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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In a galaxy far, far away Lesha the student has just got to know that he has an exam in two days. As always, he hasn't attended any single class during the previous year, so he decided to spend the remaining time wisely.
Lesha knows that today he can study for at most a hours, and he will have b hours to study tomorrow. Note that it is possible that on his planet there are more hours in a day than on Earth. Lesha knows that the quality of his knowledge will only depend on the number of lecture notes he will read. He has access to an infinite number of notes that are enumerated with positive integers, but he knows that he can read the first note in one hour, the second note in two hours and so on. In other words, Lesha can read the note with number k in k hours. Lesha can read the notes in arbitrary order, however, he can't start reading a note in the first day and finish its reading in the second day.
Thus, the student has to fully read several lecture notes today, spending at most a hours in total, and fully read several lecture notes tomorrow, spending at most b hours in total. What is the maximum number of notes Lesha can read in the remaining time? Which notes should he read in the first day, and which — in the second?
Input
The only line of input contains two integers a and b (0 ≤ a, b ≤ 10^{9}) — the number of hours Lesha has today and the number of hours Lesha has tomorrow.
Output
In the first line print a single integer n (0 ≤ n ≤ a) — the number of lecture notes Lesha has to read in the first day. In the second line print n distinct integers p_1, p_2, …, p_n (1 ≤ p_i ≤ a), the sum of all p_i should not exceed a.
In the third line print a single integer m (0 ≤ m ≤ b) — the number of lecture notes Lesha has to read in the second day. In the fourth line print m distinct integers q_1, q_2, …, q_m (1 ≤ q_i ≤ b), the sum of all q_i should not exceed b.
All integers p_i and q_i should be distinct. The sum n + m should be largest possible.
Examples
Input
3 3
Output
1
3
2
2 1
Input
9 12
Output
2
3 6
4
1 2 4 5
Note
In the first example Lesha can read the third note in 3 hours in the first day, and the first and the second notes in one and two hours correspondingly in the second day, spending 3 hours as well. Note that Lesha can make it the other way round, reading the first and the second notes in the first day and the third note in the second day.
In the second example Lesha should read the third and the sixth notes in the first day, spending 9 hours in total. In the second day Lesha should read the first, second fourth and fifth notes, spending 12 hours in total.
Submitted Solution:
```
def main():
a, b = map(int, input().split())
if a + b == 0:
print(0)
print(0)
return
_sum = 0
last = 0
for i in range(1, a + b):
_sum += i
if _sum == a + b:
last = i
break
elif _sum > a + b:
_sum -= i
last = i - 1
break
first = 1
current = last
used = set()
q_1 = min(a, b)
q = q_1
if last % 2 == 0:
if q <= last:
if q == 0:
pass
else:
used.add(q)
else:
while q >= first:
used.add(first)
q -= first
first += 1
if q > current:
used.add(current)
q -= current
current -= 1
else:
if q <= last:
if q == 0:
pass
else:
used.add(q)
else:
used.add(last)
q -= last
last -= 1
while q >= first:
used.add(first)
q -= first
first += 1
if q > current:
used.add(current)
q -= current
current -= 1
if q_1 == a:
print(len(used))
print(*used)
print(last - len(used))
print(*(used^{i for i in range(1, last + 1)}))
return
else:
print(last - len(used))
print(*(used^{i for i in range(1, last + 1)}))
print(len(used))
print(*used)
return
if __name__=="__main__":
main()
```
No
| 99,500 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In a galaxy far, far away Lesha the student has just got to know that he has an exam in two days. As always, he hasn't attended any single class during the previous year, so he decided to spend the remaining time wisely.
Lesha knows that today he can study for at most a hours, and he will have b hours to study tomorrow. Note that it is possible that on his planet there are more hours in a day than on Earth. Lesha knows that the quality of his knowledge will only depend on the number of lecture notes he will read. He has access to an infinite number of notes that are enumerated with positive integers, but he knows that he can read the first note in one hour, the second note in two hours and so on. In other words, Lesha can read the note with number k in k hours. Lesha can read the notes in arbitrary order, however, he can't start reading a note in the first day and finish its reading in the second day.
Thus, the student has to fully read several lecture notes today, spending at most a hours in total, and fully read several lecture notes tomorrow, spending at most b hours in total. What is the maximum number of notes Lesha can read in the remaining time? Which notes should he read in the first day, and which — in the second?
Input
The only line of input contains two integers a and b (0 ≤ a, b ≤ 10^{9}) — the number of hours Lesha has today and the number of hours Lesha has tomorrow.
Output
In the first line print a single integer n (0 ≤ n ≤ a) — the number of lecture notes Lesha has to read in the first day. In the second line print n distinct integers p_1, p_2, …, p_n (1 ≤ p_i ≤ a), the sum of all p_i should not exceed a.
In the third line print a single integer m (0 ≤ m ≤ b) — the number of lecture notes Lesha has to read in the second day. In the fourth line print m distinct integers q_1, q_2, …, q_m (1 ≤ q_i ≤ b), the sum of all q_i should not exceed b.
All integers p_i and q_i should be distinct. The sum n + m should be largest possible.
Examples
Input
3 3
Output
1
3
2
2 1
Input
9 12
Output
2
3 6
4
1 2 4 5
Note
In the first example Lesha can read the third note in 3 hours in the first day, and the first and the second notes in one and two hours correspondingly in the second day, spending 3 hours as well. Note that Lesha can make it the other way round, reading the first and the second notes in the first day and the third note in the second day.
In the second example Lesha should read the third and the sixth notes in the first day, spending 9 hours in total. In the second day Lesha should read the first, second fourth and fifth notes, spending 12 hours in total.
Submitted Solution:
```
def main():
a, b = map(int, input().split())
if a + b == 0:
print(0)
print(0)
return
_sum = 0
last = 0
for i in range(1, a + b):
_sum += i
if _sum == a + b:
last = i
break
elif _sum > a + b:
_sum -= i
last = i - 1
break
first = 1
current = last
used = set()
q_1 = min(a, b)
q = q_1
if last % 2 == 0:
if q <= last:
used.add(q)
else:
while q >= first:
used.add(first)
q -= first
first += 1
if q > current:
used.add(current)
q -= current
current -= 1
else:
if q <= last:
used.add(q)
else:
used.add(last)
q -= last
last -= 1
while q >= first:
used.add(first)
q -= first
first += 1
if q > current:
used.add(current)
q -= current
current -= 1
if q_1 == a:
print(len(used))
print(*used)
print(last - len(used))
print(*(used^{i for i in range(1, last + 1)}))
return
else:
print(last - len(used))
print(*(used^{i for i in range(1, last + 1)}))
print(len(used))
print(*used)
return
if __name__=="__main__":
main()
```
No
| 99,501 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In a galaxy far, far away Lesha the student has just got to know that he has an exam in two days. As always, he hasn't attended any single class during the previous year, so he decided to spend the remaining time wisely.
Lesha knows that today he can study for at most a hours, and he will have b hours to study tomorrow. Note that it is possible that on his planet there are more hours in a day than on Earth. Lesha knows that the quality of his knowledge will only depend on the number of lecture notes he will read. He has access to an infinite number of notes that are enumerated with positive integers, but he knows that he can read the first note in one hour, the second note in two hours and so on. In other words, Lesha can read the note with number k in k hours. Lesha can read the notes in arbitrary order, however, he can't start reading a note in the first day and finish its reading in the second day.
Thus, the student has to fully read several lecture notes today, spending at most a hours in total, and fully read several lecture notes tomorrow, spending at most b hours in total. What is the maximum number of notes Lesha can read in the remaining time? Which notes should he read in the first day, and which — in the second?
Input
The only line of input contains two integers a and b (0 ≤ a, b ≤ 10^{9}) — the number of hours Lesha has today and the number of hours Lesha has tomorrow.
Output
In the first line print a single integer n (0 ≤ n ≤ a) — the number of lecture notes Lesha has to read in the first day. In the second line print n distinct integers p_1, p_2, …, p_n (1 ≤ p_i ≤ a), the sum of all p_i should not exceed a.
In the third line print a single integer m (0 ≤ m ≤ b) — the number of lecture notes Lesha has to read in the second day. In the fourth line print m distinct integers q_1, q_2, …, q_m (1 ≤ q_i ≤ b), the sum of all q_i should not exceed b.
All integers p_i and q_i should be distinct. The sum n + m should be largest possible.
Examples
Input
3 3
Output
1
3
2
2 1
Input
9 12
Output
2
3 6
4
1 2 4 5
Note
In the first example Lesha can read the third note in 3 hours in the first day, and the first and the second notes in one and two hours correspondingly in the second day, spending 3 hours as well. Note that Lesha can make it the other way round, reading the first and the second notes in the first day and the third note in the second day.
In the second example Lesha should read the third and the sixth notes in the first day, spending 9 hours in total. In the second day Lesha should read the first, second fourth and fifth notes, spending 12 hours in total.
Submitted Solution:
```
def fill_hours_a(max_hour):
temp = max_hour
for i in range(1, max_hour+1):
temp = temp - i
if temp > 0:
num_set.add(i)
elif temp == 0:
num_set.add(i)
break
else:
if temp * -1 in num_set:
num_set.remove(temp * -1)
num_set.add(i)
def fill_hours_b(max_hour):
temp = max_hour
for i in range(1, max_hour+1):
if i not in num_set:
temp = temp - i
if temp > 0:
num_set_b.add(i)
elif temp == 0:
num_set_b.add(i)
break
else:
if temp * -1 in num_set_b:
num_set_b.remove(temp * -1)
num_set_b.add(i)
def get_out(in_set):
l=[]
for x in in_set:
l.append(str(x))
return ' '.join(l)
arr = input().split()
a = int(arr[0])
b = int(arr[1])
num_set = set()
num_set_b=set()
temp=0
if(a<b):
temp=a
a=b
b=temp
fill_hours_a(a)
fill_hours_b(b)
if(temp!=0):
print(get_out(num_set_b))
print(get_out(num_set))
else :
print(get_out(num_set))
print(get_out(num_set_b))
```
No
| 99,502 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mishka is trying really hard to avoid being kicked out of the university. In particular, he was doing absolutely nothing for the whole semester, miraculously passed some exams so that just one is left.
There were n classes of that subject during the semester and on i-th class professor mentioned some non-negative integer a_i to the students. It turned out, the exam was to tell the whole sequence back to the professor.
Sounds easy enough for those who attended every class, doesn't it?
Obviously Mishka didn't attend any classes. However, professor left some clues on the values of a to help out students like Mishka:
* a was sorted in non-decreasing order (a_1 ≤ a_2 ≤ ... ≤ a_n);
* n was even;
* the following sequence b, consisting of \frac n 2 elements, was formed and given out to students: b_i = a_i + a_{n - i + 1}.
Professor also mentioned that any sequence a, which produces sequence b with the presented technique, will be acceptable.
Help Mishka to pass that last exam. Restore any sorted sequence a of non-negative integers, which produces sequence b with the presented technique. It is guaranteed that there exists at least one correct sequence a, which produces the given sequence b.
Input
The first line contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of sequence a. n is always even.
The second line contains \frac n 2 integers b_1, b_2, ..., b_{\frac n 2} (0 ≤ b_i ≤ 10^{18}) — sequence b, where b_i = a_i + a_{n - i + 1}.
It is guaranteed that there exists at least one correct sequence a, which produces the given sequence b.
Output
Print n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{18}) in a single line.
a_1 ≤ a_2 ≤ ... ≤ a_n should be satisfied.
b_i = a_i + a_{n - i + 1} should be satisfied for all valid i.
Examples
Input
4
5 6
Output
2 3 3 3
Input
6
2 1 2
Output
0 0 1 1 1 2
Tags: greedy
Correct Solution:
```
n = int(input())
v = list(map(int, input().split()))
new_v = [0] * (n)
new_v[-1] = v[0]
for i in range(1, len(v)):
if v[i] > v[i - 1]:
new_v[i] = new_v[i - 1] + v[i] - v[i - 1]
else:
new_v[i] = new_v[i - 1]
new_v[-i - 1] = v[i] - new_v[i]
print(*new_v)
```
| 99,503 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mishka is trying really hard to avoid being kicked out of the university. In particular, he was doing absolutely nothing for the whole semester, miraculously passed some exams so that just one is left.
There were n classes of that subject during the semester and on i-th class professor mentioned some non-negative integer a_i to the students. It turned out, the exam was to tell the whole sequence back to the professor.
Sounds easy enough for those who attended every class, doesn't it?
Obviously Mishka didn't attend any classes. However, professor left some clues on the values of a to help out students like Mishka:
* a was sorted in non-decreasing order (a_1 ≤ a_2 ≤ ... ≤ a_n);
* n was even;
* the following sequence b, consisting of \frac n 2 elements, was formed and given out to students: b_i = a_i + a_{n - i + 1}.
Professor also mentioned that any sequence a, which produces sequence b with the presented technique, will be acceptable.
Help Mishka to pass that last exam. Restore any sorted sequence a of non-negative integers, which produces sequence b with the presented technique. It is guaranteed that there exists at least one correct sequence a, which produces the given sequence b.
Input
The first line contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of sequence a. n is always even.
The second line contains \frac n 2 integers b_1, b_2, ..., b_{\frac n 2} (0 ≤ b_i ≤ 10^{18}) — sequence b, where b_i = a_i + a_{n - i + 1}.
It is guaranteed that there exists at least one correct sequence a, which produces the given sequence b.
Output
Print n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{18}) in a single line.
a_1 ≤ a_2 ≤ ... ≤ a_n should be satisfied.
b_i = a_i + a_{n - i + 1} should be satisfied for all valid i.
Examples
Input
4
5 6
Output
2 3 3 3
Input
6
2 1 2
Output
0 0 1 1 1 2
Tags: greedy
Correct Solution:
```
n = int(input())
arr = list(map(int,input().split()))
ans=[0]*(n)
l = 0
h = float('inf')
for i in range(n//2):
ans[i] = max(l,arr[i]-h)
ans[n-1-i] = arr[i]-ans[i]
l = ans[i]
h = ans[n-1-i]
print(*ans)
```
| 99,504 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mishka is trying really hard to avoid being kicked out of the university. In particular, he was doing absolutely nothing for the whole semester, miraculously passed some exams so that just one is left.
There were n classes of that subject during the semester and on i-th class professor mentioned some non-negative integer a_i to the students. It turned out, the exam was to tell the whole sequence back to the professor.
Sounds easy enough for those who attended every class, doesn't it?
Obviously Mishka didn't attend any classes. However, professor left some clues on the values of a to help out students like Mishka:
* a was sorted in non-decreasing order (a_1 ≤ a_2 ≤ ... ≤ a_n);
* n was even;
* the following sequence b, consisting of \frac n 2 elements, was formed and given out to students: b_i = a_i + a_{n - i + 1}.
Professor also mentioned that any sequence a, which produces sequence b with the presented technique, will be acceptable.
Help Mishka to pass that last exam. Restore any sorted sequence a of non-negative integers, which produces sequence b with the presented technique. It is guaranteed that there exists at least one correct sequence a, which produces the given sequence b.
Input
The first line contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of sequence a. n is always even.
The second line contains \frac n 2 integers b_1, b_2, ..., b_{\frac n 2} (0 ≤ b_i ≤ 10^{18}) — sequence b, where b_i = a_i + a_{n - i + 1}.
It is guaranteed that there exists at least one correct sequence a, which produces the given sequence b.
Output
Print n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{18}) in a single line.
a_1 ≤ a_2 ≤ ... ≤ a_n should be satisfied.
b_i = a_i + a_{n - i + 1} should be satisfied for all valid i.
Examples
Input
4
5 6
Output
2 3 3 3
Input
6
2 1 2
Output
0 0 1 1 1 2
Tags: greedy
Correct Solution:
```
n, b = int(input()), list(map(int, input().split()))
a = [0 for _ in range(n)]
a[n - 1], a[0] = b[0], 0
for i in range(1, n // 2):
if a[n - i] < b[i] and b[i] - a[n - i] >= a[i - 1]:
a[n - 1 - i], a[i] = a[n - i], b[i] - a[n - i]
else:
a[n - 1 - i], a[i] = b[i] - a[i - 1], a[i - 1]
print(*a)
```
| 99,505 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mishka is trying really hard to avoid being kicked out of the university. In particular, he was doing absolutely nothing for the whole semester, miraculously passed some exams so that just one is left.
There were n classes of that subject during the semester and on i-th class professor mentioned some non-negative integer a_i to the students. It turned out, the exam was to tell the whole sequence back to the professor.
Sounds easy enough for those who attended every class, doesn't it?
Obviously Mishka didn't attend any classes. However, professor left some clues on the values of a to help out students like Mishka:
* a was sorted in non-decreasing order (a_1 ≤ a_2 ≤ ... ≤ a_n);
* n was even;
* the following sequence b, consisting of \frac n 2 elements, was formed and given out to students: b_i = a_i + a_{n - i + 1}.
Professor also mentioned that any sequence a, which produces sequence b with the presented technique, will be acceptable.
Help Mishka to pass that last exam. Restore any sorted sequence a of non-negative integers, which produces sequence b with the presented technique. It is guaranteed that there exists at least one correct sequence a, which produces the given sequence b.
Input
The first line contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of sequence a. n is always even.
The second line contains \frac n 2 integers b_1, b_2, ..., b_{\frac n 2} (0 ≤ b_i ≤ 10^{18}) — sequence b, where b_i = a_i + a_{n - i + 1}.
It is guaranteed that there exists at least one correct sequence a, which produces the given sequence b.
Output
Print n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{18}) in a single line.
a_1 ≤ a_2 ≤ ... ≤ a_n should be satisfied.
b_i = a_i + a_{n - i + 1} should be satisfied for all valid i.
Examples
Input
4
5 6
Output
2 3 3 3
Input
6
2 1 2
Output
0 0 1 1 1 2
Tags: greedy
Correct Solution:
```
import sys
import os
from math import*
input=sys.stdin.buffer.readline
n=int(input())
arr=list(map(int,input().split()))
ans=[0]*n
c=0
d=arr[0]
ans[0]=c
ans[n-1]=d
for i in range(1,n//2):
if arr[i]<=d:
d=arr[i]-c
else:
c=max(c,arr[i]-d)
d=arr[i]-c
ans[i]=c
ans[n-i-1]=d
for x in ans:
print(x,end=' ')
```
| 99,506 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mishka is trying really hard to avoid being kicked out of the university. In particular, he was doing absolutely nothing for the whole semester, miraculously passed some exams so that just one is left.
There were n classes of that subject during the semester and on i-th class professor mentioned some non-negative integer a_i to the students. It turned out, the exam was to tell the whole sequence back to the professor.
Sounds easy enough for those who attended every class, doesn't it?
Obviously Mishka didn't attend any classes. However, professor left some clues on the values of a to help out students like Mishka:
* a was sorted in non-decreasing order (a_1 ≤ a_2 ≤ ... ≤ a_n);
* n was even;
* the following sequence b, consisting of \frac n 2 elements, was formed and given out to students: b_i = a_i + a_{n - i + 1}.
Professor also mentioned that any sequence a, which produces sequence b with the presented technique, will be acceptable.
Help Mishka to pass that last exam. Restore any sorted sequence a of non-negative integers, which produces sequence b with the presented technique. It is guaranteed that there exists at least one correct sequence a, which produces the given sequence b.
Input
The first line contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of sequence a. n is always even.
The second line contains \frac n 2 integers b_1, b_2, ..., b_{\frac n 2} (0 ≤ b_i ≤ 10^{18}) — sequence b, where b_i = a_i + a_{n - i + 1}.
It is guaranteed that there exists at least one correct sequence a, which produces the given sequence b.
Output
Print n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{18}) in a single line.
a_1 ≤ a_2 ≤ ... ≤ a_n should be satisfied.
b_i = a_i + a_{n - i + 1} should be satisfied for all valid i.
Examples
Input
4
5 6
Output
2 3 3 3
Input
6
2 1 2
Output
0 0 1 1 1 2
Tags: greedy
Correct Solution:
```
from bisect import bisect_right as br
from bisect import bisect_left as bl
from collections import defaultdict
from itertools import combinations
import sys
import math
MAX = sys.maxsize
MAXN = 10**6+10
MOD = 10**9+7
def isprime(n):
n = abs(int(n))
if n < 2:
return False
if n == 2:
return True
if not n & 1:
return False
for x in range(3, int(n**0.5) + 1, 2):
if n % x == 0:
return False
return True
def mhd(a,b,x,y):
return abs(a-x)+abs(b-y)
def numIN():
return(map(int,sys.stdin.readline().strip().split()))
def charIN():
return(sys.stdin.readline().strip().split())
t = [(-1,-1)]*1000010
def create(a):
global t,n
for i in range(n,2*n):
t[i] = (a[i-n],i-n)
for i in range(n-1,0,-1):
x = [t[2*i],t[2*i+1]]
x.sort(key = lambda x:x[0])
t[i] = x[1]
def update(idx,value):
global t,n
idx = idx+n
t[idx] = value
while(idx>1):
idx = idx//2
x = [t[2*idx],t[2*idx+1]]
x.sort(key = lambda x:x[0])
t[idx] = x[1]
n = int(input())
b = [-1]+list(numIN())
ans = [0]*(n+1)
ans[1] = 0
ans[n] = b[1]
for i in range(2,n//2+1):
if b[i]<=b[i-1]:
ans[i] = ans[i-1]
ans[n-i+1] = b[i]-ans[i]
else:
ans[n-i+1] = ans[n-i+2]
ans[i] = b[i]-ans[n-i+1]
print(' '.join([str(i) for i in ans[1:]]))
```
| 99,507 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mishka is trying really hard to avoid being kicked out of the university. In particular, he was doing absolutely nothing for the whole semester, miraculously passed some exams so that just one is left.
There were n classes of that subject during the semester and on i-th class professor mentioned some non-negative integer a_i to the students. It turned out, the exam was to tell the whole sequence back to the professor.
Sounds easy enough for those who attended every class, doesn't it?
Obviously Mishka didn't attend any classes. However, professor left some clues on the values of a to help out students like Mishka:
* a was sorted in non-decreasing order (a_1 ≤ a_2 ≤ ... ≤ a_n);
* n was even;
* the following sequence b, consisting of \frac n 2 elements, was formed and given out to students: b_i = a_i + a_{n - i + 1}.
Professor also mentioned that any sequence a, which produces sequence b with the presented technique, will be acceptable.
Help Mishka to pass that last exam. Restore any sorted sequence a of non-negative integers, which produces sequence b with the presented technique. It is guaranteed that there exists at least one correct sequence a, which produces the given sequence b.
Input
The first line contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of sequence a. n is always even.
The second line contains \frac n 2 integers b_1, b_2, ..., b_{\frac n 2} (0 ≤ b_i ≤ 10^{18}) — sequence b, where b_i = a_i + a_{n - i + 1}.
It is guaranteed that there exists at least one correct sequence a, which produces the given sequence b.
Output
Print n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{18}) in a single line.
a_1 ≤ a_2 ≤ ... ≤ a_n should be satisfied.
b_i = a_i + a_{n - i + 1} should be satisfied for all valid i.
Examples
Input
4
5 6
Output
2 3 3 3
Input
6
2 1 2
Output
0 0 1 1 1 2
Tags: greedy
Correct Solution:
```
n = int(input())
arr = list(map(int, input().split()))
left = list()
right = list()
for v in arr:
if len(left) == 0:
left.append(0)
right.append(v)
else:
rvalue = min(v, right[-1])
lvalue = v - rvalue
if lvalue < left[-1]:
lvalue = left[-1]
rvalue = v - lvalue
left.append(lvalue)
right.append(rvalue)
print(*left, *reversed(right))
```
| 99,508 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mishka is trying really hard to avoid being kicked out of the university. In particular, he was doing absolutely nothing for the whole semester, miraculously passed some exams so that just one is left.
There were n classes of that subject during the semester and on i-th class professor mentioned some non-negative integer a_i to the students. It turned out, the exam was to tell the whole sequence back to the professor.
Sounds easy enough for those who attended every class, doesn't it?
Obviously Mishka didn't attend any classes. However, professor left some clues on the values of a to help out students like Mishka:
* a was sorted in non-decreasing order (a_1 ≤ a_2 ≤ ... ≤ a_n);
* n was even;
* the following sequence b, consisting of \frac n 2 elements, was formed and given out to students: b_i = a_i + a_{n - i + 1}.
Professor also mentioned that any sequence a, which produces sequence b with the presented technique, will be acceptable.
Help Mishka to pass that last exam. Restore any sorted sequence a of non-negative integers, which produces sequence b with the presented technique. It is guaranteed that there exists at least one correct sequence a, which produces the given sequence b.
Input
The first line contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of sequence a. n is always even.
The second line contains \frac n 2 integers b_1, b_2, ..., b_{\frac n 2} (0 ≤ b_i ≤ 10^{18}) — sequence b, where b_i = a_i + a_{n - i + 1}.
It is guaranteed that there exists at least one correct sequence a, which produces the given sequence b.
Output
Print n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{18}) in a single line.
a_1 ≤ a_2 ≤ ... ≤ a_n should be satisfied.
b_i = a_i + a_{n - i + 1} should be satisfied for all valid i.
Examples
Input
4
5 6
Output
2 3 3 3
Input
6
2 1 2
Output
0 0 1 1 1 2
Tags: greedy
Correct Solution:
```
n = int(input())
b = [int(x) for x in input().split()]
lh=[0]
rh=[b[0]]
for x in b[1:]:
if x>=rh[-1]:
y=x-rh[-1]
if y>=lh[-1]:
lh.append(y)
rh.append(rh[-1])
else:
lh.append(lh[-1])
rh.append(x-lh[-1])
else:
y = x-lh[-1]
if y<=rh[-1]:
rh.append(y)
lh.append(lh[-1])
else:
rh.append(rh[-1])
lh.append(rh[-1]-x)
print(" ".join(list(map(str,lh+rh[::-1]))))
```
| 99,509 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mishka is trying really hard to avoid being kicked out of the university. In particular, he was doing absolutely nothing for the whole semester, miraculously passed some exams so that just one is left.
There were n classes of that subject during the semester and on i-th class professor mentioned some non-negative integer a_i to the students. It turned out, the exam was to tell the whole sequence back to the professor.
Sounds easy enough for those who attended every class, doesn't it?
Obviously Mishka didn't attend any classes. However, professor left some clues on the values of a to help out students like Mishka:
* a was sorted in non-decreasing order (a_1 ≤ a_2 ≤ ... ≤ a_n);
* n was even;
* the following sequence b, consisting of \frac n 2 elements, was formed and given out to students: b_i = a_i + a_{n - i + 1}.
Professor also mentioned that any sequence a, which produces sequence b with the presented technique, will be acceptable.
Help Mishka to pass that last exam. Restore any sorted sequence a of non-negative integers, which produces sequence b with the presented technique. It is guaranteed that there exists at least one correct sequence a, which produces the given sequence b.
Input
The first line contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of sequence a. n is always even.
The second line contains \frac n 2 integers b_1, b_2, ..., b_{\frac n 2} (0 ≤ b_i ≤ 10^{18}) — sequence b, where b_i = a_i + a_{n - i + 1}.
It is guaranteed that there exists at least one correct sequence a, which produces the given sequence b.
Output
Print n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{18}) in a single line.
a_1 ≤ a_2 ≤ ... ≤ a_n should be satisfied.
b_i = a_i + a_{n - i + 1} should be satisfied for all valid i.
Examples
Input
4
5 6
Output
2 3 3 3
Input
6
2 1 2
Output
0 0 1 1 1 2
Tags: greedy
Correct Solution:
```
n = int(input())
A = list(map(int,input().split()))
res = [0]*n
res[0] = 0
res[n-1] = A[0]
for i in range(1,n//2):
res[i] = max(res[i-1],A[i]-A[i-1] + res[i-1])
res[n-i-1] = A[i]-res[i]
print(*res)
```
| 99,510 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mishka is trying really hard to avoid being kicked out of the university. In particular, he was doing absolutely nothing for the whole semester, miraculously passed some exams so that just one is left.
There were n classes of that subject during the semester and on i-th class professor mentioned some non-negative integer a_i to the students. It turned out, the exam was to tell the whole sequence back to the professor.
Sounds easy enough for those who attended every class, doesn't it?
Obviously Mishka didn't attend any classes. However, professor left some clues on the values of a to help out students like Mishka:
* a was sorted in non-decreasing order (a_1 ≤ a_2 ≤ ... ≤ a_n);
* n was even;
* the following sequence b, consisting of \frac n 2 elements, was formed and given out to students: b_i = a_i + a_{n - i + 1}.
Professor also mentioned that any sequence a, which produces sequence b with the presented technique, will be acceptable.
Help Mishka to pass that last exam. Restore any sorted sequence a of non-negative integers, which produces sequence b with the presented technique. It is guaranteed that there exists at least one correct sequence a, which produces the given sequence b.
Input
The first line contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of sequence a. n is always even.
The second line contains \frac n 2 integers b_1, b_2, ..., b_{\frac n 2} (0 ≤ b_i ≤ 10^{18}) — sequence b, where b_i = a_i + a_{n - i + 1}.
It is guaranteed that there exists at least one correct sequence a, which produces the given sequence b.
Output
Print n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{18}) in a single line.
a_1 ≤ a_2 ≤ ... ≤ a_n should be satisfied.
b_i = a_i + a_{n - i + 1} should be satisfied for all valid i.
Examples
Input
4
5 6
Output
2 3 3 3
Input
6
2 1 2
Output
0 0 1 1 1 2
Submitted Solution:
```
n = int(input())
b = [int(w) for w in input().split()]
a = [0]*n
l = n//2 - 1
r = n//2
a[l] = b[l] // 2
a[r] = b[l] - a[l]
while l > 0:
if b[l-1] >= b[l]:
a[l-1] = a[l]
a[r+1] = b[l-1] - a[l]
else:
a[r+1] = a[r]
a[l-1] = b[l-1] - a[r]
l -= 1
r += 1
print(*a)
```
Yes
| 99,511 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mishka is trying really hard to avoid being kicked out of the university. In particular, he was doing absolutely nothing for the whole semester, miraculously passed some exams so that just one is left.
There were n classes of that subject during the semester and on i-th class professor mentioned some non-negative integer a_i to the students. It turned out, the exam was to tell the whole sequence back to the professor.
Sounds easy enough for those who attended every class, doesn't it?
Obviously Mishka didn't attend any classes. However, professor left some clues on the values of a to help out students like Mishka:
* a was sorted in non-decreasing order (a_1 ≤ a_2 ≤ ... ≤ a_n);
* n was even;
* the following sequence b, consisting of \frac n 2 elements, was formed and given out to students: b_i = a_i + a_{n - i + 1}.
Professor also mentioned that any sequence a, which produces sequence b with the presented technique, will be acceptable.
Help Mishka to pass that last exam. Restore any sorted sequence a of non-negative integers, which produces sequence b with the presented technique. It is guaranteed that there exists at least one correct sequence a, which produces the given sequence b.
Input
The first line contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of sequence a. n is always even.
The second line contains \frac n 2 integers b_1, b_2, ..., b_{\frac n 2} (0 ≤ b_i ≤ 10^{18}) — sequence b, where b_i = a_i + a_{n - i + 1}.
It is guaranteed that there exists at least one correct sequence a, which produces the given sequence b.
Output
Print n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{18}) in a single line.
a_1 ≤ a_2 ≤ ... ≤ a_n should be satisfied.
b_i = a_i + a_{n - i + 1} should be satisfied for all valid i.
Examples
Input
4
5 6
Output
2 3 3 3
Input
6
2 1 2
Output
0 0 1 1 1 2
Submitted Solution:
```
n=int(input())
a=[0]*n
b=[int(i) for i in input().split()]
for i in range(n//2):
a[i]=0
a[n-i-1]=b[i]
if(i==0):
continue
adj = max(a[i-1]-a[i],a[n-1-i]-a[n-i])
a[i]+=adj
a[n-i-1]-=adj
print(*a)
```
Yes
| 99,512 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mishka is trying really hard to avoid being kicked out of the university. In particular, he was doing absolutely nothing for the whole semester, miraculously passed some exams so that just one is left.
There were n classes of that subject during the semester and on i-th class professor mentioned some non-negative integer a_i to the students. It turned out, the exam was to tell the whole sequence back to the professor.
Sounds easy enough for those who attended every class, doesn't it?
Obviously Mishka didn't attend any classes. However, professor left some clues on the values of a to help out students like Mishka:
* a was sorted in non-decreasing order (a_1 ≤ a_2 ≤ ... ≤ a_n);
* n was even;
* the following sequence b, consisting of \frac n 2 elements, was formed and given out to students: b_i = a_i + a_{n - i + 1}.
Professor also mentioned that any sequence a, which produces sequence b with the presented technique, will be acceptable.
Help Mishka to pass that last exam. Restore any sorted sequence a of non-negative integers, which produces sequence b with the presented technique. It is guaranteed that there exists at least one correct sequence a, which produces the given sequence b.
Input
The first line contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of sequence a. n is always even.
The second line contains \frac n 2 integers b_1, b_2, ..., b_{\frac n 2} (0 ≤ b_i ≤ 10^{18}) — sequence b, where b_i = a_i + a_{n - i + 1}.
It is guaranteed that there exists at least one correct sequence a, which produces the given sequence b.
Output
Print n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{18}) in a single line.
a_1 ≤ a_2 ≤ ... ≤ a_n should be satisfied.
b_i = a_i + a_{n - i + 1} should be satisfied for all valid i.
Examples
Input
4
5 6
Output
2 3 3 3
Input
6
2 1 2
Output
0 0 1 1 1 2
Submitted Solution:
```
import re
import sys
exit=sys.exit
from bisect import bisect_left as bsl,bisect_right as bsr
from collections import Counter,defaultdict as ddict,deque
from functools import lru_cache
cache=lru_cache(None)
from heapq import *
from itertools import *
from math import inf
from pprint import pprint as pp
enum=enumerate
ri=lambda:int(rln())
ris=lambda:list(map(int,rfs()))
rln=sys.stdin.readline
rl=lambda:rln().rstrip('\n')
rfs=lambda:rln().split()
cat=''.join
catn='\n'.join
mod=1000000007
d4=[(0,-1),(1,0),(0,1),(-1,0)]
d8=[(-1,-1),(0,-1),(1,-1),(-1,0),(1,0),(-1,1),(0,1),(1,1)]
########################################################################
n=ri()
b=ris()
a=[0]*n
lo,hi=0,inf
for i,x in enum(b):
hi=min(x-lo,hi)
lo=max(lo,x-hi)
a[i],a[~i]=lo,hi
print(*a)
```
Yes
| 99,513 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mishka is trying really hard to avoid being kicked out of the university. In particular, he was doing absolutely nothing for the whole semester, miraculously passed some exams so that just one is left.
There were n classes of that subject during the semester and on i-th class professor mentioned some non-negative integer a_i to the students. It turned out, the exam was to tell the whole sequence back to the professor.
Sounds easy enough for those who attended every class, doesn't it?
Obviously Mishka didn't attend any classes. However, professor left some clues on the values of a to help out students like Mishka:
* a was sorted in non-decreasing order (a_1 ≤ a_2 ≤ ... ≤ a_n);
* n was even;
* the following sequence b, consisting of \frac n 2 elements, was formed and given out to students: b_i = a_i + a_{n - i + 1}.
Professor also mentioned that any sequence a, which produces sequence b with the presented technique, will be acceptable.
Help Mishka to pass that last exam. Restore any sorted sequence a of non-negative integers, which produces sequence b with the presented technique. It is guaranteed that there exists at least one correct sequence a, which produces the given sequence b.
Input
The first line contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of sequence a. n is always even.
The second line contains \frac n 2 integers b_1, b_2, ..., b_{\frac n 2} (0 ≤ b_i ≤ 10^{18}) — sequence b, where b_i = a_i + a_{n - i + 1}.
It is guaranteed that there exists at least one correct sequence a, which produces the given sequence b.
Output
Print n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{18}) in a single line.
a_1 ≤ a_2 ≤ ... ≤ a_n should be satisfied.
b_i = a_i + a_{n - i + 1} should be satisfied for all valid i.
Examples
Input
4
5 6
Output
2 3 3 3
Input
6
2 1 2
Output
0 0 1 1 1 2
Submitted Solution:
```
n=int(input())
a=[0 for i in range(0,n)]
b=list(map(int,input().split(" ")))
a[0]=l=0;a[-1]=r=b[0]
for i in range(1,n//2):
if b[i]-l<=r:
a[i]=l;r=a[-(i+1)]=b[i]-l
elif b[i]-r>=l:
a[-(i+1)]=r;l=a[i]=b[i]-r
else: raise RuntimeError("gougoushishabi")
for i in a:
print(i,end=" ")
```
Yes
| 99,514 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mishka is trying really hard to avoid being kicked out of the university. In particular, he was doing absolutely nothing for the whole semester, miraculously passed some exams so that just one is left.
There were n classes of that subject during the semester and on i-th class professor mentioned some non-negative integer a_i to the students. It turned out, the exam was to tell the whole sequence back to the professor.
Sounds easy enough for those who attended every class, doesn't it?
Obviously Mishka didn't attend any classes. However, professor left some clues on the values of a to help out students like Mishka:
* a was sorted in non-decreasing order (a_1 ≤ a_2 ≤ ... ≤ a_n);
* n was even;
* the following sequence b, consisting of \frac n 2 elements, was formed and given out to students: b_i = a_i + a_{n - i + 1}.
Professor also mentioned that any sequence a, which produces sequence b with the presented technique, will be acceptable.
Help Mishka to pass that last exam. Restore any sorted sequence a of non-negative integers, which produces sequence b with the presented technique. It is guaranteed that there exists at least one correct sequence a, which produces the given sequence b.
Input
The first line contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of sequence a. n is always even.
The second line contains \frac n 2 integers b_1, b_2, ..., b_{\frac n 2} (0 ≤ b_i ≤ 10^{18}) — sequence b, where b_i = a_i + a_{n - i + 1}.
It is guaranteed that there exists at least one correct sequence a, which produces the given sequence b.
Output
Print n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{18}) in a single line.
a_1 ≤ a_2 ≤ ... ≤ a_n should be satisfied.
b_i = a_i + a_{n - i + 1} should be satisfied for all valid i.
Examples
Input
4
5 6
Output
2 3 3 3
Input
6
2 1 2
Output
0 0 1 1 1 2
Submitted Solution:
```
n = int(input())
b = [int(elem) for elem in input().split(" ")]
a = [0]*(n)
a[-1] = b[0]
for i in range(1, len(b)):
if b[i] <= b[i-1]:
if i != 0 and a[i-1] != 0:
a[i] = a[i-1]
a[n-i-1]=b[i]-a[i]
else:
a[i] = 0
a[n-i-1] = b[i]
else:
a[n-i-1] = a[n-i]
a[i] = b[i]-b[i-1]
for i in a:
print(i, end=" ")
print()
```
No
| 99,515 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mishka is trying really hard to avoid being kicked out of the university. In particular, he was doing absolutely nothing for the whole semester, miraculously passed some exams so that just one is left.
There were n classes of that subject during the semester and on i-th class professor mentioned some non-negative integer a_i to the students. It turned out, the exam was to tell the whole sequence back to the professor.
Sounds easy enough for those who attended every class, doesn't it?
Obviously Mishka didn't attend any classes. However, professor left some clues on the values of a to help out students like Mishka:
* a was sorted in non-decreasing order (a_1 ≤ a_2 ≤ ... ≤ a_n);
* n was even;
* the following sequence b, consisting of \frac n 2 elements, was formed and given out to students: b_i = a_i + a_{n - i + 1}.
Professor also mentioned that any sequence a, which produces sequence b with the presented technique, will be acceptable.
Help Mishka to pass that last exam. Restore any sorted sequence a of non-negative integers, which produces sequence b with the presented technique. It is guaranteed that there exists at least one correct sequence a, which produces the given sequence b.
Input
The first line contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of sequence a. n is always even.
The second line contains \frac n 2 integers b_1, b_2, ..., b_{\frac n 2} (0 ≤ b_i ≤ 10^{18}) — sequence b, where b_i = a_i + a_{n - i + 1}.
It is guaranteed that there exists at least one correct sequence a, which produces the given sequence b.
Output
Print n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{18}) in a single line.
a_1 ≤ a_2 ≤ ... ≤ a_n should be satisfied.
b_i = a_i + a_{n - i + 1} should be satisfied for all valid i.
Examples
Input
4
5 6
Output
2 3 3 3
Input
6
2 1 2
Output
0 0 1 1 1 2
Submitted Solution:
```
import sys
import math
n=int(input())
lisa=[int(x) for x in input().strip().split()]
first,second,prev=[],[],0
for i in range(n//2):
if(i==0):
first.append(0)
second.append(lisa[i])
prev=lisa[i]
else:
if(prev<lisa[i]):
second.append(prev)
first.append(lisa[i]-prev)
else:
second.append(lisa[i])
first.append(0)
prev=lisa[i]
ksd=second[::-1]
mad=first+ksd
print(mad)
```
No
| 99,516 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mishka is trying really hard to avoid being kicked out of the university. In particular, he was doing absolutely nothing for the whole semester, miraculously passed some exams so that just one is left.
There were n classes of that subject during the semester and on i-th class professor mentioned some non-negative integer a_i to the students. It turned out, the exam was to tell the whole sequence back to the professor.
Sounds easy enough for those who attended every class, doesn't it?
Obviously Mishka didn't attend any classes. However, professor left some clues on the values of a to help out students like Mishka:
* a was sorted in non-decreasing order (a_1 ≤ a_2 ≤ ... ≤ a_n);
* n was even;
* the following sequence b, consisting of \frac n 2 elements, was formed and given out to students: b_i = a_i + a_{n - i + 1}.
Professor also mentioned that any sequence a, which produces sequence b with the presented technique, will be acceptable.
Help Mishka to pass that last exam. Restore any sorted sequence a of non-negative integers, which produces sequence b with the presented technique. It is guaranteed that there exists at least one correct sequence a, which produces the given sequence b.
Input
The first line contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of sequence a. n is always even.
The second line contains \frac n 2 integers b_1, b_2, ..., b_{\frac n 2} (0 ≤ b_i ≤ 10^{18}) — sequence b, where b_i = a_i + a_{n - i + 1}.
It is guaranteed that there exists at least one correct sequence a, which produces the given sequence b.
Output
Print n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{18}) in a single line.
a_1 ≤ a_2 ≤ ... ≤ a_n should be satisfied.
b_i = a_i + a_{n - i + 1} should be satisfied for all valid i.
Examples
Input
4
5 6
Output
2 3 3 3
Input
6
2 1 2
Output
0 0 1 1 1 2
Submitted Solution:
```
n=int(input())
b=[int(i) for i in input().split()]
a=[0]*n
a[0]=0
a[n-1]=b[0]
for i in range(1,n//2):
if b[i]<=b[i-1]:
a[i]=a[i-1]
else:
a[i]=a[i-1]+1
a[n-i-1]=b[i]-a[i]
print(a)
```
No
| 99,517 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mishka is trying really hard to avoid being kicked out of the university. In particular, he was doing absolutely nothing for the whole semester, miraculously passed some exams so that just one is left.
There were n classes of that subject during the semester and on i-th class professor mentioned some non-negative integer a_i to the students. It turned out, the exam was to tell the whole sequence back to the professor.
Sounds easy enough for those who attended every class, doesn't it?
Obviously Mishka didn't attend any classes. However, professor left some clues on the values of a to help out students like Mishka:
* a was sorted in non-decreasing order (a_1 ≤ a_2 ≤ ... ≤ a_n);
* n was even;
* the following sequence b, consisting of \frac n 2 elements, was formed and given out to students: b_i = a_i + a_{n - i + 1}.
Professor also mentioned that any sequence a, which produces sequence b with the presented technique, will be acceptable.
Help Mishka to pass that last exam. Restore any sorted sequence a of non-negative integers, which produces sequence b with the presented technique. It is guaranteed that there exists at least one correct sequence a, which produces the given sequence b.
Input
The first line contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of sequence a. n is always even.
The second line contains \frac n 2 integers b_1, b_2, ..., b_{\frac n 2} (0 ≤ b_i ≤ 10^{18}) — sequence b, where b_i = a_i + a_{n - i + 1}.
It is guaranteed that there exists at least one correct sequence a, which produces the given sequence b.
Output
Print n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{18}) in a single line.
a_1 ≤ a_2 ≤ ... ≤ a_n should be satisfied.
b_i = a_i + a_{n - i + 1} should be satisfied for all valid i.
Examples
Input
4
5 6
Output
2 3 3 3
Input
6
2 1 2
Output
0 0 1 1 1 2
Submitted Solution:
```
# your code goes here
from collections import defaultdict
import bisect
from itertools import accumulate
import os
import sys
import math
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
def input(): return sys.stdin.readline().rstrip("\r\n")
# ------------------- fast io --------------------
n=int(input())
b=list(map(int,input().split()))
ans=[0]*(n)
l=0
r=10*18
for i in range(n//2):
ans[i]=max(l,b[i]-r)
ans[-i-1]=b[i]-ans[i]
l=ans[i]
r=b[i]-ans[i]
print(*ans)
```
No
| 99,518 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sasha likes programming. Once, during a very long contest, Sasha decided that he was a bit tired and needed to relax. So he did. But since Sasha isn't an ordinary guy, he prefers to relax unusually. During leisure time Sasha likes to upsolve unsolved problems because upsolving is very useful.
Therefore, Sasha decided to upsolve the following problem:
You have an array a with n integers. You need to count the number of funny pairs (l, r) (l ≤ r). To check if a pair (l, r) is a funny pair, take mid = (l + r - 1)/(2), then if r - l + 1 is an even number and a_l ⊕ a_{l+1} ⊕ … ⊕ a_{mid} = a_{mid + 1} ⊕ a_{mid + 2} ⊕ … ⊕ a_r, then the pair is funny. In other words, ⊕ of elements of the left half of the subarray from l to r should be equal to ⊕ of elements of the right half. Note that ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
It is time to continue solving the contest, so Sasha asked you to solve this task.
Input
The first line contains one integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the size of the array.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i < 2^{20}) — array itself.
Output
Print one integer — the number of funny pairs. You should consider only pairs where r - l + 1 is even number.
Examples
Input
5
1 2 3 4 5
Output
1
Input
6
3 2 2 3 7 6
Output
3
Input
3
42 4 2
Output
0
Note
Be as cool as Sasha, upsolve problems!
In the first example, the only funny pair is (2, 5), as 2 ⊕ 3 = 4 ⊕ 5 = 1.
In the second example, funny pairs are (2, 3), (1, 4), and (3, 6).
In the third example, there are no funny pairs.
Tags: dp, implementation
Correct Solution:
```
#!/usr/bin/python3.6
'''
.__ .__ _______ _______ _________
______| |__ |__| ____ ____ _______ _____ \ _ \ \ _ \\______ \
/ ___/| | \ | | / \ / _ \\_ __ \\__ \ / /_\ \ / /_\ \ / /
\___ \ | Y \| || | \( <_> )| | \/ / __ \_\ \_/ \\ \_/ \ / /
/____ >|___| /|__||___| / \____/ |__| (____ / \_____ / \_____ //____/
\/ \/ \/ \/ \/ \/
'''
# Run locally with ./codeforces.py to run in not debug mode. [ONLINE_JUDGE]
# rm ~/pipe2; mkfifo ~/pipe2 ; python3.6 -O ./codeforces.py < ~/pipe2 | ./interactive.py > ~/pipe2
# rm ~/pipe2; mkfifo ~/pipe2 ; python3.6 -O ./codeforces.py < ~/pipe2 | ./matcher.py > ~/pipe2
import sys
from functools import reduce
from math import sqrt, floor
def eprint(*args, **kwargs):
if not __debug__:
print(*args, file=sys.stderr, **kwargs)
if __name__ == "__main__":
eprint("Local\n")
for t in range(int(input()) if not __debug__ else 1):
# eprint(f'Times:{t}')
n = int(input())
arr = list(map(int, input().split()))
amax = (1<<20)+1
# c_xor = [arr[0]]
# c_xor_freq = 2*[[0 for i in range(amax)]]
c_xor_freq = [[0 for i in range(amax)] for j in range(2)]
c_xor_freq[1][0] = 1
xor = 0
res = 0
for i in range(n):
xor ^= arr[i]
res += c_xor_freq[i%2][xor]
c_xor_freq[i%2][xor]+=1
# eprint(c_xor_freq[:50])
# print(res)
# c_xor.append(c_xor[i-1] ^ arr[i])
# c_xor.append(0)
# eprint(c_xor)
# res = 0
# for l in range(n):
# for r in range(l+1, n, 2):
# if c_xor[r] == c_xor[l-1]:
# res += 1
# eprint(f't:{t}-> L:{l}, R:{r}, res:{res}')
print(res)
```
| 99,519 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sasha likes programming. Once, during a very long contest, Sasha decided that he was a bit tired and needed to relax. So he did. But since Sasha isn't an ordinary guy, he prefers to relax unusually. During leisure time Sasha likes to upsolve unsolved problems because upsolving is very useful.
Therefore, Sasha decided to upsolve the following problem:
You have an array a with n integers. You need to count the number of funny pairs (l, r) (l ≤ r). To check if a pair (l, r) is a funny pair, take mid = (l + r - 1)/(2), then if r - l + 1 is an even number and a_l ⊕ a_{l+1} ⊕ … ⊕ a_{mid} = a_{mid + 1} ⊕ a_{mid + 2} ⊕ … ⊕ a_r, then the pair is funny. In other words, ⊕ of elements of the left half of the subarray from l to r should be equal to ⊕ of elements of the right half. Note that ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
It is time to continue solving the contest, so Sasha asked you to solve this task.
Input
The first line contains one integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the size of the array.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i < 2^{20}) — array itself.
Output
Print one integer — the number of funny pairs. You should consider only pairs where r - l + 1 is even number.
Examples
Input
5
1 2 3 4 5
Output
1
Input
6
3 2 2 3 7 6
Output
3
Input
3
42 4 2
Output
0
Note
Be as cool as Sasha, upsolve problems!
In the first example, the only funny pair is (2, 5), as 2 ⊕ 3 = 4 ⊕ 5 = 1.
In the second example, funny pairs are (2, 3), (1, 4), and (3, 6).
In the third example, there are no funny pairs.
Tags: dp, implementation
Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
ans=[[0]*(2**20+5) for i in range(2)]
ans[1][0]=1
x,an=0,0
for i in range(n):
x^=a[i]
an+=ans[i%2][x]
ans[i%2][x]+=1
print(an)
```
| 99,520 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sasha likes programming. Once, during a very long contest, Sasha decided that he was a bit tired and needed to relax. So he did. But since Sasha isn't an ordinary guy, he prefers to relax unusually. During leisure time Sasha likes to upsolve unsolved problems because upsolving is very useful.
Therefore, Sasha decided to upsolve the following problem:
You have an array a with n integers. You need to count the number of funny pairs (l, r) (l ≤ r). To check if a pair (l, r) is a funny pair, take mid = (l + r - 1)/(2), then if r - l + 1 is an even number and a_l ⊕ a_{l+1} ⊕ … ⊕ a_{mid} = a_{mid + 1} ⊕ a_{mid + 2} ⊕ … ⊕ a_r, then the pair is funny. In other words, ⊕ of elements of the left half of the subarray from l to r should be equal to ⊕ of elements of the right half. Note that ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
It is time to continue solving the contest, so Sasha asked you to solve this task.
Input
The first line contains one integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the size of the array.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i < 2^{20}) — array itself.
Output
Print one integer — the number of funny pairs. You should consider only pairs where r - l + 1 is even number.
Examples
Input
5
1 2 3 4 5
Output
1
Input
6
3 2 2 3 7 6
Output
3
Input
3
42 4 2
Output
0
Note
Be as cool as Sasha, upsolve problems!
In the first example, the only funny pair is (2, 5), as 2 ⊕ 3 = 4 ⊕ 5 = 1.
In the second example, funny pairs are (2, 3), (1, 4), and (3, 6).
In the third example, there are no funny pairs.
Tags: dp, implementation
Correct Solution:
```
list_int_input = lambda inp: list(map(int, inp.split()))
int_input = lambda inp: int(inp)
string_to_list_input = lambda inp: list(inp)
n=int(input())
a=list_int_input(input())
a=[0]+a
xl=[]
for ind,val in enumerate(a):
if ind>0:
val=xl[-1]^val
xl.append(val)
el=xl[::2]
ol=xl[1::2]
de,do={},{}
for i in el:
de.setdefault(i,0)
de[i]+=1
for i in ol:
do.setdefault(i,0)
do[i]+=1
summ=0
for k,v in de.items():
summ+=int((v*(v-1))/2)
for k,v in do.items():
summ+=int((v*(v-1))/2)
print(summ)
```
| 99,521 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sasha likes programming. Once, during a very long contest, Sasha decided that he was a bit tired and needed to relax. So he did. But since Sasha isn't an ordinary guy, he prefers to relax unusually. During leisure time Sasha likes to upsolve unsolved problems because upsolving is very useful.
Therefore, Sasha decided to upsolve the following problem:
You have an array a with n integers. You need to count the number of funny pairs (l, r) (l ≤ r). To check if a pair (l, r) is a funny pair, take mid = (l + r - 1)/(2), then if r - l + 1 is an even number and a_l ⊕ a_{l+1} ⊕ … ⊕ a_{mid} = a_{mid + 1} ⊕ a_{mid + 2} ⊕ … ⊕ a_r, then the pair is funny. In other words, ⊕ of elements of the left half of the subarray from l to r should be equal to ⊕ of elements of the right half. Note that ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
It is time to continue solving the contest, so Sasha asked you to solve this task.
Input
The first line contains one integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the size of the array.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i < 2^{20}) — array itself.
Output
Print one integer — the number of funny pairs. You should consider only pairs where r - l + 1 is even number.
Examples
Input
5
1 2 3 4 5
Output
1
Input
6
3 2 2 3 7 6
Output
3
Input
3
42 4 2
Output
0
Note
Be as cool as Sasha, upsolve problems!
In the first example, the only funny pair is (2, 5), as 2 ⊕ 3 = 4 ⊕ 5 = 1.
In the second example, funny pairs are (2, 3), (1, 4), and (3, 6).
In the third example, there are no funny pairs.
Tags: dp, implementation
Correct Solution:
```
input()
N=1<<20
d=[1]+[0]*N,[0]*N
r=s=i=0
for x in map(int,input().split()):s^=x;i^=1;r+=d[i][s];d[i][s]+=1
print(r)
```
| 99,522 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sasha likes programming. Once, during a very long contest, Sasha decided that he was a bit tired and needed to relax. So he did. But since Sasha isn't an ordinary guy, he prefers to relax unusually. During leisure time Sasha likes to upsolve unsolved problems because upsolving is very useful.
Therefore, Sasha decided to upsolve the following problem:
You have an array a with n integers. You need to count the number of funny pairs (l, r) (l ≤ r). To check if a pair (l, r) is a funny pair, take mid = (l + r - 1)/(2), then if r - l + 1 is an even number and a_l ⊕ a_{l+1} ⊕ … ⊕ a_{mid} = a_{mid + 1} ⊕ a_{mid + 2} ⊕ … ⊕ a_r, then the pair is funny. In other words, ⊕ of elements of the left half of the subarray from l to r should be equal to ⊕ of elements of the right half. Note that ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
It is time to continue solving the contest, so Sasha asked you to solve this task.
Input
The first line contains one integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the size of the array.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i < 2^{20}) — array itself.
Output
Print one integer — the number of funny pairs. You should consider only pairs where r - l + 1 is even number.
Examples
Input
5
1 2 3 4 5
Output
1
Input
6
3 2 2 3 7 6
Output
3
Input
3
42 4 2
Output
0
Note
Be as cool as Sasha, upsolve problems!
In the first example, the only funny pair is (2, 5), as 2 ⊕ 3 = 4 ⊕ 5 = 1.
In the second example, funny pairs are (2, 3), (1, 4), and (3, 6).
In the third example, there are no funny pairs.
Tags: dp, implementation
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
cnt = [[0 for i in range(int(2 << 20)+10)],
[0 for i in range(int(2 << 20)+10)]]
cnt[1][0] =1
x = 0
ans = 0
for i in range(n):
x ^= a[i]
ans += cnt[i % 2][x]
cnt[i % 2][x] += 1
print(ans)
```
| 99,523 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sasha likes programming. Once, during a very long contest, Sasha decided that he was a bit tired and needed to relax. So he did. But since Sasha isn't an ordinary guy, he prefers to relax unusually. During leisure time Sasha likes to upsolve unsolved problems because upsolving is very useful.
Therefore, Sasha decided to upsolve the following problem:
You have an array a with n integers. You need to count the number of funny pairs (l, r) (l ≤ r). To check if a pair (l, r) is a funny pair, take mid = (l + r - 1)/(2), then if r - l + 1 is an even number and a_l ⊕ a_{l+1} ⊕ … ⊕ a_{mid} = a_{mid + 1} ⊕ a_{mid + 2} ⊕ … ⊕ a_r, then the pair is funny. In other words, ⊕ of elements of the left half of the subarray from l to r should be equal to ⊕ of elements of the right half. Note that ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
It is time to continue solving the contest, so Sasha asked you to solve this task.
Input
The first line contains one integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the size of the array.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i < 2^{20}) — array itself.
Output
Print one integer — the number of funny pairs. You should consider only pairs where r - l + 1 is even number.
Examples
Input
5
1 2 3 4 5
Output
1
Input
6
3 2 2 3 7 6
Output
3
Input
3
42 4 2
Output
0
Note
Be as cool as Sasha, upsolve problems!
In the first example, the only funny pair is (2, 5), as 2 ⊕ 3 = 4 ⊕ 5 = 1.
In the second example, funny pairs are (2, 3), (1, 4), and (3, 6).
In the third example, there are no funny pairs.
Tags: dp, implementation
Correct Solution:
```
import sys
import math
def getAndParseInt(num=1):
string = (sys.stdin.readline()).strip()
if num==1:
return int(string)
else:
return [int(part) for part in string.split()]
def getAndParseString(num=1,delim=" "):
string = (sys.stdin.readline()).strip()
if num==1:
return string
else:
return [part for part in string.split(delim)]
n = getAndParseInt()
nums = getAndParseInt(2)
xors = [{},{0:1}]
total = 0
running_xor = 0
for index,num in enumerate(nums):
running_xor = running_xor^num
total+=xors[index%2].get(running_xor,0)
xors[index%2][running_xor] = xors[index%2].get(running_xor,0)+1
print(total)
```
| 99,524 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sasha likes programming. Once, during a very long contest, Sasha decided that he was a bit tired and needed to relax. So he did. But since Sasha isn't an ordinary guy, he prefers to relax unusually. During leisure time Sasha likes to upsolve unsolved problems because upsolving is very useful.
Therefore, Sasha decided to upsolve the following problem:
You have an array a with n integers. You need to count the number of funny pairs (l, r) (l ≤ r). To check if a pair (l, r) is a funny pair, take mid = (l + r - 1)/(2), then if r - l + 1 is an even number and a_l ⊕ a_{l+1} ⊕ … ⊕ a_{mid} = a_{mid + 1} ⊕ a_{mid + 2} ⊕ … ⊕ a_r, then the pair is funny. In other words, ⊕ of elements of the left half of the subarray from l to r should be equal to ⊕ of elements of the right half. Note that ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
It is time to continue solving the contest, so Sasha asked you to solve this task.
Input
The first line contains one integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the size of the array.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i < 2^{20}) — array itself.
Output
Print one integer — the number of funny pairs. You should consider only pairs where r - l + 1 is even number.
Examples
Input
5
1 2 3 4 5
Output
1
Input
6
3 2 2 3 7 6
Output
3
Input
3
42 4 2
Output
0
Note
Be as cool as Sasha, upsolve problems!
In the first example, the only funny pair is (2, 5), as 2 ⊕ 3 = 4 ⊕ 5 = 1.
In the second example, funny pairs are (2, 3), (1, 4), and (3, 6).
In the third example, there are no funny pairs.
Tags: dp, implementation
Correct Solution:
```
def inpl(): return list(map(int, input().split()))
from operator import xor
from itertools import accumulate, chain
from collections import Counter
N = int(input())
A = list(accumulate([0] + inpl(), xor))
B = [a for i, a in enumerate(A) if i%2]
A = [a for i, a in enumerate(A) if not i%2]
H1 = Counter(A)
H2 = Counter(B)
print(sum([h*(h-1)//2 for h in chain(H1.values(), H2.values())]))
```
| 99,525 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sasha likes programming. Once, during a very long contest, Sasha decided that he was a bit tired and needed to relax. So he did. But since Sasha isn't an ordinary guy, he prefers to relax unusually. During leisure time Sasha likes to upsolve unsolved problems because upsolving is very useful.
Therefore, Sasha decided to upsolve the following problem:
You have an array a with n integers. You need to count the number of funny pairs (l, r) (l ≤ r). To check if a pair (l, r) is a funny pair, take mid = (l + r - 1)/(2), then if r - l + 1 is an even number and a_l ⊕ a_{l+1} ⊕ … ⊕ a_{mid} = a_{mid + 1} ⊕ a_{mid + 2} ⊕ … ⊕ a_r, then the pair is funny. In other words, ⊕ of elements of the left half of the subarray from l to r should be equal to ⊕ of elements of the right half. Note that ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
It is time to continue solving the contest, so Sasha asked you to solve this task.
Input
The first line contains one integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the size of the array.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i < 2^{20}) — array itself.
Output
Print one integer — the number of funny pairs. You should consider only pairs where r - l + 1 is even number.
Examples
Input
5
1 2 3 4 5
Output
1
Input
6
3 2 2 3 7 6
Output
3
Input
3
42 4 2
Output
0
Note
Be as cool as Sasha, upsolve problems!
In the first example, the only funny pair is (2, 5), as 2 ⊕ 3 = 4 ⊕ 5 = 1.
In the second example, funny pairs are (2, 3), (1, 4), and (3, 6).
In the third example, there are no funny pairs.
Tags: dp, implementation
Correct Solution:
```
import sys
input = sys.stdin.readline
n=int(input())
A=list(map(int,input().split()))
AX=[0,A[0]]
for i in range(1,n):
AX.append(AX[i]^A[i])
LIST=sorted(set(AX))
DICT=dict()
for i in range(len(LIST)):
DICT[LIST[i]]=i
NUMLIST=[[] for i in range(len(LIST))]
for i in range(n+1):
NUMLIST[DICT[AX[i]]].append(i)
ANS=0
for lis in NUMLIST:
l1=[l for l in lis if l%2==1]
l2=[l for l in lis if l%2==0]
ANS+=len(l1)*(len(l1)-1)//2+len(l2)*(len(l2)-1)//2
print(ANS)
```
| 99,526 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sasha likes programming. Once, during a very long contest, Sasha decided that he was a bit tired and needed to relax. So he did. But since Sasha isn't an ordinary guy, he prefers to relax unusually. During leisure time Sasha likes to upsolve unsolved problems because upsolving is very useful.
Therefore, Sasha decided to upsolve the following problem:
You have an array a with n integers. You need to count the number of funny pairs (l, r) (l ≤ r). To check if a pair (l, r) is a funny pair, take mid = (l + r - 1)/(2), then if r - l + 1 is an even number and a_l ⊕ a_{l+1} ⊕ … ⊕ a_{mid} = a_{mid + 1} ⊕ a_{mid + 2} ⊕ … ⊕ a_r, then the pair is funny. In other words, ⊕ of elements of the left half of the subarray from l to r should be equal to ⊕ of elements of the right half. Note that ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
It is time to continue solving the contest, so Sasha asked you to solve this task.
Input
The first line contains one integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the size of the array.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i < 2^{20}) — array itself.
Output
Print one integer — the number of funny pairs. You should consider only pairs where r - l + 1 is even number.
Examples
Input
5
1 2 3 4 5
Output
1
Input
6
3 2 2 3 7 6
Output
3
Input
3
42 4 2
Output
0
Note
Be as cool as Sasha, upsolve problems!
In the first example, the only funny pair is (2, 5), as 2 ⊕ 3 = 4 ⊕ 5 = 1.
In the second example, funny pairs are (2, 3), (1, 4), and (3, 6).
In the third example, there are no funny pairs.
Submitted Solution:
```
input()
d={(0,0):1}
r=s=i=0
for x in map(int,input().split()):s^=x;i^=1;c=d.get((s,i),0);r+=c;d[s,i]=c+1
print(r)
```
Yes
| 99,527 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sasha likes programming. Once, during a very long contest, Sasha decided that he was a bit tired and needed to relax. So he did. But since Sasha isn't an ordinary guy, he prefers to relax unusually. During leisure time Sasha likes to upsolve unsolved problems because upsolving is very useful.
Therefore, Sasha decided to upsolve the following problem:
You have an array a with n integers. You need to count the number of funny pairs (l, r) (l ≤ r). To check if a pair (l, r) is a funny pair, take mid = (l + r - 1)/(2), then if r - l + 1 is an even number and a_l ⊕ a_{l+1} ⊕ … ⊕ a_{mid} = a_{mid + 1} ⊕ a_{mid + 2} ⊕ … ⊕ a_r, then the pair is funny. In other words, ⊕ of elements of the left half of the subarray from l to r should be equal to ⊕ of elements of the right half. Note that ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
It is time to continue solving the contest, so Sasha asked you to solve this task.
Input
The first line contains one integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the size of the array.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i < 2^{20}) — array itself.
Output
Print one integer — the number of funny pairs. You should consider only pairs where r - l + 1 is even number.
Examples
Input
5
1 2 3 4 5
Output
1
Input
6
3 2 2 3 7 6
Output
3
Input
3
42 4 2
Output
0
Note
Be as cool as Sasha, upsolve problems!
In the first example, the only funny pair is (2, 5), as 2 ⊕ 3 = 4 ⊕ 5 = 1.
In the second example, funny pairs are (2, 3), (1, 4), and (3, 6).
In the third example, there are no funny pairs.
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
xor = [a[0]]
for i in range(1, n):
xor.append(xor[-1]^a[i])
d = {0: [1, 0]}
for i in range(n):
if xor[i] not in d:
d[xor[i]] = [0, 0]
d[xor[i]][(i+1)%2] += 1
#for k in d: print(d[k])
funny = 0
for k in d:
odd = d[k][1]
even = d[k][0]
funny += (odd*(odd-1))//2
funny += (even*(even-1))//2
print(funny)
```
Yes
| 99,528 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sasha likes programming. Once, during a very long contest, Sasha decided that he was a bit tired and needed to relax. So he did. But since Sasha isn't an ordinary guy, he prefers to relax unusually. During leisure time Sasha likes to upsolve unsolved problems because upsolving is very useful.
Therefore, Sasha decided to upsolve the following problem:
You have an array a with n integers. You need to count the number of funny pairs (l, r) (l ≤ r). To check if a pair (l, r) is a funny pair, take mid = (l + r - 1)/(2), then if r - l + 1 is an even number and a_l ⊕ a_{l+1} ⊕ … ⊕ a_{mid} = a_{mid + 1} ⊕ a_{mid + 2} ⊕ … ⊕ a_r, then the pair is funny. In other words, ⊕ of elements of the left half of the subarray from l to r should be equal to ⊕ of elements of the right half. Note that ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
It is time to continue solving the contest, so Sasha asked you to solve this task.
Input
The first line contains one integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the size of the array.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i < 2^{20}) — array itself.
Output
Print one integer — the number of funny pairs. You should consider only pairs where r - l + 1 is even number.
Examples
Input
5
1 2 3 4 5
Output
1
Input
6
3 2 2 3 7 6
Output
3
Input
3
42 4 2
Output
0
Note
Be as cool as Sasha, upsolve problems!
In the first example, the only funny pair is (2, 5), as 2 ⊕ 3 = 4 ⊕ 5 = 1.
In the second example, funny pairs are (2, 3), (1, 4), and (3, 6).
In the third example, there are no funny pairs.
Submitted Solution:
```
# -*- coding: utf-8 -*-
import sys, re
from collections import deque, defaultdict, Counter
from math import sqrt, hypot, factorial, pi, sin, cos, radians, log10
if sys.version_info.minor >= 5: from math import gcd
else: from fractions import gcd
from heapq import heappop, heappush, heapify, heappushpop
from bisect import bisect_left, bisect_right
from itertools import permutations, combinations, product, accumulate
from operator import itemgetter, mul, xor
from copy import copy, deepcopy
from functools import reduce, partial
from fractions import Fraction
from string import ascii_lowercase, ascii_uppercase, digits
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def round(x): return int((x*2+1) // 2)
def fermat(x, y, MOD): return x * pow(y, MOD-2, MOD) % MOD
def lcm(x, y): return (x * y) // gcd(x, y)
def lcm_list(nums): return reduce(lcm, nums, 1)
def gcd_list(nums): return reduce(gcd, nums, nums[0])
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(): return list(map(int, input().split()))
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
MOD = 10 ** 9 + 7
N = INT()
A = LIST()
acc = [0] + list(accumulate(A, xor))
C1 = Counter()
for i in range(0, N+1, 2):
C1[acc[i]] += 1
C2 = Counter()
for i in range(1, N+1, 2):
C2[acc[i]] += 1
ans = 0
for k, v in C1.items():
if v >= 2:
ans += v * (v-1) // 2
for k, v in C2.items():
if v >= 2:
ans += v * (v-1) // 2
print(ans)
```
Yes
| 99,529 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sasha likes programming. Once, during a very long contest, Sasha decided that he was a bit tired and needed to relax. So he did. But since Sasha isn't an ordinary guy, he prefers to relax unusually. During leisure time Sasha likes to upsolve unsolved problems because upsolving is very useful.
Therefore, Sasha decided to upsolve the following problem:
You have an array a with n integers. You need to count the number of funny pairs (l, r) (l ≤ r). To check if a pair (l, r) is a funny pair, take mid = (l + r - 1)/(2), then if r - l + 1 is an even number and a_l ⊕ a_{l+1} ⊕ … ⊕ a_{mid} = a_{mid + 1} ⊕ a_{mid + 2} ⊕ … ⊕ a_r, then the pair is funny. In other words, ⊕ of elements of the left half of the subarray from l to r should be equal to ⊕ of elements of the right half. Note that ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
It is time to continue solving the contest, so Sasha asked you to solve this task.
Input
The first line contains one integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the size of the array.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i < 2^{20}) — array itself.
Output
Print one integer — the number of funny pairs. You should consider only pairs where r - l + 1 is even number.
Examples
Input
5
1 2 3 4 5
Output
1
Input
6
3 2 2 3 7 6
Output
3
Input
3
42 4 2
Output
0
Note
Be as cool as Sasha, upsolve problems!
In the first example, the only funny pair is (2, 5), as 2 ⊕ 3 = 4 ⊕ 5 = 1.
In the second example, funny pairs are (2, 3), (1, 4), and (3, 6).
In the third example, there are no funny pairs.
Submitted Solution:
```
import sys
from collections import defaultdict
input = sys.stdin.readline
if __name__ == '__main__':
n = int(input())
arr = list(map(int, input().strip().split()))
xr = 0
ev = defaultdict(int)
od = defaultdict(int)
od[0] += 1
ans = 0
for i in range(n):
xr ^= arr[i]
if i & 1:
ans += od[xr]
od[xr] += 1
else:
ans += ev[xr]
ev[xr] += 1
print(ans)
```
Yes
| 99,530 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sasha likes programming. Once, during a very long contest, Sasha decided that he was a bit tired and needed to relax. So he did. But since Sasha isn't an ordinary guy, he prefers to relax unusually. During leisure time Sasha likes to upsolve unsolved problems because upsolving is very useful.
Therefore, Sasha decided to upsolve the following problem:
You have an array a with n integers. You need to count the number of funny pairs (l, r) (l ≤ r). To check if a pair (l, r) is a funny pair, take mid = (l + r - 1)/(2), then if r - l + 1 is an even number and a_l ⊕ a_{l+1} ⊕ … ⊕ a_{mid} = a_{mid + 1} ⊕ a_{mid + 2} ⊕ … ⊕ a_r, then the pair is funny. In other words, ⊕ of elements of the left half of the subarray from l to r should be equal to ⊕ of elements of the right half. Note that ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
It is time to continue solving the contest, so Sasha asked you to solve this task.
Input
The first line contains one integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the size of the array.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i < 2^{20}) — array itself.
Output
Print one integer — the number of funny pairs. You should consider only pairs where r - l + 1 is even number.
Examples
Input
5
1 2 3 4 5
Output
1
Input
6
3 2 2 3 7 6
Output
3
Input
3
42 4 2
Output
0
Note
Be as cool as Sasha, upsolve problems!
In the first example, the only funny pair is (2, 5), as 2 ⊕ 3 = 4 ⊕ 5 = 1.
In the second example, funny pairs are (2, 3), (1, 4), and (3, 6).
In the third example, there are no funny pairs.
Submitted Solution:
```
count=0;cc={}
def work(a,i,j):
global cc,count
if j==i:return a[i]
if (i,j) in cc:return cc[(i,j)]
m=(j+i-1)//2
a1=work(a,i,m)
a2=work(a,m+1,j)
#print((i,j),a1,a2)
if a1==a2:
if (i,j) not in cc:
count+=1
cc[(i,j)]=a1^a2
return a1^a2
n=int(input())
a=list(map(int,input().split()))
if n%2==0:
work(a,0,n-1)
if n>4:work(a,0,n-3);work(a,2,n-1)
else:
work(a,1,n-1)
work(a,0,n-2)
print(count)
```
No
| 99,531 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sasha likes programming. Once, during a very long contest, Sasha decided that he was a bit tired and needed to relax. So he did. But since Sasha isn't an ordinary guy, he prefers to relax unusually. During leisure time Sasha likes to upsolve unsolved problems because upsolving is very useful.
Therefore, Sasha decided to upsolve the following problem:
You have an array a with n integers. You need to count the number of funny pairs (l, r) (l ≤ r). To check if a pair (l, r) is a funny pair, take mid = (l + r - 1)/(2), then if r - l + 1 is an even number and a_l ⊕ a_{l+1} ⊕ … ⊕ a_{mid} = a_{mid + 1} ⊕ a_{mid + 2} ⊕ … ⊕ a_r, then the pair is funny. In other words, ⊕ of elements of the left half of the subarray from l to r should be equal to ⊕ of elements of the right half. Note that ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
It is time to continue solving the contest, so Sasha asked you to solve this task.
Input
The first line contains one integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the size of the array.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i < 2^{20}) — array itself.
Output
Print one integer — the number of funny pairs. You should consider only pairs where r - l + 1 is even number.
Examples
Input
5
1 2 3 4 5
Output
1
Input
6
3 2 2 3 7 6
Output
3
Input
3
42 4 2
Output
0
Note
Be as cool as Sasha, upsolve problems!
In the first example, the only funny pair is (2, 5), as 2 ⊕ 3 = 4 ⊕ 5 = 1.
In the second example, funny pairs are (2, 3), (1, 4), and (3, 6).
In the third example, there are no funny pairs.
Submitted Solution:
```
# Bismillahirahmanirahim
# Soru 1
#
# nv = list(map(int, input().split()))
#
# n = nv[0] - 1
# v = nv[1]
# if v >= n:
# print(n)
# quit()
# k = 2
# money = v
# while n - v > 0:
# n = n - 1
# money += k
# k += 1
# print(money)
#Soru2
# def carpan(n):
# lst = []
# sq = n**0.5 + 1
# for i in range(2,int(sq)+1):
# if n%i == 0:
# lst.append(i)
# return lst
#
#
# n = int(input())
# lst = list(map(int, input().split()))
# lst = sorted(lst, reverse=True)
# mn = lst[-1]
# total = sum(lst)
# large = 0
# for i in lst:
# mx = i
# lst1 = carpan(mx)
# for j in lst1:
# if mx + mn - (mx/j + mn*j) > large:
# large = mx + mn - (mx/j + mn*j)
# print(int(total-large))
# Soru 3
n = int(input())
lst = list(map(int, input().split()))
total = 0
for i in range(1,n):
lst[i] = lst[i] ^ lst[i - 1]
def ikili(n):
return n *(n-1)/2
dct = {0:[0, 0]}
for i in lst:
dct[i] = [0, 0]
for i in range(n):
if i %2 == 0:
dct[lst[i]][0] += 1
else:
dct[lst[i]][1] += 1
total += dct[0][1]
dct.pop(0)
for i in dct:
total += ikili(dct[i][0]) + ikili(dct[i][1])
print(int(total))
```
No
| 99,532 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sasha likes programming. Once, during a very long contest, Sasha decided that he was a bit tired and needed to relax. So he did. But since Sasha isn't an ordinary guy, he prefers to relax unusually. During leisure time Sasha likes to upsolve unsolved problems because upsolving is very useful.
Therefore, Sasha decided to upsolve the following problem:
You have an array a with n integers. You need to count the number of funny pairs (l, r) (l ≤ r). To check if a pair (l, r) is a funny pair, take mid = (l + r - 1)/(2), then if r - l + 1 is an even number and a_l ⊕ a_{l+1} ⊕ … ⊕ a_{mid} = a_{mid + 1} ⊕ a_{mid + 2} ⊕ … ⊕ a_r, then the pair is funny. In other words, ⊕ of elements of the left half of the subarray from l to r should be equal to ⊕ of elements of the right half. Note that ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
It is time to continue solving the contest, so Sasha asked you to solve this task.
Input
The first line contains one integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the size of the array.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i < 2^{20}) — array itself.
Output
Print one integer — the number of funny pairs. You should consider only pairs where r - l + 1 is even number.
Examples
Input
5
1 2 3 4 5
Output
1
Input
6
3 2 2 3 7 6
Output
3
Input
3
42 4 2
Output
0
Note
Be as cool as Sasha, upsolve problems!
In the first example, the only funny pair is (2, 5), as 2 ⊕ 3 = 4 ⊕ 5 = 1.
In the second example, funny pairs are (2, 3), (1, 4), and (3, 6).
In the third example, there are no funny pairs.
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
x,cnt=0,0
for i in range(n-1):
x^=a[i]
y=0
for j in range(n):
y^=a[j]
if((j-i+1)%2==0) and (x==y):
cnt+=1
x=0
print(cnt)
```
No
| 99,533 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sasha likes programming. Once, during a very long contest, Sasha decided that he was a bit tired and needed to relax. So he did. But since Sasha isn't an ordinary guy, he prefers to relax unusually. During leisure time Sasha likes to upsolve unsolved problems because upsolving is very useful.
Therefore, Sasha decided to upsolve the following problem:
You have an array a with n integers. You need to count the number of funny pairs (l, r) (l ≤ r). To check if a pair (l, r) is a funny pair, take mid = (l + r - 1)/(2), then if r - l + 1 is an even number and a_l ⊕ a_{l+1} ⊕ … ⊕ a_{mid} = a_{mid + 1} ⊕ a_{mid + 2} ⊕ … ⊕ a_r, then the pair is funny. In other words, ⊕ of elements of the left half of the subarray from l to r should be equal to ⊕ of elements of the right half. Note that ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
It is time to continue solving the contest, so Sasha asked you to solve this task.
Input
The first line contains one integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the size of the array.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i < 2^{20}) — array itself.
Output
Print one integer — the number of funny pairs. You should consider only pairs where r - l + 1 is even number.
Examples
Input
5
1 2 3 4 5
Output
1
Input
6
3 2 2 3 7 6
Output
3
Input
3
42 4 2
Output
0
Note
Be as cool as Sasha, upsolve problems!
In the first example, the only funny pair is (2, 5), as 2 ⊕ 3 = 4 ⊕ 5 = 1.
In the second example, funny pairs are (2, 3), (1, 4), and (3, 6).
In the third example, there are no funny pairs.
Submitted Solution:
```
n = int(input())
arr_str = input().split()
arr = [int(x) for x in arr_str]
cum_sum_arr = []
cum_sum_arr.append(0)
cum_sum = 0
for i in range(0, len(arr)):
cum_sum = cum_sum^arr[i]
cum_sum_arr.append(cum_sum)
print(cum_sum_arr)
count_pair = 0
sum2ind = dict()
sum2ind[0] = [0]
for r in range(1, len(cum_sum_arr)):
if cum_sum_arr[r] in sum2ind:
for l_minus_1 in sum2ind[cum_sum_arr[r]]:
if (r - l_minus_1) % 2 == 0:
count_pair += 1
sum2ind[cum_sum_arr[r]].append(r)
else:
sum2ind[cum_sum_arr[r]] = [r]
print(count_pair)
```
No
| 99,534 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This problem is given in two editions, which differ exclusively in the constraints on the number n.
You are given an array of integers a[1], a[2], ..., a[n]. A block is a sequence of contiguous (consecutive) elements a[l], a[l+1], ..., a[r] (1 ≤ l ≤ r ≤ n). Thus, a block is defined by a pair of indices (l, r).
Find a set of blocks (l_1, r_1), (l_2, r_2), ..., (l_k, r_k) such that:
* They do not intersect (i.e. they are disjoint). Formally, for each pair of blocks (l_i, r_i) and (l_j, r_j) where i ≠ j either r_i < l_j or r_j < l_i.
* For each block the sum of its elements is the same. Formally, $$$a[l_1]+a[l_1+1]+...+a[r_1]=a[l_2]+a[l_2+1]+...+a[r_2]= ... = a[l_k]+a[l_k+1]+...+a[r_k].$$$
* The number of the blocks in the set is maximum. Formally, there does not exist a set of blocks (l_1', r_1'), (l_2', r_2'), ..., (l_{k'}', r_{k'}') satisfying the above two requirements with k' > k.
<image> The picture corresponds to the first example. Blue boxes illustrate blocks.
Write a program to find such a set of blocks.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the length of the given array. The second line contains the sequence of elements a[1], a[2], ..., a[n] (-10^5 ≤ a_i ≤ 10^5).
Output
In the first line print the integer k (1 ≤ k ≤ n). The following k lines should contain blocks, one per line. In each line print a pair of indices l_i, r_i (1 ≤ l_i ≤ r_i ≤ n) — the bounds of the i-th block. You can print blocks in any order. If there are multiple answers, print any of them.
Examples
Input
7
4 1 2 2 1 5 3
Output
3
7 7
2 3
4 5
Input
11
-5 -4 -3 -2 -1 0 1 2 3 4 5
Output
2
3 4
1 1
Input
4
1 1 1 1
Output
4
4 4
1 1
2 2
3 3
Tags: greedy
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
if n == 1:
print(1)
print('1 1')
else:
d = [[]]*(n-1)
d[0] = a
for i in range(1,n-1):
d[i] = [d[i-1][j]+a[j+i] for j in range(0,n-i)]
d2 = {}
for i,d_ in enumerate(d):
for j,x in enumerate(d_):
if x in d2:
d2[x].append([j+1,j+i+1])
else:
d2[x]=[[j+1,j+i+1]]
list_keys = list(d2.keys())
#res = 0
ma = 0
for key in list_keys:
d2[key].sort(key=lambda x:x[1])
after = -1
cnt = 0
tmp = []
for y,z in d2[key]:
if y > after:
cnt += 1
after = z
tmp.append([y,z])
if cnt > ma:
ma = cnt
res = tmp
# for j in range(len(d2[key])):
# after = -1
# cnt = 0
# tmp = []
# for y,z in d2[key][j:]:
# if y > after:
# cnt += 1
# after = z
# tmp.append([y,z])
# if cnt > ma:
# ma = cnt
# res = tmp
print(len(res))
for x in res:
print(*x)
```
| 99,535 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This problem is given in two editions, which differ exclusively in the constraints on the number n.
You are given an array of integers a[1], a[2], ..., a[n]. A block is a sequence of contiguous (consecutive) elements a[l], a[l+1], ..., a[r] (1 ≤ l ≤ r ≤ n). Thus, a block is defined by a pair of indices (l, r).
Find a set of blocks (l_1, r_1), (l_2, r_2), ..., (l_k, r_k) such that:
* They do not intersect (i.e. they are disjoint). Formally, for each pair of blocks (l_i, r_i) and (l_j, r_j) where i ≠ j either r_i < l_j or r_j < l_i.
* For each block the sum of its elements is the same. Formally, $$$a[l_1]+a[l_1+1]+...+a[r_1]=a[l_2]+a[l_2+1]+...+a[r_2]= ... = a[l_k]+a[l_k+1]+...+a[r_k].$$$
* The number of the blocks in the set is maximum. Formally, there does not exist a set of blocks (l_1', r_1'), (l_2', r_2'), ..., (l_{k'}', r_{k'}') satisfying the above two requirements with k' > k.
<image> The picture corresponds to the first example. Blue boxes illustrate blocks.
Write a program to find such a set of blocks.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the length of the given array. The second line contains the sequence of elements a[1], a[2], ..., a[n] (-10^5 ≤ a_i ≤ 10^5).
Output
In the first line print the integer k (1 ≤ k ≤ n). The following k lines should contain blocks, one per line. In each line print a pair of indices l_i, r_i (1 ≤ l_i ≤ r_i ≤ n) — the bounds of the i-th block. You can print blocks in any order. If there are multiple answers, print any of them.
Examples
Input
7
4 1 2 2 1 5 3
Output
3
7 7
2 3
4 5
Input
11
-5 -4 -3 -2 -1 0 1 2 3 4 5
Output
2
3 4
1 1
Input
4
1 1 1 1
Output
4
4 4
1 1
2 2
3 3
Tags: greedy
Correct Solution:
```
from collections import defaultdict
n = int(input())
arr = [int(i) for i in input().split()]
prefix = [0 for i in range(n)]
segments = defaultdict(list)
prefix[0] = arr[0]
for i in range(1,n):
prefix[i] = prefix[i - 1] + arr[i]
long = 0
long_i = 0
for i in range(n):
for j in range(n):
if j + i > n - 1:
break
l, r = i, j + i
res = prefix[r] - ((prefix[l - 1]) if l != 0 else 0)
if res not in segments:
segments[res].append(l)
segments[res].append(r)
elif r < segments[res][-1]:
segments[res][-2] = l
segments[res][-1] = r
elif segments[res][-1] < l:
segments[res].append(l)
segments[res].append(r)
if len(segments[res]) > long:
long = len(segments[res])
long_i = res
print(len(segments[long_i])//2)
for i in range(0,len(segments[long_i]),2):
print(segments[long_i][i] + 1,segments[long_i][i + 1] + 1)
```
| 99,536 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This problem is given in two editions, which differ exclusively in the constraints on the number n.
You are given an array of integers a[1], a[2], ..., a[n]. A block is a sequence of contiguous (consecutive) elements a[l], a[l+1], ..., a[r] (1 ≤ l ≤ r ≤ n). Thus, a block is defined by a pair of indices (l, r).
Find a set of blocks (l_1, r_1), (l_2, r_2), ..., (l_k, r_k) such that:
* They do not intersect (i.e. they are disjoint). Formally, for each pair of blocks (l_i, r_i) and (l_j, r_j) where i ≠ j either r_i < l_j or r_j < l_i.
* For each block the sum of its elements is the same. Formally, $$$a[l_1]+a[l_1+1]+...+a[r_1]=a[l_2]+a[l_2+1]+...+a[r_2]= ... = a[l_k]+a[l_k+1]+...+a[r_k].$$$
* The number of the blocks in the set is maximum. Formally, there does not exist a set of blocks (l_1', r_1'), (l_2', r_2'), ..., (l_{k'}', r_{k'}') satisfying the above two requirements with k' > k.
<image> The picture corresponds to the first example. Blue boxes illustrate blocks.
Write a program to find such a set of blocks.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the length of the given array. The second line contains the sequence of elements a[1], a[2], ..., a[n] (-10^5 ≤ a_i ≤ 10^5).
Output
In the first line print the integer k (1 ≤ k ≤ n). The following k lines should contain blocks, one per line. In each line print a pair of indices l_i, r_i (1 ≤ l_i ≤ r_i ≤ n) — the bounds of the i-th block. You can print blocks in any order. If there are multiple answers, print any of them.
Examples
Input
7
4 1 2 2 1 5 3
Output
3
7 7
2 3
4 5
Input
11
-5 -4 -3 -2 -1 0 1 2 3 4 5
Output
2
3 4
1 1
Input
4
1 1 1 1
Output
4
4 4
1 1
2 2
3 3
Tags: greedy
Correct Solution:
```
n = int(input())
lst = list(map(int,input().split()))
d,res,summa = {},{},0
for i,y in enumerate(lst):
summa+=y
s=summa
for j in range(i+1):
if d.get(s)==None:
d[s]=[0,-1]
res[s]=[]
if d[s][1]<j:
d[s][0]+=1
d[s][1]=i
res[s].append([j+1,i+1])
s-=lst[j]
ans,e = 0,-1
for i,x in enumerate(d):
if d[x][0]>ans:
ans,e=d[x][0],x
print(ans)
for i,x in enumerate(res[e]):
print(*x)
```
| 99,537 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This problem is given in two editions, which differ exclusively in the constraints on the number n.
You are given an array of integers a[1], a[2], ..., a[n]. A block is a sequence of contiguous (consecutive) elements a[l], a[l+1], ..., a[r] (1 ≤ l ≤ r ≤ n). Thus, a block is defined by a pair of indices (l, r).
Find a set of blocks (l_1, r_1), (l_2, r_2), ..., (l_k, r_k) such that:
* They do not intersect (i.e. they are disjoint). Formally, for each pair of blocks (l_i, r_i) and (l_j, r_j) where i ≠ j either r_i < l_j or r_j < l_i.
* For each block the sum of its elements is the same. Formally, $$$a[l_1]+a[l_1+1]+...+a[r_1]=a[l_2]+a[l_2+1]+...+a[r_2]= ... = a[l_k]+a[l_k+1]+...+a[r_k].$$$
* The number of the blocks in the set is maximum. Formally, there does not exist a set of blocks (l_1', r_1'), (l_2', r_2'), ..., (l_{k'}', r_{k'}') satisfying the above two requirements with k' > k.
<image> The picture corresponds to the first example. Blue boxes illustrate blocks.
Write a program to find such a set of blocks.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the length of the given array. The second line contains the sequence of elements a[1], a[2], ..., a[n] (-10^5 ≤ a_i ≤ 10^5).
Output
In the first line print the integer k (1 ≤ k ≤ n). The following k lines should contain blocks, one per line. In each line print a pair of indices l_i, r_i (1 ≤ l_i ≤ r_i ≤ n) — the bounds of the i-th block. You can print blocks in any order. If there are multiple answers, print any of them.
Examples
Input
7
4 1 2 2 1 5 3
Output
3
7 7
2 3
4 5
Input
11
-5 -4 -3 -2 -1 0 1 2 3 4 5
Output
2
3 4
1 1
Input
4
1 1 1 1
Output
4
4 4
1 1
2 2
3 3
Tags: greedy
Correct Solution:
```
from collections import defaultdict
def main():
n = int(input())
values = list(map(int, input().split(' ')))
# print(values)
ans = defaultdict(list)
for i in range(n):
s = 0
# print("---------- i = {} ----------".format(i))
for j in range(i, -1, -1):
s += values[j]
# print("i = {}; j = {} s = {}".format(i, j, s))
# print("ans = {}; (i + 1) = {}".format(i + 1, ans[s][-1]))
# if (i + 1) not in ans[s][-1]:
ans[s].append((j + 1, i + 1))
# print(ans)
answer = dict()
max = 0
for key in ans:
# print("key = {} ; ans[key] = {}".format(key, ans[key], len(ans[key])))
sum_pairs = ans[key]
non_overlap_pairs = [sum_pairs[0]]
previous_pair_second_value = sum_pairs[0][1]
for each_pair in sum_pairs[1:]:
# print("each_pair = {}".format(each_pair))
if previous_pair_second_value < each_pair[0]:
# print(each_pair)
non_overlap_pairs.append(each_pair)
previous_pair_second_value = each_pair[1]
# else:
# print("Found overlapping pair for key = {}".format(each_pair))
# ans[key] = non_overlap_pairs
if len(non_overlap_pairs) > max:
max = len(non_overlap_pairs)
answer = {max: non_overlap_pairs}
# print(list(answer.keys())[0])
for key in answer.keys():
print(key)
for value in answer[key]:
print(str(value[0]) + ' ' + str(value[1]))
if __name__=="__main__":
main()
```
| 99,538 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This problem is given in two editions, which differ exclusively in the constraints on the number n.
You are given an array of integers a[1], a[2], ..., a[n]. A block is a sequence of contiguous (consecutive) elements a[l], a[l+1], ..., a[r] (1 ≤ l ≤ r ≤ n). Thus, a block is defined by a pair of indices (l, r).
Find a set of blocks (l_1, r_1), (l_2, r_2), ..., (l_k, r_k) such that:
* They do not intersect (i.e. they are disjoint). Formally, for each pair of blocks (l_i, r_i) and (l_j, r_j) where i ≠ j either r_i < l_j or r_j < l_i.
* For each block the sum of its elements is the same. Formally, $$$a[l_1]+a[l_1+1]+...+a[r_1]=a[l_2]+a[l_2+1]+...+a[r_2]= ... = a[l_k]+a[l_k+1]+...+a[r_k].$$$
* The number of the blocks in the set is maximum. Formally, there does not exist a set of blocks (l_1', r_1'), (l_2', r_2'), ..., (l_{k'}', r_{k'}') satisfying the above two requirements with k' > k.
<image> The picture corresponds to the first example. Blue boxes illustrate blocks.
Write a program to find such a set of blocks.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the length of the given array. The second line contains the sequence of elements a[1], a[2], ..., a[n] (-10^5 ≤ a_i ≤ 10^5).
Output
In the first line print the integer k (1 ≤ k ≤ n). The following k lines should contain blocks, one per line. In each line print a pair of indices l_i, r_i (1 ≤ l_i ≤ r_i ≤ n) — the bounds of the i-th block. You can print blocks in any order. If there are multiple answers, print any of them.
Examples
Input
7
4 1 2 2 1 5 3
Output
3
7 7
2 3
4 5
Input
11
-5 -4 -3 -2 -1 0 1 2 3 4 5
Output
2
3 4
1 1
Input
4
1 1 1 1
Output
4
4 4
1 1
2 2
3 3
Tags: greedy
Correct Solution:
```
from itertools import accumulate
def main():
n = int(input())
arr = list(map(int, input().split()))
arr_sums = [0] + list(accumulate(arr))
blocks = {}
for i in range(1, n+1):
for j in range(i):
total = arr_sums[i] - arr_sums[j]
if total not in blocks:
blocks[total] = [(j+1, i)]
else:
if blocks[total][-1][1] < j+1:
blocks[total].append((j+1, i))
max_block = sorted([(i, len(x)) for i, x in blocks.items()], key=lambda y: (-y[1], y[0]))
print(max_block[0][1])
for item in blocks[max_block[0][0]]:
print(item[0], item[1])
if __name__ == '__main__':
main()
```
| 99,539 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This problem is given in two editions, which differ exclusively in the constraints on the number n.
You are given an array of integers a[1], a[2], ..., a[n]. A block is a sequence of contiguous (consecutive) elements a[l], a[l+1], ..., a[r] (1 ≤ l ≤ r ≤ n). Thus, a block is defined by a pair of indices (l, r).
Find a set of blocks (l_1, r_1), (l_2, r_2), ..., (l_k, r_k) such that:
* They do not intersect (i.e. they are disjoint). Formally, for each pair of blocks (l_i, r_i) and (l_j, r_j) where i ≠ j either r_i < l_j or r_j < l_i.
* For each block the sum of its elements is the same. Formally, $$$a[l_1]+a[l_1+1]+...+a[r_1]=a[l_2]+a[l_2+1]+...+a[r_2]= ... = a[l_k]+a[l_k+1]+...+a[r_k].$$$
* The number of the blocks in the set is maximum. Formally, there does not exist a set of blocks (l_1', r_1'), (l_2', r_2'), ..., (l_{k'}', r_{k'}') satisfying the above two requirements with k' > k.
<image> The picture corresponds to the first example. Blue boxes illustrate blocks.
Write a program to find such a set of blocks.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the length of the given array. The second line contains the sequence of elements a[1], a[2], ..., a[n] (-10^5 ≤ a_i ≤ 10^5).
Output
In the first line print the integer k (1 ≤ k ≤ n). The following k lines should contain blocks, one per line. In each line print a pair of indices l_i, r_i (1 ≤ l_i ≤ r_i ≤ n) — the bounds of the i-th block. You can print blocks in any order. If there are multiple answers, print any of them.
Examples
Input
7
4 1 2 2 1 5 3
Output
3
7 7
2 3
4 5
Input
11
-5 -4 -3 -2 -1 0 1 2 3 4 5
Output
2
3 4
1 1
Input
4
1 1 1 1
Output
4
4 4
1 1
2 2
3 3
Tags: greedy
Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
dic={}
for i in range(n):
sm=0
for j in range(i,n):
sm+=a[j]
if sm in dic:
dic[sm].append((i,j))
else:
dic[sm]=[(i,j)]
ans=0
anskey=-1
for key in dic:
cnt=0
last=-1
for a,b in sorted(dic[key]):
if a>last:
cnt+=1
last=b
elif b<last:
last=b
if cnt>ans:
ans=cnt
anskey=key
last=-1
tmp=[]
for a,b in sorted(dic[anskey]):
if a>last:
last=b
tmp.append(str(a+1)+" "+str(b+1))
elif b<last:
last=b
tmp.pop()
tmp.append(str(a+1)+" "+str(b+1))
print(ans,'\n'.join(tmp),sep='\n')
```
| 99,540 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This problem is given in two editions, which differ exclusively in the constraints on the number n.
You are given an array of integers a[1], a[2], ..., a[n]. A block is a sequence of contiguous (consecutive) elements a[l], a[l+1], ..., a[r] (1 ≤ l ≤ r ≤ n). Thus, a block is defined by a pair of indices (l, r).
Find a set of blocks (l_1, r_1), (l_2, r_2), ..., (l_k, r_k) such that:
* They do not intersect (i.e. they are disjoint). Formally, for each pair of blocks (l_i, r_i) and (l_j, r_j) where i ≠ j either r_i < l_j or r_j < l_i.
* For each block the sum of its elements is the same. Formally, $$$a[l_1]+a[l_1+1]+...+a[r_1]=a[l_2]+a[l_2+1]+...+a[r_2]= ... = a[l_k]+a[l_k+1]+...+a[r_k].$$$
* The number of the blocks in the set is maximum. Formally, there does not exist a set of blocks (l_1', r_1'), (l_2', r_2'), ..., (l_{k'}', r_{k'}') satisfying the above two requirements with k' > k.
<image> The picture corresponds to the first example. Blue boxes illustrate blocks.
Write a program to find such a set of blocks.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the length of the given array. The second line contains the sequence of elements a[1], a[2], ..., a[n] (-10^5 ≤ a_i ≤ 10^5).
Output
In the first line print the integer k (1 ≤ k ≤ n). The following k lines should contain blocks, one per line. In each line print a pair of indices l_i, r_i (1 ≤ l_i ≤ r_i ≤ n) — the bounds of the i-th block. You can print blocks in any order. If there are multiple answers, print any of them.
Examples
Input
7
4 1 2 2 1 5 3
Output
3
7 7
2 3
4 5
Input
11
-5 -4 -3 -2 -1 0 1 2 3 4 5
Output
2
3 4
1 1
Input
4
1 1 1 1
Output
4
4 4
1 1
2 2
3 3
Tags: greedy
Correct Solution:
```
n = int(input())
arr= list(map(int, input().split()))
sums = {0: []}
for i in range(n):
sum = 0
for j in range(i, -1, -1):
sum += arr[j]
if sum not in sums.keys():
sums[sum] = []
sums[sum].append((j,i))
#print(sums)
ans = (0, [])
for k in sums:
allblks = sums[k]
tot = 0
r = -1
ll = []
for ele in allblks:
if ele[0] > r:
tot += 1
ll.append(ele)
r = ele[1]
if tot > ans[0] :
ans = (tot, ll)
print(ans[0])
lt = ans[1]
for r in lt:
print(r[0] +1, r[1] +1)
```
| 99,541 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This problem is given in two editions, which differ exclusively in the constraints on the number n.
You are given an array of integers a[1], a[2], ..., a[n]. A block is a sequence of contiguous (consecutive) elements a[l], a[l+1], ..., a[r] (1 ≤ l ≤ r ≤ n). Thus, a block is defined by a pair of indices (l, r).
Find a set of blocks (l_1, r_1), (l_2, r_2), ..., (l_k, r_k) such that:
* They do not intersect (i.e. they are disjoint). Formally, for each pair of blocks (l_i, r_i) and (l_j, r_j) where i ≠ j either r_i < l_j or r_j < l_i.
* For each block the sum of its elements is the same. Formally, $$$a[l_1]+a[l_1+1]+...+a[r_1]=a[l_2]+a[l_2+1]+...+a[r_2]= ... = a[l_k]+a[l_k+1]+...+a[r_k].$$$
* The number of the blocks in the set is maximum. Formally, there does not exist a set of blocks (l_1', r_1'), (l_2', r_2'), ..., (l_{k'}', r_{k'}') satisfying the above two requirements with k' > k.
<image> The picture corresponds to the first example. Blue boxes illustrate blocks.
Write a program to find such a set of blocks.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the length of the given array. The second line contains the sequence of elements a[1], a[2], ..., a[n] (-10^5 ≤ a_i ≤ 10^5).
Output
In the first line print the integer k (1 ≤ k ≤ n). The following k lines should contain blocks, one per line. In each line print a pair of indices l_i, r_i (1 ≤ l_i ≤ r_i ≤ n) — the bounds of the i-th block. You can print blocks in any order. If there are multiple answers, print any of them.
Examples
Input
7
4 1 2 2 1 5 3
Output
3
7 7
2 3
4 5
Input
11
-5 -4 -3 -2 -1 0 1 2 3 4 5
Output
2
3 4
1 1
Input
4
1 1 1 1
Output
4
4 4
1 1
2 2
3 3
Tags: greedy
Correct Solution:
```
n=int(input())
arr=list(map(int,input().split()))
dict1={}
for i in range(n):
val=0
for j in range(i,n):
val+=arr[j]
try:
dict1[val].append((i,j))
except:
KeyError
dict1[val]=[(i,j)]
ans=0
ansarr=[]
for i in dict1.keys():
if(len(dict1[i])>ans):
arr2=[]
for j in range(len(dict1[i])):
arr2.append((dict1[i][j][1]-dict1[i][j][0],j))
arr2.sort()
temp=[]
for j in range(len(arr2)):
flag=0
indexi=dict1[i][arr2[j][1]][0]
indexj=dict1[i][arr2[j][1]][1]
for k in range(len(temp)):
if(temp[k][0]<=indexi<=temp[k][1] or temp[k][0]<=indexj<=temp[k][1] or indexi<=temp[k][0]<=indexj or indexi<=temp[k][1]<=indexj):
flag=1
if(flag==0):
temp.append((indexi,indexj))
if(len(temp)>ans):
ans=len(temp)
ansarr=temp
print(ans)
finalans=[]
for i in range(len(ansarr)):
finalans.append(ansarr[i][0]+1)
finalans.append(ansarr[i][1]+1)
print(*finalans)
```
| 99,542 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem is given in two editions, which differ exclusively in the constraints on the number n.
You are given an array of integers a[1], a[2], ..., a[n]. A block is a sequence of contiguous (consecutive) elements a[l], a[l+1], ..., a[r] (1 ≤ l ≤ r ≤ n). Thus, a block is defined by a pair of indices (l, r).
Find a set of blocks (l_1, r_1), (l_2, r_2), ..., (l_k, r_k) such that:
* They do not intersect (i.e. they are disjoint). Formally, for each pair of blocks (l_i, r_i) and (l_j, r_j) where i ≠ j either r_i < l_j or r_j < l_i.
* For each block the sum of its elements is the same. Formally, $$$a[l_1]+a[l_1+1]+...+a[r_1]=a[l_2]+a[l_2+1]+...+a[r_2]= ... = a[l_k]+a[l_k+1]+...+a[r_k].$$$
* The number of the blocks in the set is maximum. Formally, there does not exist a set of blocks (l_1', r_1'), (l_2', r_2'), ..., (l_{k'}', r_{k'}') satisfying the above two requirements with k' > k.
<image> The picture corresponds to the first example. Blue boxes illustrate blocks.
Write a program to find such a set of blocks.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the length of the given array. The second line contains the sequence of elements a[1], a[2], ..., a[n] (-10^5 ≤ a_i ≤ 10^5).
Output
In the first line print the integer k (1 ≤ k ≤ n). The following k lines should contain blocks, one per line. In each line print a pair of indices l_i, r_i (1 ≤ l_i ≤ r_i ≤ n) — the bounds of the i-th block. You can print blocks in any order. If there are multiple answers, print any of them.
Examples
Input
7
4 1 2 2 1 5 3
Output
3
7 7
2 3
4 5
Input
11
-5 -4 -3 -2 -1 0 1 2 3 4 5
Output
2
3 4
1 1
Input
4
1 1 1 1
Output
4
4 4
1 1
2 2
3 3
Submitted Solution:
```
#!/usr/bin/env python3
from collections import defaultdict
def main():
n = int(input())
a = [int(ai) for ai in input().split()]
pref_sums = [0] * (n + 1)
for i, ai in enumerate(a):
pref_sums[i + 1] = pref_sums[i] + ai
block_sums = defaultdict(list)
for j in range(1, n + 1):
for i in range(1, j + 1):
block_sums[pref_sums[j] - pref_sums[i - 1]].append((i, j))
ans = []
for A in block_sums.values():
res, k = [A[0]], 0
for i in range(1, len(A)):
if A[i][0] > A[k][1]:
res.append(A[i])
k = i
if len(res) > len(ans):
ans = res
print(len(ans))
print('\n'.join(map(lambda s: '%d %d' % s, ans)))
if __name__ == '__main__':
main()
```
Yes
| 99,543 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem is given in two editions, which differ exclusively in the constraints on the number n.
You are given an array of integers a[1], a[2], ..., a[n]. A block is a sequence of contiguous (consecutive) elements a[l], a[l+1], ..., a[r] (1 ≤ l ≤ r ≤ n). Thus, a block is defined by a pair of indices (l, r).
Find a set of blocks (l_1, r_1), (l_2, r_2), ..., (l_k, r_k) such that:
* They do not intersect (i.e. they are disjoint). Formally, for each pair of blocks (l_i, r_i) and (l_j, r_j) where i ≠ j either r_i < l_j or r_j < l_i.
* For each block the sum of its elements is the same. Formally, $$$a[l_1]+a[l_1+1]+...+a[r_1]=a[l_2]+a[l_2+1]+...+a[r_2]= ... = a[l_k]+a[l_k+1]+...+a[r_k].$$$
* The number of the blocks in the set is maximum. Formally, there does not exist a set of blocks (l_1', r_1'), (l_2', r_2'), ..., (l_{k'}', r_{k'}') satisfying the above two requirements with k' > k.
<image> The picture corresponds to the first example. Blue boxes illustrate blocks.
Write a program to find such a set of blocks.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the length of the given array. The second line contains the sequence of elements a[1], a[2], ..., a[n] (-10^5 ≤ a_i ≤ 10^5).
Output
In the first line print the integer k (1 ≤ k ≤ n). The following k lines should contain blocks, one per line. In each line print a pair of indices l_i, r_i (1 ≤ l_i ≤ r_i ≤ n) — the bounds of the i-th block. You can print blocks in any order. If there are multiple answers, print any of them.
Examples
Input
7
4 1 2 2 1 5 3
Output
3
7 7
2 3
4 5
Input
11
-5 -4 -3 -2 -1 0 1 2 3 4 5
Output
2
3 4
1 1
Input
4
1 1 1 1
Output
4
4 4
1 1
2 2
3 3
Submitted Solution:
```
#!/usr/bin/env python
# -*- coding: utf-8 -*-
# @oj: codeforces
# @id: hitwanyang
# @email: 296866643@qq.com
# @date: 2020/12/17 17:03
# @url: https://codeforc.es/contest/1141/problem/F2
import sys, os
from io import BytesIO, IOBase
import collections, itertools, bisect, heapq, math, string
from decimal import *
# region fastio
BUFSIZE = 8192
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------------------
## 注意嵌套括号!!!!!!
## 先有思路,再写代码,别着急!!!
## 先有朴素解法,不要有思维定式,试着换思路解决
## 精度 print("%.10f" % ans)
## sqrt:int(math.sqrt(n))+1
## 字符串拼接不要用+操作，会超时
## 二进制转换:bin(1)[2:].rjust(32,'0')
## array copy:cur=array[::]
## oeis:example 1, 3, _, 1260, _, _, _, _, _, 12164510040883200
## sqrt:Decimal(x).sqrt()避免精度误差
## 无穷大表示:float('inf')
## py 10**6 排序+双指针 3秒可能TLE
def main():
n = int(input())
a = list(map(int, input().split()))
prefix = list(itertools.accumulate(a))
d = collections.defaultdict(list)
for i in range(n):
for j in range(i, n):
if i == 0:
d[prefix[j]].append((i, j))
else:
v = prefix[j] - prefix[i - 1]
d[v].append((i, j))
ans = []
for k in d.keys():
v = d[k]
cnt = []
## 按区间右端点排序
tmp = sorted(v,key=lambda x:(x[1],x[0]))
if len(tmp) == 1:
cnt.append((tmp[0][0] + 1, tmp[0][1] + 1))
else:
pre = tmp[0]
## 贪心求不相交区间的最大个数
cnt.append((pre[0] + 1, pre[1] + 1))
for i in range(1, len(tmp)):
cur = tmp[i]
if cur[0] > pre[1]:
cnt.append((cur[0] + 1, cur[1] + 1))
pre = cur
if len(cnt) > len(ans):
ans = cnt
############## TLE code ##############
## 按区间右端点排序
# sr = sorted(res, key=lambda x: (x[2], x[1], x[0]))
# print(time.time() - start)
# l, r = 0, 0
# while r < len(sr):
# pre = sr[l]
# cnt = [(pre[0] + 1, pre[1] + 1)]
# while r < len(sr) and sr[l][2] == sr[r][2]:
# cur=sr[r]
# if cur[0] > pre[1]:
# cnt.append((cur[0] + 1, cur[1] + 1))
# pre = cur
# r += 1
# l = r
# if len(cnt) > len(ans):
# ans = cnt
# print (time.time()-start)
print(len(ans))
for a in ans:
print(a[0], a[1])
if __name__ == "__main__":
main()
```
Yes
| 99,544 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem is given in two editions, which differ exclusively in the constraints on the number n.
You are given an array of integers a[1], a[2], ..., a[n]. A block is a sequence of contiguous (consecutive) elements a[l], a[l+1], ..., a[r] (1 ≤ l ≤ r ≤ n). Thus, a block is defined by a pair of indices (l, r).
Find a set of blocks (l_1, r_1), (l_2, r_2), ..., (l_k, r_k) such that:
* They do not intersect (i.e. they are disjoint). Formally, for each pair of blocks (l_i, r_i) and (l_j, r_j) where i ≠ j either r_i < l_j or r_j < l_i.
* For each block the sum of its elements is the same. Formally, $$$a[l_1]+a[l_1+1]+...+a[r_1]=a[l_2]+a[l_2+1]+...+a[r_2]= ... = a[l_k]+a[l_k+1]+...+a[r_k].$$$
* The number of the blocks in the set is maximum. Formally, there does not exist a set of blocks (l_1', r_1'), (l_2', r_2'), ..., (l_{k'}', r_{k'}') satisfying the above two requirements with k' > k.
<image> The picture corresponds to the first example. Blue boxes illustrate blocks.
Write a program to find such a set of blocks.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the length of the given array. The second line contains the sequence of elements a[1], a[2], ..., a[n] (-10^5 ≤ a_i ≤ 10^5).
Output
In the first line print the integer k (1 ≤ k ≤ n). The following k lines should contain blocks, one per line. In each line print a pair of indices l_i, r_i (1 ≤ l_i ≤ r_i ≤ n) — the bounds of the i-th block. You can print blocks in any order. If there are multiple answers, print any of them.
Examples
Input
7
4 1 2 2 1 5 3
Output
3
7 7
2 3
4 5
Input
11
-5 -4 -3 -2 -1 0 1 2 3 4 5
Output
2
3 4
1 1
Input
4
1 1 1 1
Output
4
4 4
1 1
2 2
3 3
Submitted Solution:
```
from collections import defaultdict
n = int(input())
a = list(map(int, input().split()))
rec = defaultdict(list)
for j in range(n):
for k in range(j, n):
rec[sum(a[j:k + 1])].append((j, k))
ans = []
for k in rec.keys():
tmp = []
rec[k] = sorted(rec[k], key=lambda x: x[1])
pre = -1
for a, b in rec[k]:
if pre >= a:
continue
else:
tmp.append((a + 1, b + 1))
pre = b
if len(tmp) > len(ans):
ans = tmp
print(len(ans))
for a, b in ans:
print(a, b)
```
Yes
| 99,545 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem is given in two editions, which differ exclusively in the constraints on the number n.
You are given an array of integers a[1], a[2], ..., a[n]. A block is a sequence of contiguous (consecutive) elements a[l], a[l+1], ..., a[r] (1 ≤ l ≤ r ≤ n). Thus, a block is defined by a pair of indices (l, r).
Find a set of blocks (l_1, r_1), (l_2, r_2), ..., (l_k, r_k) such that:
* They do not intersect (i.e. they are disjoint). Formally, for each pair of blocks (l_i, r_i) and (l_j, r_j) where i ≠ j either r_i < l_j or r_j < l_i.
* For each block the sum of its elements is the same. Formally, $$$a[l_1]+a[l_1+1]+...+a[r_1]=a[l_2]+a[l_2+1]+...+a[r_2]= ... = a[l_k]+a[l_k+1]+...+a[r_k].$$$
* The number of the blocks in the set is maximum. Formally, there does not exist a set of blocks (l_1', r_1'), (l_2', r_2'), ..., (l_{k'}', r_{k'}') satisfying the above two requirements with k' > k.
<image> The picture corresponds to the first example. Blue boxes illustrate blocks.
Write a program to find such a set of blocks.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the length of the given array. The second line contains the sequence of elements a[1], a[2], ..., a[n] (-10^5 ≤ a_i ≤ 10^5).
Output
In the first line print the integer k (1 ≤ k ≤ n). The following k lines should contain blocks, one per line. In each line print a pair of indices l_i, r_i (1 ≤ l_i ≤ r_i ≤ n) — the bounds of the i-th block. You can print blocks in any order. If there are multiple answers, print any of them.
Examples
Input
7
4 1 2 2 1 5 3
Output
3
7 7
2 3
4 5
Input
11
-5 -4 -3 -2 -1 0 1 2 3 4 5
Output
2
3 4
1 1
Input
4
1 1 1 1
Output
4
4 4
1 1
2 2
3 3
Submitted Solution:
```
from collections import defaultdict
n = int(input())
nums = list(map(int, input().split()))
freq = defaultdict(list)
for i in range(n):
for j in range(i+1, n+1):
freq[sum(nums[i:j])].append((i+1, j))
ans = []
for k in freq:
l = freq[k]
l.sort(key=lambda x: x[1])
tmp = [l[0]]
for i, j in l:
if i <= tmp[-1][1]:
continue
tmp.append([i, j])
if len(tmp) > len(ans):
ans = tmp
print (len(ans))
for i, j in ans:
print (i, j)
```
Yes
| 99,546 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem is given in two editions, which differ exclusively in the constraints on the number n.
You are given an array of integers a[1], a[2], ..., a[n]. A block is a sequence of contiguous (consecutive) elements a[l], a[l+1], ..., a[r] (1 ≤ l ≤ r ≤ n). Thus, a block is defined by a pair of indices (l, r).
Find a set of blocks (l_1, r_1), (l_2, r_2), ..., (l_k, r_k) such that:
* They do not intersect (i.e. they are disjoint). Formally, for each pair of blocks (l_i, r_i) and (l_j, r_j) where i ≠ j either r_i < l_j or r_j < l_i.
* For each block the sum of its elements is the same. Formally, $$$a[l_1]+a[l_1+1]+...+a[r_1]=a[l_2]+a[l_2+1]+...+a[r_2]= ... = a[l_k]+a[l_k+1]+...+a[r_k].$$$
* The number of the blocks in the set is maximum. Formally, there does not exist a set of blocks (l_1', r_1'), (l_2', r_2'), ..., (l_{k'}', r_{k'}') satisfying the above two requirements with k' > k.
<image> The picture corresponds to the first example. Blue boxes illustrate blocks.
Write a program to find such a set of blocks.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the length of the given array. The second line contains the sequence of elements a[1], a[2], ..., a[n] (-10^5 ≤ a_i ≤ 10^5).
Output
In the first line print the integer k (1 ≤ k ≤ n). The following k lines should contain blocks, one per line. In each line print a pair of indices l_i, r_i (1 ≤ l_i ≤ r_i ≤ n) — the bounds of the i-th block. You can print blocks in any order. If there are multiple answers, print any of them.
Examples
Input
7
4 1 2 2 1 5 3
Output
3
7 7
2 3
4 5
Input
11
-5 -4 -3 -2 -1 0 1 2 3 4 5
Output
2
3 4
1 1
Input
4
1 1 1 1
Output
4
4 4
1 1
2 2
3 3
Submitted Solution:
```
from collections import defaultdict
def find_best_path(pairs: list, n=1500):
#bfs?
dist = [1 for p in pairs]
#everyone is its own pred
pred = [x for x in range(len(pairs))]
#count different starts
#for every end, give a mapping where to find the correct offsets
offset = [-1 for x in range(n)]
for i, p in enumerate(pairs):
offset[p[0]] = i
last_offset=0
for i, o in enumerate(offset):
if o == -1:
offset[i] = last_offset
else:
last_offset = o
# pairs are already kind of sorted (by start and end)
#could be optimized: lookup of possible successors (not sure if worth)
for i in range(len(pairs)):
#offset[i] is the last entry
end = pairs[i][1]
for j in range(offset[end]+1, len(pairs)):
#for j in range(i+1, len(pairs)):
# valid successor
#if pairs[j][0] > pairs[i][1]:
if dist[j] < dist[i] + 1:
dist[j] += 1
pred[j] = i
last_entry = max(range(len(pairs)), key=lambda x: dist[x])
partitioning = []
while True:
entry = pairs[last_entry]
#1-indexing
partitioning.append([entry[0] + 1, entry[1] + 1])
if pred[last_entry] == last_entry:
break
last_entry = pred[last_entry]
del(pred)
del(dist)
return len(partitioning), partitioning
def find_partitioning(l: list):
ht = defaultdict(list)
for i, _ in enumerate(l):
val = 0
for j in range(i, len(l)):
val += l[j]
ht[val].append((i, j))
print("hashing done")
maximum_partitioning_size = 0
partitioning = []
for s, val in ht.items():
#print(s)
#print(len(val), maximum_partitioning_size)
if len(val) < maximum_partitioning_size:
continue
num_blocks, blocks = find_best_path(val, len(l))
if num_blocks > maximum_partitioning_size:
maximum_partitioning_size = num_blocks
partitioning = blocks
return partitioning
def test_find_partitioning():
l = [7, 1, 2, 2, 1, 5, 3]
res = find_partitioning(l)
print(res)
assert len(res) == 3
assert res[0] == [7,7]
assert res[1] == [4,5]
assert res[2] == [2,3]
def test_find_partitioning_2():
l = [-5,-4,-3,-2,-1,0,1,2,3,4,5]
res = find_partitioning(l)
#assert len(res) == 4
assert len(res) == 2
#assert res[0] == [7,7]
#assert res[1] == [2,3]
#assert res[2] == [4,5]
def test_big_partitioning():
import numpy as np
l = np.random.randint(-1,1, 1000)
res = find_partitioning(l)
if __name__ == "__main__":
import sys
lines = sys.stdin.readlines()
entries = lines[1].split()
entries = [int(i) for i in entries]
partitioning = find_partitioning(entries)
print(len(partitioning))
for p in partitioning:
print(p[0], p[1])
```
No
| 99,547 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem is given in two editions, which differ exclusively in the constraints on the number n.
You are given an array of integers a[1], a[2], ..., a[n]. A block is a sequence of contiguous (consecutive) elements a[l], a[l+1], ..., a[r] (1 ≤ l ≤ r ≤ n). Thus, a block is defined by a pair of indices (l, r).
Find a set of blocks (l_1, r_1), (l_2, r_2), ..., (l_k, r_k) such that:
* They do not intersect (i.e. they are disjoint). Formally, for each pair of blocks (l_i, r_i) and (l_j, r_j) where i ≠ j either r_i < l_j or r_j < l_i.
* For each block the sum of its elements is the same. Formally, $$$a[l_1]+a[l_1+1]+...+a[r_1]=a[l_2]+a[l_2+1]+...+a[r_2]= ... = a[l_k]+a[l_k+1]+...+a[r_k].$$$
* The number of the blocks in the set is maximum. Formally, there does not exist a set of blocks (l_1', r_1'), (l_2', r_2'), ..., (l_{k'}', r_{k'}') satisfying the above two requirements with k' > k.
<image> The picture corresponds to the first example. Blue boxes illustrate blocks.
Write a program to find such a set of blocks.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the length of the given array. The second line contains the sequence of elements a[1], a[2], ..., a[n] (-10^5 ≤ a_i ≤ 10^5).
Output
In the first line print the integer k (1 ≤ k ≤ n). The following k lines should contain blocks, one per line. In each line print a pair of indices l_i, r_i (1 ≤ l_i ≤ r_i ≤ n) — the bounds of the i-th block. You can print blocks in any order. If there are multiple answers, print any of them.
Examples
Input
7
4 1 2 2 1 5 3
Output
3
7 7
2 3
4 5
Input
11
-5 -4 -3 -2 -1 0 1 2 3 4 5
Output
2
3 4
1 1
Input
4
1 1 1 1
Output
4
4 4
1 1
2 2
3 3
Submitted Solution:
```
import time
def index(key, item, index):
if key in index:
index[key].append(item)
else:
index[key] = [item]
def is_in_order(l):
if not l:
return True
for i in range(1, len(l)):
old, new = l[i - 1], l[i]
if new[0] < old[0]:
return False
elif new[0] == old[0]:
if new[1] < old[1]:
return False
return True
def schedule2(times):
# print(times)
assert is_in_order(times)
result = []
a_min = 0
for s, e in times:
if s >= a_min:
result.append((s, e))
a_min = e
return result
def schedule(times):
index_by_b = {}
for time in times:
index(time[1], time, index_by_b)
b_keys = sorted(list(index_by_b.keys()))
result = []
a_min = 0
# Get interval with minimun end time whose start time >= a_min.
for end_time in b_keys:
start = index_by_b[end_time].pop()[0]
if start >= a_min:
result.append((start, end_time))
a_min = end_time
return result
def solve(n, a_l):
index_by_sum = {}
for i in range(n):
sum_ = 0
for j in range(i + 1, n + 1):
sum_ += a_l[j - 1]
if sum_ in index_by_sum:
index_by_sum[sum_].append((i, j))
else:
index_by_sum[sum_] = [(i, j)]
result = []
for sum_, times in index_by_sum.items():
sub_result = schedule2(times)
if len(sub_result) > len(result):
result = sub_result
return result
def test():
n = 7
a_l = [4, 1, 2, 2, 1, 5, 3]
# n = 1500
# a_l = list(range(1, n + 1))
tick = time.time()
result = solve(n, a_l)
tock = time.time()
print(len(result))
for a, b in result:
print(a + 1, b)
print("T:", round(tock - tick, 5))
def main():
n = int(input())
a_l = list(map(int, input().split()))
result = solve(n, a_l)
print(len(result))
for a, b in result:
print(a + 1, b)
if __name__ == "__main__":
main()
```
No
| 99,548 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem is given in two editions, which differ exclusively in the constraints on the number n.
You are given an array of integers a[1], a[2], ..., a[n]. A block is a sequence of contiguous (consecutive) elements a[l], a[l+1], ..., a[r] (1 ≤ l ≤ r ≤ n). Thus, a block is defined by a pair of indices (l, r).
Find a set of blocks (l_1, r_1), (l_2, r_2), ..., (l_k, r_k) such that:
* They do not intersect (i.e. they are disjoint). Formally, for each pair of blocks (l_i, r_i) and (l_j, r_j) where i ≠ j either r_i < l_j or r_j < l_i.
* For each block the sum of its elements is the same. Formally, $$$a[l_1]+a[l_1+1]+...+a[r_1]=a[l_2]+a[l_2+1]+...+a[r_2]= ... = a[l_k]+a[l_k+1]+...+a[r_k].$$$
* The number of the blocks in the set is maximum. Formally, there does not exist a set of blocks (l_1', r_1'), (l_2', r_2'), ..., (l_{k'}', r_{k'}') satisfying the above two requirements with k' > k.
<image> The picture corresponds to the first example. Blue boxes illustrate blocks.
Write a program to find such a set of blocks.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the length of the given array. The second line contains the sequence of elements a[1], a[2], ..., a[n] (-10^5 ≤ a_i ≤ 10^5).
Output
In the first line print the integer k (1 ≤ k ≤ n). The following k lines should contain blocks, one per line. In each line print a pair of indices l_i, r_i (1 ≤ l_i ≤ r_i ≤ n) — the bounds of the i-th block. You can print blocks in any order. If there are multiple answers, print any of them.
Examples
Input
7
4 1 2 2 1 5 3
Output
3
7 7
2 3
4 5
Input
11
-5 -4 -3 -2 -1 0 1 2 3 4 5
Output
2
3 4
1 1
Input
4
1 1 1 1
Output
4
4 4
1 1
2 2
3 3
Submitted Solution:
```
n=int(input())
arr=list(map(int,input().split()))
dict1={}
for i in range(n):
val=0
for j in range(i,n):
val+=arr[j]
try:
dict1[val].append((i,j))
except:
KeyError
dict1[val]=[(i,j)]
ans=0
ansarr=[]
for i in dict1.keys():
if(len(dict1[i])>ans):
arr2=[]
for j in range(len(dict1[i])):
arr2.append((dict1[i][j][1]-dict1[i][j][0],j))
arr2.sort()
temp=[]
for j in range(len(arr2)):
flag=0
indexi=dict1[i][arr2[j][1]][0]
indexj=dict1[i][arr2[j][1]][1]
for k in range(len(temp)):
if(temp[k][0]<=indexi<=temp[k][1] or temp[k][0]<=indexj<=temp[k][1] or indexi<=temp[k][0]<=indexj or indexi<=temp[k][1]<=indexj):
flag=1
if(flag==1):
break
else:
temp.append((indexi,indexj))
if(len(temp)>ans):
ans=len(temp)
ansarr=temp
print(ans)
finalans=[]
for i in range(len(ansarr)):
finalans.append(ansarr[i][0]+1)
finalans.append(ansarr[i][1]+1)
print(*finalans)
```
No
| 99,549 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem is given in two editions, which differ exclusively in the constraints on the number n.
You are given an array of integers a[1], a[2], ..., a[n]. A block is a sequence of contiguous (consecutive) elements a[l], a[l+1], ..., a[r] (1 ≤ l ≤ r ≤ n). Thus, a block is defined by a pair of indices (l, r).
Find a set of blocks (l_1, r_1), (l_2, r_2), ..., (l_k, r_k) such that:
* They do not intersect (i.e. they are disjoint). Formally, for each pair of blocks (l_i, r_i) and (l_j, r_j) where i ≠ j either r_i < l_j or r_j < l_i.
* For each block the sum of its elements is the same. Formally, $$$a[l_1]+a[l_1+1]+...+a[r_1]=a[l_2]+a[l_2+1]+...+a[r_2]= ... = a[l_k]+a[l_k+1]+...+a[r_k].$$$
* The number of the blocks in the set is maximum. Formally, there does not exist a set of blocks (l_1', r_1'), (l_2', r_2'), ..., (l_{k'}', r_{k'}') satisfying the above two requirements with k' > k.
<image> The picture corresponds to the first example. Blue boxes illustrate blocks.
Write a program to find such a set of blocks.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the length of the given array. The second line contains the sequence of elements a[1], a[2], ..., a[n] (-10^5 ≤ a_i ≤ 10^5).
Output
In the first line print the integer k (1 ≤ k ≤ n). The following k lines should contain blocks, one per line. In each line print a pair of indices l_i, r_i (1 ≤ l_i ≤ r_i ≤ n) — the bounds of the i-th block. You can print blocks in any order. If there are multiple answers, print any of them.
Examples
Input
7
4 1 2 2 1 5 3
Output
3
7 7
2 3
4 5
Input
11
-5 -4 -3 -2 -1 0 1 2 3 4 5
Output
2
3 4
1 1
Input
4
1 1 1 1
Output
4
4 4
1 1
2 2
3 3
Submitted Solution:
```
from collections import defaultdict
n = int(input())
nums = [int(x) for x in input().split()]
mapas = [defaultdict(list) for _ in range(n)]
for i, nu in enumerate(nums):
if i:
mapas[i] = defaultdict(list, mapas[i-1])
# print(mapas[i])
mapas[i][nu].append((i+1,i+1))
su = nu
for j in reversed(range(1, i)):
su += nums[j]
if len(mapas[j-1][su])+1 > len(mapas[i][su]):
mapas[i][su] = mapas[j-1][su]+ [(j+1, i+1)]
# print(su)
if i:
su += nums[0]
# print(su, mapas[i])
if len(mapas[i][su]) == 0:
mapas[i][su] = [(1, i+1)]
# print(mapas[i])
# print()
# mapas[i][su] = max(1, mapas[i][su])
mx, my = -1,-1
# print(mapas[-1])
for x, y in mapas[-1].items():
if len(y)> my:
mx, my = x, len(y)
print(my)
# print(mapas[-1][mx])
for a, b in mapas[-1][mx]:
print(a, b)
```
No
| 99,550 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let f_{x} = c^{2x-6} ⋅ f_{x-1} ⋅ f_{x-2} ⋅ f_{x-3} for x ≥ 4.
You have given integers n, f_{1}, f_{2}, f_{3}, and c. Find f_{n} mod (10^{9}+7).
Input
The only line contains five integers n, f_{1}, f_{2}, f_{3}, and c (4 ≤ n ≤ 10^{18}, 1 ≤ f_{1}, f_{2}, f_{3}, c ≤ 10^{9}).
Output
Print f_{n} mod (10^{9} + 7).
Examples
Input
5 1 2 5 3
Output
72900
Input
17 97 41 37 11
Output
317451037
Note
In the first example, f_{4} = 90, f_{5} = 72900.
In the second example, f_{17} ≈ 2.28 × 10^{29587}.
Tags: dp, math, matrices, number theory
Correct Solution:
```
def mat_dot(A, B, mod):
assert len(A[0]) == len(B), 'invalid_size'
L = len(A)
M = len(A[0])
N = len(B[0])
res = [[0]*N for _ in range(L)]
for i in range(L):
for j in range(N):
a = 0
for k in range(M):
a = (a+A[i][k]*B[k][j]) % mod
res[i][j] = a
return res
def mat_pow(A, x, mod):
N = len(A)
res = [[0]*N for _ in range(N)]
for i in range(N):
res[i][i] = 1
for i in range(x.bit_length()):
if 2**i & x:
res = mat_dot(res, A, mod)
A = mat_dot(A, A, mod)
return res
#if there exists K such that X**K %mod == Y % mod, return K % (mod-1)
#Otherwise, return -1
def bsgs(X, Y, mod):
Y %= mod
X %= mod
rm = int(mod**(0.5)) + 2
R = pow(pow(X, rm ,mod), mod-2, mod)
D = {Y: 0}
p = Y
for a in range(1, rm):
p = p*R % mod
D[p] = a
p = 1
b = 0
if p in D:
return (D[p]*rm + b)%(mod-1)
for b in range(1, rm):
p = (p*X) % mod
if p in D:
return (D[p]*rm + b)%(mod-1)
return -1
n, f1, f2, f3, c = map(int, input().split())
mod = 10**9+7
a1 = bsgs(5, f1, mod)
a2 = bsgs(5, f2, mod)
a3 = bsgs(5, f3, mod)
d = bsgs(5, c, mod)
A = [[1, 1, 1, 2*d, -6*d],
[1, 0, 0, 0, 0],
[0, 1, 0, 0, 0],
[0, 0, 0, 1, 1],
[0, 0, 0, 0, 1]]
B = mat_pow(A, n-3, mod - 1)
Ans = mat_dot(B, [[a3], [a2], [a1], [4], [1]], mod - 1)
print(pow(5, Ans[0][0], mod))
```
| 99,551 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let f_{x} = c^{2x-6} ⋅ f_{x-1} ⋅ f_{x-2} ⋅ f_{x-3} for x ≥ 4.
You have given integers n, f_{1}, f_{2}, f_{3}, and c. Find f_{n} mod (10^{9}+7).
Input
The only line contains five integers n, f_{1}, f_{2}, f_{3}, and c (4 ≤ n ≤ 10^{18}, 1 ≤ f_{1}, f_{2}, f_{3}, c ≤ 10^{9}).
Output
Print f_{n} mod (10^{9} + 7).
Examples
Input
5 1 2 5 3
Output
72900
Input
17 97 41 37 11
Output
317451037
Note
In the first example, f_{4} = 90, f_{5} = 72900.
In the second example, f_{17} ≈ 2.28 × 10^{29587}.
Tags: dp, math, matrices, number theory
Correct Solution:
```
def mat_mult(A, B, MOD):
n, m, p = len(A), len(A[0]), len(B[0])
assert (len(B) == m)
C = [[0] * p for _ in range(n)]
for i in range(n):
for k in range(m):
Aik = A[i][k]
for j in range(p):
C[i][j] = (C[i][j] + Aik * B[k][j]) % MOD
return C
def ksm(A, n, MOD):
if (n == 0):
E = [[0 for i in range(len(A))] for j in range(len(A))]
for i in range(len(A)):
E[i][i] = 1
return E
if (n == 1):
return A
k = ksm(A, n//2, MOD)
z = mat_mult(k, k, MOD)
if (n&1):
return (mat_mult(z, A, MOD))
else:
return z
def Fenjie(n):
k = {}
if (n==1):
return {}
a = 2
while (n>=2):
b = n%a
if (a*a > n):
k[n] = 1
break
if (b==0):
if (a in k):
k[a] += 1
else:
k[a] = 1
n = n//a
else:
a += 1
return k
def Euler(n):
if (n==1):
return 1
k = Fenjie(n)
m = n
for i in k:
m = m // i * (i-1)
return m
MOD = 10**9 + 7
n, b, c, d, e = list(map(int, input().split()))
l1 = [[0],[0],[1]]
l2 = [[0],[1],[0]]
l3 = [[1],[0],[0]]
l4 = [[6],[2],[0],[0],[0]]
a1 = [[1,1,1],[1,0,0],[0,1,0]]
a2 = [[3,-2,0,-1,1],[1,0,0,0,0],[0,1,0,0,0],[0,0,1,0,0],[0,0,0,1,0]]
if (n == 4):
print(e*e*b*c*d % MOD)
else:
a1 = ksm(a1, n-3, MOD-1)
a2 = ksm(a2, n-5, MOD-1)
b1 = mat_mult(a1, l1, MOD-1)
p1 = pow(b, b1[0][0], MOD)
c1 = mat_mult(a1, l2, MOD-1)
p2 = pow(c, c1[0][0], MOD)
d1 = mat_mult(a1, l3, MOD-1)
p3 = pow(d, d1[0][0], MOD)
n1 = mat_mult(a2, l4, MOD-1)[0][0]
p = pow(e, n1, MOD)
p = p1*p%MOD
p = p2*p%MOD
p = p3*p%MOD
print(p)
```
| 99,552 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let f_{x} = c^{2x-6} ⋅ f_{x-1} ⋅ f_{x-2} ⋅ f_{x-3} for x ≥ 4.
You have given integers n, f_{1}, f_{2}, f_{3}, and c. Find f_{n} mod (10^{9}+7).
Input
The only line contains five integers n, f_{1}, f_{2}, f_{3}, and c (4 ≤ n ≤ 10^{18}, 1 ≤ f_{1}, f_{2}, f_{3}, c ≤ 10^{9}).
Output
Print f_{n} mod (10^{9} + 7).
Examples
Input
5 1 2 5 3
Output
72900
Input
17 97 41 37 11
Output
317451037
Note
In the first example, f_{4} = 90, f_{5} = 72900.
In the second example, f_{17} ≈ 2.28 × 10^{29587}.
Tags: dp, math, matrices, number theory
Correct Solution:
```
def baby_step_giant_step(g, y, p):
"""y = g^x (mod p)を満たすxを求める"""
m = int(p**0.5) + 1
# Baby-step
baby = {}
b = 1
for i in range(m):
baby[b] = i
b = (b * g) % p
# Giant-step
gm = pow(b, p-2, p)
giant = y
for i in range(m):
if giant in baby:
x = i*m + baby[giant]
return x
giant = (giant * gm) % p
return -1
def _mul(A, B, MOD):
C = [[0] * len(B[0]) for i in range(len(A))]
for i in range(len(A)):
for k in range(len(B)):
for j in range(len(B[0])):
C[i][j] = (C[i][j] + A[i][k]*B[k][j]) % MOD
return C
def pow_matrix(A, n, MOD):
"""A**nをダブリングによって求める"""
B = [[0] * len(A) for i in range(len(A))]
for i in range(len(A)):
B[i][i] = 1
while n > 0:
if n & 1:
B = _mul(A, B, MOD)
A = _mul(A, A, MOD)
n = n // 2
return B
n, f1, f2, f3, c = map(int, input().split())
MOD = 10**9 + 7
log_f1 = baby_step_giant_step(5, f1, MOD)
log_f2 = baby_step_giant_step(5, f2, MOD)
log_f3 = baby_step_giant_step(5, f3, MOD)
log_c = baby_step_giant_step(5, c, MOD)
matrix = [[0]*5 for i in range(5)]
matrix[0][0] = 1
matrix[0][1] = 1
matrix[0][2] = 1
matrix[0][3] = 2 * log_c
matrix[0][4] = -6 * log_c
matrix[1][0] = 1
matrix[2][1] = 1
matrix[3][3] = 1
matrix[3][4] = 1
matrix[4][4] = 1
matrix_n = pow_matrix(matrix, n - 3, MOD - 1)
ans = log_f3 * matrix_n[0][0] + log_f2 * matrix_n[0][1] + log_f1 * matrix_n[0][2] \
+ 4 * matrix_n[0][3] + matrix_n[0][4]
ans = ans % (MOD-1)
print(pow(5, ans, MOD))
```
| 99,553 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let f_{x} = c^{2x-6} ⋅ f_{x-1} ⋅ f_{x-2} ⋅ f_{x-3} for x ≥ 4.
You have given integers n, f_{1}, f_{2}, f_{3}, and c. Find f_{n} mod (10^{9}+7).
Input
The only line contains five integers n, f_{1}, f_{2}, f_{3}, and c (4 ≤ n ≤ 10^{18}, 1 ≤ f_{1}, f_{2}, f_{3}, c ≤ 10^{9}).
Output
Print f_{n} mod (10^{9} + 7).
Examples
Input
5 1 2 5 3
Output
72900
Input
17 97 41 37 11
Output
317451037
Note
In the first example, f_{4} = 90, f_{5} = 72900.
In the second example, f_{17} ≈ 2.28 × 10^{29587}.
Tags: dp, math, matrices, number theory
Correct Solution:
```
n, f1, f2, f3, c = list(map(int,input().split()))
mat = [[1,1,1],[1,0,0],[0,1,0]]
final = [[1,0,0],[0,1,0],[0,0,1]]
nn = n - 3
N = 10**9 + 6
def prod(a, b):
m = [[0,0,0],[0,0,0],[0,0,0]]
for i in range(3):
for j in range(3):
m[i][j] = (a[i][0]*b[0][j] + a[i][1]*b[1][j]+a[i][2]*b[2][j]) % N
return m
while nn > 0:
if nn % 2 == 1:
final = prod(final, mat)
mat = prod(mat,mat)
nn //= 2
q = (final[0][0] * 3 + final[0][1] * 2 + final[0][2] * 1) % N
p = q - (n%N) + N
# p to potega c
ef3 = (final[0][0] * 1) % N
ef2 = (final[0][1] * 1) % N
ef1 = (final[0][2] * 1) % N
# print f1^ef1 *f2^ef2*f3^ef3 * c^p
def pot(a,w):
wyn = 1
while w > 0:
if w%2 == 1:
wyn = (wyn * a) % (N+1)
a = (a * a) % (N+1)
w //= 2
return wyn
l1 = pot(f1, ef1)
l2 = pot(f2, ef2)
l3 = pot(f3, ef3)
l4 = pot(c, p)
c = (l1*l2*l3*l4)%(N+1)
print(c)
```
| 99,554 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let f_{x} = c^{2x-6} ⋅ f_{x-1} ⋅ f_{x-2} ⋅ f_{x-3} for x ≥ 4.
You have given integers n, f_{1}, f_{2}, f_{3}, and c. Find f_{n} mod (10^{9}+7).
Input
The only line contains five integers n, f_{1}, f_{2}, f_{3}, and c (4 ≤ n ≤ 10^{18}, 1 ≤ f_{1}, f_{2}, f_{3}, c ≤ 10^{9}).
Output
Print f_{n} mod (10^{9} + 7).
Examples
Input
5 1 2 5 3
Output
72900
Input
17 97 41 37 11
Output
317451037
Note
In the first example, f_{4} = 90, f_{5} = 72900.
In the second example, f_{17} ≈ 2.28 × 10^{29587}.
Tags: dp, math, matrices, number theory
Correct Solution:
```
MOD = int(1e9 + 7)
def matMOD(A):
f = lambda x: x%(MOD-1)
ret = [ list(map(f, i)) for i in A ]
return ret
def matmul(A, B):
a, b = len(A), len(A[0])
c = len(B[0])
ret = [ [0] * c for i in range(a) ]
for i in range(a):
for j in range(c):
for k in range(b):
ret[i][j] += A[i][k] * B[k][j]
return ret
def matmul_log(A, m):
if m == 1:
return A
B = matmul_log(A, m//2)
if m % 2 == 0:
return matMOD(matmul(B, B))
else:
return matMOD(matmul(A, matmul(B, B)))
n, f1, f2, f3, c = map(int, input().split())
ini = [
[1, 0, 0, 1],
[0, 1, 0, 1],
[0, 0, 1, 1],
[4, 2, 0, 0, 0, 2]
]
m = [
[0, 0, 0, 0],
[1, 0, 0, 1],
[0, 1, 0, 1],
[0, 0, 1, 1]
]
m2 = [
[1, 0, 0, 0, 0, 1],
[1, 1, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0],
[0, 0, 1, 0, 0, 1],
[0, 0, 0, 1, 0, 1],
[0, 0, 0, 0, 1, 1]
]
M = matmul_log(m, n-1)
M2 = matmul_log(m2, n-1)
x, y, z = matmul([ini[0]], M)[0][0], matmul([ini[1]], M)[0][0], matmul([ini[2]], M)[0][0]
w = matmul([ini[3]], M2)[0][2]
#print(x, y, z)
#print(ini[3])
#print(w)
ans = (pow(c, w, MOD) * pow(f1, x, MOD) * pow(f2, y, MOD) * pow(f3, z, MOD)) % MOD
print(ans)
```
| 99,555 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let f_{x} = c^{2x-6} ⋅ f_{x-1} ⋅ f_{x-2} ⋅ f_{x-3} for x ≥ 4.
You have given integers n, f_{1}, f_{2}, f_{3}, and c. Find f_{n} mod (10^{9}+7).
Input
The only line contains five integers n, f_{1}, f_{2}, f_{3}, and c (4 ≤ n ≤ 10^{18}, 1 ≤ f_{1}, f_{2}, f_{3}, c ≤ 10^{9}).
Output
Print f_{n} mod (10^{9} + 7).
Examples
Input
5 1 2 5 3
Output
72900
Input
17 97 41 37 11
Output
317451037
Note
In the first example, f_{4} = 90, f_{5} = 72900.
In the second example, f_{17} ≈ 2.28 × 10^{29587}.
Tags: dp, math, matrices, number theory
Correct Solution:
```
import os
from io import BytesIO, StringIO
#input = BytesIO(os.read(0, os.fstat(0).st_size)).readline
DEBUG = False
debug_print = print if DEBUG else lambda *x,**y: None
def input_as_list():
return list(map(int, input().split()))
def array_of(f, *dim):
return [array_of(f, *dim[1:]) for _ in range(dim[0])] if dim else f()
def main():
mod = 1000000007
def modmul(x, y):
return (x * y) % mod
def row(m, i, n=3):
return m[n * i:n * (i + 1)]
def col(m, i, n=3):
return m[i::n]
def vecmul(u, v):
s = 0
for i, j in zip(u, v):
s += (i * j) % (mod-1)
return s % (mod-1)
def matmul(a, b, n=3):
out = []
for i in range(n * n):
r, c = divmod(i, n)
out.append(vecmul(row(a, r), col(b, c)))
return out
def matadd(a, b, n=3):
out = []
for i in range(n * n):
out.append((a[i] + b[i]) % (mod-1))
return out
def matpow(m, p, n=3):
bs = str(bin(p)).lstrip('0b')
out = [0] * (n * n)
out[0::n + 1] = [1] * n
for b in reversed(bs):
if b == '1':
out = matmul(out, m)
m = matmul(m, m)
return out
def brute_force(n, f1, f2, f3, c):
i = 3
while i < n:
i += 1
f4 = pow(c, 2*i-6, mod) * f1 * f2 * f3 % mod
f1, f2, f3 = f2, f3, f4
return f3
def solve(n, f1, f2, f3, c):
f = [f1, f2, f3]
g = [modmul(c, f[0]), modmul(c * c, f[1]), modmul(c * c * c, f[2])]
mat = [1, 1, 1, 1, 0, 0, 0, 1, 0]
mat_n = matpow(mat, n - 3)
g_n = (pow(g[2], mat_n[0], mod)
* pow(g[1], mat_n[1], mod)
* pow(g[0], mat_n[2], mod)) % mod
c_n = pow(c, n, mod)
c_n_inv = pow(c_n, mod - 2, mod)
f_n = modmul(g_n, c_n_inv)
return f_n
def solve_from_stdin():
i = input_as_list()
print(solve(*i))
def test():
import random
sample = [random.randrange(4, 99) for _ in range(5)]
sample = [2, 2, 2, 2]
print(*sample)
for i in range(4,200):
print(i, *sample)
print(solve(i, *sample))
print(brute_force(i, *sample))
print()
solve_from_stdin()
main()
```
| 99,556 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let f_{x} = c^{2x-6} ⋅ f_{x-1} ⋅ f_{x-2} ⋅ f_{x-3} for x ≥ 4.
You have given integers n, f_{1}, f_{2}, f_{3}, and c. Find f_{n} mod (10^{9}+7).
Input
The only line contains five integers n, f_{1}, f_{2}, f_{3}, and c (4 ≤ n ≤ 10^{18}, 1 ≤ f_{1}, f_{2}, f_{3}, c ≤ 10^{9}).
Output
Print f_{n} mod (10^{9} + 7).
Examples
Input
5 1 2 5 3
Output
72900
Input
17 97 41 37 11
Output
317451037
Note
In the first example, f_{4} = 90, f_{5} = 72900.
In the second example, f_{17} ≈ 2.28 × 10^{29587}.
Tags: dp, math, matrices, number theory
Correct Solution:
```
fib_matrix = [[1,1,1],
[1,0,0],
[0,1,0]]
def matrix_square(A, mod):
return mat_mult(A,A,mod)
def mat_mult(A,B, mod):
if mod is not None:
return [[(A[0][0]*B[0][0] + A[0][1]*B[1][0]+A[0][2]*B[2][0])%mod, (A[0][0]*B[0][1] + A[0][1]*B[1][1]+A[0][2]*B[2][1])%mod,(A[0][0]*B[0][2] + A[0][1]*B[1][2]+A[0][2]*B[2][2])%mod],
[(A[1][0]*B[0][0] + A[1][1]*B[1][0]+A[1][2]*B[2][0])%mod, (A[1][0]*B[0][1] + A[1][1]*B[1][1]+A[1][2]*B[2][1])%mod,(A[1][0]*B[0][2] + A[1][1]*B[1][2]+A[1][2]*B[2][2])%mod],
[(A[2][0]*B[0][0] + A[2][1]*B[1][0]+A[2][2]*B[2][0])%mod, (A[2][0]*B[0][1] + A[2][1]*B[1][1]+A[2][2]*B[2][1])%mod,(A[2][0]*B[0][2] + A[2][1]*B[1][2]+A[2][2]*B[2][2])%mod]]
def matrix_pow(M, power, mod):
#Special definition for power=0:
if power <= 0:
return M
powers = list(reversed([True if i=="1" else False for i in bin(power)[2:]])) #Order is 1,2,4,8,16,...
matrices = [None for _ in powers]
matrices[0] = M
for i in range(1,len(powers)):
matrices[i] = matrix_square(matrices[i-1], mod)
result = None
for matrix, power in zip(matrices, powers):
if power:
if result is None:
result = matrix
else:
result = mat_mult(result, matrix, mod)
return result
fib_matrix2 = [[1,1,1,1,0],
[1,0,0,0,0],
[0,1,0,0,0],
[0,0,0,1,1],
[0,0,0,0,1]]
def matrix_square2(A, mod):
return mat_mult2(A,A,mod)
def mat_mult2(A,B, mod):
if mod is not None:
m=[]
for i in range(5):
m.append([])
for j in range(5):
sums=0
for k in range(5):
sums+=A[i][k]*B[k][j]
m[-1].append(sums%mod)
return m
def matrix_pow2(M, power, mod):
#Special definition for power=0:
if power <= 0:
return M
powers = list(reversed([True if i=="1" else False for i in bin(power)[2:]])) #Order is 1,2,4,8,16,...
matrices = [None for _ in powers]
matrices[0] = M
for i in range(1,len(powers)):
matrices[i] = matrix_square2(matrices[i-1], mod)
result = None
for matrix, power in zip(matrices, powers):
if power:
if result is None:
result = matrix
else:
result = mat_mult2(result, matrix, mod)
return result
n,f1,f2,f3,c=map(int,input().split())
f3pow=matrix_pow(fib_matrix, n-2, 1000000006)[0][2]
f1pow=matrix_pow(fib_matrix, n-2, 1000000006)[1][2]
f2pow=matrix_pow(fib_matrix, n-2, 1000000006)[1][1]
cpow=2*matrix_pow2(fib_matrix2, n, 1000000006)[2][4]
ans=pow(c,cpow,1000000007)
ans=(ans*pow(f1,f1pow,1000000007))%1000000007
ans=(ans*pow(f2,f2pow,1000000007))%1000000007
ans=(ans*pow(f3,f3pow,1000000007))%1000000007
print(ans)
```
| 99,557 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let f_{x} = c^{2x-6} ⋅ f_{x-1} ⋅ f_{x-2} ⋅ f_{x-3} for x ≥ 4.
You have given integers n, f_{1}, f_{2}, f_{3}, and c. Find f_{n} mod (10^{9}+7).
Input
The only line contains five integers n, f_{1}, f_{2}, f_{3}, and c (4 ≤ n ≤ 10^{18}, 1 ≤ f_{1}, f_{2}, f_{3}, c ≤ 10^{9}).
Output
Print f_{n} mod (10^{9} + 7).
Examples
Input
5 1 2 5 3
Output
72900
Input
17 97 41 37 11
Output
317451037
Note
In the first example, f_{4} = 90, f_{5} = 72900.
In the second example, f_{17} ≈ 2.28 × 10^{29587}.
Tags: dp, math, matrices, number theory
Correct Solution:
```
import math
md = 10**9+7
def multiply(M,N):
md2 = 10**9+6
R = [[0 for i in range(3)] for j in range(3)]
for i in range(0, 3):
for j in range(0, 3):
for k in range(0, 3):
R[i][j] += (M[i][k] * N[k][j])%md2
R[i][j] %= md2
return R
def power(mat, n):
res = [[1,0,0],[0,1,0],[0,0,1]]
while n:
if n&1: res = multiply(res, mat)
mat = multiply(mat, mat)
n//=2
return res
n, f1, f2, f3, c = map(int, input().split())
f1 = (f1*c)%md
f2 = (f2*c**2)%md
f3 = (f3*c**3)%md
#print(f1, f2, f3)
mat = [[1,1,1],[1,0,0],[0,1,0]]
res = power(mat, n-3)
#print(res)
pw1, pw2, pw3 = res[0][2], res[0][1], res[0][0]
f1 = pow(f1, pw1, md)
f2 = pow(f2, pw2, md)
f3 = pow(f3, pw3, md)
ans = ((f1 * f2)%md * f3)%md
c = pow(c, md-2, md)
ans *= pow(c, n%(md-1), md)
ans %= md
print(ans)
```
| 99,558 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let f_{x} = c^{2x-6} ⋅ f_{x-1} ⋅ f_{x-2} ⋅ f_{x-3} for x ≥ 4.
You have given integers n, f_{1}, f_{2}, f_{3}, and c. Find f_{n} mod (10^{9}+7).
Input
The only line contains five integers n, f_{1}, f_{2}, f_{3}, and c (4 ≤ n ≤ 10^{18}, 1 ≤ f_{1}, f_{2}, f_{3}, c ≤ 10^{9}).
Output
Print f_{n} mod (10^{9} + 7).
Examples
Input
5 1 2 5 3
Output
72900
Input
17 97 41 37 11
Output
317451037
Note
In the first example, f_{4} = 90, f_{5} = 72900.
In the second example, f_{17} ≈ 2.28 × 10^{29587}.
Submitted Solution:
```
#!/usr/bin/env python
class Mat:
def __init__(self, x):
if x == 1: # identity
self.m = [
[1, 0, 0],
[0, 1, 0],
[0, 0, 1],
]
else: # x should be 3x3
self.m = [[
x[i][j]
for j in range(3)] for i in range(3)]
def __mul__(self, other):
return Mat([[
(sum(self.m[i][k] * other.m[k][j] for k in range(3))) % (mod - 1)
for j in range(3)] for i in range(3)])
def __rmul__(self, other):
return Mat([[
(sum(other.m[i][k] * self.m[k][j] for k in range(3))) % (mod - 1)
for j in range(3)] for i in range(3)])
def __mod__(self, _mod):
return Mat([[
self.m[i][j] % _mod
for j in range(3)] for i in range(3)])
def mat_bin_mod_pow(_mat, _mod, _pow):
if _pow == 0:
return Mat(1)
return (mat_bin_mod_pow((_mat * _mat) % _mod, _mod, _pow >> 1) *
(_mat if _pow & 1 else Mat(1))) % _mod
def bin_mod_pow(_num, _mod, _pow):
if _pow == 0:
return 1
return (bin_mod_pow(_num**2 % _mod, _mod, _pow >> 1) *
(_num if _pow & 1 else 1)) % _mod
def bin_mod_inv(_num, _mod):
return bin_mod_pow(_num, _mod, _mod - 2) % _mod
if __name__ == "__main__":
mod = 10**9 + 7
n, f1, f2, f3, c = map(int, input().split())
f = [f1, f2, f3]
g = [(f[i] * c**(i + 1)) % mod for i in range(3)]
mat = Mat([
[0, 1, 0],
[0, 0, 1],
[1, 1, 1],
])
p = mat_bin_mod_pow(_mat=mat, _mod=mod-1, _pow=n-3).m[-1]
print((
bin_mod_pow(_num=g[0], _mod=mod, _pow=p[0]) *
bin_mod_pow(_num=g[1], _mod=mod, _pow=p[1]) *
bin_mod_pow(_num=g[2], _mod=mod, _pow=p[2]) *
bin_mod_pow(_num=bin_mod_inv(_num=c, _mod=mod), _mod=mod, _pow=n)
) % mod)
```
Yes
| 99,559 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let f_{x} = c^{2x-6} ⋅ f_{x-1} ⋅ f_{x-2} ⋅ f_{x-3} for x ≥ 4.
You have given integers n, f_{1}, f_{2}, f_{3}, and c. Find f_{n} mod (10^{9}+7).
Input
The only line contains five integers n, f_{1}, f_{2}, f_{3}, and c (4 ≤ n ≤ 10^{18}, 1 ≤ f_{1}, f_{2}, f_{3}, c ≤ 10^{9}).
Output
Print f_{n} mod (10^{9} + 7).
Examples
Input
5 1 2 5 3
Output
72900
Input
17 97 41 37 11
Output
317451037
Note
In the first example, f_{4} = 90, f_{5} = 72900.
In the second example, f_{17} ≈ 2.28 × 10^{29587}.
Submitted Solution:
```
def _mul(A, B, MOD):
C = [[0] * len(B[0]) for i in range(len(A))]
for i in range(len(A)):
for k in range(len(B)):
for j in range(len(B[0])):
C[i][j] = (C[i][j] + A[i][k]*B[k][j]) % MOD
return C
def pow_matrix(A, n, MOD):
"""A**nをダブリングによって求める"""
B = [[0] * len(A) for i in range(len(A))]
for i in range(len(A)):
B[i][i] = 1
while n > 0:
if n & 1:
B = _mul(A, B, MOD)
A = _mul(A, A, MOD)
n = n // 2
return B
n, f1, f2, f3, c = map(int, input().split())
MOD = 10**9 + 7
ans = 1
matrix = [[0]*3 for i in range(3)]
matrix[0][0] = matrix[0][1] = matrix[0][2] =1
matrix[1][0] = 1
matrix[2][1] = 1
f_matrix = pow_matrix(matrix, n - 3, MOD - 1)
ans *= pow(f3, f_matrix[0][0], MOD)
ans *= pow(f2, f_matrix[0][1], MOD)
ans *= pow(f1, f_matrix[0][2], MOD)
matrix = [[0]*5 for i in range(5)]
matrix[0][0] = matrix[0][1] = matrix[0][2] =1
matrix[0][3] = 2
matrix[0][4] = -6
matrix[1][0] = matrix[2][1] = 1
matrix[3][3] = matrix[3][4] = 1
matrix[4][4] = 1
c_matrix = pow_matrix(matrix, n - 3, MOD - 1)
ans *= pow(c, c_matrix[0][3] * 4 + c_matrix[0][4], MOD)
print(ans % MOD)
```
Yes
| 99,560 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let f_{x} = c^{2x-6} ⋅ f_{x-1} ⋅ f_{x-2} ⋅ f_{x-3} for x ≥ 4.
You have given integers n, f_{1}, f_{2}, f_{3}, and c. Find f_{n} mod (10^{9}+7).
Input
The only line contains five integers n, f_{1}, f_{2}, f_{3}, and c (4 ≤ n ≤ 10^{18}, 1 ≤ f_{1}, f_{2}, f_{3}, c ≤ 10^{9}).
Output
Print f_{n} mod (10^{9} + 7).
Examples
Input
5 1 2 5 3
Output
72900
Input
17 97 41 37 11
Output
317451037
Note
In the first example, f_{4} = 90, f_{5} = 72900.
In the second example, f_{17} ≈ 2.28 × 10^{29587}.
Submitted Solution:
```
MD = 10**9+7
def mult(A,B):
n = len(A)
C = [[0] * n for _ in range(n)]
for i in range(len(A)):
for j in range(len(A)):
for k in range(len(A)):
C[i][j] += A[i][k]*B[k][j]
C[i][j] %= MD-1
return C
def ex(A,k):
n = len(A)
R = [[0] * n for _ in range(n)]
for i in range(n):
R[i][i] = 1
while k > 0:
if k & 1:
R = mult(R,A)
k //= 2
A = mult(A,A)
return R
cma = [
[1,1,1,1,0],
[1,0,0,0,0],
[0,1,0,0,0],
[0,0,0,1,2],
[0,0,0,0,1]
]
tri = [
[1,1,1],
[1,0,0],
[0,1,0]
]
n,f1,f2,f3,c = map(int,input().split())
ce = ex(cma,n-3)
cp = ce[0][3]*2+ce[0][4]
f = ex(tri,n-3)
fp1 = f[0][2]
fp2 = f[0][1]
fp3 = f[0][0]
r = pow(f1,fp1,MD)*pow(f2,fp2,MD)*pow(f3,fp3,MD)*pow(c,cp,MD)
#print(fp1,fp2,fp3,cp)
print(r%MD)
```
Yes
| 99,561 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let f_{x} = c^{2x-6} ⋅ f_{x-1} ⋅ f_{x-2} ⋅ f_{x-3} for x ≥ 4.
You have given integers n, f_{1}, f_{2}, f_{3}, and c. Find f_{n} mod (10^{9}+7).
Input
The only line contains five integers n, f_{1}, f_{2}, f_{3}, and c (4 ≤ n ≤ 10^{18}, 1 ≤ f_{1}, f_{2}, f_{3}, c ≤ 10^{9}).
Output
Print f_{n} mod (10^{9} + 7).
Examples
Input
5 1 2 5 3
Output
72900
Input
17 97 41 37 11
Output
317451037
Note
In the first example, f_{4} = 90, f_{5} = 72900.
In the second example, f_{17} ≈ 2.28 × 10^{29587}.
Submitted Solution:
```
# @author
import sys
class EProductOrientedRecurrence:
def solve(self):
MOD = 10 ** 9 + 7
D = 3
I = [
[1 if i == j else 0 for j in range(D)]
for i in range(D)
]
def mat_mult(A, B):
n, m, p = len(A), len(A[0]), len(B[0])
assert (len(B) == m)
C = [[0] * p for _ in range(n)]
for i in range(n):
for k in range(m):
Aik = A[i][k]
for j in range(p):
C[i][j] = (C[i][j] + Aik * B[k][j]) % (MOD - 1)
return C
def mat_pow(A, p):
if p == 0:
return I
if p == 1:
return A
if p % 2 == 0:
return mat_pow(mat_mult(A, A), p // 2)
else:
return mat_mult(A, mat_pow(A, p - 1))
n, f1, f2, f3, c = [int(_) for _ in input().split()]
M = [[0, 1, 0],
[0, 0, 1],
[1, 1, 1]]
Rf1 = mat_mult(mat_pow(M, n - 1), [[1], [0], [0]])[0][0]
Rf2 = mat_mult(mat_pow(M, n - 1), [[0], [1], [0]])[0][0]
Rf3 = mat_mult(mat_pow(M, n - 1), [[0], [0], [1]])[0][0]
Rpower = (mat_mult(mat_pow(M, n - 1), [[1], [2], [3]])[0][0] - n) % (MOD - 1)
ans = pow(f1, Rf1, MOD) * pow(f2, Rf2, MOD) * pow(f3, Rf3, MOD) * pow(c, Rpower, MOD)
ans %= MOD
print(ans)
solver = EProductOrientedRecurrence()
input = sys.stdin.readline
solver.solve()
```
Yes
| 99,562 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let f_{x} = c^{2x-6} ⋅ f_{x-1} ⋅ f_{x-2} ⋅ f_{x-3} for x ≥ 4.
You have given integers n, f_{1}, f_{2}, f_{3}, and c. Find f_{n} mod (10^{9}+7).
Input
The only line contains five integers n, f_{1}, f_{2}, f_{3}, and c (4 ≤ n ≤ 10^{18}, 1 ≤ f_{1}, f_{2}, f_{3}, c ≤ 10^{9}).
Output
Print f_{n} mod (10^{9} + 7).
Examples
Input
5 1 2 5 3
Output
72900
Input
17 97 41 37 11
Output
317451037
Note
In the first example, f_{4} = 90, f_{5} = 72900.
In the second example, f_{17} ≈ 2.28 × 10^{29587}.
Submitted Solution:
```
import math
md = 10**9+7
def multiply(M,N):
md = 10**9+7
R = [[0 for i in range(3)] for j in range(3)]
for i in range(0, 3):
for j in range(0, 3):
for k in range(0, 3):
R[i][j] += (M[i][k] * N[k][j])%md
R[i][j] %= md
return R
def power(mat, n):
res = [[1,0,0],[0,1,0],[0,0,1]]
while n:
if n&1: res = multiply(res, mat)
mat = multiply(mat, mat)
n//=2
return res
n, f1, f2, f3, c = map(int, input().split())
f1 = (f1*c)%md
f2 = (f2*c**2)%md
f3 = (f3*c**3)%md
mat = [[1,1,1],[1,0,0],[0,1,0]]
res = power(mat, n-3)
pw1, pw2, pw3 = res[0][2], res[0][1], res[0][0]
f1 = pow(f1, pw1, md)
f2 = pow(f2, pw2, md)
f3 = pow(f3, pw3, md)
ans = ((f1 * f2)%md * f3)%md
c = pow(c, md-2, md)
ans *= pow(c, n, md)
ans %= md
print(ans)
```
No
| 99,563 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let f_{x} = c^{2x-6} ⋅ f_{x-1} ⋅ f_{x-2} ⋅ f_{x-3} for x ≥ 4.
You have given integers n, f_{1}, f_{2}, f_{3}, and c. Find f_{n} mod (10^{9}+7).
Input
The only line contains five integers n, f_{1}, f_{2}, f_{3}, and c (4 ≤ n ≤ 10^{18}, 1 ≤ f_{1}, f_{2}, f_{3}, c ≤ 10^{9}).
Output
Print f_{n} mod (10^{9} + 7).
Examples
Input
5 1 2 5 3
Output
72900
Input
17 97 41 37 11
Output
317451037
Note
In the first example, f_{4} = 90, f_{5} = 72900.
In the second example, f_{17} ≈ 2.28 × 10^{29587}.
Submitted Solution:
```
import sys
# from bisect import bisect_right
# gcd
# from fractions import gcd
# from math import ceil, floor
# from copy import deepcopy
# from itertools import accumulate
# l = ['a', 'b', 'b', 'c', 'b', 'a', 'c', 'c', 'b', 'c', 'b', 'a']
# print(S.most_common(2)) # [('b', 5), ('c', 4)]
# print(S.keys()) # dict_keys(['a', 'b', 'c'])
# print(S.values()) # dict_values([3, 5, 4])
# print(S.items()) # dict_items([('a', 3), ('b', 5), ('c', 4)])
# from collections import Counter
# import math
# from functools import reduce
#
# fin = open('in_1.txt', 'r')
# sys.stdin = fin
input = sys.stdin.readline
def ii(): return int(input())
def mi(): return map(int, input().rstrip().split())
def lmi(): return list(map(int, input().rstrip().split()))
def li(): return list(input().rstrip())
# template
# def solve(L, A, B, M):
# ret, digits = 0, 0
# for d in range(18, 0, -1):
# lo = 10 ** (d - 1)
# hi = 10 ** d - 1
# if hi < A or A + B * (L - 1) < lo:
# continue
# imin = 0 if lo <= A else (lo - A + B - 1) // B
# init = A + B * imin
# if init > hi:
# continue
# imax = L - 1 if A + B * (L - 1) <= hi else imin + (hi - init) // B
# n = imax - imin + 1
# p = 10 ** d % M
# a = matpow([[0,1,0], [0,0,1], [1,1,1]], n, M)
# # a[0][3] = sum p^i for i in [0, n-1]
# # a[1][3] = sum i * p^i for i in [0, n-1]
# # sum (A + B * (imax - i)) * p^i for i in [0, n-1]
# sub = (A + B * imax) % M * a[1][3] % M + M - (B * a[0][3] % M)
# ret += sub % M * pow10(digits, M) % M
# digits += d * (imax - imin + 1)
# return ret % M
def pow10(p, mod):
if p % 2 == 1:
return 10 * pow10(p - 1, mod) % mod
elif p > 0:
sub = pow10(p // 2, mod)
return sub * sub % mod
else:
return 1
def matpow(a, p, mod):
if p % 2 == 1:
return matmul(a, matpow(a, p - 1, mod), mod)
elif p > 0:
b = matpow(a, p // 2, mod)
return matmul(b, b, mod)
else:
n = len(a)
return [[1 if i == j else 0 for j in range(n)] for i in range(n)]
def matmul(a, b, mod):
n = len(a)
ret = [[0 for j in range(n)] for i in range(n)]
for i in range(n):
for k in range(n):
for j in range(n):
ret[i][j] += a[i][k] * b[k][j]
for j in range(n):
ret[i][j] %= mod
return ret
if __name__ == '__main__':
# write code
mod = 10**9+7
n, a1, a2, a3, c = mi()
a = matpow([[0, 1, 0], [0, 0, 1], [1, 1, 1]], n - 2, mod)
# print(a)
ans = pow(a1, a[1][0], mod) * pow(a2, a[1][1], mod) * pow(a3, a[1][2], mod) % mod
# print(ans)
cmat = matpow([[0, 1, 0, 0, 0], [0, 0, 1, 0, 0], [1, 1, 1, 2, 0], [0, 0, 0, 1, 1], [0, 0, 0, 0, 1]], n - 3, mod)
# print(c)
ans *= pow(c, (cmat[2][3] + cmat[2][4]), mod)
ans %= mod
print(ans)
```
No
| 99,564 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let f_{x} = c^{2x-6} ⋅ f_{x-1} ⋅ f_{x-2} ⋅ f_{x-3} for x ≥ 4.
You have given integers n, f_{1}, f_{2}, f_{3}, and c. Find f_{n} mod (10^{9}+7).
Input
The only line contains five integers n, f_{1}, f_{2}, f_{3}, and c (4 ≤ n ≤ 10^{18}, 1 ≤ f_{1}, f_{2}, f_{3}, c ≤ 10^{9}).
Output
Print f_{n} mod (10^{9} + 7).
Examples
Input
5 1 2 5 3
Output
72900
Input
17 97 41 37 11
Output
317451037
Note
In the first example, f_{4} = 90, f_{5} = 72900.
In the second example, f_{17} ≈ 2.28 × 10^{29587}.
Submitted Solution:
```
from math import ceil
c33 = 33 ** (1 / 2)
q = (586 + 102 * c33) ** (2 / 3) + 4 - 2 * ((586 + 102 * c33) ** (1 / 3))
lol = ((586 + 102 * c33) ** (1 / 3))
M = 10 ** 9 + 7
def trib(n):
c33 = 33 ** 0.5
p = (int((((19 + 3*c33) ** (1 / 3) + (19 - 3 * c33) ** (1 / 3) + 1) / 3) ** n) % M) * lol
return ceil(3 * p / q) - (n == 5)
n, f1, f2, f3, c = map(int, input().split())
if n == 17:
print(317451037)
exit(0)
ans = 1
ans *= pow(c, (pow(2, (n - 2), M - 1) - 2), M)
a = trib(n - 4)
b = trib(n - 3)
d = trib(n - 2)
ans *= pow(f3, d, M)
ans *= pow(f2, a + b, M)
ans *= pow(f1, b, M)
print(ans % M)
```
No
| 99,565 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let f_{x} = c^{2x-6} ⋅ f_{x-1} ⋅ f_{x-2} ⋅ f_{x-3} for x ≥ 4.
You have given integers n, f_{1}, f_{2}, f_{3}, and c. Find f_{n} mod (10^{9}+7).
Input
The only line contains five integers n, f_{1}, f_{2}, f_{3}, and c (4 ≤ n ≤ 10^{18}, 1 ≤ f_{1}, f_{2}, f_{3}, c ≤ 10^{9}).
Output
Print f_{n} mod (10^{9} + 7).
Examples
Input
5 1 2 5 3
Output
72900
Input
17 97 41 37 11
Output
317451037
Note
In the first example, f_{4} = 90, f_{5} = 72900.
In the second example, f_{17} ≈ 2.28 × 10^{29587}.
Submitted Solution:
```
def mat_mul(a, b):
n, m, p = len(a), len(b), len(b[0])
res = [[0]*p for _ in range(n)]
for i in range(n):
for j in range(p):
for k in range(m):
res[i][j] += a[i][k]*b[k][j]
res[i][j] %= 1000000007
return res
def mat_pow(a, n):
if n == 1: return a
if n%2 == 1: return mat_mul(mat_pow(a, n-1), a)
t = mat_pow(a, n//2)
return mat_mul(t, t)
n, f1, f2, f3, c = map(int, input().split())
m1 = [[3, 1000000005, 0, 1000000006, 1], [1, 0, 0, 0, 0], [0, 1, 0, 0, 0], [0, 0, 1, 0, 0], [0, 0, 0, 1, 0]]
m2 = [[2], [0], [0], [0], [0]]
t1 = pow(c, mat_mul(mat_pow(m1, n), m2)[-1][0], 1000000007)
m1 = [[0, 0, 1], [1, 0, 1], [0, 1, 1]]
m2 = [[1], [0], [0]]
m3 = mat_mul(mat_pow(m1, n-1), m2)
t2 = pow(f1, m3[0][0], 1000000007)
t3 = pow(f2, m3[1][0], 1000000007)
t4 = pow(f3, m3[2][0], 1000000007)
print(t1*t2*t3*t4%1000000007)
```
No
| 99,566 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jack has become a soldier now. Unfortunately, he has trouble with the drill. Instead of marching beginning with the left foot and then changing legs with each step, as ordered, he keeps repeating a sequence of steps, in which he sometimes makes the wrong steps or — horror of horrors! — stops for a while. For example, if Jack uses the sequence 'right, left, break', when the sergeant yells: 'Left! Right! Left! Right! Left! Right!', Jack first makes a step with the right foot, then one with the left foot, then he is confused and stops for a moment, then again - this time according to the order - starts with the right foot, then uses the left foot, then - to the sergeant's irritation - he stops to catch his breath, to incorrectly start with the right foot again... Marching this way, Jack will make the step that he is supposed to in the given moment in only one third of cases.
When the officers convinced him he should do something about it, Jack decided to modify the basic sequence of steps that he repeats. However, in order not to get too tired, he has decided that the only thing he'll do is adding any number of breaks in any positions of the original sequence (a break corresponds to stopping for the duration of one step). Of course, Jack can't make a step on the same foot twice in a row, if there is no pause between these steps. It is, however, not impossible that the sequence of steps he used so far is incorrect (it would explain a lot, actually).
Help Private Jack! Given the sequence of steps he keeps repeating, calculate the maximal percentage of time that he can spend marching correctly after adding some breaks to his scheme.
Input
The first line of input contains a sequence consisting only of characters 'L', 'R' and 'X', where 'L' corresponds to a step with the left foot, 'R' — with the right foot, and 'X' — to a break. The length of the sequence will not exceed 106.
Output
Output the maximum percentage of time that Jack can spend marching correctly, rounded down to exactly six digits after the decimal point.
Examples
Input
X
Output
0.000000
Input
LXRR
Output
50.000000
Note
In the second example, if we add two breaks to receive LXXRXR, Jack will march: LXXRXRLXXRXRL... instead of LRLRLRLRLRLRL... and will make the correct step in half the cases. If we didn't add any breaks, the sequence would be incorrect — Jack can't step on his right foot twice in a row.
Tags: binary search, dp, greedy
Correct Solution:
```
def solve():
original = input()
temp = [original[0]]
for i in range(1, len(original)):
if original[i] == original[i - 1] != 'X':
temp.append('X')
temp.append(original[i])
augmented = ''.join(temp)
answer = 0
if augmented[0] == augmented[-1] != 'X':
answer = max(rate(augmented + 'X'), rate('X' + augmented))
else:
answer = rate(augmented)
print('%d.%06d' % (answer / 1000000, answer % 1000000))
def rate(seq):
correct, total, unknown, indicator = 0, 0, 0, 0
left_step = True
for action in seq:
if action == 'X':
total += 1
left_step = not left_step
else:
if left_step and action == 'L' or not left_step and action == 'R':
correct += 1
total += 1
indicator = 0
left_step = not left_step
else:
correct += 1
total += 2
unknown += indicator
indicator = 1 - indicator
if total % 2 == 1:
total += 1
unknown += indicator
if correct * 2 > total:
correct -= unknown
total -= unknown * 2
return correct * 100000000 // total
solve()
```
| 99,567 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jack has become a soldier now. Unfortunately, he has trouble with the drill. Instead of marching beginning with the left foot and then changing legs with each step, as ordered, he keeps repeating a sequence of steps, in which he sometimes makes the wrong steps or — horror of horrors! — stops for a while. For example, if Jack uses the sequence 'right, left, break', when the sergeant yells: 'Left! Right! Left! Right! Left! Right!', Jack first makes a step with the right foot, then one with the left foot, then he is confused and stops for a moment, then again - this time according to the order - starts with the right foot, then uses the left foot, then - to the sergeant's irritation - he stops to catch his breath, to incorrectly start with the right foot again... Marching this way, Jack will make the step that he is supposed to in the given moment in only one third of cases.
When the officers convinced him he should do something about it, Jack decided to modify the basic sequence of steps that he repeats. However, in order not to get too tired, he has decided that the only thing he'll do is adding any number of breaks in any positions of the original sequence (a break corresponds to stopping for the duration of one step). Of course, Jack can't make a step on the same foot twice in a row, if there is no pause between these steps. It is, however, not impossible that the sequence of steps he used so far is incorrect (it would explain a lot, actually).
Help Private Jack! Given the sequence of steps he keeps repeating, calculate the maximal percentage of time that he can spend marching correctly after adding some breaks to his scheme.
Input
The first line of input contains a sequence consisting only of characters 'L', 'R' and 'X', where 'L' corresponds to a step with the left foot, 'R' — with the right foot, and 'X' — to a break. The length of the sequence will not exceed 106.
Output
Output the maximum percentage of time that Jack can spend marching correctly, rounded down to exactly six digits after the decimal point.
Examples
Input
X
Output
0.000000
Input
LXRR
Output
50.000000
Note
In the second example, if we add two breaks to receive LXXRXR, Jack will march: LXXRXRLXXRXRL... instead of LRLRLRLRLRLRL... and will make the correct step in half the cases. If we didn't add any breaks, the sequence would be incorrect — Jack can't step on his right foot twice in a row.
Tags: binary search, dp, greedy
Correct Solution:
```
__author__ = 'Darren'
def solve():
original = input()
temp = [original[0]]
for i in range(1, len(original)):
if original[i] == original[i-1] != 'X':
temp.append('X')
temp.append(original[i])
augmented = ''.join(temp)
answer = 0
if augmented[0] == augmented[-1] != 'X':
answer = max(rate(augmented+'X'), rate('X'+augmented))
else:
answer = rate(augmented)
print('%d.%06d' % (answer / 1000000, answer % 1000000))
def rate(seq):
correct, total, unknown, indicator = 0, 0, 0, 0
left_step = True
for action in seq:
if action == 'X':
total += 1
left_step = not left_step
else:
if left_step and action == 'L' or not left_step and action == 'R':
correct += 1
total += 1
indicator = 0
left_step = not left_step
else:
correct += 1
total += 2
unknown += indicator
indicator = 1 - indicator
if total % 2 == 1:
total += 1
unknown += indicator
if correct * 2 > total:
correct -= unknown
total -= unknown * 2
return correct * 100000000 // total
if __name__ == '__main__':
solve()
```
| 99,568 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jack has become a soldier now. Unfortunately, he has trouble with the drill. Instead of marching beginning with the left foot and then changing legs with each step, as ordered, he keeps repeating a sequence of steps, in which he sometimes makes the wrong steps or — horror of horrors! — stops for a while. For example, if Jack uses the sequence 'right, left, break', when the sergeant yells: 'Left! Right! Left! Right! Left! Right!', Jack first makes a step with the right foot, then one with the left foot, then he is confused and stops for a moment, then again - this time according to the order - starts with the right foot, then uses the left foot, then - to the sergeant's irritation - he stops to catch his breath, to incorrectly start with the right foot again... Marching this way, Jack will make the step that he is supposed to in the given moment in only one third of cases.
When the officers convinced him he should do something about it, Jack decided to modify the basic sequence of steps that he repeats. However, in order not to get too tired, he has decided that the only thing he'll do is adding any number of breaks in any positions of the original sequence (a break corresponds to stopping for the duration of one step). Of course, Jack can't make a step on the same foot twice in a row, if there is no pause between these steps. It is, however, not impossible that the sequence of steps he used so far is incorrect (it would explain a lot, actually).
Help Private Jack! Given the sequence of steps he keeps repeating, calculate the maximal percentage of time that he can spend marching correctly after adding some breaks to his scheme.
Input
The first line of input contains a sequence consisting only of characters 'L', 'R' and 'X', where 'L' corresponds to a step with the left foot, 'R' — with the right foot, and 'X' — to a break. The length of the sequence will not exceed 106.
Output
Output the maximum percentage of time that Jack can spend marching correctly, rounded down to exactly six digits after the decimal point.
Examples
Input
X
Output
0.000000
Input
LXRR
Output
50.000000
Note
In the second example, if we add two breaks to receive LXXRXR, Jack will march: LXXRXRLXXRXRL... instead of LRLRLRLRLRLRL... and will make the correct step in half the cases. If we didn't add any breaks, the sequence would be incorrect — Jack can't step on his right foot twice in a row.
Submitted Solution:
```
l1=[x for x in input()]
l2=[]
c=0
if len(l1)==1:
if l1[0]=="X" or l1[0]=="R":
print("{:.6f}".format(0))
else:
print("{:.6f}".format(100))
else:
for a in range(int(len(l1)/2)):
l2.extend(["L","R"])
for a in range(len(l1)):
if l1[a]==l2[a]:
c+=1
print("{:.6f}".format((c/len(l1))*100))
```
No
| 99,569 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jack has become a soldier now. Unfortunately, he has trouble with the drill. Instead of marching beginning with the left foot and then changing legs with each step, as ordered, he keeps repeating a sequence of steps, in which he sometimes makes the wrong steps or — horror of horrors! — stops for a while. For example, if Jack uses the sequence 'right, left, break', when the sergeant yells: 'Left! Right! Left! Right! Left! Right!', Jack first makes a step with the right foot, then one with the left foot, then he is confused and stops for a moment, then again - this time according to the order - starts with the right foot, then uses the left foot, then - to the sergeant's irritation - he stops to catch his breath, to incorrectly start with the right foot again... Marching this way, Jack will make the step that he is supposed to in the given moment in only one third of cases.
When the officers convinced him he should do something about it, Jack decided to modify the basic sequence of steps that he repeats. However, in order not to get too tired, he has decided that the only thing he'll do is adding any number of breaks in any positions of the original sequence (a break corresponds to stopping for the duration of one step). Of course, Jack can't make a step on the same foot twice in a row, if there is no pause between these steps. It is, however, not impossible that the sequence of steps he used so far is incorrect (it would explain a lot, actually).
Help Private Jack! Given the sequence of steps he keeps repeating, calculate the maximal percentage of time that he can spend marching correctly after adding some breaks to his scheme.
Input
The first line of input contains a sequence consisting only of characters 'L', 'R' and 'X', where 'L' corresponds to a step with the left foot, 'R' — with the right foot, and 'X' — to a break. The length of the sequence will not exceed 106.
Output
Output the maximum percentage of time that Jack can spend marching correctly, rounded down to exactly six digits after the decimal point.
Examples
Input
X
Output
0.000000
Input
LXRR
Output
50.000000
Note
In the second example, if we add two breaks to receive LXXRXR, Jack will march: LXXRXRLXXRXRL... instead of LRLRLRLRLRLRL... and will make the correct step in half the cases. If we didn't add any breaks, the sequence would be incorrect — Jack can't step on his right foot twice in a row.
Submitted Solution:
```
d={'L':'R','R':'L'}
s=input()
cur='L'
long=0
steps=0
for i in s:
long+=1
if i!='X':
steps+=1
if i!=cur:
long+=1
cur=d[cur]
cur=d[cur]
if long%2:long+=1
t=(steps*10**8+long//2)//long
print(t//10**6,'.','0'*(6-len(str(t%10**6))),t%10**6,sep='')
```
No
| 99,570 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jack has become a soldier now. Unfortunately, he has trouble with the drill. Instead of marching beginning with the left foot and then changing legs with each step, as ordered, he keeps repeating a sequence of steps, in which he sometimes makes the wrong steps or — horror of horrors! — stops for a while. For example, if Jack uses the sequence 'right, left, break', when the sergeant yells: 'Left! Right! Left! Right! Left! Right!', Jack first makes a step with the right foot, then one with the left foot, then he is confused and stops for a moment, then again - this time according to the order - starts with the right foot, then uses the left foot, then - to the sergeant's irritation - he stops to catch his breath, to incorrectly start with the right foot again... Marching this way, Jack will make the step that he is supposed to in the given moment in only one third of cases.
When the officers convinced him he should do something about it, Jack decided to modify the basic sequence of steps that he repeats. However, in order not to get too tired, he has decided that the only thing he'll do is adding any number of breaks in any positions of the original sequence (a break corresponds to stopping for the duration of one step). Of course, Jack can't make a step on the same foot twice in a row, if there is no pause between these steps. It is, however, not impossible that the sequence of steps he used so far is incorrect (it would explain a lot, actually).
Help Private Jack! Given the sequence of steps he keeps repeating, calculate the maximal percentage of time that he can spend marching correctly after adding some breaks to his scheme.
Input
The first line of input contains a sequence consisting only of characters 'L', 'R' and 'X', where 'L' corresponds to a step with the left foot, 'R' — with the right foot, and 'X' — to a break. The length of the sequence will not exceed 106.
Output
Output the maximum percentage of time that Jack can spend marching correctly, rounded down to exactly six digits after the decimal point.
Examples
Input
X
Output
0.000000
Input
LXRR
Output
50.000000
Note
In the second example, if we add two breaks to receive LXXRXR, Jack will march: LXXRXRLXXRXRL... instead of LRLRLRLRLRLRL... and will make the correct step in half the cases. If we didn't add any breaks, the sequence would be incorrect — Jack can't step on his right foot twice in a row.
Submitted Solution:
```
d={'L':'R','R':'L'}
s=input()
cur='L'
long=len(s)
steps=0
for i in s:
if i!='X':
steps+=1
if i!=cur:
long+=1
cur=d[cur]
cur=d[cur]
if long%2:long+=1
t=steps*10**8//long
print(t//10**6,'.','0'*(6-len(str(t%10**6))),t%10**6,sep='')
```
No
| 99,571 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Suppose you are stuck on a desert island. The only way to save yourself is to craft a wooden raft and go to the sea. Fortunately, you have a hand-made saw and a forest nearby. Moreover, you've already cut several trees and prepared it to the point that now you have n logs and the i-th log has length a_i.
The wooden raft you'd like to build has the following structure: 2 logs of length x and x logs of length y. Such raft would have the area equal to x ⋅ y. Both x and y must be integers since it's the only way you can measure the lengths while being on a desert island. And both x and y must be at least 2 since the raft that is one log wide is unstable.
You can cut logs in pieces but you can't merge two logs in one. What is the maximum area of the raft you can craft?
Input
The first line contains the only integer n (1 ≤ n ≤ 5 ⋅ 10^5) — the number of logs you have.
The second line contains n integers a_1, a_2, ..., a_n (2 ≤ a_i ≤ 5 ⋅ 10^5) — the corresponding lengths of the logs.
It's guaranteed that you can always craft at least 2 × 2 raft.
Output
Print the only integer — the maximum area of the raft you can craft.
Examples
Input
1
9
Output
4
Input
9
9 10 9 18 9 9 9 28 9
Output
90
Note
In the first example, you can cut the log of the length 9 in 5 parts: 2 + 2 + 2 + 2 + 1. Now you can build 2 × 2 raft using 2 logs of length x = 2 and x = 2 logs of length y = 2.
In the second example, you can cut a_4 = 18 into two pieces 9 + 9 and a_8 = 28 in three pieces 10 + 9 + 9. Now you can make 10 × 9 raft using 2 logs of length 10 and 10 logs of length 9.
Submitted Solution:
```
n = int(input())
a = list(map(int,input().split(" ")))
summ = sum(a)
x = 2
y = 2
s = 2*x + x*y
b = [x*y]
while True:
if(y > summ//10): break
while True:
if(s > summ): break
x += 1
s = 2*x + x*y
x -= 1
b.append(x*y)
x = 2
y += 1
s = 2*x + x*y
print(max(b))
```
No
| 99,572 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Suppose you are stuck on a desert island. The only way to save yourself is to craft a wooden raft and go to the sea. Fortunately, you have a hand-made saw and a forest nearby. Moreover, you've already cut several trees and prepared it to the point that now you have n logs and the i-th log has length a_i.
The wooden raft you'd like to build has the following structure: 2 logs of length x and x logs of length y. Such raft would have the area equal to x ⋅ y. Both x and y must be integers since it's the only way you can measure the lengths while being on a desert island. And both x and y must be at least 2 since the raft that is one log wide is unstable.
You can cut logs in pieces but you can't merge two logs in one. What is the maximum area of the raft you can craft?
Input
The first line contains the only integer n (1 ≤ n ≤ 5 ⋅ 10^5) — the number of logs you have.
The second line contains n integers a_1, a_2, ..., a_n (2 ≤ a_i ≤ 5 ⋅ 10^5) — the corresponding lengths of the logs.
It's guaranteed that you can always craft at least 2 × 2 raft.
Output
Print the only integer — the maximum area of the raft you can craft.
Examples
Input
1
9
Output
4
Input
9
9 10 9 18 9 9 9 28 9
Output
90
Note
In the first example, you can cut the log of the length 9 in 5 parts: 2 + 2 + 2 + 2 + 1. Now you can build 2 × 2 raft using 2 logs of length x = 2 and x = 2 logs of length y = 2.
In the second example, you can cut a_4 = 18 into two pieces 9 + 9 and a_8 = 28 in three pieces 10 + 9 + 9. Now you can make 10 × 9 raft using 2 logs of length 10 and 10 logs of length 9.
Submitted Solution:
```
a=int(input())
b=input().split()
if a==1:
print(4)
elif a==9:
print(90)
```
No
| 99,573 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Suppose you are stuck on a desert island. The only way to save yourself is to craft a wooden raft and go to the sea. Fortunately, you have a hand-made saw and a forest nearby. Moreover, you've already cut several trees and prepared it to the point that now you have n logs and the i-th log has length a_i.
The wooden raft you'd like to build has the following structure: 2 logs of length x and x logs of length y. Such raft would have the area equal to x ⋅ y. Both x and y must be integers since it's the only way you can measure the lengths while being on a desert island. And both x and y must be at least 2 since the raft that is one log wide is unstable.
You can cut logs in pieces but you can't merge two logs in one. What is the maximum area of the raft you can craft?
Input
The first line contains the only integer n (1 ≤ n ≤ 5 ⋅ 10^5) — the number of logs you have.
The second line contains n integers a_1, a_2, ..., a_n (2 ≤ a_i ≤ 5 ⋅ 10^5) — the corresponding lengths of the logs.
It's guaranteed that you can always craft at least 2 × 2 raft.
Output
Print the only integer — the maximum area of the raft you can craft.
Examples
Input
1
9
Output
4
Input
9
9 10 9 18 9 9 9 28 9
Output
90
Note
In the first example, you can cut the log of the length 9 in 5 parts: 2 + 2 + 2 + 2 + 1. Now you can build 2 × 2 raft using 2 logs of length x = 2 and x = 2 logs of length y = 2.
In the second example, you can cut a_4 = 18 into two pieces 9 + 9 and a_8 = 28 in three pieces 10 + 9 + 9. Now you can make 10 × 9 raft using 2 logs of length 10 and 10 logs of length 9.
Submitted Solution:
```
n = int(input())
a = list(map(int,input().split(" ")))
summ = sum(a)
x = 2
y = 2
s = 2*x + x*y
b = [s]
while True:
if(y > summ//10): break
while True:
if(s > summ): break
x += 1
s = 2*x + x*y
x -= 1
b.append(x*y)
x = 2
y += 1
s = 2*x + x*y
print(max(b))
```
No
| 99,574 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You're given a tree with n vertices. The color of the i-th vertex is h_{i}.
The value of the tree is defined as ∑_{h_{i} = h_{j}, 1 ≤ i < j ≤ n}{dis(i,j)}, where dis(i,j) is the number of edges on the shortest path between i and j.
The color of each vertex is lost, you only remember that h_{i} can be any integer from [l_{i}, r_{i}](inclusive). You want to calculate the sum of values of all trees meeting these conditions modulo 10^9 + 7 (the set of edges is fixed, but each color is unknown, so there are ∏_{i = 1}^{n} (r_{i} - l_{i} + 1) different trees).
Input
The first line contains one integer n (2 ≤ n ≤ 10^5) — the number of vertices.
Then n lines follow, each line contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ 10^5) denoting the range of possible colors of vertex i.
Then n - 1 lines follow, each containing two integers u and v (1 ≤ u, v ≤ n, u ≠ v) denoting an edge of the tree. It is guaranteed that these edges form a tree.
Output
Print one integer — the sum of values of all possible trees, taken modulo 10^9 + 7.
Example
Input
4
1 1
1 2
1 1
1 2
1 2
1 3
3 4
Output
22
Note
In the first example there are four different ways to color the tree (so, there are four different trees):
* a tree with vertices colored as follows: { 1,1,1,1 }. The value of this tree is dis(1,2)+dis(1,3)+dis(1,4)+dis(2,3)+dis(2,4)+dis(3,4) = 10;
* a tree with vertices colored as follows: { 1,2,1,1 }. The value of this tree is dis(1,3)+dis(1,4)+dis(3,4)=4;
* a tree with vertices colored as follows: { 1,1,1,2 }. The value of this tree is dis(1,2)+dis(1,3)+dis(2,3)=4;
* a tree with vertices colored as follows: { 1,2,1,2 }. The value of this tree is dis(1,3)+dis(2,4)=4.
Overall the sum of all values is 10+4+4+4=22.
Submitted Solution:
```
import sys
input = lambda: sys.stdin.readline().rstrip()
NN = 18
BIT1=[0]*(2**NN+1)
BIT2=[0]*(2**NN+1)
def addrange(l0, r0, x=1):
l, r = l0, r0
while l <= 2**NN:
BIT1[l] -= x * l0
BIT2[l] -= x
l += l & (-l)
while r <= 2**NN:
BIT1[r] += x * r0
BIT2[r] += x
r += r & (-r)
def rangesum(l, r):
l0, r0, a, b, c = l, r, 0, 0, 0
while l != 0:
a -= BIT1[l]
b += BIT2[l]
l -= l&(-l)
while r != 0:
a += BIT1[r]
c += BIT2[r]
r -= r&(-r)
return a + b * l0 - c * r0
def disp():
tmp = []
for i in range(30):
tmp.append(rangesum(i, i+1))
print(tmp)
N = int(input())
Y = [[] for _ in range(N)]
MO = 10**9+7
s = 1
for i in range(N):
a, b = map(int, input().split())
Y[i] = [a, b+1, pow(b-a+1, MO-2, MO)]
s *= b-a+1
s %= MO
X = [[] for _ in range(N)]
for i in range(N-1):
a, b = map(int, input().split())
X[a-1].append(b-1)
X[b-1].append(a-1)
# print("X =", X)
ans = 0
def EulerTour(n, X):
global ans
P = [-1] * n
Q = [-1, 0]
ct = -1
ET = []
ET1 = [0] * n
ET2 = [0] * n
DE = [0] * n
de = -1
while Q:
i = Q.pop()
if i < 0:
# �� �߂�������𑫂��ꍇ�͂�����g��
# ct += 1
# �� �߂��ET�ɓ����ꍇ�͂�����g��
# ET.append(P[~i])
# print("END i =", ~i)
a, b, k = Y[i]
# addrange(a, b, -k)
ET2[~i] = ct
de -= 1
continue
if i >= 0:
# print("i =", i)
a, b, k = Y[i]
ans += rangesum(a, b)
# print("ans =", ans * s % MO)
addrange(a, b, k)
ET.append(i)
ct += 1
if ET1[i] == 0: ET1[i] = ct
de += 1
DE[i] = de
for a in X[i][::-1]:
if a != P[i]:
P[a] = i
for k in range(len(X[a])):
if X[a][k] == i:
del X[a][k]
break
Q.append(~a)
Q.append(a)
return (ET, ET1, ET2)
ET, ET1, ET2 = EulerTour(N, X)
print(ans * s % MO)
```
No
| 99,575 |
Provide tags and a correct Python 3 solution for this coding contest problem.
New Year is coming and you are excited to know how many minutes remain before the New Year. You know that currently the clock shows h hours and m minutes, where 0 ≤ hh < 24 and 0 ≤ mm < 60. We use 24-hour time format!
Your task is to find the number of minutes before the New Year. You know that New Year comes when the clock shows 0 hours and 0 minutes.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1439) — the number of test cases.
The following t lines describe test cases. The i-th line contains the time as two integers h and m (0 ≤ h < 24, 0 ≤ m < 60). It is guaranteed that this time is not a midnight, i.e. the following two conditions can't be met at the same time: h=0 and m=0. It is guaranteed that both h and m are given without leading zeros.
Output
For each test case, print the answer on it — the number of minutes before the New Year.
Example
Input
5
23 55
23 0
0 1
4 20
23 59
Output
5
60
1439
1180
1
Tags: math
Correct Solution:
```
num_of_q = int(input())
d = 24 * 60
for i in range(num_of_q):
h, m = [int(i) for i in input().split()]
m += h * 60
print(d - m)
```
| 99,576 |
Provide tags and a correct Python 3 solution for this coding contest problem.
New Year is coming and you are excited to know how many minutes remain before the New Year. You know that currently the clock shows h hours and m minutes, where 0 ≤ hh < 24 and 0 ≤ mm < 60. We use 24-hour time format!
Your task is to find the number of minutes before the New Year. You know that New Year comes when the clock shows 0 hours and 0 minutes.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1439) — the number of test cases.
The following t lines describe test cases. The i-th line contains the time as two integers h and m (0 ≤ h < 24, 0 ≤ m < 60). It is guaranteed that this time is not a midnight, i.e. the following two conditions can't be met at the same time: h=0 and m=0. It is guaranteed that both h and m are given without leading zeros.
Output
For each test case, print the answer on it — the number of minutes before the New Year.
Example
Input
5
23 55
23 0
0 1
4 20
23 59
Output
5
60
1439
1180
1
Tags: math
Correct Solution:
```
#!/usr/bin/env python
from __future__ import division, print_function
import os
import sys
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
def main():
t = int(input())
for _ in range(t):
h, m = map(int, input().split())
print((23 - h) * 60 + (60 - m))
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
```
| 99,577 |
Provide tags and a correct Python 3 solution for this coding contest problem.
New Year is coming and you are excited to know how many minutes remain before the New Year. You know that currently the clock shows h hours and m minutes, where 0 ≤ hh < 24 and 0 ≤ mm < 60. We use 24-hour time format!
Your task is to find the number of minutes before the New Year. You know that New Year comes when the clock shows 0 hours and 0 minutes.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1439) — the number of test cases.
The following t lines describe test cases. The i-th line contains the time as two integers h and m (0 ≤ h < 24, 0 ≤ m < 60). It is guaranteed that this time is not a midnight, i.e. the following two conditions can't be met at the same time: h=0 and m=0. It is guaranteed that both h and m are given without leading zeros.
Output
For each test case, print the answer on it — the number of minutes before the New Year.
Example
Input
5
23 55
23 0
0 1
4 20
23 59
Output
5
60
1439
1180
1
Tags: math
Correct Solution:
```
q = int(input())
for _ in range(q):
h, m = map(int, input().split())
print((24 - h) * 60 - m)
```
| 99,578 |
Provide tags and a correct Python 3 solution for this coding contest problem.
New Year is coming and you are excited to know how many minutes remain before the New Year. You know that currently the clock shows h hours and m minutes, where 0 ≤ hh < 24 and 0 ≤ mm < 60. We use 24-hour time format!
Your task is to find the number of minutes before the New Year. You know that New Year comes when the clock shows 0 hours and 0 minutes.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1439) — the number of test cases.
The following t lines describe test cases. The i-th line contains the time as two integers h and m (0 ≤ h < 24, 0 ≤ m < 60). It is guaranteed that this time is not a midnight, i.e. the following two conditions can't be met at the same time: h=0 and m=0. It is guaranteed that both h and m are given without leading zeros.
Output
For each test case, print the answer on it — the number of minutes before the New Year.
Example
Input
5
23 55
23 0
0 1
4 20
23 59
Output
5
60
1439
1180
1
Tags: math
Correct Solution:
```
n=int(input())
for i in range(n):
h,m=map(int,input().split())
hh=(23-h)*60
mm=60-m
print(hh+mm)
```
| 99,579 |
Provide tags and a correct Python 3 solution for this coding contest problem.
New Year is coming and you are excited to know how many minutes remain before the New Year. You know that currently the clock shows h hours and m minutes, where 0 ≤ hh < 24 and 0 ≤ mm < 60. We use 24-hour time format!
Your task is to find the number of minutes before the New Year. You know that New Year comes when the clock shows 0 hours and 0 minutes.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1439) — the number of test cases.
The following t lines describe test cases. The i-th line contains the time as two integers h and m (0 ≤ h < 24, 0 ≤ m < 60). It is guaranteed that this time is not a midnight, i.e. the following two conditions can't be met at the same time: h=0 and m=0. It is guaranteed that both h and m are given without leading zeros.
Output
For each test case, print the answer on it — the number of minutes before the New Year.
Example
Input
5
23 55
23 0
0 1
4 20
23 59
Output
5
60
1439
1180
1
Tags: math
Correct Solution:
```
t=int(input())
while(t!=0):
a,b=input().split()
a=int(a)
b=int(b)
print((((24-a)*60))-b)
t-=1
```
| 99,580 |
Provide tags and a correct Python 3 solution for this coding contest problem.
New Year is coming and you are excited to know how many minutes remain before the New Year. You know that currently the clock shows h hours and m minutes, where 0 ≤ hh < 24 and 0 ≤ mm < 60. We use 24-hour time format!
Your task is to find the number of minutes before the New Year. You know that New Year comes when the clock shows 0 hours and 0 minutes.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1439) — the number of test cases.
The following t lines describe test cases. The i-th line contains the time as two integers h and m (0 ≤ h < 24, 0 ≤ m < 60). It is guaranteed that this time is not a midnight, i.e. the following two conditions can't be met at the same time: h=0 and m=0. It is guaranteed that both h and m are given without leading zeros.
Output
For each test case, print the answer on it — the number of minutes before the New Year.
Example
Input
5
23 55
23 0
0 1
4 20
23 59
Output
5
60
1439
1180
1
Tags: math
Correct Solution:
```
for _ in range(int(input())):
h, m = [int(i) for i in input().split()]
temp = h * 60 + m
print(24 * 60 - temp)
```
| 99,581 |
Provide tags and a correct Python 3 solution for this coding contest problem.
New Year is coming and you are excited to know how many minutes remain before the New Year. You know that currently the clock shows h hours and m minutes, where 0 ≤ hh < 24 and 0 ≤ mm < 60. We use 24-hour time format!
Your task is to find the number of minutes before the New Year. You know that New Year comes when the clock shows 0 hours and 0 minutes.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1439) — the number of test cases.
The following t lines describe test cases. The i-th line contains the time as two integers h and m (0 ≤ h < 24, 0 ≤ m < 60). It is guaranteed that this time is not a midnight, i.e. the following two conditions can't be met at the same time: h=0 and m=0. It is guaranteed that both h and m are given without leading zeros.
Output
For each test case, print the answer on it — the number of minutes before the New Year.
Example
Input
5
23 55
23 0
0 1
4 20
23 59
Output
5
60
1439
1180
1
Tags: math
Correct Solution:
```
for i in range(int(input())):
h,m=[int(s) for s in input().split()]
print((24-h)*60-m)
```
| 99,582 |
Provide tags and a correct Python 3 solution for this coding contest problem.
New Year is coming and you are excited to know how many minutes remain before the New Year. You know that currently the clock shows h hours and m minutes, where 0 ≤ hh < 24 and 0 ≤ mm < 60. We use 24-hour time format!
Your task is to find the number of minutes before the New Year. You know that New Year comes when the clock shows 0 hours and 0 minutes.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1439) — the number of test cases.
The following t lines describe test cases. The i-th line contains the time as two integers h and m (0 ≤ h < 24, 0 ≤ m < 60). It is guaranteed that this time is not a midnight, i.e. the following two conditions can't be met at the same time: h=0 and m=0. It is guaranteed that both h and m are given without leading zeros.
Output
For each test case, print the answer on it — the number of minutes before the New Year.
Example
Input
5
23 55
23 0
0 1
4 20
23 59
Output
5
60
1439
1180
1
Tags: math
Correct Solution:
```
t = int(input(''))
for v in range(t):
a = input('').split(' ')
h = int(a[0])
m = int(a[1])
ans = (24-h-1)*60 + 60-m
print(ans)
```
| 99,583 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
New Year is coming and you are excited to know how many minutes remain before the New Year. You know that currently the clock shows h hours and m minutes, where 0 ≤ hh < 24 and 0 ≤ mm < 60. We use 24-hour time format!
Your task is to find the number of minutes before the New Year. You know that New Year comes when the clock shows 0 hours and 0 minutes.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1439) — the number of test cases.
The following t lines describe test cases. The i-th line contains the time as two integers h and m (0 ≤ h < 24, 0 ≤ m < 60). It is guaranteed that this time is not a midnight, i.e. the following two conditions can't be met at the same time: h=0 and m=0. It is guaranteed that both h and m are given without leading zeros.
Output
For each test case, print the answer on it — the number of minutes before the New Year.
Example
Input
5
23 55
23 0
0 1
4 20
23 59
Output
5
60
1439
1180
1
Submitted Solution:
```
a=input()
for i in range (int(a)) :
b=input()
q=b.split(" ")
h=int(q[0])
m=int(q[1])
mins=((24-h)*60)+ (60-m) -60
print (mins)
```
Yes
| 99,584 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
New Year is coming and you are excited to know how many minutes remain before the New Year. You know that currently the clock shows h hours and m minutes, where 0 ≤ hh < 24 and 0 ≤ mm < 60. We use 24-hour time format!
Your task is to find the number of minutes before the New Year. You know that New Year comes when the clock shows 0 hours and 0 minutes.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1439) — the number of test cases.
The following t lines describe test cases. The i-th line contains the time as two integers h and m (0 ≤ h < 24, 0 ≤ m < 60). It is guaranteed that this time is not a midnight, i.e. the following two conditions can't be met at the same time: h=0 and m=0. It is guaranteed that both h and m are given without leading zeros.
Output
For each test case, print the answer on it — the number of minutes before the New Year.
Example
Input
5
23 55
23 0
0 1
4 20
23 59
Output
5
60
1439
1180
1
Submitted Solution:
```
import sys
# import bisect
# from collections import deque
Ri = lambda : [int(x) for x in sys.stdin.readline().split()]
ri = lambda : sys.stdin.readline().strip()
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
INF = 10 ** 18
MOD = 998244353
for _ in range(int(ri())):
h,m = Ri()
m1 = (60-m)
m2 = (24-h-1)*60
print(m1+m2)
```
Yes
| 99,585 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
New Year is coming and you are excited to know how many minutes remain before the New Year. You know that currently the clock shows h hours and m minutes, where 0 ≤ hh < 24 and 0 ≤ mm < 60. We use 24-hour time format!
Your task is to find the number of minutes before the New Year. You know that New Year comes when the clock shows 0 hours and 0 minutes.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1439) — the number of test cases.
The following t lines describe test cases. The i-th line contains the time as two integers h and m (0 ≤ h < 24, 0 ≤ m < 60). It is guaranteed that this time is not a midnight, i.e. the following two conditions can't be met at the same time: h=0 and m=0. It is guaranteed that both h and m are given without leading zeros.
Output
For each test case, print the answer on it — the number of minutes before the New Year.
Example
Input
5
23 55
23 0
0 1
4 20
23 59
Output
5
60
1439
1180
1
Submitted Solution:
```
for t in range(int(input())):
h,m = map(int,input().split())
su = h*60+m
print(24*60-su)
```
Yes
| 99,586 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
New Year is coming and you are excited to know how many minutes remain before the New Year. You know that currently the clock shows h hours and m minutes, where 0 ≤ hh < 24 and 0 ≤ mm < 60. We use 24-hour time format!
Your task is to find the number of minutes before the New Year. You know that New Year comes when the clock shows 0 hours and 0 minutes.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1439) — the number of test cases.
The following t lines describe test cases. The i-th line contains the time as two integers h and m (0 ≤ h < 24, 0 ≤ m < 60). It is guaranteed that this time is not a midnight, i.e. the following two conditions can't be met at the same time: h=0 and m=0. It is guaranteed that both h and m are given without leading zeros.
Output
For each test case, print the answer on it — the number of minutes before the New Year.
Example
Input
5
23 55
23 0
0 1
4 20
23 59
Output
5
60
1439
1180
1
Submitted Solution:
```
from bisect import bisect_left as bl
from bisect import bisect_right as br
import heapq
import math
from collections import *
from functools import reduce,cmp_to_key
import sys
input = sys.stdin.readline
# M = mod = 998244353
def factors(n):return sorted(list(set(reduce(list.__add__,([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))))
def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip('\n').split(' ')]
def st():return input().rstrip('\n')
def val():return int(input().rstrip('\n'))
def li2():return [i for i in input().rstrip('\n').split(' ')]
def li3():return [int(i) for i in input().rstrip('\n')]
for _ in range(val()):
h, m = li()
tot = 60 - m
tot += (24 - (h + 1))*60
print(tot)
```
Yes
| 99,587 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
New Year is coming and you are excited to know how many minutes remain before the New Year. You know that currently the clock shows h hours and m minutes, where 0 ≤ hh < 24 and 0 ≤ mm < 60. We use 24-hour time format!
Your task is to find the number of minutes before the New Year. You know that New Year comes when the clock shows 0 hours and 0 minutes.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1439) — the number of test cases.
The following t lines describe test cases. The i-th line contains the time as two integers h and m (0 ≤ h < 24, 0 ≤ m < 60). It is guaranteed that this time is not a midnight, i.e. the following two conditions can't be met at the same time: h=0 and m=0. It is guaranteed that both h and m are given without leading zeros.
Output
For each test case, print the answer on it — the number of minutes before the New Year.
Example
Input
5
23 55
23 0
0 1
4 20
23 59
Output
5
60
1439
1180
1
Submitted Solution:
```
def solve(T):
h, m = map(int, input().split())
risposta = 0 # memorizza qui la risposta
if m == 0:
return (24 - h) * 60
else:
return (23 - h) * 60 + 60 - m
T = int(input())
for t in range(1, T+1):
print("Case #" + str(t) + ":", solve(T))
```
No
| 99,588 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
New Year is coming and you are excited to know how many minutes remain before the New Year. You know that currently the clock shows h hours and m minutes, where 0 ≤ hh < 24 and 0 ≤ mm < 60. We use 24-hour time format!
Your task is to find the number of minutes before the New Year. You know that New Year comes when the clock shows 0 hours and 0 minutes.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1439) — the number of test cases.
The following t lines describe test cases. The i-th line contains the time as two integers h and m (0 ≤ h < 24, 0 ≤ m < 60). It is guaranteed that this time is not a midnight, i.e. the following two conditions can't be met at the same time: h=0 and m=0. It is guaranteed that both h and m are given without leading zeros.
Output
For each test case, print the answer on it — the number of minutes before the New Year.
Example
Input
5
23 55
23 0
0 1
4 20
23 59
Output
5
60
1439
1180
1
Submitted Solution:
```
n=int(input())
for i in range(n):
a,b=map(int,input().split())
if(a<23 and b<59):
o=23-a
p=o*60
q=60-b
r=p+q
print(r)
else:
if(a==23):
e=60-b
print(e)
```
No
| 99,589 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
New Year is coming and you are excited to know how many minutes remain before the New Year. You know that currently the clock shows h hours and m minutes, where 0 ≤ hh < 24 and 0 ≤ mm < 60. We use 24-hour time format!
Your task is to find the number of minutes before the New Year. You know that New Year comes when the clock shows 0 hours and 0 minutes.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1439) — the number of test cases.
The following t lines describe test cases. The i-th line contains the time as two integers h and m (0 ≤ h < 24, 0 ≤ m < 60). It is guaranteed that this time is not a midnight, i.e. the following two conditions can't be met at the same time: h=0 and m=0. It is guaranteed that both h and m are given without leading zeros.
Output
For each test case, print the answer on it — the number of minutes before the New Year.
Example
Input
5
23 55
23 0
0 1
4 20
23 59
Output
5
60
1439
1180
1
Submitted Solution:
```
def main(lst):
res_lst = []
for i in lst:
res_lst.append((24*60) - (i[0] * 60 + i[1]))
return res_lst
if __name__=='__main__':
n = int(input())
lst = []
for i in range(n):
lst.append(list(map(int,input().split())))
print(main(lst))
```
No
| 99,590 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
New Year is coming and you are excited to know how many minutes remain before the New Year. You know that currently the clock shows h hours and m minutes, where 0 ≤ hh < 24 and 0 ≤ mm < 60. We use 24-hour time format!
Your task is to find the number of minutes before the New Year. You know that New Year comes when the clock shows 0 hours and 0 minutes.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1439) — the number of test cases.
The following t lines describe test cases. The i-th line contains the time as two integers h and m (0 ≤ h < 24, 0 ≤ m < 60). It is guaranteed that this time is not a midnight, i.e. the following two conditions can't be met at the same time: h=0 and m=0. It is guaranteed that both h and m are given without leading zeros.
Output
For each test case, print the answer on it — the number of minutes before the New Year.
Example
Input
5
23 55
23 0
0 1
4 20
23 59
Output
5
60
1439
1180
1
Submitted Solution:
```
t=int(input())
for i in range(t):
h,m=map(int,input().split())
h=abs(h-24)
if(h==1):
m=abs(m-60)
print(m)
if(h==24):
h=24*60
print(h-m)
if(1<h<23):
h=h-1
h=h*60
print(h+abs(m-60))
```
No
| 99,591 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of the problem. The difference is constraints on the number of wise men and the time limit. You can make hacks only if all versions of this task are solved.
n wise men live in a beautiful city. Some of them know each other.
For each of the n! possible permutations p_1, p_2, …, p_n of the wise men, let's generate a binary string of length n-1: for each 1 ≤ i < n set s_i=1 if p_i and p_{i+1} know each other, and s_i=0 otherwise.
For all possible 2^{n-1} binary strings, find the number of permutations that produce this binary string.
Input
The first line of input contains one integer n (2 ≤ n ≤ 14) — the number of wise men in the city.
The next n lines contain a binary string of length n each, such that the j-th character of the i-th string is equal to '1' if wise man i knows wise man j, and equals '0' otherwise.
It is guaranteed that if the i-th man knows the j-th man, then the j-th man knows i-th man and no man knows himself.
Output
Print 2^{n-1} space-separated integers. For each 0 ≤ x < 2^{n-1}:
* Let's consider a string s of length n-1, such that s_i = ⌊ \frac{x}{2^{i-1}} ⌋ mod 2 for all 1 ≤ i ≤ n - 1.
* The (x+1)-th number should be equal to the required answer for s.
Examples
Input
3
011
101
110
Output
0 0 0 6
Input
4
0101
1000
0001
1010
Output
2 2 6 2 2 6 2 2
Note
In the first test, each wise man knows each other, so every permutation will produce the string 11.
In the second test:
* If p = \{1, 2, 3, 4\}, the produced string is 101, because wise men 1 and 2 know each other, 2 and 3 don't know each other, and 3 and 4 know each other;
* If p = \{4, 1, 2, 3\}, the produced string is 110, because wise men 1 and 4 know each other, 1 and 2 know each other and 2, and 3 don't know each other;
* If p = \{1, 3, 2, 4\}, the produced string is 000, because wise men 1 and 3 don't know each other, 3 and 2 don't know each other, and 2 and 4 don't know each other.
Tags: bitmasks, brute force, dp, meet-in-the-middle
Correct Solution:
```
from sys import stdout
n = int(input())
class Person:
num = n - 1
def __init__(self, rel):
self.relationship = int(rel, 2)
def __getitem__(self, k):
return (self.relationship >> Person.num - k) & 1
rel = [Person(input()) for _ in range(n)]
dp = [[0] * n for _ in range(1 << n)]
for people in range(1, 1 << n):
ones = [i for i in range(n) if people & (1 << i)]
# print(f'ones: {ones}')
one_num = len(ones)
if one_num == 1:
dp[people][ones[0]] = [1]
continue
for i in ones:
dp[people][i] = [0] * (1 << one_num - 1)
pre_people = people ^ (1 << i)
for j in ones:
if j == i:
continue
for pre_s, times in enumerate(dp[pre_people][j]):
s = pre_s | (rel[j][i] << one_num - 2)
# print(f'dp[{people}][{i}][{s}]: {dp[people][i][s]}')
dp[people][i][s] += times
people = (1 << n) - 1
for s in range(1 << (n-1)):
ans = 0
for i in range(n):
ans += dp[people][i][s]
print(ans, end=' ')
```
| 99,592 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Uh oh! Applications to tech companies are due soon, and you've been procrastinating by doing contests instead! (Let's pretend for now that it is actually possible to get a job in these uncertain times.)
You have completed many programming projects. In fact, there are exactly n types of programming projects, and you have completed a_i projects of type i. Your résumé has limited space, but you want to carefully choose them in such a way that maximizes your chances of getting hired.
You want to include several projects of the same type to emphasize your expertise, but you also don't want to include so many that the low-quality projects start slipping in. Specifically, you determine the following quantity to be a good indicator of your chances of getting hired:
$$$ f(b_1,…,b_n)=∑_{i=1}^n b_i(a_i-b_i^2). $$$
Here, b_i denotes the number of projects of type i you include in your résumé. Of course, you cannot include more projects than you have completed, so you require 0≤ b_i ≤ a_i for all i.
Your résumé only has enough room for k projects, and you will absolutely not be hired if your résumé has empty space, so you require ∑_{i=1}^n b_i=k.
Find values for b_1,…, b_n that maximize the value of f(b_1,…,b_n) while satisfying the above two constraints.
Input
The first line contains two integers n and k (1≤ n≤ 10^5, 1≤ k≤ ∑_{i=1}^n a_i) — the number of types of programming projects and the résumé size, respectively.
The next line contains n integers a_1,…,a_n (1≤ a_i≤ 10^9) — a_i is equal to the number of completed projects of type i.
Output
In a single line, output n integers b_1,…, b_n that achieve the maximum value of f(b_1,…,b_n), while satisfying the requirements 0≤ b_i≤ a_i and ∑_{i=1}^n b_i=k. If there are multiple solutions, output any.
Note that you do not have to output the value f(b_1,…,b_n).
Examples
Input
10 32
1 2 3 4 5 5 5 5 5 5
Output
1 2 3 3 3 4 4 4 4 4
Input
5 8
4 4 8 2 1
Output
2 2 2 1 1
Note
For the first test, the optimal answer is f=-269. Note that a larger f value is possible if we ignored the constraint ∑_{i=1}^n b_i=k.
For the second test, the optimal answer is f=9.
Tags: binary search, greedy, math
Correct Solution:
```
import sys
import heapq as hq
readline = sys.stdin.readline
ns = lambda: readline().rstrip()
ni = lambda: int(readline().rstrip())
nm = lambda: map(int, readline().split())
nl = lambda: list(map(int, readline().split()))
eps = 10**-7
def solve():
n, k = nm()
a = nl()
ans = [0]*n
ok = 10**9; ng = -4*10**18
while ok - ng > 1:
mid = (ok + ng) // 2
ck = 0
for i in range(n):
d = 9 - 12 * (mid + 1 - a[i])
if d < 0:
continue
ck += min(a[i], int((3 + d**.5) / 6 + eps))
# print(mid, ck)
if ck > k:
ng = mid
else:
ok = mid
for i in range(n):
d = 9 - 12 * (ok + 1 - a[i])
if d < 0:
continue
ans[i] = min(a[i], int((3 + d**.5) / 6 + eps))
# print(ans)
rk = k - sum(ans)
l = list()
for i in range(n):
if ans[i] < a[i]:
hq.heappush(l, (-a[i] + 3 * ans[i]**2 - 3 * ans[i] + 1, i))
for _ in range(rk):
v, i = hq.heappop(l)
ans[i] += 1
if ans[i] < a[i]:
hq.heappush(l, (-a[i] + 3 * ans[i]**2 - 3 * ans[i] + 1, i))
print(*ans)
return
solve()
```
| 99,593 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The store sells n beads. The color of each bead is described by a lowercase letter of the English alphabet ("a"–"z"). You want to buy some beads to assemble a necklace from them.
A necklace is a set of beads connected in a circle.
For example, if the store sells beads "a", "b", "c", "a", "c", "c", then you can assemble the following necklaces (these are not all possible options):
<image>
And the following necklaces cannot be assembled from beads sold in the store:
<image> The first necklace cannot be assembled because it has three beads "a" (of the two available). The second necklace cannot be assembled because it contains a bead "d", which is not sold in the store.
We call a necklace k-beautiful if, when it is turned clockwise by k beads, the necklace remains unchanged. For example, here is a sequence of three turns of a necklace.
<image> As you can see, this necklace is, for example, 3-beautiful, 6-beautiful, 9-beautiful, and so on, but it is not 1-beautiful or 2-beautiful.
In particular, a necklace of length 1 is k-beautiful for any integer k. A necklace that consists of beads of the same color is also beautiful for any k.
You are given the integers n and k, and also the string s containing n lowercase letters of the English alphabet — each letter defines a bead in the store. You can buy any subset of beads and connect them in any order. Find the maximum length of a k-beautiful necklace you can assemble.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the test. Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ n, k ≤ 2000).
The second line of each test case contains the string s containing n lowercase English letters — the beads in the store.
It is guaranteed that the sum of n for all test cases does not exceed 2000.
Output
Output t answers to the test cases. Each answer is a positive integer — the maximum length of the k-beautiful necklace you can assemble.
Example
Input
6
6 3
abcbac
3 6
aaa
7 1000
abczgyo
5 4
ababa
20 10
aaebdbabdbbddaadaadc
20 5
ecbedececacbcbccbdec
Output
6
3
5
4
15
10
Note
The first test case is explained in the statement.
In the second test case, a 6-beautiful necklace can be assembled from all the letters.
In the third test case, a 1000-beautiful necklace can be assembled, for example, from beads "abzyo".
Tags: brute force, dfs and similar, dp, graphs, greedy, number theory
Correct Solution:
```
import math
def f(n):
c=[]
d=[]
i = 1
while i <= math.sqrt(n):
if (n % i == 0):
if (n // i == i):
c.append(i)
else:
c.append(i)
d.append(n//i)
i+=1
c+=d[::-1]
return c
t=int(input())
for i in range(t):
n,k=map(int,input().split())
s=input()
b=f(k)
ans=0
c=[0]*(26)
for j in range(n):
c[ord(s[j])-97]+=1
k=[]
for j in range(26):
p=c[j]
i=1
while((p//i)!=0):
k.append(p//i)
i+=1
k.sort(reverse=True)
j=0
r=len(k)
while(j<len(b)):
p=b[j]
if (p-1)<r:
ans=max(ans,p*k[p-1])
j+=1
print(ans)
```
| 99,594 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The store sells n beads. The color of each bead is described by a lowercase letter of the English alphabet ("a"–"z"). You want to buy some beads to assemble a necklace from them.
A necklace is a set of beads connected in a circle.
For example, if the store sells beads "a", "b", "c", "a", "c", "c", then you can assemble the following necklaces (these are not all possible options):
<image>
And the following necklaces cannot be assembled from beads sold in the store:
<image> The first necklace cannot be assembled because it has three beads "a" (of the two available). The second necklace cannot be assembled because it contains a bead "d", which is not sold in the store.
We call a necklace k-beautiful if, when it is turned clockwise by k beads, the necklace remains unchanged. For example, here is a sequence of three turns of a necklace.
<image> As you can see, this necklace is, for example, 3-beautiful, 6-beautiful, 9-beautiful, and so on, but it is not 1-beautiful or 2-beautiful.
In particular, a necklace of length 1 is k-beautiful for any integer k. A necklace that consists of beads of the same color is also beautiful for any k.
You are given the integers n and k, and also the string s containing n lowercase letters of the English alphabet — each letter defines a bead in the store. You can buy any subset of beads and connect them in any order. Find the maximum length of a k-beautiful necklace you can assemble.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the test. Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ n, k ≤ 2000).
The second line of each test case contains the string s containing n lowercase English letters — the beads in the store.
It is guaranteed that the sum of n for all test cases does not exceed 2000.
Output
Output t answers to the test cases. Each answer is a positive integer — the maximum length of the k-beautiful necklace you can assemble.
Example
Input
6
6 3
abcbac
3 6
aaa
7 1000
abczgyo
5 4
ababa
20 10
aaebdbabdbbddaadaadc
20 5
ecbedececacbcbccbdec
Output
6
3
5
4
15
10
Note
The first test case is explained in the statement.
In the second test case, a 6-beautiful necklace can be assembled from all the letters.
In the third test case, a 1000-beautiful necklace can be assembled, for example, from beads "abzyo".
Tags: brute force, dfs and similar, dp, graphs, greedy, number theory
Correct Solution:
```
def binary_search(item,start,end,l):
if (start>end):
return max(end,0)
if (start>len(l)-1):
return (len(l)-1)
mid=(start+end)//2
if l[mid]>item:
return binary_search(item,start,mid-1,l)
elif(l[mid]<item):
return binary_search(item,mid+1,end,l)
else:
return mid
def f():
n,k=map(int,input().split(" "))
multiplesofk=[]
for i in range(1,k+1):
if (k%i==0):
multiplesofk.append(i)
string=input()
counter=[0]*200
for i in range(0,len(string)):
counter[ord(string[i])]+=1
listofcount=[]
for i in range(ord('a'),ord('z')+1):
if counter[i]!=0:
listofcount.append(counter[i])
maximum=0
for i in range(1,max(listofcount)+1):
s=0
for j in range(0,len(listofcount)):
s+=listofcount[j]//i
element=binary_search(s,0,len(multiplesofk)-1,multiplesofk)
maximum=max(maximum,multiplesofk[element]*i)
print (maximum)
for i in range(int(input())):
f()
'''200 559
luzrgywfemprizaxktszmugkfdakyjfdwxrfcjouklnxjzypgdlxkyplaornwpktkedqimlvzpcaivmtavwskvelctnvyziytxudzxudfueyukgagjpwkltpfmowhcwjdfxzzjziuaegmwzsisthasicflftxfkfukywxndsdfdbjwhncmjfmbevlkybpxesyrgnssow
'''
```
| 99,595 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The store sells n beads. The color of each bead is described by a lowercase letter of the English alphabet ("a"–"z"). You want to buy some beads to assemble a necklace from them.
A necklace is a set of beads connected in a circle.
For example, if the store sells beads "a", "b", "c", "a", "c", "c", then you can assemble the following necklaces (these are not all possible options):
<image>
And the following necklaces cannot be assembled from beads sold in the store:
<image> The first necklace cannot be assembled because it has three beads "a" (of the two available). The second necklace cannot be assembled because it contains a bead "d", which is not sold in the store.
We call a necklace k-beautiful if, when it is turned clockwise by k beads, the necklace remains unchanged. For example, here is a sequence of three turns of a necklace.
<image> As you can see, this necklace is, for example, 3-beautiful, 6-beautiful, 9-beautiful, and so on, but it is not 1-beautiful or 2-beautiful.
In particular, a necklace of length 1 is k-beautiful for any integer k. A necklace that consists of beads of the same color is also beautiful for any k.
You are given the integers n and k, and also the string s containing n lowercase letters of the English alphabet — each letter defines a bead in the store. You can buy any subset of beads and connect them in any order. Find the maximum length of a k-beautiful necklace you can assemble.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the test. Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ n, k ≤ 2000).
The second line of each test case contains the string s containing n lowercase English letters — the beads in the store.
It is guaranteed that the sum of n for all test cases does not exceed 2000.
Output
Output t answers to the test cases. Each answer is a positive integer — the maximum length of the k-beautiful necklace you can assemble.
Example
Input
6
6 3
abcbac
3 6
aaa
7 1000
abczgyo
5 4
ababa
20 10
aaebdbabdbbddaadaadc
20 5
ecbedececacbcbccbdec
Output
6
3
5
4
15
10
Note
The first test case is explained in the statement.
In the second test case, a 6-beautiful necklace can be assembled from all the letters.
In the third test case, a 1000-beautiful necklace can be assembled, for example, from beads "abzyo".
Tags: brute force, dfs and similar, dp, graphs, greedy, number theory
Correct Solution:
```
letters = 26
for t in range(int(input())):
n, k = map(int, input().split(' '))
s = input()
counts = [0] * (letters + 1)
for c in s:
counts[ord(c) - ord('a')] += 1
counts.sort(reverse=True)
zeroi = counts.index(0)
if zeroi >= 0:
counts = counts[:zeroi]
#print('s:', s)
#print('counts:', counts)
ans = 1
for d in range(1, min(k,n)+1):
if k % d != 0:
continue
l = 1
r = n // d
while l < r:
m = (l + r + 1) // 2
inGroup = 0
for count in counts:
inGroup += count // m
if inGroup >= d:
l = m
else:
r = m - 1
#print(f"d={d} groups={l} len={l*d}")
ans = max(ans, l*d)
print(ans)
```
| 99,596 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The store sells n beads. The color of each bead is described by a lowercase letter of the English alphabet ("a"–"z"). You want to buy some beads to assemble a necklace from them.
A necklace is a set of beads connected in a circle.
For example, if the store sells beads "a", "b", "c", "a", "c", "c", then you can assemble the following necklaces (these are not all possible options):
<image>
And the following necklaces cannot be assembled from beads sold in the store:
<image> The first necklace cannot be assembled because it has three beads "a" (of the two available). The second necklace cannot be assembled because it contains a bead "d", which is not sold in the store.
We call a necklace k-beautiful if, when it is turned clockwise by k beads, the necklace remains unchanged. For example, here is a sequence of three turns of a necklace.
<image> As you can see, this necklace is, for example, 3-beautiful, 6-beautiful, 9-beautiful, and so on, but it is not 1-beautiful or 2-beautiful.
In particular, a necklace of length 1 is k-beautiful for any integer k. A necklace that consists of beads of the same color is also beautiful for any k.
You are given the integers n and k, and also the string s containing n lowercase letters of the English alphabet — each letter defines a bead in the store. You can buy any subset of beads and connect them in any order. Find the maximum length of a k-beautiful necklace you can assemble.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the test. Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ n, k ≤ 2000).
The second line of each test case contains the string s containing n lowercase English letters — the beads in the store.
It is guaranteed that the sum of n for all test cases does not exceed 2000.
Output
Output t answers to the test cases. Each answer is a positive integer — the maximum length of the k-beautiful necklace you can assemble.
Example
Input
6
6 3
abcbac
3 6
aaa
7 1000
abczgyo
5 4
ababa
20 10
aaebdbabdbbddaadaadc
20 5
ecbedececacbcbccbdec
Output
6
3
5
4
15
10
Note
The first test case is explained in the statement.
In the second test case, a 6-beautiful necklace can be assembled from all the letters.
In the third test case, a 1000-beautiful necklace can be assembled, for example, from beads "abzyo".
Tags: brute force, dfs and similar, dp, graphs, greedy, number theory
Correct Solution:
```
import sys
input = sys.stdin.readline
import math
import copy
import bisect
import collections
from collections import deque
import heapq
from collections import defaultdict
t = int(input())
for f in range(t):
n,k = map(int,input().split())
div = []
for i in range(1,2001):
if k%i == 0:
div.append(i)
s = list(input().rstrip())
c = collections.Counter(s)
a = c.most_common()
freq = []
for i in a:
freq.append(i[1])
ans = 0
for i in div:
j = 1
tmp = 0
while j*i <=n:
memo = 0
for l in range(len(freq)):
memo += freq[l]//j
if memo >= i:
tmp = j*i
j += 1
else:
break
ans = max(ans,tmp)
print(ans)
```
| 99,597 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The store sells n beads. The color of each bead is described by a lowercase letter of the English alphabet ("a"–"z"). You want to buy some beads to assemble a necklace from them.
A necklace is a set of beads connected in a circle.
For example, if the store sells beads "a", "b", "c", "a", "c", "c", then you can assemble the following necklaces (these are not all possible options):
<image>
And the following necklaces cannot be assembled from beads sold in the store:
<image> The first necklace cannot be assembled because it has three beads "a" (of the two available). The second necklace cannot be assembled because it contains a bead "d", which is not sold in the store.
We call a necklace k-beautiful if, when it is turned clockwise by k beads, the necklace remains unchanged. For example, here is a sequence of three turns of a necklace.
<image> As you can see, this necklace is, for example, 3-beautiful, 6-beautiful, 9-beautiful, and so on, but it is not 1-beautiful or 2-beautiful.
In particular, a necklace of length 1 is k-beautiful for any integer k. A necklace that consists of beads of the same color is also beautiful for any k.
You are given the integers n and k, and also the string s containing n lowercase letters of the English alphabet — each letter defines a bead in the store. You can buy any subset of beads and connect them in any order. Find the maximum length of a k-beautiful necklace you can assemble.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the test. Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ n, k ≤ 2000).
The second line of each test case contains the string s containing n lowercase English letters — the beads in the store.
It is guaranteed that the sum of n for all test cases does not exceed 2000.
Output
Output t answers to the test cases. Each answer is a positive integer — the maximum length of the k-beautiful necklace you can assemble.
Example
Input
6
6 3
abcbac
3 6
aaa
7 1000
abczgyo
5 4
ababa
20 10
aaebdbabdbbddaadaadc
20 5
ecbedececacbcbccbdec
Output
6
3
5
4
15
10
Note
The first test case is explained in the statement.
In the second test case, a 6-beautiful necklace can be assembled from all the letters.
In the third test case, a 1000-beautiful necklace can be assembled, for example, from beads "abzyo".
Tags: brute force, dfs and similar, dp, graphs, greedy, number theory
Correct Solution:
```
import os
### START FAST IO ###
inp = os.read(0, int(1e7)).split()
inp_pos = -1
def cin():
global inp_pos
inp_pos += 1
return inp[inp_pos]
# cout: os.write(1, "\n".encode().join(res))
#### END FAST IO ####
from itertools import groupby
from math import gcd
T = int(cin())
res = []
while T:
T -= 1
n = int(cin())
k = int(cin())
a = [len(list(g)) for k, g in groupby(sorted(cin()))]
ans = 0
for v in range(1, n+1):
d = gcd(v, k)
f = v//d
if sum([x//f for x in a]) >= d:
ans = v
res.append(str(ans).encode())
os.write(1, "\n".encode().join(res))
```
| 99,598 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The store sells n beads. The color of each bead is described by a lowercase letter of the English alphabet ("a"–"z"). You want to buy some beads to assemble a necklace from them.
A necklace is a set of beads connected in a circle.
For example, if the store sells beads "a", "b", "c", "a", "c", "c", then you can assemble the following necklaces (these are not all possible options):
<image>
And the following necklaces cannot be assembled from beads sold in the store:
<image> The first necklace cannot be assembled because it has three beads "a" (of the two available). The second necklace cannot be assembled because it contains a bead "d", which is not sold in the store.
We call a necklace k-beautiful if, when it is turned clockwise by k beads, the necklace remains unchanged. For example, here is a sequence of three turns of a necklace.
<image> As you can see, this necklace is, for example, 3-beautiful, 6-beautiful, 9-beautiful, and so on, but it is not 1-beautiful or 2-beautiful.
In particular, a necklace of length 1 is k-beautiful for any integer k. A necklace that consists of beads of the same color is also beautiful for any k.
You are given the integers n and k, and also the string s containing n lowercase letters of the English alphabet — each letter defines a bead in the store. You can buy any subset of beads and connect them in any order. Find the maximum length of a k-beautiful necklace you can assemble.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the test. Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ n, k ≤ 2000).
The second line of each test case contains the string s containing n lowercase English letters — the beads in the store.
It is guaranteed that the sum of n for all test cases does not exceed 2000.
Output
Output t answers to the test cases. Each answer is a positive integer — the maximum length of the k-beautiful necklace you can assemble.
Example
Input
6
6 3
abcbac
3 6
aaa
7 1000
abczgyo
5 4
ababa
20 10
aaebdbabdbbddaadaadc
20 5
ecbedececacbcbccbdec
Output
6
3
5
4
15
10
Note
The first test case is explained in the statement.
In the second test case, a 6-beautiful necklace can be assembled from all the letters.
In the third test case, a 1000-beautiful necklace can be assembled, for example, from beads "abzyo".
Tags: brute force, dfs and similar, dp, graphs, greedy, number theory
Correct Solution:
```
import atexit
import io
import sys
import math
from collections import defaultdict,Counter
# _INPUT_LINES = sys.stdin.read().splitlines()
# input = iter(_INPUT_LINES).__next__
# _OUTPUT_BUFFER = io.StringIO()
# sys.stdout = _OUTPUT_BUFFER
# @atexit.register
# def write():
# sys.__stdout__.write(_OUTPUT_BUFFER.getvalue())
# sys.stdout=open("CP2/output.txt",'w')
# sys.stdin=open("CP2/input.txt",'r')
# m=pow(10,9)+7
t=int(input())
for i in range(t):
n,k=map(int,input().split())
s=input()
if k%n==0:
print(n)
continue
c2=list(Counter(s).values())
# print(c)
for j in range(n,0,-1):
c=c2[:]
d=defaultdict(list)
for kk in range(j):
d[kk]=(kk+k)%j
visit=[0]*j
l=[]
for kk in range(j):
if visit[kk]==0:
c1=1
visit[kk]=1
p=d[kk]
while visit[p]==0:
c1+=1
visit[p]=1
p=d[p]
l.append(c1)
l.sort(reverse=True)
# print(l)
for kk in l:
m=max(c)
if m>=kk:
ind=c.index(m)
c[ind]-=kk
else:
break
else:
print(j)
break
```
| 99,599 |
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