message stringlengths 2 67k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 463 109k | cluster float64 19 19 | __index_level_0__ int64 926 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are b boys and g girls participating in Olympiad of Metropolises. There will be a board games tournament in the evening and n participants have accepted the invitation. The organizers do not know how many boys and girls are among them.
Organizers are preparing red badges for girls and blue ones for boys.
Vasya prepared n+1 decks of badges. The i-th (where i is from 0 to n, inclusive) deck contains i blue badges and n-i red ones. The total number of badges in any deck is exactly n.
Determine the minimum number of decks among these n+1 that Vasya should take, so that there will be a suitable deck no matter how many girls and boys there will be among the participants of the tournament.
Input
The first line contains an integer b (1 ≤ b ≤ 300), the number of boys.
The second line contains an integer g (1 ≤ g ≤ 300), the number of girls.
The third line contains an integer n (1 ≤ n ≤ b + g), the number of the board games tournament participants.
Output
Output the only integer, the minimum number of badge decks that Vasya could take.
Examples
Input
5
6
3
Output
4
Input
5
3
5
Output
4
Note
In the first example, each of 4 decks should be taken: (0 blue, 3 red), (1 blue, 2 red), (2 blue, 1 red), (3 blue, 0 red).
In the second example, 4 decks should be taken: (2 blue, 3 red), (3 blue, 2 red), (4 blue, 1 red), (5 blue, 0 red). Piles (0 blue, 5 red) and (1 blue, 4 red) can not be used.
Submitted Solution:
```
b=int(input())
g=int(input())
n=int(input())
print(min(min(b,g)-(n-min(max(b,g),n)),min(max(b,g),n))+1)
``` | instruction | 0 | 75,089 | 19 | 150,178 |
Yes | output | 1 | 75,089 | 19 | 150,179 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are b boys and g girls participating in Olympiad of Metropolises. There will be a board games tournament in the evening and n participants have accepted the invitation. The organizers do not know how many boys and girls are among them.
Organizers are preparing red badges for girls and blue ones for boys.
Vasya prepared n+1 decks of badges. The i-th (where i is from 0 to n, inclusive) deck contains i blue badges and n-i red ones. The total number of badges in any deck is exactly n.
Determine the minimum number of decks among these n+1 that Vasya should take, so that there will be a suitable deck no matter how many girls and boys there will be among the participants of the tournament.
Input
The first line contains an integer b (1 ≤ b ≤ 300), the number of boys.
The second line contains an integer g (1 ≤ g ≤ 300), the number of girls.
The third line contains an integer n (1 ≤ n ≤ b + g), the number of the board games tournament participants.
Output
Output the only integer, the minimum number of badge decks that Vasya could take.
Examples
Input
5
6
3
Output
4
Input
5
3
5
Output
4
Note
In the first example, each of 4 decks should be taken: (0 blue, 3 red), (1 blue, 2 red), (2 blue, 1 red), (3 blue, 0 red).
In the second example, 4 decks should be taken: (2 blue, 3 red), (3 blue, 2 red), (4 blue, 1 red), (5 blue, 0 red). Piles (0 blue, 5 red) and (1 blue, 4 red) can not be used.
Submitted Solution:
```
b = int(input())
g = int(input())
n = int(input())
if (b+g>n):
r = min(b,g,n)+1
else : r = 1
print(r)
``` | instruction | 0 | 75,090 | 19 | 150,180 |
No | output | 1 | 75,090 | 19 | 150,181 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are b boys and g girls participating in Olympiad of Metropolises. There will be a board games tournament in the evening and n participants have accepted the invitation. The organizers do not know how many boys and girls are among them.
Organizers are preparing red badges for girls and blue ones for boys.
Vasya prepared n+1 decks of badges. The i-th (where i is from 0 to n, inclusive) deck contains i blue badges and n-i red ones. The total number of badges in any deck is exactly n.
Determine the minimum number of decks among these n+1 that Vasya should take, so that there will be a suitable deck no matter how many girls and boys there will be among the participants of the tournament.
Input
The first line contains an integer b (1 ≤ b ≤ 300), the number of boys.
The second line contains an integer g (1 ≤ g ≤ 300), the number of girls.
The third line contains an integer n (1 ≤ n ≤ b + g), the number of the board games tournament participants.
Output
Output the only integer, the minimum number of badge decks that Vasya could take.
Examples
Input
5
6
3
Output
4
Input
5
3
5
Output
4
Note
In the first example, each of 4 decks should be taken: (0 blue, 3 red), (1 blue, 2 red), (2 blue, 1 red), (3 blue, 0 red).
In the second example, 4 decks should be taken: (2 blue, 3 red), (3 blue, 2 red), (4 blue, 1 red), (5 blue, 0 red). Piles (0 blue, 5 red) and (1 blue, 4 red) can not be used.
Submitted Solution:
```
a=int(input())
b=int(input())
c=int(input())
d=[]
e=[]
ans=0
s=0
s1=0
for i in range(0,c+1):
d.append(i)
e.append(c-i)
for i in range(0,len(d)):
s=s+d[i]
s1=s1+e[i]
ans=ans+1
if(s>=a and s1>=b):
break
print(ans)
``` | instruction | 0 | 75,091 | 19 | 150,182 |
No | output | 1 | 75,091 | 19 | 150,183 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are b boys and g girls participating in Olympiad of Metropolises. There will be a board games tournament in the evening and n participants have accepted the invitation. The organizers do not know how many boys and girls are among them.
Organizers are preparing red badges for girls and blue ones for boys.
Vasya prepared n+1 decks of badges. The i-th (where i is from 0 to n, inclusive) deck contains i blue badges and n-i red ones. The total number of badges in any deck is exactly n.
Determine the minimum number of decks among these n+1 that Vasya should take, so that there will be a suitable deck no matter how many girls and boys there will be among the participants of the tournament.
Input
The first line contains an integer b (1 ≤ b ≤ 300), the number of boys.
The second line contains an integer g (1 ≤ g ≤ 300), the number of girls.
The third line contains an integer n (1 ≤ n ≤ b + g), the number of the board games tournament participants.
Output
Output the only integer, the minimum number of badge decks that Vasya could take.
Examples
Input
5
6
3
Output
4
Input
5
3
5
Output
4
Note
In the first example, each of 4 decks should be taken: (0 blue, 3 red), (1 blue, 2 red), (2 blue, 1 red), (3 blue, 0 red).
In the second example, 4 decks should be taken: (2 blue, 3 red), (3 blue, 2 red), (4 blue, 1 red), (5 blue, 0 red). Piles (0 blue, 5 red) and (1 blue, 4 red) can not be used.
Submitted Solution:
```
a=int(input())
b=int(input())
c=int(input())
List=[a,b,c]
List.sort()
if c<a<b:
print(c+1)
elif (a+b)==c:
print(1)
else:
print(List[0]+1)
``` | instruction | 0 | 75,092 | 19 | 150,184 |
No | output | 1 | 75,092 | 19 | 150,185 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are b boys and g girls participating in Olympiad of Metropolises. There will be a board games tournament in the evening and n participants have accepted the invitation. The organizers do not know how many boys and girls are among them.
Organizers are preparing red badges for girls and blue ones for boys.
Vasya prepared n+1 decks of badges. The i-th (where i is from 0 to n, inclusive) deck contains i blue badges and n-i red ones. The total number of badges in any deck is exactly n.
Determine the minimum number of decks among these n+1 that Vasya should take, so that there will be a suitable deck no matter how many girls and boys there will be among the participants of the tournament.
Input
The first line contains an integer b (1 ≤ b ≤ 300), the number of boys.
The second line contains an integer g (1 ≤ g ≤ 300), the number of girls.
The third line contains an integer n (1 ≤ n ≤ b + g), the number of the board games tournament participants.
Output
Output the only integer, the minimum number of badge decks that Vasya could take.
Examples
Input
5
6
3
Output
4
Input
5
3
5
Output
4
Note
In the first example, each of 4 decks should be taken: (0 blue, 3 red), (1 blue, 2 red), (2 blue, 1 red), (3 blue, 0 red).
In the second example, 4 decks should be taken: (2 blue, 3 red), (3 blue, 2 red), (4 blue, 1 red), (5 blue, 0 red). Piles (0 blue, 5 red) and (1 blue, 4 red) can not be used.
Submitted Solution:
```
b=int(input())
g=int(input())
n=int(input())
x=min(b,g)
if(b==g and x<n):
print(1)
elif(x<n):
print(x+1)
else:
print(n+1)
``` | instruction | 0 | 75,093 | 19 | 150,186 |
No | output | 1 | 75,093 | 19 | 150,187 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You've got a rectangular table with length a and width b and the infinite number of plates of radius r. Two players play the following game: they take turns to put the plates on the table so that the plates don't lie on each other (but they can touch each other), and so that any point on any plate is located within the table's border. During the game one cannot move the plates that already lie on the table. The player who cannot make another move loses. Determine which player wins, the one who moves first or the one who moves second, provided that both players play optimally well.
Input
A single line contains three space-separated integers a, b, r (1 ≤ a, b, r ≤ 100) — the table sides and the plates' radius, correspondingly.
Output
If wins the player who moves first, print "First" (without the quotes). Otherwise print "Second" (without the quotes).
Examples
Input
5 5 2
Output
First
Input
6 7 4
Output
Second
Note
In the first sample the table has place for only one plate. The first player puts a plate on the table, the second player can't do that and loses.
<image>
In the second sample the table is so small that it doesn't have enough place even for one plate. So the first player loses without making a single move.
<image> | instruction | 0 | 75,303 | 19 | 150,606 |
Tags: constructive algorithms, games, math
Correct Solution:
```
data = input()
datalist = data.split()
a = int(datalist[0])
b = int(datalist[1])
r = int(datalist[2])
if a >= 2*r and b >= 2*r:
print("First")
else:
print("Second")
``` | output | 1 | 75,303 | 19 | 150,607 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You've got a rectangular table with length a and width b and the infinite number of plates of radius r. Two players play the following game: they take turns to put the plates on the table so that the plates don't lie on each other (but they can touch each other), and so that any point on any plate is located within the table's border. During the game one cannot move the plates that already lie on the table. The player who cannot make another move loses. Determine which player wins, the one who moves first or the one who moves second, provided that both players play optimally well.
Input
A single line contains three space-separated integers a, b, r (1 ≤ a, b, r ≤ 100) — the table sides and the plates' radius, correspondingly.
Output
If wins the player who moves first, print "First" (without the quotes). Otherwise print "Second" (without the quotes).
Examples
Input
5 5 2
Output
First
Input
6 7 4
Output
Second
Note
In the first sample the table has place for only one plate. The first player puts a plate on the table, the second player can't do that and loses.
<image>
In the second sample the table is so small that it doesn't have enough place even for one plate. So the first player loses without making a single move.
<image> | instruction | 0 | 75,304 | 19 | 150,608 |
Tags: constructive algorithms, games, math
Correct Solution:
```
x=list(map(int,input().split()))
c=min(x[1],x[0])
v="Second"
if c>=x[2]*2:
v="First"
print(v)
``` | output | 1 | 75,304 | 19 | 150,609 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You've got a rectangular table with length a and width b and the infinite number of plates of radius r. Two players play the following game: they take turns to put the plates on the table so that the plates don't lie on each other (but they can touch each other), and so that any point on any plate is located within the table's border. During the game one cannot move the plates that already lie on the table. The player who cannot make another move loses. Determine which player wins, the one who moves first or the one who moves second, provided that both players play optimally well.
Input
A single line contains three space-separated integers a, b, r (1 ≤ a, b, r ≤ 100) — the table sides and the plates' radius, correspondingly.
Output
If wins the player who moves first, print "First" (without the quotes). Otherwise print "Second" (without the quotes).
Examples
Input
5 5 2
Output
First
Input
6 7 4
Output
Second
Note
In the first sample the table has place for only one plate. The first player puts a plate on the table, the second player can't do that and loses.
<image>
In the second sample the table is so small that it doesn't have enough place even for one plate. So the first player loses without making a single move.
<image> | instruction | 0 | 75,305 | 19 | 150,610 |
Tags: constructive algorithms, games, math
Correct Solution:
```
a, b, r = map(int, input().split())
if a < 2 * r or b < 2 * r:
print("Second")
else:
print("First")
``` | output | 1 | 75,305 | 19 | 150,611 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You've got a rectangular table with length a and width b and the infinite number of plates of radius r. Two players play the following game: they take turns to put the plates on the table so that the plates don't lie on each other (but they can touch each other), and so that any point on any plate is located within the table's border. During the game one cannot move the plates that already lie on the table. The player who cannot make another move loses. Determine which player wins, the one who moves first or the one who moves second, provided that both players play optimally well.
Input
A single line contains three space-separated integers a, b, r (1 ≤ a, b, r ≤ 100) — the table sides and the plates' radius, correspondingly.
Output
If wins the player who moves first, print "First" (without the quotes). Otherwise print "Second" (without the quotes).
Examples
Input
5 5 2
Output
First
Input
6 7 4
Output
Second
Note
In the first sample the table has place for only one plate. The first player puts a plate on the table, the second player can't do that and loses.
<image>
In the second sample the table is so small that it doesn't have enough place even for one plate. So the first player loses without making a single move.
<image> | instruction | 0 | 75,306 | 19 | 150,612 |
Tags: constructive algorithms, games, math
Correct Solution:
```
a,b,c=map(int,input().split())
if min(a,b)>=2*c:
print('First')
else:
print('Second')
``` | output | 1 | 75,306 | 19 | 150,613 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You've got a rectangular table with length a and width b and the infinite number of plates of radius r. Two players play the following game: they take turns to put the plates on the table so that the plates don't lie on each other (but they can touch each other), and so that any point on any plate is located within the table's border. During the game one cannot move the plates that already lie on the table. The player who cannot make another move loses. Determine which player wins, the one who moves first or the one who moves second, provided that both players play optimally well.
Input
A single line contains three space-separated integers a, b, r (1 ≤ a, b, r ≤ 100) — the table sides and the plates' radius, correspondingly.
Output
If wins the player who moves first, print "First" (without the quotes). Otherwise print "Second" (without the quotes).
Examples
Input
5 5 2
Output
First
Input
6 7 4
Output
Second
Note
In the first sample the table has place for only one plate. The first player puts a plate on the table, the second player can't do that and loses.
<image>
In the second sample the table is so small that it doesn't have enough place even for one plate. So the first player loses without making a single move.
<image> | instruction | 0 | 75,307 | 19 | 150,614 |
Tags: constructive algorithms, games, math
Correct Solution:
```
s = input().split()
a = int(s[0])
b = int(s[1])
r = int(s[2])
if 2*r>min(a,b):
print("Second")
else:
print("First")
``` | output | 1 | 75,307 | 19 | 150,615 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You've got a rectangular table with length a and width b and the infinite number of plates of radius r. Two players play the following game: they take turns to put the plates on the table so that the plates don't lie on each other (but they can touch each other), and so that any point on any plate is located within the table's border. During the game one cannot move the plates that already lie on the table. The player who cannot make another move loses. Determine which player wins, the one who moves first or the one who moves second, provided that both players play optimally well.
Input
A single line contains three space-separated integers a, b, r (1 ≤ a, b, r ≤ 100) — the table sides and the plates' radius, correspondingly.
Output
If wins the player who moves first, print "First" (without the quotes). Otherwise print "Second" (without the quotes).
Examples
Input
5 5 2
Output
First
Input
6 7 4
Output
Second
Note
In the first sample the table has place for only one plate. The first player puts a plate on the table, the second player can't do that and loses.
<image>
In the second sample the table is so small that it doesn't have enough place even for one plate. So the first player loses without making a single move.
<image> | instruction | 0 | 75,308 | 19 | 150,616 |
Tags: constructive algorithms, games, math
Correct Solution:
```
a,b,r=map(int,input().split())
if a>=r*2 and b>=2*r:
print("First")
else:
print("Second")
``` | output | 1 | 75,308 | 19 | 150,617 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You've got a rectangular table with length a and width b and the infinite number of plates of radius r. Two players play the following game: they take turns to put the plates on the table so that the plates don't lie on each other (but they can touch each other), and so that any point on any plate is located within the table's border. During the game one cannot move the plates that already lie on the table. The player who cannot make another move loses. Determine which player wins, the one who moves first or the one who moves second, provided that both players play optimally well.
Input
A single line contains three space-separated integers a, b, r (1 ≤ a, b, r ≤ 100) — the table sides and the plates' radius, correspondingly.
Output
If wins the player who moves first, print "First" (without the quotes). Otherwise print "Second" (without the quotes).
Examples
Input
5 5 2
Output
First
Input
6 7 4
Output
Second
Note
In the first sample the table has place for only one plate. The first player puts a plate on the table, the second player can't do that and loses.
<image>
In the second sample the table is so small that it doesn't have enough place even for one plate. So the first player loses without making a single move.
<image> | instruction | 0 | 75,309 | 19 | 150,618 |
Tags: constructive algorithms, games, math
Correct Solution:
```
A, B, R = map(int, input().split())
if 2 * R > min(A, B) :
print("Second")
else :
print("First")
``` | output | 1 | 75,309 | 19 | 150,619 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You've got a rectangular table with length a and width b and the infinite number of plates of radius r. Two players play the following game: they take turns to put the plates on the table so that the plates don't lie on each other (but they can touch each other), and so that any point on any plate is located within the table's border. During the game one cannot move the plates that already lie on the table. The player who cannot make another move loses. Determine which player wins, the one who moves first or the one who moves second, provided that both players play optimally well.
Input
A single line contains three space-separated integers a, b, r (1 ≤ a, b, r ≤ 100) — the table sides and the plates' radius, correspondingly.
Output
If wins the player who moves first, print "First" (without the quotes). Otherwise print "Second" (without the quotes).
Examples
Input
5 5 2
Output
First
Input
6 7 4
Output
Second
Note
In the first sample the table has place for only one plate. The first player puts a plate on the table, the second player can't do that and loses.
<image>
In the second sample the table is so small that it doesn't have enough place even for one plate. So the first player loses without making a single move.
<image> | instruction | 0 | 75,310 | 19 | 150,620 |
Tags: constructive algorithms, games, math
Correct Solution:
```
import math
a , b , r = map(int , input().split())
if r * 2 <= min(a , b):
print ('First')
else:
print ('Second')
``` | output | 1 | 75,310 | 19 | 150,621 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've got a rectangular table with length a and width b and the infinite number of plates of radius r. Two players play the following game: they take turns to put the plates on the table so that the plates don't lie on each other (but they can touch each other), and so that any point on any plate is located within the table's border. During the game one cannot move the plates that already lie on the table. The player who cannot make another move loses. Determine which player wins, the one who moves first or the one who moves second, provided that both players play optimally well.
Input
A single line contains three space-separated integers a, b, r (1 ≤ a, b, r ≤ 100) — the table sides and the plates' radius, correspondingly.
Output
If wins the player who moves first, print "First" (without the quotes). Otherwise print "Second" (without the quotes).
Examples
Input
5 5 2
Output
First
Input
6 7 4
Output
Second
Note
In the first sample the table has place for only one plate. The first player puts a plate on the table, the second player can't do that and loses.
<image>
In the second sample the table is so small that it doesn't have enough place even for one plate. So the first player loses without making a single move.
<image>
Submitted Solution:
```
from sys import stdin, stdout
a, b, r = map(int, stdin.readline().split())
if r * 2 <= a and r * 2 <= b:
stdout.write('First')
else:
stdout.write('Second')
``` | instruction | 0 | 75,311 | 19 | 150,622 |
Yes | output | 1 | 75,311 | 19 | 150,623 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've got a rectangular table with length a and width b and the infinite number of plates of radius r. Two players play the following game: they take turns to put the plates on the table so that the plates don't lie on each other (but they can touch each other), and so that any point on any plate is located within the table's border. During the game one cannot move the plates that already lie on the table. The player who cannot make another move loses. Determine which player wins, the one who moves first or the one who moves second, provided that both players play optimally well.
Input
A single line contains three space-separated integers a, b, r (1 ≤ a, b, r ≤ 100) — the table sides and the plates' radius, correspondingly.
Output
If wins the player who moves first, print "First" (without the quotes). Otherwise print "Second" (without the quotes).
Examples
Input
5 5 2
Output
First
Input
6 7 4
Output
Second
Note
In the first sample the table has place for only one plate. The first player puts a plate on the table, the second player can't do that and loses.
<image>
In the second sample the table is so small that it doesn't have enough place even for one plate. So the first player loses without making a single move.
<image>
Submitted Solution:
```
a,b,r=map(int,input().split())
if 2*r>a or 2*r>b:
print("Second")
else :
print("First")
``` | instruction | 0 | 75,312 | 19 | 150,624 |
Yes | output | 1 | 75,312 | 19 | 150,625 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've got a rectangular table with length a and width b and the infinite number of plates of radius r. Two players play the following game: they take turns to put the plates on the table so that the plates don't lie on each other (but they can touch each other), and so that any point on any plate is located within the table's border. During the game one cannot move the plates that already lie on the table. The player who cannot make another move loses. Determine which player wins, the one who moves first or the one who moves second, provided that both players play optimally well.
Input
A single line contains three space-separated integers a, b, r (1 ≤ a, b, r ≤ 100) — the table sides and the plates' radius, correspondingly.
Output
If wins the player who moves first, print "First" (without the quotes). Otherwise print "Second" (without the quotes).
Examples
Input
5 5 2
Output
First
Input
6 7 4
Output
Second
Note
In the first sample the table has place for only one plate. The first player puts a plate on the table, the second player can't do that and loses.
<image>
In the second sample the table is so small that it doesn't have enough place even for one plate. So the first player loses without making a single move.
<image>
Submitted Solution:
```
a, b, r = map(int, input().split())
if a >= 2*r and b >= 2* r :
print('First')
else :
print('Second')
``` | instruction | 0 | 75,313 | 19 | 150,626 |
Yes | output | 1 | 75,313 | 19 | 150,627 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've got a rectangular table with length a and width b and the infinite number of plates of radius r. Two players play the following game: they take turns to put the plates on the table so that the plates don't lie on each other (but they can touch each other), and so that any point on any plate is located within the table's border. During the game one cannot move the plates that already lie on the table. The player who cannot make another move loses. Determine which player wins, the one who moves first or the one who moves second, provided that both players play optimally well.
Input
A single line contains three space-separated integers a, b, r (1 ≤ a, b, r ≤ 100) — the table sides and the plates' radius, correspondingly.
Output
If wins the player who moves first, print "First" (without the quotes). Otherwise print "Second" (without the quotes).
Examples
Input
5 5 2
Output
First
Input
6 7 4
Output
Second
Note
In the first sample the table has place for only one plate. The first player puts a plate on the table, the second player can't do that and loses.
<image>
In the second sample the table is so small that it doesn't have enough place even for one plate. So the first player loses without making a single move.
<image>
Submitted Solution:
```
a,b,r=list(map(int,input().split()))
print('Second'if r+r>min(a,b)else'First')
``` | instruction | 0 | 75,314 | 19 | 150,628 |
Yes | output | 1 | 75,314 | 19 | 150,629 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've got a rectangular table with length a and width b and the infinite number of plates of radius r. Two players play the following game: they take turns to put the plates on the table so that the plates don't lie on each other (but they can touch each other), and so that any point on any plate is located within the table's border. During the game one cannot move the plates that already lie on the table. The player who cannot make another move loses. Determine which player wins, the one who moves first or the one who moves second, provided that both players play optimally well.
Input
A single line contains three space-separated integers a, b, r (1 ≤ a, b, r ≤ 100) — the table sides and the plates' radius, correspondingly.
Output
If wins the player who moves first, print "First" (without the quotes). Otherwise print "Second" (without the quotes).
Examples
Input
5 5 2
Output
First
Input
6 7 4
Output
Second
Note
In the first sample the table has place for only one plate. The first player puts a plate on the table, the second player can't do that and loses.
<image>
In the second sample the table is so small that it doesn't have enough place even for one plate. So the first player loses without making a single move.
<image>
Submitted Solution:
```
from sys import stdin,stdout
from math import pi,ceil
nmbr = lambda: int(stdin.readline())
lst = lambda: list(map(int,stdin.readline().split()))
for _ in range(1):#nmbr()):
l,b,r=lst()
area1=l*b
area2=ceil(pi*r*r)
n=area1//area2
print('First' if n&1 else 'Second')
``` | instruction | 0 | 75,315 | 19 | 150,630 |
No | output | 1 | 75,315 | 19 | 150,631 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've got a rectangular table with length a and width b and the infinite number of plates of radius r. Two players play the following game: they take turns to put the plates on the table so that the plates don't lie on each other (but they can touch each other), and so that any point on any plate is located within the table's border. During the game one cannot move the plates that already lie on the table. The player who cannot make another move loses. Determine which player wins, the one who moves first or the one who moves second, provided that both players play optimally well.
Input
A single line contains three space-separated integers a, b, r (1 ≤ a, b, r ≤ 100) — the table sides and the plates' radius, correspondingly.
Output
If wins the player who moves first, print "First" (without the quotes). Otherwise print "Second" (without the quotes).
Examples
Input
5 5 2
Output
First
Input
6 7 4
Output
Second
Note
In the first sample the table has place for only one plate. The first player puts a plate on the table, the second player can't do that and loses.
<image>
In the second sample the table is so small that it doesn't have enough place even for one plate. So the first player loses without making a single move.
<image>
Submitted Solution:
```
a,b,r=map(int,input().split())
s=a*b
s1=3.14*r*r
d=0
if s>=s1:
while s>=s1:
s=s-s1
d=d+1
if d%2==0:
print("Second")
else:
print("First")
else:
print("Second")
``` | instruction | 0 | 75,316 | 19 | 150,632 |
No | output | 1 | 75,316 | 19 | 150,633 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've got a rectangular table with length a and width b and the infinite number of plates of radius r. Two players play the following game: they take turns to put the plates on the table so that the plates don't lie on each other (but they can touch each other), and so that any point on any plate is located within the table's border. During the game one cannot move the plates that already lie on the table. The player who cannot make another move loses. Determine which player wins, the one who moves first or the one who moves second, provided that both players play optimally well.
Input
A single line contains three space-separated integers a, b, r (1 ≤ a, b, r ≤ 100) — the table sides and the plates' radius, correspondingly.
Output
If wins the player who moves first, print "First" (without the quotes). Otherwise print "Second" (without the quotes).
Examples
Input
5 5 2
Output
First
Input
6 7 4
Output
Second
Note
In the first sample the table has place for only one plate. The first player puts a plate on the table, the second player can't do that and loses.
<image>
In the second sample the table is so small that it doesn't have enough place even for one plate. So the first player loses without making a single move.
<image>
Submitted Solution:
```
a,b,r=map(int,input().split())
(print("First")if (a>2*r and b>2*r) else print("Second"))
``` | instruction | 0 | 75,317 | 19 | 150,634 |
No | output | 1 | 75,317 | 19 | 150,635 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've got a rectangular table with length a and width b and the infinite number of plates of radius r. Two players play the following game: they take turns to put the plates on the table so that the plates don't lie on each other (but they can touch each other), and so that any point on any plate is located within the table's border. During the game one cannot move the plates that already lie on the table. The player who cannot make another move loses. Determine which player wins, the one who moves first or the one who moves second, provided that both players play optimally well.
Input
A single line contains three space-separated integers a, b, r (1 ≤ a, b, r ≤ 100) — the table sides and the plates' radius, correspondingly.
Output
If wins the player who moves first, print "First" (without the quotes). Otherwise print "Second" (without the quotes).
Examples
Input
5 5 2
Output
First
Input
6 7 4
Output
Second
Note
In the first sample the table has place for only one plate. The first player puts a plate on the table, the second player can't do that and loses.
<image>
In the second sample the table is so small that it doesn't have enough place even for one plate. So the first player loses without making a single move.
<image>
Submitted Solution:
```
# Enter your code here. Read input from STDIN. Print output to STDOUT# ===============================================================================================
# importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from math import *
from io import BytesIO, IOBase
from itertools import *
import bisect
from heapq import *
from math import ceil, floor
from copy import *
from collections import deque, defaultdict
from collections import Counter as counter # Counter(list) return a dict with {key: count}
from itertools import combinations # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
from operator import *
# If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect
# If the element is already present in the list,
# the right most position where element has to be inserted is returned
# ==============================================================================================
# fast I/O region
BUFSIZE = 8192
from sys import stderr
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "A" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for A in args:
if not at_start:
file.write(sep)
file.write(str(A))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
# ===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
#### END ITERATE RECURSION ####
###########################
# Sorted list
class SortedList:
def __init__(self, iterable=[], _load=200):
"""Initialize sorted list instance."""
values = sorted(iterable)
self._len = _len = len(values)
self._load = _load
self._lists = _lists = [values[start:start + _load] for start in range(0, _len, _load)]
self._list_lens = [len(_list) for _list in _lists]
self._mins = [_list[0] for _list in _lists]
self._fen_tree = []
self._rebuild = True
def _fen_build(self):
"""Build a fenwick tree instance."""
self._fen_tree[:] = self._list_lens
_fen_tree = self._fen_tree
for start in range(len(_fen_tree)):
if start | start + 1 < len(_fen_tree):
_fen_tree[start | start + 1] += _fen_tree[start]
self._rebuild = False
def _fen_update(self, index, value):
"""Update `fen_tree[index] += value`."""
if not self._rebuild:
_fen_tree = self._fen_tree
while index < len(_fen_tree):
_fen_tree[index] += value
index |= index + 1
def _fen_query(self, end):
"""Return `sum(_fen_tree[:end])`."""
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
A = 0
while end:
A += _fen_tree[end - 1]
end &= end - 1
return A
def _fen_findkth(self, k):
"""Return a pair of (the largest `idx` such that `sum(_fen_tree[:idx]) <= k`, `k - sum(_fen_tree[:idx])`)."""
_list_lens = self._list_lens
if k < _list_lens[0]:
return 0, k
if k >= self._len - _list_lens[-1]:
return len(_list_lens) - 1, k + _list_lens[-1] - self._len
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
idx = -1
for d in reversed(range(len(_fen_tree).bit_length())):
right_idx = idx + (1 << d)
if right_idx < len(_fen_tree) and k >= _fen_tree[right_idx]:
idx = right_idx
k -= _fen_tree[idx]
return idx + 1, k
def _delete(self, pos, idx):
"""Delete value at the given `(pos, idx)`."""
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len -= 1
self._fen_update(pos, -1)
del _lists[pos][idx]
_list_lens[pos] -= 1
if _list_lens[pos]:
_mins[pos] = _lists[pos][0]
else:
del _lists[pos]
del _list_lens[pos]
del _mins[pos]
self._rebuild = True
def _loc_left(self, value):
"""Return an index pair that corresponds to the first position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
lo, pos = -1, len(_lists) - 1
while lo + 1 < pos:
mi = (lo + pos) >> 1
if value <= _mins[mi]:
pos = mi
else:
lo = mi
if pos and value <= _lists[pos - 1][-1]:
pos -= 1
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value <= _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def _loc_right(self, value):
"""Return an index pair that corresponds to the last position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
pos, hi = 0, len(_lists)
while pos + 1 < hi:
mi = (pos + hi) >> 1
if value < _mins[mi]:
hi = mi
else:
pos = mi
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value < _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def add(self, value):
"""Add `value` to sorted list."""
_load = self._load
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len += 1
if _lists:
pos, idx = self._loc_right(value)
self._fen_update(pos, 1)
_list = _lists[pos]
_list.insert(idx, value)
_list_lens[pos] += 1
_mins[pos] = _list[0]
if _load + _load < len(_list):
_lists.insert(pos + 1, _list[_load:])
_list_lens.insert(pos + 1, len(_list) - _load)
_mins.insert(pos + 1, _list[_load])
_list_lens[pos] = _load
del _list[_load:]
self._rebuild = True
else:
_lists.append([value])
_mins.append(value)
_list_lens.append(1)
self._rebuild = True
def discard(self, value):
"""Remove `value` from sorted list if it is a member."""
_lists = self._lists
if _lists:
pos, idx = self._loc_right(value)
if idx and _lists[pos][idx - 1] == value:
self._delete(pos, idx - 1)
def remove(self, value):
"""Remove `value` from sorted list; `value` must be a member."""
_len = self._len
self.discard(value)
if _len == self._len:
raise ValueError('{0!r} not in list'.format(value))
def pop(self, index=-1):
"""Remove and return value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
value = self._lists[pos][idx]
self._delete(pos, idx)
return value
def bisect_left(self, value):
"""Return the first index to insert `value` in the sorted list."""
pos, idx = self._loc_left(value)
return self._fen_query(pos) + idx
def bisect_right(self, value):
"""Return the last index to insert `value` in the sorted list."""
pos, idx = self._loc_right(value)
return self._fen_query(pos) + idx
def count(self, value):
"""Return number of occurrences of `value` in the sorted list."""
return self.bisect_right(value) - self.bisect_left(value)
def __len__(self):
"""Return the size of the sorted list."""
return self._len
def __getitem__(self, index):
"""Lookup value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
return self._lists[pos][idx]
def __delitem__(self, index):
"""Remove value at `index` from sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
self._delete(pos, idx)
def __contains__(self, value):
"""Return true if `value` is an element of the sorted list."""
_lists = self._lists
if _lists:
pos, idx = self._loc_left(value)
return idx < len(_lists[pos]) and _lists[pos][idx] == value
return False
def __iter__(self):
"""Return an iterator over the sorted list."""
return (value for _list in self._lists for value in _list)
def __reversed__(self):
"""Return a reverse iterator over the sorted list."""
return (value for _list in reversed(self._lists) for value in reversed(_list))
def __repr__(self):
"""Return string representation of sorted list."""
return 'SortedList({0})'.format(list(self))
# ===============================================================================================
# some shortcuts
mod = 1000000007
def testcase(t):
for p in range(t):
solve()
def pow(A, B, p):
res = 1 # Initialize result
A = A % p # Update A if it is more , than or equal to p
if (A == 0):
return 0
while (B > 0):
if ((B & 1) == 1): # If B is odd, multiply, A with result
res = (res * A) % p
B = B >> 1 # B = B/2
A = (A * A) % p
return res
from functools import reduce
def factors(n):
return set(reduce(list.__add__,
([start, n // start] for start in range(1, int(n ** 0.5) + 1) if n % start == 0)))
def gcd(a, b):
if a == b: return a
while b > 0: a, b = b, a % b
return a
# discrete binary search
# minimise:
# def search():
# l = 0
# r = 10 ** 15
#
# for start in range(200):
# if isvalid(l):
# return l
# if l == r:
# return l
# m = (l + r) // 2
# if isvalid(m) and not isvalid(m - 1):
# return m
# if isvalid(m):
# r = m + 1
# else:
# l = m
# return m
# maximise:
# def search():
# l = 0
# r = 10 ** 15
#
# for start in range(200):
# # print(l,r)
# if isvalid(r):
# return r
# if l == r:
# return l
# m = (l + r) // 2
# if isvalid(m) and not isvalid(m + 1):
# return m
# if isvalid(m):
# l = m
# else:
# r = m - 1
# return m
##############Find sum of product of subsets of size k in a array
# ar=[0,1,2,3]
# k=3
# n=len(ar)-1
# dp=[0]*(n+1)
# dp[0]=1
# for pos in range(1,n+1):
# dp[pos]=0
# l=max(1,k+pos-n-1)
# for j in range(min(pos,k),l-1,-1):
# dp[j]=dp[j]+ar[pos]*dp[j-1]
# print(dp[k])
def prefix_sum(ar): # [1,2,3,4]->[1,3,6,10]
return list(accumulate(ar))
def suffix_sum(ar): # [1,2,3,4]->[10,9,7,4]
return list(accumulate(ar[::-1]))[::-1]
def N():
return int(inp())
dx = [0, 0, 1, -1]
dy = [1, -1, 0, 0]
def YES():
print("YES")
def NO():
print("NO")
def yes():
print("yes")
def no():
print("no")
def Yes():
print("Yes")
def No():
print("No")
# =========================================================================================
from collections import defaultdict
def numberOfSetBits(start):
start = start - ((start >> 1) & 0x55555555)
start = (start & 0x33333333) + ((start >> 2) & 0x33333333)
return (((start + (start >> 4) & 0xF0F0F0F) * 0x1010101) & 0xffffffff) >> 24
class MergeFind:
def __init__(self, n):
self.parent = list(range(n))
self.size = [1] * n
self.num_sets = n
# self.lista = [[_] for _ in range(n)]
def find(self, a):
to_update = []
while a != self.parent[a]:
to_update.append(a)
a = self.parent[a]
for b in to_update:
self.parent[b] = a
return self.parent[a]
def merge(self, a, b):
a = self.find(a)
b = self.find(b)
if a == b:
return
if self.size[a] < self.size[b]:
a, b = b, a
self.num_sets -= 1
self.parent[b] = a
self.size[a] += self.size[b]
# self.lista[a] += self.lista[b]
# self.lista[b] = []
def set_size(self, a):
return self.size[self.find(a)]
def __len__(self):
return self.num_sets
def lcm(a, b):
return abs((a // gcd(a, b)) * b)
# #
# to find factorial and ncr
# tot = 100005
# mod = 10**9 + 7
# fac = [1, 1]
# finv = [1, 1]
# inv = [0, 1]
#
# for start in range(2, tot + 1):
# fac.append((fac[-1] * start) % mod)
# inv.append(mod - (inv[mod % start] * (mod // start) % mod))
# finv.append(finv[-1] * inv[-1] % mod)
def inp(): return sys.stdin.readline().rstrip("\r\n") # for fast input
def out(var): sys.stdout.write(str(var)) # for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def fsep(): return map(float, inp().split())
def nextline(): out("\n") # as stdout.write always print sring.
def arr1d(n, v):
return [v] * n
def arr2d(n, m, v):
return [[v] * m for _ in range(n)]
def arr3d(n, m, p, v):
return [[[v] * p for _ in range(m)] for start in range(n)]
def ceil(a, b):
return (a + b - 1) // b
"""
def main():
t = int(input())
for _ in range(t):
n,m = map(int,input().split())
l=[]
ans=[]
p=[]
for i in range(m):
arr = list(map(int,input().split()))
arr.pop(0)
p.append(arr)
for j in arr:
l.append(j)
f = (m)//2
#print(p)
l = sorted(l)
sett = set(l)
sett = list(sett)
#print(sett)
for i in sett:
if l.count(i)<=f:
ans+=[i]*l.count(i)
else:
ans+=[i]*(f)
# print(ans)
ans1=[]
for i in range(m):
for j in range(len(p[i])):
if p[i][j] in ans:
ans1.append(p[i][j])
ans.remove(p[i][j])
break
if len(ans1)>=m:
print("YES")
print(*ans1)
else:
print("NO")
if __name__ == '__main__':
main()
"""
"""
def main():
t = int(input())
for _ in range(t):
s = input()
l=[]
ss = list(s)
for i in range(len(s)-4):
if ss[i]=="p" and ss[i+1]=="a" and ss[i+2]=="r" and ss[i+3]=="t" and ss[i+4]=="y":
ss[i],ss[i+1],ss[i+2],ss[i+3],ss[i+4]="p","a","w","r","i"
ans=""
for i in ss:
ans+=i
print(ans)
if __name__ == '__main__':
main()
"""
def main():
a,b,r = sep()
rect = a*b
plate = pi*r*r
ans = rect/plate
ans = floor(ans)
if ans%2==0:
print("second")
else:
print("first")
if __name__ == '__main__':
main()
``` | instruction | 0 | 75,318 | 19 | 150,636 |
No | output | 1 | 75,318 | 19 | 150,637 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gosha is hunting. His goal is to catch as many Pokemons as possible. Gosha has a Poke Balls and b Ultra Balls. There are n Pokemons. They are numbered 1 through n. Gosha knows that if he throws a Poke Ball at the i-th Pokemon he catches it with probability pi. If he throws an Ultra Ball at the i-th Pokemon he catches it with probability ui. He can throw at most one Ball of each type at any Pokemon.
The hunting proceeds as follows: at first, Gosha chooses no more than a Pokemons at which he will throw Poke Balls and no more than b Pokemons at which he will throw Ultra Balls. After that, he throws the chosen Balls at the chosen Pokemons. If he throws both Ultra Ball and Poke Ball at some Pokemon, he is caught if and only if he is caught by any of these Balls. The outcome of a throw doesn't depend on the other throws.
Gosha would like to know what is the expected number of the Pokemons he catches if he acts in an optimal way. In other words, he would like to know the maximum possible expected number of Pokemons can catch.
Input
The first line contains three integers n, a and b (2 ≤ n ≤ 2000, 0 ≤ a, b ≤ n) — the number of Pokemons, the number of Poke Balls and the number of Ultra Balls.
The second line contains n real values p1, p2, ..., pn (0 ≤ pi ≤ 1), where pi is the probability of catching the i-th Pokemon if Gosha throws a Poke Ball to it.
The third line contains n real values u1, u2, ..., un (0 ≤ ui ≤ 1), where ui is the probability of catching the i-th Pokemon if Gosha throws an Ultra Ball to it.
All the probabilities are given with exactly three digits after the decimal separator.
Output
Print the maximum possible expected number of Pokemons Gosha can catch. The answer is considered correct if it's absolute or relative error doesn't exceed 10 - 4.
Examples
Input
3 2 2
1.000 0.000 0.500
0.000 1.000 0.500
Output
2.75
Input
4 1 3
0.100 0.500 0.500 0.600
0.100 0.500 0.900 0.400
Output
2.16
Input
3 2 0
0.412 0.198 0.599
0.612 0.987 0.443
Output
1.011
Submitted Solution:
```
n, a, b = map(int, input().split())
pa = [float(i) for i in input().split()]
pb = [float(i) for i in input().split()]
pab = [pa[i] + pb[i] - pa[i] * pb[i] for i in range(n)]
def solve(ca, cb):
dp = [0.0] * (n + 1)
cntA = [0] * (n + 1)
cntB = [0] * (n + 1)
for i in range(n):
dp[i+1] = dp[i]
cntA[i+1] = cntA[i]
cntB[i+1] = cntB[i]
if dp[i+1] < dp[i] + pa[i] - ca:
dp[i+1] = dp[i] + pa[i] - ca
cntA[i+1] = cntA[i] + 1
cntB[i+1] = cntB[i]
if dp[i+1] < dp[i] + pb[i] - cb:
dp[i+1] = dp[i] + pb[i] - cb
cntA[i+1] = cntA[i]
cntB[i+1] = cntB[i] + 1
if dp[i+1] < dp[i] + pab[i] - ca - cb:
dp[i+1] = dp[i] + pab[i] - ca - cb
cntA[i+1] = cntA[i] + 1
cntB[i+1] = cntB[i] + 1
return (dp[n] + cntA[n] * ca + cntB[n] * cb, cntA[n], cntB[n])
la = 0.0
ra = 1.0
for i in range(50):
ma = (la + ra) / 2
lb = 0.0
rb = 1.0
for j in range(50):
mb = (lb + rb) / 2
if solve(ma, mb)[2] > b:
lb = mb
else:
rb = mb
if solve(ma, rb)[1] > a:
la = ma
else:
ra = ma
lb = 0.0
rb = 1.0
for j in range(50):
mb = (lb + rb) / 2
if solve(ra, mb)[2] > b:
lb = mb
else:
rb = mb
print(f'{solve(ra, rb)[0]:.10f}')
``` | instruction | 0 | 75,546 | 19 | 151,092 |
No | output | 1 | 75,546 | 19 | 151,093 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gosha is hunting. His goal is to catch as many Pokemons as possible. Gosha has a Poke Balls and b Ultra Balls. There are n Pokemons. They are numbered 1 through n. Gosha knows that if he throws a Poke Ball at the i-th Pokemon he catches it with probability pi. If he throws an Ultra Ball at the i-th Pokemon he catches it with probability ui. He can throw at most one Ball of each type at any Pokemon.
The hunting proceeds as follows: at first, Gosha chooses no more than a Pokemons at which he will throw Poke Balls and no more than b Pokemons at which he will throw Ultra Balls. After that, he throws the chosen Balls at the chosen Pokemons. If he throws both Ultra Ball and Poke Ball at some Pokemon, he is caught if and only if he is caught by any of these Balls. The outcome of a throw doesn't depend on the other throws.
Gosha would like to know what is the expected number of the Pokemons he catches if he acts in an optimal way. In other words, he would like to know the maximum possible expected number of Pokemons can catch.
Input
The first line contains three integers n, a and b (2 ≤ n ≤ 2000, 0 ≤ a, b ≤ n) — the number of Pokemons, the number of Poke Balls and the number of Ultra Balls.
The second line contains n real values p1, p2, ..., pn (0 ≤ pi ≤ 1), where pi is the probability of catching the i-th Pokemon if Gosha throws a Poke Ball to it.
The third line contains n real values u1, u2, ..., un (0 ≤ ui ≤ 1), where ui is the probability of catching the i-th Pokemon if Gosha throws an Ultra Ball to it.
All the probabilities are given with exactly three digits after the decimal separator.
Output
Print the maximum possible expected number of Pokemons Gosha can catch. The answer is considered correct if it's absolute or relative error doesn't exceed 10 - 4.
Examples
Input
3 2 2
1.000 0.000 0.500
0.000 1.000 0.500
Output
2.75
Input
4 1 3
0.100 0.500 0.500 0.600
0.100 0.500 0.900 0.400
Output
2.16
Input
3 2 0
0.412 0.198 0.599
0.612 0.987 0.443
Output
1.011
Submitted Solution:
```
n,a,b=map(int,input().split())
p1={}
u1={}
p=list(map(float,input().split()))
s=0
i=0
while i!=a:
z=max(p)
while z==max(p):
p1[p.index(max(p))]=max(p)
p[p.index(max(p))]=0
i+=1
if i==a:
break
p=list(map(float,input().split()))
i=0
while i!=b:
z=max(p)
while z==max(p):
u1[p.index(max(p))]=max(p)
p[p.index(max(p))]=0
i+=1
if i==b:
break
for j in range(n):
try:
s+=(p1[j]+u1[j]-p1[j]*u1[j])
except:
try:
s+=p1[j]
except:
try:
s+=u1[j]
except:
s+=0
print(s)
``` | instruction | 0 | 75,547 | 19 | 151,094 |
No | output | 1 | 75,547 | 19 | 151,095 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob begin their day with a quick game. They first choose a starting number X0 ≥ 3 and try to reach one million by the process described below.
Alice goes first and then they take alternating turns. In the i-th turn, the player whose turn it is selects a prime number smaller than the current number, and announces the smallest multiple of this prime number that is not smaller than the current number.
Formally, he or she selects a prime p < Xi - 1 and then finds the minimum Xi ≥ Xi - 1 such that p divides Xi. Note that if the selected prime p already divides Xi - 1, then the number does not change.
Eve has witnessed the state of the game after two turns. Given X2, help her determine what is the smallest possible starting number X0. Note that the players don't necessarily play optimally. You should consider all possible game evolutions.
Input
The input contains a single integer X2 (4 ≤ X2 ≤ 106). It is guaranteed that the integer X2 is composite, that is, is not prime.
Output
Output a single integer — the minimum possible X0.
Examples
Input
14
Output
6
Input
20
Output
15
Input
8192
Output
8191
Note
In the first test, the smallest possible starting number is X0 = 6. One possible course of the game is as follows:
* Alice picks prime 5 and announces X1 = 10
* Bob picks prime 7 and announces X2 = 14.
In the second case, let X0 = 15.
* Alice picks prime 2 and announces X1 = 16
* Bob picks prime 5 and announces X2 = 20. | instruction | 0 | 75,595 | 19 | 151,190 |
Tags: math, number theory
Correct Solution:
```
def testPrimal(n):
siv = [0 for _ in range(n+1)]
for i in range(2, n+1):
if siv[i] == 0:
for j in range(i+i, n+1, i):
siv[j] = i
siv[i] = i - siv[i] + 1
result = n
for i in range(siv[n], n + 1):
result = min(result, siv[i])
return result
n = int(input())
print(testPrimal(n))
``` | output | 1 | 75,595 | 19 | 151,191 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob begin their day with a quick game. They first choose a starting number X0 ≥ 3 and try to reach one million by the process described below.
Alice goes first and then they take alternating turns. In the i-th turn, the player whose turn it is selects a prime number smaller than the current number, and announces the smallest multiple of this prime number that is not smaller than the current number.
Formally, he or she selects a prime p < Xi - 1 and then finds the minimum Xi ≥ Xi - 1 such that p divides Xi. Note that if the selected prime p already divides Xi - 1, then the number does not change.
Eve has witnessed the state of the game after two turns. Given X2, help her determine what is the smallest possible starting number X0. Note that the players don't necessarily play optimally. You should consider all possible game evolutions.
Input
The input contains a single integer X2 (4 ≤ X2 ≤ 106). It is guaranteed that the integer X2 is composite, that is, is not prime.
Output
Output a single integer — the minimum possible X0.
Examples
Input
14
Output
6
Input
20
Output
15
Input
8192
Output
8191
Note
In the first test, the smallest possible starting number is X0 = 6. One possible course of the game is as follows:
* Alice picks prime 5 and announces X1 = 10
* Bob picks prime 7 and announces X2 = 14.
In the second case, let X0 = 15.
* Alice picks prime 2 and announces X1 = 16
* Bob picks prime 5 and announces X2 = 20. | instruction | 0 | 75,596 | 19 | 151,192 |
Tags: math, number theory
Correct Solution:
```
x = int(input())
n = x
vis = [False for i in range(n+1)]
flag = [0 for i in range(n+1)]
for i in range(2, n+1):
if not vis[i]:
for j in range(i+i, n+1, i):
vis[j] = True
flag[j] = i
ans = x
for i in range(x - flag[x] + 1, x+1):
ans = min(ans, i - flag[i] + 1)
print(ans)
``` | output | 1 | 75,596 | 19 | 151,193 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob begin their day with a quick game. They first choose a starting number X0 ≥ 3 and try to reach one million by the process described below.
Alice goes first and then they take alternating turns. In the i-th turn, the player whose turn it is selects a prime number smaller than the current number, and announces the smallest multiple of this prime number that is not smaller than the current number.
Formally, he or she selects a prime p < Xi - 1 and then finds the minimum Xi ≥ Xi - 1 such that p divides Xi. Note that if the selected prime p already divides Xi - 1, then the number does not change.
Eve has witnessed the state of the game after two turns. Given X2, help her determine what is the smallest possible starting number X0. Note that the players don't necessarily play optimally. You should consider all possible game evolutions.
Input
The input contains a single integer X2 (4 ≤ X2 ≤ 106). It is guaranteed that the integer X2 is composite, that is, is not prime.
Output
Output a single integer — the minimum possible X0.
Examples
Input
14
Output
6
Input
20
Output
15
Input
8192
Output
8191
Note
In the first test, the smallest possible starting number is X0 = 6. One possible course of the game is as follows:
* Alice picks prime 5 and announces X1 = 10
* Bob picks prime 7 and announces X2 = 14.
In the second case, let X0 = 15.
* Alice picks prime 2 and announces X1 = 16
* Bob picks prime 5 and announces X2 = 20. | instruction | 0 | 75,597 | 19 | 151,194 |
Tags: math, number theory
Correct Solution:
```
from sys import stdin
maxn = 1000006
prime = [0] * maxn
def getPrimes():
n = 2
while(n < maxn):
num = n*2
while(num < maxn):
prime[num] = n
num += n
num = n + 1
while( num < maxn and prime[num]!= 0 ):
num += 1
n = num
def main():
getPrimes()
n = int(stdin.readline().strip())
a = 0
b = prime[n]
x0 = 9999999
for i in range(n-b+1,n+1):
a = prime[i]
x0 = min(i-a+1,x0)
#print("i",i,"prime[i]",prime[i],"x0",x0)
print(x0)
main()
``` | output | 1 | 75,597 | 19 | 151,195 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob begin their day with a quick game. They first choose a starting number X0 ≥ 3 and try to reach one million by the process described below.
Alice goes first and then they take alternating turns. In the i-th turn, the player whose turn it is selects a prime number smaller than the current number, and announces the smallest multiple of this prime number that is not smaller than the current number.
Formally, he or she selects a prime p < Xi - 1 and then finds the minimum Xi ≥ Xi - 1 such that p divides Xi. Note that if the selected prime p already divides Xi - 1, then the number does not change.
Eve has witnessed the state of the game after two turns. Given X2, help her determine what is the smallest possible starting number X0. Note that the players don't necessarily play optimally. You should consider all possible game evolutions.
Input
The input contains a single integer X2 (4 ≤ X2 ≤ 106). It is guaranteed that the integer X2 is composite, that is, is not prime.
Output
Output a single integer — the minimum possible X0.
Examples
Input
14
Output
6
Input
20
Output
15
Input
8192
Output
8191
Note
In the first test, the smallest possible starting number is X0 = 6. One possible course of the game is as follows:
* Alice picks prime 5 and announces X1 = 10
* Bob picks prime 7 and announces X2 = 14.
In the second case, let X0 = 15.
* Alice picks prime 2 and announces X1 = 16
* Bob picks prime 5 and announces X2 = 20. | instruction | 0 | 75,598 | 19 | 151,196 |
Tags: math, number theory
Correct Solution:
```
def gf(n):
d = 2
while d * d <= n:
f = 1
while n % d is 0:
if f:
yield d
f = 0
n //= d
d += 1
if n > 1:
yield n
x2 = int(input())
print(min(x1 - p + 1 for x1 in range(x2 - max(gf(x2)) + 1, x2 + 1) for p in gf(x1) if p != x1))
``` | output | 1 | 75,598 | 19 | 151,197 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob begin their day with a quick game. They first choose a starting number X0 ≥ 3 and try to reach one million by the process described below.
Alice goes first and then they take alternating turns. In the i-th turn, the player whose turn it is selects a prime number smaller than the current number, and announces the smallest multiple of this prime number that is not smaller than the current number.
Formally, he or she selects a prime p < Xi - 1 and then finds the minimum Xi ≥ Xi - 1 such that p divides Xi. Note that if the selected prime p already divides Xi - 1, then the number does not change.
Eve has witnessed the state of the game after two turns. Given X2, help her determine what is the smallest possible starting number X0. Note that the players don't necessarily play optimally. You should consider all possible game evolutions.
Input
The input contains a single integer X2 (4 ≤ X2 ≤ 106). It is guaranteed that the integer X2 is composite, that is, is not prime.
Output
Output a single integer — the minimum possible X0.
Examples
Input
14
Output
6
Input
20
Output
15
Input
8192
Output
8191
Note
In the first test, the smallest possible starting number is X0 = 6. One possible course of the game is as follows:
* Alice picks prime 5 and announces X1 = 10
* Bob picks prime 7 and announces X2 = 14.
In the second case, let X0 = 15.
* Alice picks prime 2 and announces X1 = 16
* Bob picks prime 5 and announces X2 = 20. | instruction | 0 | 75,599 | 19 | 151,198 |
Tags: math, number theory
Correct Solution:
```
x2 = int(input())
s = [-1] * (x2 + 1)
for i in range(2, x2+1):
if s[i] == -1:
for j in range(i, x2+1, i):
s[j] = i
ans = x2
for x1 in range(x2 - s[x2] + 1, x2 + 1):
x0 = x1 - s[x1] + 1
if x0 > 1:
ans = min(ans, x0)
print(ans)
``` | output | 1 | 75,599 | 19 | 151,199 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob begin their day with a quick game. They first choose a starting number X0 ≥ 3 and try to reach one million by the process described below.
Alice goes first and then they take alternating turns. In the i-th turn, the player whose turn it is selects a prime number smaller than the current number, and announces the smallest multiple of this prime number that is not smaller than the current number.
Formally, he or she selects a prime p < Xi - 1 and then finds the minimum Xi ≥ Xi - 1 such that p divides Xi. Note that if the selected prime p already divides Xi - 1, then the number does not change.
Eve has witnessed the state of the game after two turns. Given X2, help her determine what is the smallest possible starting number X0. Note that the players don't necessarily play optimally. You should consider all possible game evolutions.
Input
The input contains a single integer X2 (4 ≤ X2 ≤ 106). It is guaranteed that the integer X2 is composite, that is, is not prime.
Output
Output a single integer — the minimum possible X0.
Examples
Input
14
Output
6
Input
20
Output
15
Input
8192
Output
8191
Note
In the first test, the smallest possible starting number is X0 = 6. One possible course of the game is as follows:
* Alice picks prime 5 and announces X1 = 10
* Bob picks prime 7 and announces X2 = 14.
In the second case, let X0 = 15.
* Alice picks prime 2 and announces X1 = 16
* Bob picks prime 5 and announces X2 = 20. | instruction | 0 | 75,600 | 19 | 151,200 |
Tags: math, number theory
Correct Solution:
```
x2 = int(input())
def maxPrimeDevisor(v):
result = 0
while v % 2 == 0:
result = 2
v //= 2
i = 3
while i * i <= v:
if v % i == 0:
while v % i == 0:
v //= i
result = max(result, i)
else:
i += 2
if v > 1:
result = max(result, v)
return result
x1 = [max(3, x2 - maxPrimeDevisor(x2) + 1), x2]
x0 = x1[1]
for i in range(x1[0], x1[1] + 1):
mpd = maxPrimeDevisor(i)
if mpd != i:
x0 = max(3, min(x0, i - maxPrimeDevisor(i) + 1))
print(x0)
``` | output | 1 | 75,600 | 19 | 151,201 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob begin their day with a quick game. They first choose a starting number X0 ≥ 3 and try to reach one million by the process described below.
Alice goes first and then they take alternating turns. In the i-th turn, the player whose turn it is selects a prime number smaller than the current number, and announces the smallest multiple of this prime number that is not smaller than the current number.
Formally, he or she selects a prime p < Xi - 1 and then finds the minimum Xi ≥ Xi - 1 such that p divides Xi. Note that if the selected prime p already divides Xi - 1, then the number does not change.
Eve has witnessed the state of the game after two turns. Given X2, help her determine what is the smallest possible starting number X0. Note that the players don't necessarily play optimally. You should consider all possible game evolutions.
Input
The input contains a single integer X2 (4 ≤ X2 ≤ 106). It is guaranteed that the integer X2 is composite, that is, is not prime.
Output
Output a single integer — the minimum possible X0.
Examples
Input
14
Output
6
Input
20
Output
15
Input
8192
Output
8191
Note
In the first test, the smallest possible starting number is X0 = 6. One possible course of the game is as follows:
* Alice picks prime 5 and announces X1 = 10
* Bob picks prime 7 and announces X2 = 14.
In the second case, let X0 = 15.
* Alice picks prime 2 and announces X1 = 16
* Bob picks prime 5 and announces X2 = 20. | instruction | 0 | 75,602 | 19 | 151,204 |
Tags: math, number theory
Correct Solution:
```
n = int(input())
ans = n
f = [0]*(n+1)
for i in range(2, n+1):
if f[i]==0:
for j in range(i*2, n+1, i):
f[j] = i
f[i] = i-f[i]+1
for i in range(f[n], n+1):
ans = min(ans, f[i])
print(ans)
# Made By Mostafa_Khaled
``` | output | 1 | 75,602 | 19 | 151,205 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob begin their day with a quick game. They first choose a starting number X0 ≥ 3 and try to reach one million by the process described below.
Alice goes first and then they take alternating turns. In the i-th turn, the player whose turn it is selects a prime number smaller than the current number, and announces the smallest multiple of this prime number that is not smaller than the current number.
Formally, he or she selects a prime p < Xi - 1 and then finds the minimum Xi ≥ Xi - 1 such that p divides Xi. Note that if the selected prime p already divides Xi - 1, then the number does not change.
Eve has witnessed the state of the game after two turns. Given X2, help her determine what is the smallest possible starting number X0. Note that the players don't necessarily play optimally. You should consider all possible game evolutions.
Input
The input contains a single integer X2 (4 ≤ X2 ≤ 106). It is guaranteed that the integer X2 is composite, that is, is not prime.
Output
Output a single integer — the minimum possible X0.
Examples
Input
14
Output
6
Input
20
Output
15
Input
8192
Output
8191
Note
In the first test, the smallest possible starting number is X0 = 6. One possible course of the game is as follows:
* Alice picks prime 5 and announces X1 = 10
* Bob picks prime 7 and announces X2 = 14.
In the second case, let X0 = 15.
* Alice picks prime 2 and announces X1 = 16
* Bob picks prime 5 and announces X2 = 20.
Submitted Solution:
```
x2 = int(input())
n = x2
vis = [False for i in range(n+1)]
f = [0 for i in range(n+1)]
for i in range(2, n+1):
if not vis[i]:
for j in range(i+i, n+1, i):
vis[j] = True
f[j] = i
ans = x2
for i in range(x2 - f[x2] + 1, x2+1):
ans = min(ans, i - f[i] + 1)
print(ans)
``` | instruction | 0 | 75,603 | 19 | 151,206 |
Yes | output | 1 | 75,603 | 19 | 151,207 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob begin their day with a quick game. They first choose a starting number X0 ≥ 3 and try to reach one million by the process described below.
Alice goes first and then they take alternating turns. In the i-th turn, the player whose turn it is selects a prime number smaller than the current number, and announces the smallest multiple of this prime number that is not smaller than the current number.
Formally, he or she selects a prime p < Xi - 1 and then finds the minimum Xi ≥ Xi - 1 such that p divides Xi. Note that if the selected prime p already divides Xi - 1, then the number does not change.
Eve has witnessed the state of the game after two turns. Given X2, help her determine what is the smallest possible starting number X0. Note that the players don't necessarily play optimally. You should consider all possible game evolutions.
Input
The input contains a single integer X2 (4 ≤ X2 ≤ 106). It is guaranteed that the integer X2 is composite, that is, is not prime.
Output
Output a single integer — the minimum possible X0.
Examples
Input
14
Output
6
Input
20
Output
15
Input
8192
Output
8191
Note
In the first test, the smallest possible starting number is X0 = 6. One possible course of the game is as follows:
* Alice picks prime 5 and announces X1 = 10
* Bob picks prime 7 and announces X2 = 14.
In the second case, let X0 = 15.
* Alice picks prime 2 and announces X1 = 16
* Bob picks prime 5 and announces X2 = 20.
Submitted Solution:
```
if __name__ == '__main__':
x = int(input())
pSieve = [True] * x
for i in range(2, x):
if pSieve[i]:
for j in range(i * i, x, i):
pSieve[j] = False
primos = []
for i in range(2,x):
if pSieve[i]:
primos.append(i)
y = x-1
for p in primos:
if x % p == 0:
y = x - p
z = y + 1
for i in primos:
if i > y:
break
d = y - y % i
if d + i <= x:
z = min(z, d+1)
print(z)
``` | instruction | 0 | 75,604 | 19 | 151,208 |
Yes | output | 1 | 75,604 | 19 | 151,209 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob begin their day with a quick game. They first choose a starting number X0 ≥ 3 and try to reach one million by the process described below.
Alice goes first and then they take alternating turns. In the i-th turn, the player whose turn it is selects a prime number smaller than the current number, and announces the smallest multiple of this prime number that is not smaller than the current number.
Formally, he or she selects a prime p < Xi - 1 and then finds the minimum Xi ≥ Xi - 1 such that p divides Xi. Note that if the selected prime p already divides Xi - 1, then the number does not change.
Eve has witnessed the state of the game after two turns. Given X2, help her determine what is the smallest possible starting number X0. Note that the players don't necessarily play optimally. You should consider all possible game evolutions.
Input
The input contains a single integer X2 (4 ≤ X2 ≤ 106). It is guaranteed that the integer X2 is composite, that is, is not prime.
Output
Output a single integer — the minimum possible X0.
Examples
Input
14
Output
6
Input
20
Output
15
Input
8192
Output
8191
Note
In the first test, the smallest possible starting number is X0 = 6. One possible course of the game is as follows:
* Alice picks prime 5 and announces X1 = 10
* Bob picks prime 7 and announces X2 = 14.
In the second case, let X0 = 15.
* Alice picks prime 2 and announces X1 = 16
* Bob picks prime 5 and announces X2 = 20.
Submitted Solution:
```
numero = int(input())
lista = [0]*(numero+1)
for i in range(2, numero+1):
if lista[i]==0:
for j in range(2*i, numero+1, i):
lista[j] = i
lista[i] = i-lista[i]+1
p = numero
for i in range(lista[numero], numero+1):
p = min(p, lista[i])
print(p)
``` | instruction | 0 | 75,605 | 19 | 151,210 |
Yes | output | 1 | 75,605 | 19 | 151,211 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob begin their day with a quick game. They first choose a starting number X0 ≥ 3 and try to reach one million by the process described below.
Alice goes first and then they take alternating turns. In the i-th turn, the player whose turn it is selects a prime number smaller than the current number, and announces the smallest multiple of this prime number that is not smaller than the current number.
Formally, he or she selects a prime p < Xi - 1 and then finds the minimum Xi ≥ Xi - 1 such that p divides Xi. Note that if the selected prime p already divides Xi - 1, then the number does not change.
Eve has witnessed the state of the game after two turns. Given X2, help her determine what is the smallest possible starting number X0. Note that the players don't necessarily play optimally. You should consider all possible game evolutions.
Input
The input contains a single integer X2 (4 ≤ X2 ≤ 106). It is guaranteed that the integer X2 is composite, that is, is not prime.
Output
Output a single integer — the minimum possible X0.
Examples
Input
14
Output
6
Input
20
Output
15
Input
8192
Output
8191
Note
In the first test, the smallest possible starting number is X0 = 6. One possible course of the game is as follows:
* Alice picks prime 5 and announces X1 = 10
* Bob picks prime 7 and announces X2 = 14.
In the second case, let X0 = 15.
* Alice picks prime 2 and announces X1 = 16
* Bob picks prime 5 and announces X2 = 20.
Submitted Solution:
```
n = int(input())
siv = [0 for _ in range(n+1)]
for i in range(2, n+1):
if siv[i] == 0:
for j in range(i+i, n+1, i):
siv[j] = i
siv[i] = i - siv[i] + 1;
result = n
for i in range(siv[n], n + 1):
result = min(result, siv[i])
print(result)
``` | instruction | 0 | 75,606 | 19 | 151,212 |
Yes | output | 1 | 75,606 | 19 | 151,213 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob begin their day with a quick game. They first choose a starting number X0 ≥ 3 and try to reach one million by the process described below.
Alice goes first and then they take alternating turns. In the i-th turn, the player whose turn it is selects a prime number smaller than the current number, and announces the smallest multiple of this prime number that is not smaller than the current number.
Formally, he or she selects a prime p < Xi - 1 and then finds the minimum Xi ≥ Xi - 1 such that p divides Xi. Note that if the selected prime p already divides Xi - 1, then the number does not change.
Eve has witnessed the state of the game after two turns. Given X2, help her determine what is the smallest possible starting number X0. Note that the players don't necessarily play optimally. You should consider all possible game evolutions.
Input
The input contains a single integer X2 (4 ≤ X2 ≤ 106). It is guaranteed that the integer X2 is composite, that is, is not prime.
Output
Output a single integer — the minimum possible X0.
Examples
Input
14
Output
6
Input
20
Output
15
Input
8192
Output
8191
Note
In the first test, the smallest possible starting number is X0 = 6. One possible course of the game is as follows:
* Alice picks prime 5 and announces X1 = 10
* Bob picks prime 7 and announces X2 = 14.
In the second case, let X0 = 15.
* Alice picks prime 2 and announces X1 = 16
* Bob picks prime 5 and announces X2 = 20.
Submitted Solution:
```
if __name__ == '__main__':
x = int(input())
pSieve = [True] * x
for i in range(2, x):
if pSieve[i]:
for j in range(i * i, x, i):
pSieve[j] = False
primos = []
for i in range(2,x):
if pSieve[i]:
primos.append(i)
y = x-1
for p in primos:
if x % p == 0:
y = x - p
z = y + 1
for i in primos:
if i > y:
break
d = y - y % i
if d + i <= x:
z = min(z, d+1)
print(d)
``` | instruction | 0 | 75,607 | 19 | 151,214 |
No | output | 1 | 75,607 | 19 | 151,215 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob begin their day with a quick game. They first choose a starting number X0 ≥ 3 and try to reach one million by the process described below.
Alice goes first and then they take alternating turns. In the i-th turn, the player whose turn it is selects a prime number smaller than the current number, and announces the smallest multiple of this prime number that is not smaller than the current number.
Formally, he or she selects a prime p < Xi - 1 and then finds the minimum Xi ≥ Xi - 1 such that p divides Xi. Note that if the selected prime p already divides Xi - 1, then the number does not change.
Eve has witnessed the state of the game after two turns. Given X2, help her determine what is the smallest possible starting number X0. Note that the players don't necessarily play optimally. You should consider all possible game evolutions.
Input
The input contains a single integer X2 (4 ≤ X2 ≤ 106). It is guaranteed that the integer X2 is composite, that is, is not prime.
Output
Output a single integer — the minimum possible X0.
Examples
Input
14
Output
6
Input
20
Output
15
Input
8192
Output
8191
Note
In the first test, the smallest possible starting number is X0 = 6. One possible course of the game is as follows:
* Alice picks prime 5 and announces X1 = 10
* Bob picks prime 7 and announces X2 = 14.
In the second case, let X0 = 15.
* Alice picks prime 2 and announces X1 = 16
* Bob picks prime 5 and announces X2 = 20.
Submitted Solution:
```
x2 = int(input())
from math import sqrt
def get_del(x):
dels = []
for i in range(2,int(sqrt(x) +3)):
if x % i == 0:
dels.append(i)
return dels
def is_simple(x):
if len(get_del(x)) == 0:
return True
else:
False
simple = [1871, 1873, 1877, 1879, 1889, 1901, 1907, 1913, 1931, 1933, 1949, 1951, 1973, 1979, 1987, 1993, 1997, 1999, 2003, 2011, 2017, 2027, 2029, 2039, 2053, 2063, 2069, 2081, 2083, 2087, 2089, 2099, 2111, 2113, 2129, 2131, 2137, 2141, 2143, 2153, 2161, 2179, 2203, 2207, 2213, 2221, 2237, 2239, 2243, 2251, 2267, 2269, 2273, 2281, 2287, 2293, 2297, 2309, 2311, 2333, 2339, 2341, 2347, 2351, 2357, 2371, 2377, 2381, 2383, 2389, 2393, 2399, 2411, 2417, 2423, 2437, 2441, 2447, 2459, 2467, 2473, 2477, 2503, 2521, 2531, 2539, 2543, 2549, 2551, 2557, 2579, 2591, 2593, 2609, 2617, 2621, 2633, 2647, 2657, 2659, 2663, 2671, 2677, 2683, 2687, 2689, 2693, 2699, 2707, 2711, 2713, 2719, 2729, 2731, 2741, 2749, 2753, 2767, 2777, 2789, 2791, 2797, 2801, 2803, 2819, 2833, 2837, 2843, 2851, 2857, 2861, 2879, 2887, 2897, 2903, 2909, 2917, 2927, 2939, 2953, 2957, 2963, 2969, 2971, 2999, 3001, 3011, 3019, 3023, 3037, 3041, 3049, 3061, 3067, 3079, 3083, 3089, 3109, 3119, 3121, 3137, 3163, 3167, 3169, 3181, 3187, 3191, 3203, 3209, 3217, 3221, 3229, 3251, 3253, 3257, 3259, 3271, 3299, 3301, 3307, 3313, 3319, 3323, 3329, 3331, 3343, 3347, 3359, 3361, 3371, 3373, 3389, 3391, 3407, 3413, 3433, 3449, 3457, 3461, 3463, 3467, 3469, 3491, 3499, 3511, 3517, 3527, 3529, 3533, 3539, 3541, 3547, 3557, 3559, 3571, 3581, 3583, 3593, 3607, 3613, 3617, 3623, 3631, 3637, 3643, 3659, 3671, 3673, 3677, 3691, 3697, 3701, 3709, 3719, 3727, 3733, 3739, 3761, 3767, 3769, 3779, 3793, 3797, 3803, 3821, 3823, 3833, 3847, 3851, 3853, 3863, 3877, 3881, 3889, 3907, 3911, 3917, 3919, 3923, 3929, 3931, 3943, 3947, 3967, 3989, 4001, 4003, 4007, 4013, 4019, 4021, 4027, 4049, 4051, 4057, 4073, 4079, 4091, 4093, 4099, 4111, 4127, 4129, 4133, 4139, 4153, 4157, 4159, 4177, 4201, 4211, 4217, 4219, 4229, 4231, 4241, 4243, 4253, 4259, 4261, 4271, 4273, 4283, 4289, 4297, 4327, 4337, 4339, 4349, 4357, 4363, 4373, 4391, 4397, 4409, 4421, 4423, 4441, 4447, 4451, 4457, 4463, 4481, 4483, 4493, 4507, 4513, 4517, 4519, 4523, 4547, 4549, 4561, 4567, 4583, 4591, 4597, 4603, 4621, 4637, 4639, 4643, 4649, 4651, 4657, 4663, 4673, 4679, 4691, 4703, 4721, 4723, 4729, 4733, 4751, 4759, 4783, 4787, 4789, 4793, 4799, 4801, 4813, 4817, 4831, 4861, 4871, 4877, 4889, 4903, 4909, 4919, 4931, 4933, 4937, 4943, 4951, 4957, 4967, 4969, 4973, 4987, 4993, 4999, 5003, 5009, 5011, 5021, 5023, 5039, 5051, 5059, 5077, 5081, 5087, 5099, 5101, 5107, 5113, 5119, 5147, 5153, 5167, 5171, 5179, 5189, 5197, 5209, 5227, 5231, 5233, 5237, 5261, 5273, 5279, 5281, 5297, 5303, 5309, 5323, 5333, 5347, 5351, 5381, 5387, 5393, 5399, 5407, 5413, 5417, 5419, 5431, 5437, 5441, 5443, 5449, 5471, 5477, 5479, 5483, 5501, 5503, 5507, 5519, 5521, 5527, 5531, 5557, 5563, 5569, 5573, 5581, 5591, 5623, 5639, 5641, 5647, 5651, 5653, 5657, 5659, 5669, 5683, 5689, 5693, 5701, 5711, 5717, 5737, 5741, 5743, 5749, 5779, 5783, 5791, 5801, 5807, 5813, 5821, 5827, 5839, 5843, 5849, 5851, 5857, 5861, 5867, 5869, 5879, 5881, 5897, 5903, 5923, 5927, 5939, 5953, 5981, 5987, 6007, 6011, 6029, 6037, 6043, 6047, 6053, 6067, 6073, 6079, 6089, 6091, 6101, 6113, 6121, 6131, 6133, 6143, 6151, 6163, 6173, 6197, 6199, 6203, 6211, 6217, 6221, 6229, 6247, 6257, 6263, 6269, 6271, 6277, 6287, 6299, 6301, 6311, 6317, 6323, 6329, 6337, 6343, 6353, 6359, 6361, 6367, 6373, 6379, 6389, 6397, 6421, 6427, 6449, 6451, 6469, 6473, 6481, 6491, 6521, 6529, 6547, 6551, 6553, 6563, 6569, 6571, 6577, 6581, 6599, 6607, 6619, 6637, 6653, 6659, 6661, 6673, 6679, 6689, 6691, 6701, 6703, 6709, 6719, 6733, 6737, 6761, 6763, 6779, 6781, 6791, 6793, 6803, 6823, 6827, 6829, 6833, 6841, 6857, 6863, 6869, 6871, 6883, 6899, 6907, 6911, 6917, 6947, 6949, 6959, 6961, 6967, 6971, 6977, 6983, 6991, 6997, 7001, 7013, 7019, 7027, 7039, 7043, 7057, 7069, 7079, 7103, 7109, 7121, 7127, 7129, 7151, 7159, 7177, 7187, 7193, 7207, 7211, 7213, 7219, 7229, 7237, 7243, 7247, 7253, 7283, 7297, 7307, 7309, 7321, 7331, 7333, 7349, 7351, 7369, 7393, 7411, 7417, 7433, 7451, 7457, 7459, 7477, 7481, 7487, 7489, 7499, 7507, 7517, 7523, 7529, 7537, 7541, 7547, 7549, 7559, 7561, 7573, 7577, 7583, 7589, 7591, 7603, 7607, 7621, 7639, 7643, 7649, 7669, 7673, 7681, 7687, 7691, 7699, 7703, 7717, 7723, 7727, 7741, 7753, 7757, 7759, 7789, 7793, 7817, 7823, 7829, 7841, 7853, 7867, 7873, 7877, 7879, 7883, 7901, 7907, 7919, 7927, 7933, 7937, 7949, 7951, 7963, 7993, 8009, 8011, 8017, 8039, 8053, 8059, 8069, 8081, 8087, 8089, 8093, 8101, 8111, 8117, 8123, 8147, 8161, 8167, 8171, 8179, 8191, 8209, 8219, 8221, 8231, 8233, 8237, 8243, 8263, 8269, 8273, 8287, 8291, 8293, 8297, 8311, 8317, 8329, 8353, 8363, 8369, 8377, 8387, 8389, 8419, 8423, 8429, 8431, 8443, 8447, 8461, 8467, 8501, 8513, 8521, 8527, 8537, 8539, 8543, 8563, 8573, 8581, 8597, 8599, 8609, 8623, 8627, 8629, 8641, 8647, 8663, 8669, 8677, 8681, 8689, 8693, 8699, 8707, 8713, 8719, 8731, 8737, 8741, 8747, 8753, 8761, 8779, 8783, 8803, 8807, 8819, 8821, 8831, 8837, 8839, 8849, 8861, 8863, 8867, 8887, 8893, 8923, 8929, 8933, 8941, 8951, 8963, 8969, 8971, 8999, 9001, 9007, 9011, 9013, 9029, 9041, 9043, 9049, 9059, 9067, 9091, 9103, 9109, 9127, 9133, 9137, 9151, 9157, 9161, 9173, 9181, 9187, 9199, 9203, 9209, 9221, 9227, 9239, 9241, 9257, 9277, 9281, 9283, 9293, 9311, 9319, 9323, 9337, 9341, 9343, 9349, 9371, 9377, 9391, 9397, 9403, 9413, 9419, 9421, 9431, 9433, 9437, 9439, 9461, 9463, 9467, 9473, 9479, 9491, 9497, 9511, 9521, 9533, 9539, 9547, 9551, 9587, 9601, 9613, 9619, 9623, 9629, 9631, 9643, 9649, 9661, 9677, 9679, 9689, 9697, 9719, 9721, 9733, 9739, 9743, 9749, 9767, 9769, 9781, 9787, 9791, 9803, 9811, 9817, 9829, 9833, 9839, 9851, 9857, 9859, 9871, 9883, 9887, 9901, 9907, 9923, 9929, 9931, 9941, 9949, 9967, 9973, 10007, 10009, 10037, 10039, 10061, 10067, 10069, 10079, 10091, 10093, 10099, 10103, 10111, 10133, 10139, 10141, 10151, 10159, 10163, 10169, 10177, 10181, 10193, 10211, 10223, 10243, 10247, 10253, 10259, 10267, 10271, 10273, 10289, 10301, 10303, 10313, 10321, 10331, 10333, 10337, 10343, 10357, 10369, 10391, 10399, 10427, 10429, 10433, 10453, 10457, 10459, 10463, 10477, 10487, 10499, 10501, 10513, 10529, 10531, 10559, 10567, 10589, 10597, 10601, 10607, 10613, 10627, 10631, 10639, 10651, 10657, 10663, 10667, 10687, 10691, 10709, 10711, 10723, 10729, 10733, 10739, 10753, 10771, 10781, 10789, 10799, 10831, 10837, 10847, 10853, 10859, 10861, 10867, 10883, 10889, 10891, 10903, 10909, 10937, 10939, 10949, 10957, 10973, 10979, 10987, 10993, 11003, 11027, 11047, 11057, 11059, 11069, 11071, 11083, 11087, 11093, 11113, 11117, 11119, 11131, 11149, 11159, 11161, 11171, 11173, 11177, 11197, 11213, 11239, 11243, 11251, 11257, 11261, 11273, 11279, 11287, 11299, 11311, 11317, 11321, 11329, 11351, 11353, 11369, 11383, 11393, 11399, 11411, 11423, 11437, 11443, 11447, 11467, 11471, 11483, 11489, 11491, 11497, 11503, 11519, 11527, 11549, 11551, 11579, 11587, 11593, 11597, 11617, 11621, 11633, 11657, 11677, 11681, 11689, 11699, 11701, 11717, 11719, 11731, 11743, 11777, 11779, 11783, 11789, 11801, 11807, 11813, 11821, 11827, 11831, 11833, 11839, 11863, 11867, 11887, 11897, 11903, 11909, 11923, 11927, 11933, 11939, 11941, 11953, 11959, 11969, 11971, 11981, 11987, 12007, 12011, 12037, 12041, 12043, 12049, 12071, 12073, 12097, 12101, 12107, 12109, 12113, 12119, 12143, 12149, 12157, 12161, 12163, 12197, 12203, 12211, 12227, 12239, 12241, 12251, 12253, 12263, 12269, 12277, 12281, 12289, 12301, 12323, 12329, 12343, 12347, 12373, 12377, 12379, 12391, 12401, 12409, 12413, 12421, 12433, 12437, 12451, 12457, 12473, 12479, 12487, 12491, 12497, 12503, 12511, 12517, 12527, 12539, 12541, 12547, 12553, 12569, 12577, 12583, 12589, 12601, 12611, 12613, 12619, 12637, 12641, 12647, 12653, 12659, 12671, 12689, 12697, 12703, 12713, 12721, 12739, 12743, 12757, 12763, 12781, 12791, 12799, 12809, 12821, 12823, 12829, 12841, 12853, 12889, 12893, 12899, 12907, 12911, 12917, 12919, 12923, 12941, 12953, 12959, 12967, 12973, 12979, 12983, 13001, 13003, 13007, 13009, 13033, 13037, 13043, 13049, 13063, 13093, 13099, 13103, 13109, 13121, 13127, 13147, 13151, 13159, 13163, 13171, 13177, 13183, 13187, 13217, 13219, 13229, 13241, 13249, 13259, 13267, 13291, 13297, 13309, 13313, 13327, 13331, 13337, 13339, 13367, 13381, 13397, 13399, 13411, 13417, 13421, 13441, 13451, 13457, 13463, 13469, 13477, 13487, 13499, 13513, 13523, 13537, 13553, 13567, 13577, 13591, 13597, 13613, 13619, 13627, 13633, 13649, 13669, 13679, 13681, 13687, 13691, 13693, 13697, 13709, 13711, 13721, 13723, 13729, 13751, 13757, 13759, 13763, 13781, 13789, 13799, 13807, 13829, 13831, 13841, 13859, 13873, 13877, 13879, 13883, 13901, 13903, 13907, 13913, 13921, 13931, 13933, 13963, 13967, 13997, 13999, 14009, 14011, 14029, 14033, 14051, 14057, 14071, 14081, 14083, 14087, 14107, 14143, 14149, 14153, 14159, 14173, 14177, 14197, 14207, 14221, 14243, 14249, 14251, 14281, 14293, 14303, 14321, 14323, 14327, 14341, 14347, 14369, 14387, 14389, 14401, 14407, 14411, 14419, 14423, 14431, 14437, 14447, 14449, 14461, 14479, 14489, 14503, 14519, 14533, 14537, 14543, 14549, 14551, 14557, 14561, 14563, 14591, 14593, 14621, 14627, 14629, 14633, 14639, 14653, 14657, 14669, 14683, 14699, 14713, 14717, 14723, 14731, 14737, 14741, 14747, 14753, 14759, 14767, 14771, 14779, 14783, 14797, 14813, 14821, 14827, 14831, 14843, 14851, 14867, 14869, 14879, 14887, 14891, 14897, 14923, 14929, 14939, 14947, 14951, 14957, 14969, 14983, 15013, 15017, 15031, 15053, 15061, 15073, 15077, 15083, 15091, 15101, 15107, 15121, 15131, 15137, 15139, 15149, 15161, 15173, 15187, 15193, 15199, 15217, 15227, 15233, 15241, 15259, 15263, 15269, 15271, 15277, 15287, 15289, 15299, 15307, 15313, 15319, 15329, 15331, 15349, 15359, 15361, 15373, 15377, 15383, 15391, 15401, 15413, 15427, 15439, 15443, 15451, 15461, 15467, 15473, 15493, 15497, 15511, 15527, 15541, 15551, 15559, 15569, 15581, 15583, 15601, 15607, 15619, 15629, 15641, 15643, 15647, 15649, 15661, 15667, 15671, 15679, 15683, 15727, 15731, 15733, 15737, 15739, 15749, 15761, 15767, 15773, 15787, 15791, 15797, 15803, 15809, 15817, 15823, 15859, 15877, 15881, 15887, 15889, 15901, 15907, 15913, 15919, 15923, 15937, 15959, 15971, 15973, 15991, 16001, 16007, 16033, 16057, 16061, 16063, 16067, 16069, 16073, 16087, 16091, 16097, 16103, 16111, 16127, 16139, 16141, 16183, 16187, 16189, 16193, 16217, 16223, 16229, 16231, 16249, 16253, 16267, 16273, 16301, 16319, 16333, 16339, 16349, 16361, 16363, 16369, 16381, 16411, 16417, 16421, 16427, 16433, 16447, 16451, 16453, 16477, 16481, 16487, 16493, 16519, 16529, 16547, 16553, 16561, 16567, 16573, 16603, 16607, 16619, 16631, 16633, 16649, 16651, 16657, 16661, 16673, 16691, 16693, 16699, 16703, 16729, 16741, 16747, 16759, 16763, 16787, 16811, 16823, 16829, 16831, 16843, 16871, 16879, 16883, 16889, 16901, 16903, 16921, 16927, 16931, 16937, 16943, 16963, 16979, 16981, 16987, 16993, 17011, 17021, 17027, 17029, 17033, 17041, 17047, 17053, 17077, 17093, 17099, 17107, 17117, 17123, 17137, 17159, 17167, 17183, 17189, 17191, 17203, 17207, 17209, 17231, 17239, 17257, 17291, 17293, 17299, 17317, 17321, 17327, 17333, 17341, 17351, 17359, 17377, 17383, 17387, 17389, 17393, 17401, 17417, 17419, 17431, 17443, 17449, 17467, 17471, 17477, 17483, 17489, 17491, 17497, 17509, 17519, 17539, 17551, 17569, 17573, 17579, 17581, 17597, 17599, 17609, 17623, 17627, 17657, 17659, 17669, 17681, 17683, 17707, 17713, 17729, 17737, 17747, 17749, 17761, 17783, 17789, 17791, 17807, 17827, 17837, 17839, 17851, 17863, 17881, 17891, 17903, 17909, 17911, 17921, 17923, 17929, 17939, 17957, 17959, 17971, 17977, 17981, 17987, 17989, 18013, 18041, 18043, 18047, 18049, 18059, 18061, 18077, 18089, 18097, 18119, 18121, 18127, 18131, 18133, 18143, 18149, 18169, 18181, 18191, 18199, 18211, 18217, 18223, 18229, 18233, 18251, 18253, 18257, 18269, 18287, 18289, 18301, 18307, 18311, 18313, 18329, 18341, 18353, 18367, 18371, 18379, 18397, 18401, 18413, 18427, 18433, 18439, 18443, 18451, 18457, 18461, 18481, 18493, 18503, 18517, 18521, 18523, 18539, 18541, 18553, 18583, 18587, 18593, 18617, 18637, 18661, 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93133, 93139, 93151, 93169, 93179, 93187, 93199, 93229, 93239, 93241, 93251, 93253, 93257, 93263, 93281, 93283, 93287, 93307, 93319, 93323, 93329, 93337, 93371, 93377, 93383, 93407, 93419, 93427, 93463, 93479, 93481, 93487, 93491, 93493, 93497, 93503, 93523, 93529, 93553, 93557, 93559, 93563, 93581, 93601, 93607, 93629, 93637, 93683, 93701, 93703, 93719, 93739, 93761, 93763, 93787, 93809, 93811, 93827, 93851, 93871, 93887, 93889, 93893, 93901, 93911, 93913, 93923, 93937, 93941, 93949, 93967, 93971, 93979, 93983, 93997, 94007, 94009, 94033, 94049, 94057, 94063, 94079, 94099, 94109, 94111, 94117, 94121, 94151, 94153, 94169, 94201, 94207, 94219, 94229, 94253, 94261, 94273, 94291, 94307, 94309, 94321, 94327, 94331, 94343, 94349, 94351, 94379, 94397, 94399, 94421, 94427, 94433, 94439, 94441, 94447, 94463, 94477, 94483, 94513, 94529, 94531, 94541, 94543, 94547, 94559, 94561, 94573, 94583, 94597, 94603, 94613, 94621, 94649, 94651, 94687, 94693, 94709, 94723, 94727, 94747, 94771, 94777, 94781, 94789, 94793, 94811, 94819, 94823, 94837, 94841, 94847, 94849, 94873, 94889, 94903, 94907, 94933, 94949, 94951, 94961, 94993, 94999, 95003, 95009, 95021, 95027, 95063, 95071, 95083, 95087, 95089, 95093, 95101, 95107, 95111, 95131, 95143, 95153, 95177, 95189, 95191, 95203, 95213, 95219, 95231, 95233, 95239, 95257, 95261, 95267, 95273, 95279, 95287, 95311, 95317, 95327, 95339, 95369, 95383, 95393, 95401, 95413, 95419, 95429, 95441, 95443, 95461, 95467, 95471, 95479, 95483, 95507, 95527, 95531, 95539, 95549, 95561, 95569, 95581, 95597, 95603, 95617, 95621, 95629, 95633, 95651, 95701, 95707, 95713, 95717, 95723, 95731, 95737, 95747, 95773, 95783, 95789, 95791, 95801, 95803, 95813, 95819, 95857, 95869, 95873, 95881, 95891, 95911, 95917, 95923, 95929, 95947, 95957, 95959, 95971, 95987, 95989, 96001, 96013, 96017, 96043, 96053, 96059, 96079, 96097, 96137, 96149, 96157, 96167, 96179, 96181, 96199, 96211, 96221, 96223, 96233, 96259, 96263, 96269, 96281, 96289, 96293, 96323, 96329, 96331, 96337, 96353, 96377, 96401, 96419, 96431, 96443, 96451, 96457, 96461, 96469, 96479, 96487, 96493, 96497, 96517, 96527, 96553, 96557, 96581, 96587, 96589, 96601, 96643, 96661, 96667, 96671, 96697, 96703, 96731, 96737, 96739, 96749, 96757, 96763, 96769, 96779, 96787, 96797, 96799, 96821, 96823, 96827, 96847, 96851, 96857, 96893, 96907, 96911, 96931, 96953, 96959, 96973, 96979, 96989, 96997, 97001, 97003, 97007, 97021, 97039, 97073, 97081, 97103, 97117, 97127, 97151, 97157, 97159, 97169, 97171, 97177, 97187, 97213, 97231, 97241, 97259, 97283, 97301, 97303, 97327, 97367, 97369, 97373, 97379, 97381, 97387, 97397, 97423, 97429, 97441, 97453, 97459, 97463, 97499, 97501, 97511, 97523, 97547, 97549, 97553, 97561, 97571, 97577, 97579, 97583, 97607, 97609, 97613, 97649, 97651, 97673, 97687, 97711, 97729, 97771, 97777, 97787, 97789, 97813, 97829, 97841, 97843, 97847, 97849, 97859, 97861, 97871, 97879, 97883, 97919, 97927, 97931, 97943, 97961, 97967, 97973, 97987, 98009, 98011, 98017, 98041, 98047, 98057, 98081, 98101, 98123, 98129, 98143, 98179, 98207, 98213, 98221, 98227, 98251, 98257, 98269, 98297, 98299, 98317, 98321, 98323, 98327, 98347, 98369, 98377, 98387, 98389, 98407, 98411, 98419, 98429, 98443, 98453, 98459, 98467, 98473, 98479, 98491, 98507, 98519, 98533, 98543, 98561, 98563, 98573, 98597, 98621, 98627, 98639, 98641, 98663, 98669, 98689, 98711, 98713, 98717, 98729, 98731, 98737, 98773, 98779, 98801, 98807, 98809, 98837, 98849, 98867, 98869, 98873, 98887, 98893, 98897, 98899, 98909, 98911, 98927, 98929, 98939, 98947, 98953, 98963, 98981, 98993, 98999, 99013, 99017, 99023, 99041, 99053, 99079, 99083, 99089, 99103, 99109, 99119, 99131, 99133, 99137, 99139, 99149, 99173, 99181, 99191, 99223, 99233, 99241, 99251, 99257, 99259, 99277, 99289, 99317, 99347, 99349, 99367, 99371, 99377, 99391, 99397, 99401, 99409, 99431, 99439, 99469, 99487, 99497, 99523, 99527, 99529, 99551, 99559, 99563, 99571, 99577, 99581, 99607, 99611, 99623, 99643, 99661, 99667, 99679, 99689, 99707, 99709, 99713, 99719, 99721, 99733, 99761, 99767, 99787, 99793, 99809, 99817, 99823, 99829, 99833, 99839, 99859, 99871, 99877, 99881, 99901, 99907, 99923, 99929, 99961, 99971, 99989, 99991]
for i in range(1870,-1,-1):
if is_simple(i):
# print('Hello')
simple.insert(0,i)
simple.insert(1,2)
simple.insert(2,3)
simple = list(filter(lambda x:x<=x2,simple))
answer = []
for i in range(len(simple)-1,-1,-1):
if x2 % simple[i] == 0:
chisla = [j for j in range(x2 -simple[i]+1,x2)]
if x2 == simple[i]:
answer = [simple[i]]
break
if len(answer) > 0:
print(answer[0])
else:
answer = []
# print()
for i in chisla:
small_simple = list(filter(lambda x: x <= i, simple))
for j in range(len(small_simple) - 1, -1, -1):
if i % small_simple[j] == 0:
if i == small_simple[j]:
answer.append(i)
else:
answer.append(i-small_simple[j]+1)
break
print(min(answer))
``` | instruction | 0 | 75,608 | 19 | 151,216 |
No | output | 1 | 75,608 | 19 | 151,217 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob begin their day with a quick game. They first choose a starting number X0 ≥ 3 and try to reach one million by the process described below.
Alice goes first and then they take alternating turns. In the i-th turn, the player whose turn it is selects a prime number smaller than the current number, and announces the smallest multiple of this prime number that is not smaller than the current number.
Formally, he or she selects a prime p < Xi - 1 and then finds the minimum Xi ≥ Xi - 1 such that p divides Xi. Note that if the selected prime p already divides Xi - 1, then the number does not change.
Eve has witnessed the state of the game after two turns. Given X2, help her determine what is the smallest possible starting number X0. Note that the players don't necessarily play optimally. You should consider all possible game evolutions.
Input
The input contains a single integer X2 (4 ≤ X2 ≤ 106). It is guaranteed that the integer X2 is composite, that is, is not prime.
Output
Output a single integer — the minimum possible X0.
Examples
Input
14
Output
6
Input
20
Output
15
Input
8192
Output
8191
Note
In the first test, the smallest possible starting number is X0 = 6. One possible course of the game is as follows:
* Alice picks prime 5 and announces X1 = 10
* Bob picks prime 7 and announces X2 = 14.
In the second case, let X0 = 15.
* Alice picks prime 2 and announces X1 = 16
* Bob picks prime 5 and announces X2 = 20.
Submitted Solution:
```
import sys
from math import floor, sqrt
#comment these out later
#sys.stdin = open("in.in", "r")
#sys.stdout = open("out.out", "w")
from collections import Counter
def prime_sieve(n):
"""returns a sieve of primes >= 5 and < n"""
flag = n % 6 == 2
sieve = bytearray((n // 3 + flag >> 3) + 1)
for i in range(1, int(n**0.5) // 3 + 1):
if not (sieve[i >> 3] >> (i & 7)) & 1:
k = (3 * i + 1) | 1
for j in range(k * k // 3, n // 3 + flag, 2 * k):
sieve[j >> 3] |= 1 << (j & 7)
for j in range(k * (k - 2 * (i & 1) + 4) // 3, n // 3 + flag, 2 * k):
sieve[j >> 3] |= 1 << (j & 7)
return sieve
def prime_list(n):
"""returns a list of primes <= n"""
res = []
if n > 1:
res.append(2)
if n > 2:
res.append(3)
if n > 4:
sieve = prime_sieve(n + 1)
res.extend(3 * i + 1 | 1 for i in range(1, (n + 1) // 3 + (n % 6 == 1)) if not (sieve[i >> 3] >> (i & 7)) & 1)
return res
def main():
mxs = [0]*1000005
ps = prime_list(1000)
for x in ps:
for y in range(x, 1000005, x):
mxs[y] = x
n = int(sys.stdin.readline())
mxp = mxs[n]
ans = 10000000
for y in range(n-mxp+1, n):
mxp2 = mxs[y]
if mxp2 == 0:
c = y
else:
c = y - mxp2 + 1
ans = min(ans, c)
print(ans)
main()
``` | instruction | 0 | 75,609 | 19 | 151,218 |
No | output | 1 | 75,609 | 19 | 151,219 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob begin their day with a quick game. They first choose a starting number X0 ≥ 3 and try to reach one million by the process described below.
Alice goes first and then they take alternating turns. In the i-th turn, the player whose turn it is selects a prime number smaller than the current number, and announces the smallest multiple of this prime number that is not smaller than the current number.
Formally, he or she selects a prime p < Xi - 1 and then finds the minimum Xi ≥ Xi - 1 such that p divides Xi. Note that if the selected prime p already divides Xi - 1, then the number does not change.
Eve has witnessed the state of the game after two turns. Given X2, help her determine what is the smallest possible starting number X0. Note that the players don't necessarily play optimally. You should consider all possible game evolutions.
Input
The input contains a single integer X2 (4 ≤ X2 ≤ 106). It is guaranteed that the integer X2 is composite, that is, is not prime.
Output
Output a single integer — the minimum possible X0.
Examples
Input
14
Output
6
Input
20
Output
15
Input
8192
Output
8191
Note
In the first test, the smallest possible starting number is X0 = 6. One possible course of the game is as follows:
* Alice picks prime 5 and announces X1 = 10
* Bob picks prime 7 and announces X2 = 14.
In the second case, let X0 = 15.
* Alice picks prime 2 and announces X1 = 16
* Bob picks prime 5 and announces X2 = 20.
Submitted Solution:
```
n = int(input())
primes = [2,3]
for i in range(4,1001):
pr=True
for p in primes:
if i%p==0:
pr = False
if pr:
primes.append(i)
def lp(x,primes):
if x in primes:
return 1
res = 1
for p in primes:
if x%p==0:
res = p
return res
lpn = lp(n,primes)
res = 10**19
for i in range(n-lpn+1,n+1):
res = min(res, i-lp(i,primes)+1)
print(res)
``` | instruction | 0 | 75,610 | 19 | 151,220 |
No | output | 1 | 75,610 | 19 | 151,221 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Don't you tell me what you think that I can be
If you say that Arkady is a bit old-fashioned playing checkers, you won't be right. There is also a modern computer game Arkady and his friends are keen on. We won't discuss its rules, the only feature important to this problem is that each player has to pick a distinct hero in the beginning of the game.
There are 2 teams each having n players and 2n heroes to distribute between the teams. The teams take turns picking heroes: at first, the first team chooses a hero in its team, after that the second team chooses a hero and so on. Note that after a hero is chosen it becomes unavailable to both teams.
The friends estimate the power of the i-th of the heroes as p_i. Each team wants to maximize the total power of its heroes. However, there is one exception: there are m pairs of heroes that are especially strong against each other, so when any team chooses a hero from such a pair, the other team must choose the other one on its turn. Each hero is in at most one such pair.
This is an interactive problem. You are to write a program that will optimally choose the heroes for one team, while the jury's program will play for the other team. Note that the jury's program may behave inefficiently, in this case you have to take the opportunity and still maximize the total power of your team. Formally, if you ever have chance to reach the total power of q or greater regardless of jury's program choices, you must get q or greater to pass a test.
Input
The first line contains two integers n and m (1 ≤ n ≤ 10^3, 0 ≤ m ≤ n) — the number of players in one team and the number of special pairs of heroes.
The second line contains 2n integers p_1, p_2, …, p_{2n} (1 ≤ p_i ≤ 10^3) — the powers of the heroes.
Each of the next m lines contains two integer a and b (1 ≤ a, b ≤ 2n, a ≠ b) — a pair of heroes that are especially strong against each other. It is guaranteed that each hero appears at most once in this list.
The next line contains a single integer t (1 ≤ t ≤ 2) — the team you are to play for. If t = 1, the first turn is yours, otherwise you have the second turn.
Hacks
In order to hack, use the format described above with one additional line. In this line output 2n distinct integers from 1 to 2n — the priority order for the jury's team. The jury's team will on each turn select the first possible hero from this list. Here possible means that it is not yet taken and does not contradict the rules about special pair of heroes.
Interaction
When it is your turn, print a single integer x (1 ≤ x ≤ 2n) — the index of the hero chosen by you. Note that you can't choose a hero previously chosen by either you of the other player, and you must follow the rules about special pairs of heroes.
When it is the other team's turn, read a line containing a single integer x (1 ≤ x ≤ 2n) — the index of the hero chosen by the other team. It is guaranteed that this index is not chosen before and that the other team also follows the rules about special pairs of heroes.
After the last turn you should terminate without printing anything.
After printing your choice do not forget to output end of line and flush the output. Otherwise you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Jury's answer -1 instead of a valid choice means that you made an invalid turn. Exit immediately after receiving -1 and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.
Examples
Input
3 1
1 2 3 4 5 6
2 6
1
2
4
1
Output
6
5
3
Input
3 1
1 2 3 4 5 6
1 5
2
6
1
3
Output
5
4
2
Note
In the first example the first turn is yours. In example, you choose 6, the other team is forced to reply with 2. You choose 5, the other team chooses 4. Finally, you choose 3 and the other team choose 1.
In the second example you have the second turn. The other team chooses 6, you choose 5, forcing the other team to choose 1. Now you choose 4, the other team chooses 3 and you choose 2. | instruction | 0 | 75,892 | 19 | 151,784 |
Tags: greedy, implementation, interactive, sortings
Correct Solution:
```
# -*- coding:utf-8 -*-
"""
created by shuangquan.huang at 12/14/18
"""
import collections
import time
import os
import sys
import bisect
import heapq
N, M = map(int, input().split())
P = [0] + [int(x) for x in input().split()]
A = [0 for _ in range(2*N+1)]
pairs = []
for i in range(M):
a, b = map(int, input().split())
A[a] = b
A[b] = a
pairs.append((a, b))
sp = [(P[i], i) for i in range(1, 2*N+1)]
sp.sort(reverse=True)
vis = [False] * (2*N+1)
T = int(input())
if T == 1:
for a, b in pairs:
if P[a] >= P[b]:
print(a)
sys.stdout.flush()
input()
else:
print(b)
sys.stdout.flush()
input()
vis[a] = True
vis[b] = True
for p, i in sp:
if not vis[i]:
print(i)
sys.stdout.flush()
b = int(input())
vis[b] = True
else:
selected = 0
for _ in range(N):
b = int(input())
selected += 1
vis[b] = True
if A[b] > 0 and not vis[A[b]]:
print(A[b])
selected += 1
vis[A[b]] = True
sys.stdout.flush()
else:
break
for a, b in pairs:
if vis[a] or vis[b]:
continue
selected += 2
if P[a] >= P[b]:
print(a)
sys.stdout.flush()
input()
else:
print(b)
sys.stdout.flush()
input()
vis[a] = True
vis[b] = True
for p, i in sp:
if not vis[i]:
print(i)
sys.stdout.flush()
selected += 1
if selected >= 2*N:
break
b = int(input())
vis[b] = True
selected += 1
``` | output | 1 | 75,892 | 19 | 151,785 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Don't you tell me what you think that I can be
If you say that Arkady is a bit old-fashioned playing checkers, you won't be right. There is also a modern computer game Arkady and his friends are keen on. We won't discuss its rules, the only feature important to this problem is that each player has to pick a distinct hero in the beginning of the game.
There are 2 teams each having n players and 2n heroes to distribute between the teams. The teams take turns picking heroes: at first, the first team chooses a hero in its team, after that the second team chooses a hero and so on. Note that after a hero is chosen it becomes unavailable to both teams.
The friends estimate the power of the i-th of the heroes as p_i. Each team wants to maximize the total power of its heroes. However, there is one exception: there are m pairs of heroes that are especially strong against each other, so when any team chooses a hero from such a pair, the other team must choose the other one on its turn. Each hero is in at most one such pair.
This is an interactive problem. You are to write a program that will optimally choose the heroes for one team, while the jury's program will play for the other team. Note that the jury's program may behave inefficiently, in this case you have to take the opportunity and still maximize the total power of your team. Formally, if you ever have chance to reach the total power of q or greater regardless of jury's program choices, you must get q or greater to pass a test.
Input
The first line contains two integers n and m (1 ≤ n ≤ 10^3, 0 ≤ m ≤ n) — the number of players in one team and the number of special pairs of heroes.
The second line contains 2n integers p_1, p_2, …, p_{2n} (1 ≤ p_i ≤ 10^3) — the powers of the heroes.
Each of the next m lines contains two integer a and b (1 ≤ a, b ≤ 2n, a ≠ b) — a pair of heroes that are especially strong against each other. It is guaranteed that each hero appears at most once in this list.
The next line contains a single integer t (1 ≤ t ≤ 2) — the team you are to play for. If t = 1, the first turn is yours, otherwise you have the second turn.
Hacks
In order to hack, use the format described above with one additional line. In this line output 2n distinct integers from 1 to 2n — the priority order for the jury's team. The jury's team will on each turn select the first possible hero from this list. Here possible means that it is not yet taken and does not contradict the rules about special pair of heroes.
Interaction
When it is your turn, print a single integer x (1 ≤ x ≤ 2n) — the index of the hero chosen by you. Note that you can't choose a hero previously chosen by either you of the other player, and you must follow the rules about special pairs of heroes.
When it is the other team's turn, read a line containing a single integer x (1 ≤ x ≤ 2n) — the index of the hero chosen by the other team. It is guaranteed that this index is not chosen before and that the other team also follows the rules about special pairs of heroes.
After the last turn you should terminate without printing anything.
After printing your choice do not forget to output end of line and flush the output. Otherwise you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Jury's answer -1 instead of a valid choice means that you made an invalid turn. Exit immediately after receiving -1 and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.
Examples
Input
3 1
1 2 3 4 5 6
2 6
1
2
4
1
Output
6
5
3
Input
3 1
1 2 3 4 5 6
1 5
2
6
1
3
Output
5
4
2
Note
In the first example the first turn is yours. In example, you choose 6, the other team is forced to reply with 2. You choose 5, the other team chooses 4. Finally, you choose 3 and the other team choose 1.
In the second example you have the second turn. The other team chooses 6, you choose 5, forcing the other team to choose 1. Now you choose 4, the other team chooses 3 and you choose 2.
Submitted Solution:
```
from sys import stdout
n, m = map(int, input().split())
powers = list(map(int, input().split()))
agr = {}
for elz in range(m):
k1, k2 = map(int, input().split())
agr[k1] = k2
agr[k2] = k1
team = int(input())
if team == 1:
z = max(powers)
print(z)
stdout.flush()
powers[powers.index(z)] = 0
for b in range(n-3+team):
pick = int(input())
powers[powers.index(pick)] = 0
if pick in agr and agr[pick] in set(powers):
print(agr[pick])
stdout.flush()
powers[powers.index(agr[pick])] = 0
else:
z = max(powers)
print(z)
stdout.flush()
powers[powers.index(z)] = 0
if n > 1 and team==1:
pick = int(input())
powers[powers.index(pick)] = 0
if pick in agr and agr[pick] in set(powers):
print(agr[pick])
stdout.flush()
powers[powers.index(agr[pick])] = 0
else:
z = max(powers)
print(z)
stdout.flush()
powers[powers.index(z)] = 0
if team == 1:
__ = input()
``` | instruction | 0 | 75,893 | 19 | 151,786 |
No | output | 1 | 75,893 | 19 | 151,787 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Don't you tell me what you think that I can be
If you say that Arkady is a bit old-fashioned playing checkers, you won't be right. There is also a modern computer game Arkady and his friends are keen on. We won't discuss its rules, the only feature important to this problem is that each player has to pick a distinct hero in the beginning of the game.
There are 2 teams each having n players and 2n heroes to distribute between the teams. The teams take turns picking heroes: at first, the first team chooses a hero in its team, after that the second team chooses a hero and so on. Note that after a hero is chosen it becomes unavailable to both teams.
The friends estimate the power of the i-th of the heroes as p_i. Each team wants to maximize the total power of its heroes. However, there is one exception: there are m pairs of heroes that are especially strong against each other, so when any team chooses a hero from such a pair, the other team must choose the other one on its turn. Each hero is in at most one such pair.
This is an interactive problem. You are to write a program that will optimally choose the heroes for one team, while the jury's program will play for the other team. Note that the jury's program may behave inefficiently, in this case you have to take the opportunity and still maximize the total power of your team. Formally, if you ever have chance to reach the total power of q or greater regardless of jury's program choices, you must get q or greater to pass a test.
Input
The first line contains two integers n and m (1 ≤ n ≤ 10^3, 0 ≤ m ≤ n) — the number of players in one team and the number of special pairs of heroes.
The second line contains 2n integers p_1, p_2, …, p_{2n} (1 ≤ p_i ≤ 10^3) — the powers of the heroes.
Each of the next m lines contains two integer a and b (1 ≤ a, b ≤ 2n, a ≠ b) — a pair of heroes that are especially strong against each other. It is guaranteed that each hero appears at most once in this list.
The next line contains a single integer t (1 ≤ t ≤ 2) — the team you are to play for. If t = 1, the first turn is yours, otherwise you have the second turn.
Hacks
In order to hack, use the format described above with one additional line. In this line output 2n distinct integers from 1 to 2n — the priority order for the jury's team. The jury's team will on each turn select the first possible hero from this list. Here possible means that it is not yet taken and does not contradict the rules about special pair of heroes.
Interaction
When it is your turn, print a single integer x (1 ≤ x ≤ 2n) — the index of the hero chosen by you. Note that you can't choose a hero previously chosen by either you of the other player, and you must follow the rules about special pairs of heroes.
When it is the other team's turn, read a line containing a single integer x (1 ≤ x ≤ 2n) — the index of the hero chosen by the other team. It is guaranteed that this index is not chosen before and that the other team also follows the rules about special pairs of heroes.
After the last turn you should terminate without printing anything.
After printing your choice do not forget to output end of line and flush the output. Otherwise you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Jury's answer -1 instead of a valid choice means that you made an invalid turn. Exit immediately after receiving -1 and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.
Examples
Input
3 1
1 2 3 4 5 6
2 6
1
2
4
1
Output
6
5
3
Input
3 1
1 2 3 4 5 6
1 5
2
6
1
3
Output
5
4
2
Note
In the first example the first turn is yours. In example, you choose 6, the other team is forced to reply with 2. You choose 5, the other team chooses 4. Finally, you choose 3 and the other team choose 1.
In the second example you have the second turn. The other team chooses 6, you choose 5, forcing the other team to choose 1. Now you choose 4, the other team chooses 3 and you choose 2.
Submitted Solution:
```
n, m = map(int, input().split())
heroes = [0] + [int(x) for x in input().split()]
forces = {}
for i, v in enumerate(heroes[1:], 1):
forces[int(v)] = forces.get(int(v), []) + [i]
links = {}
for _ in range(m):
a, b = map(int, input().split())
links[a] = b
links[b] = a
order = int(input())
if order == 1:
n -= 1
choice = max(forces)
x = forces[choice].pop()
if len(forces[choice]) == 0:
forces.pop(choice)
print(x)
for _ in range(n):
choice = int(input())
val = heroes[choice]
forces[val].remove(choice)
if len(forces[val]) == 0:
forces.pop(val)
if choice in links:
b = links[choice]
links.pop(choice)
links.pop(b)
if heroes[b] in forces and b in forces[heroes[b]]:
forces[heroes[b]].remove(b)
if len(forces[heroes[b]]) == 0:
forces.pop(heroes[b])
print(b)
else:
choice = max(forces)
x = forces[choice].pop()
if len(forces[choice]) == 0:
forces.pop(choice)
print(x)
else:
choice = max(forces)
x = forces[choice].pop()
if len(forces[choice]) == 0:
forces.pop(choice)
print(x)
if order == 1:
choice = int(input())
``` | instruction | 0 | 75,894 | 19 | 151,788 |
No | output | 1 | 75,894 | 19 | 151,789 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Don't you tell me what you think that I can be
If you say that Arkady is a bit old-fashioned playing checkers, you won't be right. There is also a modern computer game Arkady and his friends are keen on. We won't discuss its rules, the only feature important to this problem is that each player has to pick a distinct hero in the beginning of the game.
There are 2 teams each having n players and 2n heroes to distribute between the teams. The teams take turns picking heroes: at first, the first team chooses a hero in its team, after that the second team chooses a hero and so on. Note that after a hero is chosen it becomes unavailable to both teams.
The friends estimate the power of the i-th of the heroes as p_i. Each team wants to maximize the total power of its heroes. However, there is one exception: there are m pairs of heroes that are especially strong against each other, so when any team chooses a hero from such a pair, the other team must choose the other one on its turn. Each hero is in at most one such pair.
This is an interactive problem. You are to write a program that will optimally choose the heroes for one team, while the jury's program will play for the other team. Note that the jury's program may behave inefficiently, in this case you have to take the opportunity and still maximize the total power of your team. Formally, if you ever have chance to reach the total power of q or greater regardless of jury's program choices, you must get q or greater to pass a test.
Input
The first line contains two integers n and m (1 ≤ n ≤ 10^3, 0 ≤ m ≤ n) — the number of players in one team and the number of special pairs of heroes.
The second line contains 2n integers p_1, p_2, …, p_{2n} (1 ≤ p_i ≤ 10^3) — the powers of the heroes.
Each of the next m lines contains two integer a and b (1 ≤ a, b ≤ 2n, a ≠ b) — a pair of heroes that are especially strong against each other. It is guaranteed that each hero appears at most once in this list.
The next line contains a single integer t (1 ≤ t ≤ 2) — the team you are to play for. If t = 1, the first turn is yours, otherwise you have the second turn.
Hacks
In order to hack, use the format described above with one additional line. In this line output 2n distinct integers from 1 to 2n — the priority order for the jury's team. The jury's team will on each turn select the first possible hero from this list. Here possible means that it is not yet taken and does not contradict the rules about special pair of heroes.
Interaction
When it is your turn, print a single integer x (1 ≤ x ≤ 2n) — the index of the hero chosen by you. Note that you can't choose a hero previously chosen by either you of the other player, and you must follow the rules about special pairs of heroes.
When it is the other team's turn, read a line containing a single integer x (1 ≤ x ≤ 2n) — the index of the hero chosen by the other team. It is guaranteed that this index is not chosen before and that the other team also follows the rules about special pairs of heroes.
After the last turn you should terminate without printing anything.
After printing your choice do not forget to output end of line and flush the output. Otherwise you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Jury's answer -1 instead of a valid choice means that you made an invalid turn. Exit immediately after receiving -1 and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.
Examples
Input
3 1
1 2 3 4 5 6
2 6
1
2
4
1
Output
6
5
3
Input
3 1
1 2 3 4 5 6
1 5
2
6
1
3
Output
5
4
2
Note
In the first example the first turn is yours. In example, you choose 6, the other team is forced to reply with 2. You choose 5, the other team chooses 4. Finally, you choose 3 and the other team choose 1.
In the second example you have the second turn. The other team chooses 6, you choose 5, forcing the other team to choose 1. Now you choose 4, the other team chooses 3 and you choose 2.
Submitted Solution:
```
import sys
a, b = map(int, input().split())
powers = list(map(int, input().split()))
P = []
for i in range(len(powers)):
P.append((powers[i], i))
P.sort(reverse=True)
heroes = [1 for i in range(2 * a)]
E = [[] for _ in range(2 * a)]
for i in range(b):
k, l = map(int, input().split())
E[k - 1].append((powers[l - 1], l - 1))
E[l - 1].append((powers[k - 1], k - 1))
for i in range(len(E)):
E[i].sort()
t = int(input())
hero_max = 0
all_hero = a * 2
if t == 1:
print(P[hero_max][1] + 1)
heroes[P[hero_max][1]] = 0
hero_max += 1
all_hero -= 1
while all_hero > 0:
stop = 0
enemy_pick = int(input()) - 1
all_hero -= 1
heroes[enemy_pick] = 0
if all_hero >= 1:
if E[enemy_pick]:
for i in range(len(E[enemy_pick])):
if heroes[E[enemy_pick][i][1]]:
heroes[E[enemy_pick][i][1]] = 0
print(E[enemy_pick][i][1] + 1)
stop = 1
break
if not stop:
while not heroes[P[hero_max][1]]:
hero_max += 1
print(P[hero_max][1] + 1)
heroes[P[hero_max][1]] = 0
hero_max += 1
all_hero -= 1
sys.stdout.flush()
``` | instruction | 0 | 75,895 | 19 | 151,790 |
No | output | 1 | 75,895 | 19 | 151,791 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Don't you tell me what you think that I can be
If you say that Arkady is a bit old-fashioned playing checkers, you won't be right. There is also a modern computer game Arkady and his friends are keen on. We won't discuss its rules, the only feature important to this problem is that each player has to pick a distinct hero in the beginning of the game.
There are 2 teams each having n players and 2n heroes to distribute between the teams. The teams take turns picking heroes: at first, the first team chooses a hero in its team, after that the second team chooses a hero and so on. Note that after a hero is chosen it becomes unavailable to both teams.
The friends estimate the power of the i-th of the heroes as p_i. Each team wants to maximize the total power of its heroes. However, there is one exception: there are m pairs of heroes that are especially strong against each other, so when any team chooses a hero from such a pair, the other team must choose the other one on its turn. Each hero is in at most one such pair.
This is an interactive problem. You are to write a program that will optimally choose the heroes for one team, while the jury's program will play for the other team. Note that the jury's program may behave inefficiently, in this case you have to take the opportunity and still maximize the total power of your team. Formally, if you ever have chance to reach the total power of q or greater regardless of jury's program choices, you must get q or greater to pass a test.
Input
The first line contains two integers n and m (1 ≤ n ≤ 10^3, 0 ≤ m ≤ n) — the number of players in one team and the number of special pairs of heroes.
The second line contains 2n integers p_1, p_2, …, p_{2n} (1 ≤ p_i ≤ 10^3) — the powers of the heroes.
Each of the next m lines contains two integer a and b (1 ≤ a, b ≤ 2n, a ≠ b) — a pair of heroes that are especially strong against each other. It is guaranteed that each hero appears at most once in this list.
The next line contains a single integer t (1 ≤ t ≤ 2) — the team you are to play for. If t = 1, the first turn is yours, otherwise you have the second turn.
Hacks
In order to hack, use the format described above with one additional line. In this line output 2n distinct integers from 1 to 2n — the priority order for the jury's team. The jury's team will on each turn select the first possible hero from this list. Here possible means that it is not yet taken and does not contradict the rules about special pair of heroes.
Interaction
When it is your turn, print a single integer x (1 ≤ x ≤ 2n) — the index of the hero chosen by you. Note that you can't choose a hero previously chosen by either you of the other player, and you must follow the rules about special pairs of heroes.
When it is the other team's turn, read a line containing a single integer x (1 ≤ x ≤ 2n) — the index of the hero chosen by the other team. It is guaranteed that this index is not chosen before and that the other team also follows the rules about special pairs of heroes.
After the last turn you should terminate without printing anything.
After printing your choice do not forget to output end of line and flush the output. Otherwise you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Jury's answer -1 instead of a valid choice means that you made an invalid turn. Exit immediately after receiving -1 and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.
Examples
Input
3 1
1 2 3 4 5 6
2 6
1
2
4
1
Output
6
5
3
Input
3 1
1 2 3 4 5 6
1 5
2
6
1
3
Output
5
4
2
Note
In the first example the first turn is yours. In example, you choose 6, the other team is forced to reply with 2. You choose 5, the other team chooses 4. Finally, you choose 3 and the other team choose 1.
In the second example you have the second turn. The other team chooses 6, you choose 5, forcing the other team to choose 1. Now you choose 4, the other team chooses 3 and you choose 2.
Submitted Solution:
```
n,m = list(map(int, input().split()))
heroes = []
h = list(map(int, input().split()))
for u in range(len(h)):
heroes.append((h[u], u))
heroes.sort()
heroes.reverse()
pairs = []
for q in range(m):
pairs.append(list(map(int, input().split())))
t = int(input())
for za in range(2*n):
if t == 1:
er = 0
for y in range(len(pairs)):
if pairs[y][0] == (u-1):
print(pairs[y][1])
pairs.pop(y)
for t in range(len(heroes)):
if heroes[t][1] == (u-1):
heroes.pop(t)
break
else:
er += 1
if er != 0:
print(heroes[0][1]+1)
heroes.pop(0)
t += 1
else:
u = int(input())
if u == -1:
break
for t in range(len(heroes)):
if heroes[t][1] == (u - 1):
heroes.pop(t)
break
t = 1
``` | instruction | 0 | 75,896 | 19 | 151,792 |
No | output | 1 | 75,896 | 19 | 151,793 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Everyone knows that agents in Valorant decide, who will play as attackers, and who will play as defenders. To do that Raze and Breach decided to play t matches of a digit game...
In each of t matches of the digit game, a positive integer is generated. It consists of n digits. The digits of this integer are numerated from 1 to n from the highest-order digit to the lowest-order digit. After this integer is announced, the match starts.
Agents play in turns. Raze starts. In one turn an agent can choose any unmarked digit and mark it. Raze can choose digits on odd positions, but can not choose digits on even positions. Breach can choose digits on even positions, but can not choose digits on odd positions. The match ends, when there is only one unmarked digit left. If the single last digit is odd, then Raze wins, else Breach wins.
It can be proved, that before the end of the match (for every initial integer with n digits) each agent has an ability to make a turn, i.e. there is at least one unmarked digit, that stands on a position of required parity.
For each of t matches find out, which agent wins, if both of them want to win and play optimally.
Input
First line of input contains an integer t (1 ≤ t ≤ 100) — the number of matches.
The first line of each match description contains an integer n (1 ≤ n ≤ 10^3) — the number of digits of the generated number.
The second line of each match description contains an n-digit positive integer without leading zeros.
Output
For each match print 1, if Raze wins, and 2, if Breach wins.
Example
Input
4
1
2
1
3
3
102
4
2069
Output
2
1
1
2
Note
In the first match no one can make a turn, the only digit left is 2, it's even, so Breach wins.
In the second match the only digit left is 3, it's odd, so Raze wins.
In the third match Raze can mark the last digit, after that Breach can only mark 0. 1 will be the last digit left, it's odd, so Raze wins.
In the fourth match no matter how Raze plays, Breach can mark 9, and in the end there will be digit 0. It's even, so Breach wins. | instruction | 0 | 76,056 | 19 | 152,112 |
Tags: games, greedy, implementation
Correct Solution:
```
# import sys
# # For getting input from input.txt file
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
t = int(input())
while t>0:
n = int(input())
number = input()
number_ls = [int(i) for i in number]
start,last = 0,n-1
for i in range(n):
if start>=last:
break
if i%2 == 0:
if start%2==0:
if last%2==0:
if number_ls[last]%2==0:
last-=1
else:
start+=1
else:
start+=1
else:
last-=1
else:
if start%2 != 0:
if last%2 != 0:
if number_ls[last]%2 !=0:
last-=1
else:
start+=1
else:
start+=1
else:
last-=1
if number_ls[start]%2==0:
print(2)
else:
print(1)
t-=1
``` | output | 1 | 76,056 | 19 | 152,113 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Everyone knows that agents in Valorant decide, who will play as attackers, and who will play as defenders. To do that Raze and Breach decided to play t matches of a digit game...
In each of t matches of the digit game, a positive integer is generated. It consists of n digits. The digits of this integer are numerated from 1 to n from the highest-order digit to the lowest-order digit. After this integer is announced, the match starts.
Agents play in turns. Raze starts. In one turn an agent can choose any unmarked digit and mark it. Raze can choose digits on odd positions, but can not choose digits on even positions. Breach can choose digits on even positions, but can not choose digits on odd positions. The match ends, when there is only one unmarked digit left. If the single last digit is odd, then Raze wins, else Breach wins.
It can be proved, that before the end of the match (for every initial integer with n digits) each agent has an ability to make a turn, i.e. there is at least one unmarked digit, that stands on a position of required parity.
For each of t matches find out, which agent wins, if both of them want to win and play optimally.
Input
First line of input contains an integer t (1 ≤ t ≤ 100) — the number of matches.
The first line of each match description contains an integer n (1 ≤ n ≤ 10^3) — the number of digits of the generated number.
The second line of each match description contains an n-digit positive integer without leading zeros.
Output
For each match print 1, if Raze wins, and 2, if Breach wins.
Example
Input
4
1
2
1
3
3
102
4
2069
Output
2
1
1
2
Note
In the first match no one can make a turn, the only digit left is 2, it's even, so Breach wins.
In the second match the only digit left is 3, it's odd, so Raze wins.
In the third match Raze can mark the last digit, after that Breach can only mark 0. 1 will be the last digit left, it's odd, so Raze wins.
In the fourth match no matter how Raze plays, Breach can mark 9, and in the end there will be digit 0. It's even, so Breach wins. | instruction | 0 | 76,057 | 19 | 152,114 |
Tags: games, greedy, implementation
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
s=input()
if n%2==1:
a=[int(i)%2 for i in s[::2]]
if 1 in a:print(1)
else:print(2)
else:
a=[int(i)%2 for i in s[1::2]]
if 0 in a:print(2)
else:print(1)
``` | output | 1 | 76,057 | 19 | 152,115 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Everyone knows that agents in Valorant decide, who will play as attackers, and who will play as defenders. To do that Raze and Breach decided to play t matches of a digit game...
In each of t matches of the digit game, a positive integer is generated. It consists of n digits. The digits of this integer are numerated from 1 to n from the highest-order digit to the lowest-order digit. After this integer is announced, the match starts.
Agents play in turns. Raze starts. In one turn an agent can choose any unmarked digit and mark it. Raze can choose digits on odd positions, but can not choose digits on even positions. Breach can choose digits on even positions, but can not choose digits on odd positions. The match ends, when there is only one unmarked digit left. If the single last digit is odd, then Raze wins, else Breach wins.
It can be proved, that before the end of the match (for every initial integer with n digits) each agent has an ability to make a turn, i.e. there is at least one unmarked digit, that stands on a position of required parity.
For each of t matches find out, which agent wins, if both of them want to win and play optimally.
Input
First line of input contains an integer t (1 ≤ t ≤ 100) — the number of matches.
The first line of each match description contains an integer n (1 ≤ n ≤ 10^3) — the number of digits of the generated number.
The second line of each match description contains an n-digit positive integer without leading zeros.
Output
For each match print 1, if Raze wins, and 2, if Breach wins.
Example
Input
4
1
2
1
3
3
102
4
2069
Output
2
1
1
2
Note
In the first match no one can make a turn, the only digit left is 2, it's even, so Breach wins.
In the second match the only digit left is 3, it's odd, so Raze wins.
In the third match Raze can mark the last digit, after that Breach can only mark 0. 1 will be the last digit left, it's odd, so Raze wins.
In the fourth match no matter how Raze plays, Breach can mark 9, and in the end there will be digit 0. It's even, so Breach wins. | instruction | 0 | 76,058 | 19 | 152,116 |
Tags: games, greedy, implementation
Correct Solution:
```
#*******************************************
#* CODER : ANIRBAN DEY *
#* NICK : nonstop-baban(since 2001) *
#* INSTITUTION : IIEST, SHIBPUR(2019-2023) *
#*******************************************
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
import collections
from itertools import permutations
from collections import defaultdict
import threading
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")
#------------------------------main code------------------------------------------#
t=int(input())
while t!=0:
t-=1
n=int(input())
s=input()
odd1=0
evn2=0
for i in range(n):
if((ord(s[i])-ord('0'))%2!=0 and i%2==0):
odd1+=1
elif((ord(s[i])-ord('0'))%2==0 and i%2!=0):
evn2+=1
if(n%2==0):
if(evn2>0):
print(2)
else:
print(1)
else:
if(odd1>0):
print(1)
else:
print(2)
``` | output | 1 | 76,058 | 19 | 152,117 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Everyone knows that agents in Valorant decide, who will play as attackers, and who will play as defenders. To do that Raze and Breach decided to play t matches of a digit game...
In each of t matches of the digit game, a positive integer is generated. It consists of n digits. The digits of this integer are numerated from 1 to n from the highest-order digit to the lowest-order digit. After this integer is announced, the match starts.
Agents play in turns. Raze starts. In one turn an agent can choose any unmarked digit and mark it. Raze can choose digits on odd positions, but can not choose digits on even positions. Breach can choose digits on even positions, but can not choose digits on odd positions. The match ends, when there is only one unmarked digit left. If the single last digit is odd, then Raze wins, else Breach wins.
It can be proved, that before the end of the match (for every initial integer with n digits) each agent has an ability to make a turn, i.e. there is at least one unmarked digit, that stands on a position of required parity.
For each of t matches find out, which agent wins, if both of them want to win and play optimally.
Input
First line of input contains an integer t (1 ≤ t ≤ 100) — the number of matches.
The first line of each match description contains an integer n (1 ≤ n ≤ 10^3) — the number of digits of the generated number.
The second line of each match description contains an n-digit positive integer without leading zeros.
Output
For each match print 1, if Raze wins, and 2, if Breach wins.
Example
Input
4
1
2
1
3
3
102
4
2069
Output
2
1
1
2
Note
In the first match no one can make a turn, the only digit left is 2, it's even, so Breach wins.
In the second match the only digit left is 3, it's odd, so Raze wins.
In the third match Raze can mark the last digit, after that Breach can only mark 0. 1 will be the last digit left, it's odd, so Raze wins.
In the fourth match no matter how Raze plays, Breach can mark 9, and in the end there will be digit 0. It's even, so Breach wins. | instruction | 0 | 76,059 | 19 | 152,118 |
Tags: games, greedy, implementation
Correct Solution:
```
for _ in [0]*int(input()):
n=int(input())
s=input()
if n%2:
for c in s[::2]:
if c in '13579':
print(1)
break
else:
print(2)
else:
for c in s[1::2]:
if c in '24680':
print(2)
break
else:
print(1)
``` | output | 1 | 76,059 | 19 | 152,119 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Everyone knows that agents in Valorant decide, who will play as attackers, and who will play as defenders. To do that Raze and Breach decided to play t matches of a digit game...
In each of t matches of the digit game, a positive integer is generated. It consists of n digits. The digits of this integer are numerated from 1 to n from the highest-order digit to the lowest-order digit. After this integer is announced, the match starts.
Agents play in turns. Raze starts. In one turn an agent can choose any unmarked digit and mark it. Raze can choose digits on odd positions, but can not choose digits on even positions. Breach can choose digits on even positions, but can not choose digits on odd positions. The match ends, when there is only one unmarked digit left. If the single last digit is odd, then Raze wins, else Breach wins.
It can be proved, that before the end of the match (for every initial integer with n digits) each agent has an ability to make a turn, i.e. there is at least one unmarked digit, that stands on a position of required parity.
For each of t matches find out, which agent wins, if both of them want to win and play optimally.
Input
First line of input contains an integer t (1 ≤ t ≤ 100) — the number of matches.
The first line of each match description contains an integer n (1 ≤ n ≤ 10^3) — the number of digits of the generated number.
The second line of each match description contains an n-digit positive integer without leading zeros.
Output
For each match print 1, if Raze wins, and 2, if Breach wins.
Example
Input
4
1
2
1
3
3
102
4
2069
Output
2
1
1
2
Note
In the first match no one can make a turn, the only digit left is 2, it's even, so Breach wins.
In the second match the only digit left is 3, it's odd, so Raze wins.
In the third match Raze can mark the last digit, after that Breach can only mark 0. 1 will be the last digit left, it's odd, so Raze wins.
In the fourth match no matter how Raze plays, Breach can mark 9, and in the end there will be digit 0. It's even, so Breach wins. | instruction | 0 | 76,060 | 19 | 152,120 |
Tags: games, greedy, implementation
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
ara = str(input())
if n == 1:
if int(ara[0]) % 2 == 0:
print(2)
else:
print(1)
elif n % 2 == 0:
l = 0
for i in range(1,n,2):
if int(ara[i]) % 2 == 0:
print('2')
l += 1
break
if l == 0:
print('1')
elif n % 2 != 0:
l = 0
for i in range(0,n,2):
if int(ara[i]) % 2 != 0:
print('1')
l += 1
break
if l == 0:
print('2')
``` | output | 1 | 76,060 | 19 | 152,121 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Everyone knows that agents in Valorant decide, who will play as attackers, and who will play as defenders. To do that Raze and Breach decided to play t matches of a digit game...
In each of t matches of the digit game, a positive integer is generated. It consists of n digits. The digits of this integer are numerated from 1 to n from the highest-order digit to the lowest-order digit. After this integer is announced, the match starts.
Agents play in turns. Raze starts. In one turn an agent can choose any unmarked digit and mark it. Raze can choose digits on odd positions, but can not choose digits on even positions. Breach can choose digits on even positions, but can not choose digits on odd positions. The match ends, when there is only one unmarked digit left. If the single last digit is odd, then Raze wins, else Breach wins.
It can be proved, that before the end of the match (for every initial integer with n digits) each agent has an ability to make a turn, i.e. there is at least one unmarked digit, that stands on a position of required parity.
For each of t matches find out, which agent wins, if both of them want to win and play optimally.
Input
First line of input contains an integer t (1 ≤ t ≤ 100) — the number of matches.
The first line of each match description contains an integer n (1 ≤ n ≤ 10^3) — the number of digits of the generated number.
The second line of each match description contains an n-digit positive integer without leading zeros.
Output
For each match print 1, if Raze wins, and 2, if Breach wins.
Example
Input
4
1
2
1
3
3
102
4
2069
Output
2
1
1
2
Note
In the first match no one can make a turn, the only digit left is 2, it's even, so Breach wins.
In the second match the only digit left is 3, it's odd, so Raze wins.
In the third match Raze can mark the last digit, after that Breach can only mark 0. 1 will be the last digit left, it's odd, so Raze wins.
In the fourth match no matter how Raze plays, Breach can mark 9, and in the end there will be digit 0. It's even, so Breach wins. | instruction | 0 | 76,061 | 19 | 152,122 |
Tags: games, greedy, implementation
Correct Solution:
```
try:
t=int(input())
for i in range(t):
n=int(input())
s=input()
l=list(s)
l=[int(x) for x in l]
flag,flag2=0,0
for i in range(0,n,2):
if(l[i]%2!=0):
flag=1
break
for i in range(1,n,2):
if(l[i]%2==0):
flag2=1
break
if(n%2!=0):
if(flag==1):
print(1)
else:
print(2)
else:
if(flag2==1):
print(2)
else:
print(1)
except:
pass
``` | output | 1 | 76,061 | 19 | 152,123 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Everyone knows that agents in Valorant decide, who will play as attackers, and who will play as defenders. To do that Raze and Breach decided to play t matches of a digit game...
In each of t matches of the digit game, a positive integer is generated. It consists of n digits. The digits of this integer are numerated from 1 to n from the highest-order digit to the lowest-order digit. After this integer is announced, the match starts.
Agents play in turns. Raze starts. In one turn an agent can choose any unmarked digit and mark it. Raze can choose digits on odd positions, but can not choose digits on even positions. Breach can choose digits on even positions, but can not choose digits on odd positions. The match ends, when there is only one unmarked digit left. If the single last digit is odd, then Raze wins, else Breach wins.
It can be proved, that before the end of the match (for every initial integer with n digits) each agent has an ability to make a turn, i.e. there is at least one unmarked digit, that stands on a position of required parity.
For each of t matches find out, which agent wins, if both of them want to win and play optimally.
Input
First line of input contains an integer t (1 ≤ t ≤ 100) — the number of matches.
The first line of each match description contains an integer n (1 ≤ n ≤ 10^3) — the number of digits of the generated number.
The second line of each match description contains an n-digit positive integer without leading zeros.
Output
For each match print 1, if Raze wins, and 2, if Breach wins.
Example
Input
4
1
2
1
3
3
102
4
2069
Output
2
1
1
2
Note
In the first match no one can make a turn, the only digit left is 2, it's even, so Breach wins.
In the second match the only digit left is 3, it's odd, so Raze wins.
In the third match Raze can mark the last digit, after that Breach can only mark 0. 1 will be the last digit left, it's odd, so Raze wins.
In the fourth match no matter how Raze plays, Breach can mark 9, and in the end there will be digit 0. It's even, so Breach wins. | instruction | 0 | 76,062 | 19 | 152,124 |
Tags: games, greedy, implementation
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
s=input()
d={
"opos_e":0,
"opos_o": 0,
"epos_e": 0,
"epos_o": 0,
}
for i in range(n):
tmp=int(s[i])
if (i+1)%2==1:
if tmp%2==0:
d["opos_e"]+=1
else:
d["opos_o"]+=1
else:
if tmp%2==0:
d["epos_e"]+=1
else:
d["epos_o"]+=1
for i in range(n-1):
if (i+1)%2==1:
if d["opos_e"]-1 >=0:
d["opos_e"] -= 1
else:
d["opos_o"] -= 1
else:
if d["epos_o"]-1>=0:
d["epos_o"] -= 1
else:
d["epos_e"] -= 1
if d["opos_e"]==1 or d["epos_e"]==1:
print(2)
else:
print(1)
``` | output | 1 | 76,062 | 19 | 152,125 |
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