message stringlengths 2 30.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 237 109k | cluster float64 10 10 | __index_level_0__ int64 474 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya has a scales for weighing loads and weights of masses w0, w1, w2, ..., w100 grams where w is some integer not less than 2 (exactly one weight of each nominal value). Vanya wonders whether he can weight an item with mass m using the given weights, if the weights can be put on both pans of the scales. Formally speaking, your task is to determine whether it is possible to place an item of mass m and some weights on the left pan of the scales, and some weights on the right pan of the scales so that the pans of the scales were in balance.
Input
The first line contains two integers w, m (2 β€ w β€ 109, 1 β€ m β€ 109) β the number defining the masses of the weights and the mass of the item.
Output
Print word 'YES' if the item can be weighted and 'NO' if it cannot.
Examples
Input
3 7
Output
YES
Input
100 99
Output
YES
Input
100 50
Output
NO
Note
Note to the first sample test. One pan can have an item of mass 7 and a weight of mass 3, and the second pan can have two weights of masses 9 and 1, correspondingly. Then 7 + 3 = 9 + 1.
Note to the second sample test. One pan of the scales can have an item of mass 99 and the weight of mass 1, and the second pan can have the weight of mass 100.
Note to the third sample test. It is impossible to measure the weight of the item in the manner described in the input.
Submitted Solution:
```
w, m = map(int, input().split())
for i in range(100, -1, -1):
if abs(m - w**i) < m:
m = abs(m - w**i)
print('YES' if m == 0 else 'NO')
``` | instruction | 0 | 12,886 | 10 | 25,772 |
Yes | output | 1 | 12,886 | 10 | 25,773 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya has a scales for weighing loads and weights of masses w0, w1, w2, ..., w100 grams where w is some integer not less than 2 (exactly one weight of each nominal value). Vanya wonders whether he can weight an item with mass m using the given weights, if the weights can be put on both pans of the scales. Formally speaking, your task is to determine whether it is possible to place an item of mass m and some weights on the left pan of the scales, and some weights on the right pan of the scales so that the pans of the scales were in balance.
Input
The first line contains two integers w, m (2 β€ w β€ 109, 1 β€ m β€ 109) β the number defining the masses of the weights and the mass of the item.
Output
Print word 'YES' if the item can be weighted and 'NO' if it cannot.
Examples
Input
3 7
Output
YES
Input
100 99
Output
YES
Input
100 50
Output
NO
Note
Note to the first sample test. One pan can have an item of mass 7 and a weight of mass 3, and the second pan can have two weights of masses 9 and 1, correspondingly. Then 7 + 3 = 9 + 1.
Note to the second sample test. One pan of the scales can have an item of mass 99 and the weight of mass 1, and the second pan can have the weight of mass 100.
Note to the third sample test. It is impossible to measure the weight of the item in the manner described in the input.
Submitted Solution:
```
w,m=map(int,input().split())
if m>1:
while m:
if 1<m%w<w-1: break
m=(m+1)%w
print("NO" if m%w>1 else "YES")
``` | instruction | 0 | 12,887 | 10 | 25,774 |
No | output | 1 | 12,887 | 10 | 25,775 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya has a scales for weighing loads and weights of masses w0, w1, w2, ..., w100 grams where w is some integer not less than 2 (exactly one weight of each nominal value). Vanya wonders whether he can weight an item with mass m using the given weights, if the weights can be put on both pans of the scales. Formally speaking, your task is to determine whether it is possible to place an item of mass m and some weights on the left pan of the scales, and some weights on the right pan of the scales so that the pans of the scales were in balance.
Input
The first line contains two integers w, m (2 β€ w β€ 109, 1 β€ m β€ 109) β the number defining the masses of the weights and the mass of the item.
Output
Print word 'YES' if the item can be weighted and 'NO' if it cannot.
Examples
Input
3 7
Output
YES
Input
100 99
Output
YES
Input
100 50
Output
NO
Note
Note to the first sample test. One pan can have an item of mass 7 and a weight of mass 3, and the second pan can have two weights of masses 9 and 1, correspondingly. Then 7 + 3 = 9 + 1.
Note to the second sample test. One pan of the scales can have an item of mass 99 and the weight of mass 1, and the second pan can have the weight of mass 100.
Note to the third sample test. It is impossible to measure the weight of the item in the manner described in the input.
Submitted Solution:
```
import math
while(1):
a,b=input().split()
a=int(a)
b=int(b)
k=math.log(b)//math.log(a)
diff=min(abs(b-a**k),abs(b-a**(k+1)))
if(diff>=2):
if(math.log(diff)/math.log(a)==math.log(diff)//math.log(a) or math.log(diff+1)/math.log(a)==math.log(diff+1)//math.log(a) or math.log(diff-1)/math.log(a)==math.log(diff-1)//math.log(a)):
print("YES")
break
if(diff==abs(b-a**k)):
k=k
else:
k=k+1
m=a**k
n=m+1
t=b
for x in range (0,1000000000):
if(b==m or b==n):
print("YES")
break
elif(b>n):
print("NO")
break
b=t
b=b+3**x
break
``` | instruction | 0 | 12,888 | 10 | 25,776 |
No | output | 1 | 12,888 | 10 | 25,777 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya has a scales for weighing loads and weights of masses w0, w1, w2, ..., w100 grams where w is some integer not less than 2 (exactly one weight of each nominal value). Vanya wonders whether he can weight an item with mass m using the given weights, if the weights can be put on both pans of the scales. Formally speaking, your task is to determine whether it is possible to place an item of mass m and some weights on the left pan of the scales, and some weights on the right pan of the scales so that the pans of the scales were in balance.
Input
The first line contains two integers w, m (2 β€ w β€ 109, 1 β€ m β€ 109) β the number defining the masses of the weights and the mass of the item.
Output
Print word 'YES' if the item can be weighted and 'NO' if it cannot.
Examples
Input
3 7
Output
YES
Input
100 99
Output
YES
Input
100 50
Output
NO
Note
Note to the first sample test. One pan can have an item of mass 7 and a weight of mass 3, and the second pan can have two weights of masses 9 and 1, correspondingly. Then 7 + 3 = 9 + 1.
Note to the second sample test. One pan of the scales can have an item of mass 99 and the weight of mass 1, and the second pan can have the weight of mass 100.
Note to the third sample test. It is impossible to measure the weight of the item in the manner described in the input.
Submitted Solution:
```
base,num = map(int,input().split())
if base==3 or base==2 :
print("YES")
quit()
newNum = []
while num > 0:
newNum=[num % base] + newNum
num //= base
for i in range(len(newNum)-1,-1,-1) :
if newNum[i]==base-1 :
for j in range(i-1,-1,-1) :
if newNum[j]==0 :
break
else :
if newNum[j]!=base-1 :
print("NO")
quit()
else :
if newNum[i]!=0 and newNum[i]!=1 :
print("NO")
quit()
print("YES")
``` | instruction | 0 | 12,889 | 10 | 25,778 |
No | output | 1 | 12,889 | 10 | 25,779 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya has a scales for weighing loads and weights of masses w0, w1, w2, ..., w100 grams where w is some integer not less than 2 (exactly one weight of each nominal value). Vanya wonders whether he can weight an item with mass m using the given weights, if the weights can be put on both pans of the scales. Formally speaking, your task is to determine whether it is possible to place an item of mass m and some weights on the left pan of the scales, and some weights on the right pan of the scales so that the pans of the scales were in balance.
Input
The first line contains two integers w, m (2 β€ w β€ 109, 1 β€ m β€ 109) β the number defining the masses of the weights and the mass of the item.
Output
Print word 'YES' if the item can be weighted and 'NO' if it cannot.
Examples
Input
3 7
Output
YES
Input
100 99
Output
YES
Input
100 50
Output
NO
Note
Note to the first sample test. One pan can have an item of mass 7 and a weight of mass 3, and the second pan can have two weights of masses 9 and 1, correspondingly. Then 7 + 3 = 9 + 1.
Note to the second sample test. One pan of the scales can have an item of mass 99 and the weight of mass 1, and the second pan can have the weight of mass 100.
Note to the third sample test. It is impossible to measure the weight of the item in the manner described in the input.
Submitted Solution:
```
w, m = map(int, input().split())
for i in range(101):
tmp = m+0
base = 1
for j in range(i+1):
tmp += base
base *= w
while base > 0:
if tmp >= base*2:
tmp -= base*2
base //= w
if tmp == 0:
print('YES')
exit()
print('NO')
``` | instruction | 0 | 12,890 | 10 | 25,780 |
No | output | 1 | 12,890 | 10 | 25,781 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, Dima met with Sasha in a philatelic store, and since then they are collecting coins together. Their favorite occupation is to sort collections of coins. Sasha likes having things in order, that is why he wants his coins to be arranged in a row in such a way that firstly come coins out of circulation, and then come coins still in circulation.
For arranging coins Dima uses the following algorithm. One step of his algorithm looks like the following:
1. He looks through all the coins from left to right;
2. If he sees that the i-th coin is still in circulation, and (i + 1)-th coin is already out of circulation, he exchanges these two coins and continues watching coins from (i + 1)-th.
Dima repeats the procedure above until it happens that no two coins were exchanged during this procedure. Dima calls hardness of ordering the number of steps required for him according to the algorithm above to sort the sequence, e.g. the number of times he looks through the coins from the very beginning. For example, for the ordered sequence hardness of ordering equals one.
Today Sasha invited Dima and proposed him a game. First he puts n coins in a row, all of them are out of circulation. Then Sasha chooses one of the coins out of circulation and replaces it with a coin in circulation for n times. During this process Sasha constantly asks Dima what is the hardness of ordering of the sequence.
The task is more complicated because Dima should not touch the coins and he should determine hardness of ordering in his mind. Help Dima with this task.
Input
The first line contains single integer n (1 β€ n β€ 300 000) β number of coins that Sasha puts behind Dima.
Second line contains n distinct integers p1, p2, ..., pn (1 β€ pi β€ n) β positions that Sasha puts coins in circulation to. At first Sasha replaces coin located at position p1, then coin located at position p2 and so on. Coins are numbered from left to right.
Output
Print n + 1 numbers a0, a1, ..., an, where a0 is a hardness of ordering at the beginning, a1 is a hardness of ordering after the first replacement and so on.
Examples
Input
4
1 3 4 2
Output
1 2 3 2 1
Input
8
6 8 3 4 7 2 1 5
Output
1 2 2 3 4 3 4 5 1
Note
Let's denote as O coin out of circulation, and as X β coin is circulation.
At the first sample, initially in row there are coins that are not in circulation, so Dima will look through them from left to right and won't make any exchanges.
After replacement of the first coin with a coin in circulation, Dima will exchange this coin with next three times and after that he will finally look through the coins and finish the process.
XOOO β OOOX
After replacement of the third coin, Dima's actions look this way:
XOXO β OXOX β OOXX
After replacement of the fourth coin, Dima's actions look this way:
XOXX β OXXX
Finally, after replacement of the second coin, row becomes consisting of coins that are in circulation and Dima will look through coins from left to right without any exchanges. | instruction | 0 | 13,035 | 10 | 26,070 |
Tags: dsu, implementation, sortings, two pointers
Correct Solution:
```
n = int(input());
p = [int(x) for x in input().split()];
st = [1] * n;
r = n-1;
print(1, end = " ");
for i in range(0, n, 1):
st[p[i]-1] = 0;
j = r;
while ( j>=0 and st[j] == 0 ):
j -= 1;
r = j;
print(i+2 - (n - 1 - r), end = " ");
``` | output | 1 | 13,035 | 10 | 26,071 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, Dima met with Sasha in a philatelic store, and since then they are collecting coins together. Their favorite occupation is to sort collections of coins. Sasha likes having things in order, that is why he wants his coins to be arranged in a row in such a way that firstly come coins out of circulation, and then come coins still in circulation.
For arranging coins Dima uses the following algorithm. One step of his algorithm looks like the following:
1. He looks through all the coins from left to right;
2. If he sees that the i-th coin is still in circulation, and (i + 1)-th coin is already out of circulation, he exchanges these two coins and continues watching coins from (i + 1)-th.
Dima repeats the procedure above until it happens that no two coins were exchanged during this procedure. Dima calls hardness of ordering the number of steps required for him according to the algorithm above to sort the sequence, e.g. the number of times he looks through the coins from the very beginning. For example, for the ordered sequence hardness of ordering equals one.
Today Sasha invited Dima and proposed him a game. First he puts n coins in a row, all of them are out of circulation. Then Sasha chooses one of the coins out of circulation and replaces it with a coin in circulation for n times. During this process Sasha constantly asks Dima what is the hardness of ordering of the sequence.
The task is more complicated because Dima should not touch the coins and he should determine hardness of ordering in his mind. Help Dima with this task.
Input
The first line contains single integer n (1 β€ n β€ 300 000) β number of coins that Sasha puts behind Dima.
Second line contains n distinct integers p1, p2, ..., pn (1 β€ pi β€ n) β positions that Sasha puts coins in circulation to. At first Sasha replaces coin located at position p1, then coin located at position p2 and so on. Coins are numbered from left to right.
Output
Print n + 1 numbers a0, a1, ..., an, where a0 is a hardness of ordering at the beginning, a1 is a hardness of ordering after the first replacement and so on.
Examples
Input
4
1 3 4 2
Output
1 2 3 2 1
Input
8
6 8 3 4 7 2 1 5
Output
1 2 2 3 4 3 4 5 1
Note
Let's denote as O coin out of circulation, and as X β coin is circulation.
At the first sample, initially in row there are coins that are not in circulation, so Dima will look through them from left to right and won't make any exchanges.
After replacement of the first coin with a coin in circulation, Dima will exchange this coin with next three times and after that he will finally look through the coins and finish the process.
XOOO β OOOX
After replacement of the third coin, Dima's actions look this way:
XOXO β OXOX β OOXX
After replacement of the fourth coin, Dima's actions look this way:
XOXX β OXXX
Finally, after replacement of the second coin, row becomes consisting of coins that are in circulation and Dima will look through coins from left to right without any exchanges. | instruction | 0 | 13,036 | 10 | 26,072 |
Tags: dsu, implementation, sortings, two pointers
Correct Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
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
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------game starts now-----------------------------------------------------
n=int(input())
a=list(map(int,input().split()))
r=[1]
c=1
b=[0]*n
for i in range (n):
b[a[i]-1]=1
temp=c
for j in range (n-temp,-1,-1):
if b[j]==1:
c+=1
else:
break
r.append(i+2-c+1)
print(*r)
``` | output | 1 | 13,036 | 10 | 26,073 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, Dima met with Sasha in a philatelic store, and since then they are collecting coins together. Their favorite occupation is to sort collections of coins. Sasha likes having things in order, that is why he wants his coins to be arranged in a row in such a way that firstly come coins out of circulation, and then come coins still in circulation.
For arranging coins Dima uses the following algorithm. One step of his algorithm looks like the following:
1. He looks through all the coins from left to right;
2. If he sees that the i-th coin is still in circulation, and (i + 1)-th coin is already out of circulation, he exchanges these two coins and continues watching coins from (i + 1)-th.
Dima repeats the procedure above until it happens that no two coins were exchanged during this procedure. Dima calls hardness of ordering the number of steps required for him according to the algorithm above to sort the sequence, e.g. the number of times he looks through the coins from the very beginning. For example, for the ordered sequence hardness of ordering equals one.
Today Sasha invited Dima and proposed him a game. First he puts n coins in a row, all of them are out of circulation. Then Sasha chooses one of the coins out of circulation and replaces it with a coin in circulation for n times. During this process Sasha constantly asks Dima what is the hardness of ordering of the sequence.
The task is more complicated because Dima should not touch the coins and he should determine hardness of ordering in his mind. Help Dima with this task.
Input
The first line contains single integer n (1 β€ n β€ 300 000) β number of coins that Sasha puts behind Dima.
Second line contains n distinct integers p1, p2, ..., pn (1 β€ pi β€ n) β positions that Sasha puts coins in circulation to. At first Sasha replaces coin located at position p1, then coin located at position p2 and so on. Coins are numbered from left to right.
Output
Print n + 1 numbers a0, a1, ..., an, where a0 is a hardness of ordering at the beginning, a1 is a hardness of ordering after the first replacement and so on.
Examples
Input
4
1 3 4 2
Output
1 2 3 2 1
Input
8
6 8 3 4 7 2 1 5
Output
1 2 2 3 4 3 4 5 1
Note
Let's denote as O coin out of circulation, and as X β coin is circulation.
At the first sample, initially in row there are coins that are not in circulation, so Dima will look through them from left to right and won't make any exchanges.
After replacement of the first coin with a coin in circulation, Dima will exchange this coin with next three times and after that he will finally look through the coins and finish the process.
XOOO β OOOX
After replacement of the third coin, Dima's actions look this way:
XOXO β OXOX β OOXX
After replacement of the fourth coin, Dima's actions look this way:
XOXX β OXXX
Finally, after replacement of the second coin, row becomes consisting of coins that are in circulation and Dima will look through coins from left to right without any exchanges. | instruction | 0 | 13,037 | 10 | 26,074 |
Tags: dsu, implementation, sortings, two pointers
Correct Solution:
```
n = int(input())
l = list(map(int,input().split()))
print(1,end = " ")
ptr = n-1
v = [0]*n
for i in range(n):
v[l[i]-1] = 1
while(ptr>=0 and v[ptr]==1):ptr-=1
print(i+1-(n-1-ptr)+1,end = " ")
``` | output | 1 | 13,037 | 10 | 26,075 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, Dima met with Sasha in a philatelic store, and since then they are collecting coins together. Their favorite occupation is to sort collections of coins. Sasha likes having things in order, that is why he wants his coins to be arranged in a row in such a way that firstly come coins out of circulation, and then come coins still in circulation.
For arranging coins Dima uses the following algorithm. One step of his algorithm looks like the following:
1. He looks through all the coins from left to right;
2. If he sees that the i-th coin is still in circulation, and (i + 1)-th coin is already out of circulation, he exchanges these two coins and continues watching coins from (i + 1)-th.
Dima repeats the procedure above until it happens that no two coins were exchanged during this procedure. Dima calls hardness of ordering the number of steps required for him according to the algorithm above to sort the sequence, e.g. the number of times he looks through the coins from the very beginning. For example, for the ordered sequence hardness of ordering equals one.
Today Sasha invited Dima and proposed him a game. First he puts n coins in a row, all of them are out of circulation. Then Sasha chooses one of the coins out of circulation and replaces it with a coin in circulation for n times. During this process Sasha constantly asks Dima what is the hardness of ordering of the sequence.
The task is more complicated because Dima should not touch the coins and he should determine hardness of ordering in his mind. Help Dima with this task.
Input
The first line contains single integer n (1 β€ n β€ 300 000) β number of coins that Sasha puts behind Dima.
Second line contains n distinct integers p1, p2, ..., pn (1 β€ pi β€ n) β positions that Sasha puts coins in circulation to. At first Sasha replaces coin located at position p1, then coin located at position p2 and so on. Coins are numbered from left to right.
Output
Print n + 1 numbers a0, a1, ..., an, where a0 is a hardness of ordering at the beginning, a1 is a hardness of ordering after the first replacement and so on.
Examples
Input
4
1 3 4 2
Output
1 2 3 2 1
Input
8
6 8 3 4 7 2 1 5
Output
1 2 2 3 4 3 4 5 1
Note
Let's denote as O coin out of circulation, and as X β coin is circulation.
At the first sample, initially in row there are coins that are not in circulation, so Dima will look through them from left to right and won't make any exchanges.
After replacement of the first coin with a coin in circulation, Dima will exchange this coin with next three times and after that he will finally look through the coins and finish the process.
XOOO β OOOX
After replacement of the third coin, Dima's actions look this way:
XOXO β OXOX β OOXX
After replacement of the fourth coin, Dima's actions look this way:
XOXX β OXXX
Finally, after replacement of the second coin, row becomes consisting of coins that are in circulation and Dima will look through coins from left to right without any exchanges. | instruction | 0 | 13,038 | 10 | 26,076 |
Tags: dsu, implementation, sortings, two pointers
Correct Solution:
```
n = int(input())
p = list(map(int, input().split()))
lp = n+1
ans = [1]
vis = [0 for i in range(n)]
ans = [1]
top = n
hardness = 1
for i in range(len(p)):
vis[p[i]-1] = 1
hardness += 1
while vis[top-1] == 1 and top > 0:
top -= 1
hardness -=1
ans.append(hardness)
print(' '.join([str(i) for i in ans]))
``` | output | 1 | 13,038 | 10 | 26,077 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, Dima met with Sasha in a philatelic store, and since then they are collecting coins together. Their favorite occupation is to sort collections of coins. Sasha likes having things in order, that is why he wants his coins to be arranged in a row in such a way that firstly come coins out of circulation, and then come coins still in circulation.
For arranging coins Dima uses the following algorithm. One step of his algorithm looks like the following:
1. He looks through all the coins from left to right;
2. If he sees that the i-th coin is still in circulation, and (i + 1)-th coin is already out of circulation, he exchanges these two coins and continues watching coins from (i + 1)-th.
Dima repeats the procedure above until it happens that no two coins were exchanged during this procedure. Dima calls hardness of ordering the number of steps required for him according to the algorithm above to sort the sequence, e.g. the number of times he looks through the coins from the very beginning. For example, for the ordered sequence hardness of ordering equals one.
Today Sasha invited Dima and proposed him a game. First he puts n coins in a row, all of them are out of circulation. Then Sasha chooses one of the coins out of circulation and replaces it with a coin in circulation for n times. During this process Sasha constantly asks Dima what is the hardness of ordering of the sequence.
The task is more complicated because Dima should not touch the coins and he should determine hardness of ordering in his mind. Help Dima with this task.
Input
The first line contains single integer n (1 β€ n β€ 300 000) β number of coins that Sasha puts behind Dima.
Second line contains n distinct integers p1, p2, ..., pn (1 β€ pi β€ n) β positions that Sasha puts coins in circulation to. At first Sasha replaces coin located at position p1, then coin located at position p2 and so on. Coins are numbered from left to right.
Output
Print n + 1 numbers a0, a1, ..., an, where a0 is a hardness of ordering at the beginning, a1 is a hardness of ordering after the first replacement and so on.
Examples
Input
4
1 3 4 2
Output
1 2 3 2 1
Input
8
6 8 3 4 7 2 1 5
Output
1 2 2 3 4 3 4 5 1
Note
Let's denote as O coin out of circulation, and as X β coin is circulation.
At the first sample, initially in row there are coins that are not in circulation, so Dima will look through them from left to right and won't make any exchanges.
After replacement of the first coin with a coin in circulation, Dima will exchange this coin with next three times and after that he will finally look through the coins and finish the process.
XOOO β OOOX
After replacement of the third coin, Dima's actions look this way:
XOXO β OXOX β OOXX
After replacement of the fourth coin, Dima's actions look this way:
XOXX β OXXX
Finally, after replacement of the second coin, row becomes consisting of coins that are in circulation and Dima will look through coins from left to right without any exchanges. | instruction | 0 | 13,039 | 10 | 26,078 |
Tags: dsu, implementation, sortings, two pointers
Correct Solution:
```
n = int(input())
input_ = list(map(int, input().split(' ')))
pos = n
a = [0 for i in range(n+1)]
res = 1
ans = [1]
for x in input_:
a[x] = 1
res += 1
while a[pos]==1:
pos -= 1
res -= 1
ans.append(res)
print (' '.join(map(str, ans)))
``` | output | 1 | 13,039 | 10 | 26,079 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, Dima met with Sasha in a philatelic store, and since then they are collecting coins together. Their favorite occupation is to sort collections of coins. Sasha likes having things in order, that is why he wants his coins to be arranged in a row in such a way that firstly come coins out of circulation, and then come coins still in circulation.
For arranging coins Dima uses the following algorithm. One step of his algorithm looks like the following:
1. He looks through all the coins from left to right;
2. If he sees that the i-th coin is still in circulation, and (i + 1)-th coin is already out of circulation, he exchanges these two coins and continues watching coins from (i + 1)-th.
Dima repeats the procedure above until it happens that no two coins were exchanged during this procedure. Dima calls hardness of ordering the number of steps required for him according to the algorithm above to sort the sequence, e.g. the number of times he looks through the coins from the very beginning. For example, for the ordered sequence hardness of ordering equals one.
Today Sasha invited Dima and proposed him a game. First he puts n coins in a row, all of them are out of circulation. Then Sasha chooses one of the coins out of circulation and replaces it with a coin in circulation for n times. During this process Sasha constantly asks Dima what is the hardness of ordering of the sequence.
The task is more complicated because Dima should not touch the coins and he should determine hardness of ordering in his mind. Help Dima with this task.
Input
The first line contains single integer n (1 β€ n β€ 300 000) β number of coins that Sasha puts behind Dima.
Second line contains n distinct integers p1, p2, ..., pn (1 β€ pi β€ n) β positions that Sasha puts coins in circulation to. At first Sasha replaces coin located at position p1, then coin located at position p2 and so on. Coins are numbered from left to right.
Output
Print n + 1 numbers a0, a1, ..., an, where a0 is a hardness of ordering at the beginning, a1 is a hardness of ordering after the first replacement and so on.
Examples
Input
4
1 3 4 2
Output
1 2 3 2 1
Input
8
6 8 3 4 7 2 1 5
Output
1 2 2 3 4 3 4 5 1
Note
Let's denote as O coin out of circulation, and as X β coin is circulation.
At the first sample, initially in row there are coins that are not in circulation, so Dima will look through them from left to right and won't make any exchanges.
After replacement of the first coin with a coin in circulation, Dima will exchange this coin with next three times and after that he will finally look through the coins and finish the process.
XOOO β OOOX
After replacement of the third coin, Dima's actions look this way:
XOXO β OXOX β OOXX
After replacement of the fourth coin, Dima's actions look this way:
XOXX β OXXX
Finally, after replacement of the second coin, row becomes consisting of coins that are in circulation and Dima will look through coins from left to right without any exchanges. | instruction | 0 | 13,040 | 10 | 26,080 |
Tags: dsu, implementation, sortings, two pointers
Correct Solution:
```
n=int(input())
a=[0]*n
p=list(map(lambda x:int(x)-1,input().split()))
print(1,end=' ')
x=n-1
for i in range(n-1):
a[p[i]]=1
if p[i]==x:
while a[x]:
x-=1
print(i-n+x+3,end=' ')
print(1)
``` | output | 1 | 13,040 | 10 | 26,081 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, Dima met with Sasha in a philatelic store, and since then they are collecting coins together. Their favorite occupation is to sort collections of coins. Sasha likes having things in order, that is why he wants his coins to be arranged in a row in such a way that firstly come coins out of circulation, and then come coins still in circulation.
For arranging coins Dima uses the following algorithm. One step of his algorithm looks like the following:
1. He looks through all the coins from left to right;
2. If he sees that the i-th coin is still in circulation, and (i + 1)-th coin is already out of circulation, he exchanges these two coins and continues watching coins from (i + 1)-th.
Dima repeats the procedure above until it happens that no two coins were exchanged during this procedure. Dima calls hardness of ordering the number of steps required for him according to the algorithm above to sort the sequence, e.g. the number of times he looks through the coins from the very beginning. For example, for the ordered sequence hardness of ordering equals one.
Today Sasha invited Dima and proposed him a game. First he puts n coins in a row, all of them are out of circulation. Then Sasha chooses one of the coins out of circulation and replaces it with a coin in circulation for n times. During this process Sasha constantly asks Dima what is the hardness of ordering of the sequence.
The task is more complicated because Dima should not touch the coins and he should determine hardness of ordering in his mind. Help Dima with this task.
Input
The first line contains single integer n (1 β€ n β€ 300 000) β number of coins that Sasha puts behind Dima.
Second line contains n distinct integers p1, p2, ..., pn (1 β€ pi β€ n) β positions that Sasha puts coins in circulation to. At first Sasha replaces coin located at position p1, then coin located at position p2 and so on. Coins are numbered from left to right.
Output
Print n + 1 numbers a0, a1, ..., an, where a0 is a hardness of ordering at the beginning, a1 is a hardness of ordering after the first replacement and so on.
Examples
Input
4
1 3 4 2
Output
1 2 3 2 1
Input
8
6 8 3 4 7 2 1 5
Output
1 2 2 3 4 3 4 5 1
Note
Let's denote as O coin out of circulation, and as X β coin is circulation.
At the first sample, initially in row there are coins that are not in circulation, so Dima will look through them from left to right and won't make any exchanges.
After replacement of the first coin with a coin in circulation, Dima will exchange this coin with next three times and after that he will finally look through the coins and finish the process.
XOOO β OOOX
After replacement of the third coin, Dima's actions look this way:
XOXO β OXOX β OOXX
After replacement of the fourth coin, Dima's actions look this way:
XOXX β OXXX
Finally, after replacement of the second coin, row becomes consisting of coins that are in circulation and Dima will look through coins from left to right without any exchanges. | instruction | 0 | 13,041 | 10 | 26,082 |
Tags: dsu, implementation, sortings, two pointers
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
p = [0] * (n + 1)
ans = [1] * (n + 1)
ind = n
for i in range(n):
p[a[i] - 1] = 1
while ind > 0 and p[ind - 1] == 1:
ind -= 1
ans[i + 1] = 1 + (i + 1) - (n - ind)
print(' '.join(map(str, ans)))
``` | output | 1 | 13,041 | 10 | 26,083 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, Dima met with Sasha in a philatelic store, and since then they are collecting coins together. Their favorite occupation is to sort collections of coins. Sasha likes having things in order, that is why he wants his coins to be arranged in a row in such a way that firstly come coins out of circulation, and then come coins still in circulation.
For arranging coins Dima uses the following algorithm. One step of his algorithm looks like the following:
1. He looks through all the coins from left to right;
2. If he sees that the i-th coin is still in circulation, and (i + 1)-th coin is already out of circulation, he exchanges these two coins and continues watching coins from (i + 1)-th.
Dima repeats the procedure above until it happens that no two coins were exchanged during this procedure. Dima calls hardness of ordering the number of steps required for him according to the algorithm above to sort the sequence, e.g. the number of times he looks through the coins from the very beginning. For example, for the ordered sequence hardness of ordering equals one.
Today Sasha invited Dima and proposed him a game. First he puts n coins in a row, all of them are out of circulation. Then Sasha chooses one of the coins out of circulation and replaces it with a coin in circulation for n times. During this process Sasha constantly asks Dima what is the hardness of ordering of the sequence.
The task is more complicated because Dima should not touch the coins and he should determine hardness of ordering in his mind. Help Dima with this task.
Input
The first line contains single integer n (1 β€ n β€ 300 000) β number of coins that Sasha puts behind Dima.
Second line contains n distinct integers p1, p2, ..., pn (1 β€ pi β€ n) β positions that Sasha puts coins in circulation to. At first Sasha replaces coin located at position p1, then coin located at position p2 and so on. Coins are numbered from left to right.
Output
Print n + 1 numbers a0, a1, ..., an, where a0 is a hardness of ordering at the beginning, a1 is a hardness of ordering after the first replacement and so on.
Examples
Input
4
1 3 4 2
Output
1 2 3 2 1
Input
8
6 8 3 4 7 2 1 5
Output
1 2 2 3 4 3 4 5 1
Note
Let's denote as O coin out of circulation, and as X β coin is circulation.
At the first sample, initially in row there are coins that are not in circulation, so Dima will look through them from left to right and won't make any exchanges.
After replacement of the first coin with a coin in circulation, Dima will exchange this coin with next three times and after that he will finally look through the coins and finish the process.
XOOO β OOOX
After replacement of the third coin, Dima's actions look this way:
XOXO β OXOX β OOXX
After replacement of the fourth coin, Dima's actions look this way:
XOXX β OXXX
Finally, after replacement of the second coin, row becomes consisting of coins that are in circulation and Dima will look through coins from left to right without any exchanges. | instruction | 0 | 13,042 | 10 | 26,084 |
Tags: dsu, implementation, sortings, two pointers
Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
end=1
c=[]
s='1'
for i in range(n):
c.append(False)
for i in range(n):
c[a[i]-1]=True
while(n>=end and c[n-end]):
end+=1
s+=' ' + str(3+i-end)
print(s)
``` | output | 1 | 13,042 | 10 | 26,085 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, Dima met with Sasha in a philatelic store, and since then they are collecting coins together. Their favorite occupation is to sort collections of coins. Sasha likes having things in order, that is why he wants his coins to be arranged in a row in such a way that firstly come coins out of circulation, and then come coins still in circulation.
For arranging coins Dima uses the following algorithm. One step of his algorithm looks like the following:
1. He looks through all the coins from left to right;
2. If he sees that the i-th coin is still in circulation, and (i + 1)-th coin is already out of circulation, he exchanges these two coins and continues watching coins from (i + 1)-th.
Dima repeats the procedure above until it happens that no two coins were exchanged during this procedure. Dima calls hardness of ordering the number of steps required for him according to the algorithm above to sort the sequence, e.g. the number of times he looks through the coins from the very beginning. For example, for the ordered sequence hardness of ordering equals one.
Today Sasha invited Dima and proposed him a game. First he puts n coins in a row, all of them are out of circulation. Then Sasha chooses one of the coins out of circulation and replaces it with a coin in circulation for n times. During this process Sasha constantly asks Dima what is the hardness of ordering of the sequence.
The task is more complicated because Dima should not touch the coins and he should determine hardness of ordering in his mind. Help Dima with this task.
Input
The first line contains single integer n (1 β€ n β€ 300 000) β number of coins that Sasha puts behind Dima.
Second line contains n distinct integers p1, p2, ..., pn (1 β€ pi β€ n) β positions that Sasha puts coins in circulation to. At first Sasha replaces coin located at position p1, then coin located at position p2 and so on. Coins are numbered from left to right.
Output
Print n + 1 numbers a0, a1, ..., an, where a0 is a hardness of ordering at the beginning, a1 is a hardness of ordering after the first replacement and so on.
Examples
Input
4
1 3 4 2
Output
1 2 3 2 1
Input
8
6 8 3 4 7 2 1 5
Output
1 2 2 3 4 3 4 5 1
Note
Let's denote as O coin out of circulation, and as X β coin is circulation.
At the first sample, initially in row there are coins that are not in circulation, so Dima will look through them from left to right and won't make any exchanges.
After replacement of the first coin with a coin in circulation, Dima will exchange this coin with next three times and after that he will finally look through the coins and finish the process.
XOOO β OOOX
After replacement of the third coin, Dima's actions look this way:
XOXO β OXOX β OOXX
After replacement of the fourth coin, Dima's actions look this way:
XOXX β OXXX
Finally, after replacement of the second coin, row becomes consisting of coins that are in circulation and Dima will look through coins from left to right without any exchanges.
Submitted Solution:
```
n = int(input())
x = [0]*n
a = 0
p = list(map(int, input().split()))
z = n-1
ans = ['1']
for i in range(n):
x[p[i]-1] = 1
a += 1
while z> -1 and x[z] == 1:
z-=1
a-=1
ans.append(str(a+1))
print(' '.join(ans))
``` | instruction | 0 | 13,043 | 10 | 26,086 |
Yes | output | 1 | 13,043 | 10 | 26,087 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, Dima met with Sasha in a philatelic store, and since then they are collecting coins together. Their favorite occupation is to sort collections of coins. Sasha likes having things in order, that is why he wants his coins to be arranged in a row in such a way that firstly come coins out of circulation, and then come coins still in circulation.
For arranging coins Dima uses the following algorithm. One step of his algorithm looks like the following:
1. He looks through all the coins from left to right;
2. If he sees that the i-th coin is still in circulation, and (i + 1)-th coin is already out of circulation, he exchanges these two coins and continues watching coins from (i + 1)-th.
Dima repeats the procedure above until it happens that no two coins were exchanged during this procedure. Dima calls hardness of ordering the number of steps required for him according to the algorithm above to sort the sequence, e.g. the number of times he looks through the coins from the very beginning. For example, for the ordered sequence hardness of ordering equals one.
Today Sasha invited Dima and proposed him a game. First he puts n coins in a row, all of them are out of circulation. Then Sasha chooses one of the coins out of circulation and replaces it with a coin in circulation for n times. During this process Sasha constantly asks Dima what is the hardness of ordering of the sequence.
The task is more complicated because Dima should not touch the coins and he should determine hardness of ordering in his mind. Help Dima with this task.
Input
The first line contains single integer n (1 β€ n β€ 300 000) β number of coins that Sasha puts behind Dima.
Second line contains n distinct integers p1, p2, ..., pn (1 β€ pi β€ n) β positions that Sasha puts coins in circulation to. At first Sasha replaces coin located at position p1, then coin located at position p2 and so on. Coins are numbered from left to right.
Output
Print n + 1 numbers a0, a1, ..., an, where a0 is a hardness of ordering at the beginning, a1 is a hardness of ordering after the first replacement and so on.
Examples
Input
4
1 3 4 2
Output
1 2 3 2 1
Input
8
6 8 3 4 7 2 1 5
Output
1 2 2 3 4 3 4 5 1
Note
Let's denote as O coin out of circulation, and as X β coin is circulation.
At the first sample, initially in row there are coins that are not in circulation, so Dima will look through them from left to right and won't make any exchanges.
After replacement of the first coin with a coin in circulation, Dima will exchange this coin with next three times and after that he will finally look through the coins and finish the process.
XOOO β OOOX
After replacement of the third coin, Dima's actions look this way:
XOXO β OXOX β OOXX
After replacement of the fourth coin, Dima's actions look this way:
XOXX β OXXX
Finally, after replacement of the second coin, row becomes consisting of coins that are in circulation and Dima will look through coins from left to right without any exchanges.
Submitted Solution:
```
n=int(input())
l=list(map(int,input().split()))
p=n
iss=set()
ans=[1]
out=1
for i in range(n) :
if l[i]==p :
p-=1
while p in iss :
p-=1
out-=1
ans.append(out)
else :
iss.add(l[i])
out+=1
ans.append(out)
print(*ans)
``` | instruction | 0 | 13,044 | 10 | 26,088 |
Yes | output | 1 | 13,044 | 10 | 26,089 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, Dima met with Sasha in a philatelic store, and since then they are collecting coins together. Their favorite occupation is to sort collections of coins. Sasha likes having things in order, that is why he wants his coins to be arranged in a row in such a way that firstly come coins out of circulation, and then come coins still in circulation.
For arranging coins Dima uses the following algorithm. One step of his algorithm looks like the following:
1. He looks through all the coins from left to right;
2. If he sees that the i-th coin is still in circulation, and (i + 1)-th coin is already out of circulation, he exchanges these two coins and continues watching coins from (i + 1)-th.
Dima repeats the procedure above until it happens that no two coins were exchanged during this procedure. Dima calls hardness of ordering the number of steps required for him according to the algorithm above to sort the sequence, e.g. the number of times he looks through the coins from the very beginning. For example, for the ordered sequence hardness of ordering equals one.
Today Sasha invited Dima and proposed him a game. First he puts n coins in a row, all of them are out of circulation. Then Sasha chooses one of the coins out of circulation and replaces it with a coin in circulation for n times. During this process Sasha constantly asks Dima what is the hardness of ordering of the sequence.
The task is more complicated because Dima should not touch the coins and he should determine hardness of ordering in his mind. Help Dima with this task.
Input
The first line contains single integer n (1 β€ n β€ 300 000) β number of coins that Sasha puts behind Dima.
Second line contains n distinct integers p1, p2, ..., pn (1 β€ pi β€ n) β positions that Sasha puts coins in circulation to. At first Sasha replaces coin located at position p1, then coin located at position p2 and so on. Coins are numbered from left to right.
Output
Print n + 1 numbers a0, a1, ..., an, where a0 is a hardness of ordering at the beginning, a1 is a hardness of ordering after the first replacement and so on.
Examples
Input
4
1 3 4 2
Output
1 2 3 2 1
Input
8
6 8 3 4 7 2 1 5
Output
1 2 2 3 4 3 4 5 1
Note
Let's denote as O coin out of circulation, and as X β coin is circulation.
At the first sample, initially in row there are coins that are not in circulation, so Dima will look through them from left to right and won't make any exchanges.
After replacement of the first coin with a coin in circulation, Dima will exchange this coin with next three times and after that he will finally look through the coins and finish the process.
XOOO β OOOX
After replacement of the third coin, Dima's actions look this way:
XOXO β OXOX β OOXX
After replacement of the fourth coin, Dima's actions look this way:
XOXX β OXXX
Finally, after replacement of the second coin, row becomes consisting of coins that are in circulation and Dima will look through coins from left to right without any exchanges.
Submitted Solution:
```
n=int(input())
a=[0]*n
p=list(map(int,input().split()))
tmp=n-1
ans=['1']
curr=1
for i in range(n):
a[p[i]-1]=1
if p[i]-1==tmp:
tmp-=1
while tmp>=0 and a[tmp]==1:
tmp-=1
curr-=1
else:
curr+=1
ans.append(str(curr))
print(' '.join(ans))
``` | instruction | 0 | 13,045 | 10 | 26,090 |
Yes | output | 1 | 13,045 | 10 | 26,091 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, Dima met with Sasha in a philatelic store, and since then they are collecting coins together. Their favorite occupation is to sort collections of coins. Sasha likes having things in order, that is why he wants his coins to be arranged in a row in such a way that firstly come coins out of circulation, and then come coins still in circulation.
For arranging coins Dima uses the following algorithm. One step of his algorithm looks like the following:
1. He looks through all the coins from left to right;
2. If he sees that the i-th coin is still in circulation, and (i + 1)-th coin is already out of circulation, he exchanges these two coins and continues watching coins from (i + 1)-th.
Dima repeats the procedure above until it happens that no two coins were exchanged during this procedure. Dima calls hardness of ordering the number of steps required for him according to the algorithm above to sort the sequence, e.g. the number of times he looks through the coins from the very beginning. For example, for the ordered sequence hardness of ordering equals one.
Today Sasha invited Dima and proposed him a game. First he puts n coins in a row, all of them are out of circulation. Then Sasha chooses one of the coins out of circulation and replaces it with a coin in circulation for n times. During this process Sasha constantly asks Dima what is the hardness of ordering of the sequence.
The task is more complicated because Dima should not touch the coins and he should determine hardness of ordering in his mind. Help Dima with this task.
Input
The first line contains single integer n (1 β€ n β€ 300 000) β number of coins that Sasha puts behind Dima.
Second line contains n distinct integers p1, p2, ..., pn (1 β€ pi β€ n) β positions that Sasha puts coins in circulation to. At first Sasha replaces coin located at position p1, then coin located at position p2 and so on. Coins are numbered from left to right.
Output
Print n + 1 numbers a0, a1, ..., an, where a0 is a hardness of ordering at the beginning, a1 is a hardness of ordering after the first replacement and so on.
Examples
Input
4
1 3 4 2
Output
1 2 3 2 1
Input
8
6 8 3 4 7 2 1 5
Output
1 2 2 3 4 3 4 5 1
Note
Let's denote as O coin out of circulation, and as X β coin is circulation.
At the first sample, initially in row there are coins that are not in circulation, so Dima will look through them from left to right and won't make any exchanges.
After replacement of the first coin with a coin in circulation, Dima will exchange this coin with next three times and after that he will finally look through the coins and finish the process.
XOOO β OOOX
After replacement of the third coin, Dima's actions look this way:
XOXO β OXOX β OOXX
After replacement of the fourth coin, Dima's actions look this way:
XOXX β OXXX
Finally, after replacement of the second coin, row becomes consisting of coins that are in circulation and Dima will look through coins from left to right without any exchanges.
Submitted Solution:
```
n = int(input())
a = [int(i) for i in input().split()]
last = n
count = 1
state = [1 for i in range(n+1)]
b=['1']
for i in a:
if i<=n:
state[i]=0
count+=1
if i==n:
while state[n]==0:
count-=1
n-=1
b.append(str(count))
print(" ".join(b))
``` | instruction | 0 | 13,046 | 10 | 26,092 |
Yes | output | 1 | 13,046 | 10 | 26,093 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, Dima met with Sasha in a philatelic store, and since then they are collecting coins together. Their favorite occupation is to sort collections of coins. Sasha likes having things in order, that is why he wants his coins to be arranged in a row in such a way that firstly come coins out of circulation, and then come coins still in circulation.
For arranging coins Dima uses the following algorithm. One step of his algorithm looks like the following:
1. He looks through all the coins from left to right;
2. If he sees that the i-th coin is still in circulation, and (i + 1)-th coin is already out of circulation, he exchanges these two coins and continues watching coins from (i + 1)-th.
Dima repeats the procedure above until it happens that no two coins were exchanged during this procedure. Dima calls hardness of ordering the number of steps required for him according to the algorithm above to sort the sequence, e.g. the number of times he looks through the coins from the very beginning. For example, for the ordered sequence hardness of ordering equals one.
Today Sasha invited Dima and proposed him a game. First he puts n coins in a row, all of them are out of circulation. Then Sasha chooses one of the coins out of circulation and replaces it with a coin in circulation for n times. During this process Sasha constantly asks Dima what is the hardness of ordering of the sequence.
The task is more complicated because Dima should not touch the coins and he should determine hardness of ordering in his mind. Help Dima with this task.
Input
The first line contains single integer n (1 β€ n β€ 300 000) β number of coins that Sasha puts behind Dima.
Second line contains n distinct integers p1, p2, ..., pn (1 β€ pi β€ n) β positions that Sasha puts coins in circulation to. At first Sasha replaces coin located at position p1, then coin located at position p2 and so on. Coins are numbered from left to right.
Output
Print n + 1 numbers a0, a1, ..., an, where a0 is a hardness of ordering at the beginning, a1 is a hardness of ordering after the first replacement and so on.
Examples
Input
4
1 3 4 2
Output
1 2 3 2 1
Input
8
6 8 3 4 7 2 1 5
Output
1 2 2 3 4 3 4 5 1
Note
Let's denote as O coin out of circulation, and as X β coin is circulation.
At the first sample, initially in row there are coins that are not in circulation, so Dima will look through them from left to right and won't make any exchanges.
After replacement of the first coin with a coin in circulation, Dima will exchange this coin with next three times and after that he will finally look through the coins and finish the process.
XOOO β OOOX
After replacement of the third coin, Dima's actions look this way:
XOXO β OXOX β OOXX
After replacement of the fourth coin, Dima's actions look this way:
XOXX β OXXX
Finally, after replacement of the second coin, row becomes consisting of coins that are in circulation and Dima will look through coins from left to right without any exchanges.
Submitted Solution:
```
n = int(input())
input_ = list(map(int, input().split()))
pos = n
a = [0 for i in range(n+1)]
res = 1
ans = [1]
print(1, end=" ")
for x in input_:
a[x] = 1
res += 1
while a[pos]==1:
pos -= 1
res -= 1
ans.append(res)
print(' '.join(map(str, ans)))
``` | instruction | 0 | 13,047 | 10 | 26,094 |
No | output | 1 | 13,047 | 10 | 26,095 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, Dima met with Sasha in a philatelic store, and since then they are collecting coins together. Their favorite occupation is to sort collections of coins. Sasha likes having things in order, that is why he wants his coins to be arranged in a row in such a way that firstly come coins out of circulation, and then come coins still in circulation.
For arranging coins Dima uses the following algorithm. One step of his algorithm looks like the following:
1. He looks through all the coins from left to right;
2. If he sees that the i-th coin is still in circulation, and (i + 1)-th coin is already out of circulation, he exchanges these two coins and continues watching coins from (i + 1)-th.
Dima repeats the procedure above until it happens that no two coins were exchanged during this procedure. Dima calls hardness of ordering the number of steps required for him according to the algorithm above to sort the sequence, e.g. the number of times he looks through the coins from the very beginning. For example, for the ordered sequence hardness of ordering equals one.
Today Sasha invited Dima and proposed him a game. First he puts n coins in a row, all of them are out of circulation. Then Sasha chooses one of the coins out of circulation and replaces it with a coin in circulation for n times. During this process Sasha constantly asks Dima what is the hardness of ordering of the sequence.
The task is more complicated because Dima should not touch the coins and he should determine hardness of ordering in his mind. Help Dima with this task.
Input
The first line contains single integer n (1 β€ n β€ 300 000) β number of coins that Sasha puts behind Dima.
Second line contains n distinct integers p1, p2, ..., pn (1 β€ pi β€ n) β positions that Sasha puts coins in circulation to. At first Sasha replaces coin located at position p1, then coin located at position p2 and so on. Coins are numbered from left to right.
Output
Print n + 1 numbers a0, a1, ..., an, where a0 is a hardness of ordering at the beginning, a1 is a hardness of ordering after the first replacement and so on.
Examples
Input
4
1 3 4 2
Output
1 2 3 2 1
Input
8
6 8 3 4 7 2 1 5
Output
1 2 2 3 4 3 4 5 1
Note
Let's denote as O coin out of circulation, and as X β coin is circulation.
At the first sample, initially in row there are coins that are not in circulation, so Dima will look through them from left to right and won't make any exchanges.
After replacement of the first coin with a coin in circulation, Dima will exchange this coin with next three times and after that he will finally look through the coins and finish the process.
XOOO β OOOX
After replacement of the third coin, Dima's actions look this way:
XOXO β OXOX β OOXX
After replacement of the fourth coin, Dima's actions look this way:
XOXX β OXXX
Finally, after replacement of the second coin, row becomes consisting of coins that are in circulation and Dima will look through coins from left to right without any exchanges.
Submitted Solution:
```
num = int(input())
ints = input()
ints = ints.split(' ')
ints = list(map(lambda x: int(x), ints))
set_of_coins = set()
list_resuts = []
def result(iterable, i, num, set):
set.add(iterable[i - 1])
score = 0
for j in range(num, num - i, -1):
if j not in set_of_coins:
score = j - (num - i) + 1
break
list_resuts.append(score)
list_resuts.append(1)
for i in range(1, num + 1):
result(ints, i, num, set_of_coins)
string = ''
for i in list_resuts:
string += str(i) + ' '
print(string)
``` | instruction | 0 | 13,048 | 10 | 26,096 |
No | output | 1 | 13,048 | 10 | 26,097 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, Dima met with Sasha in a philatelic store, and since then they are collecting coins together. Their favorite occupation is to sort collections of coins. Sasha likes having things in order, that is why he wants his coins to be arranged in a row in such a way that firstly come coins out of circulation, and then come coins still in circulation.
For arranging coins Dima uses the following algorithm. One step of his algorithm looks like the following:
1. He looks through all the coins from left to right;
2. If he sees that the i-th coin is still in circulation, and (i + 1)-th coin is already out of circulation, he exchanges these two coins and continues watching coins from (i + 1)-th.
Dima repeats the procedure above until it happens that no two coins were exchanged during this procedure. Dima calls hardness of ordering the number of steps required for him according to the algorithm above to sort the sequence, e.g. the number of times he looks through the coins from the very beginning. For example, for the ordered sequence hardness of ordering equals one.
Today Sasha invited Dima and proposed him a game. First he puts n coins in a row, all of them are out of circulation. Then Sasha chooses one of the coins out of circulation and replaces it with a coin in circulation for n times. During this process Sasha constantly asks Dima what is the hardness of ordering of the sequence.
The task is more complicated because Dima should not touch the coins and he should determine hardness of ordering in his mind. Help Dima with this task.
Input
The first line contains single integer n (1 β€ n β€ 300 000) β number of coins that Sasha puts behind Dima.
Second line contains n distinct integers p1, p2, ..., pn (1 β€ pi β€ n) β positions that Sasha puts coins in circulation to. At first Sasha replaces coin located at position p1, then coin located at position p2 and so on. Coins are numbered from left to right.
Output
Print n + 1 numbers a0, a1, ..., an, where a0 is a hardness of ordering at the beginning, a1 is a hardness of ordering after the first replacement and so on.
Examples
Input
4
1 3 4 2
Output
1 2 3 2 1
Input
8
6 8 3 4 7 2 1 5
Output
1 2 2 3 4 3 4 5 1
Note
Let's denote as O coin out of circulation, and as X β coin is circulation.
At the first sample, initially in row there are coins that are not in circulation, so Dima will look through them from left to right and won't make any exchanges.
After replacement of the first coin with a coin in circulation, Dima will exchange this coin with next three times and after that he will finally look through the coins and finish the process.
XOOO β OOOX
After replacement of the third coin, Dima's actions look this way:
XOXO β OXOX β OOXX
After replacement of the fourth coin, Dima's actions look this way:
XOXX β OXXX
Finally, after replacement of the second coin, row becomes consisting of coins that are in circulation and Dima will look through coins from left to right without any exchanges.
Submitted Solution:
```
n = int(input())
numbers=[int(i)-1 for i in input().split(' ')]
data= [0 for i in range(n)]
print(1,end=" ")
gloabal_count = 0
for i in numbers:
flag = False
data[i] = 1
count = 0
breakPoint = -1
for d in range(len(data)-1,0, -1):
count +=data[d]
if count != len(data) - d:
breakPoint = d
gloabal_count+=len(data) - breakPoint
data = data[0: breakPoint+1]
break
if breakPoint != -1:
print(i - gloabal_count, end=" ")
else:
print(1)
print()
``` | instruction | 0 | 13,049 | 10 | 26,098 |
No | output | 1 | 13,049 | 10 | 26,099 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, Dima met with Sasha in a philatelic store, and since then they are collecting coins together. Their favorite occupation is to sort collections of coins. Sasha likes having things in order, that is why he wants his coins to be arranged in a row in such a way that firstly come coins out of circulation, and then come coins still in circulation.
For arranging coins Dima uses the following algorithm. One step of his algorithm looks like the following:
1. He looks through all the coins from left to right;
2. If he sees that the i-th coin is still in circulation, and (i + 1)-th coin is already out of circulation, he exchanges these two coins and continues watching coins from (i + 1)-th.
Dima repeats the procedure above until it happens that no two coins were exchanged during this procedure. Dima calls hardness of ordering the number of steps required for him according to the algorithm above to sort the sequence, e.g. the number of times he looks through the coins from the very beginning. For example, for the ordered sequence hardness of ordering equals one.
Today Sasha invited Dima and proposed him a game. First he puts n coins in a row, all of them are out of circulation. Then Sasha chooses one of the coins out of circulation and replaces it with a coin in circulation for n times. During this process Sasha constantly asks Dima what is the hardness of ordering of the sequence.
The task is more complicated because Dima should not touch the coins and he should determine hardness of ordering in his mind. Help Dima with this task.
Input
The first line contains single integer n (1 β€ n β€ 300 000) β number of coins that Sasha puts behind Dima.
Second line contains n distinct integers p1, p2, ..., pn (1 β€ pi β€ n) β positions that Sasha puts coins in circulation to. At first Sasha replaces coin located at position p1, then coin located at position p2 and so on. Coins are numbered from left to right.
Output
Print n + 1 numbers a0, a1, ..., an, where a0 is a hardness of ordering at the beginning, a1 is a hardness of ordering after the first replacement and so on.
Examples
Input
4
1 3 4 2
Output
1 2 3 2 1
Input
8
6 8 3 4 7 2 1 5
Output
1 2 2 3 4 3 4 5 1
Note
Let's denote as O coin out of circulation, and as X β coin is circulation.
At the first sample, initially in row there are coins that are not in circulation, so Dima will look through them from left to right and won't make any exchanges.
After replacement of the first coin with a coin in circulation, Dima will exchange this coin with next three times and after that he will finally look through the coins and finish the process.
XOOO β OOOX
After replacement of the third coin, Dima's actions look this way:
XOXO β OXOX β OOXX
After replacement of the fourth coin, Dima's actions look this way:
XOXX β OXXX
Finally, after replacement of the second coin, row becomes consisting of coins that are in circulation and Dima will look through coins from left to right without any exchanges.
Submitted Solution:
```
n=int(input())
p=list(map(int,input().split()))
s=[1]
b=[0 for i in range(n)]
for i in range(n):
b[n-p[i]]=1
step=0
for j in range(i+1):
if b[j]==0:
step=step+1
elif b[j]==1 and step!=0:
step=step+1
s.append(step+1)
print(s)
``` | instruction | 0 | 13,050 | 10 | 26,100 |
No | output | 1 | 13,050 | 10 | 26,101 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties β n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study β you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 β€ n, m β€ 3000) β the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 β€ p_i β€ m, 1 β€ c_i β€ 10^9) β the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number β the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party. | instruction | 0 | 13,307 | 10 | 26,614 |
Tags: brute force, greedy
Correct Solution:
```
n, m = map(int, input().split())
parties = [0 for _ in range(m)]
pairs = []
for i in range(n):
p, c = map(int, input().split())
parties[p - 1] += 1
pairs.append((c, p - 1))
pairs.sort()
min_sum = 10 ** 31
for i in range(n, parties[0] - 1, -1):
cur_sum = 0
cur_amount = parties[0]
colours = [0 for _ in range(n)]
need_to_buy = [max(parties[j] + 1 - i, 0) for j in range(m)]
need_to_buy[0] = 0
tmp = sum(need_to_buy)
if tmp + parties[0] > i:
print(min_sum)
exit()
for j in range(n):
if need_to_buy[pairs[j][1]] > 0:
need_to_buy[pairs[j][1]] -= 1
colours[j] = 1
cur_amount += 1
cur_sum += pairs[j][0]
j = 0
while cur_amount < i:
if colours[j] == 0 and pairs[j][1] != 0:
cur_amount += 1
cur_sum += pairs[j][0]
j += 1
min_sum = min(min_sum, cur_sum)
print(min_sum)
``` | output | 1 | 13,307 | 10 | 26,615 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties β n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study β you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 β€ n, m β€ 3000) β the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 β€ p_i β€ m, 1 β€ c_i β€ 10^9) β the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number β the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party. | instruction | 0 | 13,308 | 10 | 26,616 |
Tags: brute force, greedy
Correct Solution:
```
#!/usr/bin/python3
def solve(N, M, A):
pv = [0] * M
pm = [[] for _ in range(M)]
allm = {}
for (p, c) in A:
p -= 1
pv[p] += 1
if p == 0:
continue
pm[p].append(c)
if c not in allm:
allm[c] = 0
allm[c] += 1
for m in pm:
m.sort()
allml = list(allm)
allml.sort()
maxv = max(pv)
best = 10 ** 9 * 3000 + 1
for iv in range(1, min(maxv + 2, N + 1)):
allmrest = dict(allm)
cost = 0
reqd = 0
if pv[0] < iv:
reqd = iv - pv[0]
got = 0
for i in range(1, M):
if pv[i] >= iv:
ad = pv[i] - iv + 1
for j in range(ad):
cost += pm[i][j]
allmrest[pm[i][j]] -= 1
got += ad
if reqd > got:
for m in allml:
c = allmrest[m]
toget = min(reqd - got, c)
cost += toget * m
got += toget
if reqd == got:
break
if reqd > got:
continue
best = min(best, cost)
return best
def main():
N, M = [int(e) for e in input().split(' ')]
A = [[int(e) for e in input().split(' ')] for _ in range(N)]
print(solve(N, M, A))
if __name__ == '__main__':
main()
``` | output | 1 | 13,308 | 10 | 26,617 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties β n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study β you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 β€ n, m β€ 3000) β the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 β€ p_i β€ m, 1 β€ c_i β€ 10^9) β the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number β the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party. | instruction | 0 | 13,309 | 10 | 26,618 |
Tags: brute force, greedy
Correct Solution:
```
import sys
n,m=map(int,input().split())
V=[None]*n
for i in range(n):
V[i]=tuple(map(int,input().split()))
V.sort(key=lambda x:x[1])
if n==1:
if V[0][0]==1:
print(0)
else:
print(V[0][1])
sys.exit()
Voterlist=[[] for i in range(m+1)]
Partylist=[0 for i in range(m+1)]
Vnot1=[]
for i in range(n):
Voterlist[V[i][0]]+=[V[i][1]]
Partylist[V[i][0]]+=1
if V[i][0]!=1:
Vnot1.append(V[i])
Vnot1.sort(key=lambda x:x[1])
maxvote=max(Partylist)
Votenumberlist=[[] for i in range(maxvote+1)]
for i in range(1,m+1):
Votenumberlist[Partylist[i]].append(i)
if Votenumberlist[maxvote]==[1]:
print(0)
sys.exit()
#print(Votenumberlist,maxvote,Votenumberlist[maxvote])
ANS=0
for i in range(maxvote+1-Partylist[1]):
ANS+=Vnot1[i][1]
maxvoteminus=1
while True:
checkparty=[0 for i in range(m+1)]
money=0
for i in range(maxvoteminus):
for j in Votenumberlist[maxvote-i]:
checkparty[j]+=maxvoteminus-i
#print(checkparty)
for i in range(2,m+1):
money+=sum(Voterlist[i][:checkparty[i]])
if money>ANS:
break
neednumber=maxvote+1-maxvoteminus-Partylist[1]
for i in range(maxvoteminus):
neednumber-=len(Votenumberlist[maxvote-i])*(maxvoteminus-i)
#print(checkparty,neednumber,money)
for pm in Vnot1:
if checkparty[pm[0]]!=0:
checkparty[pm[0]]-=1
else:
money+=pm[1]
neednumber-=1
if neednumber==0:
break
#print(money,ANS)
if money<ANS:
ANS=money
maxvoteminus+=1
print(ANS)
``` | output | 1 | 13,309 | 10 | 26,619 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties β n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study β you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 β€ n, m β€ 3000) β the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 β€ p_i β€ m, 1 β€ c_i β€ 10^9) β the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number β the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party. | instruction | 0 | 13,310 | 10 | 26,620 |
Tags: brute force, greedy
Correct Solution:
```
import sys
#sys.stdin=open("data.txt")
input=sys.stdin.readline
n,m=map(int,input().split())
party=[[] for _ in range(m+5)]
pc=sorted([list(map(int,input().split())) for _ in range(n)],key=lambda x:x[1])
choose=[0]*n
for i in range(n):
party[pc[i][0]].append(i)
want=10**18
for i in range(1,n+1):
p1=len(party[1])
# want all other parties to have <i voters
for j in range(2,m+5):
if len(party[j])<i: continue
for k in range(len(party[j])-i+1):
p1+=1
choose[party[j][k]]=1
# want party 1 to have >=i voters
want2=0
for j in range(n):
if p1<i and choose[j]==0 and pc[j][0]!=1:
choose[j]=1
p1+=1
if choose[j]==1:
want2+=pc[j][1]
if want>want2:
want=want2
#print(i,want2)
# reset
choose=[0]*n
print(want)
``` | output | 1 | 13,310 | 10 | 26,621 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties β n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study β you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 β€ n, m β€ 3000) β the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 β€ p_i β€ m, 1 β€ c_i β€ 10^9) β the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number β the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party. | instruction | 0 | 13,311 | 10 | 26,622 |
Tags: brute force, greedy
Correct Solution:
```
import sys
import os
def solve(m, candidates):
n = len(candidates)
candidates.sort(key=lambda x: x[1])
party = dict()
granted = 0
for i in range(len(candidates)):
p = candidates[i][0]
c = candidates[i][1]
if p == 1:
granted += 1
continue
if p in party:
party[p].append((i, c))
else:
party[p] = [(i, c)]
result = None
for t in range(1, n // 2 + 2):
total = 0
chosen = set()
for k, v in party.items():
if len(v) >= t:
for i in range(len(v) + 1 - t):
chosen.add(v[i][0])
total += v[i][1]
for i in range(n):
if len(chosen) + granted >= t:
break
if i not in chosen and candidates[i][0] != 1:
chosen.add(i)
total += candidates[i][1]
if result is None:
result = total
else:
result = min(result, total)
return result
def main():
n, m = map(int, input().split())
candidates = []
for i in range(n):
p, c = map(int, input().split())
candidates.append((p, c))
print(solve(m, candidates))
if __name__ == '__main__':
main()
``` | output | 1 | 13,311 | 10 | 26,623 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties β n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study β you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 β€ n, m β€ 3000) β the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 β€ p_i β€ m, 1 β€ c_i β€ 10^9) β the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number β the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party. | instruction | 0 | 13,312 | 10 | 26,624 |
Tags: brute force, greedy
Correct Solution:
```
class Solver:
def solve(self):
self.num_voters, self.num_parties = (int(x) for x in input().split())
self.votes_per_party = [[] for _ in range(self.num_parties)]
for _ in range(self.num_voters):
party, price = (int(x) for x in input().split())
party -= 1
self.votes_per_party[party].append(price)
for party in range(self.num_parties):
self.votes_per_party[party].sort()
max_necessary_votes = self.num_voters//2 + 1
cost = lambda min_votes : self.conversion_price_with_fixed_votes(min_votes)
return self.ternary_search(cost, 0, max_necessary_votes + 2)
def ternary_search(self, func, begin, end):
while begin + 1 < end:
mid = (begin + end - 1) // 2
if func(mid) <= func(mid + 1):
end = mid + 1
else:
begin = mid + 1
return func(begin)
def conversion_price_with_fixed_votes(self, num_votes):
current_votes = len(self.votes_per_party[0])
total_cost = 0
for votes in self.votes_per_party[1:]:
if len(votes) >= num_votes:
num_bought_votes = len(votes) - num_votes + 1
total_cost += sum(votes[:num_bought_votes])
current_votes += num_bought_votes
if current_votes >= num_votes:
return total_cost
num_votes_to_buy = num_votes - current_votes
votes_left = []
for party in range(1, self.num_parties):
votes_left += self.votes_per_party[party][-(num_votes-1):]
votes_left.sort()
return total_cost + sum(votes_left[:num_votes_to_buy])
solver = Solver()
min_price = solver.solve()
print(min_price)
``` | output | 1 | 13,312 | 10 | 26,625 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties β n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study β you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 β€ n, m β€ 3000) β the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 β€ p_i β€ m, 1 β€ c_i β€ 10^9) β the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number β the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party. | instruction | 0 | 13,313 | 10 | 26,626 |
Tags: brute force, greedy
Correct Solution:
```
from sys import stdin
from collections import deque
n,m = map(int, stdin.readline().split())
pc=dict()
ph=dict()
costs=[]
class Cost:
def __init__(self, id, cost, party) -> None:
self.id=id
self.cost=cost
self.party=party
self.removed=False
for i in range(n):
p, c = map(int,stdin.readline().split())
if p != 1:
cost = Cost(i, c, p)
costs.append(cost)
pc.setdefault(p,[]).append(cost)
ph[p] = ph.setdefault(p, 0) + 1
max_h=0
for p, pa in pc.items():
pa.sort(reverse=True, key=lambda x: x.cost)
max_h = max(max_h, len(pa))
hp = [[] for _ in range(max_h+2)]
for p, h in ph.items():
if p != 1:
hp[h].append(p)
ans=[0]*(max_h+2)
costs.sort(key=lambda x:x.cost)
dq = deque(costs)
p_set = set()
height_p1 = ph[1] if 1 in ph else 0
top_sum = 0
top_count = 0
for i in range(max_h+1, -1, -1):
if len(costs) < i:
ans[i] = float('inf')
continue
for p in hp[i]:
p_set.add(p)
for p in p_set:
if len(pc[p]) > 0:
min_cost = pc[p].pop()
min_cost.removed=True
top_sum += min_cost.cost
top_count += 1
ans[i] += top_sum
if height_p1+top_count < i:
for j in range(i-height_p1-top_count):
while dq[0].removed: dq.popleft()
ans[i] += dq[0].cost
dq.rotate(-1)
dq.rotate(i-height_p1-top_count)
print(min(ans))
``` | output | 1 | 13,313 | 10 | 26,627 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties β n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study β you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 β€ n, m β€ 3000) β the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 β€ p_i β€ m, 1 β€ c_i β€ 10^9) β the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number β the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party. | instruction | 0 | 13,314 | 10 | 26,628 |
Tags: brute force, greedy
Correct Solution:
```
n,m=map(int,input().split())
men=[]
for i in range(n):
x,y=map(int,input().split())
men.append((y,x-1))
def Calc(lim):
cnt=[0]*m
vis=[False]*n
for i in range(n):
cnt[men[i][1]]+=1
cost=0
for i in range(n):
if men[i][1]!=0 and cnt[men[i][1]]>=lim:
cnt[men[i][1]]-=1
cost+=men[i][0]
vis[i]=True
cnt[0]+=1
for i in range(n):
if cnt[0]<lim and vis[i] == False and men[i][1]!=0:
cnt[0]+=1
cost+=men[i][0]
return cost
men.sort()
ans = 10**18
for i in range(n):
ans=min(ans,Calc(i))
print(ans)
``` | output | 1 | 13,314 | 10 | 26,629 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties β n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study β you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 β€ n, m β€ 3000) β the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 β€ p_i β€ m, 1 β€ c_i β€ 10^9) β the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number β the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party.
Submitted Solution:
```
import sys
from collections import defaultdict
n, m = map(int, sys.stdin.readline().rstrip('\n').split(' '))
p = defaultdict(list)
for _ in range(n):
x, y = map(int, sys.stdin.readline().rstrip('\n').split(' '))
p[x].append(y)
for key in p:
p[key] = sorted(p[key])
ans = 10**100
for k in range(1, n + 1):
cur = 0
r = []
for key in p:
if key == 1:
continue
a = p[key]
cnt = max(0, len(a) - (k - 1))
cur += sum(a[:cnt])
r += a[cnt:]
r = sorted(r)
cnt = max(0, len(r) - (n - k))
cur += sum(r[:cnt])
ans = min(ans, cur)
print(ans)
``` | instruction | 0 | 13,315 | 10 | 26,630 |
Yes | output | 1 | 13,315 | 10 | 26,631 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties β n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study β you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 β€ n, m β€ 3000) β the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 β€ p_i β€ m, 1 β€ c_i β€ 10^9) β the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number β the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party.
Submitted Solution:
```
import sys
def pro():
return sys.stdin.readline().strip()
def rop():
return map(int, pro().split())
a, s = rop()
q = [0] * a
w = [0] * s
for i in range(a):
n, p = rop()
q[i] = (p, n-1)
w[n - 1] += 1
q.sort()
t = 1e100
for k in range(1, a + 1):
p = w[::]
r = [True] * a
m = 0
for i in range(a):
po, rty = q[i]
if rty != 0 and p[rty] >= k:
p[0] += 1
p[rty] -= 1
m += po
r[i] = False
for i in range(a):
po, rty = q[i]
if rty != 0 and r[i] and p[0] < k:
p[0] += 1
p[rty] -= 1
m += po
r[i] = False
t = min(t, m)
print(t)
``` | instruction | 0 | 13,316 | 10 | 26,632 |
Yes | output | 1 | 13,316 | 10 | 26,633 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties β n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study β you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 β€ n, m β€ 3000) β the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 β€ p_i β€ m, 1 β€ c_i β€ 10^9) β the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number β the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party.
Submitted Solution:
```
import sys
input=sys.stdin.readline
n,m=map(int,input().split())
men=[]
for i in range(n):
x,y=map(int,input().split())
men.append((y,x-1))
def solve(lim):
cnt=[0]*m
vis=[0]*n
cost=0
for i in range(n):
cnt[men[i][1]]+=1
for i in range(n):
if(men[i][1]!=0 and cnt[men[i][1]]>=lim):
cnt[men[i][1]]-=1
cnt[0]+=1
cost+=men[i][0]
vis[i]=1
for i in range(n):
if cnt[0]<lim and vis[i] == False and men[i][1]!=0:
cnt[0]+=1
cost+=men[i][0]
return cost
men.sort()
ans=10**18
for i in range(n):
ans=min(ans,solve(i))
print(ans)
``` | instruction | 0 | 13,317 | 10 | 26,634 |
Yes | output | 1 | 13,317 | 10 | 26,635 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties β n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study β you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 β€ n, m β€ 3000) β the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 β€ p_i β€ m, 1 β€ c_i β€ 10^9) β the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number β the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party.
Submitted Solution:
```
# -*- coding:utf-8 -*-
"""
created by shuangquan.huang at 1/29/19
Let's iterate over final number of votes for The United Party of Berland.
We can see that all opponents should get less votes than our party,
and our party should get at least our chosen number of votes.
We can sort all voters by their costs, and solve the problem in two passes.
First, if we need to get π₯ votes, we should definitely buy all cheap votes for parties that have at least π₯ votes.
Second, if we don't have π₯ votes yet, we should by the cheapest votes to get π₯ votes.
We can see that this solution is optimal: consider the optimal answer, and see how many votes The United Party got.
We tried such number of votes, and we tried to achieve this number of votes by cheapest way, so we couldn't miss the
optimal answer. This can be implemented in π(π2logπ) or even π(πlogπ).
"""
import collections
import time
import os
import sys
import bisect
import heapq
def solve(N, M, A):
votes = [[] for _ in range(M+1)]
for p, v in A:
votes[p].append(v)
for p in range(1, M+1):
votes[p].sort()
own = len(votes[1])
votes = votes[2:]
votes.sort(reverse=True, key=len)
size = [len(v) for v in votes]
if not size or own > size[0]:
return 0
nvotes = len(votes)
ans = float('inf')
for buy in range((size[0]-own)//2+1, min(N, (N+1) // 2 + 1) + 1):
cost = 0
target = own + buy
done = 0
for p in range(nvotes):
if size[p] >= target:
t = size[p] - target + 1
cost += sum(votes[p][: t] or [0])
done += t
else:
break
if done >= buy:
ans = min(ans, cost)
else:
more = buy - done
q = []
for p in range(nvotes):
t = max(size[p] - target + 1, 0)
q.extend(votes[p][t: t+more])
q.sort()
cost += sum(q[:more])
ans = min(ans, cost)
return ans
N, M = map(int, input().split())
A = []
for i in range(N):
p, v = map(int, input().split())
A.append((p, v))
print(solve(N, M, A))
``` | instruction | 0 | 13,318 | 10 | 26,636 |
Yes | output | 1 | 13,318 | 10 | 26,637 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties β n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study β you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 β€ n, m β€ 3000) β the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 β€ p_i β€ m, 1 β€ c_i β€ 10^9) β the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number β the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party.
Submitted Solution:
```
# https://codeforces.com/contest/1019/problem/A
n, m = map(int, input().split())
d = {}
count = 0
for _ in range(n):
p, c = map(int, input().split())
if p == 1:
count += 1
continue
if p not in d:
d[p] = []
d[p].append(c)
for k in d:
d[k] = sorted(d[k])
Min = float('inf')
for i in range(n-count+1):
cur = count + i
remain = []
c = 0
s = 0
for k in d:
if len(d[k]) >= cur:
for j, x in enumerate(d[k]):
if j <= len(d[k]) - cur:
s += x
c += 1
else:
remain.append(x)
else:
remain.extend(d[k])
if c < cur:
remain = sorted(remain)
for x in remain[:cur-c]:
s += x
Min = min(Min, s)
print(Min)
#5 5
#2 100
#3 200
#4 300
#5 800
#5 900
``` | instruction | 0 | 13,319 | 10 | 26,638 |
No | output | 1 | 13,319 | 10 | 26,639 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties β n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study β you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 β€ n, m β€ 3000) β the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 β€ p_i β€ m, 1 β€ c_i β€ 10^9) β the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number β the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party.
Submitted Solution:
```
import math
n,m = map(int,(input().split()))
arr = []
for _ in range(n):
arr.append(list(map(int,input().split())))
arr.sort(key = lambda x : x[0])
arrC= []
flag = -1
for i in range(n):
if (arr[i][0] != 1):
flag = i
break
if(flag == -1):
print(0)
else:
ans = 0
arrC= arr[flag:]
arrC.sort(key = lambda x:x[1])
count = [0]*m
count[0] = flag
for num in arr[flag:]:
count[num[0]-1] += 1
j = 0
while(count[0] <= max(count) ):
maxV = max(count)
valV = 0
for i in range(j,n-flag):
if count[arrC[i][0] -1] == maxV :
valV = arrC[i][1]
break
if(maxV == count[0]):
if(valV == 0 ):
break
else:
ans += arrC[j][1]
break
if(valV > arrC[j][1] + arrC[j+1][1]):
ans += arrC[j][1]
count[arrC[j][0] -1] -= 1
j += 1
count[0] += 1
else:
ans += valV
j += 1
count[arrC[i][0] -1] -= 1
count[0] += 1
print(ans)
``` | instruction | 0 | 13,320 | 10 | 26,640 |
No | output | 1 | 13,320 | 10 | 26,641 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties β n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study β you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 β€ n, m β€ 3000) β the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 β€ p_i β€ m, 1 β€ c_i β€ 10^9) β the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number β the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party.
Submitted Solution:
```
import math
n,m = map(int,(input().split()))
arr = []
for _ in range(n):
arr.append(list(map(int,input().split())))
arr.sort(key = lambda x : x[0])
arrC= []
flag = -1
for i in range(n):
if (arr[i][0] != 1):
flag = i
break
if(flag == -1):
print(0)
else:
ans = 0
arrC= arr[flag:]
arrC.sort(key = lambda x:x[1])
count = [0]*m
count[0] = flag
for num in arr[flag:]:
count[num[0]-1] += 1
j = 0
while(count[0] < max(count) - 1):
ans += arrC[j][1]
count[arrC[j][0] -1] -= 1
count[0] += 1
j += 1
val = 0
maxV = max(count)
pos = -1
valV = 0
for i in range(1,m):
if maxV == count[i]:
pos = i
val += 1
if(maxV != count[0]):
if(val == 1):
for i in range(j,n):
if arrC[i][0] == pos+1 :
valV = arr[i][1]
break
if(valV < arrC[j][1] + arr[j+1][1]):
ans += valV
else:
ans += arrC[j][1] + arr[j+1][1]
else:
ans += arrC[j][1]+arr[j+1][1]
print(ans)
else:
if(val == 0):
print(0)
else:
print(arrC[j][1])
``` | instruction | 0 | 13,321 | 10 | 26,642 |
No | output | 1 | 13,321 | 10 | 26,643 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties β n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study β you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 β€ n, m β€ 3000) β the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 β€ p_i β€ m, 1 β€ c_i β€ 10^9) β the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number β the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party.
Submitted Solution:
```
n, m = map(int, input().split())
cnt = {}
un = 0
arr = [[] for _ in range(m+1)]
for i in range(n):
p, c = map(int, input().split())
if p != 1:
if p not in cnt:
cnt[p] = 1
else:
cnt[p] += 1
arr[p].append(c)
else:
un += 1
for i in range(m+1):
arr[i].sort(reverse=True)
total = 0
while cnt:
if un > max(cnt.values()):
break
mn1 = 1000000001
mn1_index = -1
for i in range(1, m+1):
if arr[i] and arr[i][-1] < mn1:
mn1 = arr[i][-1]
mn1_index = i
mn2 = 1000000001
mn2_index = -1
for i in range(1, m+1):
if arr[i] and i != mn1_index and arr[i][-1] < mn2:
mn2 = arr[i][-1]
mn2_index = i
if len(arr[mn1_index]) > 1 and arr[mn1_index][-2] < mn2:
mn2 = arr[mn1_index][-2]
mn2_index = mn1_index
mx = 0
mx_index = -1
for x in cnt:
if cnt[x] > mx:
mx = cnt[x]
mx_index = x
# print(mn1,mn1_index,mn2,mx_index,mn2_index,arr[mx_index][-1])
if cnt[mx_index] > un:
if mn1+mn2 < arr[mx_index][-1]:
total += mn1
arr[mn1_index].pop()
cnt[mn1_index] -= 1
else:
total += arr[mx_index][-1]
arr[mx_index].pop()
cnt[mx_index] -= 1
else:
if mn1 < arr[mx_index][-1]:
total += mn1
arr[mn1_index].pop()
cnt[mn1_index] -= 1
else:
total += arr[mx_index][-1]
arr[mx_index].pop()
cnt[mx_index] -= 1
un += 1
print(total)
``` | instruction | 0 | 13,322 | 10 | 26,644 |
No | output | 1 | 13,322 | 10 | 26,645 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs a_i coins.
You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.
However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.
So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 100) β the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 β€ n β€ 100) β the number of goods. The second line of the query contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^7), where a_i is the price of the i-th good.
Output
For each query, print the answer for it β the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
Example
Input
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
Output
3
2
1 | instruction | 0 | 13,395 | 10 | 26,790 |
Tags: math
Correct Solution:
```
import math
test=int(input())
for i in range(test):
n=int(input())
l=list(map(int,input().split()))
suml=sum(l)
print(math.ceil(suml/n))
``` | output | 1 | 13,395 | 10 | 26,791 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs a_i coins.
You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.
However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.
So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 100) β the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 β€ n β€ 100) β the number of goods. The second line of the query contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^7), where a_i is the price of the i-th good.
Output
For each query, print the answer for it β the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
Example
Input
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
Output
3
2
1 | instruction | 0 | 13,396 | 10 | 26,792 |
Tags: math
Correct Solution:
```
t = int(input())
import math
for _ in range(t):
n = int(input())
A = list(map(int,input().split()))
s =sum(A)
print(math.ceil(s/n))
``` | output | 1 | 13,396 | 10 | 26,793 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs a_i coins.
You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.
However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.
So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 100) β the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 β€ n β€ 100) β the number of goods. The second line of the query contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^7), where a_i is the price of the i-th good.
Output
For each query, print the answer for it β the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
Example
Input
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
Output
3
2
1 | instruction | 0 | 13,397 | 10 | 26,794 |
Tags: math
Correct Solution:
```
import math
q=int(input())
for i in range(q):
n=int(input())
l=list(map(int,input().split()))
p=sum(l)
p=math.ceil(p/n)
print(p)
``` | output | 1 | 13,397 | 10 | 26,795 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs a_i coins.
You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.
However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.
So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 100) β the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 β€ n β€ 100) β the number of goods. The second line of the query contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^7), where a_i is the price of the i-th good.
Output
For each query, print the answer for it β the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
Example
Input
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
Output
3
2
1 | instruction | 0 | 13,398 | 10 | 26,796 |
Tags: math
Correct Solution:
```
from collections import Counter
if __name__ == '__main__':
q = int(input())
for t in range(q):
n = int(input())
l = [int(i) for i in input().split(" ")]
s = sum(l)
if s%n==0:
print(s//n)
else:
print(s//n+1)
``` | output | 1 | 13,398 | 10 | 26,797 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs a_i coins.
You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.
However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.
So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 100) β the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 β€ n β€ 100) β the number of goods. The second line of the query contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^7), where a_i is the price of the i-th good.
Output
For each query, print the answer for it β the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
Example
Input
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
Output
3
2
1 | instruction | 0 | 13,399 | 10 | 26,798 |
Tags: math
Correct Solution:
```
q=int(input())
for i in range(q):
n=int(input())
total=sum([int(x) for x in input().split()])
if total%n>0:
print(int(total/n)+1)
else:
print(int(total/n))
``` | output | 1 | 13,399 | 10 | 26,799 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs a_i coins.
You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.
However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.
So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 100) β the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 β€ n β€ 100) β the number of goods. The second line of the query contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^7), where a_i is the price of the i-th good.
Output
For each query, print the answer for it β the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
Example
Input
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
Output
3
2
1 | instruction | 0 | 13,400 | 10 | 26,800 |
Tags: math
Correct Solution:
```
from math import ceil
q = int(input())
for i in range(q):
n = int(input())
x = [int(i) for i in input().split()]
print(ceil(sum(x)/n))
``` | output | 1 | 13,400 | 10 | 26,801 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs a_i coins.
You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.
However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.
So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 100) β the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 β€ n β€ 100) β the number of goods. The second line of the query contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^7), where a_i is the price of the i-th good.
Output
For each query, print the answer for it β the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
Example
Input
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
Output
3
2
1 | instruction | 0 | 13,401 | 10 | 26,802 |
Tags: math
Correct Solution:
```
import math
t=int(input())
for i in range(t):
n=int(input())
sum=0
ar=list(map(int,input().split()))
for j in range(n):
sum=sum+ar[j]
if(sum%n==0):
print(math.floor(sum/n))
else:
print(math.ceil(sum/n))
``` | output | 1 | 13,401 | 10 | 26,803 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs a_i coins.
You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.
However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.
So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 100) β the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 β€ n β€ 100) β the number of goods. The second line of the query contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^7), where a_i is the price of the i-th good.
Output
For each query, print the answer for it β the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
Example
Input
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
Output
3
2
1 | instruction | 0 | 13,402 | 10 | 26,804 |
Tags: math
Correct Solution:
```
for t in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
sum=0
for k in a:
sum+=k
i=0
while(True):
if((sum+i)%n==0):
print((sum+i)//n)
break;
else:
i+=1
``` | output | 1 | 13,402 | 10 | 26,805 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs a_i coins.
You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.
However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.
So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 100) β the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 β€ n β€ 100) β the number of goods. The second line of the query contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^7), where a_i is the price of the i-th good.
Output
For each query, print the answer for it β the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
Example
Input
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
Output
3
2
1
Submitted Solution:
```
import math
t=int(input())
for _ in range(t):
n=int(input())
l=list(map(int,input().split()))
l.sort()
print(math.ceil((sum(l))/n))
``` | instruction | 0 | 13,403 | 10 | 26,806 |
Yes | output | 1 | 13,403 | 10 | 26,807 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs a_i coins.
You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.
However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.
So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 100) β the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 β€ n β€ 100) β the number of goods. The second line of the query contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^7), where a_i is the price of the i-th good.
Output
For each query, print the answer for it β the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
Example
Input
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
Output
3
2
1
Submitted Solution:
```
q = int(input())
for _ in range(q):
n = int(input())
dat = list(map(int, input().split()))
import math
print(math.ceil(sum(dat)/n))
``` | instruction | 0 | 13,404 | 10 | 26,808 |
Yes | output | 1 | 13,404 | 10 | 26,809 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs a_i coins.
You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.
However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.
So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 100) β the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 β€ n β€ 100) β the number of goods. The second line of the query contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^7), where a_i is the price of the i-th good.
Output
For each query, print the answer for it β the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
Example
Input
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
Output
3
2
1
Submitted Solution:
```
import math
for i in range(int(input())):
n = int(input())
print(math.ceil(sum(list(map(int, input().split(' ')))) / n))
``` | instruction | 0 | 13,405 | 10 | 26,810 |
Yes | output | 1 | 13,405 | 10 | 26,811 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs a_i coins.
You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.
However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.
So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 100) β the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 β€ n β€ 100) β the number of goods. The second line of the query contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^7), where a_i is the price of the i-th good.
Output
For each query, print the answer for it β the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
Example
Input
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
Output
3
2
1
Submitted Solution:
```
q = int (input ())
q1 = q;
while q1 > 0:
n = int (input ())
l = map(int, input().split(" "))
sumatotal = sum(l)
rpta = (int) (sumatotal/n);
if (sumatotal % n != 0):
rpta = rpta +1
print (rpta)
q1 = q1 - 1
``` | instruction | 0 | 13,406 | 10 | 26,812 |
Yes | output | 1 | 13,406 | 10 | 26,813 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs a_i coins.
You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.
However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.
So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 100) β the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 β€ n β€ 100) β the number of goods. The second line of the query contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^7), where a_i is the price of the i-th good.
Output
For each query, print the answer for it β the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
Example
Input
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
Output
3
2
1
Submitted Solution:
```
import sys
T = int(sys.stdin.readline())
for i in range(T):
N = int(sys.stdin.readline())
nums = list(map(int, sys.stdin.readline().split()))
print("%.f" %(sum(nums)/N))
``` | instruction | 0 | 13,407 | 10 | 26,814 |
No | output | 1 | 13,407 | 10 | 26,815 |
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