message stringlengths 2 20.1k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 1.95k 109k | cluster float64 17 17 | __index_level_0__ int64 3.91k 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A programming coach has n students to teach. We know that n is divisible by 3. Let's assume that all students are numbered from 1 to n, inclusive.
Before the university programming championship the coach wants to split all students into groups of three. For some pairs of students we know that they want to be on the same team. Besides, if the i-th student wants to be on the same team with the j-th one, then the j-th student wants to be on the same team with the i-th one. The coach wants the teams to show good results, so he wants the following condition to hold: if the i-th student wants to be on the same team with the j-th, then the i-th and the j-th students must be on the same team. Also, it is obvious that each student must be on exactly one team.
Help the coach and divide the teams the way he wants.
Input
The first line of the input contains integers n and m (3 ≤ n ≤ 48, <image>. Then follow m lines, each contains a pair of integers ai, bi (1 ≤ ai < bi ≤ n) — the pair ai, bi means that students with numbers ai and bi want to be on the same team.
It is guaranteed that n is divisible by 3. It is guaranteed that each pair ai, bi occurs in the input at most once.
Output
If the required division into teams doesn't exist, print number -1. Otherwise, print <image> lines. In each line print three integers xi, yi, zi (1 ≤ xi, yi, zi ≤ n) — the i-th team.
If there are multiple answers, you are allowed to print any of them.
Examples
Input
3 0
Output
3 2 1
Input
6 4
1 2
2 3
3 4
5 6
Output
-1
Input
3 3
1 2
2 3
1 3
Output
3 2 1
Submitted Solution:
```
from collections import defaultdict
n,m=map(int,input().split())
d=defaultdict(list)
f=[False]*(n+1)
v=[]
def dfs(i,s):
s.add(i)
f[i]=True
for k in d[i]:
if not f[k]:
dfs(k,s)
for j in range(m):
x,y=map(int,input().split())
d[x].append(y)
d[y].append(x)
for i in range(1,n+1):
if d[i] and not f[i]:
s=set()
dfs(i,s)
if len(s)>3:
print(-1)
exit()
v.append(list(s))
if len(v)>(n//3):
print(-1)
exit()
while len(v)<(n//3):
v.append([])
j=0
for i in range(1, n + 1):
if not f[i]:
while len(v[j]) == 3:
j += 1
v[j].append(i)
for i in v:
print(i[0],i[1],i[2])
``` | instruction | 0 | 82,882 | 17 | 165,764 |
Yes | output | 1 | 82,882 | 17 | 165,765 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A programming coach has n students to teach. We know that n is divisible by 3. Let's assume that all students are numbered from 1 to n, inclusive.
Before the university programming championship the coach wants to split all students into groups of three. For some pairs of students we know that they want to be on the same team. Besides, if the i-th student wants to be on the same team with the j-th one, then the j-th student wants to be on the same team with the i-th one. The coach wants the teams to show good results, so he wants the following condition to hold: if the i-th student wants to be on the same team with the j-th, then the i-th and the j-th students must be on the same team. Also, it is obvious that each student must be on exactly one team.
Help the coach and divide the teams the way he wants.
Input
The first line of the input contains integers n and m (3 ≤ n ≤ 48, <image>. Then follow m lines, each contains a pair of integers ai, bi (1 ≤ ai < bi ≤ n) — the pair ai, bi means that students with numbers ai and bi want to be on the same team.
It is guaranteed that n is divisible by 3. It is guaranteed that each pair ai, bi occurs in the input at most once.
Output
If the required division into teams doesn't exist, print number -1. Otherwise, print <image> lines. In each line print three integers xi, yi, zi (1 ≤ xi, yi, zi ≤ n) — the i-th team.
If there are multiple answers, you are allowed to print any of them.
Examples
Input
3 0
Output
3 2 1
Input
6 4
1 2
2 3
3 4
5 6
Output
-1
Input
3 3
1 2
2 3
1 3
Output
3 2 1
Submitted Solution:
```
'http://codeforces.com/problemset/problem/300/b'
import sys
sys.setrecursionlimit(9000)
#grup numarası gibi dusun
komsu=[0]*48
'kurallilari bulmak icin'
def bul(x):
if (komsu[x] == x):
return x
komsu[x] = bul(komsu[x])
return komsu[x]
'graphta node ve edgeleri yerlestiriyormus gibi'
def yerlestir(x,y):
x = bul(x)
y = bul(y)
if (x == y):
return
komsu[x] = y
def main():
n, m=map(int, sys.stdin.readline().strip().split(' '))
x,y,z,p,r = 0,0,0,0,0
a=[0]*48
'gruplar 16x3 gtup'
b=[[0 for _ in range(3)] for __ in range(16)]
'graph gibi dusun kac tane komsu ekliyom'
c=[0]*48
for i in range(n):
komsu[i]=i
for i in range(m):
l,o = map(int, sys.stdin.readline().strip().split(' '))
yerlestir(l-1,o-1)
for i in range(n):
c[bul(i)]+=1
for i in range(n):
if (c[i] > 3):
print(-1)
exit()
if (c[i] == 1):
x+=1
elif (c[i] == 2):
y+=1
elif (c[i] == 3):
z+=1
if ((y + z) > (n / 3)):
print(-1)
exit()
'print(komsu,"komsu")'
'kurallilari gruplara yerlesrine'
for i in range(n):
'3 lu hazirlar'
if (c[i] == 3):
q = 0
for j in range(n):
if (bul(j) == i):
a[j] = 1
b[p][q] = j + 1
q+=1
p+=1
'2 li grubu olup 3.sunu bekleyenler'
for i in range(n):
if (c[i] == 2):
q = 0
for j in range(n):
if (bul(j) == i):
a[j] = 1
b[p][q] = j + 1
q+=1
for j in range(n):
if (a[j] == 0 and c[bul(j)] == 1):
a[j] = 1
b[p][q] = j + 1
q+=1
break
p+=1
'print(b)'
'kuralsızları gruplara yerlestirme'
for i in range(n):
if (a[i] == 0):
b[p][r] = i + 1
r+=1
if (r == 3):
p+=1
r = 0
for i in range(n//3):
print(b[i][0], b[i][1], b[i][2])
"""
print(b)
print(komsu," komsu")
print(c,"ccccccc")
print(a,"aaaaaaa")
"""
if __name__ == "__main__":
main()
``` | instruction | 0 | 82,883 | 17 | 165,766 |
Yes | output | 1 | 82,883 | 17 | 165,767 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A programming coach has n students to teach. We know that n is divisible by 3. Let's assume that all students are numbered from 1 to n, inclusive.
Before the university programming championship the coach wants to split all students into groups of three. For some pairs of students we know that they want to be on the same team. Besides, if the i-th student wants to be on the same team with the j-th one, then the j-th student wants to be on the same team with the i-th one. The coach wants the teams to show good results, so he wants the following condition to hold: if the i-th student wants to be on the same team with the j-th, then the i-th and the j-th students must be on the same team. Also, it is obvious that each student must be on exactly one team.
Help the coach and divide the teams the way he wants.
Input
The first line of the input contains integers n and m (3 ≤ n ≤ 48, <image>. Then follow m lines, each contains a pair of integers ai, bi (1 ≤ ai < bi ≤ n) — the pair ai, bi means that students with numbers ai and bi want to be on the same team.
It is guaranteed that n is divisible by 3. It is guaranteed that each pair ai, bi occurs in the input at most once.
Output
If the required division into teams doesn't exist, print number -1. Otherwise, print <image> lines. In each line print three integers xi, yi, zi (1 ≤ xi, yi, zi ≤ n) — the i-th team.
If there are multiple answers, you are allowed to print any of them.
Examples
Input
3 0
Output
3 2 1
Input
6 4
1 2
2 3
3 4
5 6
Output
-1
Input
3 3
1 2
2 3
1 3
Output
3 2 1
Submitted Solution:
```
N = 64
g = [[False]*N for _ in range(N)]
a = [[],[],[],[]]
c = set()
rd = lambda: list(map(int,input().split()))
def dfs(u):
x = [u]
c.add(u)
for v in I:
if g[u][v] and v not in c:
x += dfs(v)
return x
def fail():
print(-1)
exit()
n, m = rd()
for _ in range(m):
u, v = rd()
g[u][v] = g[v][u] = True
I = range(1, n+1)
for i in I:
if i not in c:
x = dfs(i)
l = len(x)
if l > 3: fail()
a[l] += [x]
for x in a[2]:
if a[1]: a[3] += [x+a[1].pop()]
else: fail()
if len(a[1])%3: fail()
while a[1]: a[3] += [a[1].pop()+a[1].pop()+a[1].pop()]
for x in a[3]: print(' '.join(map(str, x)))
``` | instruction | 0 | 82,884 | 17 | 165,768 |
Yes | output | 1 | 82,884 | 17 | 165,769 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A programming coach has n students to teach. We know that n is divisible by 3. Let's assume that all students are numbered from 1 to n, inclusive.
Before the university programming championship the coach wants to split all students into groups of three. For some pairs of students we know that they want to be on the same team. Besides, if the i-th student wants to be on the same team with the j-th one, then the j-th student wants to be on the same team with the i-th one. The coach wants the teams to show good results, so he wants the following condition to hold: if the i-th student wants to be on the same team with the j-th, then the i-th and the j-th students must be on the same team. Also, it is obvious that each student must be on exactly one team.
Help the coach and divide the teams the way he wants.
Input
The first line of the input contains integers n and m (3 ≤ n ≤ 48, <image>. Then follow m lines, each contains a pair of integers ai, bi (1 ≤ ai < bi ≤ n) — the pair ai, bi means that students with numbers ai and bi want to be on the same team.
It is guaranteed that n is divisible by 3. It is guaranteed that each pair ai, bi occurs in the input at most once.
Output
If the required division into teams doesn't exist, print number -1. Otherwise, print <image> lines. In each line print three integers xi, yi, zi (1 ≤ xi, yi, zi ≤ n) — the i-th team.
If there are multiple answers, you are allowed to print any of them.
Examples
Input
3 0
Output
3 2 1
Input
6 4
1 2
2 3
3 4
5 6
Output
-1
Input
3 3
1 2
2 3
1 3
Output
3 2 1
Submitted Solution:
```
def dfs(node):
global cnt,i
seen[node] = 1
cnt += 1
teams[i].append(node)
for each in arr[node]:
if seen[each] == 0:
dfs(each)
n,m = map(int,input().split())
arr = [list() for x in range(49)]
seen = [0]*49
teams = [list() for k in range(49)]
teams1 = list()
teams2 = list()
for i in range(m):
a,b = map(int,input().split())
arr[a].append(b)
arr[b].append(a)
for i in range(1,n+1):
if seen[i] == 0:
cnt = 0
dfs(i)
if cnt > 3:
print(-1)
break
if cnt == 2:
teams2.append(teams[i])
teams[i] = []
elif cnt == 1:
teams1.append(teams[i])
teams[i] =[]
else:
if len(teams2) > len(teams1):
print(-1)
else:
for each in teams:
if each:
x,y,z = each
print(x,y,z)
for i in range(min(len(teams2),len(teams1))):
x,y = teams2.pop()
z = teams1.pop()
print(x,y,z[0])
cnt = 0
for i in teams1:
cnt += 1
if i:
print(i[0],end = " ")
if cnt % 3 == 0:
print()
``` | instruction | 0 | 82,885 | 17 | 165,770 |
Yes | output | 1 | 82,885 | 17 | 165,771 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A programming coach has n students to teach. We know that n is divisible by 3. Let's assume that all students are numbered from 1 to n, inclusive.
Before the university programming championship the coach wants to split all students into groups of three. For some pairs of students we know that they want to be on the same team. Besides, if the i-th student wants to be on the same team with the j-th one, then the j-th student wants to be on the same team with the i-th one. The coach wants the teams to show good results, so he wants the following condition to hold: if the i-th student wants to be on the same team with the j-th, then the i-th and the j-th students must be on the same team. Also, it is obvious that each student must be on exactly one team.
Help the coach and divide the teams the way he wants.
Input
The first line of the input contains integers n and m (3 ≤ n ≤ 48, <image>. Then follow m lines, each contains a pair of integers ai, bi (1 ≤ ai < bi ≤ n) — the pair ai, bi means that students with numbers ai and bi want to be on the same team.
It is guaranteed that n is divisible by 3. It is guaranteed that each pair ai, bi occurs in the input at most once.
Output
If the required division into teams doesn't exist, print number -1. Otherwise, print <image> lines. In each line print three integers xi, yi, zi (1 ≤ xi, yi, zi ≤ n) — the i-th team.
If there are multiple answers, you are allowed to print any of them.
Examples
Input
3 0
Output
3 2 1
Input
6 4
1 2
2 3
3 4
5 6
Output
-1
Input
3 3
1 2
2 3
1 3
Output
3 2 1
Submitted Solution:
```
n, m=list(input().split())
n=int(n)
m=int(m)
l=[{i} for i in range(1, n+1)]
for i in range(m):
a, b=list(map(int, input().split()))
fa=False
fb=False
for j in l:
if a in j:
fa=True
x=l.index(j)
l[l.index(j)]|={a, b}
if b in j:
y=l.index(j)
fb=True
l[l.index(j)]|={a, b}
if fb and fa and y!=x:
l[x]|=l[y]
l.remove(l[y])
print(l)
ll=[len(i) for i in l]
f=True
for i in ll:
if i>3:
print(-1)
f=False
break
if f:
p=[]
if ll.count(1)<ll.count(2) or (ll.count(1)-ll.count(2))%3!=0: print(-1)
else:
while len(l)>0:
if all (len(i)==1 for i in l):
c=0
for i in range(len(l)):
if c in {0, 1}:
print(' '.join(map(str, l[i])), end=' ')
c+=1
else:
print(' '.join(map(str, l[i])))
c=0
break
else:
for i in l:
if len(i)==3:
print(' '.join(map(str, i)))
l.remove(i)
elif len(i)==2:
if len(p)==0:
p+=list(i)
l.remove(i)
if len(p)==1:
print(' '.join(map(str, p+list(i))))
l.remove(i)
p=[]
elif len(i)==1:
if len(p)==0:
p+=list(i)
l.remove(i)
elif len(p)==2:
print(' '.join(map(str, p+list(i))))
p=[]
l.remove(i)
``` | instruction | 0 | 82,886 | 17 | 165,772 |
No | output | 1 | 82,886 | 17 | 165,773 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A programming coach has n students to teach. We know that n is divisible by 3. Let's assume that all students are numbered from 1 to n, inclusive.
Before the university programming championship the coach wants to split all students into groups of three. For some pairs of students we know that they want to be on the same team. Besides, if the i-th student wants to be on the same team with the j-th one, then the j-th student wants to be on the same team with the i-th one. The coach wants the teams to show good results, so he wants the following condition to hold: if the i-th student wants to be on the same team with the j-th, then the i-th and the j-th students must be on the same team. Also, it is obvious that each student must be on exactly one team.
Help the coach and divide the teams the way he wants.
Input
The first line of the input contains integers n and m (3 ≤ n ≤ 48, <image>. Then follow m lines, each contains a pair of integers ai, bi (1 ≤ ai < bi ≤ n) — the pair ai, bi means that students with numbers ai and bi want to be on the same team.
It is guaranteed that n is divisible by 3. It is guaranteed that each pair ai, bi occurs in the input at most once.
Output
If the required division into teams doesn't exist, print number -1. Otherwise, print <image> lines. In each line print three integers xi, yi, zi (1 ≤ xi, yi, zi ≤ n) — the i-th team.
If there are multiple answers, you are allowed to print any of them.
Examples
Input
3 0
Output
3 2 1
Input
6 4
1 2
2 3
3 4
5 6
Output
-1
Input
3 3
1 2
2 3
1 3
Output
3 2 1
Submitted Solution:
```
n, m = [int(i) for i in input().split()]
students = {i: -1 for i in range(1, n + 1)}
teams = []
have_solution = True
for i in range(m):
a, b = [int(i) for i in input().split()]
if students[a] == -1 and students[b] == -1:
teams.append(set([a, b]))
students[a] = students[b] = len(teams) - 1
elif students[a] != -1 and students[b] != -1:
have_solution = False
break
else:
students[a] = students[b] = max(students[a], students[b])
teams[students[a]].add(a)
teams[students[a]].add(b)
if len(teams[students[a]]) > 3:
have_solution = False
break
if not have_solution:
print(-1)
else:
not_teams = []
for key, value in students.items():
if value == -1:
not_teams.append(key)
for i in teams:
tmp = list(i)
if len(i) != 3:
tmp.append(not_teams.pop())
tmp.sort()
print(' '.join(str(j) for j in tmp))
while not_teams:
print(not_teams.pop(), not_teams.pop(), not_teams.pop())
``` | instruction | 0 | 82,887 | 17 | 165,774 |
No | output | 1 | 82,887 | 17 | 165,775 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A programming coach has n students to teach. We know that n is divisible by 3. Let's assume that all students are numbered from 1 to n, inclusive.
Before the university programming championship the coach wants to split all students into groups of three. For some pairs of students we know that they want to be on the same team. Besides, if the i-th student wants to be on the same team with the j-th one, then the j-th student wants to be on the same team with the i-th one. The coach wants the teams to show good results, so he wants the following condition to hold: if the i-th student wants to be on the same team with the j-th, then the i-th and the j-th students must be on the same team. Also, it is obvious that each student must be on exactly one team.
Help the coach and divide the teams the way he wants.
Input
The first line of the input contains integers n and m (3 ≤ n ≤ 48, <image>. Then follow m lines, each contains a pair of integers ai, bi (1 ≤ ai < bi ≤ n) — the pair ai, bi means that students with numbers ai and bi want to be on the same team.
It is guaranteed that n is divisible by 3. It is guaranteed that each pair ai, bi occurs in the input at most once.
Output
If the required division into teams doesn't exist, print number -1. Otherwise, print <image> lines. In each line print three integers xi, yi, zi (1 ≤ xi, yi, zi ≤ n) — the i-th team.
If there are multiple answers, you are allowed to print any of them.
Examples
Input
3 0
Output
3 2 1
Input
6 4
1 2
2 3
3 4
5 6
Output
-1
Input
3 3
1 2
2 3
1 3
Output
3 2 1
Submitted Solution:
```
n, m = [int(i) for i in input().split()]
students = {i: -1 for i in range(1, n)}
teams = []
have_solution = True
for i in range(m):
a, b = [int(i) for i in input().split()]
if students[a] == -1 and students[b] == -1:
teams.append(set([a, b]))
students[a] = students[b] = len(teams) - 1
elif students[a] != -1 and students[b] != -1:
have_solution = False
break
else:
students[a] = students[b] = max(students[a], students[b])
teams[students[a]].add(a)
teams[students[a]].add(b)
if len(teams[students[a]]) > 3:
have_solution = False
break
if not have_solution:
print(-1)
else:
not_teams = []
for key, value in students.items():
if value != -1:
not_teams.append(key)
for i in teams:
if len(i) != 3:
print(not_teams.pop(), end=' ')
print(' '.join(str(j) for j in i))
``` | instruction | 0 | 82,888 | 17 | 165,776 |
No | output | 1 | 82,888 | 17 | 165,777 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A programming coach has n students to teach. We know that n is divisible by 3. Let's assume that all students are numbered from 1 to n, inclusive.
Before the university programming championship the coach wants to split all students into groups of three. For some pairs of students we know that they want to be on the same team. Besides, if the i-th student wants to be on the same team with the j-th one, then the j-th student wants to be on the same team with the i-th one. The coach wants the teams to show good results, so he wants the following condition to hold: if the i-th student wants to be on the same team with the j-th, then the i-th and the j-th students must be on the same team. Also, it is obvious that each student must be on exactly one team.
Help the coach and divide the teams the way he wants.
Input
The first line of the input contains integers n and m (3 ≤ n ≤ 48, <image>. Then follow m lines, each contains a pair of integers ai, bi (1 ≤ ai < bi ≤ n) — the pair ai, bi means that students with numbers ai and bi want to be on the same team.
It is guaranteed that n is divisible by 3. It is guaranteed that each pair ai, bi occurs in the input at most once.
Output
If the required division into teams doesn't exist, print number -1. Otherwise, print <image> lines. In each line print three integers xi, yi, zi (1 ≤ xi, yi, zi ≤ n) — the i-th team.
If there are multiple answers, you are allowed to print any of them.
Examples
Input
3 0
Output
3 2 1
Input
6 4
1 2
2 3
3 4
5 6
Output
-1
Input
3 3
1 2
2 3
1 3
Output
3 2 1
Submitted Solution:
```
n, m=list(input().split())
n=int(n)
m=int(m)
l=[{i} for i in range(1, n+1)]
for i in range(m):
a, b=list(map(int, input().split()))
fa=False
fb=False
for j in l:
if a in j:
fa=True
x=l.index(j)
l[l.index(j)]|={a, b}
if b in j:
y=l.index(j)
fb=True
l[l.index(j)]|={a, b}
if fb and fa and y!=x:
l[x]|=l[y]
l.remove(l[y])
y={tuple(i) for i in l}
l=[set(i) for i in y]
ll=[len(i) for i in l]
f=True
for i in ll:
if i>3:
print(-1)
f=False
break
print(l)
if f:
p=[]
if ll.count(1)<ll.count(2) or (ll.count(1)-ll.count(2))%3!=0: print(-1)
else:
while len(l)>0:
for i in l:
if len(i)==3:
print(' '.join(map(str, i)))
l.remove(i)
elif len(i)==1:
if len(p)==2:
print(' '.join(map(str, p+list(i))))
p=[]
l.remove(i)
elif len(i)==2:
if len(p)==0:
p+=list(i)
l.remove(i)
if all (len(i)==1 for i in l) and len(l)%3==0:
c=0
for i in range(len(l)):
if c in {0, 1}:
print(' '.join(map(str, l[i])), end=' ')
c+=1
else:
print(' '.join(map(str, l[i])))
c=0
break
``` | instruction | 0 | 82,889 | 17 | 165,778 |
No | output | 1 | 82,889 | 17 | 165,779 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One tradition of ACM-ICPC contests is that a team gets a balloon for every solved problem. We assume that the submission time doesn't matter and teams are sorted only by the number of balloons they have. It means that one's place is equal to the number of teams with more balloons, increased by 1. For example, if there are seven teams with more balloons, you get the eight place. Ties are allowed.
You should know that it's important to eat before a contest. If the number of balloons of a team is greater than the weight of this team, the team starts to float in the air together with their workstation. They eventually touch the ceiling, what is strictly forbidden by the rules. The team is then disqualified and isn't considered in the standings.
A contest has just finished. There are n teams, numbered 1 through n. The i-th team has ti balloons and weight wi. It's guaranteed that ti doesn't exceed wi so nobody floats initially.
Limak is a member of the first team. He doesn't like cheating and he would never steal balloons from other teams. Instead, he can give his balloons away to other teams, possibly making them float. Limak can give away zero or more balloons of his team. Obviously, he can't give away more balloons than his team initially has.
What is the best place Limak can get?
Input
The first line of the standard input contains one integer n (2 ≤ n ≤ 300 000) — the number of teams.
The i-th of n following lines contains two integers ti and wi (0 ≤ ti ≤ wi ≤ 1018) — respectively the number of balloons and the weight of the i-th team. Limak is a member of the first team.
Output
Print one integer denoting the best place Limak can get.
Examples
Input
8
20 1000
32 37
40 1000
45 50
16 16
16 16
14 1000
2 1000
Output
3
Input
7
4 4
4 4
4 4
4 4
4 4
4 4
5 5
Output
2
Input
7
14000000003 1000000000000000000
81000000000 88000000000
5000000000 7000000000
15000000000 39000000000
46000000000 51000000000
0 1000000000
0 0
Output
2
Note
In the first sample, Limak has 20 balloons initially. There are three teams with more balloons (32, 40 and 45 balloons), so Limak has the fourth place initially. One optimal strategy is:
1. Limak gives 6 balloons away to a team with 32 balloons and weight 37, which is just enough to make them fly. Unfortunately, Limak has only 14 balloons now and he would get the fifth place.
2. Limak gives 6 balloons away to a team with 45 balloons. Now they have 51 balloons and weight 50 so they fly and get disqualified.
3. Limak gives 1 balloon to each of two teams with 16 balloons initially.
4. Limak has 20 - 6 - 6 - 1 - 1 = 6 balloons.
5. There are three other teams left and their numbers of balloons are 40, 14 and 2.
6. Limak gets the third place because there are two teams with more balloons.
In the second sample, Limak has the second place and he can't improve it.
In the third sample, Limak has just enough balloons to get rid of teams 2, 3 and 5 (the teams with 81 000 000 000, 5 000 000 000 and 46 000 000 000 balloons respectively). With zero balloons left, he will get the second place (ex-aequo with team 6 and team 7). | instruction | 0 | 83,030 | 17 | 166,060 |
Tags: data structures, greedy
Correct Solution:
```
import heapq
from bisect import bisect_right
from collections import defaultdict
import os
import sys
from io import BytesIO, IOBase
from collections import defaultdict
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")
n=int(input())
res=[]
curr,we=map(int,input().split())
for j in range(n-1):
p,q=map(int,input().split())
res.append([p,q-p])
res.sort(reverse=True)
li=[]
j=0
ans=n
tot=0
while(j<n-1):
if res[j][0]>curr:
heapq.heappush(li,res[j][1]+1)
j+=1
else:
ans=min(ans,j+1-tot)
try:
p=li[0]
except:
p=float("inf")
if curr>=p:
curr-=p
heapq.heappop(li)
tot+=1
else:
break
while(curr>=0):
ans = min(ans, j + 1 - tot)
try:
p = li[0]
except:
p = float("inf")
if curr >= p:
curr -= p
heapq.heappop(li)
tot += 1
else:
break
print(ans)
``` | output | 1 | 83,030 | 17 | 166,061 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One tradition of ACM-ICPC contests is that a team gets a balloon for every solved problem. We assume that the submission time doesn't matter and teams are sorted only by the number of balloons they have. It means that one's place is equal to the number of teams with more balloons, increased by 1. For example, if there are seven teams with more balloons, you get the eight place. Ties are allowed.
You should know that it's important to eat before a contest. If the number of balloons of a team is greater than the weight of this team, the team starts to float in the air together with their workstation. They eventually touch the ceiling, what is strictly forbidden by the rules. The team is then disqualified and isn't considered in the standings.
A contest has just finished. There are n teams, numbered 1 through n. The i-th team has ti balloons and weight wi. It's guaranteed that ti doesn't exceed wi so nobody floats initially.
Limak is a member of the first team. He doesn't like cheating and he would never steal balloons from other teams. Instead, he can give his balloons away to other teams, possibly making them float. Limak can give away zero or more balloons of his team. Obviously, he can't give away more balloons than his team initially has.
What is the best place Limak can get?
Input
The first line of the standard input contains one integer n (2 ≤ n ≤ 300 000) — the number of teams.
The i-th of n following lines contains two integers ti and wi (0 ≤ ti ≤ wi ≤ 1018) — respectively the number of balloons and the weight of the i-th team. Limak is a member of the first team.
Output
Print one integer denoting the best place Limak can get.
Examples
Input
8
20 1000
32 37
40 1000
45 50
16 16
16 16
14 1000
2 1000
Output
3
Input
7
4 4
4 4
4 4
4 4
4 4
4 4
5 5
Output
2
Input
7
14000000003 1000000000000000000
81000000000 88000000000
5000000000 7000000000
15000000000 39000000000
46000000000 51000000000
0 1000000000
0 0
Output
2
Note
In the first sample, Limak has 20 balloons initially. There are three teams with more balloons (32, 40 and 45 balloons), so Limak has the fourth place initially. One optimal strategy is:
1. Limak gives 6 balloons away to a team with 32 balloons and weight 37, which is just enough to make them fly. Unfortunately, Limak has only 14 balloons now and he would get the fifth place.
2. Limak gives 6 balloons away to a team with 45 balloons. Now they have 51 balloons and weight 50 so they fly and get disqualified.
3. Limak gives 1 balloon to each of two teams with 16 balloons initially.
4. Limak has 20 - 6 - 6 - 1 - 1 = 6 balloons.
5. There are three other teams left and their numbers of balloons are 40, 14 and 2.
6. Limak gets the third place because there are two teams with more balloons.
In the second sample, Limak has the second place and he can't improve it.
In the third sample, Limak has just enough balloons to get rid of teams 2, 3 and 5 (the teams with 81 000 000 000, 5 000 000 000 and 46 000 000 000 balloons respectively). With zero balloons left, he will get the second place (ex-aequo with team 6 and team 7). | instruction | 0 | 83,031 | 17 | 166,062 |
Tags: data structures, greedy
Correct Solution:
```
#!/usr/bin/env python3
from sys import stdin,stdout
from bisect import *
from heapq import *
def ri():
return map(int, input().split())
n = int(input())
abw = [list(ri()) for i in range(n)]
a1 = abw[0][0]
a = [abw[i][0] for i in range(1,n)]
a.sort()
abw = abw[1:]
abw.sort(key=lambda e: e[0])
j = bisect(a,a1)
f = 0
mins = n - j
pq = [abw[i][1]-abw[i][0] for i in range(j, n-1)]
heapify(pq)
iiprv = j
while len(pq) and a1 > 0:
g = heappop(pq)
if a1 > g:
f += 1
a1 -= g+1
ii = bisect(a,a1)
for k in range(ii, iiprv):
heappush(pq, abw[k][1]-abw[k][0])
iiprv = ii
mins = min(mins, n-ii-f)
print(mins)
``` | output | 1 | 83,031 | 17 | 166,063 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One tradition of ACM-ICPC contests is that a team gets a balloon for every solved problem. We assume that the submission time doesn't matter and teams are sorted only by the number of balloons they have. It means that one's place is equal to the number of teams with more balloons, increased by 1. For example, if there are seven teams with more balloons, you get the eight place. Ties are allowed.
You should know that it's important to eat before a contest. If the number of balloons of a team is greater than the weight of this team, the team starts to float in the air together with their workstation. They eventually touch the ceiling, what is strictly forbidden by the rules. The team is then disqualified and isn't considered in the standings.
A contest has just finished. There are n teams, numbered 1 through n. The i-th team has ti balloons and weight wi. It's guaranteed that ti doesn't exceed wi so nobody floats initially.
Limak is a member of the first team. He doesn't like cheating and he would never steal balloons from other teams. Instead, he can give his balloons away to other teams, possibly making them float. Limak can give away zero or more balloons of his team. Obviously, he can't give away more balloons than his team initially has.
What is the best place Limak can get?
Input
The first line of the standard input contains one integer n (2 ≤ n ≤ 300 000) — the number of teams.
The i-th of n following lines contains two integers ti and wi (0 ≤ ti ≤ wi ≤ 1018) — respectively the number of balloons and the weight of the i-th team. Limak is a member of the first team.
Output
Print one integer denoting the best place Limak can get.
Examples
Input
8
20 1000
32 37
40 1000
45 50
16 16
16 16
14 1000
2 1000
Output
3
Input
7
4 4
4 4
4 4
4 4
4 4
4 4
5 5
Output
2
Input
7
14000000003 1000000000000000000
81000000000 88000000000
5000000000 7000000000
15000000000 39000000000
46000000000 51000000000
0 1000000000
0 0
Output
2
Note
In the first sample, Limak has 20 balloons initially. There are three teams with more balloons (32, 40 and 45 balloons), so Limak has the fourth place initially. One optimal strategy is:
1. Limak gives 6 balloons away to a team with 32 balloons and weight 37, which is just enough to make them fly. Unfortunately, Limak has only 14 balloons now and he would get the fifth place.
2. Limak gives 6 balloons away to a team with 45 balloons. Now they have 51 balloons and weight 50 so they fly and get disqualified.
3. Limak gives 1 balloon to each of two teams with 16 balloons initially.
4. Limak has 20 - 6 - 6 - 1 - 1 = 6 balloons.
5. There are three other teams left and their numbers of balloons are 40, 14 and 2.
6. Limak gets the third place because there are two teams with more balloons.
In the second sample, Limak has the second place and he can't improve it.
In the third sample, Limak has just enough balloons to get rid of teams 2, 3 and 5 (the teams with 81 000 000 000, 5 000 000 000 and 46 000 000 000 balloons respectively). With zero balloons left, he will get the second place (ex-aequo with team 6 and team 7). | instruction | 0 | 83,032 | 17 | 166,064 |
Tags: data structures, greedy
Correct Solution:
```
'''input
8
20 1000
32 37
40 1000
45 50
16 16
16 16
14 1000
2 1000
'''
import heapq
from bisect import bisect
inf = 10**18 + 2
def rints():
return list(map(int, input().split()))
def ri():
return int(input())
def bin_search(arr, pred, lo=0, hi = None):
if hi is None:
hi = len(arr)
while lo < hi :
mid = (lo+hi) // 2
if pred(arr[mid]):
hi = mid
else:
lo = mid + 1
return lo
n = ri()
score, _ = rints()
teams = []
for _ in range(n-1):
t, w = rints()
teams.append((w-t + 1, t, w))
teams.sort(key = lambda x : x[1])
# print(teams)
def solve(score):
idx = bin_search(teams, lambda x : x[1] > score)
best = pos = len(teams) - idx + 1
# print(teams[idx:], pos)
ahead = teams[idx:]
behind = teams[:idx]
heapq.heapify(ahead)
# print(ahead)
while ahead and score >= ahead[0][0]:
score -= heapq.heappop(ahead)[0]
pos -= 1
while behind and behind[-1][1] > score:
heapq.heappush(ahead, behind.pop())
pos += 1
# print(score, pos)
best = min(best, pos)
return best
print(solve(score))
``` | output | 1 | 83,032 | 17 | 166,065 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One tradition of ACM-ICPC contests is that a team gets a balloon for every solved problem. We assume that the submission time doesn't matter and teams are sorted only by the number of balloons they have. It means that one's place is equal to the number of teams with more balloons, increased by 1. For example, if there are seven teams with more balloons, you get the eight place. Ties are allowed.
You should know that it's important to eat before a contest. If the number of balloons of a team is greater than the weight of this team, the team starts to float in the air together with their workstation. They eventually touch the ceiling, what is strictly forbidden by the rules. The team is then disqualified and isn't considered in the standings.
A contest has just finished. There are n teams, numbered 1 through n. The i-th team has ti balloons and weight wi. It's guaranteed that ti doesn't exceed wi so nobody floats initially.
Limak is a member of the first team. He doesn't like cheating and he would never steal balloons from other teams. Instead, he can give his balloons away to other teams, possibly making them float. Limak can give away zero or more balloons of his team. Obviously, he can't give away more balloons than his team initially has.
What is the best place Limak can get?
Input
The first line of the standard input contains one integer n (2 ≤ n ≤ 300 000) — the number of teams.
The i-th of n following lines contains two integers ti and wi (0 ≤ ti ≤ wi ≤ 1018) — respectively the number of balloons and the weight of the i-th team. Limak is a member of the first team.
Output
Print one integer denoting the best place Limak can get.
Examples
Input
8
20 1000
32 37
40 1000
45 50
16 16
16 16
14 1000
2 1000
Output
3
Input
7
4 4
4 4
4 4
4 4
4 4
4 4
5 5
Output
2
Input
7
14000000003 1000000000000000000
81000000000 88000000000
5000000000 7000000000
15000000000 39000000000
46000000000 51000000000
0 1000000000
0 0
Output
2
Note
In the first sample, Limak has 20 balloons initially. There are three teams with more balloons (32, 40 and 45 balloons), so Limak has the fourth place initially. One optimal strategy is:
1. Limak gives 6 balloons away to a team with 32 balloons and weight 37, which is just enough to make them fly. Unfortunately, Limak has only 14 balloons now and he would get the fifth place.
2. Limak gives 6 balloons away to a team with 45 balloons. Now they have 51 balloons and weight 50 so they fly and get disqualified.
3. Limak gives 1 balloon to each of two teams with 16 balloons initially.
4. Limak has 20 - 6 - 6 - 1 - 1 = 6 balloons.
5. There are three other teams left and their numbers of balloons are 40, 14 and 2.
6. Limak gets the third place because there are two teams with more balloons.
In the second sample, Limak has the second place and he can't improve it.
In the third sample, Limak has just enough balloons to get rid of teams 2, 3 and 5 (the teams with 81 000 000 000, 5 000 000 000 and 46 000 000 000 balloons respectively). With zero balloons left, he will get the second place (ex-aequo with team 6 and team 7). | instruction | 0 | 83,033 | 17 | 166,066 |
Tags: data structures, greedy
Correct Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import random
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
import threading
from collections import defaultdict
#threading.stack_size(10**8)
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
#sys.setrecursionlimit(300000)
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")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=0, func=lambda a, b: max(a , b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b:a + b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] < key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] > k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
n=int(input())
a,b=map(int,input().split())
l=[]
bal=[]
rev=[]
for i in range(1,n):
a1,b1=map(int,input().split())
if a1>a:
l.append((b1-a1+1,a1,b1))
bal.append(a1)
else:
rev.append((a1,b1))
bal.sort()
rev.sort(reverse=True)
heapq.heapify(l)
ans=len(bal)+1
ma=ans
#print(ma,bal)
t=0
f=0
while(a>0):
if len(l)==0:
break
e=heapq.heappop(l)
#print(e,bal,a)
if a<e[0]:
break
else:
a+=-e[0]
t+=1
while(f<len(rev) and rev[f][0]>a):
heapq.heappush(l,(rev[f][1]-rev[f][0]+1,rev[f][0],rev[f][1]))
bal.append(rev[f][0])
f+=1
ans = len(bal)+ 1-t
ma=min(ma,ans)
print(ma)
``` | output | 1 | 83,033 | 17 | 166,067 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One tradition of ACM-ICPC contests is that a team gets a balloon for every solved problem. We assume that the submission time doesn't matter and teams are sorted only by the number of balloons they have. It means that one's place is equal to the number of teams with more balloons, increased by 1. For example, if there are seven teams with more balloons, you get the eight place. Ties are allowed.
You should know that it's important to eat before a contest. If the number of balloons of a team is greater than the weight of this team, the team starts to float in the air together with their workstation. They eventually touch the ceiling, what is strictly forbidden by the rules. The team is then disqualified and isn't considered in the standings.
A contest has just finished. There are n teams, numbered 1 through n. The i-th team has ti balloons and weight wi. It's guaranteed that ti doesn't exceed wi so nobody floats initially.
Limak is a member of the first team. He doesn't like cheating and he would never steal balloons from other teams. Instead, he can give his balloons away to other teams, possibly making them float. Limak can give away zero or more balloons of his team. Obviously, he can't give away more balloons than his team initially has.
What is the best place Limak can get?
Input
The first line of the standard input contains one integer n (2 ≤ n ≤ 300 000) — the number of teams.
The i-th of n following lines contains two integers ti and wi (0 ≤ ti ≤ wi ≤ 1018) — respectively the number of balloons and the weight of the i-th team. Limak is a member of the first team.
Output
Print one integer denoting the best place Limak can get.
Examples
Input
8
20 1000
32 37
40 1000
45 50
16 16
16 16
14 1000
2 1000
Output
3
Input
7
4 4
4 4
4 4
4 4
4 4
4 4
5 5
Output
2
Input
7
14000000003 1000000000000000000
81000000000 88000000000
5000000000 7000000000
15000000000 39000000000
46000000000 51000000000
0 1000000000
0 0
Output
2
Note
In the first sample, Limak has 20 balloons initially. There are three teams with more balloons (32, 40 and 45 balloons), so Limak has the fourth place initially. One optimal strategy is:
1. Limak gives 6 balloons away to a team with 32 balloons and weight 37, which is just enough to make them fly. Unfortunately, Limak has only 14 balloons now and he would get the fifth place.
2. Limak gives 6 balloons away to a team with 45 balloons. Now they have 51 balloons and weight 50 so they fly and get disqualified.
3. Limak gives 1 balloon to each of two teams with 16 balloons initially.
4. Limak has 20 - 6 - 6 - 1 - 1 = 6 balloons.
5. There are three other teams left and their numbers of balloons are 40, 14 and 2.
6. Limak gets the third place because there are two teams with more balloons.
In the second sample, Limak has the second place and he can't improve it.
In the third sample, Limak has just enough balloons to get rid of teams 2, 3 and 5 (the teams with 81 000 000 000, 5 000 000 000 and 46 000 000 000 balloons respectively). With zero balloons left, he will get the second place (ex-aequo with team 6 and team 7). | instruction | 0 | 83,034 | 17 | 166,068 |
Tags: data structures, greedy
Correct Solution:
```
import os
import io
import heapq
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
#inp = sys.stdin.readlines()
ii = 0
out = []
n = int(input())
ii += 1
pq = [] # PriorityQueue()
diffs = []
arr = []
limak = int(input().split()[0])
ii += 1
for team in range(n - 1):
b, w = map(int, input().split())
ii += 1
# print(b, w, w - b + 1)
arr.append(b)
diffs.append((w - b) + 1)
arr1 = sorted(arr, reverse=True)
sortingbelike = zip(arr, diffs)
howtobeorz = sorted(sortingbelike)
#print(howtobeorz)
diffs = [x for _, x in howtobeorz]
diffs1 = diffs[::-1]
#print(diffs1, arr1)
# print(arr, arr1, diffs, diffs1)
# pos_limak = binsearch(arr, limak)
# last = pos_limak
#
# best = n - pos_limak
place = 0
removed = 0
best = 300000
#
# for i in range(n-best,n-1):
# pq.put(diffs[i])
# # print(diffs[i])
while True:
while place < n - 1 and arr1[place] > limak:
# print(diffs1[place - 1])
place += 1
heapq.heappush(pq, diffs1[place - 1])
# print(place, removed, arr1, diffs1)
# print(best, pq)
best = min(best, place - removed + 1)
if len(pq) == 0:
break
curr = heapq.heappop(pq)
if curr > limak:
# print(curr, limak)
break
limak -= curr
removed += 1
print(best)
# sys.stdout.write(str(best))
``` | output | 1 | 83,034 | 17 | 166,069 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One tradition of ACM-ICPC contests is that a team gets a balloon for every solved problem. We assume that the submission time doesn't matter and teams are sorted only by the number of balloons they have. It means that one's place is equal to the number of teams with more balloons, increased by 1. For example, if there are seven teams with more balloons, you get the eight place. Ties are allowed.
You should know that it's important to eat before a contest. If the number of balloons of a team is greater than the weight of this team, the team starts to float in the air together with their workstation. They eventually touch the ceiling, what is strictly forbidden by the rules. The team is then disqualified and isn't considered in the standings.
A contest has just finished. There are n teams, numbered 1 through n. The i-th team has ti balloons and weight wi. It's guaranteed that ti doesn't exceed wi so nobody floats initially.
Limak is a member of the first team. He doesn't like cheating and he would never steal balloons from other teams. Instead, he can give his balloons away to other teams, possibly making them float. Limak can give away zero or more balloons of his team. Obviously, he can't give away more balloons than his team initially has.
What is the best place Limak can get?
Input
The first line of the standard input contains one integer n (2 ≤ n ≤ 300 000) — the number of teams.
The i-th of n following lines contains two integers ti and wi (0 ≤ ti ≤ wi ≤ 1018) — respectively the number of balloons and the weight of the i-th team. Limak is a member of the first team.
Output
Print one integer denoting the best place Limak can get.
Examples
Input
8
20 1000
32 37
40 1000
45 50
16 16
16 16
14 1000
2 1000
Output
3
Input
7
4 4
4 4
4 4
4 4
4 4
4 4
5 5
Output
2
Input
7
14000000003 1000000000000000000
81000000000 88000000000
5000000000 7000000000
15000000000 39000000000
46000000000 51000000000
0 1000000000
0 0
Output
2
Note
In the first sample, Limak has 20 balloons initially. There are three teams with more balloons (32, 40 and 45 balloons), so Limak has the fourth place initially. One optimal strategy is:
1. Limak gives 6 balloons away to a team with 32 balloons and weight 37, which is just enough to make them fly. Unfortunately, Limak has only 14 balloons now and he would get the fifth place.
2. Limak gives 6 balloons away to a team with 45 balloons. Now they have 51 balloons and weight 50 so they fly and get disqualified.
3. Limak gives 1 balloon to each of two teams with 16 balloons initially.
4. Limak has 20 - 6 - 6 - 1 - 1 = 6 balloons.
5. There are three other teams left and their numbers of balloons are 40, 14 and 2.
6. Limak gets the third place because there are two teams with more balloons.
In the second sample, Limak has the second place and he can't improve it.
In the third sample, Limak has just enough balloons to get rid of teams 2, 3 and 5 (the teams with 81 000 000 000, 5 000 000 000 and 46 000 000 000 balloons respectively). With zero balloons left, he will get the second place (ex-aequo with team 6 and team 7). | instruction | 0 | 83,035 | 17 | 166,070 |
Tags: data structures, greedy
Correct Solution:
```
#!/usr/bin/env python3
from sys import stdin,stdout
from bisect import *
from heapq import *
def ri():
return map(int, input().split())
n = int(input())
abw = [list(ri()) for i in range(n)]
a1 = abw[0][0]
a = [abw[i][0] for i in range(1,n)]
a.sort()
abw = abw[1:]
abw.sort(key=lambda e: e[0])
for j in range(len(abw)):
if abw[j][0] > a1:
break
else:
j += 1
f = 0
mins = n - j
pq = [abw[i][1]-abw[i][0] for i in range(j, n-1)]
heapify(pq)
iiprv = j
while len(pq) and a1 > 0:
g = heappop(pq)
if a1 > g:
f += 1
a1 -= g+1
ii = bisect(a,a1)
for k in range(ii, iiprv):
heappush(pq, abw[k][1]-abw[k][0])
iiprv = ii
mins = min(mins, n-ii-f)
print(mins)
``` | output | 1 | 83,035 | 17 | 166,071 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One tradition of ACM-ICPC contests is that a team gets a balloon for every solved problem. We assume that the submission time doesn't matter and teams are sorted only by the number of balloons they have. It means that one's place is equal to the number of teams with more balloons, increased by 1. For example, if there are seven teams with more balloons, you get the eight place. Ties are allowed.
You should know that it's important to eat before a contest. If the number of balloons of a team is greater than the weight of this team, the team starts to float in the air together with their workstation. They eventually touch the ceiling, what is strictly forbidden by the rules. The team is then disqualified and isn't considered in the standings.
A contest has just finished. There are n teams, numbered 1 through n. The i-th team has ti balloons and weight wi. It's guaranteed that ti doesn't exceed wi so nobody floats initially.
Limak is a member of the first team. He doesn't like cheating and he would never steal balloons from other teams. Instead, he can give his balloons away to other teams, possibly making them float. Limak can give away zero or more balloons of his team. Obviously, he can't give away more balloons than his team initially has.
What is the best place Limak can get?
Input
The first line of the standard input contains one integer n (2 ≤ n ≤ 300 000) — the number of teams.
The i-th of n following lines contains two integers ti and wi (0 ≤ ti ≤ wi ≤ 1018) — respectively the number of balloons and the weight of the i-th team. Limak is a member of the first team.
Output
Print one integer denoting the best place Limak can get.
Examples
Input
8
20 1000
32 37
40 1000
45 50
16 16
16 16
14 1000
2 1000
Output
3
Input
7
4 4
4 4
4 4
4 4
4 4
4 4
5 5
Output
2
Input
7
14000000003 1000000000000000000
81000000000 88000000000
5000000000 7000000000
15000000000 39000000000
46000000000 51000000000
0 1000000000
0 0
Output
2
Note
In the first sample, Limak has 20 balloons initially. There are three teams with more balloons (32, 40 and 45 balloons), so Limak has the fourth place initially. One optimal strategy is:
1. Limak gives 6 balloons away to a team with 32 balloons and weight 37, which is just enough to make them fly. Unfortunately, Limak has only 14 balloons now and he would get the fifth place.
2. Limak gives 6 balloons away to a team with 45 balloons. Now they have 51 balloons and weight 50 so they fly and get disqualified.
3. Limak gives 1 balloon to each of two teams with 16 balloons initially.
4. Limak has 20 - 6 - 6 - 1 - 1 = 6 balloons.
5. There are three other teams left and their numbers of balloons are 40, 14 and 2.
6. Limak gets the third place because there are two teams with more balloons.
In the second sample, Limak has the second place and he can't improve it.
In the third sample, Limak has just enough balloons to get rid of teams 2, 3 and 5 (the teams with 81 000 000 000, 5 000 000 000 and 46 000 000 000 balloons respectively). With zero balloons left, he will get the second place (ex-aequo with team 6 and team 7). | instruction | 0 | 83,036 | 17 | 166,072 |
Tags: data structures, greedy
Correct Solution:
```
import os
import sys
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)
input = lambda: sys.stdin.readline().rstrip("\r\n")
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 Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
INF = 10 ** 18
MOD = 10**9+7
from fractions import Fraction
import heapq as hp
Ri = lambda : [int(x) for x in sys.stdin.readline().split()]
ri = lambda : sys.stdin.readline().strip()
n = int(ri())
lis = []
a,b = -1,-1
for i in range(n):
ta,tb = Ri()
if i == 0:
a,b = ta,tb
else:
lis.append([ta,tb, i])
lis.sort(key = lambda x : x[0])
cnt = 0
i = 0
for i in range(len(lis)):
if lis[i][0] <= a:
cnt+=1
else:
break
if lis[i][0] <= a:
i+=1
curplace = n-(cnt+1)+1
place = curplace
# print(place)
h = []; hp.heapify(h)
for j in range(i, len(lis)):
hp.heappush(h, lis[j][1]- lis[j][0] + 1)
i-=1
while len(h) > 0:
temp = hp.heappop(h)
if temp <= a:
a-=temp
curplace-=1
else:
break
while i >=0 and lis[i][0] > a:
hp.heappush(h, lis[i][1]-lis[i][0] + 1)
curplace+=1
i-=1
# print(curplace)
place = min(place, curplace)
print(place)
``` | output | 1 | 83,036 | 17 | 166,073 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One tradition of ACM-ICPC contests is that a team gets a balloon for every solved problem. We assume that the submission time doesn't matter and teams are sorted only by the number of balloons they have. It means that one's place is equal to the number of teams with more balloons, increased by 1. For example, if there are seven teams with more balloons, you get the eight place. Ties are allowed.
You should know that it's important to eat before a contest. If the number of balloons of a team is greater than the weight of this team, the team starts to float in the air together with their workstation. They eventually touch the ceiling, what is strictly forbidden by the rules. The team is then disqualified and isn't considered in the standings.
A contest has just finished. There are n teams, numbered 1 through n. The i-th team has ti balloons and weight wi. It's guaranteed that ti doesn't exceed wi so nobody floats initially.
Limak is a member of the first team. He doesn't like cheating and he would never steal balloons from other teams. Instead, he can give his balloons away to other teams, possibly making them float. Limak can give away zero or more balloons of his team. Obviously, he can't give away more balloons than his team initially has.
What is the best place Limak can get?
Input
The first line of the standard input contains one integer n (2 ≤ n ≤ 300 000) — the number of teams.
The i-th of n following lines contains two integers ti and wi (0 ≤ ti ≤ wi ≤ 1018) — respectively the number of balloons and the weight of the i-th team. Limak is a member of the first team.
Output
Print one integer denoting the best place Limak can get.
Examples
Input
8
20 1000
32 37
40 1000
45 50
16 16
16 16
14 1000
2 1000
Output
3
Input
7
4 4
4 4
4 4
4 4
4 4
4 4
5 5
Output
2
Input
7
14000000003 1000000000000000000
81000000000 88000000000
5000000000 7000000000
15000000000 39000000000
46000000000 51000000000
0 1000000000
0 0
Output
2
Note
In the first sample, Limak has 20 balloons initially. There are three teams with more balloons (32, 40 and 45 balloons), so Limak has the fourth place initially. One optimal strategy is:
1. Limak gives 6 balloons away to a team with 32 balloons and weight 37, which is just enough to make them fly. Unfortunately, Limak has only 14 balloons now and he would get the fifth place.
2. Limak gives 6 balloons away to a team with 45 balloons. Now they have 51 balloons and weight 50 so they fly and get disqualified.
3. Limak gives 1 balloon to each of two teams with 16 balloons initially.
4. Limak has 20 - 6 - 6 - 1 - 1 = 6 balloons.
5. There are three other teams left and their numbers of balloons are 40, 14 and 2.
6. Limak gets the third place because there are two teams with more balloons.
In the second sample, Limak has the second place and he can't improve it.
In the third sample, Limak has just enough balloons to get rid of teams 2, 3 and 5 (the teams with 81 000 000 000, 5 000 000 000 and 46 000 000 000 balloons respectively). With zero balloons left, he will get the second place (ex-aequo with team 6 and team 7). | instruction | 0 | 83,037 | 17 | 166,074 |
Tags: data structures, greedy
Correct Solution:
```
import os
import sys
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)
input = lambda: sys.stdin.readline().rstrip("\r\n")
##########################################################
#print('%d %d' %ans)
from collections import Counter
import math
#for _ in range(int(input())):
#for i in range(int(input())):
import math
#for i in range(int(input())):
import heapq
import sys
from math import gcd
n=int(input())
b,w=map(int,input().split())
heap=[]
heap1=[]
for _ in range(n-1):
ba,w=map(int,input().split())
if ba>b:
heap.append(w-ba+1)
else:
heap1.append((-1*ba,w))
heapq.heapify(heap)
heapq.heapify(heap1)
ans=len(heap)+1
while heap:
need=heap[0]
if need>b:
break
b-=need
heapq.heappop(heap)
while heap1 and -1*heap1[0][0]>b:
t,w=heapq.heappop(heap1)
t=-1*t
w=w
heapq.heappush(heap,w+1-t)
ans=min(ans,len(heap)+1)
print(ans)
``` | output | 1 | 83,037 | 17 | 166,075 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One tradition of ACM-ICPC contests is that a team gets a balloon for every solved problem. We assume that the submission time doesn't matter and teams are sorted only by the number of balloons they have. It means that one's place is equal to the number of teams with more balloons, increased by 1. For example, if there are seven teams with more balloons, you get the eight place. Ties are allowed.
You should know that it's important to eat before a contest. If the number of balloons of a team is greater than the weight of this team, the team starts to float in the air together with their workstation. They eventually touch the ceiling, what is strictly forbidden by the rules. The team is then disqualified and isn't considered in the standings.
A contest has just finished. There are n teams, numbered 1 through n. The i-th team has ti balloons and weight wi. It's guaranteed that ti doesn't exceed wi so nobody floats initially.
Limak is a member of the first team. He doesn't like cheating and he would never steal balloons from other teams. Instead, he can give his balloons away to other teams, possibly making them float. Limak can give away zero or more balloons of his team. Obviously, he can't give away more balloons than his team initially has.
What is the best place Limak can get?
Input
The first line of the standard input contains one integer n (2 ≤ n ≤ 300 000) — the number of teams.
The i-th of n following lines contains two integers ti and wi (0 ≤ ti ≤ wi ≤ 1018) — respectively the number of balloons and the weight of the i-th team. Limak is a member of the first team.
Output
Print one integer denoting the best place Limak can get.
Examples
Input
8
20 1000
32 37
40 1000
45 50
16 16
16 16
14 1000
2 1000
Output
3
Input
7
4 4
4 4
4 4
4 4
4 4
4 4
5 5
Output
2
Input
7
14000000003 1000000000000000000
81000000000 88000000000
5000000000 7000000000
15000000000 39000000000
46000000000 51000000000
0 1000000000
0 0
Output
2
Note
In the first sample, Limak has 20 balloons initially. There are three teams with more balloons (32, 40 and 45 balloons), so Limak has the fourth place initially. One optimal strategy is:
1. Limak gives 6 balloons away to a team with 32 balloons and weight 37, which is just enough to make them fly. Unfortunately, Limak has only 14 balloons now and he would get the fifth place.
2. Limak gives 6 balloons away to a team with 45 balloons. Now they have 51 balloons and weight 50 so they fly and get disqualified.
3. Limak gives 1 balloon to each of two teams with 16 balloons initially.
4. Limak has 20 - 6 - 6 - 1 - 1 = 6 balloons.
5. There are three other teams left and their numbers of balloons are 40, 14 and 2.
6. Limak gets the third place because there are two teams with more balloons.
In the second sample, Limak has the second place and he can't improve it.
In the third sample, Limak has just enough balloons to get rid of teams 2, 3 and 5 (the teams with 81 000 000 000, 5 000 000 000 and 46 000 000 000 balloons respectively). With zero balloons left, he will get the second place (ex-aequo with team 6 and team 7).
Submitted Solution:
```
import sys
input = sys.stdin.readline
from heapq import heappop, heappush, heapify
n = int(input())
b, _ = map(int, input().split())
heap = []
heap2 = []
for _ in range(n-1):
t, w = map(int, input().split())
if t > b:
heap.append(w + 1 - t)
else:
heap2.append((-t, -w))
heapify(heap)
heapify(heap2)
ans = len(heap) + 1
while heap:
need = heap[0]
if need > b:
break
b -= need
heappop(heap)
while heap2 and -heap2[0][0] > b:
t, w = heappop(heap2)
t, w = -t, -w
heappush(heap, w + 1 - t)
cur = len(heap) + 1
ans = min(ans, cur)
print(ans)
``` | instruction | 0 | 83,038 | 17 | 166,076 |
Yes | output | 1 | 83,038 | 17 | 166,077 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One tradition of ACM-ICPC contests is that a team gets a balloon for every solved problem. We assume that the submission time doesn't matter and teams are sorted only by the number of balloons they have. It means that one's place is equal to the number of teams with more balloons, increased by 1. For example, if there are seven teams with more balloons, you get the eight place. Ties are allowed.
You should know that it's important to eat before a contest. If the number of balloons of a team is greater than the weight of this team, the team starts to float in the air together with their workstation. They eventually touch the ceiling, what is strictly forbidden by the rules. The team is then disqualified and isn't considered in the standings.
A contest has just finished. There are n teams, numbered 1 through n. The i-th team has ti balloons and weight wi. It's guaranteed that ti doesn't exceed wi so nobody floats initially.
Limak is a member of the first team. He doesn't like cheating and he would never steal balloons from other teams. Instead, he can give his balloons away to other teams, possibly making them float. Limak can give away zero or more balloons of his team. Obviously, he can't give away more balloons than his team initially has.
What is the best place Limak can get?
Input
The first line of the standard input contains one integer n (2 ≤ n ≤ 300 000) — the number of teams.
The i-th of n following lines contains two integers ti and wi (0 ≤ ti ≤ wi ≤ 1018) — respectively the number of balloons and the weight of the i-th team. Limak is a member of the first team.
Output
Print one integer denoting the best place Limak can get.
Examples
Input
8
20 1000
32 37
40 1000
45 50
16 16
16 16
14 1000
2 1000
Output
3
Input
7
4 4
4 4
4 4
4 4
4 4
4 4
5 5
Output
2
Input
7
14000000003 1000000000000000000
81000000000 88000000000
5000000000 7000000000
15000000000 39000000000
46000000000 51000000000
0 1000000000
0 0
Output
2
Note
In the first sample, Limak has 20 balloons initially. There are three teams with more balloons (32, 40 and 45 balloons), so Limak has the fourth place initially. One optimal strategy is:
1. Limak gives 6 balloons away to a team with 32 balloons and weight 37, which is just enough to make them fly. Unfortunately, Limak has only 14 balloons now and he would get the fifth place.
2. Limak gives 6 balloons away to a team with 45 balloons. Now they have 51 balloons and weight 50 so they fly and get disqualified.
3. Limak gives 1 balloon to each of two teams with 16 balloons initially.
4. Limak has 20 - 6 - 6 - 1 - 1 = 6 balloons.
5. There are three other teams left and their numbers of balloons are 40, 14 and 2.
6. Limak gets the third place because there are two teams with more balloons.
In the second sample, Limak has the second place and he can't improve it.
In the third sample, Limak has just enough balloons to get rid of teams 2, 3 and 5 (the teams with 81 000 000 000, 5 000 000 000 and 46 000 000 000 balloons respectively). With zero balloons left, he will get the second place (ex-aequo with team 6 and team 7).
Submitted Solution:
```
import sys
input = sys.stdin.readline
n = int(input())
limak = list(map(int, input().split()))
import heapq
pq = []
arr = []
for i in range(1, n):
ti, wi = map(int, input().split())
if ti > limak[0]:
heapq.heappush(pq, wi - ti + 1)
else:
arr.append((ti, wi - ti + 1))
arr.sort(reverse=True)
rem = limak[0]
best = len(pq) + 1
cur = len(pq) + 1
idx = 0
while rem and pq:
nxt = pq[0]
heapq.heappop(pq)
if rem < nxt:
break
rem -= nxt
cur -= 1
while idx < len(arr) and arr[idx][0] > rem:
heapq.heappush(pq, arr[idx][1])
idx += 1
cur += 1
best = min(best, cur)
print(best)
``` | instruction | 0 | 83,039 | 17 | 166,078 |
Yes | output | 1 | 83,039 | 17 | 166,079 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One tradition of ACM-ICPC contests is that a team gets a balloon for every solved problem. We assume that the submission time doesn't matter and teams are sorted only by the number of balloons they have. It means that one's place is equal to the number of teams with more balloons, increased by 1. For example, if there are seven teams with more balloons, you get the eight place. Ties are allowed.
You should know that it's important to eat before a contest. If the number of balloons of a team is greater than the weight of this team, the team starts to float in the air together with their workstation. They eventually touch the ceiling, what is strictly forbidden by the rules. The team is then disqualified and isn't considered in the standings.
A contest has just finished. There are n teams, numbered 1 through n. The i-th team has ti balloons and weight wi. It's guaranteed that ti doesn't exceed wi so nobody floats initially.
Limak is a member of the first team. He doesn't like cheating and he would never steal balloons from other teams. Instead, he can give his balloons away to other teams, possibly making them float. Limak can give away zero or more balloons of his team. Obviously, he can't give away more balloons than his team initially has.
What is the best place Limak can get?
Input
The first line of the standard input contains one integer n (2 ≤ n ≤ 300 000) — the number of teams.
The i-th of n following lines contains two integers ti and wi (0 ≤ ti ≤ wi ≤ 1018) — respectively the number of balloons and the weight of the i-th team. Limak is a member of the first team.
Output
Print one integer denoting the best place Limak can get.
Examples
Input
8
20 1000
32 37
40 1000
45 50
16 16
16 16
14 1000
2 1000
Output
3
Input
7
4 4
4 4
4 4
4 4
4 4
4 4
5 5
Output
2
Input
7
14000000003 1000000000000000000
81000000000 88000000000
5000000000 7000000000
15000000000 39000000000
46000000000 51000000000
0 1000000000
0 0
Output
2
Note
In the first sample, Limak has 20 balloons initially. There are three teams with more balloons (32, 40 and 45 balloons), so Limak has the fourth place initially. One optimal strategy is:
1. Limak gives 6 balloons away to a team with 32 balloons and weight 37, which is just enough to make them fly. Unfortunately, Limak has only 14 balloons now and he would get the fifth place.
2. Limak gives 6 balloons away to a team with 45 balloons. Now they have 51 balloons and weight 50 so they fly and get disqualified.
3. Limak gives 1 balloon to each of two teams with 16 balloons initially.
4. Limak has 20 - 6 - 6 - 1 - 1 = 6 balloons.
5. There are three other teams left and their numbers of balloons are 40, 14 and 2.
6. Limak gets the third place because there are two teams with more balloons.
In the second sample, Limak has the second place and he can't improve it.
In the third sample, Limak has just enough balloons to get rid of teams 2, 3 and 5 (the teams with 81 000 000 000, 5 000 000 000 and 46 000 000 000 balloons respectively). With zero balloons left, he will get the second place (ex-aequo with team 6 and team 7).
Submitted Solution:
```
from bisect import bisect_right
import heapq
n = int(input())
l = []
ti, wi = map(int, input().split())
bal = ti
pos = 1
for _ in range(n - 1):
ti, wi = map(int, input().split())
if ti > bal:
pos += 1
l.append((ti, wi - ti + 1))
l.sort()
best_pos = pos
op = bisect_right(l, (bal, float('inf')))
#print(l)
w = []
for i, v in l[op:]:
heapq.heappush(w, v)
op -= 1
while w:
head = heapq.heappop(w)
if bal < head:
break
bal -= head
pos -= 1
#print(w, op)
while op >= 0 and l[op][0] > bal:
heapq.heappush(w, l[op][1])
op -= 1
pos += 1
best_pos = min(best_pos, pos)
print(best_pos)
``` | instruction | 0 | 83,040 | 17 | 166,080 |
Yes | output | 1 | 83,040 | 17 | 166,081 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One tradition of ACM-ICPC contests is that a team gets a balloon for every solved problem. We assume that the submission time doesn't matter and teams are sorted only by the number of balloons they have. It means that one's place is equal to the number of teams with more balloons, increased by 1. For example, if there are seven teams with more balloons, you get the eight place. Ties are allowed.
You should know that it's important to eat before a contest. If the number of balloons of a team is greater than the weight of this team, the team starts to float in the air together with their workstation. They eventually touch the ceiling, what is strictly forbidden by the rules. The team is then disqualified and isn't considered in the standings.
A contest has just finished. There are n teams, numbered 1 through n. The i-th team has ti balloons and weight wi. It's guaranteed that ti doesn't exceed wi so nobody floats initially.
Limak is a member of the first team. He doesn't like cheating and he would never steal balloons from other teams. Instead, he can give his balloons away to other teams, possibly making them float. Limak can give away zero or more balloons of his team. Obviously, he can't give away more balloons than his team initially has.
What is the best place Limak can get?
Input
The first line of the standard input contains one integer n (2 ≤ n ≤ 300 000) — the number of teams.
The i-th of n following lines contains two integers ti and wi (0 ≤ ti ≤ wi ≤ 1018) — respectively the number of balloons and the weight of the i-th team. Limak is a member of the first team.
Output
Print one integer denoting the best place Limak can get.
Examples
Input
8
20 1000
32 37
40 1000
45 50
16 16
16 16
14 1000
2 1000
Output
3
Input
7
4 4
4 4
4 4
4 4
4 4
4 4
5 5
Output
2
Input
7
14000000003 1000000000000000000
81000000000 88000000000
5000000000 7000000000
15000000000 39000000000
46000000000 51000000000
0 1000000000
0 0
Output
2
Note
In the first sample, Limak has 20 balloons initially. There are three teams with more balloons (32, 40 and 45 balloons), so Limak has the fourth place initially. One optimal strategy is:
1. Limak gives 6 balloons away to a team with 32 balloons and weight 37, which is just enough to make them fly. Unfortunately, Limak has only 14 balloons now and he would get the fifth place.
2. Limak gives 6 balloons away to a team with 45 balloons. Now they have 51 balloons and weight 50 so they fly and get disqualified.
3. Limak gives 1 balloon to each of two teams with 16 balloons initially.
4. Limak has 20 - 6 - 6 - 1 - 1 = 6 balloons.
5. There are three other teams left and their numbers of balloons are 40, 14 and 2.
6. Limak gets the third place because there are two teams with more balloons.
In the second sample, Limak has the second place and he can't improve it.
In the third sample, Limak has just enough balloons to get rid of teams 2, 3 and 5 (the teams with 81 000 000 000, 5 000 000 000 and 46 000 000 000 balloons respectively). With zero balloons left, he will get the second place (ex-aequo with team 6 and team 7).
Submitted Solution:
```
import heapq
n = int(input())
num_balloons = int(input().split()[0])
better_teams = []
current_place = 1
worse_teams = []
for i in range(1, n):
inp = input().split()
team_balloons = int(inp[0])
team_size = int(inp[1])
if team_balloons>num_balloons:
current_place+=1
better_teams.append(team_size-team_balloons+1)
else:
worse_teams.append([team_balloons, team_size])
heapq.heapify(better_teams)
best_place = current_place
worse_teams = sorted(worse_teams,key=lambda x:-x[0])
ind = 0
while True:
if len(better_teams)==0:
break
removed_team = heapq.heappop(better_teams)
num_balloons -= removed_team
current_place-=1
if num_balloons<0:
break
while ind<len(worse_teams) and worse_teams[ind][0]>num_balloons:
heapq.heappush(better_teams,worse_teams[ind][1]-worse_teams[ind][0]+1)
current_place+=1
ind+=1
if current_place<best_place:
best_place=current_place
print(best_place)
``` | instruction | 0 | 83,041 | 17 | 166,082 |
Yes | output | 1 | 83,041 | 17 | 166,083 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One tradition of ACM-ICPC contests is that a team gets a balloon for every solved problem. We assume that the submission time doesn't matter and teams are sorted only by the number of balloons they have. It means that one's place is equal to the number of teams with more balloons, increased by 1. For example, if there are seven teams with more balloons, you get the eight place. Ties are allowed.
You should know that it's important to eat before a contest. If the number of balloons of a team is greater than the weight of this team, the team starts to float in the air together with their workstation. They eventually touch the ceiling, what is strictly forbidden by the rules. The team is then disqualified and isn't considered in the standings.
A contest has just finished. There are n teams, numbered 1 through n. The i-th team has ti balloons and weight wi. It's guaranteed that ti doesn't exceed wi so nobody floats initially.
Limak is a member of the first team. He doesn't like cheating and he would never steal balloons from other teams. Instead, he can give his balloons away to other teams, possibly making them float. Limak can give away zero or more balloons of his team. Obviously, he can't give away more balloons than his team initially has.
What is the best place Limak can get?
Input
The first line of the standard input contains one integer n (2 ≤ n ≤ 300 000) — the number of teams.
The i-th of n following lines contains two integers ti and wi (0 ≤ ti ≤ wi ≤ 1018) — respectively the number of balloons and the weight of the i-th team. Limak is a member of the first team.
Output
Print one integer denoting the best place Limak can get.
Examples
Input
8
20 1000
32 37
40 1000
45 50
16 16
16 16
14 1000
2 1000
Output
3
Input
7
4 4
4 4
4 4
4 4
4 4
4 4
5 5
Output
2
Input
7
14000000003 1000000000000000000
81000000000 88000000000
5000000000 7000000000
15000000000 39000000000
46000000000 51000000000
0 1000000000
0 0
Output
2
Note
In the first sample, Limak has 20 balloons initially. There are three teams with more balloons (32, 40 and 45 balloons), so Limak has the fourth place initially. One optimal strategy is:
1. Limak gives 6 balloons away to a team with 32 balloons and weight 37, which is just enough to make them fly. Unfortunately, Limak has only 14 balloons now and he would get the fifth place.
2. Limak gives 6 balloons away to a team with 45 balloons. Now they have 51 balloons and weight 50 so they fly and get disqualified.
3. Limak gives 1 balloon to each of two teams with 16 balloons initially.
4. Limak has 20 - 6 - 6 - 1 - 1 = 6 balloons.
5. There are three other teams left and their numbers of balloons are 40, 14 and 2.
6. Limak gets the third place because there are two teams with more balloons.
In the second sample, Limak has the second place and he can't improve it.
In the third sample, Limak has just enough balloons to get rid of teams 2, 3 and 5 (the teams with 81 000 000 000, 5 000 000 000 and 46 000 000 000 balloons respectively). With zero balloons left, he will get the second place (ex-aequo with team 6 and team 7).
Submitted Solution:
```
import heapq
from bisect import bisect_right
from collections import defaultdict
import os
import sys
from io import BytesIO, IOBase
from collections import defaultdict
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")
n=int(input())
res=[]
curr,we=map(int,input().split())
for j in range(n-1):
p,q=map(int,input().split())
res.append([p,q-p])
res.sort(reverse=True)
li=[]
j=0
ans=n
tot=0
while(j<n-1):
if res[j][0]>curr:
heapq.heappush(li,res[j][1]+1)
j+=1
else:
ans=min(ans,j+1-tot)
try:
p=li[0]
except:
p=float("inf")
if curr>=p:
curr-=p
heapq.heappop(li)
tot+=1
else:
break
print(ans)
``` | instruction | 0 | 83,042 | 17 | 166,084 |
No | output | 1 | 83,042 | 17 | 166,085 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One tradition of ACM-ICPC contests is that a team gets a balloon for every solved problem. We assume that the submission time doesn't matter and teams are sorted only by the number of balloons they have. It means that one's place is equal to the number of teams with more balloons, increased by 1. For example, if there are seven teams with more balloons, you get the eight place. Ties are allowed.
You should know that it's important to eat before a contest. If the number of balloons of a team is greater than the weight of this team, the team starts to float in the air together with their workstation. They eventually touch the ceiling, what is strictly forbidden by the rules. The team is then disqualified and isn't considered in the standings.
A contest has just finished. There are n teams, numbered 1 through n. The i-th team has ti balloons and weight wi. It's guaranteed that ti doesn't exceed wi so nobody floats initially.
Limak is a member of the first team. He doesn't like cheating and he would never steal balloons from other teams. Instead, he can give his balloons away to other teams, possibly making them float. Limak can give away zero or more balloons of his team. Obviously, he can't give away more balloons than his team initially has.
What is the best place Limak can get?
Input
The first line of the standard input contains one integer n (2 ≤ n ≤ 300 000) — the number of teams.
The i-th of n following lines contains two integers ti and wi (0 ≤ ti ≤ wi ≤ 1018) — respectively the number of balloons and the weight of the i-th team. Limak is a member of the first team.
Output
Print one integer denoting the best place Limak can get.
Examples
Input
8
20 1000
32 37
40 1000
45 50
16 16
16 16
14 1000
2 1000
Output
3
Input
7
4 4
4 4
4 4
4 4
4 4
4 4
5 5
Output
2
Input
7
14000000003 1000000000000000000
81000000000 88000000000
5000000000 7000000000
15000000000 39000000000
46000000000 51000000000
0 1000000000
0 0
Output
2
Note
In the first sample, Limak has 20 balloons initially. There are three teams with more balloons (32, 40 and 45 balloons), so Limak has the fourth place initially. One optimal strategy is:
1. Limak gives 6 balloons away to a team with 32 balloons and weight 37, which is just enough to make them fly. Unfortunately, Limak has only 14 balloons now and he would get the fifth place.
2. Limak gives 6 balloons away to a team with 45 balloons. Now they have 51 balloons and weight 50 so they fly and get disqualified.
3. Limak gives 1 balloon to each of two teams with 16 balloons initially.
4. Limak has 20 - 6 - 6 - 1 - 1 = 6 balloons.
5. There are three other teams left and their numbers of balloons are 40, 14 and 2.
6. Limak gets the third place because there are two teams with more balloons.
In the second sample, Limak has the second place and he can't improve it.
In the third sample, Limak has just enough balloons to get rid of teams 2, 3 and 5 (the teams with 81 000 000 000, 5 000 000 000 and 46 000 000 000 balloons respectively). With zero balloons left, he will get the second place (ex-aequo with team 6 and team 7).
Submitted Solution:
```
import bisect
import sys
input=sys.stdin.readline
n=int(input())
ar=[]
for i in range(n):
t,w=map(int,input().split())
if(i==0):
mai=[t,w]
else:
ar.append([w-t+1,w,t])
ar.sort(key=lambda x:(x[2],-x[0]))
br=[]
cr=[]
for i in range(n-1):
br.append(ar[i][2])
cr.append([ar[i][0],i])
cr.sort(key=lambda x:x[0])
pos=bisect.bisect(br,mai[0])
ext=0
ans=[pos]
for i in range(n):
ke=cr[i][1]
if(mai[0]-cr[i][0]>=0):
mai[0]-=cr[i][0]
jj=pos
br[i]=-1
while(jj>=0 and mai[0]<br[jj]):
if(br[jj]==-1):
ext-=1
jj-=1
if(pos>=ke):
ext+=1
pos=jj
ans.append(pos+ext)
else:
break
print(n-max(ans))
``` | instruction | 0 | 83,043 | 17 | 166,086 |
No | output | 1 | 83,043 | 17 | 166,087 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One tradition of ACM-ICPC contests is that a team gets a balloon for every solved problem. We assume that the submission time doesn't matter and teams are sorted only by the number of balloons they have. It means that one's place is equal to the number of teams with more balloons, increased by 1. For example, if there are seven teams with more balloons, you get the eight place. Ties are allowed.
You should know that it's important to eat before a contest. If the number of balloons of a team is greater than the weight of this team, the team starts to float in the air together with their workstation. They eventually touch the ceiling, what is strictly forbidden by the rules. The team is then disqualified and isn't considered in the standings.
A contest has just finished. There are n teams, numbered 1 through n. The i-th team has ti balloons and weight wi. It's guaranteed that ti doesn't exceed wi so nobody floats initially.
Limak is a member of the first team. He doesn't like cheating and he would never steal balloons from other teams. Instead, he can give his balloons away to other teams, possibly making them float. Limak can give away zero or more balloons of his team. Obviously, he can't give away more balloons than his team initially has.
What is the best place Limak can get?
Input
The first line of the standard input contains one integer n (2 ≤ n ≤ 300 000) — the number of teams.
The i-th of n following lines contains two integers ti and wi (0 ≤ ti ≤ wi ≤ 1018) — respectively the number of balloons and the weight of the i-th team. Limak is a member of the first team.
Output
Print one integer denoting the best place Limak can get.
Examples
Input
8
20 1000
32 37
40 1000
45 50
16 16
16 16
14 1000
2 1000
Output
3
Input
7
4 4
4 4
4 4
4 4
4 4
4 4
5 5
Output
2
Input
7
14000000003 1000000000000000000
81000000000 88000000000
5000000000 7000000000
15000000000 39000000000
46000000000 51000000000
0 1000000000
0 0
Output
2
Note
In the first sample, Limak has 20 balloons initially. There are three teams with more balloons (32, 40 and 45 balloons), so Limak has the fourth place initially. One optimal strategy is:
1. Limak gives 6 balloons away to a team with 32 balloons and weight 37, which is just enough to make them fly. Unfortunately, Limak has only 14 balloons now and he would get the fifth place.
2. Limak gives 6 balloons away to a team with 45 balloons. Now they have 51 balloons and weight 50 so they fly and get disqualified.
3. Limak gives 1 balloon to each of two teams with 16 balloons initially.
4. Limak has 20 - 6 - 6 - 1 - 1 = 6 balloons.
5. There are three other teams left and their numbers of balloons are 40, 14 and 2.
6. Limak gets the third place because there are two teams with more balloons.
In the second sample, Limak has the second place and he can't improve it.
In the third sample, Limak has just enough balloons to get rid of teams 2, 3 and 5 (the teams with 81 000 000 000, 5 000 000 000 and 46 000 000 000 balloons respectively). With zero balloons left, he will get the second place (ex-aequo with team 6 and team 7).
Submitted Solution:
```
import bisect
n = int(input())
my, _ = list(map(int, input().strip().split()))
#print(my)
competitors = []
for i in range(n - 1):
b, w = list(map(int, input().strip().split()))
competitors.append([b, w - b + 1])
competitors.sort(key = lambda x: x[0])
#print(competitors)
bs = [c[0] for c in competitors]
#print(bs)
pos = bisect.bisect_right(bs, my)
#print(pos)
cur_rank = len(competitors) - pos + 1
#print(cur_rank)
best_rank = cur_rank
overs = competitors[pos: ]
overs.sort(key = lambda x: x[1])
#print(overs)
nearest_lower = pos - 1
#print('nearest_lower %d' % nearest_lower)
for i in range(len(overs)):
#print('my=%d overs[%d][1]=%d' % (my, i, overs[i][1]))
if my >= overs[i][1]:
my -= overs[i][1]
cur_rank -= 1
#print('my=%d' % my)
while nearest_lower >= 0 and competitors[nearest_lower][0] > my:
cur_rank += 1
#print('nearest_lower %d' % nearest_lower)
if my >= competitors[nearest_lower][1]:
my -= competitors[nearest_lower][1]
nearest_lower -= 1
cur_rank -= 1
#print('my=%d' % my)
else:
nearest_lower -= 1
best_rank = min(best_rank, cur_rank)
else:
break
print(best_rank)
``` | instruction | 0 | 83,044 | 17 | 166,088 |
No | output | 1 | 83,044 | 17 | 166,089 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One tradition of ACM-ICPC contests is that a team gets a balloon for every solved problem. We assume that the submission time doesn't matter and teams are sorted only by the number of balloons they have. It means that one's place is equal to the number of teams with more balloons, increased by 1. For example, if there are seven teams with more balloons, you get the eight place. Ties are allowed.
You should know that it's important to eat before a contest. If the number of balloons of a team is greater than the weight of this team, the team starts to float in the air together with their workstation. They eventually touch the ceiling, what is strictly forbidden by the rules. The team is then disqualified and isn't considered in the standings.
A contest has just finished. There are n teams, numbered 1 through n. The i-th team has ti balloons and weight wi. It's guaranteed that ti doesn't exceed wi so nobody floats initially.
Limak is a member of the first team. He doesn't like cheating and he would never steal balloons from other teams. Instead, he can give his balloons away to other teams, possibly making them float. Limak can give away zero or more balloons of his team. Obviously, he can't give away more balloons than his team initially has.
What is the best place Limak can get?
Input
The first line of the standard input contains one integer n (2 ≤ n ≤ 300 000) — the number of teams.
The i-th of n following lines contains two integers ti and wi (0 ≤ ti ≤ wi ≤ 1018) — respectively the number of balloons and the weight of the i-th team. Limak is a member of the first team.
Output
Print one integer denoting the best place Limak can get.
Examples
Input
8
20 1000
32 37
40 1000
45 50
16 16
16 16
14 1000
2 1000
Output
3
Input
7
4 4
4 4
4 4
4 4
4 4
4 4
5 5
Output
2
Input
7
14000000003 1000000000000000000
81000000000 88000000000
5000000000 7000000000
15000000000 39000000000
46000000000 51000000000
0 1000000000
0 0
Output
2
Note
In the first sample, Limak has 20 balloons initially. There are three teams with more balloons (32, 40 and 45 balloons), so Limak has the fourth place initially. One optimal strategy is:
1. Limak gives 6 balloons away to a team with 32 balloons and weight 37, which is just enough to make them fly. Unfortunately, Limak has only 14 balloons now and he would get the fifth place.
2. Limak gives 6 balloons away to a team with 45 balloons. Now they have 51 balloons and weight 50 so they fly and get disqualified.
3. Limak gives 1 balloon to each of two teams with 16 balloons initially.
4. Limak has 20 - 6 - 6 - 1 - 1 = 6 balloons.
5. There are three other teams left and their numbers of balloons are 40, 14 and 2.
6. Limak gets the third place because there are two teams with more balloons.
In the second sample, Limak has the second place and he can't improve it.
In the third sample, Limak has just enough balloons to get rid of teams 2, 3 and 5 (the teams with 81 000 000 000, 5 000 000 000 and 46 000 000 000 balloons respectively). With zero balloons left, he will get the second place (ex-aequo with team 6 and team 7).
Submitted Solution:
```
#!/usr/bin/env python3
from sys import stdin,stdout
def ri():
return map(int, input().split())
#stdin.readline().strip()
``` | instruction | 0 | 83,045 | 17 | 166,090 |
No | output | 1 | 83,045 | 17 | 166,091 |
Provide a correct Python 3 solution for this coding contest problem.
problem
You have to play a darts game with the following rules.
You can throw up to four arrows towards the target. You don't have to throw all four, you don't have to throw one. The target is divided into N parts. The score P1, ..., PN is written in each part. The total score S of the place where the arrow is stuck is the basis of your score. S is a predetermined score M or less. In the case of, S will be your score as it is. However, if S exceeds M, your score will be 0 points.
Create a program that finds the maximum number of points you can get given the written points and the value of M.
input
The input consists of multiple datasets. Each dataset is given in the following format.
On the first line, the integers N (1 ≤ N ≤ 1000) and M (1 ≤ M ≤ 200000000 = 2 × 108) are written in this order, separated by blanks. In (1 ≤ i ≤ N), Pi (1 ≤ Pi ≤ 100000000 = 108) is written.
Of the scoring data, 20% of the points satisfy N ≤ 100 and 50% of the points satisfy N ≤ 300.
When both N and M are 0, the input is completed. The number of data sets does not exceed 10.
output
Outputs the maximum number of points you can get for each dataset on one line.
Examples
Input
4 50
3
14
15
9
3 21
16
11
2
0 0
Output
48
20
Input
None
Output
None | instruction | 0 | 83,306 | 17 | 166,612 |
"Correct Solution:
```
from bisect import bisect_right
while True:
N, M = [int(x) for x in input().split()]
if N == 0 and M == 0:
break
_P = [0] * N
for i in range(N):
_P[i] = int(input())
_P.append(0)
P = list(set(x + y for x in _P for y in _P))
P.sort()
ans = 0
for p in P:
if p > M // 2:
break
i = bisect_right(P, M - p)
ans = max(ans, p + P[i - 1])
print(ans)
``` | output | 1 | 83,306 | 17 | 166,613 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
You have to play a darts game with the following rules.
You can throw up to four arrows towards the target. You don't have to throw all four, you don't have to throw one. The target is divided into N parts. The score P1, ..., PN is written in each part. The total score S of the place where the arrow is stuck is the basis of your score. S is a predetermined score M or less. In the case of, S will be your score as it is. However, if S exceeds M, your score will be 0 points.
Create a program that finds the maximum number of points you can get given the written points and the value of M.
input
The input consists of multiple datasets. Each dataset is given in the following format.
On the first line, the integers N (1 ≤ N ≤ 1000) and M (1 ≤ M ≤ 200000000 = 2 × 108) are written in this order, separated by blanks. In (1 ≤ i ≤ N), Pi (1 ≤ Pi ≤ 100000000 = 108) is written.
Of the scoring data, 20% of the points satisfy N ≤ 100 and 50% of the points satisfy N ≤ 300.
When both N and M are 0, the input is completed. The number of data sets does not exceed 10.
output
Outputs the maximum number of points you can get for each dataset on one line.
Examples
Input
4 50
3
14
15
9
3 21
16
11
2
0 0
Output
48
20
Input
None
Output
None
Submitted Solution:
```
import bisect
while True:
max_S = 0
p = [0]
pp = []
n, m = map(int, input().split())
if n == 0 and m == 0:
break
for _ in range(n):
p.append(int(input()))
# 最初にダーツ2本を投げた状態にする
for i in range(n+1):
for j in range(i, n+1):
pp.append(p[i] + p[j])
pp.sort()
# t: ppの中でm以下の数字の位置
t = bisect.bisect_left(pp, m)
# h: ppの中でm/2以下の数字の位置
h = bisect.bisect_left(pp, m//2, 0, t)
if t < len(pp):
if pp[t] == m:
print(m)
continue
pp = pp[:t]
max_S = 0
for i in range(len(pp)):
t = bisect.bisect_left(pp, m-pp[i])
if pp[i] + pp[t-1] > max_S:
max_S = pp[i] + pp[t-1]
print(max_S)
``` | instruction | 0 | 83,310 | 17 | 166,620 |
Yes | output | 1 | 83,310 | 17 | 166,621 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
You have to play a darts game with the following rules.
You can throw up to four arrows towards the target. You don't have to throw all four, you don't have to throw one. The target is divided into N parts. The score P1, ..., PN is written in each part. The total score S of the place where the arrow is stuck is the basis of your score. S is a predetermined score M or less. In the case of, S will be your score as it is. However, if S exceeds M, your score will be 0 points.
Create a program that finds the maximum number of points you can get given the written points and the value of M.
input
The input consists of multiple datasets. Each dataset is given in the following format.
On the first line, the integers N (1 ≤ N ≤ 1000) and M (1 ≤ M ≤ 200000000 = 2 × 108) are written in this order, separated by blanks. In (1 ≤ i ≤ N), Pi (1 ≤ Pi ≤ 100000000 = 108) is written.
Of the scoring data, 20% of the points satisfy N ≤ 100 and 50% of the points satisfy N ≤ 300.
When both N and M are 0, the input is completed. The number of data sets does not exceed 10.
output
Outputs the maximum number of points you can get for each dataset on one line.
Examples
Input
4 50
3
14
15
9
3 21
16
11
2
0 0
Output
48
20
Input
None
Output
None
Submitted Solution:
```
import bisect
def main():
while True:
n,m=map(int,input().split())
if n==0 and m==0:
break
p=list()
for i in range(n):
p.append(int(input()))
p.append(0)
ls=list()
for i in p:
for j in p:
if i+j<=m:
ls.append(i+j)
ans=0
for x in ls:
idx=bisect.bisect_right(ls,m-x)
if idx==0:
continue
idx-=1
ans=max(ans,x+ls[idx])
print(ans)
if __name__=="__main__":
main()
``` | instruction | 0 | 83,314 | 17 | 166,628 |
No | output | 1 | 83,314 | 17 | 166,629 |
Provide a correct Python 3 solution for this coding contest problem.
Yui Hirasawa, who attends private Sakuragaoka Girls' High School, has to make a career hope by the day after tomorrow, but the trouble is that she hasn't decided anything yet. When I consulted with my friend Wa, I knew that my first choice was K University, so I consulted with my career guidance teacher to see if I could join K University.
As a career guidance teacher who was consulted, you decided to refer to Yui's final exam results in order to predict whether Yui would be able to enter K University. However, Yui's grades are greatly affected by whether or not he wins the exam, so he decided to check the best and worst scores of the final exams in the past. The data of the past final exams only remained for each score of the five subjects, and the total score of the exam did not remain. Therefore, your job is to write a program that takes the score data of each exam as input and outputs the score of the best time and the score of the worst time in the past final exams.
Notes on Test Cases
Multiple datasets are given in the above input format. Create a program that outputs each data set in the above output format.
When n is 0, it indicates the end of input.
<!-
Input
n
s11 s12 s13 s14 s15
...
sn1 sn2 sn3 sn4 sn5
n is the number of tests taken so far. The number of tests is 1 or more and 100 or less. The next n lines are given scores for 5 subjects in each exam. The range of points is an integer between 0 and 100.
Output
Output the highest and lowest points in one line. Separate the highest and lowest points with a single space and do not include any other characters.
Examples
Input
3
49 50 87 78 41
27 61 100 45 84
28 88 40 95 66
2
100 100 100 100 100
0 0 0 0 0
1
89 90 85 93 82
0
Output
317 305
500 0
439 439
Input
3
49 50 87 78 41
27 61 100 45 84
28 88 40 95 66
Output
317 305
Input
2
100 100 100 100 100
0 0 0 0 0
Output
500 0
Input
1
89 90 85 93 82
Output
439 439 | instruction | 0 | 83,327 | 17 | 166,654 |
"Correct Solution:
```
while 1:
n = int(input())
if n == 0:
break
a = []
for _ in range(n):
a.append(sum(map(int, input().split())))
print(max(a), min(a))
``` | output | 1 | 83,327 | 17 | 166,655 |
Provide a correct Python 3 solution for this coding contest problem.
Yui Hirasawa, who attends private Sakuragaoka Girls' High School, has to make a career hope by the day after tomorrow, but the trouble is that she hasn't decided anything yet. When I consulted with my friend Wa, I knew that my first choice was K University, so I consulted with my career guidance teacher to see if I could join K University.
As a career guidance teacher who was consulted, you decided to refer to Yui's final exam results in order to predict whether Yui would be able to enter K University. However, Yui's grades are greatly affected by whether or not he wins the exam, so he decided to check the best and worst scores of the final exams in the past. The data of the past final exams only remained for each score of the five subjects, and the total score of the exam did not remain. Therefore, your job is to write a program that takes the score data of each exam as input and outputs the score of the best time and the score of the worst time in the past final exams.
Notes on Test Cases
Multiple datasets are given in the above input format. Create a program that outputs each data set in the above output format.
When n is 0, it indicates the end of input.
<!-
Input
n
s11 s12 s13 s14 s15
...
sn1 sn2 sn3 sn4 sn5
n is the number of tests taken so far. The number of tests is 1 or more and 100 or less. The next n lines are given scores for 5 subjects in each exam. The range of points is an integer between 0 and 100.
Output
Output the highest and lowest points in one line. Separate the highest and lowest points with a single space and do not include any other characters.
Examples
Input
3
49 50 87 78 41
27 61 100 45 84
28 88 40 95 66
2
100 100 100 100 100
0 0 0 0 0
1
89 90 85 93 82
0
Output
317 305
500 0
439 439
Input
3
49 50 87 78 41
27 61 100 45 84
28 88 40 95 66
Output
317 305
Input
2
100 100 100 100 100
0 0 0 0 0
Output
500 0
Input
1
89 90 85 93 82
Output
439 439 | instruction | 0 | 83,328 | 17 | 166,656 |
"Correct Solution:
```
d = {0: lambda x, y: max(x, y),
1: lambda x, y: min(x, y)}
while True:
n = int(input())
if n == 0:
break
ans = [0, 500]
for _ in range(n):
score = sum(map(int, input().split()))
ans = [d[i](ans[i], score) for i in range(2)]
print(*ans)
``` | output | 1 | 83,328 | 17 | 166,657 |
Provide a correct Python 3 solution for this coding contest problem.
Yui Hirasawa, who attends private Sakuragaoka Girls' High School, has to make a career hope by the day after tomorrow, but the trouble is that she hasn't decided anything yet. When I consulted with my friend Wa, I knew that my first choice was K University, so I consulted with my career guidance teacher to see if I could join K University.
As a career guidance teacher who was consulted, you decided to refer to Yui's final exam results in order to predict whether Yui would be able to enter K University. However, Yui's grades are greatly affected by whether or not he wins the exam, so he decided to check the best and worst scores of the final exams in the past. The data of the past final exams only remained for each score of the five subjects, and the total score of the exam did not remain. Therefore, your job is to write a program that takes the score data of each exam as input and outputs the score of the best time and the score of the worst time in the past final exams.
Notes on Test Cases
Multiple datasets are given in the above input format. Create a program that outputs each data set in the above output format.
When n is 0, it indicates the end of input.
<!-
Input
n
s11 s12 s13 s14 s15
...
sn1 sn2 sn3 sn4 sn5
n is the number of tests taken so far. The number of tests is 1 or more and 100 or less. The next n lines are given scores for 5 subjects in each exam. The range of points is an integer between 0 and 100.
Output
Output the highest and lowest points in one line. Separate the highest and lowest points with a single space and do not include any other characters.
Examples
Input
3
49 50 87 78 41
27 61 100 45 84
28 88 40 95 66
2
100 100 100 100 100
0 0 0 0 0
1
89 90 85 93 82
0
Output
317 305
500 0
439 439
Input
3
49 50 87 78 41
27 61 100 45 84
28 88 40 95 66
Output
317 305
Input
2
100 100 100 100 100
0 0 0 0 0
Output
500 0
Input
1
89 90 85 93 82
Output
439 439 | instruction | 0 | 83,329 | 17 | 166,658 |
"Correct Solution:
```
while 1:
n = int(input())
if n == 0:
break
total = []
for i in range(n):
score = input().split()
sum = 0
for s in score:
sum += int(s)
total.append(sum)
print(max(total), min(total))
``` | output | 1 | 83,329 | 17 | 166,659 |
Provide a correct Python 3 solution for this coding contest problem.
Yui Hirasawa, who attends private Sakuragaoka Girls' High School, has to make a career hope by the day after tomorrow, but the trouble is that she hasn't decided anything yet. When I consulted with my friend Wa, I knew that my first choice was K University, so I consulted with my career guidance teacher to see if I could join K University.
As a career guidance teacher who was consulted, you decided to refer to Yui's final exam results in order to predict whether Yui would be able to enter K University. However, Yui's grades are greatly affected by whether or not he wins the exam, so he decided to check the best and worst scores of the final exams in the past. The data of the past final exams only remained for each score of the five subjects, and the total score of the exam did not remain. Therefore, your job is to write a program that takes the score data of each exam as input and outputs the score of the best time and the score of the worst time in the past final exams.
Notes on Test Cases
Multiple datasets are given in the above input format. Create a program that outputs each data set in the above output format.
When n is 0, it indicates the end of input.
<!-
Input
n
s11 s12 s13 s14 s15
...
sn1 sn2 sn3 sn4 sn5
n is the number of tests taken so far. The number of tests is 1 or more and 100 or less. The next n lines are given scores for 5 subjects in each exam. The range of points is an integer between 0 and 100.
Output
Output the highest and lowest points in one line. Separate the highest and lowest points with a single space and do not include any other characters.
Examples
Input
3
49 50 87 78 41
27 61 100 45 84
28 88 40 95 66
2
100 100 100 100 100
0 0 0 0 0
1
89 90 85 93 82
0
Output
317 305
500 0
439 439
Input
3
49 50 87 78 41
27 61 100 45 84
28 88 40 95 66
Output
317 305
Input
2
100 100 100 100 100
0 0 0 0 0
Output
500 0
Input
1
89 90 85 93 82
Output
439 439 | instruction | 0 | 83,330 | 17 | 166,660 |
"Correct Solution:
```
while 1:
n=int(input())
if n==0:break
s=[]
for i in range(n):
s.append(sum(map(int,input().split())))
print(max(s),min(s))
``` | output | 1 | 83,330 | 17 | 166,661 |
Provide a correct Python 3 solution for this coding contest problem.
Yui Hirasawa, who attends private Sakuragaoka Girls' High School, has to make a career hope by the day after tomorrow, but the trouble is that she hasn't decided anything yet. When I consulted with my friend Wa, I knew that my first choice was K University, so I consulted with my career guidance teacher to see if I could join K University.
As a career guidance teacher who was consulted, you decided to refer to Yui's final exam results in order to predict whether Yui would be able to enter K University. However, Yui's grades are greatly affected by whether or not he wins the exam, so he decided to check the best and worst scores of the final exams in the past. The data of the past final exams only remained for each score of the five subjects, and the total score of the exam did not remain. Therefore, your job is to write a program that takes the score data of each exam as input and outputs the score of the best time and the score of the worst time in the past final exams.
Notes on Test Cases
Multiple datasets are given in the above input format. Create a program that outputs each data set in the above output format.
When n is 0, it indicates the end of input.
<!-
Input
n
s11 s12 s13 s14 s15
...
sn1 sn2 sn3 sn4 sn5
n is the number of tests taken so far. The number of tests is 1 or more and 100 or less. The next n lines are given scores for 5 subjects in each exam. The range of points is an integer between 0 and 100.
Output
Output the highest and lowest points in one line. Separate the highest and lowest points with a single space and do not include any other characters.
Examples
Input
3
49 50 87 78 41
27 61 100 45 84
28 88 40 95 66
2
100 100 100 100 100
0 0 0 0 0
1
89 90 85 93 82
0
Output
317 305
500 0
439 439
Input
3
49 50 87 78 41
27 61 100 45 84
28 88 40 95 66
Output
317 305
Input
2
100 100 100 100 100
0 0 0 0 0
Output
500 0
Input
1
89 90 85 93 82
Output
439 439 | instruction | 0 | 83,331 | 17 | 166,662 |
"Correct Solution:
```
while True:
n = int(input())
if n == 0:break
p = [sum(map(int, input().split())) for _ in range(n)]
print("{} {}".format(max(p), min(p)))
``` | output | 1 | 83,331 | 17 | 166,663 |
Provide a correct Python 3 solution for this coding contest problem.
Yui Hirasawa, who attends private Sakuragaoka Girls' High School, has to make a career hope by the day after tomorrow, but the trouble is that she hasn't decided anything yet. When I consulted with my friend Wa, I knew that my first choice was K University, so I consulted with my career guidance teacher to see if I could join K University.
As a career guidance teacher who was consulted, you decided to refer to Yui's final exam results in order to predict whether Yui would be able to enter K University. However, Yui's grades are greatly affected by whether or not he wins the exam, so he decided to check the best and worst scores of the final exams in the past. The data of the past final exams only remained for each score of the five subjects, and the total score of the exam did not remain. Therefore, your job is to write a program that takes the score data of each exam as input and outputs the score of the best time and the score of the worst time in the past final exams.
Notes on Test Cases
Multiple datasets are given in the above input format. Create a program that outputs each data set in the above output format.
When n is 0, it indicates the end of input.
<!-
Input
n
s11 s12 s13 s14 s15
...
sn1 sn2 sn3 sn4 sn5
n is the number of tests taken so far. The number of tests is 1 or more and 100 or less. The next n lines are given scores for 5 subjects in each exam. The range of points is an integer between 0 and 100.
Output
Output the highest and lowest points in one line. Separate the highest and lowest points with a single space and do not include any other characters.
Examples
Input
3
49 50 87 78 41
27 61 100 45 84
28 88 40 95 66
2
100 100 100 100 100
0 0 0 0 0
1
89 90 85 93 82
0
Output
317 305
500 0
439 439
Input
3
49 50 87 78 41
27 61 100 45 84
28 88 40 95 66
Output
317 305
Input
2
100 100 100 100 100
0 0 0 0 0
Output
500 0
Input
1
89 90 85 93 82
Output
439 439 | instruction | 0 | 83,332 | 17 | 166,664 |
"Correct Solution:
```
while True :
n = int(input())
if n == 0 :
break
max_sum = 0
min_sum = 500
for i in range(n) :
m_sum = 0
m = list(map(int, input().split()))
for j in range(5) :
m_sum += m[j]
if max_sum < m_sum :
max_sum = m_sum
if min_sum > m_sum :
min_sum = m_sum
print(max_sum, min_sum)
``` | output | 1 | 83,332 | 17 | 166,665 |
Provide a correct Python 3 solution for this coding contest problem.
Yui Hirasawa, who attends private Sakuragaoka Girls' High School, has to make a career hope by the day after tomorrow, but the trouble is that she hasn't decided anything yet. When I consulted with my friend Wa, I knew that my first choice was K University, so I consulted with my career guidance teacher to see if I could join K University.
As a career guidance teacher who was consulted, you decided to refer to Yui's final exam results in order to predict whether Yui would be able to enter K University. However, Yui's grades are greatly affected by whether or not he wins the exam, so he decided to check the best and worst scores of the final exams in the past. The data of the past final exams only remained for each score of the five subjects, and the total score of the exam did not remain. Therefore, your job is to write a program that takes the score data of each exam as input and outputs the score of the best time and the score of the worst time in the past final exams.
Notes on Test Cases
Multiple datasets are given in the above input format. Create a program that outputs each data set in the above output format.
When n is 0, it indicates the end of input.
<!-
Input
n
s11 s12 s13 s14 s15
...
sn1 sn2 sn3 sn4 sn5
n is the number of tests taken so far. The number of tests is 1 or more and 100 or less. The next n lines are given scores for 5 subjects in each exam. The range of points is an integer between 0 and 100.
Output
Output the highest and lowest points in one line. Separate the highest and lowest points with a single space and do not include any other characters.
Examples
Input
3
49 50 87 78 41
27 61 100 45 84
28 88 40 95 66
2
100 100 100 100 100
0 0 0 0 0
1
89 90 85 93 82
0
Output
317 305
500 0
439 439
Input
3
49 50 87 78 41
27 61 100 45 84
28 88 40 95 66
Output
317 305
Input
2
100 100 100 100 100
0 0 0 0 0
Output
500 0
Input
1
89 90 85 93 82
Output
439 439 | instruction | 0 | 83,333 | 17 | 166,666 |
"Correct Solution:
```
while 1:
n = int(input())
if n == 0:
break
score = list(map(int, input().split()))
top = sum(score)
bottom = sum(score)
for _ in range(n-1):
score = list(map(int, input().split()))
total = sum(score)
if top < total:
top = total
elif bottom > total:
bottom = total
print(top, bottom)
``` | output | 1 | 83,333 | 17 | 166,667 |
Provide a correct Python 3 solution for this coding contest problem.
Yui Hirasawa, who attends private Sakuragaoka Girls' High School, has to make a career hope by the day after tomorrow, but the trouble is that she hasn't decided anything yet. When I consulted with my friend Wa, I knew that my first choice was K University, so I consulted with my career guidance teacher to see if I could join K University.
As a career guidance teacher who was consulted, you decided to refer to Yui's final exam results in order to predict whether Yui would be able to enter K University. However, Yui's grades are greatly affected by whether or not he wins the exam, so he decided to check the best and worst scores of the final exams in the past. The data of the past final exams only remained for each score of the five subjects, and the total score of the exam did not remain. Therefore, your job is to write a program that takes the score data of each exam as input and outputs the score of the best time and the score of the worst time in the past final exams.
Notes on Test Cases
Multiple datasets are given in the above input format. Create a program that outputs each data set in the above output format.
When n is 0, it indicates the end of input.
<!-
Input
n
s11 s12 s13 s14 s15
...
sn1 sn2 sn3 sn4 sn5
n is the number of tests taken so far. The number of tests is 1 or more and 100 or less. The next n lines are given scores for 5 subjects in each exam. The range of points is an integer between 0 and 100.
Output
Output the highest and lowest points in one line. Separate the highest and lowest points with a single space and do not include any other characters.
Examples
Input
3
49 50 87 78 41
27 61 100 45 84
28 88 40 95 66
2
100 100 100 100 100
0 0 0 0 0
1
89 90 85 93 82
0
Output
317 305
500 0
439 439
Input
3
49 50 87 78 41
27 61 100 45 84
28 88 40 95 66
Output
317 305
Input
2
100 100 100 100 100
0 0 0 0 0
Output
500 0
Input
1
89 90 85 93 82
Output
439 439 | instruction | 0 | 83,334 | 17 | 166,668 |
"Correct Solution:
```
while 1:
n,a,b=int(input()),0,500
if n==0:break
for _ in range(n):
s=sum(map(int,input().split()))
a=max(a,s)
b=min(b,s)
print(a,b)
``` | output | 1 | 83,334 | 17 | 166,669 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, Duff has been practicing weight lifting. As a hard practice, Malek gave her a task. He gave her a sequence of weights. Weight of i-th of them is 2wi pounds. In each step, Duff can lift some of the remaining weights and throw them away. She does this until there's no more weight left. Malek asked her to minimize the number of steps.
<image>
Duff is a competitive programming fan. That's why in each step, she can only lift and throw away a sequence of weights 2a1, ..., 2ak if and only if there exists a non-negative integer x such that 2a1 + 2a2 + ... + 2ak = 2x, i. e. the sum of those numbers is a power of two.
Duff is a competitive programming fan, but not a programmer. That's why she asked for your help. Help her minimize the number of steps.
Input
The first line of input contains integer n (1 ≤ n ≤ 106), the number of weights.
The second line contains n integers w1, ..., wn separated by spaces (0 ≤ wi ≤ 106 for each 1 ≤ i ≤ n), the powers of two forming the weights values.
Output
Print the minimum number of steps in a single line.
Examples
Input
5
1 1 2 3 3
Output
2
Input
4
0 1 2 3
Output
4
Note
In the first sample case: One optimal way would be to throw away the first three in the first step and the rest in the second step. Also, it's not possible to do it in one step because their sum is not a power of two.
In the second sample case: The only optimal way is to throw away one weight in each step. It's not possible to do it in less than 4 steps because there's no subset of weights with more than one weight and sum equal to a power of two. | instruction | 0 | 83,796 | 17 | 167,592 |
Tags: greedy
Correct Solution:
```
#INTNOF
n = input()
weights = input()
bits = [0 for i in range(1001015)]
#print sorted(weights)
for w in weights.split():
bits[int(w)]+=1
ans = 0
#print bits[0:n]
for i in range(1001010):
bits[i+1] += bits[i]>>1
ans+= (bits[i]&1)
print(ans)
#print 1000000
#print ' '.join(map(str, range(1000000)))
``` | output | 1 | 83,796 | 17 | 167,593 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, Duff has been practicing weight lifting. As a hard practice, Malek gave her a task. He gave her a sequence of weights. Weight of i-th of them is 2wi pounds. In each step, Duff can lift some of the remaining weights and throw them away. She does this until there's no more weight left. Malek asked her to minimize the number of steps.
<image>
Duff is a competitive programming fan. That's why in each step, she can only lift and throw away a sequence of weights 2a1, ..., 2ak if and only if there exists a non-negative integer x such that 2a1 + 2a2 + ... + 2ak = 2x, i. e. the sum of those numbers is a power of two.
Duff is a competitive programming fan, but not a programmer. That's why she asked for your help. Help her minimize the number of steps.
Input
The first line of input contains integer n (1 ≤ n ≤ 106), the number of weights.
The second line contains n integers w1, ..., wn separated by spaces (0 ≤ wi ≤ 106 for each 1 ≤ i ≤ n), the powers of two forming the weights values.
Output
Print the minimum number of steps in a single line.
Examples
Input
5
1 1 2 3 3
Output
2
Input
4
0 1 2 3
Output
4
Note
In the first sample case: One optimal way would be to throw away the first three in the first step and the rest in the second step. Also, it's not possible to do it in one step because their sum is not a power of two.
In the second sample case: The only optimal way is to throw away one weight in each step. It's not possible to do it in less than 4 steps because there's no subset of weights with more than one weight and sum equal to a power of two. | instruction | 0 | 83,798 | 17 | 167,596 |
Tags: greedy
Correct Solution:
```
n=int(input())
A=list(map(int,input().split()))
d={}
for i in range(n):
if A[i] in d:
k=A[i]
while k in d:
del d[k]
k+=1
d[k]=1
else:
d[A[i]]=1
print(len(d))
``` | output | 1 | 83,798 | 17 | 167,597 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, Duff has been practicing weight lifting. As a hard practice, Malek gave her a task. He gave her a sequence of weights. Weight of i-th of them is 2wi pounds. In each step, Duff can lift some of the remaining weights and throw them away. She does this until there's no more weight left. Malek asked her to minimize the number of steps.
<image>
Duff is a competitive programming fan. That's why in each step, she can only lift and throw away a sequence of weights 2a1, ..., 2ak if and only if there exists a non-negative integer x such that 2a1 + 2a2 + ... + 2ak = 2x, i. e. the sum of those numbers is a power of two.
Duff is a competitive programming fan, but not a programmer. That's why she asked for your help. Help her minimize the number of steps.
Input
The first line of input contains integer n (1 ≤ n ≤ 106), the number of weights.
The second line contains n integers w1, ..., wn separated by spaces (0 ≤ wi ≤ 106 for each 1 ≤ i ≤ n), the powers of two forming the weights values.
Output
Print the minimum number of steps in a single line.
Examples
Input
5
1 1 2 3 3
Output
2
Input
4
0 1 2 3
Output
4
Note
In the first sample case: One optimal way would be to throw away the first three in the first step and the rest in the second step. Also, it's not possible to do it in one step because their sum is not a power of two.
In the second sample case: The only optimal way is to throw away one weight in each step. It's not possible to do it in less than 4 steps because there's no subset of weights with more than one weight and sum equal to a power of two. | instruction | 0 | 83,799 | 17 | 167,598 |
Tags: greedy
Correct Solution:
```
n=int(input())
a=list(map(int, input().split()))
di=[0]*(10**6+699)
for i in a:
di[i]+=1
ans=0
for i in range(10**6+698):
di[i], di[i+1] = di[i] % 2, di[i+1] + di[i] // 2
if di[i] % 2 == 1: ans += 1
#print(di[:3])
print(ans)
``` | output | 1 | 83,799 | 17 | 167,599 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, Duff has been practicing weight lifting. As a hard practice, Malek gave her a task. He gave her a sequence of weights. Weight of i-th of them is 2wi pounds. In each step, Duff can lift some of the remaining weights and throw them away. She does this until there's no more weight left. Malek asked her to minimize the number of steps.
<image>
Duff is a competitive programming fan. That's why in each step, she can only lift and throw away a sequence of weights 2a1, ..., 2ak if and only if there exists a non-negative integer x such that 2a1 + 2a2 + ... + 2ak = 2x, i. e. the sum of those numbers is a power of two.
Duff is a competitive programming fan, but not a programmer. That's why she asked for your help. Help her minimize the number of steps.
Input
The first line of input contains integer n (1 ≤ n ≤ 106), the number of weights.
The second line contains n integers w1, ..., wn separated by spaces (0 ≤ wi ≤ 106 for each 1 ≤ i ≤ n), the powers of two forming the weights values.
Output
Print the minimum number of steps in a single line.
Examples
Input
5
1 1 2 3 3
Output
2
Input
4
0 1 2 3
Output
4
Note
In the first sample case: One optimal way would be to throw away the first three in the first step and the rest in the second step. Also, it's not possible to do it in one step because their sum is not a power of two.
In the second sample case: The only optimal way is to throw away one weight in each step. It's not possible to do it in less than 4 steps because there's no subset of weights with more than one weight and sum equal to a power of two. | instruction | 0 | 83,800 | 17 | 167,600 |
Tags: greedy
Correct Solution:
```
n = int(input())
a = [0 for i in range(10**6+1)]
for x in input().split():
a[int(x)] += 1
c = 0
prev = 0
for el in a:
prev += el
if prev%2 == 1:
c+=1
prev //= 2
while prev>0:
c += prev%2
prev //= 2
print(c)
``` | output | 1 | 83,800 | 17 | 167,601 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, Duff has been practicing weight lifting. As a hard practice, Malek gave her a task. He gave her a sequence of weights. Weight of i-th of them is 2wi pounds. In each step, Duff can lift some of the remaining weights and throw them away. She does this until there's no more weight left. Malek asked her to minimize the number of steps.
<image>
Duff is a competitive programming fan. That's why in each step, she can only lift and throw away a sequence of weights 2a1, ..., 2ak if and only if there exists a non-negative integer x such that 2a1 + 2a2 + ... + 2ak = 2x, i. e. the sum of those numbers is a power of two.
Duff is a competitive programming fan, but not a programmer. That's why she asked for your help. Help her minimize the number of steps.
Input
The first line of input contains integer n (1 ≤ n ≤ 106), the number of weights.
The second line contains n integers w1, ..., wn separated by spaces (0 ≤ wi ≤ 106 for each 1 ≤ i ≤ n), the powers of two forming the weights values.
Output
Print the minimum number of steps in a single line.
Examples
Input
5
1 1 2 3 3
Output
2
Input
4
0 1 2 3
Output
4
Note
In the first sample case: One optimal way would be to throw away the first three in the first step and the rest in the second step. Also, it's not possible to do it in one step because their sum is not a power of two.
In the second sample case: The only optimal way is to throw away one weight in each step. It's not possible to do it in less than 4 steps because there's no subset of weights with more than one weight and sum equal to a power of two. | instruction | 0 | 83,801 | 17 | 167,602 |
Tags: greedy
Correct Solution:
```
n = int(input())
MAXN = 10**6 + 20
bit = [0]*MAXN
for s in input().split(' '):
ss = int(s)
bit[ss] += 1
while bit[ss] > 1:
bit[ss+1] += 1
bit[ss] = bit[ss] & 1
ss += 1
print(sum(bit))
``` | output | 1 | 83,801 | 17 | 167,603 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The last stage of Football World Cup is played using the play-off system.
There are n teams left in this stage, they are enumerated from 1 to n. Several rounds are held, in each round the remaining teams are sorted in the order of their ids, then the first in this order plays with the second, the third — with the fourth, the fifth — with the sixth, and so on. It is guaranteed that in each round there is even number of teams. The winner of each game advances to the next round, the loser is eliminated from the tournament, there are no draws. In the last round there is the only game with two remaining teams: the round is called the Final, the winner is called the champion, and the tournament is over.
Arkady wants his two favorite teams to play in the Final. Unfortunately, the team ids are already determined, and it may happen that it is impossible for teams to meet in the Final, because they are to meet in some earlier stage, if they are strong enough. Determine, in which round the teams with ids a and b can meet.
Input
The only line contains three integers n, a and b (2 ≤ n ≤ 256, 1 ≤ a, b ≤ n) — the total number of teams, and the ids of the teams that Arkady is interested in.
It is guaranteed that n is such that in each round an even number of team advance, and that a and b are not equal.
Output
In the only line print "Final!" (without quotes), if teams a and b can meet in the Final.
Otherwise, print a single integer — the number of the round in which teams a and b can meet. The round are enumerated from 1.
Examples
Input
4 1 2
Output
1
Input
8 2 6
Output
Final!
Input
8 7 5
Output
2
Note
In the first example teams 1 and 2 meet in the first round.
In the second example teams 2 and 6 can only meet in the third round, which is the Final, if they win all their opponents in earlier rounds.
In the third example the teams with ids 7 and 5 can meet in the second round, if they win their opponents in the first round. | instruction | 0 | 83,916 | 17 | 167,832 |
Tags: constructive algorithms, implementation
Correct Solution:
```
n, a, b = list(map(int, input().split()))
ls = list(range(1, n + 1))
k = 0
rounds = 0
m = n
while m > 1:
m >>= 1
rounds += 1
# print(rounds)
while len(ls) > 1:
# print(ls)
k += 1
newls = []
for i in range(0, len(ls), 2):
if (ls[i] == a or ls[i] == b) and (ls[i + 1] == a or ls[i + 1] == b):
if k == rounds:
print("Final!")
else:
print(k)
newls = []
break
elif ls[i] == a or ls[i + 1] == a:
newls.append(a)
elif ls[i] == b or ls[i + 1] == b:
newls.append(b)
else:
newls.append(ls[i])
ls = newls
``` | output | 1 | 83,916 | 17 | 167,833 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The last stage of Football World Cup is played using the play-off system.
There are n teams left in this stage, they are enumerated from 1 to n. Several rounds are held, in each round the remaining teams are sorted in the order of their ids, then the first in this order plays with the second, the third — with the fourth, the fifth — with the sixth, and so on. It is guaranteed that in each round there is even number of teams. The winner of each game advances to the next round, the loser is eliminated from the tournament, there are no draws. In the last round there is the only game with two remaining teams: the round is called the Final, the winner is called the champion, and the tournament is over.
Arkady wants his two favorite teams to play in the Final. Unfortunately, the team ids are already determined, and it may happen that it is impossible for teams to meet in the Final, because they are to meet in some earlier stage, if they are strong enough. Determine, in which round the teams with ids a and b can meet.
Input
The only line contains three integers n, a and b (2 ≤ n ≤ 256, 1 ≤ a, b ≤ n) — the total number of teams, and the ids of the teams that Arkady is interested in.
It is guaranteed that n is such that in each round an even number of team advance, and that a and b are not equal.
Output
In the only line print "Final!" (without quotes), if teams a and b can meet in the Final.
Otherwise, print a single integer — the number of the round in which teams a and b can meet. The round are enumerated from 1.
Examples
Input
4 1 2
Output
1
Input
8 2 6
Output
Final!
Input
8 7 5
Output
2
Note
In the first example teams 1 and 2 meet in the first round.
In the second example teams 2 and 6 can only meet in the third round, which is the Final, if they win all their opponents in earlier rounds.
In the third example the teams with ids 7 and 5 can meet in the second round, if they win their opponents in the first round. | instruction | 0 | 83,917 | 17 | 167,834 |
Tags: constructive algorithms, implementation
Correct Solution:
```
#!/usr/bin/env python3
import sys
[n, a, b] = map(int, sys.stdin.readline().strip().split())
def log2(n):
if n == 1:
return 0
else:
return log2(n // 2) + 1
def zf(x, d):
return bin(x)[2:].zfill(d)
def common_suffix_len(a, b):
i = 0
for ca, cb in zip(a, b):
if ca != cb:
break
i += 1
return i
d = log2(n)
sl = common_suffix_len(zf(a - 1, d), zf(b - 1, d))
if sl == 0:
print ('Final!')
else:
print (d - sl)
``` | output | 1 | 83,917 | 17 | 167,835 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The last stage of Football World Cup is played using the play-off system.
There are n teams left in this stage, they are enumerated from 1 to n. Several rounds are held, in each round the remaining teams are sorted in the order of their ids, then the first in this order plays with the second, the third — with the fourth, the fifth — with the sixth, and so on. It is guaranteed that in each round there is even number of teams. The winner of each game advances to the next round, the loser is eliminated from the tournament, there are no draws. In the last round there is the only game with two remaining teams: the round is called the Final, the winner is called the champion, and the tournament is over.
Arkady wants his two favorite teams to play in the Final. Unfortunately, the team ids are already determined, and it may happen that it is impossible for teams to meet in the Final, because they are to meet in some earlier stage, if they are strong enough. Determine, in which round the teams with ids a and b can meet.
Input
The only line contains three integers n, a and b (2 ≤ n ≤ 256, 1 ≤ a, b ≤ n) — the total number of teams, and the ids of the teams that Arkady is interested in.
It is guaranteed that n is such that in each round an even number of team advance, and that a and b are not equal.
Output
In the only line print "Final!" (without quotes), if teams a and b can meet in the Final.
Otherwise, print a single integer — the number of the round in which teams a and b can meet. The round are enumerated from 1.
Examples
Input
4 1 2
Output
1
Input
8 2 6
Output
Final!
Input
8 7 5
Output
2
Note
In the first example teams 1 and 2 meet in the first round.
In the second example teams 2 and 6 can only meet in the third round, which is the Final, if they win all their opponents in earlier rounds.
In the third example the teams with ids 7 and 5 can meet in the second round, if they win their opponents in the first round. | instruction | 0 | 83,918 | 17 | 167,836 |
Tags: constructive algorithms, implementation
Correct Solution:
```
from math import log
n, a, b = map(int, input().split())
a -= 1
b -= 1
max_r = int(log(n, 2)) #n would be power of 2 , bcz there will alaways be even no of matches
#n n//2 n/4......1
c = 0
while a // 2 != b // 2: #if a//2==b//2 thwy will be next to each other and then they will fight and one of them will win
a //= 2
b //= 2
n //= 2
c += 1
c += 1
#our target is that these 2 teams would meet in final,so whatever match they play they will win,,untill unless they both
#are opponent to each other
print("Final!" if c == max_r else c)
``` | output | 1 | 83,918 | 17 | 167,837 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The last stage of Football World Cup is played using the play-off system.
There are n teams left in this stage, they are enumerated from 1 to n. Several rounds are held, in each round the remaining teams are sorted in the order of their ids, then the first in this order plays with the second, the third — with the fourth, the fifth — with the sixth, and so on. It is guaranteed that in each round there is even number of teams. The winner of each game advances to the next round, the loser is eliminated from the tournament, there are no draws. In the last round there is the only game with two remaining teams: the round is called the Final, the winner is called the champion, and the tournament is over.
Arkady wants his two favorite teams to play in the Final. Unfortunately, the team ids are already determined, and it may happen that it is impossible for teams to meet in the Final, because they are to meet in some earlier stage, if they are strong enough. Determine, in which round the teams with ids a and b can meet.
Input
The only line contains three integers n, a and b (2 ≤ n ≤ 256, 1 ≤ a, b ≤ n) — the total number of teams, and the ids of the teams that Arkady is interested in.
It is guaranteed that n is such that in each round an even number of team advance, and that a and b are not equal.
Output
In the only line print "Final!" (without quotes), if teams a and b can meet in the Final.
Otherwise, print a single integer — the number of the round in which teams a and b can meet. The round are enumerated from 1.
Examples
Input
4 1 2
Output
1
Input
8 2 6
Output
Final!
Input
8 7 5
Output
2
Note
In the first example teams 1 and 2 meet in the first round.
In the second example teams 2 and 6 can only meet in the third round, which is the Final, if they win all their opponents in earlier rounds.
In the third example the teams with ids 7 and 5 can meet in the second round, if they win their opponents in the first round. | instruction | 0 | 83,919 | 17 | 167,838 |
Tags: constructive algorithms, implementation
Correct Solution:
```
n, a, b = [int(v) for v in input().split()]
a -= 1
b -= 1
r = 1
while True:
nn = n // 2
aa = a // 2
bb = b // 2
if aa == bb:
print("Final!" if nn == 1 else r)
break
n, a, b = nn, aa, bb
r += 1
``` | output | 1 | 83,919 | 17 | 167,839 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The last stage of Football World Cup is played using the play-off system.
There are n teams left in this stage, they are enumerated from 1 to n. Several rounds are held, in each round the remaining teams are sorted in the order of their ids, then the first in this order plays with the second, the third — with the fourth, the fifth — with the sixth, and so on. It is guaranteed that in each round there is even number of teams. The winner of each game advances to the next round, the loser is eliminated from the tournament, there are no draws. In the last round there is the only game with two remaining teams: the round is called the Final, the winner is called the champion, and the tournament is over.
Arkady wants his two favorite teams to play in the Final. Unfortunately, the team ids are already determined, and it may happen that it is impossible for teams to meet in the Final, because they are to meet in some earlier stage, if they are strong enough. Determine, in which round the teams with ids a and b can meet.
Input
The only line contains three integers n, a and b (2 ≤ n ≤ 256, 1 ≤ a, b ≤ n) — the total number of teams, and the ids of the teams that Arkady is interested in.
It is guaranteed that n is such that in each round an even number of team advance, and that a and b are not equal.
Output
In the only line print "Final!" (without quotes), if teams a and b can meet in the Final.
Otherwise, print a single integer — the number of the round in which teams a and b can meet. The round are enumerated from 1.
Examples
Input
4 1 2
Output
1
Input
8 2 6
Output
Final!
Input
8 7 5
Output
2
Note
In the first example teams 1 and 2 meet in the first round.
In the second example teams 2 and 6 can only meet in the third round, which is the Final, if they win all their opponents in earlier rounds.
In the third example the teams with ids 7 and 5 can meet in the second round, if they win their opponents in the first round. | instruction | 0 | 83,920 | 17 | 167,840 |
Tags: constructive algorithms, implementation
Correct Solution:
```
import math
n,a,b=map(int,input().split())
arr=[i for i in range(1,n+1)]
i,r=0,1
#print(arr)
while True:
arr1=[]
for i in range(0,len(arr),2):
if (arr[i]==a and arr[i+1]==b) or (arr[i]==b and arr[i+1]==a):
if len(arr)==2:
print("Final!")
else:
print(r)
exit(0)
else:
if arr[i+1]==a or arr[i]==a:
arr1.append(a)
elif arr[i+1]==b or arr[i]==b:
arr1.append(b)
else:
arr1.append(arr[i])
arr=arr1
r+=1
#print(arr)
``` | output | 1 | 83,920 | 17 | 167,841 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The last stage of Football World Cup is played using the play-off system.
There are n teams left in this stage, they are enumerated from 1 to n. Several rounds are held, in each round the remaining teams are sorted in the order of their ids, then the first in this order plays with the second, the third — with the fourth, the fifth — with the sixth, and so on. It is guaranteed that in each round there is even number of teams. The winner of each game advances to the next round, the loser is eliminated from the tournament, there are no draws. In the last round there is the only game with two remaining teams: the round is called the Final, the winner is called the champion, and the tournament is over.
Arkady wants his two favorite teams to play in the Final. Unfortunately, the team ids are already determined, and it may happen that it is impossible for teams to meet in the Final, because they are to meet in some earlier stage, if they are strong enough. Determine, in which round the teams with ids a and b can meet.
Input
The only line contains three integers n, a and b (2 ≤ n ≤ 256, 1 ≤ a, b ≤ n) — the total number of teams, and the ids of the teams that Arkady is interested in.
It is guaranteed that n is such that in each round an even number of team advance, and that a and b are not equal.
Output
In the only line print "Final!" (without quotes), if teams a and b can meet in the Final.
Otherwise, print a single integer — the number of the round in which teams a and b can meet. The round are enumerated from 1.
Examples
Input
4 1 2
Output
1
Input
8 2 6
Output
Final!
Input
8 7 5
Output
2
Note
In the first example teams 1 and 2 meet in the first round.
In the second example teams 2 and 6 can only meet in the third round, which is the Final, if they win all their opponents in earlier rounds.
In the third example the teams with ids 7 and 5 can meet in the second round, if they win their opponents in the first round. | instruction | 0 | 83,921 | 17 | 167,842 |
Tags: constructive algorithms, implementation
Correct Solution:
```
import sys
from math import log
input = sys.stdin.readline
n,a,b = map(int,input().split())
x = [[i] for i in range(1,n+1)]
for i in range(int(log(n,2))):
for z in x:
if a in z and b in z:
print(i)
sys.exit(0)
new = []
for k in range(0,len(x)//2):
new.append(x[k*2]+x[k*2+1])
x = new[:]
print("Final!")
``` | output | 1 | 83,921 | 17 | 167,843 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The last stage of Football World Cup is played using the play-off system.
There are n teams left in this stage, they are enumerated from 1 to n. Several rounds are held, in each round the remaining teams are sorted in the order of their ids, then the first in this order plays with the second, the third — with the fourth, the fifth — with the sixth, and so on. It is guaranteed that in each round there is even number of teams. The winner of each game advances to the next round, the loser is eliminated from the tournament, there are no draws. In the last round there is the only game with two remaining teams: the round is called the Final, the winner is called the champion, and the tournament is over.
Arkady wants his two favorite teams to play in the Final. Unfortunately, the team ids are already determined, and it may happen that it is impossible for teams to meet in the Final, because they are to meet in some earlier stage, if they are strong enough. Determine, in which round the teams with ids a and b can meet.
Input
The only line contains three integers n, a and b (2 ≤ n ≤ 256, 1 ≤ a, b ≤ n) — the total number of teams, and the ids of the teams that Arkady is interested in.
It is guaranteed that n is such that in each round an even number of team advance, and that a and b are not equal.
Output
In the only line print "Final!" (without quotes), if teams a and b can meet in the Final.
Otherwise, print a single integer — the number of the round in which teams a and b can meet. The round are enumerated from 1.
Examples
Input
4 1 2
Output
1
Input
8 2 6
Output
Final!
Input
8 7 5
Output
2
Note
In the first example teams 1 and 2 meet in the first round.
In the second example teams 2 and 6 can only meet in the third round, which is the Final, if they win all their opponents in earlier rounds.
In the third example the teams with ids 7 and 5 can meet in the second round, if they win their opponents in the first round. | instruction | 0 | 83,922 | 17 | 167,844 |
Tags: constructive algorithms, implementation
Correct Solution:
```
from math import log2
def bin_(x):
l_ = ln
return str(bin(x))[2:].zfill(l_)
n, a, b = map(int, input().split())
ln = int(log2(n)) + 1
a_, b_ = bin_(a - 1), bin_(b - 1)
for i in range(ln):
if a_[i] != b_[i]:
ans = i
break
else:
ans = 0
print((ln - ans, 'Final!')[(ln - ans) % (ln - 1) == 0])
``` | output | 1 | 83,922 | 17 | 167,845 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The last stage of Football World Cup is played using the play-off system.
There are n teams left in this stage, they are enumerated from 1 to n. Several rounds are held, in each round the remaining teams are sorted in the order of their ids, then the first in this order plays with the second, the third — with the fourth, the fifth — with the sixth, and so on. It is guaranteed that in each round there is even number of teams. The winner of each game advances to the next round, the loser is eliminated from the tournament, there are no draws. In the last round there is the only game with two remaining teams: the round is called the Final, the winner is called the champion, and the tournament is over.
Arkady wants his two favorite teams to play in the Final. Unfortunately, the team ids are already determined, and it may happen that it is impossible for teams to meet in the Final, because they are to meet in some earlier stage, if they are strong enough. Determine, in which round the teams with ids a and b can meet.
Input
The only line contains three integers n, a and b (2 ≤ n ≤ 256, 1 ≤ a, b ≤ n) — the total number of teams, and the ids of the teams that Arkady is interested in.
It is guaranteed that n is such that in each round an even number of team advance, and that a and b are not equal.
Output
In the only line print "Final!" (without quotes), if teams a and b can meet in the Final.
Otherwise, print a single integer — the number of the round in which teams a and b can meet. The round are enumerated from 1.
Examples
Input
4 1 2
Output
1
Input
8 2 6
Output
Final!
Input
8 7 5
Output
2
Note
In the first example teams 1 and 2 meet in the first round.
In the second example teams 2 and 6 can only meet in the third round, which is the Final, if they win all their opponents in earlier rounds.
In the third example the teams with ids 7 and 5 can meet in the second round, if they win their opponents in the first round. | instruction | 0 | 83,923 | 17 | 167,846 |
Tags: constructive algorithms, implementation
Correct Solution:
```
import sys
input=sys.stdin.readline
def solve(n,a,b):
if(a%2==1 and b%2==0 and b==a+1):
return 1
return 1+solve(n//2,(a+1)//2,(b+1)//2)
l=input().split()
n=int(l[0])
a=int(l[1])
b=int(l[2])
if(a>b):
a,b=b,a
z=solve(n,a,b)
if(pow(2,z)==n):
print("Final!")
else:
print(z)
``` | output | 1 | 83,923 | 17 | 167,847 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The last stage of Football World Cup is played using the play-off system.
There are n teams left in this stage, they are enumerated from 1 to n. Several rounds are held, in each round the remaining teams are sorted in the order of their ids, then the first in this order plays with the second, the third — with the fourth, the fifth — with the sixth, and so on. It is guaranteed that in each round there is even number of teams. The winner of each game advances to the next round, the loser is eliminated from the tournament, there are no draws. In the last round there is the only game with two remaining teams: the round is called the Final, the winner is called the champion, and the tournament is over.
Arkady wants his two favorite teams to play in the Final. Unfortunately, the team ids are already determined, and it may happen that it is impossible for teams to meet in the Final, because they are to meet in some earlier stage, if they are strong enough. Determine, in which round the teams with ids a and b can meet.
Input
The only line contains three integers n, a and b (2 ≤ n ≤ 256, 1 ≤ a, b ≤ n) — the total number of teams, and the ids of the teams that Arkady is interested in.
It is guaranteed that n is such that in each round an even number of team advance, and that a and b are not equal.
Output
In the only line print "Final!" (without quotes), if teams a and b can meet in the Final.
Otherwise, print a single integer — the number of the round in which teams a and b can meet. The round are enumerated from 1.
Examples
Input
4 1 2
Output
1
Input
8 2 6
Output
Final!
Input
8 7 5
Output
2
Note
In the first example teams 1 and 2 meet in the first round.
In the second example teams 2 and 6 can only meet in the third round, which is the Final, if they win all their opponents in earlier rounds.
In the third example the teams with ids 7 and 5 can meet in the second round, if they win their opponents in the first round.
Submitted Solution:
```
import math
arr = list(map(int, input().split()))
n = arr[0]
rounds = math.log2(n)
a = arr[1]
a = int((a%2 + a)/2)
b = arr[2]
b = int((b%2 + b)/2)
roundCount = 1
while a!=b:
a=math.ceil(a/2)
b=math.ceil(b/2)
roundCount+=1
if roundCount == rounds:
print ("Final!")
else:
print (roundCount)
``` | instruction | 0 | 83,924 | 17 | 167,848 |
Yes | output | 1 | 83,924 | 17 | 167,849 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The last stage of Football World Cup is played using the play-off system.
There are n teams left in this stage, they are enumerated from 1 to n. Several rounds are held, in each round the remaining teams are sorted in the order of their ids, then the first in this order plays with the second, the third — with the fourth, the fifth — with the sixth, and so on. It is guaranteed that in each round there is even number of teams. The winner of each game advances to the next round, the loser is eliminated from the tournament, there are no draws. In the last round there is the only game with two remaining teams: the round is called the Final, the winner is called the champion, and the tournament is over.
Arkady wants his two favorite teams to play in the Final. Unfortunately, the team ids are already determined, and it may happen that it is impossible for teams to meet in the Final, because they are to meet in some earlier stage, if they are strong enough. Determine, in which round the teams with ids a and b can meet.
Input
The only line contains three integers n, a and b (2 ≤ n ≤ 256, 1 ≤ a, b ≤ n) — the total number of teams, and the ids of the teams that Arkady is interested in.
It is guaranteed that n is such that in each round an even number of team advance, and that a and b are not equal.
Output
In the only line print "Final!" (without quotes), if teams a and b can meet in the Final.
Otherwise, print a single integer — the number of the round in which teams a and b can meet. The round are enumerated from 1.
Examples
Input
4 1 2
Output
1
Input
8 2 6
Output
Final!
Input
8 7 5
Output
2
Note
In the first example teams 1 and 2 meet in the first round.
In the second example teams 2 and 6 can only meet in the third round, which is the Final, if they win all their opponents in earlier rounds.
In the third example the teams with ids 7 and 5 can meet in the second round, if they win their opponents in the first round.
Submitted Solution:
```
line = input().split(' ')
n = int(line[0])
a = int(line[1])
b = int(line[2])
teams = list(range(1, n+1))
rounds=0
while len(teams) > 2:
rounds+=1
new_teams=[]
for index in range(0, len(teams)//2):
x = teams[index*2]
y = teams[index*2 + 1]
if x==a and y==b or x==b and y==a:
break
if a in [x, y]:
new_teams.append(a)
else:
if b in [x, y]:
new_teams.append(b)
else:
new_teams.append(max(x, y))
else:
teams=new_teams
continue
break
else:
print("Final!")
if len(teams) > 2:
print(rounds)
``` | instruction | 0 | 83,925 | 17 | 167,850 |
Yes | output | 1 | 83,925 | 17 | 167,851 |
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