10. Note that the order of files is kept while transferring archives through the network, so we can't say that there are three files of sizes 1, 10 and 100.
In the third example the only possibility is that the archive contains a single file of size 4.
Tags: greedy, implementation
Correct Solution:
```
n,m = map(int,input().split(' '))
a = list(map(int,input().split(' ')))
b = list(map(int,input().split(' ')))
i = j = 0
sa = a[0]
sb = b[0]
k = 0
while i<n-1 and j<m-1:
if sa<sb:
i+=1
sa+=a[i]
elif sb<sa:
j+=1
sb+=b[j]
else:
i+=1
j+=1
sa+=a[i]
sb+=b[j]
k+=1
if i<n or j<m:
k+=1
print(k)
```
| 90,667 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Hacker Zhorik wants to decipher two secret messages he intercepted yesterday. Yeah message is a sequence of encrypted blocks, each of them consists of several bytes of information.
Zhorik knows that each of the messages is an archive containing one or more files. Zhorik knows how each of these archives was transferred through the network: if an archive consists of k files of sizes l1, l2, ..., lk bytes, then the i-th file is split to one or more blocks bi, 1, bi, 2, ..., bi, mi (here the total length of the blocks bi, 1 + bi, 2 + ... + bi, mi is equal to the length of the file li), and after that all blocks are transferred through the network, maintaining the order of files in the archive.
Zhorik thinks that the two messages contain the same archive, because their total lengths are equal. However, each file can be split in blocks in different ways in the two messages.
You are given the lengths of blocks in each of the two messages. Help Zhorik to determine what is the maximum number of files could be in the archive, if the Zhorik's assumption is correct.
Input
The first line contains two integers n, m (1 β€ n, m β€ 105) β the number of blocks in the first and in the second messages.
The second line contains n integers x1, x2, ..., xn (1 β€ xi β€ 106) β the length of the blocks that form the first message.
The third line contains m integers y1, y2, ..., ym (1 β€ yi β€ 106) β the length of the blocks that form the second message.
It is guaranteed that x1 + ... + xn = y1 + ... + ym. Also, it is guaranteed that x1 + ... + xn β€ 106.
Output
Print the maximum number of files the intercepted array could consist of.
Examples
Input
7 6
2 5 3 1 11 4 4
7 8 2 4 1 8
Output
3
Input
3 3
1 10 100
1 100 10
Output
2
Input
1 4
4
1 1 1 1
Output
1
Note
In the first example the maximum number of files in the archive is 3. For example, it is possible that in the archive are three files of sizes 2 + 5 = 7, 15 = 3 + 1 + 11 = 8 + 2 + 4 + 1 and 4 + 4 = 8.
In the second example it is possible that the archive contains two files of sizes 1 and 110 = 10 + 100 = 100 + 10. Note that the order of files is kept while transferring archives through the network, so we can't say that there are three files of sizes 1, 10 and 100.
In the third example the only possibility is that the archive contains a single file of size 4.
Tags: greedy, implementation
Correct Solution:
```
import os
import sys
debug = True
if debug and os.path.exists("input.in"):
input = open("input.in", "r").readline
else:
debug = False
input = sys.stdin.readline
def inp():
return (int(input()))
def inlt():
return (list(map(int, input().split())))
def insr():
s = input()
return s[:len(s) - 1] # Remove line char from end
def invr():
return (map(int, input().split()))
test_count = 1
if debug:
test_count = inp()
for t in range(test_count):
if debug:
print("Test Case #", t + 1)
# Start code here
ans = 0
n, m = invr()
a = inlt()
b = inlt()
i = 0
j = 0
a_sum = a[0]
b_sum = b[0]
while i < n and j < m:
if a_sum < b_sum:
i += 1
a_sum += a[i]
elif a_sum > b_sum:
j += 1
b_sum += b[j]
else:
ans += 1
i += 1
j += 1
if i == n and j == m:
break
a_sum += a[i]
b_sum += b[j]
print(ans)
```
| 90,668 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Hacker Zhorik wants to decipher two secret messages he intercepted yesterday. Yeah message is a sequence of encrypted blocks, each of them consists of several bytes of information.
Zhorik knows that each of the messages is an archive containing one or more files. Zhorik knows how each of these archives was transferred through the network: if an archive consists of k files of sizes l1, l2, ..., lk bytes, then the i-th file is split to one or more blocks bi, 1, bi, 2, ..., bi, mi (here the total length of the blocks bi, 1 + bi, 2 + ... + bi, mi is equal to the length of the file li), and after that all blocks are transferred through the network, maintaining the order of files in the archive.
Zhorik thinks that the two messages contain the same archive, because their total lengths are equal. However, each file can be split in blocks in different ways in the two messages.
You are given the lengths of blocks in each of the two messages. Help Zhorik to determine what is the maximum number of files could be in the archive, if the Zhorik's assumption is correct.
Input
The first line contains two integers n, m (1 β€ n, m β€ 105) β the number of blocks in the first and in the second messages.
The second line contains n integers x1, x2, ..., xn (1 β€ xi β€ 106) β the length of the blocks that form the first message.
The third line contains m integers y1, y2, ..., ym (1 β€ yi β€ 106) β the length of the blocks that form the second message.
It is guaranteed that x1 + ... + xn = y1 + ... + ym. Also, it is guaranteed that x1 + ... + xn β€ 106.
Output
Print the maximum number of files the intercepted array could consist of.
Examples
Input
7 6
2 5 3 1 11 4 4
7 8 2 4 1 8
Output
3
Input
3 3
1 10 100
1 100 10
Output
2
Input
1 4
4
1 1 1 1
Output
1
Note
In the first example the maximum number of files in the archive is 3. For example, it is possible that in the archive are three files of sizes 2 + 5 = 7, 15 = 3 + 1 + 11 = 8 + 2 + 4 + 1 and 4 + 4 = 8.
In the second example it is possible that the archive contains two files of sizes 1 and 110 = 10 + 100 = 100 + 10. Note that the order of files is kept while transferring archives through the network, so we can't say that there are three files of sizes 1, 10 and 100.
In the third example the only possibility is that the archive contains a single file of size 4.
Tags: greedy, implementation
Correct Solution:
```
n, m = map(int, input().split())
arr1=list(map(int,input().split()))
arr2=list(map(int,input().split()))
file=0; sum1=0; sum2=0; i=0; j=0
while i<n or j<m:
if sum1 >= sum2:
sum2 += arr2[j]
j+=1
else:
sum1 += arr1[i]
i+=1
if sum1==sum2:
file+=1
print(file)
```
| 90,669 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hacker Zhorik wants to decipher two secret messages he intercepted yesterday. Yeah message is a sequence of encrypted blocks, each of them consists of several bytes of information.
Zhorik knows that each of the messages is an archive containing one or more files. Zhorik knows how each of these archives was transferred through the network: if an archive consists of k files of sizes l1, l2, ..., lk bytes, then the i-th file is split to one or more blocks bi, 1, bi, 2, ..., bi, mi (here the total length of the blocks bi, 1 + bi, 2 + ... + bi, mi is equal to the length of the file li), and after that all blocks are transferred through the network, maintaining the order of files in the archive.
Zhorik thinks that the two messages contain the same archive, because their total lengths are equal. However, each file can be split in blocks in different ways in the two messages.
You are given the lengths of blocks in each of the two messages. Help Zhorik to determine what is the maximum number of files could be in the archive, if the Zhorik's assumption is correct.
Input
The first line contains two integers n, m (1 β€ n, m β€ 105) β the number of blocks in the first and in the second messages.
The second line contains n integers x1, x2, ..., xn (1 β€ xi β€ 106) β the length of the blocks that form the first message.
The third line contains m integers y1, y2, ..., ym (1 β€ yi β€ 106) β the length of the blocks that form the second message.
It is guaranteed that x1 + ... + xn = y1 + ... + ym. Also, it is guaranteed that x1 + ... + xn β€ 106.
Output
Print the maximum number of files the intercepted array could consist of.
Examples
Input
7 6
2 5 3 1 11 4 4
7 8 2 4 1 8
Output
3
Input
3 3
1 10 100
1 100 10
Output
2
Input
1 4
4
1 1 1 1
Output
1
Note
In the first example the maximum number of files in the archive is 3. For example, it is possible that in the archive are three files of sizes 2 + 5 = 7, 15 = 3 + 1 + 11 = 8 + 2 + 4 + 1 and 4 + 4 = 8.
In the second example it is possible that the archive contains two files of sizes 1 and 110 = 10 + 100 = 100 + 10. Note that the order of files is kept while transferring archives through the network, so we can't say that there are three files of sizes 1, 10 and 100.
In the third example the only possibility is that the archive contains a single file of size 4.
Submitted Solution:
```
n, m = map(int, input().split())
n_f = list(map(int, input().split()))
m_f = list(map(int, input().split()))
s_n = 0
s_m = 0
f = 0
i = 0
j = 0
while i < n and j < m:
if s_n > s_m:
flag = True
k = j
while flag and k < m:
s_m += m_f[k]
if s_n == s_m:
s_n = 0
s_m = 0
flag = False
f += 1
j = k + 1
if s_m > s_n:
j = k + 1
flag = False
k +=1
else:
flag1 = True
k = i
while flag1 and i < n:
s_n += n_f[k]
if s_n == s_m:
s_n = 0
s_m = 0
flag1= False
f += 1
i = k + 1
if s_n > s_m:
i = k + 1
flag1 = False
k += 1
print(f + 1)
```
Yes
| 90,670 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hacker Zhorik wants to decipher two secret messages he intercepted yesterday. Yeah message is a sequence of encrypted blocks, each of them consists of several bytes of information.
Zhorik knows that each of the messages is an archive containing one or more files. Zhorik knows how each of these archives was transferred through the network: if an archive consists of k files of sizes l1, l2, ..., lk bytes, then the i-th file is split to one or more blocks bi, 1, bi, 2, ..., bi, mi (here the total length of the blocks bi, 1 + bi, 2 + ... + bi, mi is equal to the length of the file li), and after that all blocks are transferred through the network, maintaining the order of files in the archive.
Zhorik thinks that the two messages contain the same archive, because their total lengths are equal. However, each file can be split in blocks in different ways in the two messages.
You are given the lengths of blocks in each of the two messages. Help Zhorik to determine what is the maximum number of files could be in the archive, if the Zhorik's assumption is correct.
Input
The first line contains two integers n, m (1 β€ n, m β€ 105) β the number of blocks in the first and in the second messages.
The second line contains n integers x1, x2, ..., xn (1 β€ xi β€ 106) β the length of the blocks that form the first message.
The third line contains m integers y1, y2, ..., ym (1 β€ yi β€ 106) β the length of the blocks that form the second message.
It is guaranteed that x1 + ... + xn = y1 + ... + ym. Also, it is guaranteed that x1 + ... + xn β€ 106.
Output
Print the maximum number of files the intercepted array could consist of.
Examples
Input
7 6
2 5 3 1 11 4 4
7 8 2 4 1 8
Output
3
Input
3 3
1 10 100
1 100 10
Output
2
Input
1 4
4
1 1 1 1
Output
1
Note
In the first example the maximum number of files in the archive is 3. For example, it is possible that in the archive are three files of sizes 2 + 5 = 7, 15 = 3 + 1 + 11 = 8 + 2 + 4 + 1 and 4 + 4 = 8.
In the second example it is possible that the archive contains two files of sizes 1 and 110 = 10 + 100 = 100 + 10. Note that the order of files is kept while transferring archives through the network, so we can't say that there are three files of sizes 1, 10 and 100.
In the third example the only possibility is that the archive contains a single file of size 4.
Submitted Solution:
```
a, b = map(int, input().split())
lista = list(map(int, input().split()))
listb = list(map(int, input().split()))
# print(lista)
i = 0
j = 0
cnta = 0
cntb = 0
ans = 0
while(i<len(lista) and j < len(listb)):
if(cnta == cntb):
cnta += lista[i]
cntb += listb[j]
ans += 1
i += 1
j += 1
elif cnta < cntb:
cnta += lista[i]
i += 1
elif cnta > cntb:
cntb += listb[j]
j += 1
# print("cnta==" + str(cnta) + ", cntb==" + str(cntb))
print(ans)
```
Yes
| 90,671 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hacker Zhorik wants to decipher two secret messages he intercepted yesterday. Yeah message is a sequence of encrypted blocks, each of them consists of several bytes of information.
Zhorik knows that each of the messages is an archive containing one or more files. Zhorik knows how each of these archives was transferred through the network: if an archive consists of k files of sizes l1, l2, ..., lk bytes, then the i-th file is split to one or more blocks bi, 1, bi, 2, ..., bi, mi (here the total length of the blocks bi, 1 + bi, 2 + ... + bi, mi is equal to the length of the file li), and after that all blocks are transferred through the network, maintaining the order of files in the archive.
Zhorik thinks that the two messages contain the same archive, because their total lengths are equal. However, each file can be split in blocks in different ways in the two messages.
You are given the lengths of blocks in each of the two messages. Help Zhorik to determine what is the maximum number of files could be in the archive, if the Zhorik's assumption is correct.
Input
The first line contains two integers n, m (1 β€ n, m β€ 105) β the number of blocks in the first and in the second messages.
The second line contains n integers x1, x2, ..., xn (1 β€ xi β€ 106) β the length of the blocks that form the first message.
The third line contains m integers y1, y2, ..., ym (1 β€ yi β€ 106) β the length of the blocks that form the second message.
It is guaranteed that x1 + ... + xn = y1 + ... + ym. Also, it is guaranteed that x1 + ... + xn β€ 106.
Output
Print the maximum number of files the intercepted array could consist of.
Examples
Input
7 6
2 5 3 1 11 4 4
7 8 2 4 1 8
Output
3
Input
3 3
1 10 100
1 100 10
Output
2
Input
1 4
4
1 1 1 1
Output
1
Note
In the first example the maximum number of files in the archive is 3. For example, it is possible that in the archive are three files of sizes 2 + 5 = 7, 15 = 3 + 1 + 11 = 8 + 2 + 4 + 1 and 4 + 4 = 8.
In the second example it is possible that the archive contains two files of sizes 1 and 110 = 10 + 100 = 100 + 10. Note that the order of files is kept while transferring archives through the network, so we can't say that there are three files of sizes 1, 10 and 100.
In the third example the only possibility is that the archive contains a single file of size 4.
Submitted Solution:
```
n, m = map(int,input().split())
x = list(map(int,input().split()))
y = list(map(int,input().split()))
i,j,s1,s2=0,0,0,0
count=1
while i <= n-1 and j <= m-1:
if s1==s2 and s1:
count+=1
s1=0
s2=0
elif s1>s2 or not s2:
s2+=y[j]
j+=1
elif s1<s2 or not s1:
s1+=x[i]
i+=1
print(count)
```
Yes
| 90,672 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hacker Zhorik wants to decipher two secret messages he intercepted yesterday. Yeah message is a sequence of encrypted blocks, each of them consists of several bytes of information.
Zhorik knows that each of the messages is an archive containing one or more files. Zhorik knows how each of these archives was transferred through the network: if an archive consists of k files of sizes l1, l2, ..., lk bytes, then the i-th file is split to one or more blocks bi, 1, bi, 2, ..., bi, mi (here the total length of the blocks bi, 1 + bi, 2 + ... + bi, mi is equal to the length of the file li), and after that all blocks are transferred through the network, maintaining the order of files in the archive.
Zhorik thinks that the two messages contain the same archive, because their total lengths are equal. However, each file can be split in blocks in different ways in the two messages.
You are given the lengths of blocks in each of the two messages. Help Zhorik to determine what is the maximum number of files could be in the archive, if the Zhorik's assumption is correct.
Input
The first line contains two integers n, m (1 β€ n, m β€ 105) β the number of blocks in the first and in the second messages.
The second line contains n integers x1, x2, ..., xn (1 β€ xi β€ 106) β the length of the blocks that form the first message.
The third line contains m integers y1, y2, ..., ym (1 β€ yi β€ 106) β the length of the blocks that form the second message.
It is guaranteed that x1 + ... + xn = y1 + ... + ym. Also, it is guaranteed that x1 + ... + xn β€ 106.
Output
Print the maximum number of files the intercepted array could consist of.
Examples
Input
7 6
2 5 3 1 11 4 4
7 8 2 4 1 8
Output
3
Input
3 3
1 10 100
1 100 10
Output
2
Input
1 4
4
1 1 1 1
Output
1
Note
In the first example the maximum number of files in the archive is 3. For example, it is possible that in the archive are three files of sizes 2 + 5 = 7, 15 = 3 + 1 + 11 = 8 + 2 + 4 + 1 and 4 + 4 = 8.
In the second example it is possible that the archive contains two files of sizes 1 and 110 = 10 + 100 = 100 + 10. Note that the order of files is kept while transferring archives through the network, so we can't say that there are three files of sizes 1, 10 and 100.
In the third example the only possibility is that the archive contains a single file of size 4.
Submitted Solution:
```
_, _ = list(map(int, input().strip().split()))
msg1 = list(map(int, input().strip().split()))
msg2 = list(map(int, input().strip().split()))
idx1, idx2 = 0, 0
count = 0
while idx1 < len(msg1):
sum1, sum2 = msg1[idx1], msg2[idx2]
while sum1 != sum2:
if sum1 < sum2:
idx1 += 1
sum1 += msg1[idx1]
else:
idx2 += 1
sum2 += msg2[idx2]
idx1, idx2 = idx1 + 1, idx2 + 1
count += 1
print(count)
```
Yes
| 90,673 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hacker Zhorik wants to decipher two secret messages he intercepted yesterday. Yeah message is a sequence of encrypted blocks, each of them consists of several bytes of information.
Zhorik knows that each of the messages is an archive containing one or more files. Zhorik knows how each of these archives was transferred through the network: if an archive consists of k files of sizes l1, l2, ..., lk bytes, then the i-th file is split to one or more blocks bi, 1, bi, 2, ..., bi, mi (here the total length of the blocks bi, 1 + bi, 2 + ... + bi, mi is equal to the length of the file li), and after that all blocks are transferred through the network, maintaining the order of files in the archive.
Zhorik thinks that the two messages contain the same archive, because their total lengths are equal. However, each file can be split in blocks in different ways in the two messages.
You are given the lengths of blocks in each of the two messages. Help Zhorik to determine what is the maximum number of files could be in the archive, if the Zhorik's assumption is correct.
Input
The first line contains two integers n, m (1 β€ n, m β€ 105) β the number of blocks in the first and in the second messages.
The second line contains n integers x1, x2, ..., xn (1 β€ xi β€ 106) β the length of the blocks that form the first message.
The third line contains m integers y1, y2, ..., ym (1 β€ yi β€ 106) β the length of the blocks that form the second message.
It is guaranteed that x1 + ... + xn = y1 + ... + ym. Also, it is guaranteed that x1 + ... + xn β€ 106.
Output
Print the maximum number of files the intercepted array could consist of.
Examples
Input
7 6
2 5 3 1 11 4 4
7 8 2 4 1 8
Output
3
Input
3 3
1 10 100
1 100 10
Output
2
Input
1 4
4
1 1 1 1
Output
1
Note
In the first example the maximum number of files in the archive is 3. For example, it is possible that in the archive are three files of sizes 2 + 5 = 7, 15 = 3 + 1 + 11 = 8 + 2 + 4 + 1 and 4 + 4 = 8.
In the second example it is possible that the archive contains two files of sizes 1 and 110 = 10 + 100 = 100 + 10. Note that the order of files is kept while transferring archives through the network, so we can't say that there are three files of sizes 1, 10 and 100.
In the third example the only possibility is that the archive contains a single file of size 4.
Submitted Solution:
```
n,m = map(int,input().split())
l1 = list(map(int,input().split()))
l2 = list(map(int,input().split()))
i,j = 1,1
sum1 = l1[0]
sum2 = l2[0]
count = 0
while i < len(l1) and j < len(l2):
print(sum1,sum2)
if sum1 > sum2:
sum2 = sum2 + l2[j]
j = j + 1
elif sum2 > sum1:
sum1 = sum1 + l1[i]
i = i + 1
if sum2 == sum1:
count = count + 1
if j < len(l2):
sum2 = l2[j]
j = j + 1
if i < len(l1):
sum1 = l1[i]
i = i + 1
count = count + 1
print(count)
```
No
| 90,674 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hacker Zhorik wants to decipher two secret messages he intercepted yesterday. Yeah message is a sequence of encrypted blocks, each of them consists of several bytes of information.
Zhorik knows that each of the messages is an archive containing one or more files. Zhorik knows how each of these archives was transferred through the network: if an archive consists of k files of sizes l1, l2, ..., lk bytes, then the i-th file is split to one or more blocks bi, 1, bi, 2, ..., bi, mi (here the total length of the blocks bi, 1 + bi, 2 + ... + bi, mi is equal to the length of the file li), and after that all blocks are transferred through the network, maintaining the order of files in the archive.
Zhorik thinks that the two messages contain the same archive, because their total lengths are equal. However, each file can be split in blocks in different ways in the two messages.
You are given the lengths of blocks in each of the two messages. Help Zhorik to determine what is the maximum number of files could be in the archive, if the Zhorik's assumption is correct.
Input
The first line contains two integers n, m (1 β€ n, m β€ 105) β the number of blocks in the first and in the second messages.
The second line contains n integers x1, x2, ..., xn (1 β€ xi β€ 106) β the length of the blocks that form the first message.
The third line contains m integers y1, y2, ..., ym (1 β€ yi β€ 106) β the length of the blocks that form the second message.
It is guaranteed that x1 + ... + xn = y1 + ... + ym. Also, it is guaranteed that x1 + ... + xn β€ 106.
Output
Print the maximum number of files the intercepted array could consist of.
Examples
Input
7 6
2 5 3 1 11 4 4
7 8 2 4 1 8
Output
3
Input
3 3
1 10 100
1 100 10
Output
2
Input
1 4
4
1 1 1 1
Output
1
Note
In the first example the maximum number of files in the archive is 3. For example, it is possible that in the archive are three files of sizes 2 + 5 = 7, 15 = 3 + 1 + 11 = 8 + 2 + 4 + 1 and 4 + 4 = 8.
In the second example it is possible that the archive contains two files of sizes 1 and 110 = 10 + 100 = 100 + 10. Note that the order of files is kept while transferring archives through the network, so we can't say that there are three files of sizes 1, 10 and 100.
In the third example the only possibility is that the archive contains a single file of size 4.
Submitted Solution:
```
x=[int(i) for i in input().split()]
y=[int(i) for i in input().split()]
z=[int(i) for i in input().split()]
a=1
t=1
f=0
sum1=y[0]
sum2=z[0]
if len(y)==1 or len(z)==1:
f=1
else:
for i in range(1,sum(x)+1):
if t>x[0]+1 or a>x[1]+1:
break
if sum1==sum2:
f+=1
try:
sum1=y[a]
sum2=z[t]
t+=1
a+=1
except:
break
elif sum1<sum2:
sum1+=y[a]
a+=1
elif sum1>sum2:
sum2+=z[t]
t+=1
print(f)
```
No
| 90,675 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hacker Zhorik wants to decipher two secret messages he intercepted yesterday. Yeah message is a sequence of encrypted blocks, each of them consists of several bytes of information.
Zhorik knows that each of the messages is an archive containing one or more files. Zhorik knows how each of these archives was transferred through the network: if an archive consists of k files of sizes l1, l2, ..., lk bytes, then the i-th file is split to one or more blocks bi, 1, bi, 2, ..., bi, mi (here the total length of the blocks bi, 1 + bi, 2 + ... + bi, mi is equal to the length of the file li), and after that all blocks are transferred through the network, maintaining the order of files in the archive.
Zhorik thinks that the two messages contain the same archive, because their total lengths are equal. However, each file can be split in blocks in different ways in the two messages.
You are given the lengths of blocks in each of the two messages. Help Zhorik to determine what is the maximum number of files could be in the archive, if the Zhorik's assumption is correct.
Input
The first line contains two integers n, m (1 β€ n, m β€ 105) β the number of blocks in the first and in the second messages.
The second line contains n integers x1, x2, ..., xn (1 β€ xi β€ 106) β the length of the blocks that form the first message.
The third line contains m integers y1, y2, ..., ym (1 β€ yi β€ 106) β the length of the blocks that form the second message.
It is guaranteed that x1 + ... + xn = y1 + ... + ym. Also, it is guaranteed that x1 + ... + xn β€ 106.
Output
Print the maximum number of files the intercepted array could consist of.
Examples
Input
7 6
2 5 3 1 11 4 4
7 8 2 4 1 8
Output
3
Input
3 3
1 10 100
1 100 10
Output
2
Input
1 4
4
1 1 1 1
Output
1
Note
In the first example the maximum number of files in the archive is 3. For example, it is possible that in the archive are three files of sizes 2 + 5 = 7, 15 = 3 + 1 + 11 = 8 + 2 + 4 + 1 and 4 + 4 = 8.
In the second example it is possible that the archive contains two files of sizes 1 and 110 = 10 + 100 = 100 + 10. Note that the order of files is kept while transferring archives through the network, so we can't say that there are three files of sizes 1, 10 and 100.
In the third example the only possibility is that the archive contains a single file of size 4.
Submitted Solution:
```
n,m=(map(int,input().strip().split(' ')))
a= list(map(int,input().strip().split(' ')))
b= list(map(int,input().strip().split(' ')))
index = -1
i = 0
j = 0
s1 = 0
s2 = 0
cnt=0
while(i<n and j<m):
if(index==0):
s1+=a[i]
i+=1
elif(index==1):
s2+=b[j]
j+=1
else:
s1+=a[i]
s2+=b[j]
j+=1
i+=1
if(s1==s2):
cnt+=1
s1=0
s2=0
index=-1
elif(s1>s2):
index=1
else:
index=0
print(cnt+1)
```
No
| 90,676 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hacker Zhorik wants to decipher two secret messages he intercepted yesterday. Yeah message is a sequence of encrypted blocks, each of them consists of several bytes of information.
Zhorik knows that each of the messages is an archive containing one or more files. Zhorik knows how each of these archives was transferred through the network: if an archive consists of k files of sizes l1, l2, ..., lk bytes, then the i-th file is split to one or more blocks bi, 1, bi, 2, ..., bi, mi (here the total length of the blocks bi, 1 + bi, 2 + ... + bi, mi is equal to the length of the file li), and after that all blocks are transferred through the network, maintaining the order of files in the archive.
Zhorik thinks that the two messages contain the same archive, because their total lengths are equal. However, each file can be split in blocks in different ways in the two messages.
You are given the lengths of blocks in each of the two messages. Help Zhorik to determine what is the maximum number of files could be in the archive, if the Zhorik's assumption is correct.
Input
The first line contains two integers n, m (1 β€ n, m β€ 105) β the number of blocks in the first and in the second messages.
The second line contains n integers x1, x2, ..., xn (1 β€ xi β€ 106) β the length of the blocks that form the first message.
The third line contains m integers y1, y2, ..., ym (1 β€ yi β€ 106) β the length of the blocks that form the second message.
It is guaranteed that x1 + ... + xn = y1 + ... + ym. Also, it is guaranteed that x1 + ... + xn β€ 106.
Output
Print the maximum number of files the intercepted array could consist of.
Examples
Input
7 6
2 5 3 1 11 4 4
7 8 2 4 1 8
Output
3
Input
3 3
1 10 100
1 100 10
Output
2
Input
1 4
4
1 1 1 1
Output
1
Note
In the first example the maximum number of files in the archive is 3. For example, it is possible that in the archive are three files of sizes 2 + 5 = 7, 15 = 3 + 1 + 11 = 8 + 2 + 4 + 1 and 4 + 4 = 8.
In the second example it is possible that the archive contains two files of sizes 1 and 110 = 10 + 100 = 100 + 10. Note that the order of files is kept while transferring archives through the network, so we can't say that there are three files of sizes 1, 10 and 100.
In the third example the only possibility is that the archive contains a single file of size 4.
Submitted Solution:
```
'''input
1 4
4
1 1 1 1
'''
n, m = [int(i) for i in input().split(" ")]
N = [int(i) for i in input().split(" ")]
M = [int(i) for i in input().split(" ")]
s1 = N[0]
s2 = M[0]
if s1 == s2:
ans = 1
else:
ans = 0
i, j = 1, 1
while 1:
if i == m or j == n:
break
if s1 > s2:
s2 += M[i]
i += 1
else:
s1 += N[j]
j += 1
if s1 == s2:
ans += 1
print(ans + 1)
```
No
| 90,677 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The busses in Berland are equipped with a video surveillance system. The system records information about changes in the number of passengers in a bus after stops.
If x is the number of passengers in a bus just before the current bus stop and y is the number of passengers in the bus just after current bus stop, the system records the number y-x. So the system records show how number of passengers changed.
The test run was made for single bus and n bus stops. Thus, the system recorded the sequence of integers a_1, a_2, ..., a_n (exactly one number for each bus stop), where a_i is the record for the bus stop i. The bus stops are numbered from 1 to n in chronological order.
Determine the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w (that is, at any time in the bus there should be from 0 to w passengers inclusive).
Input
The first line contains two integers n and w (1 β€ n β€ 1 000, 1 β€ w β€ 10^{9}) β the number of bus stops and the capacity of the bus.
The second line contains a sequence a_1, a_2, ..., a_n (-10^{6} β€ a_i β€ 10^{6}), where a_i equals to the number, which has been recorded by the video system after the i-th bus stop.
Output
Print the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w. If the situation is contradictory (i.e. for any initial number of passengers there will be a contradiction), print 0.
Examples
Input
3 5
2 1 -3
Output
3
Input
2 4
-1 1
Output
4
Input
4 10
2 4 1 2
Output
2
Note
In the first example initially in the bus could be 0, 1 or 2 passengers.
In the second example initially in the bus could be 1, 2, 3 or 4 passengers.
In the third example initially in the bus could be 0 or 1 passenger.
Tags: combinatorics, math
Correct Solution:
```
n,w = map(int, input().split())
a = list(map(int, input().split()))
from itertools import accumulate
b = list(accumulate(a))
c1, c2 = max(b), min(b)
if c1 < 0:
c1 = abs(c2)
elif c2 < 0:
c1, c2 = c1-c2, 0
if c1 > w:
print(0)
else:
print(w - c1+1)
```
| 90,678 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The busses in Berland are equipped with a video surveillance system. The system records information about changes in the number of passengers in a bus after stops.
If x is the number of passengers in a bus just before the current bus stop and y is the number of passengers in the bus just after current bus stop, the system records the number y-x. So the system records show how number of passengers changed.
The test run was made for single bus and n bus stops. Thus, the system recorded the sequence of integers a_1, a_2, ..., a_n (exactly one number for each bus stop), where a_i is the record for the bus stop i. The bus stops are numbered from 1 to n in chronological order.
Determine the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w (that is, at any time in the bus there should be from 0 to w passengers inclusive).
Input
The first line contains two integers n and w (1 β€ n β€ 1 000, 1 β€ w β€ 10^{9}) β the number of bus stops and the capacity of the bus.
The second line contains a sequence a_1, a_2, ..., a_n (-10^{6} β€ a_i β€ 10^{6}), where a_i equals to the number, which has been recorded by the video system after the i-th bus stop.
Output
Print the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w. If the situation is contradictory (i.e. for any initial number of passengers there will be a contradiction), print 0.
Examples
Input
3 5
2 1 -3
Output
3
Input
2 4
-1 1
Output
4
Input
4 10
2 4 1 2
Output
2
Note
In the first example initially in the bus could be 0, 1 or 2 passengers.
In the second example initially in the bus could be 1, 2, 3 or 4 passengers.
In the third example initially in the bus could be 0 or 1 passenger.
Tags: combinatorics, math
Correct Solution:
```
# your code goes here
# your code goes here
n,w = map(int,input().split())
a = list(map(int,input().split()))
maxi,mini = 0,0
total = 0
for i in range(n):
total += a[i]
if total > maxi:
maxi = total
if total < mini:
mini = total
ans = w - max(maxi,0) + 1 - max(-mini,0)
if ans < 0:
print(0)
else:
print(ans)
```
| 90,679 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The busses in Berland are equipped with a video surveillance system. The system records information about changes in the number of passengers in a bus after stops.
If x is the number of passengers in a bus just before the current bus stop and y is the number of passengers in the bus just after current bus stop, the system records the number y-x. So the system records show how number of passengers changed.
The test run was made for single bus and n bus stops. Thus, the system recorded the sequence of integers a_1, a_2, ..., a_n (exactly one number for each bus stop), where a_i is the record for the bus stop i. The bus stops are numbered from 1 to n in chronological order.
Determine the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w (that is, at any time in the bus there should be from 0 to w passengers inclusive).
Input
The first line contains two integers n and w (1 β€ n β€ 1 000, 1 β€ w β€ 10^{9}) β the number of bus stops and the capacity of the bus.
The second line contains a sequence a_1, a_2, ..., a_n (-10^{6} β€ a_i β€ 10^{6}), where a_i equals to the number, which has been recorded by the video system after the i-th bus stop.
Output
Print the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w. If the situation is contradictory (i.e. for any initial number of passengers there will be a contradiction), print 0.
Examples
Input
3 5
2 1 -3
Output
3
Input
2 4
-1 1
Output
4
Input
4 10
2 4 1 2
Output
2
Note
In the first example initially in the bus could be 0, 1 or 2 passengers.
In the second example initially in the bus could be 1, 2, 3 or 4 passengers.
In the third example initially in the bus could be 0 or 1 passenger.
Tags: combinatorics, math
Correct Solution:
```
a = input().split(" ")
a = [int(e) for e in a]
capacity = a[1]
l = input().split(" ")
l = [int(e) for e in l]
assert a[0] == len(l)
min_num = 0
max_num = 0
num = 0
for change in l:
num += change
if min_num == None or num < min_num:
min_num = num
if max_num == None or num > max_num:
max_num = num
print(max(capacity - max_num + min_num + 1, 0))
```
| 90,680 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The busses in Berland are equipped with a video surveillance system. The system records information about changes in the number of passengers in a bus after stops.
If x is the number of passengers in a bus just before the current bus stop and y is the number of passengers in the bus just after current bus stop, the system records the number y-x. So the system records show how number of passengers changed.
The test run was made for single bus and n bus stops. Thus, the system recorded the sequence of integers a_1, a_2, ..., a_n (exactly one number for each bus stop), where a_i is the record for the bus stop i. The bus stops are numbered from 1 to n in chronological order.
Determine the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w (that is, at any time in the bus there should be from 0 to w passengers inclusive).
Input
The first line contains two integers n and w (1 β€ n β€ 1 000, 1 β€ w β€ 10^{9}) β the number of bus stops and the capacity of the bus.
The second line contains a sequence a_1, a_2, ..., a_n (-10^{6} β€ a_i β€ 10^{6}), where a_i equals to the number, which has been recorded by the video system after the i-th bus stop.
Output
Print the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w. If the situation is contradictory (i.e. for any initial number of passengers there will be a contradiction), print 0.
Examples
Input
3 5
2 1 -3
Output
3
Input
2 4
-1 1
Output
4
Input
4 10
2 4 1 2
Output
2
Note
In the first example initially in the bus could be 0, 1 or 2 passengers.
In the second example initially in the bus could be 1, 2, 3 or 4 passengers.
In the third example initially in the bus could be 0 or 1 passenger.
Tags: combinatorics, math
Correct Solution:
```
z,s,a,b=input,0,0,0;n,m=map(int,z().split())
for i in map(int,z().split()):s+=i;a,b=min(a,s),max(b,s)
print(max(m-b+a+1,0))
```
| 90,681 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The busses in Berland are equipped with a video surveillance system. The system records information about changes in the number of passengers in a bus after stops.
If x is the number of passengers in a bus just before the current bus stop and y is the number of passengers in the bus just after current bus stop, the system records the number y-x. So the system records show how number of passengers changed.
The test run was made for single bus and n bus stops. Thus, the system recorded the sequence of integers a_1, a_2, ..., a_n (exactly one number for each bus stop), where a_i is the record for the bus stop i. The bus stops are numbered from 1 to n in chronological order.
Determine the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w (that is, at any time in the bus there should be from 0 to w passengers inclusive).
Input
The first line contains two integers n and w (1 β€ n β€ 1 000, 1 β€ w β€ 10^{9}) β the number of bus stops and the capacity of the bus.
The second line contains a sequence a_1, a_2, ..., a_n (-10^{6} β€ a_i β€ 10^{6}), where a_i equals to the number, which has been recorded by the video system after the i-th bus stop.
Output
Print the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w. If the situation is contradictory (i.e. for any initial number of passengers there will be a contradiction), print 0.
Examples
Input
3 5
2 1 -3
Output
3
Input
2 4
-1 1
Output
4
Input
4 10
2 4 1 2
Output
2
Note
In the first example initially in the bus could be 0, 1 or 2 passengers.
In the second example initially in the bus could be 1, 2, 3 or 4 passengers.
In the third example initially in the bus could be 0 or 1 passenger.
Tags: combinatorics, math
Correct Solution:
```
n,w = map(int,input().split())
a = [int(s) for s in input().split()]
mi,ma,s = 0,0,0
for i in a:
s += i
if s < mi:
mi = s
elif s > ma:
ma = s
if -mi <= w-ma:
print(w-ma+mi+1)
else:
print(0)
```
| 90,682 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The busses in Berland are equipped with a video surveillance system. The system records information about changes in the number of passengers in a bus after stops.
If x is the number of passengers in a bus just before the current bus stop and y is the number of passengers in the bus just after current bus stop, the system records the number y-x. So the system records show how number of passengers changed.
The test run was made for single bus and n bus stops. Thus, the system recorded the sequence of integers a_1, a_2, ..., a_n (exactly one number for each bus stop), where a_i is the record for the bus stop i. The bus stops are numbered from 1 to n in chronological order.
Determine the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w (that is, at any time in the bus there should be from 0 to w passengers inclusive).
Input
The first line contains two integers n and w (1 β€ n β€ 1 000, 1 β€ w β€ 10^{9}) β the number of bus stops and the capacity of the bus.
The second line contains a sequence a_1, a_2, ..., a_n (-10^{6} β€ a_i β€ 10^{6}), where a_i equals to the number, which has been recorded by the video system after the i-th bus stop.
Output
Print the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w. If the situation is contradictory (i.e. for any initial number of passengers there will be a contradiction), print 0.
Examples
Input
3 5
2 1 -3
Output
3
Input
2 4
-1 1
Output
4
Input
4 10
2 4 1 2
Output
2
Note
In the first example initially in the bus could be 0, 1 or 2 passengers.
In the second example initially in the bus could be 1, 2, 3 or 4 passengers.
In the third example initially in the bus could be 0 or 1 passenger.
Tags: combinatorics, math
Correct Solution:
```
"""Codeforces Round #481 (Div. 3) - Bus Video System.
http://codeforces.com/contest/978/problem/E
The busses in Berland are equipped with a video surveillance system. The
system records information about changes in the number of passengers in a bus
after stops.
If ``x`` is the number of passengers in a bus just before the current bus
stop and ``y`` is the number of passengers in the bus just after current bus
stop, the system records the number y - x. So the system records show how
number of passengers changed.
The test run was made for single bus and ``n`` bus stops. Thus, the system
recorded the sequence of integers a[1], a[2], ..., a[n] (exactly one
number for each bus stop), where a[i] is the record for the bus stop
``i``. The bus stops are numbered from 1 to n in chronological order.
Determine the number of possible ways how many people could be in the bus
before the first bus stop, if the bus has a capacity equals to ``w`` (that
is, at any time in the bus there should be from 0 to w passengers inclusive).
"""
def solve(capacity, sequence):
diff = 0
lb, ub = float('inf'), -float('inf')
for record in sequence:
diff += record
ub = max(ub, diff)
lb = min(lb, diff)
return max(0, capacity - max(0, ub) + min(0, lb) + 1)
def main():
_, capacity = [int(x) for x in input().strip().split()]
sequence = [int(x) for x in input().strip().split()]
result = solve(capacity, sequence)
print(result)
if __name__ == '__main__':
main()
```
| 90,683 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The busses in Berland are equipped with a video surveillance system. The system records information about changes in the number of passengers in a bus after stops.
If x is the number of passengers in a bus just before the current bus stop and y is the number of passengers in the bus just after current bus stop, the system records the number y-x. So the system records show how number of passengers changed.
The test run was made for single bus and n bus stops. Thus, the system recorded the sequence of integers a_1, a_2, ..., a_n (exactly one number for each bus stop), where a_i is the record for the bus stop i. The bus stops are numbered from 1 to n in chronological order.
Determine the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w (that is, at any time in the bus there should be from 0 to w passengers inclusive).
Input
The first line contains two integers n and w (1 β€ n β€ 1 000, 1 β€ w β€ 10^{9}) β the number of bus stops and the capacity of the bus.
The second line contains a sequence a_1, a_2, ..., a_n (-10^{6} β€ a_i β€ 10^{6}), where a_i equals to the number, which has been recorded by the video system after the i-th bus stop.
Output
Print the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w. If the situation is contradictory (i.e. for any initial number of passengers there will be a contradiction), print 0.
Examples
Input
3 5
2 1 -3
Output
3
Input
2 4
-1 1
Output
4
Input
4 10
2 4 1 2
Output
2
Note
In the first example initially in the bus could be 0, 1 or 2 passengers.
In the second example initially in the bus could be 1, 2, 3 or 4 passengers.
In the third example initially in the bus could be 0 or 1 passenger.
Tags: combinatorics, math
Correct Solution:
```
n, w = map(int, input().split())
a = list(map(int, input().split()))
minw, maxw = 0, w
k = 0
for e in a:
w -= e
k -= e
maxw = min(w, maxw)
minw = max(minw, k)
if minw > maxw:
print(0)
else:
print(maxw - minw + 1)
```
| 90,684 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The busses in Berland are equipped with a video surveillance system. The system records information about changes in the number of passengers in a bus after stops.
If x is the number of passengers in a bus just before the current bus stop and y is the number of passengers in the bus just after current bus stop, the system records the number y-x. So the system records show how number of passengers changed.
The test run was made for single bus and n bus stops. Thus, the system recorded the sequence of integers a_1, a_2, ..., a_n (exactly one number for each bus stop), where a_i is the record for the bus stop i. The bus stops are numbered from 1 to n in chronological order.
Determine the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w (that is, at any time in the bus there should be from 0 to w passengers inclusive).
Input
The first line contains two integers n and w (1 β€ n β€ 1 000, 1 β€ w β€ 10^{9}) β the number of bus stops and the capacity of the bus.
The second line contains a sequence a_1, a_2, ..., a_n (-10^{6} β€ a_i β€ 10^{6}), where a_i equals to the number, which has been recorded by the video system after the i-th bus stop.
Output
Print the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w. If the situation is contradictory (i.e. for any initial number of passengers there will be a contradiction), print 0.
Examples
Input
3 5
2 1 -3
Output
3
Input
2 4
-1 1
Output
4
Input
4 10
2 4 1 2
Output
2
Note
In the first example initially in the bus could be 0, 1 or 2 passengers.
In the second example initially in the bus could be 1, 2, 3 or 4 passengers.
In the third example initially in the bus could be 0 or 1 passenger.
Tags: combinatorics, math
Correct Solution:
```
n, w = map(int, input().split())
a = list(map(int, input().split()))
L = 0
R = w
c = [0] * (n + 1)
for i in range(1, n + 1):
c[i] = c[i - 1] + a[i - 1]
L = - min(c)
R = w - max(c)
if R - L + 1 < 0:
print(0)
else:
print(R - L + 1)
```
| 90,685 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The busses in Berland are equipped with a video surveillance system. The system records information about changes in the number of passengers in a bus after stops.
If x is the number of passengers in a bus just before the current bus stop and y is the number of passengers in the bus just after current bus stop, the system records the number y-x. So the system records show how number of passengers changed.
The test run was made for single bus and n bus stops. Thus, the system recorded the sequence of integers a_1, a_2, ..., a_n (exactly one number for each bus stop), where a_i is the record for the bus stop i. The bus stops are numbered from 1 to n in chronological order.
Determine the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w (that is, at any time in the bus there should be from 0 to w passengers inclusive).
Input
The first line contains two integers n and w (1 β€ n β€ 1 000, 1 β€ w β€ 10^{9}) β the number of bus stops and the capacity of the bus.
The second line contains a sequence a_1, a_2, ..., a_n (-10^{6} β€ a_i β€ 10^{6}), where a_i equals to the number, which has been recorded by the video system after the i-th bus stop.
Output
Print the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w. If the situation is contradictory (i.e. for any initial number of passengers there will be a contradiction), print 0.
Examples
Input
3 5
2 1 -3
Output
3
Input
2 4
-1 1
Output
4
Input
4 10
2 4 1 2
Output
2
Note
In the first example initially in the bus could be 0, 1 or 2 passengers.
In the second example initially in the bus could be 1, 2, 3 or 4 passengers.
In the third example initially in the bus could be 0 or 1 passenger.
Submitted Solution:
```
n, w = list(map(int, input().strip().split()))
a = list(map(int, input().strip().split()))
summ = 0
mmax = w
mmin = 0
for i in range(n):
cur = a[i]
if cur >= 0:
summ += cur
mmax = min(mmax, w - summ)
else:
cmp = 0
if summ + cur < 0:
cmp = abs(summ + cur)
mmin = max(mmin, cmp)
summ += cur
if mmax < mmin:
print(0)
else:
print(mmax - mmin + 1)
```
Yes
| 90,686 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The busses in Berland are equipped with a video surveillance system. The system records information about changes in the number of passengers in a bus after stops.
If x is the number of passengers in a bus just before the current bus stop and y is the number of passengers in the bus just after current bus stop, the system records the number y-x. So the system records show how number of passengers changed.
The test run was made for single bus and n bus stops. Thus, the system recorded the sequence of integers a_1, a_2, ..., a_n (exactly one number for each bus stop), where a_i is the record for the bus stop i. The bus stops are numbered from 1 to n in chronological order.
Determine the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w (that is, at any time in the bus there should be from 0 to w passengers inclusive).
Input
The first line contains two integers n and w (1 β€ n β€ 1 000, 1 β€ w β€ 10^{9}) β the number of bus stops and the capacity of the bus.
The second line contains a sequence a_1, a_2, ..., a_n (-10^{6} β€ a_i β€ 10^{6}), where a_i equals to the number, which has been recorded by the video system after the i-th bus stop.
Output
Print the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w. If the situation is contradictory (i.e. for any initial number of passengers there will be a contradiction), print 0.
Examples
Input
3 5
2 1 -3
Output
3
Input
2 4
-1 1
Output
4
Input
4 10
2 4 1 2
Output
2
Note
In the first example initially in the bus could be 0, 1 or 2 passengers.
In the second example initially in the bus could be 1, 2, 3 or 4 passengers.
In the third example initially in the bus could be 0 or 1 passenger.
Submitted Solution:
```
n,w = map(int,input().split())
l = list(map(int,input().split()))
ss = 0
cc = 0
if l[-1]<0:
c = w+l[-1]
c = sum(l)
for i in l:
if i!=(-(w//n)) or w%n!=0:
cc = 1
break
if l[-1]>0:
k = w
kk = l[-1]
else:
k = w+l[-1]
kk = 0
for i in l[-1::-1]:
if i<0:
if w-k<-i:
k = w
else:
k = k-i
if w-kk<-i:
kk = w
else:
kk+=(-i)
else:
if kk<i:
kk = 0
else:
kk-=i
k-=i
if k<kk or k<0 or kk>w:
ss = 1
break
if k == w and kk == w:
if cc == 1:
ss = 1
break
if ss == 0 :
print(k-kk+1)
else:
print(0)
```
Yes
| 90,687 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The busses in Berland are equipped with a video surveillance system. The system records information about changes in the number of passengers in a bus after stops.
If x is the number of passengers in a bus just before the current bus stop and y is the number of passengers in the bus just after current bus stop, the system records the number y-x. So the system records show how number of passengers changed.
The test run was made for single bus and n bus stops. Thus, the system recorded the sequence of integers a_1, a_2, ..., a_n (exactly one number for each bus stop), where a_i is the record for the bus stop i. The bus stops are numbered from 1 to n in chronological order.
Determine the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w (that is, at any time in the bus there should be from 0 to w passengers inclusive).
Input
The first line contains two integers n and w (1 β€ n β€ 1 000, 1 β€ w β€ 10^{9}) β the number of bus stops and the capacity of the bus.
The second line contains a sequence a_1, a_2, ..., a_n (-10^{6} β€ a_i β€ 10^{6}), where a_i equals to the number, which has been recorded by the video system after the i-th bus stop.
Output
Print the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w. If the situation is contradictory (i.e. for any initial number of passengers there will be a contradiction), print 0.
Examples
Input
3 5
2 1 -3
Output
3
Input
2 4
-1 1
Output
4
Input
4 10
2 4 1 2
Output
2
Note
In the first example initially in the bus could be 0, 1 or 2 passengers.
In the second example initially in the bus could be 1, 2, 3 or 4 passengers.
In the third example initially in the bus could be 0 or 1 passenger.
Submitted Solution:
```
n,w=input().split()
n,w=int(n),int(w)
A=list(map(int,input().split()))
S=[0]*len(A)
s=0
for i in range(len(A)):
s+=A[i]
S[i]=s
if w>=abs(max(S)) and w>=abs(min(S)):
if max(S)<=0:
up=w
else:
up=w-max(S)
if min(S)>=0:
down=0
else:
down=abs(min(S))
if up-down<0 or abs(min(S)-max(S))>w :
print('0')
else:
print(up-down+1)
else:
print('0')
```
Yes
| 90,688 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The busses in Berland are equipped with a video surveillance system. The system records information about changes in the number of passengers in a bus after stops.
If x is the number of passengers in a bus just before the current bus stop and y is the number of passengers in the bus just after current bus stop, the system records the number y-x. So the system records show how number of passengers changed.
The test run was made for single bus and n bus stops. Thus, the system recorded the sequence of integers a_1, a_2, ..., a_n (exactly one number for each bus stop), where a_i is the record for the bus stop i. The bus stops are numbered from 1 to n in chronological order.
Determine the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w (that is, at any time in the bus there should be from 0 to w passengers inclusive).
Input
The first line contains two integers n and w (1 β€ n β€ 1 000, 1 β€ w β€ 10^{9}) β the number of bus stops and the capacity of the bus.
The second line contains a sequence a_1, a_2, ..., a_n (-10^{6} β€ a_i β€ 10^{6}), where a_i equals to the number, which has been recorded by the video system after the i-th bus stop.
Output
Print the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w. If the situation is contradictory (i.e. for any initial number of passengers there will be a contradiction), print 0.
Examples
Input
3 5
2 1 -3
Output
3
Input
2 4
-1 1
Output
4
Input
4 10
2 4 1 2
Output
2
Note
In the first example initially in the bus could be 0, 1 or 2 passengers.
In the second example initially in the bus could be 1, 2, 3 or 4 passengers.
In the third example initially in the bus could be 0 or 1 passenger.
Submitted Solution:
```
n, w = map(int, input().split())
a = list(map(int, input().split()))
mn = 0
mx = 0
now = 0
for i in a:
now += i
mx = max(mx, now)
mn = min(mn, now)
print(max(w-(mx-mn)+1, 0))
```
Yes
| 90,689 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The busses in Berland are equipped with a video surveillance system. The system records information about changes in the number of passengers in a bus after stops.
If x is the number of passengers in a bus just before the current bus stop and y is the number of passengers in the bus just after current bus stop, the system records the number y-x. So the system records show how number of passengers changed.
The test run was made for single bus and n bus stops. Thus, the system recorded the sequence of integers a_1, a_2, ..., a_n (exactly one number for each bus stop), where a_i is the record for the bus stop i. The bus stops are numbered from 1 to n in chronological order.
Determine the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w (that is, at any time in the bus there should be from 0 to w passengers inclusive).
Input
The first line contains two integers n and w (1 β€ n β€ 1 000, 1 β€ w β€ 10^{9}) β the number of bus stops and the capacity of the bus.
The second line contains a sequence a_1, a_2, ..., a_n (-10^{6} β€ a_i β€ 10^{6}), where a_i equals to the number, which has been recorded by the video system after the i-th bus stop.
Output
Print the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w. If the situation is contradictory (i.e. for any initial number of passengers there will be a contradiction), print 0.
Examples
Input
3 5
2 1 -3
Output
3
Input
2 4
-1 1
Output
4
Input
4 10
2 4 1 2
Output
2
Note
In the first example initially in the bus could be 0, 1 or 2 passengers.
In the second example initially in the bus could be 1, 2, 3 or 4 passengers.
In the third example initially in the bus could be 0 or 1 passenger.
Submitted Solution:
```
n,w=map(int,input().split())
l=map(int,input().split())
S,s=0,0
for x in l:
S+=x
if abs(S)>s:s=abs(S)
print(w-s+1)
```
No
| 90,690 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The busses in Berland are equipped with a video surveillance system. The system records information about changes in the number of passengers in a bus after stops.
If x is the number of passengers in a bus just before the current bus stop and y is the number of passengers in the bus just after current bus stop, the system records the number y-x. So the system records show how number of passengers changed.
The test run was made for single bus and n bus stops. Thus, the system recorded the sequence of integers a_1, a_2, ..., a_n (exactly one number for each bus stop), where a_i is the record for the bus stop i. The bus stops are numbered from 1 to n in chronological order.
Determine the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w (that is, at any time in the bus there should be from 0 to w passengers inclusive).
Input
The first line contains two integers n and w (1 β€ n β€ 1 000, 1 β€ w β€ 10^{9}) β the number of bus stops and the capacity of the bus.
The second line contains a sequence a_1, a_2, ..., a_n (-10^{6} β€ a_i β€ 10^{6}), where a_i equals to the number, which has been recorded by the video system after the i-th bus stop.
Output
Print the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w. If the situation is contradictory (i.e. for any initial number of passengers there will be a contradiction), print 0.
Examples
Input
3 5
2 1 -3
Output
3
Input
2 4
-1 1
Output
4
Input
4 10
2 4 1 2
Output
2
Note
In the first example initially in the bus could be 0, 1 or 2 passengers.
In the second example initially in the bus could be 1, 2, 3 or 4 passengers.
In the third example initially in the bus could be 0 or 1 passenger.
Submitted Solution:
```
n,w=[int(x) for x in input().split()]
l=list(map(int, input().split()))
for i in range(n-1):
l[i+1]+=l[i]
mn=min(l)
mx=max(l)
x=w+1
if mx>=0:
x-=mx
if mn<=0:
x+=mn
print(x)
```
No
| 90,691 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The busses in Berland are equipped with a video surveillance system. The system records information about changes in the number of passengers in a bus after stops.
If x is the number of passengers in a bus just before the current bus stop and y is the number of passengers in the bus just after current bus stop, the system records the number y-x. So the system records show how number of passengers changed.
The test run was made for single bus and n bus stops. Thus, the system recorded the sequence of integers a_1, a_2, ..., a_n (exactly one number for each bus stop), where a_i is the record for the bus stop i. The bus stops are numbered from 1 to n in chronological order.
Determine the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w (that is, at any time in the bus there should be from 0 to w passengers inclusive).
Input
The first line contains two integers n and w (1 β€ n β€ 1 000, 1 β€ w β€ 10^{9}) β the number of bus stops and the capacity of the bus.
The second line contains a sequence a_1, a_2, ..., a_n (-10^{6} β€ a_i β€ 10^{6}), where a_i equals to the number, which has been recorded by the video system after the i-th bus stop.
Output
Print the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w. If the situation is contradictory (i.e. for any initial number of passengers there will be a contradiction), print 0.
Examples
Input
3 5
2 1 -3
Output
3
Input
2 4
-1 1
Output
4
Input
4 10
2 4 1 2
Output
2
Note
In the first example initially in the bus could be 0, 1 or 2 passengers.
In the second example initially in the bus could be 1, 2, 3 or 4 passengers.
In the third example initially in the bus could be 0 or 1 passenger.
Submitted Solution:
```
import itertools
n, w = map(int,input().split())
A = list(map(int,input().split()))
if min(A) >= 0:
print(w-sum(A)+1)
exit()
if max(A) <= 0:
print(w+sum(A)+1)
exit()
csA = list(itertools.accumulate(A))
#print(csA)
if max(csA) > w:
print(0)
exit()
if min(csA) < -w:
print(0)
exit()
low = abs(min(csA))
if max(csA) > 0:
ans = (w-max(csA)-low+1)
print(ans if ans > 0 else 0)
else:
ans = w+min(csA)+1
print(ans if ans > 0 else 0)
```
No
| 90,692 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The busses in Berland are equipped with a video surveillance system. The system records information about changes in the number of passengers in a bus after stops.
If x is the number of passengers in a bus just before the current bus stop and y is the number of passengers in the bus just after current bus stop, the system records the number y-x. So the system records show how number of passengers changed.
The test run was made for single bus and n bus stops. Thus, the system recorded the sequence of integers a_1, a_2, ..., a_n (exactly one number for each bus stop), where a_i is the record for the bus stop i. The bus stops are numbered from 1 to n in chronological order.
Determine the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w (that is, at any time in the bus there should be from 0 to w passengers inclusive).
Input
The first line contains two integers n and w (1 β€ n β€ 1 000, 1 β€ w β€ 10^{9}) β the number of bus stops and the capacity of the bus.
The second line contains a sequence a_1, a_2, ..., a_n (-10^{6} β€ a_i β€ 10^{6}), where a_i equals to the number, which has been recorded by the video system after the i-th bus stop.
Output
Print the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w. If the situation is contradictory (i.e. for any initial number of passengers there will be a contradiction), print 0.
Examples
Input
3 5
2 1 -3
Output
3
Input
2 4
-1 1
Output
4
Input
4 10
2 4 1 2
Output
2
Note
In the first example initially in the bus could be 0, 1 or 2 passengers.
In the second example initially in the bus could be 1, 2, 3 or 4 passengers.
In the third example initially in the bus could be 0 or 1 passenger.
Submitted Solution:
```
n,w=map(int,input().split())
a=[int(i) for i in input().split()]
x1=0
x2=w
for i in a:
x1=max(x1+i,0)
x2=max(min(x2+i,w),0)
k=max(0,x2-x1+1)
print(k if k<=w and k>0 else 0)
```
No
| 90,693 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's introduce a number system which is based on a roman digits. There are digits I, V, X, L which correspond to the numbers 1, 5, 10 and 50 respectively. The use of other roman digits is not allowed.
Numbers in this system are written as a sequence of one or more digits. We define the value of the sequence simply as the sum of digits in it.
For example, the number XXXV evaluates to 35 and the number IXI β to 12.
Pay attention to the difference to the traditional roman system β in our system any sequence of digits is valid, moreover the order of digits doesn't matter, for example IX means 11, not 9.
One can notice that this system is ambiguous, and some numbers can be written in many different ways. Your goal is to determine how many distinct integers can be represented by exactly n roman digits I, V, X, L.
Input
The only line of the input file contains a single integer n (1 β€ n β€ 10^9) β the number of roman digits to use.
Output
Output a single integer β the number of distinct integers which can be represented using n roman digits exactly.
Examples
Input
1
Output
4
Input
2
Output
10
Input
10
Output
244
Note
In the first sample there are exactly 4 integers which can be represented β I, V, X and L.
In the second sample it is possible to represent integers 2 (II), 6 (VI), 10 (VV), 11 (XI), 15 (XV), 20 (XX), 51 (IL), 55 (VL), 60 (XL) and 100 (LL).
Tags: brute force, combinatorics, dp, greedy, math
Correct Solution:
```
from sys import setrecursionlimit
setrecursionlimit((10**9))
n = int(input())
s = set()
arr = [1,5,10,50]
def comb(num, digit):
global n
if digit == n:
s.add(num)
else:
for i in range(4):
comb(num+arr[i], digit+1)
if n <= 12:
comb(0, 0)
print(len(s))
else:
print(49*n-247)
```
| 90,694 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's introduce a number system which is based on a roman digits. There are digits I, V, X, L which correspond to the numbers 1, 5, 10 and 50 respectively. The use of other roman digits is not allowed.
Numbers in this system are written as a sequence of one or more digits. We define the value of the sequence simply as the sum of digits in it.
For example, the number XXXV evaluates to 35 and the number IXI β to 12.
Pay attention to the difference to the traditional roman system β in our system any sequence of digits is valid, moreover the order of digits doesn't matter, for example IX means 11, not 9.
One can notice that this system is ambiguous, and some numbers can be written in many different ways. Your goal is to determine how many distinct integers can be represented by exactly n roman digits I, V, X, L.
Input
The only line of the input file contains a single integer n (1 β€ n β€ 10^9) β the number of roman digits to use.
Output
Output a single integer β the number of distinct integers which can be represented using n roman digits exactly.
Examples
Input
1
Output
4
Input
2
Output
10
Input
10
Output
244
Note
In the first sample there are exactly 4 integers which can be represented β I, V, X and L.
In the second sample it is possible to represent integers 2 (II), 6 (VI), 10 (VV), 11 (XI), 15 (XV), 20 (XX), 51 (IL), 55 (VL), 60 (XL) and 100 (LL).
Tags: brute force, combinatorics, dp, greedy, math
Correct Solution:
```
#!/usr/bin/env python
# -*- coding: utf-8 -*-
n = int(input())
if n < 12:
print([1, 4, 10, 20, 35, 56, 83, 116, 155, 198, 244, 292][n])
else:
print(n*50-n-247)
```
| 90,695 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's introduce a number system which is based on a roman digits. There are digits I, V, X, L which correspond to the numbers 1, 5, 10 and 50 respectively. The use of other roman digits is not allowed.
Numbers in this system are written as a sequence of one or more digits. We define the value of the sequence simply as the sum of digits in it.
For example, the number XXXV evaluates to 35 and the number IXI β to 12.
Pay attention to the difference to the traditional roman system β in our system any sequence of digits is valid, moreover the order of digits doesn't matter, for example IX means 11, not 9.
One can notice that this system is ambiguous, and some numbers can be written in many different ways. Your goal is to determine how many distinct integers can be represented by exactly n roman digits I, V, X, L.
Input
The only line of the input file contains a single integer n (1 β€ n β€ 10^9) β the number of roman digits to use.
Output
Output a single integer β the number of distinct integers which can be represented using n roman digits exactly.
Examples
Input
1
Output
4
Input
2
Output
10
Input
10
Output
244
Note
In the first sample there are exactly 4 integers which can be represented β I, V, X and L.
In the second sample it is possible to represent integers 2 (II), 6 (VI), 10 (VV), 11 (XI), 15 (XV), 20 (XX), 51 (IL), 55 (VL), 60 (XL) and 100 (LL).
Tags: brute force, combinatorics, dp, greedy, math
Correct Solution:
```
def check(n, b, c):
for tb in range(0, 9):
for tc in range(0, min(n - tb + 1, 9)):
s1 = 9 * b + 4 * c
s2 = 9 * tb + 4 * tc
if s1 > s2 and (s1 - s2) % 49 == 0:
return False
return True
def main():
n = int(input())
ans = 0
for b in range(0, min(n + 1, 9)):
for c in range(0, min(n - b + 1, 9)):
if check(n, b, c):
ans += (n - b - c + 1)
print(ans)
if __name__ == "__main__":
exit(main())
```
| 90,696 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's introduce a number system which is based on a roman digits. There are digits I, V, X, L which correspond to the numbers 1, 5, 10 and 50 respectively. The use of other roman digits is not allowed.
Numbers in this system are written as a sequence of one or more digits. We define the value of the sequence simply as the sum of digits in it.
For example, the number XXXV evaluates to 35 and the number IXI β to 12.
Pay attention to the difference to the traditional roman system β in our system any sequence of digits is valid, moreover the order of digits doesn't matter, for example IX means 11, not 9.
One can notice that this system is ambiguous, and some numbers can be written in many different ways. Your goal is to determine how many distinct integers can be represented by exactly n roman digits I, V, X, L.
Input
The only line of the input file contains a single integer n (1 β€ n β€ 10^9) β the number of roman digits to use.
Output
Output a single integer β the number of distinct integers which can be represented using n roman digits exactly.
Examples
Input
1
Output
4
Input
2
Output
10
Input
10
Output
244
Note
In the first sample there are exactly 4 integers which can be represented β I, V, X and L.
In the second sample it is possible to represent integers 2 (II), 6 (VI), 10 (VV), 11 (XI), 15 (XV), 20 (XX), 51 (IL), 55 (VL), 60 (XL) and 100 (LL).
Tags: brute force, combinatorics, dp, greedy, math
Correct Solution:
```
a = [0, 4, 10, 20, 35, 56, 83, 116, 155, 198, 244, 292]
n = int(input())
if n < len(a):
print (a[n])
else:
print (a[11] + 49*(n-11))
```
| 90,697 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's introduce a number system which is based on a roman digits. There are digits I, V, X, L which correspond to the numbers 1, 5, 10 and 50 respectively. The use of other roman digits is not allowed.
Numbers in this system are written as a sequence of one or more digits. We define the value of the sequence simply as the sum of digits in it.
For example, the number XXXV evaluates to 35 and the number IXI β to 12.
Pay attention to the difference to the traditional roman system β in our system any sequence of digits is valid, moreover the order of digits doesn't matter, for example IX means 11, not 9.
One can notice that this system is ambiguous, and some numbers can be written in many different ways. Your goal is to determine how many distinct integers can be represented by exactly n roman digits I, V, X, L.
Input
The only line of the input file contains a single integer n (1 β€ n β€ 10^9) β the number of roman digits to use.
Output
Output a single integer β the number of distinct integers which can be represented using n roman digits exactly.
Examples
Input
1
Output
4
Input
2
Output
10
Input
10
Output
244
Note
In the first sample there are exactly 4 integers which can be represented β I, V, X and L.
In the second sample it is possible to represent integers 2 (II), 6 (VI), 10 (VV), 11 (XI), 15 (XV), 20 (XX), 51 (IL), 55 (VL), 60 (XL) and 100 (LL).
Tags: brute force, combinatorics, dp, greedy, math
Correct Solution:
```
def ceildiv(a, b):
return -(-a // b)
def test_ceildiv():
assert ceildiv(6, 3) == 2
assert ceildiv(7, 3) == 3
assert ceildiv(5, 3) == 2
def brute_force(n, m):
"""Count numbers in range [0; m] equal to 4a + 9b + 49c
...for some non-negative a, b and c such that (a + b + c) <= n
"""
numbers = set()
max_c = min(n, m // 49)
for c in range(max_c + 1):
_49c = 49 * c
max_b = min(n - c, (m - _49c) // 9, 48)
for b in range(max_b + 1):
_9b_49c = 9 * b + _49c
max_a = min(n - b - c, (m - _9b_49c) // 4, 8)
for a in range(max_a + 1):
numbers.add(4 * a + _9b_49c)
assert 0 in numbers
for x in (4, 9, 49):
if m // x <= n:
assert m - (m % 49) in numbers
return len(numbers)
def brute_force_range(n, l, u):
"""Count numbers in range [l; u] equal to 4a + 9b + 49c
...for some non-negative a, b and c such that (a + b + c) <= n
"""
numbers = set()
min_c = max(0, ceildiv(l - 9 * n, 40))
max_c = min(n, u // 49)
for c in range(min_c, max_c + 1):
_49c = 49 * c
min_b = max(0, ceildiv(l - 4 * n - 45 * c, 5))
max_b = min(n - c, (u - _49c) // 9, 48)
for b in range(min_b, max_b + 1):
_9b_49c = 9 * b + _49c
min_a = max(0, ceildiv(l - _9b_49c, 4))
max_a = min(n - b - c, (u - _9b_49c) // 4, 8)
for a in range(min_a, max_a + 1):
numbers.add(4 * a + _9b_49c)
if numbers:
assert min(numbers) >= l
assert max(numbers) <= u
return len(numbers)
def test_brute_force_range():
for (u, l) in ((60, 80), (104, 599), (200, 777)):
for n in (10, 13, 15, 17, 20):
assert brute_force_range(n, u, l) == brute_force(n, l) - brute_force(n, u - 1)
def solid_interval(n):
assert n >= 18
return 27, (49 * n - 807) + 1
def test_solid_interval():
for n in 18, 19, 100, 200, 300, 1000, 5000:
begin, end = solid_interval(n)
assert brute_force_range(n, begin, end - 1) == end - begin
def main():
n = int(input())
if n < 18:
result = brute_force(n, 49 * n) # naΓ―ve solution
else:
begin, end = solid_interval(n)
assert begin < end < 49 * n
result = sum((
brute_force_range(n, 0, begin - 1),
end - begin, # == brute_force_range(n, begin, end - 1)
brute_force_range(n, end, 49 * n)
))
print(result)
if __name__ == '__main__':
main()
```
| 90,698 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's introduce a number system which is based on a roman digits. There are digits I, V, X, L which correspond to the numbers 1, 5, 10 and 50 respectively. The use of other roman digits is not allowed.
Numbers in this system are written as a sequence of one or more digits. We define the value of the sequence simply as the sum of digits in it.
For example, the number XXXV evaluates to 35 and the number IXI β to 12.
Pay attention to the difference to the traditional roman system β in our system any sequence of digits is valid, moreover the order of digits doesn't matter, for example IX means 11, not 9.
One can notice that this system is ambiguous, and some numbers can be written in many different ways. Your goal is to determine how many distinct integers can be represented by exactly n roman digits I, V, X, L.
Input
The only line of the input file contains a single integer n (1 β€ n β€ 10^9) β the number of roman digits to use.
Output
Output a single integer β the number of distinct integers which can be represented using n roman digits exactly.
Examples
Input
1
Output
4
Input
2
Output
10
Input
10
Output
244
Note
In the first sample there are exactly 4 integers which can be represented β I, V, X and L.
In the second sample it is possible to represent integers 2 (II), 6 (VI), 10 (VV), 11 (XI), 15 (XV), 20 (XX), 51 (IL), 55 (VL), 60 (XL) and 100 (LL).
Tags: brute force, combinatorics, dp, greedy, math
Correct Solution:
```
ans=0
def solve(nn,su,ans):
if nn==n:
s.add(su)
ans+=1
return
solve(nn+1,su+1,ans)
solve(nn+1,su+5,ans)
solve(nn+1,su+10,ans)
solve(nn+1,su+50,ans)
s = set()
n = int(input())
if n<=13:
solve(0,0,ans)
print(len(s))
else:
print(292 + 49*(n-11))
```
| 90,699 |
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