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.
The only difference between easy and hard versions is constraints.
Now elections are held in Berland and you want to win them. More precisely, you want everyone to vote for you.
There are n voters, and two ways to convince each of them to vote for you. The first way to convince the i-th voter is to pay him p_i coins. The second way is to make m_i other voters vote for you, and the i-th voter will vote for free.
Moreover, the process of such voting takes place in several steps. For example, if there are five voters with m_1 = 1, m_2 = 2, m_3 = 2, m_4 = 4, m_5 = 5, then you can buy the vote of the fifth voter, and eventually everyone will vote for you. Set of people voting for you will change as follows: {5} β {1, 5} β {1, 2, 3, 5} β {1, 2, 3, 4, 5}.
Calculate the minimum number of coins you have to spend so that everyone votes for you.
Input
The first line contains one integer t (1 β€ t β€ 5000) β the number of test cases.
The first line of each test case contains one integer n (1 β€ n β€ 5000) β the number of voters.
The next n lines contains the description of voters. i-th line contains two integers m_i and p_i (1 β€ p_i β€ 10^9, 0 β€ m_i < n).
It is guaranteed that the sum of all n over all test cases does not exceed 5000.
Output
For each test case print one integer β the minimum number of coins you have to spend so that everyone votes for you.
Example
Input
3
3
1 5
2 10
2 8
7
0 1
3 1
1 1
6 1
1 1
4 1
4 1
6
2 6
2 3
2 8
2 7
4 4
5 5
Output
8
0
7
Note
In the first test case you have to buy vote of the third voter. Then the set of people voting for you will change as follows: {3} β {1, 3} β {1, 2, 3}.
In the second example you don't need to buy votes. The set of people voting for you will change as follows: {1} β {1, 3, 5} β {1, 2, 3, 5} β {1, 2, 3, 5, 6, 7} β {1, 2, 3, 4, 5, 6, 7}.
In the third test case you have to buy votes of the second and the fifth voters. Then the set of people voting for you will change as follows: {2, 5} β {1, 2, 3, 4, 5} β {1, 2, 3, 4, 5, 6}.
Submitted Solution:
```
import sys
import heapq as hq
readline = sys.stdin.readline
read = sys.stdin.read
ns = lambda: readline().rstrip()
ni = lambda: int(readline().rstrip())
nm = lambda: map(int, readline().split())
nl = lambda: list(map(int, readline().split()))
prn = lambda x: print(*x, sep='\n')
def solve():
n = ni()
vot = [tuple(nm()) for _ in range(n)]
vot.sort(key = lambda x: (-x[0], x[1]))
q = list()
c = 0
cost = 0
for i in range(n):
hq.heappush(q, vot[i][1])
while n - i - 1 + c < vot[i][0]:
cost += hq.heappop(q)
c += 1
print(cost)
return
# solve()
T = ni()
for _ in range(T):
solve()
``` | instruction | 0 | 62,400 | 10 | 124,800 |
Yes | output | 1 | 62,400 | 10 | 124,801 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is constraints.
Now elections are held in Berland and you want to win them. More precisely, you want everyone to vote for you.
There are n voters, and two ways to convince each of them to vote for you. The first way to convince the i-th voter is to pay him p_i coins. The second way is to make m_i other voters vote for you, and the i-th voter will vote for free.
Moreover, the process of such voting takes place in several steps. For example, if there are five voters with m_1 = 1, m_2 = 2, m_3 = 2, m_4 = 4, m_5 = 5, then you can buy the vote of the fifth voter, and eventually everyone will vote for you. Set of people voting for you will change as follows: {5} β {1, 5} β {1, 2, 3, 5} β {1, 2, 3, 4, 5}.
Calculate the minimum number of coins you have to spend so that everyone votes for you.
Input
The first line contains one integer t (1 β€ t β€ 5000) β the number of test cases.
The first line of each test case contains one integer n (1 β€ n β€ 5000) β the number of voters.
The next n lines contains the description of voters. i-th line contains two integers m_i and p_i (1 β€ p_i β€ 10^9, 0 β€ m_i < n).
It is guaranteed that the sum of all n over all test cases does not exceed 5000.
Output
For each test case print one integer β the minimum number of coins you have to spend so that everyone votes for you.
Example
Input
3
3
1 5
2 10
2 8
7
0 1
3 1
1 1
6 1
1 1
4 1
4 1
6
2 6
2 3
2 8
2 7
4 4
5 5
Output
8
0
7
Note
In the first test case you have to buy vote of the third voter. Then the set of people voting for you will change as follows: {3} β {1, 3} β {1, 2, 3}.
In the second example you don't need to buy votes. The set of people voting for you will change as follows: {1} β {1, 3, 5} β {1, 2, 3, 5} β {1, 2, 3, 5, 6, 7} β {1, 2, 3, 4, 5, 6, 7}.
In the third test case you have to buy votes of the second and the fifth voters. Then the set of people voting for you will change as follows: {2, 5} β {1, 2, 3, 4, 5} β {1, 2, 3, 4, 5, 6}.
Submitted Solution:
```
import sys
from heapq import *
#sys.stdin = open('in', 'r')
t = int(input())
for ti in range(t):
n = int(input())
a = []
for i in range(n):
mi, pi = map(int, input().split())
a.append((mi, -pi))
a.sort()
c = 0
h = []
res = 0
for i in reversed(range(n)):
heappush(h, -a[i][1])
while c + i < a[i][0]:
res += heappop(h)
c += 1
print(res)
#sys.stdout.write('YES\n')
#sys.stdout.write(f'{res}\n')
#sys.stdout.write(f'{y1} {x1} {y2} {x2}\n')
``` | instruction | 0 | 62,401 | 10 | 124,802 |
Yes | output | 1 | 62,401 | 10 | 124,803 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is constraints.
Now elections are held in Berland and you want to win them. More precisely, you want everyone to vote for you.
There are n voters, and two ways to convince each of them to vote for you. The first way to convince the i-th voter is to pay him p_i coins. The second way is to make m_i other voters vote for you, and the i-th voter will vote for free.
Moreover, the process of such voting takes place in several steps. For example, if there are five voters with m_1 = 1, m_2 = 2, m_3 = 2, m_4 = 4, m_5 = 5, then you can buy the vote of the fifth voter, and eventually everyone will vote for you. Set of people voting for you will change as follows: {5} β {1, 5} β {1, 2, 3, 5} β {1, 2, 3, 4, 5}.
Calculate the minimum number of coins you have to spend so that everyone votes for you.
Input
The first line contains one integer t (1 β€ t β€ 5000) β the number of test cases.
The first line of each test case contains one integer n (1 β€ n β€ 5000) β the number of voters.
The next n lines contains the description of voters. i-th line contains two integers m_i and p_i (1 β€ p_i β€ 10^9, 0 β€ m_i < n).
It is guaranteed that the sum of all n over all test cases does not exceed 5000.
Output
For each test case print one integer β the minimum number of coins you have to spend so that everyone votes for you.
Example
Input
3
3
1 5
2 10
2 8
7
0 1
3 1
1 1
6 1
1 1
4 1
4 1
6
2 6
2 3
2 8
2 7
4 4
5 5
Output
8
0
7
Note
In the first test case you have to buy vote of the third voter. Then the set of people voting for you will change as follows: {3} β {1, 3} β {1, 2, 3}.
In the second example you don't need to buy votes. The set of people voting for you will change as follows: {1} β {1, 3, 5} β {1, 2, 3, 5} β {1, 2, 3, 5, 6, 7} β {1, 2, 3, 4, 5, 6, 7}.
In the third test case you have to buy votes of the second and the fifth voters. Then the set of people voting for you will change as follows: {2, 5} β {1, 2, 3, 4, 5} β {1, 2, 3, 4, 5, 6}.
Submitted Solution:
```
import sys
def I():
return sys.stdin.readline().rstrip()
class Heap:
def __init__( self ):
self.l = [ -1 ]
self.n = 0
def n( self ):
return self.n
def top( self ):
return self.l[ 1 ]
def ins( self, x ):
self.l.append( x )
n = len( self.l ) - 1
i = n
while i > 1:
j = i // 2
if self.l[ j ] > self.l[ i ]:
self.l[ j ], self.l[ i ] = self.l[ i ], self.l[ j ]
i = j
else:
break
def pop( self ):
r = self.l[ 1 ]
l = self.l.pop()
n = len( self.l ) - 1
if n:
self.l[ 1 ] = l
i = 1
while True:
j = i * 2
k = j + 1
if k < len( self.l ) and self.l[ i ] > max( self.l[ j ], self.l[ k ] ):
if self.l[ j ] == min( self.l[ j ], self.l[ k ] ):
self.l[ i ], self.l[ j ] = self.l[ j ], self.l[ i ]
i = j
else:
self.l[ i ], self.l[ k ] = self.l[ k ], self.l[ i ]
i = k
elif k < len( self.l ) and self.l[ i ] > self.l[ k ]:
self.l[ i ], self.l[ k ] = self.l[ k ], self.l[ i ]
i = k
elif j < len( self.l ) and self.l[ i ] > self.l[ j ]:
self.l[ i ], self.l[ j ] = self.l[ j ], self.l[ i ]
i = j
else:
break
return r
t = int( I() )
for _ in range( t ):
n = int( I() )
voter = [ list( map( int, I().split() ) ) for _ in range( n ) ]
h = Heap()
d = {}
for m, p in voter:
if m not in d:
d[ m ] = []
d[ m ].append( p )
need = {}
c = 0
sk = sorted( d.keys() )
for m in sk:
need[ m ] = max( 0, m - c )
c += len( d[ m ] )
c = 0
ans = 0
for m in sk[::-1]:
for p in d[ m ]:
h.ins( p )
while c < need[ m ]:
c += 1
ans += h.pop()
print( ans )
``` | instruction | 0 | 62,402 | 10 | 124,804 |
Yes | output | 1 | 62,402 | 10 | 124,805 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is constraints.
Now elections are held in Berland and you want to win them. More precisely, you want everyone to vote for you.
There are n voters, and two ways to convince each of them to vote for you. The first way to convince the i-th voter is to pay him p_i coins. The second way is to make m_i other voters vote for you, and the i-th voter will vote for free.
Moreover, the process of such voting takes place in several steps. For example, if there are five voters with m_1 = 1, m_2 = 2, m_3 = 2, m_4 = 4, m_5 = 5, then you can buy the vote of the fifth voter, and eventually everyone will vote for you. Set of people voting for you will change as follows: {5} β {1, 5} β {1, 2, 3, 5} β {1, 2, 3, 4, 5}.
Calculate the minimum number of coins you have to spend so that everyone votes for you.
Input
The first line contains one integer t (1 β€ t β€ 5000) β the number of test cases.
The first line of each test case contains one integer n (1 β€ n β€ 5000) β the number of voters.
The next n lines contains the description of voters. i-th line contains two integers m_i and p_i (1 β€ p_i β€ 10^9, 0 β€ m_i < n).
It is guaranteed that the sum of all n over all test cases does not exceed 5000.
Output
For each test case print one integer β the minimum number of coins you have to spend so that everyone votes for you.
Example
Input
3
3
1 5
2 10
2 8
7
0 1
3 1
1 1
6 1
1 1
4 1
4 1
6
2 6
2 3
2 8
2 7
4 4
5 5
Output
8
0
7
Note
In the first test case you have to buy vote of the third voter. Then the set of people voting for you will change as follows: {3} β {1, 3} β {1, 2, 3}.
In the second example you don't need to buy votes. The set of people voting for you will change as follows: {1} β {1, 3, 5} β {1, 2, 3, 5} β {1, 2, 3, 5, 6, 7} β {1, 2, 3, 4, 5, 6, 7}.
In the third test case you have to buy votes of the second and the fifth voters. Then the set of people voting for you will change as follows: {2, 5} β {1, 2, 3, 4, 5} β {1, 2, 3, 4, 5, 6}.
Submitted Solution:
```
from sys import stdin, stdout
import heapq
class MyHeap(object):
def __init__(self, initial=None, key=lambda x:x):
self.key = key
if initial:
self._data = [(key(item), item) for item in initial]
heapq.heapify(self._data)
else:
self._data = []
def push(self, item):
heapq.heappush(self._data, (self.key(item), item))
def pop(self):
return heapq.heappop(self._data)[1]
def print(self):
for hd in self._data:
print(hd)
print('-------------------------------')
def getminnumerofcoins(n, mpa):
res = 0
mpa.sort(key=lambda x: (x[0], -x[1]))
gap = []
cur = 0
for i in range(len(mpa)):
mp = mpa[i]
#print(mp[0])
if mp[0] > cur:
t = [i, mp[0]-cur]
gap.append(t)
#cur = mp[0]
cur += 1
#print(gap)
if len(gap) == 0:
return 0
hp = MyHeap(key=lambda x: x[1])
lidx = gap.pop()[0]
for i in range(lidx, len(mpa)):
ci = [i, mpa[i][1]]
hp.push(ci)
cur = 0
offset = 0
for i in range(len(mpa)):
mp = mpa[i]
need = mp[0] - cur
if need > 0:
for j in range(need):
if len(hp._data) == 0 and len(gap) > 0:
lg = gap.pop()
while len(gap) > 0 and lg[1] - offset <= 0:
lg = gap.pop()
for k in range(lg[0], lidx):
ci = [mpa[i][1], k]
hp.push(ci)
lidx = lg[0]
c = hp.pop()
#print(c)
if c[0] == i:
c = hp.pop()
#print(c)
res += c[1]
cur += 1
offset += 1
cur += 1
return res
if __name__ == '__main__':
t = int(stdin.readline())
for i in range(t):
n = int(stdin.readline())
mpa = []
for j in range(n):
mp = list(map(int, stdin.readline().split()))
mpa.append(mp)
res = getminnumerofcoins(n, mpa)
stdout.write(str(res) + '\n')
``` | instruction | 0 | 62,403 | 10 | 124,806 |
No | output | 1 | 62,403 | 10 | 124,807 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is constraints.
Now elections are held in Berland and you want to win them. More precisely, you want everyone to vote for you.
There are n voters, and two ways to convince each of them to vote for you. The first way to convince the i-th voter is to pay him p_i coins. The second way is to make m_i other voters vote for you, and the i-th voter will vote for free.
Moreover, the process of such voting takes place in several steps. For example, if there are five voters with m_1 = 1, m_2 = 2, m_3 = 2, m_4 = 4, m_5 = 5, then you can buy the vote of the fifth voter, and eventually everyone will vote for you. Set of people voting for you will change as follows: {5} β {1, 5} β {1, 2, 3, 5} β {1, 2, 3, 4, 5}.
Calculate the minimum number of coins you have to spend so that everyone votes for you.
Input
The first line contains one integer t (1 β€ t β€ 5000) β the number of test cases.
The first line of each test case contains one integer n (1 β€ n β€ 5000) β the number of voters.
The next n lines contains the description of voters. i-th line contains two integers m_i and p_i (1 β€ p_i β€ 10^9, 0 β€ m_i < n).
It is guaranteed that the sum of all n over all test cases does not exceed 5000.
Output
For each test case print one integer β the minimum number of coins you have to spend so that everyone votes for you.
Example
Input
3
3
1 5
2 10
2 8
7
0 1
3 1
1 1
6 1
1 1
4 1
4 1
6
2 6
2 3
2 8
2 7
4 4
5 5
Output
8
0
7
Note
In the first test case you have to buy vote of the third voter. Then the set of people voting for you will change as follows: {3} β {1, 3} β {1, 2, 3}.
In the second example you don't need to buy votes. The set of people voting for you will change as follows: {1} β {1, 3, 5} β {1, 2, 3, 5} β {1, 2, 3, 5, 6, 7} β {1, 2, 3, 4, 5, 6, 7}.
In the third test case you have to buy votes of the second and the fifth voters. Then the set of people voting for you will change as follows: {2, 5} β {1, 2, 3, 4, 5} β {1, 2, 3, 4, 5, 6}.
Submitted Solution:
```
t = int(input())
for k in range(t):
n = int(input())
arr = []
for i in range(n):
arr.append([int(c) for c in input().split(' ')])
arr = sorted(arr)
cost = 0
v = 0
while len(arr)!=0:
while(len(arr)>0 and arr[0][0]<=v):
arr.pop(0)
v+=1
if len(arr)!=0:
if v+1==arr[0][0]:
_min=len(arr)-1
for i in range(len(arr)):
if arr[i][0]==v+1:
continue
if arr[_min][1] > arr[i][1]:
_min = i
cost+=arr[_min][1]
arr.pop(i)
v+=1
else:
_min=0
for i in range(len(arr)):
if arr[_min][1] > arr[i][1]:
_min = i
cost+=arr[_min][1]
arr.pop(i)
v+=1
print(cost)
``` | instruction | 0 | 62,404 | 10 | 124,808 |
No | output | 1 | 62,404 | 10 | 124,809 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is constraints.
Now elections are held in Berland and you want to win them. More precisely, you want everyone to vote for you.
There are n voters, and two ways to convince each of them to vote for you. The first way to convince the i-th voter is to pay him p_i coins. The second way is to make m_i other voters vote for you, and the i-th voter will vote for free.
Moreover, the process of such voting takes place in several steps. For example, if there are five voters with m_1 = 1, m_2 = 2, m_3 = 2, m_4 = 4, m_5 = 5, then you can buy the vote of the fifth voter, and eventually everyone will vote for you. Set of people voting for you will change as follows: {5} β {1, 5} β {1, 2, 3, 5} β {1, 2, 3, 4, 5}.
Calculate the minimum number of coins you have to spend so that everyone votes for you.
Input
The first line contains one integer t (1 β€ t β€ 5000) β the number of test cases.
The first line of each test case contains one integer n (1 β€ n β€ 5000) β the number of voters.
The next n lines contains the description of voters. i-th line contains two integers m_i and p_i (1 β€ p_i β€ 10^9, 0 β€ m_i < n).
It is guaranteed that the sum of all n over all test cases does not exceed 5000.
Output
For each test case print one integer β the minimum number of coins you have to spend so that everyone votes for you.
Example
Input
3
3
1 5
2 10
2 8
7
0 1
3 1
1 1
6 1
1 1
4 1
4 1
6
2 6
2 3
2 8
2 7
4 4
5 5
Output
8
0
7
Note
In the first test case you have to buy vote of the third voter. Then the set of people voting for you will change as follows: {3} β {1, 3} β {1, 2, 3}.
In the second example you don't need to buy votes. The set of people voting for you will change as follows: {1} β {1, 3, 5} β {1, 2, 3, 5} β {1, 2, 3, 5, 6, 7} β {1, 2, 3, 4, 5, 6, 7}.
In the third test case you have to buy votes of the second and the fifth voters. Then the set of people voting for you will change as follows: {2, 5} β {1, 2, 3, 4, 5} β {1, 2, 3, 4, 5, 6}.
Submitted Solution:
```
ctr = 0
minsum = 0
t = int(input())
while(t<=ctr):
minsum = 5001
n = int(input())
for i in range(n):
m,p = int(input()).split()
s = (p*n)//(m+1)
if(s<minsum):
minsum = s
print(minsum)
ctr+=1
``` | instruction | 0 | 62,405 | 10 | 124,810 |
No | output | 1 | 62,405 | 10 | 124,811 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is constraints.
Now elections are held in Berland and you want to win them. More precisely, you want everyone to vote for you.
There are n voters, and two ways to convince each of them to vote for you. The first way to convince the i-th voter is to pay him p_i coins. The second way is to make m_i other voters vote for you, and the i-th voter will vote for free.
Moreover, the process of such voting takes place in several steps. For example, if there are five voters with m_1 = 1, m_2 = 2, m_3 = 2, m_4 = 4, m_5 = 5, then you can buy the vote of the fifth voter, and eventually everyone will vote for you. Set of people voting for you will change as follows: {5} β {1, 5} β {1, 2, 3, 5} β {1, 2, 3, 4, 5}.
Calculate the minimum number of coins you have to spend so that everyone votes for you.
Input
The first line contains one integer t (1 β€ t β€ 5000) β the number of test cases.
The first line of each test case contains one integer n (1 β€ n β€ 5000) β the number of voters.
The next n lines contains the description of voters. i-th line contains two integers m_i and p_i (1 β€ p_i β€ 10^9, 0 β€ m_i < n).
It is guaranteed that the sum of all n over all test cases does not exceed 5000.
Output
For each test case print one integer β the minimum number of coins you have to spend so that everyone votes for you.
Example
Input
3
3
1 5
2 10
2 8
7
0 1
3 1
1 1
6 1
1 1
4 1
4 1
6
2 6
2 3
2 8
2 7
4 4
5 5
Output
8
0
7
Note
In the first test case you have to buy vote of the third voter. Then the set of people voting for you will change as follows: {3} β {1, 3} β {1, 2, 3}.
In the second example you don't need to buy votes. The set of people voting for you will change as follows: {1} β {1, 3, 5} β {1, 2, 3, 5} β {1, 2, 3, 5, 6, 7} β {1, 2, 3, 4, 5, 6, 7}.
In the third test case you have to buy votes of the second and the fifth voters. Then the set of people voting for you will change as follows: {2, 5} β {1, 2, 3, 4, 5} β {1, 2, 3, 4, 5, 6}.
Submitted Solution:
```
print("Hello")
``` | instruction | 0 | 62,406 | 10 | 124,812 |
No | output | 1 | 62,406 | 10 | 124,813 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Phoenix wonders what it is like to rob diamonds from a jewelry store!
There are n types of diamonds. The i-th type has weight w_i and value v_i. The store initially has a_i diamonds of the i-th type.
Each day, for q days, one of the following will happen:
1. A new shipment of k_i diamonds of type d_i arrive.
2. The store sells k_i diamonds of type d_i.
3. Phoenix wonders what will happen if he robs the store using a bag that can fit diamonds with total weight not exceeding c_i. If he greedily takes diamonds of the largest value that fit, how much value would be taken? If there are multiple diamonds with the largest value, he will take the one with minimum weight. If, of the diamonds with the largest value, there are multiple with the same minimum weight, he will take any of them.
Of course, since Phoenix is a law-abiding citizen, this is all a thought experiment and he never actually robs any diamonds from the store. This means that queries of type 3 do not affect the diamonds in the store.
Input
The first line contains two integers n and q (1 β€ n β€ 2 β
10^5; 1 β€ q β€ 10^5) β the number of types of diamonds and number of days, respectively.
The next n lines describe each type of diamond. The i-th line will contain three integers a_i, w_i, and v_i (0 β€ a_i β€ 10^5; 1 β€ w_i, v_i β€ 10^5) β the initial number of diamonds of the i-th type, the weight of diamonds of the i-th type, and the value of diamonds of the i-th type, respectively.
The next q lines contain the queries. For each query, the first integer of each line is t (1 β€ t β€ 3) β the type of query.
If t=1, then two integers k_i, d_i follow (1 β€ k_i β€ 10^5; 1 β€ d_i β€ n). This means that a new shipment of k_i diamonds arrived, each of type d_i.
If t=2, then two integers k_i, d_i follow (1 β€ k_i β€ 10^5; 1 β€ d_i β€ n). This means that the store has sold k_i diamonds, each of type d_i. It is guaranteed that the store had the diamonds before they sold them.
If t=3, an integer c_i will follow (1 β€ c_i β€ 10^{18}) β the weight capacity of Phoenix's bag.
It is guaranteed that there is at least one query where t=3.
Output
Print the answer for each query of the third type (t=3).
Example
Input
3 5
2 3 4
1 5 1
0 2 4
3 6
1 3 3
3 10
2 2 3
3 30
Output
8
16
13
Note
For the first query where t=3, Phoenix can fit 2 diamonds of type 1, with total weight 6 and value 8.
For the second query where t=3, Phoenix will first fit in 3 diamonds of type 3, then one diamond of type 1 for a total weight of 9 and a value of 16. Note that diamonds of type 3 are prioritized over type 1 because type 3 has equal value but less weight.
For the final query where t=3, Phoenix can fit every diamond for a total value of 13.
Submitted Solution:
```
#import time
#t1=time.time()
a=input().split()
no_d=int(a[0])
no_day=int(a[1])
dim=[[0 for m in range(3)] for n in range(no_d)]
mat=[[0 for m in range(4)] for n in range(no_d)]
for n in range(no_d):
dim[n]=[int(i) for i in input().split()]
mat[n]=[i for i in dim[n]]
mat[n].append(n)
mat.sort(key=lambda x:x[2],reverse=True)
flag=0
for n in range(no_d-1):
if mat[n][2]==mat[n+1][2] and flag==0:
flag=1
i=n
if flag==1 and mat[n][2]!=mat[n+1][2]:
j=n
flag=0
lst=mat[i:j+1]
lst.sort(key=lambda x:x[1])
mat[i:j+1]=lst
#t2=time.time()
for n in range(no_day):
inp=[int(i) for i in input().split()]
if inp[0]==1:
dim[inp[2]-1][0]+=inp[1]
elif inp[0]==2:
dim[inp[2]-1][0]-=inp[1]
else:
wt=inp[1]
res=0
for m in mat:
if wt==0: break
dn=dim[m[3]][0]
if dn>0 :
if dn<=wt//m[1]:
wt-=dn*m[1]
res+=m[2]*dn
else:
g=4
res+=m[2]*(wt//m[1])
wt-=m[1]*(wt//m[1])
print(res)
``` | instruction | 0 | 62,527 | 10 | 125,054 |
No | output | 1 | 62,527 | 10 | 125,055 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Phoenix wonders what it is like to rob diamonds from a jewelry store!
There are n types of diamonds. The i-th type has weight w_i and value v_i. The store initially has a_i diamonds of the i-th type.
Each day, for q days, one of the following will happen:
1. A new shipment of k_i diamonds of type d_i arrive.
2. The store sells k_i diamonds of type d_i.
3. Phoenix wonders what will happen if he robs the store using a bag that can fit diamonds with total weight not exceeding c_i. If he greedily takes diamonds of the largest value that fit, how much value would be taken? If there are multiple diamonds with the largest value, he will take the one with minimum weight. If, of the diamonds with the largest value, there are multiple with the same minimum weight, he will take any of them.
Of course, since Phoenix is a law-abiding citizen, this is all a thought experiment and he never actually robs any diamonds from the store. This means that queries of type 3 do not affect the diamonds in the store.
Input
The first line contains two integers n and q (1 β€ n β€ 2 β
10^5; 1 β€ q β€ 10^5) β the number of types of diamonds and number of days, respectively.
The next n lines describe each type of diamond. The i-th line will contain three integers a_i, w_i, and v_i (0 β€ a_i β€ 10^5; 1 β€ w_i, v_i β€ 10^5) β the initial number of diamonds of the i-th type, the weight of diamonds of the i-th type, and the value of diamonds of the i-th type, respectively.
The next q lines contain the queries. For each query, the first integer of each line is t (1 β€ t β€ 3) β the type of query.
If t=1, then two integers k_i, d_i follow (1 β€ k_i β€ 10^5; 1 β€ d_i β€ n). This means that a new shipment of k_i diamonds arrived, each of type d_i.
If t=2, then two integers k_i, d_i follow (1 β€ k_i β€ 10^5; 1 β€ d_i β€ n). This means that the store has sold k_i diamonds, each of type d_i. It is guaranteed that the store had the diamonds before they sold them.
If t=3, an integer c_i will follow (1 β€ c_i β€ 10^{18}) β the weight capacity of Phoenix's bag.
It is guaranteed that there is at least one query where t=3.
Output
Print the answer for each query of the third type (t=3).
Example
Input
3 5
2 3 4
1 5 1
0 2 4
3 6
1 3 3
3 10
2 2 3
3 30
Output
8
16
13
Note
For the first query where t=3, Phoenix can fit 2 diamonds of type 1, with total weight 6 and value 8.
For the second query where t=3, Phoenix will first fit in 3 diamonds of type 3, then one diamond of type 1 for a total weight of 9 and a value of 16. Note that diamonds of type 3 are prioritized over type 1 because type 3 has equal value but less weight.
For the final query where t=3, Phoenix can fit every diamond for a total value of 13.
Submitted Solution:
```
#import time
#t1=time.time()
a=input().split()
no_d=int(a[0])
no_day=int(a[1])
dim=[[0 for m in range(3)] for n in range(no_d)]
mat=[[0 for m in range(4)] for n in range(no_d)]
for n in range(no_d):
dim[n]=[int(i) for i in input().split()]
mat[n]=[i for i in dim[n]]
mat[n].append(n)
mat.sort(key=lambda x:x[2],reverse=True)
flag=0
for n in range(no_d-1):
if mat[n][2]==mat[n+1][2] and flag==0:
flag=1
i=n
if flag==1 and mat[n][2]!=mat[n+1][2]:
j=n
flag=0
lst=mat[i:j+1]
lst.sort(key=lambda x:x[1])
mat[i:j+1]=lst
#t2=time.time()
for n in range(no_day):
inp=[int(i) for i in input().split()]
if(no_d==50 and n>295):
print(*inp)
else:
if inp[0]==1:
dim[inp[2]-1][0]+=inp[1]
elif inp[0]==2:
dim[inp[2]-1][0]-=inp[1]
else:
wt=inp[1]
res=0
for m in mat:
if wt==0: break
dn=dim[m[3]][0]
if dn>0 :
if dn<=wt//m[1]:
wt-=dn*m[1]
res+=m[2]*dn
else:
res+=m[2]*(wt//m[1])
wt-=m[1]*(wt//m[1])
if(no_d!=50):
print(res)
"""
for n in range(no_day):
inp=[int(i) for i in input().split()]
if inp[0]==1:
dim[inp[2]-1][0]+=inp[1]
elif inp[0]==2:
dim[inp[2]-1][0]-=inp[1]
else:
wt=inp[1]
res=0
for m in mat:
if wt==0: break
dn=dim[m[3]][0]
if dn>0 :
if dn<=wt//m[1]:
wt-=dn*m[1]
res+=m[2]*dn
else:
res+=m[2]*(wt//m[1])
wt-=m[1]*(wt//m[1])
print(res)"""
``` | instruction | 0 | 62,528 | 10 | 125,056 |
No | output | 1 | 62,528 | 10 | 125,057 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Phoenix wonders what it is like to rob diamonds from a jewelry store!
There are n types of diamonds. The i-th type has weight w_i and value v_i. The store initially has a_i diamonds of the i-th type.
Each day, for q days, one of the following will happen:
1. A new shipment of k_i diamonds of type d_i arrive.
2. The store sells k_i diamonds of type d_i.
3. Phoenix wonders what will happen if he robs the store using a bag that can fit diamonds with total weight not exceeding c_i. If he greedily takes diamonds of the largest value that fit, how much value would be taken? If there are multiple diamonds with the largest value, he will take the one with minimum weight. If, of the diamonds with the largest value, there are multiple with the same minimum weight, he will take any of them.
Of course, since Phoenix is a law-abiding citizen, this is all a thought experiment and he never actually robs any diamonds from the store. This means that queries of type 3 do not affect the diamonds in the store.
Input
The first line contains two integers n and q (1 β€ n β€ 2 β
10^5; 1 β€ q β€ 10^5) β the number of types of diamonds and number of days, respectively.
The next n lines describe each type of diamond. The i-th line will contain three integers a_i, w_i, and v_i (0 β€ a_i β€ 10^5; 1 β€ w_i, v_i β€ 10^5) β the initial number of diamonds of the i-th type, the weight of diamonds of the i-th type, and the value of diamonds of the i-th type, respectively.
The next q lines contain the queries. For each query, the first integer of each line is t (1 β€ t β€ 3) β the type of query.
If t=1, then two integers k_i, d_i follow (1 β€ k_i β€ 10^5; 1 β€ d_i β€ n). This means that a new shipment of k_i diamonds arrived, each of type d_i.
If t=2, then two integers k_i, d_i follow (1 β€ k_i β€ 10^5; 1 β€ d_i β€ n). This means that the store has sold k_i diamonds, each of type d_i. It is guaranteed that the store had the diamonds before they sold them.
If t=3, an integer c_i will follow (1 β€ c_i β€ 10^{18}) β the weight capacity of Phoenix's bag.
It is guaranteed that there is at least one query where t=3.
Output
Print the answer for each query of the third type (t=3).
Example
Input
3 5
2 3 4
1 5 1
0 2 4
3 6
1 3 3
3 10
2 2 3
3 30
Output
8
16
13
Note
For the first query where t=3, Phoenix can fit 2 diamonds of type 1, with total weight 6 and value 8.
For the second query where t=3, Phoenix will first fit in 3 diamonds of type 3, then one diamond of type 1 for a total weight of 9 and a value of 16. Note that diamonds of type 3 are prioritized over type 1 because type 3 has equal value but less weight.
For the final query where t=3, Phoenix can fit every diamond for a total value of 13.
Submitted Solution:
```
a=20
b=10
print(a+b)
``` | instruction | 0 | 62,529 | 10 | 125,058 |
No | output | 1 | 62,529 | 10 | 125,059 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Phoenix wonders what it is like to rob diamonds from a jewelry store!
There are n types of diamonds. The i-th type has weight w_i and value v_i. The store initially has a_i diamonds of the i-th type.
Each day, for q days, one of the following will happen:
1. A new shipment of k_i diamonds of type d_i arrive.
2. The store sells k_i diamonds of type d_i.
3. Phoenix wonders what will happen if he robs the store using a bag that can fit diamonds with total weight not exceeding c_i. If he greedily takes diamonds of the largest value that fit, how much value would be taken? If there are multiple diamonds with the largest value, he will take the one with minimum weight. If, of the diamonds with the largest value, there are multiple with the same minimum weight, he will take any of them.
Of course, since Phoenix is a law-abiding citizen, this is all a thought experiment and he never actually robs any diamonds from the store. This means that queries of type 3 do not affect the diamonds in the store.
Input
The first line contains two integers n and q (1 β€ n β€ 2 β
10^5; 1 β€ q β€ 10^5) β the number of types of diamonds and number of days, respectively.
The next n lines describe each type of diamond. The i-th line will contain three integers a_i, w_i, and v_i (0 β€ a_i β€ 10^5; 1 β€ w_i, v_i β€ 10^5) β the initial number of diamonds of the i-th type, the weight of diamonds of the i-th type, and the value of diamonds of the i-th type, respectively.
The next q lines contain the queries. For each query, the first integer of each line is t (1 β€ t β€ 3) β the type of query.
If t=1, then two integers k_i, d_i follow (1 β€ k_i β€ 10^5; 1 β€ d_i β€ n). This means that a new shipment of k_i diamonds arrived, each of type d_i.
If t=2, then two integers k_i, d_i follow (1 β€ k_i β€ 10^5; 1 β€ d_i β€ n). This means that the store has sold k_i diamonds, each of type d_i. It is guaranteed that the store had the diamonds before they sold them.
If t=3, an integer c_i will follow (1 β€ c_i β€ 10^{18}) β the weight capacity of Phoenix's bag.
It is guaranteed that there is at least one query where t=3.
Output
Print the answer for each query of the third type (t=3).
Example
Input
3 5
2 3 4
1 5 1
0 2 4
3 6
1 3 3
3 10
2 2 3
3 30
Output
8
16
13
Note
For the first query where t=3, Phoenix can fit 2 diamonds of type 1, with total weight 6 and value 8.
For the second query where t=3, Phoenix will first fit in 3 diamonds of type 3, then one diamond of type 1 for a total weight of 9 and a value of 16. Note that diamonds of type 3 are prioritized over type 1 because type 3 has equal value but less weight.
For the final query where t=3, Phoenix can fit every diamond for a total value of 13.
Submitted Solution:
```
#import time
#t1=time.time()
a=input().split()
no_d=int(a[0])
no_day=int(a[1])
dim=[[0 for m in range(3)] for n in range(no_d)]
mat=[[0 for m in range(4)] for n in range(no_d)]
for n in range(no_d):
dim[n]=[int(i) for i in input().split()]
mat[n]=[i for i in dim[n]]
mat[n].append(n)
mat.sort(key=lambda x:x[2],reverse=True)
flag=0
for n in range(no_d-1):
if mat[n][2]==mat[n+1][2] and flag==0:
flag=1
i=n
if flag==1 and mat[n][2]!=mat[n+1][2]:
j=n
flag=0
lst=mat[i:j+1]
lst.sort(key=lambda x:x[1])
mat[i:j+1]=lst
#t2=time.time()
for n in range(no_day):
inp=[int(i) for i in input().split()]
if(no_d==50 and n>104):
print(*inp)
else:
if inp[0]==1:
dim[inp[2]-1][0]+=inp[1]
elif inp[0]==2:
dim[inp[2]-1][0]-=inp[1]
else:
wt=inp[1]
res=0
for m in mat:
if wt==0: break
dn=dim[m[3]][0]
if dn>0 :
if dn<=wt//m[1]:
wt-=dn*m[1]
res+=m[2]*dn
else:
res+=m[2]*(wt//m[1])
wt-=m[1]*(wt//m[1])
if(no_d!=50):
print(res)
"""
for n in range(no_day):
inp=[int(i) for i in input().split()]
if inp[0]==1:
dim[inp[2]-1][0]+=inp[1]
elif inp[0]==2:
dim[inp[2]-1][0]-=inp[1]
else:
wt=inp[1]
res=0
for m in mat:
if wt==0: break
dn=dim[m[3]][0]
if dn>0 :
if dn<=wt//m[1]:
wt-=dn*m[1]
res+=m[2]*dn
else:
res+=m[2]*(wt//m[1])
wt-=m[1]*(wt//m[1])
print(res)"""
``` | instruction | 0 | 62,530 | 10 | 125,060 |
No | output | 1 | 62,530 | 10 | 125,061 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Toshizo is the manager of a convenience store chain in Hakodate. Every day, each of the stores in his chain sends him a table of the products that they have sold. Toshizo's job is to compile these figures and calculate how much the stores have sold in total.
The type of a table that is sent to Toshizo by the stores looks like this (all numbers represent units of products sold):
Store1 Store2 Store3 Totals
Product A -5 40 70 | 105
Product B 20 50 80 | 150
Product C 30 60 90 | 180
-----------------------------------------------------------
Store's total sales 45 150 240 435
By looking at the table, Toshizo can tell at a glance which goods are selling well or selling badly, and which stores are paying well and which stores are too empty. Sometimes, customers will bring products back, so numbers in a table can be negative as well as positive.
Toshizo reports these figures to his boss in Tokyo, and together they sometimes decide to close stores that are not doing well and sometimes decide to open new stores in places that they think will be profitable. So, the total number of stores managed by Toshizo is not fixed. Also, they often decide to discontinue selling some products that are not popular, as well as deciding to stock new items that are likely to be popular. So, the number of products that Toshizo has to monitor is also not fixed.
One New Year, it is very cold in Hakodate. A water pipe bursts in Toshizo's office and floods his desk. When Toshizo comes to work, he finds that the latest sales table is not legible. He can make out some of the figures, but not all of them. He wants to call his boss in Tokyo to tell him that the figures will be late, but he knows that his boss will expect him to reconstruct the table if at all possible.
Waiting until the next day won't help, because it is the New Year, and his shops will be on holiday. So, Toshizo decides either to work out the values for himself, or to be sure that there really is no unique solution. Only then can he call his boss and tell him what has happened.
But Toshizo also wants to be sure that a problem like this never happens again. So he decides to write a computer program that all the managers in his company can use if some of the data goes missing from their sales tables. For instance, if they have a table like:
Store 1 Store 2 Store 3 Totals
Product A ? ? 70 | 105
Product B ? 50 ? | 150
Product C 30 60 90 | 180
--------------------------------------------------------
Store's total sales 45 150 240 435
then Toshizo's program will be able to tell them the correct figures to replace the question marks. In some cases, however, even his program will not be able to replace all the question marks, in general. For instance, if a table like:
Store 1 Store 2 Totals
Product A ? ? | 40
Product B ? ? | 40
---------------------------------------------
Store's total sales 40 40 80
is given, there are infinitely many possible solutions. In this sort of case, his program will just say "NO". Toshizo's program works for any data where the totals row and column are still intact. Can you reproduce Toshizo's program?
Input
The input consists of multiple data sets, each in the following format:
p s
row1
row2
...
rowp
totals
The first line consists of two integers p and s, representing the numbers of products and stores, respectively. The former is less than 100 and the latter is less than 10.
They are separated by a blank character. Each of the subsequent lines represents a row of the table and consists of s+1 elements, each of which is either an integer or a question mark. The i-th line (1 <= i <= p) corresponds to the i-th product and the last line the totals row. The first s elements in a line represent the sales of each store and the last one the total sales of the product. Numbers or question marks in a line are separated by a blank character. There are no missing numbers in the totals column or row. There is at least one question mark in each data set.
The known numbers in the table except the totals row and column is between -1,000 and 1,000, inclusive. However, unknown numbers, represented by question marks, may not be in this range. The total sales number of each product is between -10,000 and 10,000, inclusive. The total sales number of each store is between -100,000 and 100,000, inclusive.
There is a blank line between each data set in the input, and the input is terminated by a line consisting of a zero.
Output
When there is a unique solution, your program should print the missing numbers in the occurrence order in the input data set. Otherwise, your program should print just "NO". The answers of consecutive data sets should be separated by an empty line. Each non-empty output line contains only a single number or "NO".
Example
Input
3 3
? ? 70 105
? 50 ? 150
30 60 90 180
45 150 240 435
2 2
? ? 40
? ? 40
40 40 80
2 3
? 30 40 90
50 60 70 180
70 90 110 270
0
Output
-5
40
20
80
NO
20
Submitted Solution:
```
from collections import namedtuple
from pprint import pprint
Question = namedtuple("Question", ['all_sum', 'row_sums', 'col_sums', 'cells', 'row_count', 'col_count'])
def pp_q(q):
print(q.row_count, q.col_count)
for irow in range(q.row_count):
for icol in range(q.col_count):
print(q.cells[irow, icol], end="\t")
print(q.row_sums[irow])
for icol in range(q.col_count):
print(q.col_sums[icol], end="\t")
print(q.all_sum)
def solve(q):
determined = {}
while True:
good = False
# print(equations(q))
eqns = equations(q)
if not eqns:
break
for e in equations(q):
if len(e.vars) == 1:
determined[e.vars[0]] = e.sum
# print(e.vars, e.sum)
q.cells[e.vars[0]] = e.sum
good = True
break
if not good:
return None
return [v for _, v in sorted(determined.items())]
def equations(q):
Equation = namedtuple("Equation", ["vars", "sum"])
equations = []
for irow in range(q.row_count):
s = int(q.row_sums[irow])
eq_vars = []
for icol in range(q.col_count):
x = q.cells[irow, icol]
if x is not None:
s -= int(x)
else:
eq_vars.append((irow, icol))
if not eq_vars: continue
equations.append(Equation(eq_vars, s))
for icol in range(q.col_count):
s = int(q.col_sums[icol])
eq_vars = []
for irow in range(q.row_count):
x = q.cells[irow, icol]
if x is not None:
s -= int(x)
else:
eq_vars.append((irow, icol))
if not eq_vars: continue
equations.append(Equation(eq_vars, s))
return equations
# def solve_equation(variables, ):
def main():
while True:
line = input()
if line.strip() == '0':
break
product_count, store_count = map(int, line.split())
cells = {}
row_sums = {}
col_sums = {}
for i in range(product_count):
row = [(None if x == '?' else int(x)) for x in input().split()]
row_sums[i] = row[-1]
for k in range(store_count):
cells[i, k] = row[k]
row = [(None if x == '?' else int(x)) for x in input().split()]
for k in range(store_count):
col_sums[k] = row[k]
all_sum = row[-1]
q = Question(all_sum, row_sums, col_sums, cells, product_count, store_count)
a = solve(q)
if a is None:
print("No")
else:
for x in a:
print(x)
print()
input() # ???????Β£???Β°???
main()
``` | instruction | 0 | 63,081 | 10 | 126,162 |
No | output | 1 | 63,081 | 10 | 126,163 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Toshizo is the manager of a convenience store chain in Hakodate. Every day, each of the stores in his chain sends him a table of the products that they have sold. Toshizo's job is to compile these figures and calculate how much the stores have sold in total.
The type of a table that is sent to Toshizo by the stores looks like this (all numbers represent units of products sold):
Store1 Store2 Store3 Totals
Product A -5 40 70 | 105
Product B 20 50 80 | 150
Product C 30 60 90 | 180
-----------------------------------------------------------
Store's total sales 45 150 240 435
By looking at the table, Toshizo can tell at a glance which goods are selling well or selling badly, and which stores are paying well and which stores are too empty. Sometimes, customers will bring products back, so numbers in a table can be negative as well as positive.
Toshizo reports these figures to his boss in Tokyo, and together they sometimes decide to close stores that are not doing well and sometimes decide to open new stores in places that they think will be profitable. So, the total number of stores managed by Toshizo is not fixed. Also, they often decide to discontinue selling some products that are not popular, as well as deciding to stock new items that are likely to be popular. So, the number of products that Toshizo has to monitor is also not fixed.
One New Year, it is very cold in Hakodate. A water pipe bursts in Toshizo's office and floods his desk. When Toshizo comes to work, he finds that the latest sales table is not legible. He can make out some of the figures, but not all of them. He wants to call his boss in Tokyo to tell him that the figures will be late, but he knows that his boss will expect him to reconstruct the table if at all possible.
Waiting until the next day won't help, because it is the New Year, and his shops will be on holiday. So, Toshizo decides either to work out the values for himself, or to be sure that there really is no unique solution. Only then can he call his boss and tell him what has happened.
But Toshizo also wants to be sure that a problem like this never happens again. So he decides to write a computer program that all the managers in his company can use if some of the data goes missing from their sales tables. For instance, if they have a table like:
Store 1 Store 2 Store 3 Totals
Product A ? ? 70 | 105
Product B ? 50 ? | 150
Product C 30 60 90 | 180
--------------------------------------------------------
Store's total sales 45 150 240 435
then Toshizo's program will be able to tell them the correct figures to replace the question marks. In some cases, however, even his program will not be able to replace all the question marks, in general. For instance, if a table like:
Store 1 Store 2 Totals
Product A ? ? | 40
Product B ? ? | 40
---------------------------------------------
Store's total sales 40 40 80
is given, there are infinitely many possible solutions. In this sort of case, his program will just say "NO". Toshizo's program works for any data where the totals row and column are still intact. Can you reproduce Toshizo's program?
Input
The input consists of multiple data sets, each in the following format:
p s
row1
row2
...
rowp
totals
The first line consists of two integers p and s, representing the numbers of products and stores, respectively. The former is less than 100 and the latter is less than 10.
They are separated by a blank character. Each of the subsequent lines represents a row of the table and consists of s+1 elements, each of which is either an integer or a question mark. The i-th line (1 <= i <= p) corresponds to the i-th product and the last line the totals row. The first s elements in a line represent the sales of each store and the last one the total sales of the product. Numbers or question marks in a line are separated by a blank character. There are no missing numbers in the totals column or row. There is at least one question mark in each data set.
The known numbers in the table except the totals row and column is between -1,000 and 1,000, inclusive. However, unknown numbers, represented by question marks, may not be in this range. The total sales number of each product is between -10,000 and 10,000, inclusive. The total sales number of each store is between -100,000 and 100,000, inclusive.
There is a blank line between each data set in the input, and the input is terminated by a line consisting of a zero.
Output
When there is a unique solution, your program should print the missing numbers in the occurrence order in the input data set. Otherwise, your program should print just "NO". The answers of consecutive data sets should be separated by an empty line. Each non-empty output line contains only a single number or "NO".
Example
Input
3 3
? ? 70 105
? 50 ? 150
30 60 90 180
45 150 240 435
2 2
? ? 40
? ? 40
40 40 80
2 3
? 30 40 90
50 60 70 180
70 90 110 270
0
Output
-5
40
20
80
NO
20
Submitted Solution:
```
from collections import namedtuple
from pprint import pprint
Question = namedtuple("Question", ['all_sum', 'row_sums', 'col_sums', 'cells', 'row_count', 'col_count'])
def pp_q(q):
print(q.row_count, q.col_count)
for irow in range(q.row_count):
for icol in range(q.col_count):
print(q.cells[irow, icol], end="\t")
print(q.row_sums[irow])
for icol in range(q.col_count):
print(q.col_sums[icol], end="\t")
print(q.all_sum)
def solve(q):
determined = {}
while True:
good = False
# print(equations(q))
eqns = equations(q)
if not eqns:
break
for e in equations(q):
if len(e.vars) == 1:
determined[e.vars[0]] = e.sum
# print(e.vars, e.sum)
q.cells[e.vars[0]] = e.sum
good = True
break
if not good:
return None
return [v for _, v in sorted(determined.items())]
def equations(q):
Equation = namedtuple("Equation", ["vars", "sum"])
equations = []
for irow in range(q.row_count):
s = int(q.row_sums[irow])
eq_vars = []
for icol in range(q.col_count):
x = q.cells[irow, icol]
if x is not None:
s -= int(x)
else:
eq_vars.append((irow, icol))
if not eq_vars: continue
equations.append(Equation(eq_vars, s))
for icol in range(q.col_count):
s = int(q.col_sums[icol])
eq_vars = []
for irow in range(q.row_count):
x = q.cells[irow, icol]
if x is not None:
s -= int(x)
else:
eq_vars.append((irow, icol))
if not eq_vars: continue
equations.append(Equation(eq_vars, s))
return equations
# def solve_equation(variables, ):
def main():
first = True
while True:
line = input()
if line.strip() == '0':
break
product_count, store_count = map(int, line.split())
cells = {}
row_sums = {}
col_sums = {}
for i in range(product_count):
row = [(None if x == '?' else int(x)) for x in input().split()]
row_sums[i] = row[-1]
for k in range(store_count):
cells[i, k] = row[k]
row = [(None if x == '?' else int(x)) for x in input().split()]
for k in range(store_count):
col_sums[k] = row[k]
all_sum = row[-1]
q = Question(all_sum, row_sums, col_sums, cells, product_count, store_count)
a = solve(q)
if not first:
print()
first = False
if a is None:
print("No")
else:
for x in a:
print(x)
input() # ???????Β£???Β°???
main()
``` | instruction | 0 | 63,082 | 10 | 126,164 |
No | output | 1 | 63,082 | 10 | 126,165 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Toshizo is the manager of a convenience store chain in Hakodate. Every day, each of the stores in his chain sends him a table of the products that they have sold. Toshizo's job is to compile these figures and calculate how much the stores have sold in total.
The type of a table that is sent to Toshizo by the stores looks like this (all numbers represent units of products sold):
Store1 Store2 Store3 Totals
Product A -5 40 70 | 105
Product B 20 50 80 | 150
Product C 30 60 90 | 180
-----------------------------------------------------------
Store's total sales 45 150 240 435
By looking at the table, Toshizo can tell at a glance which goods are selling well or selling badly, and which stores are paying well and which stores are too empty. Sometimes, customers will bring products back, so numbers in a table can be negative as well as positive.
Toshizo reports these figures to his boss in Tokyo, and together they sometimes decide to close stores that are not doing well and sometimes decide to open new stores in places that they think will be profitable. So, the total number of stores managed by Toshizo is not fixed. Also, they often decide to discontinue selling some products that are not popular, as well as deciding to stock new items that are likely to be popular. So, the number of products that Toshizo has to monitor is also not fixed.
One New Year, it is very cold in Hakodate. A water pipe bursts in Toshizo's office and floods his desk. When Toshizo comes to work, he finds that the latest sales table is not legible. He can make out some of the figures, but not all of them. He wants to call his boss in Tokyo to tell him that the figures will be late, but he knows that his boss will expect him to reconstruct the table if at all possible.
Waiting until the next day won't help, because it is the New Year, and his shops will be on holiday. So, Toshizo decides either to work out the values for himself, or to be sure that there really is no unique solution. Only then can he call his boss and tell him what has happened.
But Toshizo also wants to be sure that a problem like this never happens again. So he decides to write a computer program that all the managers in his company can use if some of the data goes missing from their sales tables. For instance, if they have a table like:
Store 1 Store 2 Store 3 Totals
Product A ? ? 70 | 105
Product B ? 50 ? | 150
Product C 30 60 90 | 180
--------------------------------------------------------
Store's total sales 45 150 240 435
then Toshizo's program will be able to tell them the correct figures to replace the question marks. In some cases, however, even his program will not be able to replace all the question marks, in general. For instance, if a table like:
Store 1 Store 2 Totals
Product A ? ? | 40
Product B ? ? | 40
---------------------------------------------
Store's total sales 40 40 80
is given, there are infinitely many possible solutions. In this sort of case, his program will just say "NO". Toshizo's program works for any data where the totals row and column are still intact. Can you reproduce Toshizo's program?
Input
The input consists of multiple data sets, each in the following format:
p s
row1
row2
...
rowp
totals
The first line consists of two integers p and s, representing the numbers of products and stores, respectively. The former is less than 100 and the latter is less than 10.
They are separated by a blank character. Each of the subsequent lines represents a row of the table and consists of s+1 elements, each of which is either an integer or a question mark. The i-th line (1 <= i <= p) corresponds to the i-th product and the last line the totals row. The first s elements in a line represent the sales of each store and the last one the total sales of the product. Numbers or question marks in a line are separated by a blank character. There are no missing numbers in the totals column or row. There is at least one question mark in each data set.
The known numbers in the table except the totals row and column is between -1,000 and 1,000, inclusive. However, unknown numbers, represented by question marks, may not be in this range. The total sales number of each product is between -10,000 and 10,000, inclusive. The total sales number of each store is between -100,000 and 100,000, inclusive.
There is a blank line between each data set in the input, and the input is terminated by a line consisting of a zero.
Output
When there is a unique solution, your program should print the missing numbers in the occurrence order in the input data set. Otherwise, your program should print just "NO". The answers of consecutive data sets should be separated by an empty line. Each non-empty output line contains only a single number or "NO".
Example
Input
3 3
? ? 70 105
? 50 ? 150
30 60 90 180
45 150 240 435
2 2
? ? 40
? ? 40
40 40 80
2 3
? 30 40 90
50 60 70 180
70 90 110 270
0
Output
-5
40
20
80
NO
20
Submitted Solution:
```
from collections import namedtuple
from pprint import pprint
Question = namedtuple("Question", ['all_sum', 'row_sums', 'col_sums', 'cells', 'row_count', 'col_count'])
def pp_q(q):
print(q.row_count, q.col_count)
for irow in range(q.row_count):
for icol in range(q.col_count):
print(q.cells[irow, icol], end="\t")
print(q.row_sums[irow])
for icol in range(q.col_count):
print(q.col_sums[icol], end="\t")
print(q.all_sum)
def solve(q):
determined = {}
while True:
good = False
# print(equations(q))
eqns = equations(q)
if not eqns:
break
for e in equations(q):
if len(e.vars) == 1:
determined[e.vars[0]] = e.sum
# print(e.vars, e.sum)
q.cells[e.vars[0]] = e.sum
good = True
break
if not good:
return None
return [v for _, v in sorted(determined.items())]
def equations(q):
Equation = namedtuple("Equation", ["vars", "sum"])
equations = []
for irow in range(q.row_count):
s = int(q.row_sums[irow])
eq_vars = []
for icol in range(q.col_count):
x = q.cells[irow, icol]
if x is not None:
s -= int(x)
else:
eq_vars.append((irow, icol))
if not eq_vars: continue
equations.append(Equation(eq_vars, s))
for icol in range(q.col_count):
s = int(q.col_sums[icol])
eq_vars = []
for irow in range(q.row_count):
x = q.cells[irow, icol]
if x is not None:
s -= int(x)
else:
eq_vars.append((irow, icol))
if not eq_vars: continue
equations.append(Equation(eq_vars, s))
return equations
# def solve_equation(variables, ):
def main():
first = True
while True:
line = input()
if line.strip() == '0':
break
product_count, store_count = map(int, line.split())
cells = {}
row_sums = {}
col_sums = {}
for i in range(product_count):
row = [(None if x == '?' else int(x)) for x in input().split()]
row_sums[i] = row[-1]
for k in range(store_count):
cells[i, k] = row[k]
row = [(None if x == '?' else int(x)) for x in input().split()]
for k in range(store_count):
col_sums[k] = row[k]
all_sum = row[-1]
q = Question(all_sum, row_sums, col_sums, cells, product_count, store_count)
a = solve(q)
if a is None:
print("No")
else:
for x in a:
print(x)
if not first:
print()
first = False
input() # ???????Β£???Β°???
main()
``` | instruction | 0 | 63,083 | 10 | 126,166 |
No | output | 1 | 63,083 | 10 | 126,167 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
The popular video posting site "ZouTube" is now in the midst of an unprecedented "virtual ZouTuber" boom. Among them, the one that has been attracting particular attention recently is the junior virtual ZouTuber "Aizumarim (commonly known as Azurim)".
As a big fan of Azlim, you're going to send her a "special chat" on Azlim's live stream today.
"Special chat" is a "function that viewers give points to distributors" provided by ZouTube. Viewers can spend $ 500 $, $ 1000 $, $ 5000 $, or $ 10000 $ for each $ 1 special chat, and give the distributor the same amount of points as they spend.
Given the total amount of points you have now, spend those points to find the maximum total amount of points you can give to Azlim. You can have as many special chats as you like, as long as the amount of points you hold is not less than the amount of points you consume.
Constraints
The input satisfies the following conditions.
* $ 1 \ le P \ le 10 ^ 5 $
Input
The input is given in the following format.
$ P $
An integer $ P $ representing the total amount of points you have now is given in the $ 1 $ line.
Output
Output the maximum total amount of points that can be given to Azulim on the $ 1 $ line.
Examples
Input
5700
Output
5500
Input
1333
Output
1000
Input
100000
Output
100000 | instruction | 0 | 63,118 | 10 | 126,236 |
"Correct Solution:
```
p = int(input())
sum = 0
if p >= 10000:
sum += p//10000*10000
p -= p//10000*10000
if p >= 5000:
sum += p//5000*5000
p -= p//5000*5000
if p >= 1000:
sum += p//1000*1000
p -= p//1000*1000
if p >= 500:
sum += p//500*500
p -= p//500*500
print(sum)
``` | output | 1 | 63,118 | 10 | 126,237 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
The popular video posting site "ZouTube" is now in the midst of an unprecedented "virtual ZouTuber" boom. Among them, the one that has been attracting particular attention recently is the junior virtual ZouTuber "Aizumarim (commonly known as Azurim)".
As a big fan of Azlim, you're going to send her a "special chat" on Azlim's live stream today.
"Special chat" is a "function that viewers give points to distributors" provided by ZouTube. Viewers can spend $ 500 $, $ 1000 $, $ 5000 $, or $ 10000 $ for each $ 1 special chat, and give the distributor the same amount of points as they spend.
Given the total amount of points you have now, spend those points to find the maximum total amount of points you can give to Azlim. You can have as many special chats as you like, as long as the amount of points you hold is not less than the amount of points you consume.
Constraints
The input satisfies the following conditions.
* $ 1 \ le P \ le 10 ^ 5 $
Input
The input is given in the following format.
$ P $
An integer $ P $ representing the total amount of points you have now is given in the $ 1 $ line.
Output
Output the maximum total amount of points that can be given to Azulim on the $ 1 $ line.
Examples
Input
5700
Output
5500
Input
1333
Output
1000
Input
100000
Output
100000 | instruction | 0 | 63,119 | 10 | 126,238 |
"Correct Solution:
```
n=int(input())
if n%500==0:
print(n)
else:
print((n//500)*500)
``` | output | 1 | 63,119 | 10 | 126,239 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
The popular video posting site "ZouTube" is now in the midst of an unprecedented "virtual ZouTuber" boom. Among them, the one that has been attracting particular attention recently is the junior virtual ZouTuber "Aizumarim (commonly known as Azurim)".
As a big fan of Azlim, you're going to send her a "special chat" on Azlim's live stream today.
"Special chat" is a "function that viewers give points to distributors" provided by ZouTube. Viewers can spend $ 500 $, $ 1000 $, $ 5000 $, or $ 10000 $ for each $ 1 special chat, and give the distributor the same amount of points as they spend.
Given the total amount of points you have now, spend those points to find the maximum total amount of points you can give to Azlim. You can have as many special chats as you like, as long as the amount of points you hold is not less than the amount of points you consume.
Constraints
The input satisfies the following conditions.
* $ 1 \ le P \ le 10 ^ 5 $
Input
The input is given in the following format.
$ P $
An integer $ P $ representing the total amount of points you have now is given in the $ 1 $ line.
Output
Output the maximum total amount of points that can be given to Azulim on the $ 1 $ line.
Examples
Input
5700
Output
5500
Input
1333
Output
1000
Input
100000
Output
100000 | instruction | 0 | 63,120 | 10 | 126,240 |
"Correct Solution:
```
print(int(input()) // 500 * 500)#ζ¨ζΊε
₯εγγ500γ§ε²γ£γεγ«500γγγ
``` | output | 1 | 63,120 | 10 | 126,241 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
The popular video posting site "ZouTube" is now in the midst of an unprecedented "virtual ZouTuber" boom. Among them, the one that has been attracting particular attention recently is the junior virtual ZouTuber "Aizumarim (commonly known as Azurim)".
As a big fan of Azlim, you're going to send her a "special chat" on Azlim's live stream today.
"Special chat" is a "function that viewers give points to distributors" provided by ZouTube. Viewers can spend $ 500 $, $ 1000 $, $ 5000 $, or $ 10000 $ for each $ 1 special chat, and give the distributor the same amount of points as they spend.
Given the total amount of points you have now, spend those points to find the maximum total amount of points you can give to Azlim. You can have as many special chats as you like, as long as the amount of points you hold is not less than the amount of points you consume.
Constraints
The input satisfies the following conditions.
* $ 1 \ le P \ le 10 ^ 5 $
Input
The input is given in the following format.
$ P $
An integer $ P $ representing the total amount of points you have now is given in the $ 1 $ line.
Output
Output the maximum total amount of points that can be given to Azulim on the $ 1 $ line.
Examples
Input
5700
Output
5500
Input
1333
Output
1000
Input
100000
Output
100000 | instruction | 0 | 63,121 | 10 | 126,242 |
"Correct Solution:
```
n = int(input())
ans = 0
l = [10000, 5000, 1000, 500]
for i in range(4) :
ans += (n // l[i]) * l[i]
n = n % l[i]
print(ans)
``` | output | 1 | 63,121 | 10 | 126,243 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
The popular video posting site "ZouTube" is now in the midst of an unprecedented "virtual ZouTuber" boom. Among them, the one that has been attracting particular attention recently is the junior virtual ZouTuber "Aizumarim (commonly known as Azurim)".
As a big fan of Azlim, you're going to send her a "special chat" on Azlim's live stream today.
"Special chat" is a "function that viewers give points to distributors" provided by ZouTube. Viewers can spend $ 500 $, $ 1000 $, $ 5000 $, or $ 10000 $ for each $ 1 special chat, and give the distributor the same amount of points as they spend.
Given the total amount of points you have now, spend those points to find the maximum total amount of points you can give to Azlim. You can have as many special chats as you like, as long as the amount of points you hold is not less than the amount of points you consume.
Constraints
The input satisfies the following conditions.
* $ 1 \ le P \ le 10 ^ 5 $
Input
The input is given in the following format.
$ P $
An integer $ P $ representing the total amount of points you have now is given in the $ 1 $ line.
Output
Output the maximum total amount of points that can be given to Azulim on the $ 1 $ line.
Examples
Input
5700
Output
5500
Input
1333
Output
1000
Input
100000
Output
100000 | instruction | 0 | 63,122 | 10 | 126,244 |
"Correct Solution:
```
print(int(int(input()) // 500 * 500))
``` | output | 1 | 63,122 | 10 | 126,245 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
The popular video posting site "ZouTube" is now in the midst of an unprecedented "virtual ZouTuber" boom. Among them, the one that has been attracting particular attention recently is the junior virtual ZouTuber "Aizumarim (commonly known as Azurim)".
As a big fan of Azlim, you're going to send her a "special chat" on Azlim's live stream today.
"Special chat" is a "function that viewers give points to distributors" provided by ZouTube. Viewers can spend $ 500 $, $ 1000 $, $ 5000 $, or $ 10000 $ for each $ 1 special chat, and give the distributor the same amount of points as they spend.
Given the total amount of points you have now, spend those points to find the maximum total amount of points you can give to Azlim. You can have as many special chats as you like, as long as the amount of points you hold is not less than the amount of points you consume.
Constraints
The input satisfies the following conditions.
* $ 1 \ le P \ le 10 ^ 5 $
Input
The input is given in the following format.
$ P $
An integer $ P $ representing the total amount of points you have now is given in the $ 1 $ line.
Output
Output the maximum total amount of points that can be given to Azulim on the $ 1 $ line.
Examples
Input
5700
Output
5500
Input
1333
Output
1000
Input
100000
Output
100000 | instruction | 0 | 63,123 | 10 | 126,246 |
"Correct Solution:
```
print(int(input())//500*500)
``` | output | 1 | 63,123 | 10 | 126,247 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya collects coins: he has exactly one coin for every year from 1 to n. Naturally, Vasya keeps all the coins in his collection in the order in which they were released. Once Vasya's younger brother made a change β he took all the coins whose release year dated from l to r inclusively and put them in the reverse order. That is, he took a certain segment [l, r] and reversed it. At that the segment's endpoints did not coincide. For example, if n = 8, then initially Vasya's coins were kept in the order 1 2 3 4 5 6 7 8. If Vasya's younger brother chose the segment [2, 6], then after the reversal the coin order will change to 1 6 5 4 3 2 7 8. Vasya suspects that someone else could have spoilt the permutation after his brother. Help him to find that out. Check if the given permutation can be obtained from the permutation 1 2 ... n using exactly one segment reversal. If it is possible, find the segment itself.
Input
The first line contains an integer n (1 β€ n β€ 1000) which is the number of coins in Vasya's collection. The second line contains space-separated n integers which are the spoilt sequence of coins. It is guaranteed that the given sequence is a permutation, i.e. it contains only integers from 1 to n, and every number is used exactly 1 time.
Output
If it is impossible to obtain the given permutation from the original one in exactly one action, print 0 0. Otherwise, print two numbers l r (1 β€ l < r β€ n) which are the endpoints of the segment that needs to be reversed to obtain from permutation 1 2 ... n the given one.
Examples
Input
8
1 6 5 4 3 2 7 8
Output
2 6
Input
4
2 3 4 1
Output
0 0
Input
4
1 2 3 4
Output
0 0
Submitted Solution:
```
n=int(input())
arr=list(map(int,input().split()))
s=-1
e=-1
for i in range(n):
if arr[i]!=i+1:
s=i
e=arr[i]-1
break
new_arr=arr[:s]+list(reversed(arr[s:e+1]))+arr[e+1:]
#print(s,e,new_arr)
if new_arr==sorted(arr) and s!=-1:
print(s+1,e+1)
else:
print(0,0)
``` | instruction | 0 | 63,534 | 10 | 127,068 |
Yes | output | 1 | 63,534 | 10 | 127,069 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya collects coins: he has exactly one coin for every year from 1 to n. Naturally, Vasya keeps all the coins in his collection in the order in which they were released. Once Vasya's younger brother made a change β he took all the coins whose release year dated from l to r inclusively and put them in the reverse order. That is, he took a certain segment [l, r] and reversed it. At that the segment's endpoints did not coincide. For example, if n = 8, then initially Vasya's coins were kept in the order 1 2 3 4 5 6 7 8. If Vasya's younger brother chose the segment [2, 6], then after the reversal the coin order will change to 1 6 5 4 3 2 7 8. Vasya suspects that someone else could have spoilt the permutation after his brother. Help him to find that out. Check if the given permutation can be obtained from the permutation 1 2 ... n using exactly one segment reversal. If it is possible, find the segment itself.
Input
The first line contains an integer n (1 β€ n β€ 1000) which is the number of coins in Vasya's collection. The second line contains space-separated n integers which are the spoilt sequence of coins. It is guaranteed that the given sequence is a permutation, i.e. it contains only integers from 1 to n, and every number is used exactly 1 time.
Output
If it is impossible to obtain the given permutation from the original one in exactly one action, print 0 0. Otherwise, print two numbers l r (1 β€ l < r β€ n) which are the endpoints of the segment that needs to be reversed to obtain from permutation 1 2 ... n the given one.
Examples
Input
8
1 6 5 4 3 2 7 8
Output
2 6
Input
4
2 3 4 1
Output
0 0
Input
4
1 2 3 4
Output
0 0
Submitted Solution:
```
a=int(input());flag=0;l,r=0,0;i=0
b=[int(i) for i in input().split()]
while i<a:
if b[i]!=i+1:
if sorted(b[i:b[i]])!=[*range(i+1,b[i]+1)]:
exit(print("0 0"))
else:
if flag==0:flag=1;l=i+1;r=b[i];i=b[i]
else:exit(print("0 0"))
else:i+=1
print(l,r)
``` | instruction | 0 | 63,535 | 10 | 127,070 |
Yes | output | 1 | 63,535 | 10 | 127,071 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya collects coins: he has exactly one coin for every year from 1 to n. Naturally, Vasya keeps all the coins in his collection in the order in which they were released. Once Vasya's younger brother made a change β he took all the coins whose release year dated from l to r inclusively and put them in the reverse order. That is, he took a certain segment [l, r] and reversed it. At that the segment's endpoints did not coincide. For example, if n = 8, then initially Vasya's coins were kept in the order 1 2 3 4 5 6 7 8. If Vasya's younger brother chose the segment [2, 6], then after the reversal the coin order will change to 1 6 5 4 3 2 7 8. Vasya suspects that someone else could have spoilt the permutation after his brother. Help him to find that out. Check if the given permutation can be obtained from the permutation 1 2 ... n using exactly one segment reversal. If it is possible, find the segment itself.
Input
The first line contains an integer n (1 β€ n β€ 1000) which is the number of coins in Vasya's collection. The second line contains space-separated n integers which are the spoilt sequence of coins. It is guaranteed that the given sequence is a permutation, i.e. it contains only integers from 1 to n, and every number is used exactly 1 time.
Output
If it is impossible to obtain the given permutation from the original one in exactly one action, print 0 0. Otherwise, print two numbers l r (1 β€ l < r β€ n) which are the endpoints of the segment that needs to be reversed to obtain from permutation 1 2 ... n the given one.
Examples
Input
8
1 6 5 4 3 2 7 8
Output
2 6
Input
4
2 3 4 1
Output
0 0
Input
4
1 2 3 4
Output
0 0
Submitted Solution:
```
import sys
input = sys.stdin.readline
n = int(input())
a = list(map(int, input().split()))
pos = []
i = 0
while i < n and a[i] == i+1:
i += 1
if i == n:
print(0, 0)
else:
pos.append(i+1)
sn = a[i]
j = 0
while i < n and a[i] == sn-j:
i += 1
j += 1
pos.append(i)
j = 1
while i < n:
if a[i] != sn+j:
print(0, 0)
break
i += 1
j += 1
else:
print(pos[0], pos[1])
``` | instruction | 0 | 63,536 | 10 | 127,072 |
Yes | output | 1 | 63,536 | 10 | 127,073 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya collects coins: he has exactly one coin for every year from 1 to n. Naturally, Vasya keeps all the coins in his collection in the order in which they were released. Once Vasya's younger brother made a change β he took all the coins whose release year dated from l to r inclusively and put them in the reverse order. That is, he took a certain segment [l, r] and reversed it. At that the segment's endpoints did not coincide. For example, if n = 8, then initially Vasya's coins were kept in the order 1 2 3 4 5 6 7 8. If Vasya's younger brother chose the segment [2, 6], then after the reversal the coin order will change to 1 6 5 4 3 2 7 8. Vasya suspects that someone else could have spoilt the permutation after his brother. Help him to find that out. Check if the given permutation can be obtained from the permutation 1 2 ... n using exactly one segment reversal. If it is possible, find the segment itself.
Input
The first line contains an integer n (1 β€ n β€ 1000) which is the number of coins in Vasya's collection. The second line contains space-separated n integers which are the spoilt sequence of coins. It is guaranteed that the given sequence is a permutation, i.e. it contains only integers from 1 to n, and every number is used exactly 1 time.
Output
If it is impossible to obtain the given permutation from the original one in exactly one action, print 0 0. Otherwise, print two numbers l r (1 β€ l < r β€ n) which are the endpoints of the segment that needs to be reversed to obtain from permutation 1 2 ... n the given one.
Examples
Input
8
1 6 5 4 3 2 7 8
Output
2 6
Input
4
2 3 4 1
Output
0 0
Input
4
1 2 3 4
Output
0 0
Submitted Solution:
```
##############--->>>>> Deepcoder Amit Kumar Bhuyan <<<<<---##############
"""
Perfection is achieved not when there is nothing more to add, but rather when there is nothing more to take away.
"""
from __future__ import division, print_function
import os,sys
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
def ii(): return int(input())
def si(): return input()
def mi(): return map(int,input().strip().split(" "))
def msi(): return map(str,input().strip().split(" "))
def li(): return list(mi())
def dmain():
sys.setrecursionlimit(1000000)
threading.stack_size(1024000)
thread = threading.Thread(target=main)
thread.start()
#from collections import deque, Counter, OrderedDict,defaultdict
#from heapq import nsmallest, nlargest, heapify,heappop ,heappush, heapreplace
#from math import log,sqrt,factorial,cos,tan,sin,radians
#from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
#from decimal import *
#import threading
#from itertools import permutations
#Copy 2D list m = [x[:] for x in mark] .. Avoid Using Deepcopy
import sys
input = sys.stdin.readline
scan_int = lambda: int(input())
scan_string = lambda: input().rstrip()
read_list = lambda: list(read())
read = lambda: map(int, input().split())
read_float = lambda: map(float, input().split())
# from bisect import bisect_left as lower_bound;
# from bisect import bisect_right as upper_bound;
# from math import ceil, factorial;
def ceil(x):
if x != int(x):
x = int(x) + 1
return x
def factorial(x, m):
val = 1
while x>0:
val = (val * x) % m
x -= 1
return val
def fact(x):
val = 1
while x > 0:
val *= x
x -= 1
return val
# swap_array function
def swaparr(arr, a,b):
temp = arr[a];
arr[a] = arr[b];
arr[b] = temp;
## gcd function
def gcd(a,b):
if b == 0:
return a;
return gcd(b, a % b);
## lcm function
def lcm(a, b):
return (a * b) // math.gcd(a, b)
def is_integer(n):
return math.ceil(n) == math.floor(n)
## nCr function efficient using Binomial Cofficient
def nCr(n, k):
if k > n:
return 0
if(k > n - k):
k = n - k
res = 1
for i in range(k):
res = res * (n - i)
res = res / (i + 1)
return int(res)
## upper bound function code -- such that e in a[:i] e < x;
## prime factorization
def primefs(n):
## if n == 1 ## calculating primes
primes = {}
while(n%2 == 0 and n > 0):
primes[2] = primes.get(2, 0) + 1
n = n//2
for i in range(3, int(n**0.5)+2, 2):
while(n%i == 0 and n > 0):
primes[i] = primes.get(i, 0) + 1
n = n//i
if n > 2:
primes[n] = primes.get(n, 0) + 1
## prime factoriazation of n is stored in dictionary
## primes and can be accesed. O(sqrt n)
return primes
## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p):
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
temp = a
a = b
b = temp
return a,b;
# find function with path compression included (recursive)
# def find(x, link):
# if link[x] == x:
# return x
# link[x] = find(link[x], link);
# return link[x];
# find function with path compression (ITERATIVE)
def find(x, link):
p = x;
while( p != link[p]):
p = link[p];
while( x != p):
nex = link[x];
link[x] = p;
x = nex;
return p;
# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
x = find(x, link)
y = find(y, link)
if size[x] < size[y]:
x,y = swap(x,y)
if x != y:
size[x] += size[y]
link[y] = x
## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n):
prime = [True for i in range(n+1)]
prime[0], prime[1] = False, False
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
# Euler's Toitent Function phi
def phi(n) :
result = n
p = 2
while(p * p<= n) :
if (n % p == 0) :
while (n % p == 0) :
n = n // p
result = result * (1.0 - (1.0 / (float) (p)))
p = p + 1
if (n > 1) :
result = result * (1.0 - (1.0 / (float)(n)))
return (int)(result)
def is_prime(n):
if n == 0:
return False
if n == 1:
return True
for i in range(2, int(n ** (1 / 2)) + 1):
if not n % i:
return False
return True
def next_prime(n, primes):
while primes[n] != True:
n += 1
return n
#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e5 + 5)
def spf_sieve():
spf[1] = 1;
for i in range(2, MAXN):
spf[i] = i;
for i in range(4, MAXN, 2):
spf[i] = 2;
for i in range(3, ceil(MAXN ** 0.5), 2):
if spf[i] == i:
for j in range(i*i, MAXN, i):
if spf[j] == j:
spf[j] = i;
## function for storing smallest prime factors (spf) in the array
################## un-comment below 2 lines when using factorization #################
spf = [0 for i in range(MAXN)]
# spf_sieve();
def factoriazation(x):
res = []
for i in range(2, int(x ** 0.5) + 1):
while x % i == 0:
res.append(i)
x //= i
if x != 1:
res.append(x)
return res
## this function is useful for multiple queries only, o/w use
## primefs function above. complexity O(log n)
def factors(n):
res = []
for i in range(1, int(n ** 0.5) + 1):
if n % i == 0:
res.append(i)
res.append(n // i)
return list(set(res))
## taking integer array input
def int_array():
return list(map(int, input().strip().split()));
def float_array():
return list(map(float, input().strip().split()));
## taking string array input
def str_array():
return input().strip().split();
def binary_search(low, high, w, h, n):
while low < high:
mid = low + (high - low) // 2
# print(low, mid, high)
if check(mid, w, h, n):
low = mid + 1
else:
high = mid
return low
## for checking any conditions
def check(beauty, s, n, count):
pass
#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');
################### ---------------- TEMPLATE ENDS HERE ---------------- ###################
from itertools import permutations
import math
import bisect as bis
import random
import sys
import collections as collect
def solve():
n = scan_int()
a = read_list()
start = 0
end = 0
for i in range(n):
if a[i] != i + 1:
start = i + 1
break
for j in range(n - 1, -1, -1):
if a[j] != j + 1:
end = j + 1
break
if sorted(a[start - 1: end]) != a[start - 1: end][::-1]:
print(0, 0)
return
print(start, end)
# region fastio
# template taken from https://github.com/cheran-senthil/PyRival/blob/master/templates/template.py
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
#read()
for i in range(1):
solve()
#dmain()
# Comment Read()
# fin_time = datetime.now()
# print("Execution time (for loop): ", (fin_time-init_time))
``` | instruction | 0 | 63,537 | 10 | 127,074 |
Yes | output | 1 | 63,537 | 10 | 127,075 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya collects coins: he has exactly one coin for every year from 1 to n. Naturally, Vasya keeps all the coins in his collection in the order in which they were released. Once Vasya's younger brother made a change β he took all the coins whose release year dated from l to r inclusively and put them in the reverse order. That is, he took a certain segment [l, r] and reversed it. At that the segment's endpoints did not coincide. For example, if n = 8, then initially Vasya's coins were kept in the order 1 2 3 4 5 6 7 8. If Vasya's younger brother chose the segment [2, 6], then after the reversal the coin order will change to 1 6 5 4 3 2 7 8. Vasya suspects that someone else could have spoilt the permutation after his brother. Help him to find that out. Check if the given permutation can be obtained from the permutation 1 2 ... n using exactly one segment reversal. If it is possible, find the segment itself.
Input
The first line contains an integer n (1 β€ n β€ 1000) which is the number of coins in Vasya's collection. The second line contains space-separated n integers which are the spoilt sequence of coins. It is guaranteed that the given sequence is a permutation, i.e. it contains only integers from 1 to n, and every number is used exactly 1 time.
Output
If it is impossible to obtain the given permutation from the original one in exactly one action, print 0 0. Otherwise, print two numbers l r (1 β€ l < r β€ n) which are the endpoints of the segment that needs to be reversed to obtain from permutation 1 2 ... n the given one.
Examples
Input
8
1 6 5 4 3 2 7 8
Output
2 6
Input
4
2 3 4 1
Output
0 0
Input
4
1 2 3 4
Output
0 0
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
l1=[];b=a.copy();b.reverse()
if len(a)>1:
if sorted(a)==b: print(1,len(a))
else:
for i in range(1,n):
if abs(a[i]-a[i-1])!=1: l1.append(i)
if len(l1)>2: break
if len(l1)!=2: print("0 0")
else: print(l1[0]+1,l1[1])
``` | instruction | 0 | 63,538 | 10 | 127,076 |
No | output | 1 | 63,538 | 10 | 127,077 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya collects coins: he has exactly one coin for every year from 1 to n. Naturally, Vasya keeps all the coins in his collection in the order in which they were released. Once Vasya's younger brother made a change β he took all the coins whose release year dated from l to r inclusively and put them in the reverse order. That is, he took a certain segment [l, r] and reversed it. At that the segment's endpoints did not coincide. For example, if n = 8, then initially Vasya's coins were kept in the order 1 2 3 4 5 6 7 8. If Vasya's younger brother chose the segment [2, 6], then after the reversal the coin order will change to 1 6 5 4 3 2 7 8. Vasya suspects that someone else could have spoilt the permutation after his brother. Help him to find that out. Check if the given permutation can be obtained from the permutation 1 2 ... n using exactly one segment reversal. If it is possible, find the segment itself.
Input
The first line contains an integer n (1 β€ n β€ 1000) which is the number of coins in Vasya's collection. The second line contains space-separated n integers which are the spoilt sequence of coins. It is guaranteed that the given sequence is a permutation, i.e. it contains only integers from 1 to n, and every number is used exactly 1 time.
Output
If it is impossible to obtain the given permutation from the original one in exactly one action, print 0 0. Otherwise, print two numbers l r (1 β€ l < r β€ n) which are the endpoints of the segment that needs to be reversed to obtain from permutation 1 2 ... n the given one.
Examples
Input
8
1 6 5 4 3 2 7 8
Output
2 6
Input
4
2 3 4 1
Output
0 0
Input
4
1 2 3 4
Output
0 0
Submitted Solution:
```
n=int(input())
S=str(input())
l=S.split(' ')
M='0 0'
f=0
L=[int(i) for i in l]
for i in range(n):
if L[i]!=i+1 and f==0:
j=L[i]
if L[j:]!=[i for i in range(j,n+1)]:
print('0 0')
f=2
break
else:
M=str(i+1)+' '+str(j)
f=1
if L[L[i]-1]!=i+1:
print('0 0')
f=2
break
if f!=2:
print(M)
``` | instruction | 0 | 63,539 | 10 | 127,078 |
No | output | 1 | 63,539 | 10 | 127,079 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya collects coins: he has exactly one coin for every year from 1 to n. Naturally, Vasya keeps all the coins in his collection in the order in which they were released. Once Vasya's younger brother made a change β he took all the coins whose release year dated from l to r inclusively and put them in the reverse order. That is, he took a certain segment [l, r] and reversed it. At that the segment's endpoints did not coincide. For example, if n = 8, then initially Vasya's coins were kept in the order 1 2 3 4 5 6 7 8. If Vasya's younger brother chose the segment [2, 6], then after the reversal the coin order will change to 1 6 5 4 3 2 7 8. Vasya suspects that someone else could have spoilt the permutation after his brother. Help him to find that out. Check if the given permutation can be obtained from the permutation 1 2 ... n using exactly one segment reversal. If it is possible, find the segment itself.
Input
The first line contains an integer n (1 β€ n β€ 1000) which is the number of coins in Vasya's collection. The second line contains space-separated n integers which are the spoilt sequence of coins. It is guaranteed that the given sequence is a permutation, i.e. it contains only integers from 1 to n, and every number is used exactly 1 time.
Output
If it is impossible to obtain the given permutation from the original one in exactly one action, print 0 0. Otherwise, print two numbers l r (1 β€ l < r β€ n) which are the endpoints of the segment that needs to be reversed to obtain from permutation 1 2 ... n the given one.
Examples
Input
8
1 6 5 4 3 2 7 8
Output
2 6
Input
4
2 3 4 1
Output
0 0
Input
4
1 2 3 4
Output
0 0
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
l1=[];b=a.copy();b.reverse()
if sorted(a)==b and len(a)>1: print(1,len(a))
else:
for i in range(1,n):
if abs(a[i]-a[i-1])!=1: l1.append(i)
if len(l1)>2: break
if len(l1)>0:
if l1[0]==n-2: l1.append(n)
if len(l1)!=2: print("0 0")
else: print(l1[0]+1,l1[1])
``` | instruction | 0 | 63,540 | 10 | 127,080 |
No | output | 1 | 63,540 | 10 | 127,081 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya collects coins: he has exactly one coin for every year from 1 to n. Naturally, Vasya keeps all the coins in his collection in the order in which they were released. Once Vasya's younger brother made a change β he took all the coins whose release year dated from l to r inclusively and put them in the reverse order. That is, he took a certain segment [l, r] and reversed it. At that the segment's endpoints did not coincide. For example, if n = 8, then initially Vasya's coins were kept in the order 1 2 3 4 5 6 7 8. If Vasya's younger brother chose the segment [2, 6], then after the reversal the coin order will change to 1 6 5 4 3 2 7 8. Vasya suspects that someone else could have spoilt the permutation after his brother. Help him to find that out. Check if the given permutation can be obtained from the permutation 1 2 ... n using exactly one segment reversal. If it is possible, find the segment itself.
Input
The first line contains an integer n (1 β€ n β€ 1000) which is the number of coins in Vasya's collection. The second line contains space-separated n integers which are the spoilt sequence of coins. It is guaranteed that the given sequence is a permutation, i.e. it contains only integers from 1 to n, and every number is used exactly 1 time.
Output
If it is impossible to obtain the given permutation from the original one in exactly one action, print 0 0. Otherwise, print two numbers l r (1 β€ l < r β€ n) which are the endpoints of the segment that needs to be reversed to obtain from permutation 1 2 ... n the given one.
Examples
Input
8
1 6 5 4 3 2 7 8
Output
2 6
Input
4
2 3 4 1
Output
0 0
Input
4
1 2 3 4
Output
0 0
Submitted Solution:
```
n = int(input())
arr = list(map(int, input().split()))
ans = True
for i in range(n):
if i + 1 !=arr[i]:
for j in range(arr[i],n):
if j + 1 != arr[j]:
ans = False
break
for j in range(i, arr[i]):
if arr[i]-j+1 != arr[j]:
ans = False
break
if ans:
print(i+1, arr[i])
exit(0)
break
print(0, 0)
``` | instruction | 0 | 63,541 | 10 | 127,082 |
No | output | 1 | 63,541 | 10 | 127,083 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Pupils decided to go to amusement park. Some of them were with parents. In total, n people came to the park and they all want to get to the most extreme attraction and roll on it exactly once.
Tickets for group of x people are sold on the attraction, there should be at least one adult in each group (it is possible that the group consists of one adult). The ticket price for such group is c1 + c2Β·(x - 1)2 (in particular, if the group consists of one person, then the price is c1).
All pupils who came to the park and their parents decided to split into groups in such a way that each visitor join exactly one group, and the total price of visiting the most extreme attraction is as low as possible. You are to determine this minimum possible total price. There should be at least one adult in each group.
Input
The first line contains three integers n, c1 and c2 (1 β€ n β€ 200 000, 1 β€ c1, c2 β€ 107) β the number of visitors and parameters for determining the ticket prices for a group.
The second line contains the string of length n, which consists of zeros and ones. If the i-th symbol of the string is zero, then the i-th visitor is a pupil, otherwise the i-th person is an adult. It is guaranteed that there is at least one adult. It is possible that there are no pupils.
Output
Print the minimum price of visiting the most extreme attraction for all pupils and their parents. Each of them should roll on the attraction exactly once.
Examples
Input
3 4 1
011
Output
8
Input
4 7 2
1101
Output
18
Note
In the first test one group of three people should go to the attraction. Then they have to pay 4 + 1 * (3 - 1)2 = 8.
In the second test it is better to go to the attraction in two groups. The first group should consist of two adults (for example, the first and the second person), the second should consist of one pupil and one adult (the third and the fourth person). Then each group will have a size of two and for each the price of ticket is 7 + 2 * (2 - 1)2 = 9. Thus, the total price for two groups is 18.
Submitted Solution:
```
import math
[n, c1, c2] = input().split()
people = input()
def calc(x):
return int(c1)+int(c2)*(int(x)-1)*(x-1)
pupils = people.count('0')
adults = int(n)-pupils
optimal_people = math.floor(math.sqrt(int(c1)/int(c2)) + 1)
remainder = int(math.remainder(len(people), optimal_people))
optimal_groups = int((len(people)-remainder)/optimal_people)
if adults < optimal_groups:
remainder = math.remainder(pupils, adults)
optimal_amount = ((pupils-remainder)/adults) +1
arr = adults*[optimal_amount]
arr = arr[:(adults-remainder)]
arr = arr + remainder*[optimal_amount + 1]
print(sum([calc[arg] for arg in arr]))
else:
arr = optimal_groups*[optimal_people]
arr = arr[:(optimal_groups-remainder)]
arr = arr + remainder*[optimal_people + 1]
print(sum([calc(arg) for arg in arr]))
``` | instruction | 0 | 63,596 | 10 | 127,192 |
No | output | 1 | 63,596 | 10 | 127,193 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Pupils decided to go to amusement park. Some of them were with parents. In total, n people came to the park and they all want to get to the most extreme attraction and roll on it exactly once.
Tickets for group of x people are sold on the attraction, there should be at least one adult in each group (it is possible that the group consists of one adult). The ticket price for such group is c1 + c2Β·(x - 1)2 (in particular, if the group consists of one person, then the price is c1).
All pupils who came to the park and their parents decided to split into groups in such a way that each visitor join exactly one group, and the total price of visiting the most extreme attraction is as low as possible. You are to determine this minimum possible total price. There should be at least one adult in each group.
Input
The first line contains three integers n, c1 and c2 (1 β€ n β€ 200 000, 1 β€ c1, c2 β€ 107) β the number of visitors and parameters for determining the ticket prices for a group.
The second line contains the string of length n, which consists of zeros and ones. If the i-th symbol of the string is zero, then the i-th visitor is a pupil, otherwise the i-th person is an adult. It is guaranteed that there is at least one adult. It is possible that there are no pupils.
Output
Print the minimum price of visiting the most extreme attraction for all pupils and their parents. Each of them should roll on the attraction exactly once.
Examples
Input
3 4 1
011
Output
8
Input
4 7 2
1101
Output
18
Note
In the first test one group of three people should go to the attraction. Then they have to pay 4 + 1 * (3 - 1)2 = 8.
In the second test it is better to go to the attraction in two groups. The first group should consist of two adults (for example, the first and the second person), the second should consist of one pupil and one adult (the third and the fourth person). Then each group will have a size of two and for each the price of ticket is 7 + 2 * (2 - 1)2 = 9. Thus, the total price for two groups is 18.
Submitted Solution:
```
import math
n,c1,c2=map(int,input().split())
s=input()
cnt=s.count('1')
if(cnt==1):
print(int(c1+c2*math.pow((n-1),2)))
else:
if(n%2):
l=int(c1+c2*math.pow((n-1),2))
r=int(2*c1+c2*math.pow(int((n+1)/2)-1,2)+c2*math.pow(int((n-1)/2)-1,2))
print(min(l,r))
else:
l=int(c1+c2*math.pow((n-1),2))
r=2*int(c1+c2*math.pow(int(n/2)-1,2))
print(min(l,r))
``` | instruction | 0 | 63,597 | 10 | 127,194 |
No | output | 1 | 63,597 | 10 | 127,195 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Pupils decided to go to amusement park. Some of them were with parents. In total, n people came to the park and they all want to get to the most extreme attraction and roll on it exactly once.
Tickets for group of x people are sold on the attraction, there should be at least one adult in each group (it is possible that the group consists of one adult). The ticket price for such group is c1 + c2Β·(x - 1)2 (in particular, if the group consists of one person, then the price is c1).
All pupils who came to the park and their parents decided to split into groups in such a way that each visitor join exactly one group, and the total price of visiting the most extreme attraction is as low as possible. You are to determine this minimum possible total price. There should be at least one adult in each group.
Input
The first line contains three integers n, c1 and c2 (1 β€ n β€ 200 000, 1 β€ c1, c2 β€ 107) β the number of visitors and parameters for determining the ticket prices for a group.
The second line contains the string of length n, which consists of zeros and ones. If the i-th symbol of the string is zero, then the i-th visitor is a pupil, otherwise the i-th person is an adult. It is guaranteed that there is at least one adult. It is possible that there are no pupils.
Output
Print the minimum price of visiting the most extreme attraction for all pupils and their parents. Each of them should roll on the attraction exactly once.
Examples
Input
3 4 1
011
Output
8
Input
4 7 2
1101
Output
18
Note
In the first test one group of three people should go to the attraction. Then they have to pay 4 + 1 * (3 - 1)2 = 8.
In the second test it is better to go to the attraction in two groups. The first group should consist of two adults (for example, the first and the second person), the second should consist of one pupil and one adult (the third and the fourth person). Then each group will have a size of two and for each the price of ticket is 7 + 2 * (2 - 1)2 = 9. Thus, the total price for two groups is 18.
Submitted Solution:
```
t = input()
t = t.split()
n = int(t[0])
c1 = int(t[1])
c2 = int(t[2])
t = input()
d = 0
for i in t:
if(i=="1"):
d = d+1
min = 10**1488
for i in range(1, d+1):
t = c1*i + i*c2*(((n//i)-1)**2) + (n%i)*(2*(n//i)-1)
if t<min:
min = t
print(min)
``` | instruction | 0 | 63,598 | 10 | 127,196 |
No | output | 1 | 63,598 | 10 | 127,197 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Pupils decided to go to amusement park. Some of them were with parents. In total, n people came to the park and they all want to get to the most extreme attraction and roll on it exactly once.
Tickets for group of x people are sold on the attraction, there should be at least one adult in each group (it is possible that the group consists of one adult). The ticket price for such group is c1 + c2Β·(x - 1)2 (in particular, if the group consists of one person, then the price is c1).
All pupils who came to the park and their parents decided to split into groups in such a way that each visitor join exactly one group, and the total price of visiting the most extreme attraction is as low as possible. You are to determine this minimum possible total price. There should be at least one adult in each group.
Input
The first line contains three integers n, c1 and c2 (1 β€ n β€ 200 000, 1 β€ c1, c2 β€ 107) β the number of visitors and parameters for determining the ticket prices for a group.
The second line contains the string of length n, which consists of zeros and ones. If the i-th symbol of the string is zero, then the i-th visitor is a pupil, otherwise the i-th person is an adult. It is guaranteed that there is at least one adult. It is possible that there are no pupils.
Output
Print the minimum price of visiting the most extreme attraction for all pupils and their parents. Each of them should roll on the attraction exactly once.
Examples
Input
3 4 1
011
Output
8
Input
4 7 2
1101
Output
18
Note
In the first test one group of three people should go to the attraction. Then they have to pay 4 + 1 * (3 - 1)2 = 8.
In the second test it is better to go to the attraction in two groups. The first group should consist of two adults (for example, the first and the second person), the second should consist of one pupil and one adult (the third and the fourth person). Then each group will have a size of two and for each the price of ticket is 7 + 2 * (2 - 1)2 = 9. Thus, the total price for two groups is 18.
Submitted Solution:
```
import math
[n, c1, c2] = input().split()
people = input()
def calc(x):
return int(c1)+int(c2)*(int(x)-1)*(x-1)
pupils = people.count('0')
adults = int(n)-pupils
optimal_people = math.floor(math.sqrt(int(c1)/int(c2)) + 1)
print(optimal_people)
remainder = int(math.remainder(len(people), optimal_people))
optimal_groups = int((len(people)-remainder)/optimal_people)
print(optimal_groups)
if adults < optimal_groups:
remainder = math.remainder(pupils, adults)
optimal_amount = ((pupils-remainder)/adults) +1
arr = adults*[optimal_amount]
arr = arr[:(adults-remainder)]
arr = arr + remainder*(optimal_amount + 1)
print(sum([calc[arg] for arg in arr]))
else:
arr = optimal_groups*[optimal_people]
arr = arr[:(optimal_groups-remainder)]
arr = arr + remainder*[optimal_people + 1]
print(sum([calc(arg) for arg in arr]))
``` | instruction | 0 | 63,599 | 10 | 127,198 |
No | output | 1 | 63,599 | 10 | 127,199 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 β€ A β€ 105), the desired number of ways.
Output
In the first line print integers N and M (1 β€ N β€ 106, 1 β€ M β€ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 β€ Di β€ 106), the denominations of the coins. All denominations must be distinct: for any i β j we must have Di β Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Examples
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139 | instruction | 0 | 63,640 | 10 | 127,280 |
Tags: constructive algorithms
Correct Solution:
```
print(int(input())*2-1,2,1,2)
``` | output | 1 | 63,640 | 10 | 127,281 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 β€ A β€ 105), the desired number of ways.
Output
In the first line print integers N and M (1 β€ N β€ 106, 1 β€ M β€ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 β€ Di β€ 106), the denominations of the coins. All denominations must be distinct: for any i β j we must have Di β Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Examples
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139 | instruction | 0 | 63,641 | 10 | 127,282 |
Tags: constructive algorithms
Correct Solution:
```
n=int(input())
a=2*n-1
print(a,2)
print(1,2)
``` | output | 1 | 63,641 | 10 | 127,283 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 β€ A β€ 105), the desired number of ways.
Output
In the first line print integers N and M (1 β€ N β€ 106, 1 β€ M β€ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 β€ Di β€ 106), the denominations of the coins. All denominations must be distinct: for any i β j we must have Di β Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Examples
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139 | instruction | 0 | 63,642 | 10 | 127,284 |
Tags: constructive algorithms
Correct Solution:
```
ways = int(input())
if ways == 1:
print(1, 1)
print(1)
else:
print(2*(ways-1), 2)
print(1, 2)
``` | output | 1 | 63,642 | 10 | 127,285 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 β€ A β€ 105), the desired number of ways.
Output
In the first line print integers N and M (1 β€ N β€ 106, 1 β€ M β€ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 β€ Di β€ 106), the denominations of the coins. All denominations must be distinct: for any i β j we must have Di β Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Examples
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139 | instruction | 0 | 63,643 | 10 | 127,286 |
Tags: constructive algorithms
Correct Solution:
```
#!/usr/bin/env python3
import sys
def main():
tgt = int(input())
N = (tgt -1)*2
if N == 0:
N = 1
print(N, 2)
print(1, 2)
if __name__ == '__main__':
main()
``` | output | 1 | 63,643 | 10 | 127,287 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 β€ A β€ 105), the desired number of ways.
Output
In the first line print integers N and M (1 β€ N β€ 106, 1 β€ M β€ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 β€ Di β€ 106), the denominations of the coins. All denominations must be distinct: for any i β j we must have Di β Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Examples
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139 | instruction | 0 | 63,644 | 10 | 127,288 |
Tags: constructive algorithms
Correct Solution:
```
print(2 * int(input()) - 1, '2\n1 2')
``` | output | 1 | 63,644 | 10 | 127,289 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 β€ A β€ 105), the desired number of ways.
Output
In the first line print integers N and M (1 β€ N β€ 106, 1 β€ M β€ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 β€ Di β€ 106), the denominations of the coins. All denominations must be distinct: for any i β j we must have Di β Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Examples
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139 | instruction | 0 | 63,645 | 10 | 127,290 |
Tags: constructive algorithms
Correct Solution:
```
a = int(input())
if (a == 1):
print(1, 2)
print(1, 2)
exit(0)
print(2 * a - 2, 2)
print(1, 2)
``` | output | 1 | 63,645 | 10 | 127,291 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 β€ A β€ 105), the desired number of ways.
Output
In the first line print integers N and M (1 β€ N β€ 106, 1 β€ M β€ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 β€ Di β€ 106), the denominations of the coins. All denominations must be distinct: for any i β j we must have Di β Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Examples
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139 | instruction | 0 | 63,646 | 10 | 127,292 |
Tags: constructive algorithms
Correct Solution:
```
n=int(input())
if n==1:
print(1,1)
print(1)
else:
print(n*2-2,2)
print(1,2)
``` | output | 1 | 63,646 | 10 | 127,293 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 β€ A β€ 105), the desired number of ways.
Output
In the first line print integers N and M (1 β€ N β€ 106, 1 β€ M β€ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 β€ Di β€ 106), the denominations of the coins. All denominations must be distinct: for any i β j we must have Di β Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Examples
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139 | instruction | 0 | 63,647 | 10 | 127,294 |
Tags: constructive algorithms
Correct Solution:
```
n = int(input())
print(max(1, (n - 1) * 2), 2)
print(1, 2)
``` | output | 1 | 63,647 | 10 | 127,295 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 β€ A β€ 105), the desired number of ways.
Output
In the first line print integers N and M (1 β€ N β€ 106, 1 β€ M β€ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 β€ Di β€ 106), the denominations of the coins. All denominations must be distinct: for any i β j we must have Di β Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Examples
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139
Submitted Solution:
```
print(2*int(input())-1,"2\n1 2")
``` | instruction | 0 | 63,648 | 10 | 127,296 |
Yes | output | 1 | 63,648 | 10 | 127,297 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 β€ A β€ 105), the desired number of ways.
Output
In the first line print integers N and M (1 β€ N β€ 106, 1 β€ M β€ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 β€ Di β€ 106), the denominations of the coins. All denominations must be distinct: for any i β j we must have Di β Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Examples
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139
Submitted Solution:
```
print(2*int(input())-1,"2 1 2")
``` | instruction | 0 | 63,649 | 10 | 127,298 |
Yes | output | 1 | 63,649 | 10 | 127,299 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 β€ A β€ 105), the desired number of ways.
Output
In the first line print integers N and M (1 β€ N β€ 106, 1 β€ M β€ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 β€ Di β€ 106), the denominations of the coins. All denominations must be distinct: for any i β j we must have Di β Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Examples
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139
Submitted Solution:
```
a=int(input())
if(a==1):
print(3,1)
print(3)
else:
print(2*(a-1),2)
print(1,2)
``` | instruction | 0 | 63,650 | 10 | 127,300 |
Yes | output | 1 | 63,650 | 10 | 127,301 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 β€ A β€ 105), the desired number of ways.
Output
In the first line print integers N and M (1 β€ N β€ 106, 1 β€ M β€ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 β€ Di β€ 106), the denominations of the coins. All denominations must be distinct: for any i β j we must have Di β Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Examples
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139
Submitted Solution:
```
n=int(input())
if n==1:
print("1 1")
print("1")
else:
print(2*(n-1),end="")
print(" 2")
print("1 2")
``` | instruction | 0 | 63,651 | 10 | 127,302 |
Yes | output | 1 | 63,651 | 10 | 127,303 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 β€ A β€ 105), the desired number of ways.
Output
In the first line print integers N and M (1 β€ N β€ 106, 1 β€ M β€ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 β€ Di β€ 106), the denominations of the coins. All denominations must be distinct: for any i β j we must have Di β Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Examples
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139
Submitted Solution:
```
a = int(input())
print((a - 1) * 2, 2)
print(1, 2)
``` | instruction | 0 | 63,652 | 10 | 127,304 |
No | output | 1 | 63,652 | 10 | 127,305 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 β€ A β€ 105), the desired number of ways.
Output
In the first line print integers N and M (1 β€ N β€ 106, 1 β€ M β€ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 β€ Di β€ 106), the denominations of the coins. All denominations must be distinct: for any i β j we must have Di β Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Examples
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139
Submitted Solution:
```
n = int(input())
x = (n - 1) * 2
print(x, 2)
print(1, 2)
``` | instruction | 0 | 63,653 | 10 | 127,306 |
No | output | 1 | 63,653 | 10 | 127,307 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 β€ A β€ 105), the desired number of ways.
Output
In the first line print integers N and M (1 β€ N β€ 106, 1 β€ M β€ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 β€ Di β€ 106), the denominations of the coins. All denominations must be distinct: for any i β j we must have Di β Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Examples
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139
Submitted Solution:
```
n = int(input())
number = 2*n -1
m = 2
denom = [1,2]
print(n,m)
print(*denom)
``` | instruction | 0 | 63,654 | 10 | 127,308 |
No | output | 1 | 63,654 | 10 | 127,309 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 β€ A β€ 105), the desired number of ways.
Output
In the first line print integers N and M (1 β€ N β€ 106, 1 β€ M β€ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 β€ Di β€ 106), the denominations of the coins. All denominations must be distinct: for any i β j we must have Di β Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Examples
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139
Submitted Solution:
```
print("240 6")
print("48 80 120 60 30 15")
print("4 2")
print("1 2")
print("122825 6")
print("122825 24656 4913 7225 425 25")
``` | instruction | 0 | 63,655 | 10 | 127,310 |
No | output | 1 | 63,655 | 10 | 127,311 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vladimir would like to prepare a present for his wife: they have an anniversary! He decided to buy her exactly n flowers.
Vladimir went to a flower shop, and he was amazed to see that there are m types of flowers being sold there, and there is unlimited supply of flowers of each type. Vladimir wants to choose flowers to maximize the happiness of his wife. He knows that after receiving the first flower of the i-th type happiness of his wife increases by a_i and after receiving each consecutive flower of this type her happiness increases by b_i. That is, if among the chosen flowers there are x_i > 0 flowers of type i, his wife gets a_i + (x_i - 1) β
b_i additional happiness (and if there are no flowers of type i, she gets nothing for this particular type).
Please help Vladimir to choose exactly n flowers to maximize the total happiness of his wife.
Input
The first line contains the only integer t (1 β€ t β€ 10 000), the number of test cases. It is followed by t descriptions of the test cases.
Each test case description starts with two integers n and m (1 β€ n β€ 10^9, 1 β€ m β€ 100 000), the number of flowers Vladimir needs to choose and the number of types of available flowers.
The following m lines describe the types of flowers: each line contains integers a_i and b_i (0 β€ a_i, b_i β€ 10^9) for i-th available type of flowers.
The test cases are separated by a blank line. It is guaranteed that the sum of values m among all test cases does not exceed 100 000.
Output
For each test case output a single integer: the maximum total happiness of Vladimir's wife after choosing exactly n flowers optimally.
Example
Input
2
4 3
5 0
1 4
2 2
5 3
5 2
4 2
3 1
Output
14
16
Note
In the first example case Vladimir can pick 1 flower of the first type and 3 flowers of the second type, in this case the total happiness equals 5 + (1 + 2 β
4) = 14.
In the second example Vladimir can pick 2 flowers of the first type, 2 flowers of the second type, and 1 flower of the third type, in this case the total happiness equals (5 + 1 β
2) + (4 + 1 β
2) + 3 = 16.
Submitted Solution:
```
import sys
input = sys.stdin.readline
t = int(input())
for trial in range(t):
n, m = (map(int, input().split()))
both = []
for i in range(m):
a1, b1 = map(int, input().split())
both.append([a1, b1])
if trial < t-1:
input()
ans = 0
sort_by_a = sorted(both, key=lambda x: (x[0], x[1]), reverse=True)
sort_by_b = sorted(both, key=lambda x: (x[1], x[0]), reverse=True)
sum_of_a = 0
a_index = 0
max_b = 0
for i in range(m):
while a_index < m and sort_by_a[a_index][0] > sort_by_b[i][1]:
sum_of_a += sort_by_a[a_index][0]
max_b = max(sort_by_a[a_index][1], max_b)
a_index += 1
if a_index == n or a_index == m:
break
if a_index == n:
ans = max(sum_of_a, ans)
break
if sort_by_b[i][0] > sort_by_b[i][1]:
ans = max(ans, sum_of_a + (n - a_index)*sort_by_b[i][1])
else:
ans = max(ans, sum_of_a + (n - a_index - 1) * sort_by_b[i][1] + sort_by_b[i][0])
print(ans)
``` | instruction | 0 | 64,157 | 10 | 128,314 |
Yes | output | 1 | 64,157 | 10 | 128,315 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vladimir would like to prepare a present for his wife: they have an anniversary! He decided to buy her exactly n flowers.
Vladimir went to a flower shop, and he was amazed to see that there are m types of flowers being sold there, and there is unlimited supply of flowers of each type. Vladimir wants to choose flowers to maximize the happiness of his wife. He knows that after receiving the first flower of the i-th type happiness of his wife increases by a_i and after receiving each consecutive flower of this type her happiness increases by b_i. That is, if among the chosen flowers there are x_i > 0 flowers of type i, his wife gets a_i + (x_i - 1) β
b_i additional happiness (and if there are no flowers of type i, she gets nothing for this particular type).
Please help Vladimir to choose exactly n flowers to maximize the total happiness of his wife.
Input
The first line contains the only integer t (1 β€ t β€ 10 000), the number of test cases. It is followed by t descriptions of the test cases.
Each test case description starts with two integers n and m (1 β€ n β€ 10^9, 1 β€ m β€ 100 000), the number of flowers Vladimir needs to choose and the number of types of available flowers.
The following m lines describe the types of flowers: each line contains integers a_i and b_i (0 β€ a_i, b_i β€ 10^9) for i-th available type of flowers.
The test cases are separated by a blank line. It is guaranteed that the sum of values m among all test cases does not exceed 100 000.
Output
For each test case output a single integer: the maximum total happiness of Vladimir's wife after choosing exactly n flowers optimally.
Example
Input
2
4 3
5 0
1 4
2 2
5 3
5 2
4 2
3 1
Output
14
16
Note
In the first example case Vladimir can pick 1 flower of the first type and 3 flowers of the second type, in this case the total happiness equals 5 + (1 + 2 β
4) = 14.
In the second example Vladimir can pick 2 flowers of the first type, 2 flowers of the second type, and 1 flower of the third type, in this case the total happiness equals (5 + 1 β
2) + (4 + 1 β
2) + 3 = 16.
Submitted Solution:
```
import io
import os
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
from bisect import bisect_left
def solve():
n, m = map(int, input().split())
a = []
for i in range(m):
ai, bi = map(int, input().split())
a.append((ai, bi))
a.sort(key = lambda x: x[0])
b = []
presum = [0]
for i in range(m):
b.append(a[i][1])
a[i] = a[i][0]
presum.append(presum[-1]+a[i])
ma = 0
for i in range(m):
s = a[i]
ind = bisect_left(a, b[i])
if m-ind>n-1:
s+=presum[m]-presum[m-n+1]
else:
s+=presum[m]-presum[ind]
s+=b[i]*(n-1-(m-ind))
if ind<=i:
if m-ind<=n-1:
s-=a[i]
s+=b[i]
else:
s-=a[i]
s+=a[m-n]
#print(s)
if s>ma:
ma = s
print(ma)
def main():
t = int(input())
for i in range(t):
solve()
if i!=t-1:
input()
main()
``` | instruction | 0 | 64,158 | 10 | 128,316 |
Yes | output | 1 | 64,158 | 10 | 128,317 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.