text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Each evening after the dinner the SIS's students gather together to play the game of Sport Mafia.
For the tournament, Alya puts candies into the box, which will serve as a prize for a winner. To do that, she performs n actions. The first action performed is to put a single candy into the box. For each of the remaining moves she can choose from two options:
* the first option, in case the box contains at least one candy, is to take exactly one candy out and eat it. This way the number of candies in the box decreased by 1;
* the second option is to put candies in the box. In this case, Alya will put 1 more candy, than she put in the previous time.
Thus, if the box is empty, then it can only use the second option.
For example, one possible sequence of Alya's actions look as follows:
* put one candy into the box;
* put two candies into the box;
* eat one candy from the box;
* eat one candy from the box;
* put three candies into the box;
* eat one candy from the box;
* put four candies into the box;
* eat one candy from the box;
* put five candies into the box;
This way she will perform 9 actions, the number of candies at the end will be 11, while Alya will eat 4 candies in total.
You know the total number of actions n and the number of candies at the end k. You need to find the total number of sweets Alya ate. That is the number of moves of the first option. It's guaranteed, that for the given n and k the answer always exists.
Please note, that during an action of the first option, Alya takes out and eats exactly one candy.
Input
The first line contains two integers n and k (1 β€ n β€ 10^9; 0 β€ k β€ 10^9) β the total number of moves and the number of candies in the box at the end.
It's guaranteed, that for the given n and k the answer exists.
Output
Print a single integer β the number of candies, which Alya ate. Please note, that in this problem there aren't multiple possible answers β the answer is unique for any input data.
Examples
Input
1 1
Output
0
Input
9 11
Output
4
Input
5 0
Output
3
Input
3 2
Output
1
Note
In the first example, Alya has made one move only. According to the statement, the first move is always putting one candy in the box. Hence Alya ate 0 candies.
In the second example the possible sequence of Alya's actions looks as follows:
* put 1 candy,
* put 2 candies,
* eat a candy,
* eat a candy,
* put 3 candies,
* eat a candy,
* put 4 candies,
* eat a candy,
* put 5 candies.
This way, she will make exactly n=9 actions and in the end the box will contain 1+2-1-1+3-1+4-1+5=11 candies. The answer is 4, since she ate 4 candies in total.
Submitted Solution:
```
n, k = map(int,input().split())
print(int(n-(-3+(9+8*n+8*k)**(1/2))//2))
```
Yes
| 96,200 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Each evening after the dinner the SIS's students gather together to play the game of Sport Mafia.
For the tournament, Alya puts candies into the box, which will serve as a prize for a winner. To do that, she performs n actions. The first action performed is to put a single candy into the box. For each of the remaining moves she can choose from two options:
* the first option, in case the box contains at least one candy, is to take exactly one candy out and eat it. This way the number of candies in the box decreased by 1;
* the second option is to put candies in the box. In this case, Alya will put 1 more candy, than she put in the previous time.
Thus, if the box is empty, then it can only use the second option.
For example, one possible sequence of Alya's actions look as follows:
* put one candy into the box;
* put two candies into the box;
* eat one candy from the box;
* eat one candy from the box;
* put three candies into the box;
* eat one candy from the box;
* put four candies into the box;
* eat one candy from the box;
* put five candies into the box;
This way she will perform 9 actions, the number of candies at the end will be 11, while Alya will eat 4 candies in total.
You know the total number of actions n and the number of candies at the end k. You need to find the total number of sweets Alya ate. That is the number of moves of the first option. It's guaranteed, that for the given n and k the answer always exists.
Please note, that during an action of the first option, Alya takes out and eats exactly one candy.
Input
The first line contains two integers n and k (1 β€ n β€ 10^9; 0 β€ k β€ 10^9) β the total number of moves and the number of candies in the box at the end.
It's guaranteed, that for the given n and k the answer exists.
Output
Print a single integer β the number of candies, which Alya ate. Please note, that in this problem there aren't multiple possible answers β the answer is unique for any input data.
Examples
Input
1 1
Output
0
Input
9 11
Output
4
Input
5 0
Output
3
Input
3 2
Output
1
Note
In the first example, Alya has made one move only. According to the statement, the first move is always putting one candy in the box. Hence Alya ate 0 candies.
In the second example the possible sequence of Alya's actions looks as follows:
* put 1 candy,
* put 2 candies,
* eat a candy,
* eat a candy,
* put 3 candies,
* eat a candy,
* put 4 candies,
* eat a candy,
* put 5 candies.
This way, she will make exactly n=9 actions and in the end the box will contain 1+2-1-1+3-1+4-1+5=11 candies. The answer is 4, since she ate 4 candies in total.
Submitted Solution:
```
from collections import Counter
import math
n, k = [int(x) for x in input().split()]
sum = 0
if k != 0:
for i in range(n):
sum = sum + i
if sum > k:
break
else:
for i in range(n):
sum = sum + i
if sum == n-sum+1:
break
if sum == 0:
sum = 1
print(sum - k)
```
No
| 96,201 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Each evening after the dinner the SIS's students gather together to play the game of Sport Mafia.
For the tournament, Alya puts candies into the box, which will serve as a prize for a winner. To do that, she performs n actions. The first action performed is to put a single candy into the box. For each of the remaining moves she can choose from two options:
* the first option, in case the box contains at least one candy, is to take exactly one candy out and eat it. This way the number of candies in the box decreased by 1;
* the second option is to put candies in the box. In this case, Alya will put 1 more candy, than she put in the previous time.
Thus, if the box is empty, then it can only use the second option.
For example, one possible sequence of Alya's actions look as follows:
* put one candy into the box;
* put two candies into the box;
* eat one candy from the box;
* eat one candy from the box;
* put three candies into the box;
* eat one candy from the box;
* put four candies into the box;
* eat one candy from the box;
* put five candies into the box;
This way she will perform 9 actions, the number of candies at the end will be 11, while Alya will eat 4 candies in total.
You know the total number of actions n and the number of candies at the end k. You need to find the total number of sweets Alya ate. That is the number of moves of the first option. It's guaranteed, that for the given n and k the answer always exists.
Please note, that during an action of the first option, Alya takes out and eats exactly one candy.
Input
The first line contains two integers n and k (1 β€ n β€ 10^9; 0 β€ k β€ 10^9) β the total number of moves and the number of candies in the box at the end.
It's guaranteed, that for the given n and k the answer exists.
Output
Print a single integer β the number of candies, which Alya ate. Please note, that in this problem there aren't multiple possible answers β the answer is unique for any input data.
Examples
Input
1 1
Output
0
Input
9 11
Output
4
Input
5 0
Output
3
Input
3 2
Output
1
Note
In the first example, Alya has made one move only. According to the statement, the first move is always putting one candy in the box. Hence Alya ate 0 candies.
In the second example the possible sequence of Alya's actions looks as follows:
* put 1 candy,
* put 2 candies,
* eat a candy,
* eat a candy,
* put 3 candies,
* eat a candy,
* put 4 candies,
* eat a candy,
* put 5 candies.
This way, she will make exactly n=9 actions and in the end the box will contain 1+2-1-1+3-1+4-1+5=11 candies. The answer is 4, since she ate 4 candies in total.
Submitted Solution:
```
# cook your dish here
n,k = tuple([int(x) for x in input().split()])
l=0
u=n
if k!=0:
while l<=u:
m=(l+u)//2
#print(m)
if m*(m+1)//2>=k:
u=m-1
elif m*(m+1)//2<k:
l=m+1
if k==0:
while l<=u:
m=(l+u)//2
#print(m)
if m*(m+1)//2>=n+1:
u=m-1
elif m*(m+1)//2<n+1:
l=m+1
print(m*(m-1)//2)
else:
print(m*(m+1)//2-k)
```
No
| 96,202 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Each evening after the dinner the SIS's students gather together to play the game of Sport Mafia.
For the tournament, Alya puts candies into the box, which will serve as a prize for a winner. To do that, she performs n actions. The first action performed is to put a single candy into the box. For each of the remaining moves she can choose from two options:
* the first option, in case the box contains at least one candy, is to take exactly one candy out and eat it. This way the number of candies in the box decreased by 1;
* the second option is to put candies in the box. In this case, Alya will put 1 more candy, than she put in the previous time.
Thus, if the box is empty, then it can only use the second option.
For example, one possible sequence of Alya's actions look as follows:
* put one candy into the box;
* put two candies into the box;
* eat one candy from the box;
* eat one candy from the box;
* put three candies into the box;
* eat one candy from the box;
* put four candies into the box;
* eat one candy from the box;
* put five candies into the box;
This way she will perform 9 actions, the number of candies at the end will be 11, while Alya will eat 4 candies in total.
You know the total number of actions n and the number of candies at the end k. You need to find the total number of sweets Alya ate. That is the number of moves of the first option. It's guaranteed, that for the given n and k the answer always exists.
Please note, that during an action of the first option, Alya takes out and eats exactly one candy.
Input
The first line contains two integers n and k (1 β€ n β€ 10^9; 0 β€ k β€ 10^9) β the total number of moves and the number of candies in the box at the end.
It's guaranteed, that for the given n and k the answer exists.
Output
Print a single integer β the number of candies, which Alya ate. Please note, that in this problem there aren't multiple possible answers β the answer is unique for any input data.
Examples
Input
1 1
Output
0
Input
9 11
Output
4
Input
5 0
Output
3
Input
3 2
Output
1
Note
In the first example, Alya has made one move only. According to the statement, the first move is always putting one candy in the box. Hence Alya ate 0 candies.
In the second example the possible sequence of Alya's actions looks as follows:
* put 1 candy,
* put 2 candies,
* eat a candy,
* eat a candy,
* put 3 candies,
* eat a candy,
* put 4 candies,
* eat a candy,
* put 5 candies.
This way, she will make exactly n=9 actions and in the end the box will contain 1+2-1-1+3-1+4-1+5=11 candies. The answer is 4, since she ate 4 candies in total.
Submitted Solution:
```
n, k = map(int,input().split())
s = n*(n+1)//2
print(s-k)
```
No
| 96,203 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Each evening after the dinner the SIS's students gather together to play the game of Sport Mafia.
For the tournament, Alya puts candies into the box, which will serve as a prize for a winner. To do that, she performs n actions. The first action performed is to put a single candy into the box. For each of the remaining moves she can choose from two options:
* the first option, in case the box contains at least one candy, is to take exactly one candy out and eat it. This way the number of candies in the box decreased by 1;
* the second option is to put candies in the box. In this case, Alya will put 1 more candy, than she put in the previous time.
Thus, if the box is empty, then it can only use the second option.
For example, one possible sequence of Alya's actions look as follows:
* put one candy into the box;
* put two candies into the box;
* eat one candy from the box;
* eat one candy from the box;
* put three candies into the box;
* eat one candy from the box;
* put four candies into the box;
* eat one candy from the box;
* put five candies into the box;
This way she will perform 9 actions, the number of candies at the end will be 11, while Alya will eat 4 candies in total.
You know the total number of actions n and the number of candies at the end k. You need to find the total number of sweets Alya ate. That is the number of moves of the first option. It's guaranteed, that for the given n and k the answer always exists.
Please note, that during an action of the first option, Alya takes out and eats exactly one candy.
Input
The first line contains two integers n and k (1 β€ n β€ 10^9; 0 β€ k β€ 10^9) β the total number of moves and the number of candies in the box at the end.
It's guaranteed, that for the given n and k the answer exists.
Output
Print a single integer β the number of candies, which Alya ate. Please note, that in this problem there aren't multiple possible answers β the answer is unique for any input data.
Examples
Input
1 1
Output
0
Input
9 11
Output
4
Input
5 0
Output
3
Input
3 2
Output
1
Note
In the first example, Alya has made one move only. According to the statement, the first move is always putting one candy in the box. Hence Alya ate 0 candies.
In the second example the possible sequence of Alya's actions looks as follows:
* put 1 candy,
* put 2 candies,
* eat a candy,
* eat a candy,
* put 3 candies,
* eat a candy,
* put 4 candies,
* eat a candy,
* put 5 candies.
This way, she will make exactly n=9 actions and in the end the box will contain 1+2-1-1+3-1+4-1+5=11 candies. The answer is 4, since she ate 4 candies in total.
Submitted Solution:
```
n,x=map(int,input().split())
formula=0
i=0
if(x==0):
while(n-(i+formula)!=0):
i=i+1
formula=(i*(i+1))//2
print(formula)
else:
while(formula<x):
i=i+1
formula=(i*(i+1))//2
#print(formula,i)
print(abs(i-n))
```
No
| 96,204 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of lowercase Latin letters and q queries for this string.
Recall that the substring s[l; r] of the string s is the string s_l s_{l + 1} ... s_r. For example, the substrings of "codeforces" are "code", "force", "f", "for", but not "coder" and "top".
There are two types of queries:
* 1~ pos~ c (1 β€ pos β€ |s|, c is lowercase Latin letter): replace s_{pos} with c (set s_{pos} := c);
* 2~ l~ r (1 β€ l β€ r β€ |s|): calculate the number of distinct characters in the substring s[l; r].
Input
The first line of the input contains one string s consisting of no more than 10^5 lowercase Latin letters.
The second line of the input contains one integer q (1 β€ q β€ 10^5) β the number of queries.
The next q lines contain queries, one per line. Each query is given in the format described in the problem statement. It is guaranteed that there is at least one query of the second type.
Output
For each query of the second type print the answer for it β the number of distinct characters in the required substring in this query.
Examples
Input
abacaba
5
2 1 4
1 4 b
1 5 b
2 4 6
2 1 7
Output
3
1
2
Input
dfcbbcfeeedbaea
15
1 6 e
1 4 b
2 6 14
1 7 b
1 12 c
2 6 8
2 1 6
1 7 c
1 2 f
1 10 a
2 7 9
1 10 a
1 14 b
1 1 f
2 1 11
Output
5
2
5
2
6
Tags: data structures
Correct Solution:
```
# 1234D. Distinct Characters Queries
import sys
input = sys.stdin.readline
class SegmentTree:
def __init__(
self,
init_val: "initial value: Array or str",
segfunc: "operation unique in case",
ide_ele: "identity element corresponding init_val" = 0,
):
self.segfunc = segfunc
self.size = 1 << (len(init_val) - 1).bit_length()
self.ide_ele = ide_ele
self.tree = self._build(init_val)
def _build(self, init_val: "Array or str"):
tree = [self.ide_ele] * (2 * self.size)
for idx, val in enumerate(init_val): # set
# modify val if needed (e.g. str -> ord())
val = 1 << (ord(val) - 97)
tree[idx + self.size - 1] = val
for idx in range(self.size - 2, -1, -1): # build
tree[idx] = self.segfunc(tree[2 * idx + 1], tree[2 * idx + 2])
return tree
def update(self, idx: int, val):
idx += self.size - 1
# modify val if needed as same as in _build()
val = 1 << (ord(val) - 97)
self.tree[idx] = val
while idx > 0:
idx = (idx - 1) // 2
self.tree[idx] = self.segfunc(
self.tree[2 * idx + 1], self.tree[2 * idx + 2]
)
def query(self, left: int, right: int):
if left >= right:
return self.ide_ele
left += self.size
right += self.size
ret = self.ide_ele
while left < right:
if left & 1:
ret = self.segfunc(ret, self.tree[left - 1])
left += 1
if right & 1:
right -= 1
ret = self.segfunc(ret, self.tree[right - 1])
left >>= 1
right >>= 1
return ret
def main():
S = input().rstrip()
Q = int(input())
queries = tuple(input().split() for _ in range(Q))
segfunc = lambda x, y: x | y
seg = SegmentTree(S, segfunc)
ans = []
for q_type, var1, var2 in queries:
if q_type == "1":
idx, char = int(var1) - 1, var2
seg.update(idx, char)
else:
left, right = int(var1) - 1, int(var2) # [left, right)
x = seg.query(left, right)
cnt = bin(x).count("1")
ans.append(cnt)
print("\n".join(map(str, ans)))
if __name__ == "__main__":
main()
```
| 96,205 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of lowercase Latin letters and q queries for this string.
Recall that the substring s[l; r] of the string s is the string s_l s_{l + 1} ... s_r. For example, the substrings of "codeforces" are "code", "force", "f", "for", but not "coder" and "top".
There are two types of queries:
* 1~ pos~ c (1 β€ pos β€ |s|, c is lowercase Latin letter): replace s_{pos} with c (set s_{pos} := c);
* 2~ l~ r (1 β€ l β€ r β€ |s|): calculate the number of distinct characters in the substring s[l; r].
Input
The first line of the input contains one string s consisting of no more than 10^5 lowercase Latin letters.
The second line of the input contains one integer q (1 β€ q β€ 10^5) β the number of queries.
The next q lines contain queries, one per line. Each query is given in the format described in the problem statement. It is guaranteed that there is at least one query of the second type.
Output
For each query of the second type print the answer for it β the number of distinct characters in the required substring in this query.
Examples
Input
abacaba
5
2 1 4
1 4 b
1 5 b
2 4 6
2 1 7
Output
3
1
2
Input
dfcbbcfeeedbaea
15
1 6 e
1 4 b
2 6 14
1 7 b
1 12 c
2 6 8
2 1 6
1 7 c
1 2 f
1 10 a
2 7 9
1 10 a
1 14 b
1 1 f
2 1 11
Output
5
2
5
2
6
Tags: data structures
Correct Solution:
```
import sys,math,itertools
from collections import Counter,deque,defaultdict
from bisect import bisect_left,bisect_right
from heapq import heappop,heappush,heapify
from copy import deepcopy
mod = 10**9+7
INF = float('inf')
def inp(): return int(sys.stdin.readline())
def inpl(): return list(map(int, sys.stdin.readline().split()))
def inpl_1(): return list(map(lambda x:int(x)-1, sys.stdin.readline().split()))
def inps(): return sys.stdin.readline()
def inpsl(x): tmp = sys.stdin.readline(); return list(tmp[:x])
def err(x): print(x); exit()
import sys
input = sys.stdin.readline
def solve():
ordA = ord('a')
# N = int(input())
Ss = input().rstrip()
N = len(Ss)
Q = int(input())
querys = [tuple(input().split()) for _ in range(Q)]
idEle = 0
def _binOpe(x, y):
return x | y
def makeSegTree(numEle):
numPow2 = 2 ** (numEle-1).bit_length()
data = [idEle] * (2*numPow2)
return data, numPow2
def setInit(As):
for iST, A in enumerate(As, numPow2):
data[iST] = A
for iST in reversed(range(1, numPow2)):
data[iST] = _binOpe(data[2*iST], data[2*iST+1])
def _update1(iA, A):
iST = iA + numPow2
data[iST] = A
def update(iA, A):
_update1(iA, A)
iST = iA + numPow2
while iST > 1:
iST >>= 1
data[iST] = _binOpe(data[2*iST], data[2*iST+1])
def getValue(iSt, iEn):
L = iSt + numPow2
R = iEn + numPow2
ans = idEle
while L < R:
if L & 1:
ans = _binOpe(ans, data[L])
L += 1
if R & 1:
R -= 1
ans = _binOpe(ans, data[R])
L >>= 1
R >>= 1
return ans
data, numPow2 = makeSegTree(N)
As = [(1<<(ord(S)-ordA)) for S in Ss]
setInit(As)
anss = []
for tp, *vs in querys:
if tp == '1':
i, c = vs
i = int(i)
update(i-1, 1<<(ord(c)-ordA))
else:
L, R = vs
L, R = int(L), int(R)
v = getValue(L-1, R)
anss.append(bin(v).count('1'))
print('\n'.join(map(str, anss)))
solve()
```
| 96,206 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of lowercase Latin letters and q queries for this string.
Recall that the substring s[l; r] of the string s is the string s_l s_{l + 1} ... s_r. For example, the substrings of "codeforces" are "code", "force", "f", "for", but not "coder" and "top".
There are two types of queries:
* 1~ pos~ c (1 β€ pos β€ |s|, c is lowercase Latin letter): replace s_{pos} with c (set s_{pos} := c);
* 2~ l~ r (1 β€ l β€ r β€ |s|): calculate the number of distinct characters in the substring s[l; r].
Input
The first line of the input contains one string s consisting of no more than 10^5 lowercase Latin letters.
The second line of the input contains one integer q (1 β€ q β€ 10^5) β the number of queries.
The next q lines contain queries, one per line. Each query is given in the format described in the problem statement. It is guaranteed that there is at least one query of the second type.
Output
For each query of the second type print the answer for it β the number of distinct characters in the required substring in this query.
Examples
Input
abacaba
5
2 1 4
1 4 b
1 5 b
2 4 6
2 1 7
Output
3
1
2
Input
dfcbbcfeeedbaea
15
1 6 e
1 4 b
2 6 14
1 7 b
1 12 c
2 6 8
2 1 6
1 7 c
1 2 f
1 10 a
2 7 9
1 10 a
1 14 b
1 1 f
2 1 11
Output
5
2
5
2
6
Tags: data structures
Correct Solution:
```
import heapq
import math
import sys
input = sys.stdin.readline
def li():return [int(i) for i in input().rstrip('\n').split()]
def value():return int(input())
def givesum(a,b):
c=[0]*26
for i in range(len(a)):
c[i] = a[i]+b[i]
return c
def dolist(a):
c = [0]*26
c[ord(a)-ord('a')] = 1
return c
class Node:
def __init__(self,s,e):
self.start=s
self.end=e
self.lis=[0]*26
self.left=None
self.right=None
def build(nums,l,r):
if l == r:
temp = Node(l,l)
# print(temp.start,ord(nums[l])-ord('a'),nums)
temp.lis[ord(nums[l])-ord('a')] = 1
else:
mid=(l+r)>>1
# print(mid,l,r)
temp=Node(l,r)
temp.left=build(nums,l,mid)
temp.right=build(nums,mid+1,r)
temp.lis=givesum(temp.left.lis,temp.right.lis)
return temp
def update(root,start,value):
if root.start==root.end==start:
root.lis=dolist(value)
elif root.start<=start and root.end>=start:
mid=(root.start+root.end)>>1
root.left=update(root.left,start,value)
root.right=update(root.right,start,value)
root.lis=givesum(root.left.lis,root.right.lis)
return root
def query(root,start,end):
if root.start>=start and root.end<=end:return root.lis
elif root.start>end or root.end<start:return [0]*26;
else:
mid=(start+end)>>1
return givesum(query(root.left,start,end),query(root.right,start,end))
s = input().rstrip('\n')
root = build(s,0,len(s)-1)
ansarun = []
# print('arun')
for _ in range(int(input())):
templist = list(input().split())
if templist[0] == '1':
root = update(root,int(templist[1])-1,templist[2])
else:
temp1 = query(root,int(templist[1])-1,int(templist[2])-1)
total = 0
for i in temp1:
if i:total+=1
ansarun.append(total)
for i in ansarun:print(i)
```
| 96,207 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of lowercase Latin letters and q queries for this string.
Recall that the substring s[l; r] of the string s is the string s_l s_{l + 1} ... s_r. For example, the substrings of "codeforces" are "code", "force", "f", "for", but not "coder" and "top".
There are two types of queries:
* 1~ pos~ c (1 β€ pos β€ |s|, c is lowercase Latin letter): replace s_{pos} with c (set s_{pos} := c);
* 2~ l~ r (1 β€ l β€ r β€ |s|): calculate the number of distinct characters in the substring s[l; r].
Input
The first line of the input contains one string s consisting of no more than 10^5 lowercase Latin letters.
The second line of the input contains one integer q (1 β€ q β€ 10^5) β the number of queries.
The next q lines contain queries, one per line. Each query is given in the format described in the problem statement. It is guaranteed that there is at least one query of the second type.
Output
For each query of the second type print the answer for it β the number of distinct characters in the required substring in this query.
Examples
Input
abacaba
5
2 1 4
1 4 b
1 5 b
2 4 6
2 1 7
Output
3
1
2
Input
dfcbbcfeeedbaea
15
1 6 e
1 4 b
2 6 14
1 7 b
1 12 c
2 6 8
2 1 6
1 7 c
1 2 f
1 10 a
2 7 9
1 10 a
1 14 b
1 1 f
2 1 11
Output
5
2
5
2
6
Tags: data structures
Correct Solution:
```
class SortedList:
def __init__(self, iterable=[], _load=200):
"""Initialize sorted list instance."""
values = sorted(iterable)
self._len = _len = len(values)
self._load = _load
self._lists = _lists = [values[i:i + _load] for i in range(0, _len, _load)]
self._list_lens = [len(_list) for _list in _lists]
self._mins = [_list[0] for _list in _lists]
self._fen_tree = []
self._rebuild = True
def _fen_build(self):
"""Build a fenwick tree instance."""
self._fen_tree[:] = self._list_lens
_fen_tree = self._fen_tree
for i in range(len(_fen_tree)):
if i | i + 1 < len(_fen_tree):
_fen_tree[i | i + 1] += _fen_tree[i]
self._rebuild = False
def _fen_update(self, index, value):
"""Update `fen_tree[index] += value`."""
if not self._rebuild:
_fen_tree = self._fen_tree
while index < len(_fen_tree):
_fen_tree[index] += value
index |= index + 1
def _fen_query(self, end):
"""Return `sum(_fen_tree[:end])`."""
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
x = 0
while end:
x += _fen_tree[end - 1]
end &= end - 1
return x
def _fen_findkth(self, k):
"""Return a pair of (the largest `idx` such that `sum(_fen_tree[:idx]) <= k`, `k - sum(_fen_tree[:idx])`)."""
_list_lens = self._list_lens
if k < _list_lens[0]:
return 0, k
if k >= self._len - _list_lens[-1]:
return len(_list_lens) - 1, k + _list_lens[-1] - self._len
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
idx = -1
for d in reversed(range(len(_fen_tree).bit_length())):
right_idx = idx + (1 << d)
if right_idx < len(_fen_tree) and k >= _fen_tree[right_idx]:
idx = right_idx
k -= _fen_tree[idx]
return idx + 1, k
def _delete(self, pos, idx):
"""Delete value at the given `(pos, idx)`."""
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len -= 1
self._fen_update(pos, -1)
del _lists[pos][idx]
_list_lens[pos] -= 1
if _list_lens[pos]:
_mins[pos] = _lists[pos][0]
else:
del _lists[pos]
del _list_lens[pos]
del _mins[pos]
self._rebuild = True
def _loc_left(self, value):
"""Return an index pair that corresponds to the first position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
lo, pos = -1, len(_lists) - 1
while lo + 1 < pos:
mi = (lo + pos) >> 1
if value <= _mins[mi]:
pos = mi
else:
lo = mi
if pos and value <= _lists[pos - 1][-1]:
pos -= 1
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value <= _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def _loc_right(self, value):
"""Return an index pair that corresponds to the last position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
pos, hi = 0, len(_lists)
while pos + 1 < hi:
mi = (pos + hi) >> 1
if value < _mins[mi]:
hi = mi
else:
pos = mi
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value < _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def add(self, value):
"""Add `value` to sorted list."""
_load = self._load
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len += 1
if _lists:
pos, idx = self._loc_right(value)
self._fen_update(pos, 1)
_list = _lists[pos]
_list.insert(idx, value)
_list_lens[pos] += 1
_mins[pos] = _list[0]
if _load + _load < len(_list):
_lists.insert(pos + 1, _list[_load:])
_list_lens.insert(pos + 1, len(_list) - _load)
_mins.insert(pos + 1, _list[_load])
_list_lens[pos] = _load
del _list[_load:]
self._rebuild = True
else:
_lists.append([value])
_mins.append(value)
_list_lens.append(1)
self._rebuild = True
def discard(self, value):
"""Remove `value` from sorted list if it is a member."""
_lists = self._lists
if _lists:
pos, idx = self._loc_right(value)
if idx and _lists[pos][idx - 1] == value:
self._delete(pos, idx - 1)
def remove(self, value):
"""Remove `value` from sorted list; `value` must be a member."""
_len = self._len
self.discard(value)
if _len == self._len:
raise ValueError('{0!r} not in list'.format(value))
def pop(self, index=-1):
"""Remove and return value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
value = self._lists[pos][idx]
self._delete(pos, idx)
return value
def bisect_left(self, value):
"""Return the first index to insert `value` in the sorted list."""
pos, idx = self._loc_left(value)
return self._fen_query(pos) + idx
def bisect_right(self, value):
"""Return the last index to insert `value` in the sorted list."""
pos, idx = self._loc_right(value)
return self._fen_query(pos) + idx
def count(self, value):
"""Return number of occurrences of `value` in the sorted list."""
return self.bisect_right(value) - self.bisect_left(value)
def __len__(self):
"""Return the size of the sorted list."""
return self._len
def __getitem__(self, index):
"""Lookup value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
return self._lists[pos][idx]
def __delitem__(self, index):
"""Remove value at `index` from sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
self._delete(pos, idx)
def __contains__(self, value):
"""Return true if `value` is an element of the sorted list."""
_lists = self._lists
if _lists:
pos, idx = self._loc_left(value)
return idx < len(_lists[pos]) and _lists[pos][idx] == value
return False
def __iter__(self):
"""Return an iterator over the sorted list."""
return (value for _list in self._lists for value in _list)
def __reversed__(self):
"""Return a reverse iterator over the sorted list."""
return (value for _list in reversed(self._lists) for value in reversed(_list))
def __repr__(self):
"""Return string representation of sorted list."""
return 'SortedList({0})'.format(list(self))
class OrderedList(SortedList): #Codeforces, Ordered Multiset
def __init__(self, arg):
super().__init__(arg)
def rangeCountByValue(self, leftVal, rightVal): #returns number of items in range [leftVal,rightVal] inclusive
leftCummulative = self.bisect_left(leftVal)
rightCummulative = self.bisect_left(rightVal + 1)
return rightCummulative - leftCummulative
def charToInt(c): #'a'->0
return ord(c)-ord('a')
def intToChar(x): #0->'a'
return chr(ord('a')+x)
def main():
s=input()
n=len(s)
arr=[charToInt(c) for c in s]
olArr=[OrderedList([]) for _ in range(26)]
for i in range(n):
olArr[arr[i]].add(i)
q=int(input())
allans=[]
for _ in range(q):
typee,x,y=input().split()
if typee=='1':
replaceIdx=int(x)-1
replaceChar=charToInt(y)
prevChar=arr[replaceIdx]
olArr[prevChar].remove(replaceIdx)
olArr[replaceChar].add(replaceIdx)
arr[replaceIdx]=replaceChar
else:
l=int(x)-1
r=int(y)-1
cnt=0
for i in range(26):
left=olArr[i].bisect_left(l)
right=olArr[i].bisect_right(r)
if left<right:
cnt+=1
allans.append(cnt)
multiLineArrayPrint(allans)
return
import sys
# input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.
def oneLineArrayPrint(arr):
print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
def readIntArr():
return [int(x) for x in input().split()]
# def readFloatArr():
# return [float(x) for x in input().split()]
def makeArr(defaultValFactory,dimensionArr): # eg. makeArr(lambda:0,[n,m])
dv=defaultValFactory;da=dimensionArr
if len(da)==1:return [dv() for _ in range(da[0])]
else:return [makeArr(dv,da[1:]) for _ in range(da[0])]
def queryInteractive(i,j):
print('? {} {}'.format(i,j))
sys.stdout.flush()
return int(input())
def answerInteractive(ans):
print('! {}'.format(' '.join([str(x) for x in ans])))
sys.stdout.flush()
inf=float('inf')
MOD=10**9+7
# MOD=998244353
for _abc in range(1):
main()
```
| 96,208 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of lowercase Latin letters and q queries for this string.
Recall that the substring s[l; r] of the string s is the string s_l s_{l + 1} ... s_r. For example, the substrings of "codeforces" are "code", "force", "f", "for", but not "coder" and "top".
There are two types of queries:
* 1~ pos~ c (1 β€ pos β€ |s|, c is lowercase Latin letter): replace s_{pos} with c (set s_{pos} := c);
* 2~ l~ r (1 β€ l β€ r β€ |s|): calculate the number of distinct characters in the substring s[l; r].
Input
The first line of the input contains one string s consisting of no more than 10^5 lowercase Latin letters.
The second line of the input contains one integer q (1 β€ q β€ 10^5) β the number of queries.
The next q lines contain queries, one per line. Each query is given in the format described in the problem statement. It is guaranteed that there is at least one query of the second type.
Output
For each query of the second type print the answer for it β the number of distinct characters in the required substring in this query.
Examples
Input
abacaba
5
2 1 4
1 4 b
1 5 b
2 4 6
2 1 7
Output
3
1
2
Input
dfcbbcfeeedbaea
15
1 6 e
1 4 b
2 6 14
1 7 b
1 12 c
2 6 8
2 1 6
1 7 c
1 2 f
1 10 a
2 7 9
1 10 a
1 14 b
1 1 f
2 1 11
Output
5
2
5
2
6
Tags: data structures
Correct Solution:
```
import sys
input = sys.stdin.readline
class SegmentTree:
a = ord('a')
def __init__(self):
data = [1 << ord(c) - self.a for c in input()[:-1]]
self.n = 1 << len(bin(len(data) - 1)) - 2
self.data = [0] * self.n + data + [0] * (self.n - len(data))
for k in range(self.n - 1, 0, -1):
self.data[k] = self.data[2 * k + 1] | self.data[2 * k]
def update(self, k, c):
k += self.n
self.data[k] = 1 << ord(c) - self.a
while k > 1:
k >>= 1
self.data[k] = self.data[2 * k + 1] | self.data[2 * k]
def query(self, l, r):
l += self.n
r += self.n
s = self.data[r] | self.data[l]
while l < r - 1:
if r & 1: s |= self.data[r - 1]
if not l & 1: s |= self.data[l + 1]
l >>= 1
r >>= 1
return s
tree = SegmentTree()
for i in range(int(input())):
q = input()[:-1].split()
if q[0] == "1":
tree.update(int(q[1]) - 1, q[2])
else:
s = tree.query(int(q[1]) - 1, int(q[2]) - 1)
print(bin(s)[2:].count("1"))
```
| 96,209 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of lowercase Latin letters and q queries for this string.
Recall that the substring s[l; r] of the string s is the string s_l s_{l + 1} ... s_r. For example, the substrings of "codeforces" are "code", "force", "f", "for", but not "coder" and "top".
There are two types of queries:
* 1~ pos~ c (1 β€ pos β€ |s|, c is lowercase Latin letter): replace s_{pos} with c (set s_{pos} := c);
* 2~ l~ r (1 β€ l β€ r β€ |s|): calculate the number of distinct characters in the substring s[l; r].
Input
The first line of the input contains one string s consisting of no more than 10^5 lowercase Latin letters.
The second line of the input contains one integer q (1 β€ q β€ 10^5) β the number of queries.
The next q lines contain queries, one per line. Each query is given in the format described in the problem statement. It is guaranteed that there is at least one query of the second type.
Output
For each query of the second type print the answer for it β the number of distinct characters in the required substring in this query.
Examples
Input
abacaba
5
2 1 4
1 4 b
1 5 b
2 4 6
2 1 7
Output
3
1
2
Input
dfcbbcfeeedbaea
15
1 6 e
1 4 b
2 6 14
1 7 b
1 12 c
2 6 8
2 1 6
1 7 c
1 2 f
1 10 a
2 7 9
1 10 a
1 14 b
1 1 f
2 1 11
Output
5
2
5
2
6
Tags: data structures
Correct Solution:
```
import atexit
import io
import sys
_INPUT_LINES = sys.stdin.read().splitlines()
input = iter(_INPUT_LINES).__next__
_OUTPUT_BUFFER = io.StringIO()
sys.stdout = _OUTPUT_BUFFER
@atexit.register
def write():
sys.__stdout__.write(_OUTPUT_BUFFER.getvalue())
class SegTree:
# limit for array size
N = 100002
def __init__(self):
self.tree = [0] * (2 * self.N)
# function to build the tree
def build(self, arr):
# insert leaf nodes in tree
for i in range(n):
self.tree[n + i] = arr[i]
# build the tree by calculating parents
for i in range(n - 1, 0, -1):
self.tree[i] = self.tree[i << 1] + self.tree[i << 1 | 1]
# function to update a tree node
def update_tree_node(self, p, value):
# set value at position p
self.tree[p + n] = value
p = p + n
# move upward and update parents
i = p
while i > 1:
self.tree[i >> 1] = self.tree[i] + self.tree[i ^ 1]
i >>= 1
# function to get sum on interval [l, r)
def query(self, l, r):
res = 0
# loop to find the sum in the range
l += n
r += n
while l < r:
if l & 1:
res += self.tree[l]
l += 1
if r & 1:
r -= 1
res += self.tree[r]
l >>= 1
r >>= 1
return res
if __name__ == "__main__":
s = list(input())
n = len(s)
trees = []
for i in range(26):
trees.append(SegTree())
trees[-1].build([0] * n)
for idx, c in enumerate(s):
trees[ord(c) - ord('a')].update_tree_node(idx, 1)
t = int(input())
for _ in range(t):
q, p1, p2 = input().split()
q, p1 = int(q), int(p1) - 1
if q == 1:
trees[ord(s[p1]) - ord('a')].update_tree_node(p1, 0)
trees[ord(p2) - ord('a')].update_tree_node(p1, 1)
s[p1] = p2
else:
p2 = int(p2) - 1
ans = 0
for seg in trees:
if seg.query(p1, p2 + 1) != 0:
ans += 1
print(ans)
```
| 96,210 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of lowercase Latin letters and q queries for this string.
Recall that the substring s[l; r] of the string s is the string s_l s_{l + 1} ... s_r. For example, the substrings of "codeforces" are "code", "force", "f", "for", but not "coder" and "top".
There are two types of queries:
* 1~ pos~ c (1 β€ pos β€ |s|, c is lowercase Latin letter): replace s_{pos} with c (set s_{pos} := c);
* 2~ l~ r (1 β€ l β€ r β€ |s|): calculate the number of distinct characters in the substring s[l; r].
Input
The first line of the input contains one string s consisting of no more than 10^5 lowercase Latin letters.
The second line of the input contains one integer q (1 β€ q β€ 10^5) β the number of queries.
The next q lines contain queries, one per line. Each query is given in the format described in the problem statement. It is guaranteed that there is at least one query of the second type.
Output
For each query of the second type print the answer for it β the number of distinct characters in the required substring in this query.
Examples
Input
abacaba
5
2 1 4
1 4 b
1 5 b
2 4 6
2 1 7
Output
3
1
2
Input
dfcbbcfeeedbaea
15
1 6 e
1 4 b
2 6 14
1 7 b
1 12 c
2 6 8
2 1 6
1 7 c
1 2 f
1 10 a
2 7 9
1 10 a
1 14 b
1 1 f
2 1 11
Output
5
2
5
2
6
Tags: data structures
Correct Solution:
```
class SegmentTree:
def __init__(self, n):
self.n = 2**(n-1).bit_length()
self.data = [0 for _ in range(2*self.n)]
def build(self, arr):
for i in range(len(arr)):
self.data[self.n+i] = 1<<arr[i]
for i in range(self.n-1, 0, -1):
self.data[i] = self.data[i<<1] | self.data[i<<1 | 1]
def updateTreeNode(self, p, value):
p = p+self.n
self.data[p] = 1<<value
while p > 1:
self.data[p>>1] = self.data[p] | self.data[p^1]
p>>=1
def __repr__(self):
return ' '.join(str(bin(val).count("1")) for val in self.data)
def query(self, l, r):
ans = 0
l+=self.n
r+=self.n
while l < r:
if l & 1:
ans |= self.data[l]
l+=1
if r & 1:
r-=1
ans |= self.data[r]
l >>= 1
r >>= 1
return ans
aval = ord('a')
s = [ord(c)-aval for c in input()]
segTree = SegmentTree(len(s))
segTree.build(s)
for _ in range(int(input())):
q, a, b = input().split()
#print(segTree)
if q=="1":
segTree.updateTreeNode(int(a)-1, ord(b)-aval)
else:
print(bin(segTree.query(int(a)-1, int(b))).count("1"))
```
| 96,211 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of lowercase Latin letters and q queries for this string.
Recall that the substring s[l; r] of the string s is the string s_l s_{l + 1} ... s_r. For example, the substrings of "codeforces" are "code", "force", "f", "for", but not "coder" and "top".
There are two types of queries:
* 1~ pos~ c (1 β€ pos β€ |s|, c is lowercase Latin letter): replace s_{pos} with c (set s_{pos} := c);
* 2~ l~ r (1 β€ l β€ r β€ |s|): calculate the number of distinct characters in the substring s[l; r].
Input
The first line of the input contains one string s consisting of no more than 10^5 lowercase Latin letters.
The second line of the input contains one integer q (1 β€ q β€ 10^5) β the number of queries.
The next q lines contain queries, one per line. Each query is given in the format described in the problem statement. It is guaranteed that there is at least one query of the second type.
Output
For each query of the second type print the answer for it β the number of distinct characters in the required substring in this query.
Examples
Input
abacaba
5
2 1 4
1 4 b
1 5 b
2 4 6
2 1 7
Output
3
1
2
Input
dfcbbcfeeedbaea
15
1 6 e
1 4 b
2 6 14
1 7 b
1 12 c
2 6 8
2 1 6
1 7 c
1 2 f
1 10 a
2 7 9
1 10 a
1 14 b
1 1 f
2 1 11
Output
5
2
5
2
6
Tags: data structures
Correct Solution:
```
# -*- coding: utf-8 -*-
# Baqir Khan
# Software Engineer (Backend)
from sys import stdin
inp = stdin.readline
class SegmentTree:
def __init__(self, values, default, func):
"""
Segment tree to find Value function for any segment of some array A.
Suppose we want to find some V for every slice of a given array
and that if we know values on slices [a:b] and [b:c] then it's easy to
calculate value for slice [a:c] - in other words there exists associative binary
operation ~ such that for any given a, b (where 0 <= a <= b < len(A))
V(A[a:b]) = V(A[a:c]) ~ V(A[c:b]) for any c between a and b, particularly:
V(A[a:b]) = V(A[a]) ~ V(A[a + 1]) ~ ... ~ V(A[b - 1])
And suppose that we need to change separate values in array and keep results up to date.
Example 1:
~ is min of two numbers and Value(array[a:b]) is minimum element on slice array[a:b]
Example 2:
~ is addition and Value(array[a:b]) means sum of elements in A[a:b])
Args:
values:
pre-calculated values for V function on elements of array
default:
value for empty segment, such that func(v, default) = v
func:
combining binary operation as function taking two arguments: func(a, b) means a ~ b
"""
k = 0
while 2 ** k < len(values):
k += 1
# heap_size = 2 ** (k + 1) - 1
heap = (2 ** k - 1) * [default] + list(values) + (2 ** k - len(values)) * [default]
for index in range(2 ** k - 1 - 1, -1, -1):
heap[index] = func(heap[2 * index + 1], heap[2 * index + 2])
self._length = len(values)
self.func = func
self.heap = heap
self.k = k
self.default = default
def __getitem__(self, indexer):
shift = 2 ** self.k - 1
if isinstance(indexer, int):
return self.heap[shift + indexer]
elif isinstance(indexer, slice):
left, right = shift + indexer.start, shift + indexer.stop - 1
if left > right:
return self.default
func = self.func
heap = self.heap
res = func(heap[left], heap[right])
while left < right:
if ~(left & 1):
res = func(res, heap[left])
left += 1
if right & 1:
res = func(res, heap[right])
right -= 1
left = (left + 1) // 2 - 1
right = (right + 1) // 2 - 1
if left == right:
res = func(res, heap[left])
return res
raise TypeError()
def __setitem__(self, array_index, value):
if array_index > len(self):
raise IndexError
index = (2 ** self.k) - 1 + array_index
heap = self.heap
func = self.func
heap[index] = value
while index:
# until there is an ancestor node in heap for current index
index = (index + 1) // 2 - 1
heap[index] = func(heap[2 * index + 1], heap[2 * index + 2])
def __len__(self):
return self._length
def get_binary(c):
return 1 << (ord(c) - ord('a'))
def get_ones_place(c):
place = 0
while c:
place += c & 1
c >>= 1
return place
s = list(inp()[:-1])
q = int(inp())
segment_tree = SegmentTree([get_binary(c) for c in s], 0, lambda x, y: x | y)
while q:
q -= 1
a, b, c = inp().split()
if a == "1":
segment_tree[int(b) - 1] = get_binary(c)
else:
print(get_ones_place(segment_tree[int(b) - 1: int(c)]))
```
| 96,212 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of lowercase Latin letters and q queries for this string.
Recall that the substring s[l; r] of the string s is the string s_l s_{l + 1} ... s_r. For example, the substrings of "codeforces" are "code", "force", "f", "for", but not "coder" and "top".
There are two types of queries:
* 1~ pos~ c (1 β€ pos β€ |s|, c is lowercase Latin letter): replace s_{pos} with c (set s_{pos} := c);
* 2~ l~ r (1 β€ l β€ r β€ |s|): calculate the number of distinct characters in the substring s[l; r].
Input
The first line of the input contains one string s consisting of no more than 10^5 lowercase Latin letters.
The second line of the input contains one integer q (1 β€ q β€ 10^5) β the number of queries.
The next q lines contain queries, one per line. Each query is given in the format described in the problem statement. It is guaranteed that there is at least one query of the second type.
Output
For each query of the second type print the answer for it β the number of distinct characters in the required substring in this query.
Examples
Input
abacaba
5
2 1 4
1 4 b
1 5 b
2 4 6
2 1 7
Output
3
1
2
Input
dfcbbcfeeedbaea
15
1 6 e
1 4 b
2 6 14
1 7 b
1 12 c
2 6 8
2 1 6
1 7 c
1 2 f
1 10 a
2 7 9
1 10 a
1 14 b
1 1 f
2 1 11
Output
5
2
5
2
6
Submitted Solution:
```
'''input
abacaba
5
2 1 4
1 4 b
1 5 b
2 4 6
2 1 7
'''
from sys import stdin
from collections import deque
import math
# builds the segment tree.... time complexity O(n)
def build_segment_tree(tree, string, index, start, end):
if start > end:
return
# for node element
if start == end:
tree[index][ord(string[start]) - ord('a')] += 1
# for internal node
else:
mid = (start + end) // 2
build_segment_tree(tree, string, 2 * index, start, mid)
build_segment_tree(tree, string, 2 * index + 1, mid + 1, end)
for i in range(26):
tree[index][i] = tree[2 * index][i] + tree[2 * index + 1][i]
def combine(first, second):
aux = [0] * 26
for i in range(26):
aux[i] = first[i] + second[i]
return aux
# query function. Outputs the minimum element in the given range. Time complexity is O(log(n))
def query(tree, a, index, start, end, qs, qe):
if qs > end or qe < start:
return [0] * 26
# total overlap
if qs <= start and end <= qe:
return tree[index][:]
# partial overlap
else:
mid = (start + end) // 2
left = query(tree, a, 2 * index, start, mid, qs, qe)
right = query(tree, a, 2 * index + 1, mid + 1, end, qs, qe)
return combine(left, right)
# update the given node. Time complexity is O(log(n))
def update_node(tree, string, index, start, end, n_index, c):
# No overlap
if n_index < start or n_index > end:
return
# reach the node
if start == end:
tree[index][ord(string[n_index]) - ord('a')] -= 1
tree[index][ord(c) - ord('a')] += 1
string[n_index] = c
return
# partial or complete overlap
else:
mid = (start + end) // 2
update_node(tree, string, 2 * index, start, mid, n_index, c)
update_node(tree, string, 2 * index + 1, mid + 1, end, n_index, c)
aux = combine(tree[ 2 * index], tree[ 2 * index + 1])
for i in range(26):
tree[index][i] = aux[i]
return
# main starts
string = list(stdin.readline().strip())
n = len(string)
tree = [[0 for x in range(26)] for y in range(4 * n + 1)]
index = 1
start = 0
end = n - 1
build_segment_tree(tree, string, index, start, end)
# print(tree[1])
q = int(stdin.readline().strip())
for _ in range(q):
t, x, y = list(stdin.readline().split())
if t == '1':
update_node(tree, string, index, start, end, int(x) - 1, y)
# print(tree[1])
else:
result = query(tree, string, index, start, end, int(x) - 1, int(y) - 1)
# print(result)
count = 0
for i in result:
if i > 0:
count += 1
print(count)
```
Yes
| 96,213 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of lowercase Latin letters and q queries for this string.
Recall that the substring s[l; r] of the string s is the string s_l s_{l + 1} ... s_r. For example, the substrings of "codeforces" are "code", "force", "f", "for", but not "coder" and "top".
There are two types of queries:
* 1~ pos~ c (1 β€ pos β€ |s|, c is lowercase Latin letter): replace s_{pos} with c (set s_{pos} := c);
* 2~ l~ r (1 β€ l β€ r β€ |s|): calculate the number of distinct characters in the substring s[l; r].
Input
The first line of the input contains one string s consisting of no more than 10^5 lowercase Latin letters.
The second line of the input contains one integer q (1 β€ q β€ 10^5) β the number of queries.
The next q lines contain queries, one per line. Each query is given in the format described in the problem statement. It is guaranteed that there is at least one query of the second type.
Output
For each query of the second type print the answer for it β the number of distinct characters in the required substring in this query.
Examples
Input
abacaba
5
2 1 4
1 4 b
1 5 b
2 4 6
2 1 7
Output
3
1
2
Input
dfcbbcfeeedbaea
15
1 6 e
1 4 b
2 6 14
1 7 b
1 12 c
2 6 8
2 1 6
1 7 c
1 2 f
1 10 a
2 7 9
1 10 a
1 14 b
1 1 f
2 1 11
Output
5
2
5
2
6
Submitted Solution:
```
def solve(s,qs):
n=len(s)
bits=[[0 for i in range(n+1)] for j in range(26)]
def update(i,j,val):
while j <= n:
bits[i][j] += val
j += j & (-j)
def _query(i,j):
s = 0
while j > 0:
s += bits[i][j]
j -= j & (-j)
return s
def query(i,a,b):
return _query(i,b) - _query(i,a-1)
for i,c in enumerate(s):
update(c-ord('a'),i+1,1)
for q in qs:
if q[0] == 1:
update(s[q[1]-1]-ord('a'),q[1],-1)
s[q[1]-1]=q[2]
update(q[2]-ord('a'),q[1],1)
else:
ans = 0
for i in range(26):
ans += query(i,q[1],q[2]) > 0
print(ans)
s=[ord(c) for c in input()]
qs=[input().split() for i in range(int(input()))]
for i in range(len(qs)):
q=qs[i]
if q[0] == '1': qs[i] = [int(q[0]),int(q[1]),ord(q[2])]
else: qs[i] = list(map(int,q))
solve(s,qs)
```
Yes
| 96,214 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of lowercase Latin letters and q queries for this string.
Recall that the substring s[l; r] of the string s is the string s_l s_{l + 1} ... s_r. For example, the substrings of "codeforces" are "code", "force", "f", "for", but not "coder" and "top".
There are two types of queries:
* 1~ pos~ c (1 β€ pos β€ |s|, c is lowercase Latin letter): replace s_{pos} with c (set s_{pos} := c);
* 2~ l~ r (1 β€ l β€ r β€ |s|): calculate the number of distinct characters in the substring s[l; r].
Input
The first line of the input contains one string s consisting of no more than 10^5 lowercase Latin letters.
The second line of the input contains one integer q (1 β€ q β€ 10^5) β the number of queries.
The next q lines contain queries, one per line. Each query is given in the format described in the problem statement. It is guaranteed that there is at least one query of the second type.
Output
For each query of the second type print the answer for it β the number of distinct characters in the required substring in this query.
Examples
Input
abacaba
5
2 1 4
1 4 b
1 5 b
2 4 6
2 1 7
Output
3
1
2
Input
dfcbbcfeeedbaea
15
1 6 e
1 4 b
2 6 14
1 7 b
1 12 c
2 6 8
2 1 6
1 7 c
1 2 f
1 10 a
2 7 9
1 10 a
1 14 b
1 1 f
2 1 11
Output
5
2
5
2
6
Submitted Solution:
```
class SegmentTree:
def __init__(self, data, default=0, func=max):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
s = input()
a = []
for i in s:
a += [2**(ord(i)-97)]
st = SegmentTree(a, func=lambda x, y: x | y)
for _ in range(int(input())):
p, q, r = input().split()
if p == "1":
st.__setitem__(int(q)-1, 2**(ord(r) - 97))
else:
result = st.query(int(q)-1, int(r)-1)
print(bin(result)[2:].count('1'))
```
Yes
| 96,215 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of lowercase Latin letters and q queries for this string.
Recall that the substring s[l; r] of the string s is the string s_l s_{l + 1} ... s_r. For example, the substrings of "codeforces" are "code", "force", "f", "for", but not "coder" and "top".
There are two types of queries:
* 1~ pos~ c (1 β€ pos β€ |s|, c is lowercase Latin letter): replace s_{pos} with c (set s_{pos} := c);
* 2~ l~ r (1 β€ l β€ r β€ |s|): calculate the number of distinct characters in the substring s[l; r].
Input
The first line of the input contains one string s consisting of no more than 10^5 lowercase Latin letters.
The second line of the input contains one integer q (1 β€ q β€ 10^5) β the number of queries.
The next q lines contain queries, one per line. Each query is given in the format described in the problem statement. It is guaranteed that there is at least one query of the second type.
Output
For each query of the second type print the answer for it β the number of distinct characters in the required substring in this query.
Examples
Input
abacaba
5
2 1 4
1 4 b
1 5 b
2 4 6
2 1 7
Output
3
1
2
Input
dfcbbcfeeedbaea
15
1 6 e
1 4 b
2 6 14
1 7 b
1 12 c
2 6 8
2 1 6
1 7 c
1 2 f
1 10 a
2 7 9
1 10 a
1 14 b
1 1 f
2 1 11
Output
5
2
5
2
6
Submitted Solution:
```
import sys
input = sys.stdin.readline
def update(BIT,n,i,v):
while i<=n:
BIT[i]+=v
i+=i&(-i)
#print(BIT)
def getsum(BIT,i):
s=0
while i>0:
s+=BIT[i]
i-=i&(-i)
return(s)
s=[i for i in input() if i !='\n']
#print(s)
LEN=len(s)
BIT=[[0]*(LEN+1) for i in range(26)]
for i in range(len(s)):
char=ord(s[i])-97
update(BIT[char],LEN,i+1,1)
#print(BIT)
q=int(input())
for i in range(q):
c=[i for i in input().split()]
#print(c)
if c[0]=='1':
char1=ord(s[int(c[1])-1])-97
s[int(c[1])-1]=c[2]
update(BIT[char1],LEN,int(c[1]),-1)
char2=ord(c[2])-97
update(BIT[char2],LEN,int(c[1]),1)
#print(BIT)
if c[0]=='2':
l=int(c[1])
r=int(c[2])
p=0
for i in range(len(BIT)):
p+=(min(1,getsum(BIT[i],r)-getsum(BIT[i],l-1)))
print(p)
```
Yes
| 96,216 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of lowercase Latin letters and q queries for this string.
Recall that the substring s[l; r] of the string s is the string s_l s_{l + 1} ... s_r. For example, the substrings of "codeforces" are "code", "force", "f", "for", but not "coder" and "top".
There are two types of queries:
* 1~ pos~ c (1 β€ pos β€ |s|, c is lowercase Latin letter): replace s_{pos} with c (set s_{pos} := c);
* 2~ l~ r (1 β€ l β€ r β€ |s|): calculate the number of distinct characters in the substring s[l; r].
Input
The first line of the input contains one string s consisting of no more than 10^5 lowercase Latin letters.
The second line of the input contains one integer q (1 β€ q β€ 10^5) β the number of queries.
The next q lines contain queries, one per line. Each query is given in the format described in the problem statement. It is guaranteed that there is at least one query of the second type.
Output
For each query of the second type print the answer for it β the number of distinct characters in the required substring in this query.
Examples
Input
abacaba
5
2 1 4
1 4 b
1 5 b
2 4 6
2 1 7
Output
3
1
2
Input
dfcbbcfeeedbaea
15
1 6 e
1 4 b
2 6 14
1 7 b
1 12 c
2 6 8
2 1 6
1 7 c
1 2 f
1 10 a
2 7 9
1 10 a
1 14 b
1 1 f
2 1 11
Output
5
2
5
2
6
Submitted Solution:
```
s = input()
list1 = list(s)
q = int(input())
for i in range(q):
a,b,c = input().split()
ds=0
list3 = []
if int(a) == 1:
list1[int(b)-1] = c
else:
list2 = list1[int(b)-1:int(c)-1]
for char in list2:
if char not in list3:
list3.append(char)
ds += 1
print(ds)
```
No
| 96,217 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of lowercase Latin letters and q queries for this string.
Recall that the substring s[l; r] of the string s is the string s_l s_{l + 1} ... s_r. For example, the substrings of "codeforces" are "code", "force", "f", "for", but not "coder" and "top".
There are two types of queries:
* 1~ pos~ c (1 β€ pos β€ |s|, c is lowercase Latin letter): replace s_{pos} with c (set s_{pos} := c);
* 2~ l~ r (1 β€ l β€ r β€ |s|): calculate the number of distinct characters in the substring s[l; r].
Input
The first line of the input contains one string s consisting of no more than 10^5 lowercase Latin letters.
The second line of the input contains one integer q (1 β€ q β€ 10^5) β the number of queries.
The next q lines contain queries, one per line. Each query is given in the format described in the problem statement. It is guaranteed that there is at least one query of the second type.
Output
For each query of the second type print the answer for it β the number of distinct characters in the required substring in this query.
Examples
Input
abacaba
5
2 1 4
1 4 b
1 5 b
2 4 6
2 1 7
Output
3
1
2
Input
dfcbbcfeeedbaea
15
1 6 e
1 4 b
2 6 14
1 7 b
1 12 c
2 6 8
2 1 6
1 7 c
1 2 f
1 10 a
2 7 9
1 10 a
1 14 b
1 1 f
2 1 11
Output
5
2
5
2
6
Submitted Solution:
```
n = 1101053527189095067 * 1000000000000000007
mod = 1000003
aa = [[0,0],[0,0]]
bb = [[0,0],[0,0]]
def mul():
ans = [[0,0],[0,0]]
for i in range(2):
for j in range(2):
ans[i][j] = sum(aa[i][k] * bb[k][j] % mod for k in range(2)) % mod
return ans
def mpow(a, b):
ret = [[1,0],[0,1]]
while b:
if b & 1:
ret = mul()
b >>= 1
a = mul()
return ret
print(mpow([[1,1],[1,0]], n)[0][0])
mod = 100003
n = 100003
a = 1
for i in range(10000):
a = mpow([[1,1],[1,0]], n)[0][0]
# for i in range(2, 1000009):
# a = mpow([[1,1],[1,0]], i)[0][0]
# b = mpow([[1,1],[1,0]], i+1)[0][0]
# if a == 1 and b == 1:
# print(i)
# fib = [0,1]
# for i in range(2000009):
# fib.append((fib[-1] + fib[-2])%mod)
# if fib[-1] == 1 and fib[-2] == 1:
# print(i)
```
No
| 96,218 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of lowercase Latin letters and q queries for this string.
Recall that the substring s[l; r] of the string s is the string s_l s_{l + 1} ... s_r. For example, the substrings of "codeforces" are "code", "force", "f", "for", but not "coder" and "top".
There are two types of queries:
* 1~ pos~ c (1 β€ pos β€ |s|, c is lowercase Latin letter): replace s_{pos} with c (set s_{pos} := c);
* 2~ l~ r (1 β€ l β€ r β€ |s|): calculate the number of distinct characters in the substring s[l; r].
Input
The first line of the input contains one string s consisting of no more than 10^5 lowercase Latin letters.
The second line of the input contains one integer q (1 β€ q β€ 10^5) β the number of queries.
The next q lines contain queries, one per line. Each query is given in the format described in the problem statement. It is guaranteed that there is at least one query of the second type.
Output
For each query of the second type print the answer for it β the number of distinct characters in the required substring in this query.
Examples
Input
abacaba
5
2 1 4
1 4 b
1 5 b
2 4 6
2 1 7
Output
3
1
2
Input
dfcbbcfeeedbaea
15
1 6 e
1 4 b
2 6 14
1 7 b
1 12 c
2 6 8
2 1 6
1 7 c
1 2 f
1 10 a
2 7 9
1 10 a
1 14 b
1 1 f
2 1 11
Output
5
2
5
2
6
Submitted Solution:
```
string = list(input())
answer = ""
for query in range(int(input())):
type, a, b = input().split()
if type == '1':
a = int(a)
string[a-1] = b
else:
a,b = map(int,[a,b])
amDistinct = len(set(string[a:b]))
answer += str(amDistinct) + "\n"
print(answer[:-1])
```
No
| 96,219 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of lowercase Latin letters and q queries for this string.
Recall that the substring s[l; r] of the string s is the string s_l s_{l + 1} ... s_r. For example, the substrings of "codeforces" are "code", "force", "f", "for", but not "coder" and "top".
There are two types of queries:
* 1~ pos~ c (1 β€ pos β€ |s|, c is lowercase Latin letter): replace s_{pos} with c (set s_{pos} := c);
* 2~ l~ r (1 β€ l β€ r β€ |s|): calculate the number of distinct characters in the substring s[l; r].
Input
The first line of the input contains one string s consisting of no more than 10^5 lowercase Latin letters.
The second line of the input contains one integer q (1 β€ q β€ 10^5) β the number of queries.
The next q lines contain queries, one per line. Each query is given in the format described in the problem statement. It is guaranteed that there is at least one query of the second type.
Output
For each query of the second type print the answer for it β the number of distinct characters in the required substring in this query.
Examples
Input
abacaba
5
2 1 4
1 4 b
1 5 b
2 4 6
2 1 7
Output
3
1
2
Input
dfcbbcfeeedbaea
15
1 6 e
1 4 b
2 6 14
1 7 b
1 12 c
2 6 8
2 1 6
1 7 c
1 2 f
1 10 a
2 7 9
1 10 a
1 14 b
1 1 f
2 1 11
Output
5
2
5
2
6
Submitted Solution:
```
s = input()
q = int(input())
qs = []
for _ in range(q):
qs.append(input().split())
# num_uniq = []
# uniq_count = 0
# s_set = set()
# for _s in s:
# if _s not in s_set:
# s_set.add(_s)
# uniq_count += 1
# num_uniq.append(uniq_count)
# print(num_uniq)
for t, a, b in qs:
if int(t) != 2:
continue
a, b = int(a), int(b)
print(len(set(s[a:b+1])))
```
No
| 96,220 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer a_i.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1, 2, 3, β¦). Eating the sweet i at the d-th day will cause a sugar penalty of (d β
a_i), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1 β€ m β€ n β€ 200\ 000).
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 200\ 000).
Output
You have to output n integers x_1, x_2, β¦, x_n on a single line, separed by spaces, where x_k is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let's analyze the answer for k = 5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
* Day 1: sweets 1 and 4
* Day 2: sweets 5 and 3
* Day 3 : sweet 6
Total penalty is 1 β
a_1 + 1 β
a_4 + 2 β
a_5 + 2 β
a_3 + 3 β
a_6 = 6 + 4 + 8 + 6 + 6 = 30. We can prove that it's the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x_5 = 30.
Tags: dp, greedy, math, sortings
Correct Solution:
```
N, M = map(int, input().split())
A = list(map(int, input().split()))
A.sort()
Ans = []
ans = 0
cumA = A[:]
for i in range(N):
j = i - M
if j>=0:
cumA[i] += cumA[j]
for a in cumA:
ans += a
Ans.append(ans)
print(" ".join(map(str, Ans)))
```
| 96,221 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer a_i.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1, 2, 3, β¦). Eating the sweet i at the d-th day will cause a sugar penalty of (d β
a_i), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1 β€ m β€ n β€ 200\ 000).
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 200\ 000).
Output
You have to output n integers x_1, x_2, β¦, x_n on a single line, separed by spaces, where x_k is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let's analyze the answer for k = 5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
* Day 1: sweets 1 and 4
* Day 2: sweets 5 and 3
* Day 3 : sweet 6
Total penalty is 1 β
a_1 + 1 β
a_4 + 2 β
a_5 + 2 β
a_3 + 3 β
a_6 = 6 + 4 + 8 + 6 + 6 = 30. We can prove that it's the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x_5 = 30.
Tags: dp, greedy, math, sortings
Correct Solution:
```
n, m = map(int, input().split())
a = [int(i) for i in input().split()]
a.sort()
cum = [0 for _ in range(n)]
dp1 = [0 for _ in range(n)]
dp2 = [0 for _ in range(n)]
cum[0] = a[0]
dp1[0] = a[0]
dp2[0] = a[0]
for i in range(1, n):
cum[i] = a[i] + cum[i - 1]
dp1[i] = cum[i]
if i - m >= 0:
dp1[i] -= cum[i - m]
dp1[i] += dp1[i - m] + dp2[i - m]
dp2[i] = a[i] + dp2[i - 1]
ans = " ".join([str(i) for i in dp1])
print(ans)
```
| 96,222 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer a_i.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1, 2, 3, β¦). Eating the sweet i at the d-th day will cause a sugar penalty of (d β
a_i), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1 β€ m β€ n β€ 200\ 000).
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 200\ 000).
Output
You have to output n integers x_1, x_2, β¦, x_n on a single line, separed by spaces, where x_k is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let's analyze the answer for k = 5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
* Day 1: sweets 1 and 4
* Day 2: sweets 5 and 3
* Day 3 : sweet 6
Total penalty is 1 β
a_1 + 1 β
a_4 + 2 β
a_5 + 2 β
a_3 + 3 β
a_6 = 6 + 4 + 8 + 6 + 6 = 30. We can prove that it's the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x_5 = 30.
Tags: dp, greedy, math, sortings
Correct Solution:
```
n,m = map(int,input().split())
a = list(map(int,input().split()))
a.sort()
state = [0]*n
ans = 0
for k in range(n):
state[k%m] += a[k]
ans += state[k%m]
print(ans,end=" ")
```
| 96,223 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer a_i.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1, 2, 3, β¦). Eating the sweet i at the d-th day will cause a sugar penalty of (d β
a_i), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1 β€ m β€ n β€ 200\ 000).
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 200\ 000).
Output
You have to output n integers x_1, x_2, β¦, x_n on a single line, separed by spaces, where x_k is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let's analyze the answer for k = 5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
* Day 1: sweets 1 and 4
* Day 2: sweets 5 and 3
* Day 3 : sweet 6
Total penalty is 1 β
a_1 + 1 β
a_4 + 2 β
a_5 + 2 β
a_3 + 3 β
a_6 = 6 + 4 + 8 + 6 + 6 = 30. We can prove that it's the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x_5 = 30.
Tags: dp, greedy, math, sortings
Correct Solution:
```
from collections import *
from functools import reduce
import sys
input = sys.stdin.readline
def factors(n):
return set(reduce(list.__add__,([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
def li():return [int(i) for i in input().rstrip('\n').split(' ')]
def st():return input().rstrip('\n')
def val():return int(input())
n,m = li()
a = li()
a.sort()
# print(a)
helparray = []
for i in range(m):
su = 0
for j in range(i,n,m):
su += a[j]
# print(i+1,su)
helparray.append(su)
helplen = len(helparray)
helparray = helparray[:n%m][::-1] + helparray[n%m:][::-1]
ind = -1
for i in range(m,n,1):
helparray.append(helparray[i-m] - a[ind])
ind-=1
# print(a)
# print(helparray)
helparray = [0] + helparray[::-1]
ans = []
for i in range(1,n+1):
helparray[i] += helparray[i-1]
ans.append(helparray[i])
print(*ans)
```
| 96,224 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer a_i.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1, 2, 3, β¦). Eating the sweet i at the d-th day will cause a sugar penalty of (d β
a_i), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1 β€ m β€ n β€ 200\ 000).
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 200\ 000).
Output
You have to output n integers x_1, x_2, β¦, x_n on a single line, separed by spaces, where x_k is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let's analyze the answer for k = 5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
* Day 1: sweets 1 and 4
* Day 2: sweets 5 and 3
* Day 3 : sweet 6
Total penalty is 1 β
a_1 + 1 β
a_4 + 2 β
a_5 + 2 β
a_3 + 3 β
a_6 = 6 + 4 + 8 + 6 + 6 = 30. We can prove that it's the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x_5 = 30.
Tags: dp, greedy, math, sortings
Correct Solution:
```
n, m = map(int, input().split())
a = [int(j) for j in input().split()]
a.sort()
prefix = [0] * n
suffix = [0] * n
# prefix
cur_prefix = 0
for i in range(n):
cur_prefix += a[i]
prefix[i] = cur_prefix
# suffix
suffix[m - 1] = prefix[m - 1]
for i in range(m, n):
suffix[i] = prefix[i] - prefix[i - m]
# print(a)
# print(prefix)
# print(suffix)
ans = []
for k in range(n):
k_look = k - m
if k_look < 0:
ans.append(prefix[k])
else:
ans.append(ans[k_look] + prefix[k_look] + suffix[k])
for i in ans:
print(i)
```
| 96,225 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer a_i.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1, 2, 3, β¦). Eating the sweet i at the d-th day will cause a sugar penalty of (d β
a_i), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1 β€ m β€ n β€ 200\ 000).
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 200\ 000).
Output
You have to output n integers x_1, x_2, β¦, x_n on a single line, separed by spaces, where x_k is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let's analyze the answer for k = 5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
* Day 1: sweets 1 and 4
* Day 2: sweets 5 and 3
* Day 3 : sweet 6
Total penalty is 1 β
a_1 + 1 β
a_4 + 2 β
a_5 + 2 β
a_3 + 3 β
a_6 = 6 + 4 + 8 + 6 + 6 = 30. We can prove that it's the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x_5 = 30.
Tags: dp, greedy, math, sortings
Correct Solution:
```
from sys import stdin
from collections import deque
from math import sqrt, floor, ceil, log, log2, log10, pi
def ii(): return int(stdin.readline())
def fi(): return float(stdin.readline())
def mi(): return map(int, stdin.readline().split())
def fmi(): return map(float, stdin.readline().split())
def li(): return list(mi())
def si(): return stdin.readline()
n, k=mi()
a=li()
a.sort()
#p=[]*k
for j in range(k):
for i in range(j+k, n, k):
#s+=a[i]
#p[j].append(s)
a[i]+=a[i-k]
#print(a)
x=[]
s=0
for i in range(n):
s+=a[i]
x.append(s)
print(*x)
```
| 96,226 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer a_i.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1, 2, 3, β¦). Eating the sweet i at the d-th day will cause a sugar penalty of (d β
a_i), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1 β€ m β€ n β€ 200\ 000).
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 200\ 000).
Output
You have to output n integers x_1, x_2, β¦, x_n on a single line, separed by spaces, where x_k is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let's analyze the answer for k = 5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
* Day 1: sweets 1 and 4
* Day 2: sweets 5 and 3
* Day 3 : sweet 6
Total penalty is 1 β
a_1 + 1 β
a_4 + 2 β
a_5 + 2 β
a_3 + 3 β
a_6 = 6 + 4 + 8 + 6 + 6 = 30. We can prove that it's the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x_5 = 30.
Tags: dp, greedy, math, sortings
Correct Solution:
```
for _ in range(1):
n,m=list(map(int,input().split()))
a=list(map(int,input().split()))
a.sort()
ans=[]
su=[a[0]]
for i in range(1,m):
su.append(a[i]+su[-1])
for j in range(m,n):
su.append(su[-1]+a[j])
#print(a[i-m])
d=1
#print(su)
for i in range(n):
s=su[i]
d=1
if i-m>=0:
#print(s,ans[i-m],su[i])
s+=ans[i-m]
ans.append(s)
#del(ans[0])
print(*ans)
```
| 96,227 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer a_i.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1, 2, 3, β¦). Eating the sweet i at the d-th day will cause a sugar penalty of (d β
a_i), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1 β€ m β€ n β€ 200\ 000).
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 200\ 000).
Output
You have to output n integers x_1, x_2, β¦, x_n on a single line, separed by spaces, where x_k is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let's analyze the answer for k = 5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
* Day 1: sweets 1 and 4
* Day 2: sweets 5 and 3
* Day 3 : sweet 6
Total penalty is 1 β
a_1 + 1 β
a_4 + 2 β
a_5 + 2 β
a_3 + 3 β
a_6 = 6 + 4 + 8 + 6 + 6 = 30. We can prove that it's the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x_5 = 30.
Tags: dp, greedy, math, sortings
Correct Solution:
```
n,m=map(int,input().split())
z=list(map(int,input().split()))
z.sort()
ans=[z[0]]
for i in range(1,len(z)):
ans.append(ans[-1]+z[i])
fin=[]
for i in range(len(z)):
if(i<=m-1):
fin.append(ans[i])
else:
fin.append(ans[i]+fin[i-m])
print(*fin)
```
| 96,228 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer a_i.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1, 2, 3, β¦). Eating the sweet i at the d-th day will cause a sugar penalty of (d β
a_i), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1 β€ m β€ n β€ 200\ 000).
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 200\ 000).
Output
You have to output n integers x_1, x_2, β¦, x_n on a single line, separed by spaces, where x_k is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let's analyze the answer for k = 5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
* Day 1: sweets 1 and 4
* Day 2: sweets 5 and 3
* Day 3 : sweet 6
Total penalty is 1 β
a_1 + 1 β
a_4 + 2 β
a_5 + 2 β
a_3 + 3 β
a_6 = 6 + 4 + 8 + 6 + 6 = 30. We can prove that it's the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x_5 = 30.
Submitted Solution:
```
n, m = [int(i) for i in input().split()]
d = [int(i) for i in input().split()]
d.sort()
ans = []
tot = 0
psum = [0]*m
for i in range(n):
md = i % m
psum[md] += d[i]
tot += psum[md]
ans.append(tot)
print(' '.join(map(str,ans)))
```
Yes
| 96,229 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer a_i.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1, 2, 3, β¦). Eating the sweet i at the d-th day will cause a sugar penalty of (d β
a_i), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1 β€ m β€ n β€ 200\ 000).
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 200\ 000).
Output
You have to output n integers x_1, x_2, β¦, x_n on a single line, separed by spaces, where x_k is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let's analyze the answer for k = 5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
* Day 1: sweets 1 and 4
* Day 2: sweets 5 and 3
* Day 3 : sweet 6
Total penalty is 1 β
a_1 + 1 β
a_4 + 2 β
a_5 + 2 β
a_3 + 3 β
a_6 = 6 + 4 + 8 + 6 + 6 = 30. We can prove that it's the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x_5 = 30.
Submitted Solution:
```
import os
import sys
from bisect import bisect_right
from fractions import Fraction
def main():
n , m = map(int , input().split())
arr=list(map(int , input().split()))
tot=[0]*(n)
cur=0
arr.sort()
for i in range(n):
cur=cur+arr[i]
tot[i]=cur
if(i>=m):
tot[i]+=tot[i-m]
print(*tot)
if __name__ == "__main__":
main()
```
Yes
| 96,230 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer a_i.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1, 2, 3, β¦). Eating the sweet i at the d-th day will cause a sugar penalty of (d β
a_i), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1 β€ m β€ n β€ 200\ 000).
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 200\ 000).
Output
You have to output n integers x_1, x_2, β¦, x_n on a single line, separed by spaces, where x_k is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let's analyze the answer for k = 5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
* Day 1: sweets 1 and 4
* Day 2: sweets 5 and 3
* Day 3 : sweet 6
Total penalty is 1 β
a_1 + 1 β
a_4 + 2 β
a_5 + 2 β
a_3 + 3 β
a_6 = 6 + 4 + 8 + 6 + 6 = 30. We can prove that it's the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x_5 = 30.
Submitted Solution:
```
import sys
input = lambda: sys.stdin.readline().strip("\r\n")
n, m = map(int, input().split())
a = sorted(list(map(int, input().split())))
dp = [0] * n
ans = 0
for i in range(n):
ans += a[i]
if i < m:
dp[i] = ans
else:
dp[i] = ans + dp[i-m]
print(*dp)
```
Yes
| 96,231 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer a_i.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1, 2, 3, β¦). Eating the sweet i at the d-th day will cause a sugar penalty of (d β
a_i), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1 β€ m β€ n β€ 200\ 000).
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 200\ 000).
Output
You have to output n integers x_1, x_2, β¦, x_n on a single line, separed by spaces, where x_k is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let's analyze the answer for k = 5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
* Day 1: sweets 1 and 4
* Day 2: sweets 5 and 3
* Day 3 : sweet 6
Total penalty is 1 β
a_1 + 1 β
a_4 + 2 β
a_5 + 2 β
a_3 + 3 β
a_6 = 6 + 4 + 8 + 6 + 6 = 30. We can prove that it's the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x_5 = 30.
Submitted Solution:
```
#Ashish_Sagar
n,m=map(int,input().split())
l=list(map(int,input().split()))
l.sort()
ans=[]
prefix=[0]
for i in range(1,n+1):
prefix.append(prefix[-1]+l[i-1])
for i in range(n):
if i<m:
ans.append(prefix[i+1])
else:
ans.append(prefix[i+1]+ans[i-m])
print(*ans)
```
Yes
| 96,232 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer a_i.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1, 2, 3, β¦). Eating the sweet i at the d-th day will cause a sugar penalty of (d β
a_i), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1 β€ m β€ n β€ 200\ 000).
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 200\ 000).
Output
You have to output n integers x_1, x_2, β¦, x_n on a single line, separed by spaces, where x_k is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let's analyze the answer for k = 5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
* Day 1: sweets 1 and 4
* Day 2: sweets 5 and 3
* Day 3 : sweet 6
Total penalty is 1 β
a_1 + 1 β
a_4 + 2 β
a_5 + 2 β
a_3 + 3 β
a_6 = 6 + 4 + 8 + 6 + 6 = 30. We can prove that it's the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x_5 = 30.
Submitted Solution:
```
# from sys import stdout
n, m = tuple(map(int, input().split()))
lis = list(map(int, input().split()))
if n==m and m==1:
print(*lis)
exit()
# en = n-1
# di = dict()
# k = 1; eaten = [[2]]
# di[k] = eaten; k = 2; eaten = [[3, 2]]
# di[k] = eaten; k = 3; eaten = [[4, 3], [2]]
# di[k] = eaten; k = 4; eaten = [[4, 4], [3, 2]]
# di[k] = eaten; k = 5; eaten = [[6, 4], [4, 3], [2]]
# di[k] = eaten; k = 6; eaten = [[6, 6], [4, 4], [3, 2]]
# di[k] = eaten; k = 7; eaten = [[7, 6], [6, 4], [4, 3], [2]]
# di[k] = eaten; k = 8; eaten = [[8, 7], [6, 6], [4, 4], [3, 2]]
# di[k] = eaten; k = 9; eaten = [[19, 8], [7, 6], [6, 4], [4, 3], [2]]
# di[k] = eaten
# for k in di.keys():
# #k = [1,2, .., m-1]
# # st = en-k
# li = di[k]
# for i in range(len(li)):
# posi = i+1
# for j in li[i]:
# # print(j,'*',posi, sep='', end = ' + ')
# # print()
# eaten = [[]]
lis.sort(reverse=True)
acc = [sum(lis)]
for i in range(1, n):
acc.append(acc[-1] - lis[i-1])
ans = acc[n-1:n-m-1:-1]
# print('ans =', ans)
# print('lis =', lis)
# print('acc =', acc)
for i in range(m, n):
ans.append(acc[-i-1] + ans[i-m])
# print(acc[-i-1], ans[i-m])
print(*ans)
```
No
| 96,233 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer a_i.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1, 2, 3, β¦). Eating the sweet i at the d-th day will cause a sugar penalty of (d β
a_i), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1 β€ m β€ n β€ 200\ 000).
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 200\ 000).
Output
You have to output n integers x_1, x_2, β¦, x_n on a single line, separed by spaces, where x_k is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let's analyze the answer for k = 5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
* Day 1: sweets 1 and 4
* Day 2: sweets 5 and 3
* Day 3 : sweet 6
Total penalty is 1 β
a_1 + 1 β
a_4 + 2 β
a_5 + 2 β
a_3 + 3 β
a_6 = 6 + 4 + 8 + 6 + 6 = 30. We can prove that it's the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x_5 = 30.
Submitted Solution:
```
n, m = map(int, input().split())
l = list(map(int, input().split()))
l.sort()
mass = [0]
for i in range(n):
mass += [mass[-1] + l[i]]
for i in range(1, n + 1):
if i <= m:
print(mass[i])
else:
k = i // m
r = i % m
res = 0
if r > 0:
res += (mass[r]) * (k + 1)
for y in range(k):
res += (mass[y * m + m + r] - mass[y * m + r]) * (k - y)
print(res, i, k, r)
```
No
| 96,234 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer a_i.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1, 2, 3, β¦). Eating the sweet i at the d-th day will cause a sugar penalty of (d β
a_i), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1 β€ m β€ n β€ 200\ 000).
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 200\ 000).
Output
You have to output n integers x_1, x_2, β¦, x_n on a single line, separed by spaces, where x_k is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let's analyze the answer for k = 5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
* Day 1: sweets 1 and 4
* Day 2: sweets 5 and 3
* Day 3 : sweet 6
Total penalty is 1 β
a_1 + 1 β
a_4 + 2 β
a_5 + 2 β
a_3 + 3 β
a_6 = 6 + 4 + 8 + 6 + 6 = 30. We can prove that it's the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x_5 = 30.
Submitted Solution:
```
n, m = map(int, input().split())
l1 = list(map(int, input().split()))
l1.sort()
dp = [0] * m
answer = []
for i in range(n):
dp[i % m]=dp[i%m] + l1[i]
answer.append(sum(dp))
for i in range(n):
print(answer[i], end=" ")
```
No
| 96,235 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer a_i.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1, 2, 3, β¦). Eating the sweet i at the d-th day will cause a sugar penalty of (d β
a_i), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1 β€ m β€ n β€ 200\ 000).
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 200\ 000).
Output
You have to output n integers x_1, x_2, β¦, x_n on a single line, separed by spaces, where x_k is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let's analyze the answer for k = 5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
* Day 1: sweets 1 and 4
* Day 2: sweets 5 and 3
* Day 3 : sweet 6
Total penalty is 1 β
a_1 + 1 β
a_4 + 2 β
a_5 + 2 β
a_3 + 3 β
a_6 = 6 + 4 + 8 + 6 + 6 = 30. We can prove that it's the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x_5 = 30.
Submitted Solution:
```
f = lambda: list(map(int, input().split()))
n, max_sweet_per_days = f()
sweets = f()
sweets.sort()
sweet_penalty = []
sweet_penalty.append(sweets[0])
for k in range(2, n + 1):
sweet_penalty.append(sweet_penalty[-1] + ((k // max_sweet_per_days) + 1) * sweets[k - 1])
print(" ".join(list(map(str, sweet_penalty))))
```
No
| 96,236 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n positive integers a_1, a_2, ..., a_n. For the one move you can choose any even value c and divide by two all elements that equal c.
For example, if a=[6,8,12,6,3,12] and you choose c=6, and a is transformed into a=[3,8,12,3,3,12] after the move.
You need to find the minimal number of moves for transforming a to an array of only odd integers (each element shouldn't be divisible by 2).
Input
The first line of the input contains one integer t (1 β€ t β€ 10^4) β the number of test cases in the input. Then t test cases follow.
The first line of a test case contains n (1 β€ n β€ 2β
10^5) β the number of integers in the sequence a. The second line contains positive integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9).
The sum of n for all test cases in the input doesn't exceed 2β
10^5.
Output
For t test cases print the answers in the order of test cases in the input. The answer for the test case is the minimal number of moves needed to make all numbers in the test case odd (i.e. not divisible by 2).
Example
Input
4
6
40 6 40 3 20 1
1
1024
4
2 4 8 16
3
3 1 7
Output
4
10
4
0
Note
In the first test case of the example, the optimal sequence of moves can be as follows:
* before making moves a=[40, 6, 40, 3, 20, 1];
* choose c=6;
* now a=[40, 3, 40, 3, 20, 1];
* choose c=40;
* now a=[20, 3, 20, 3, 20, 1];
* choose c=20;
* now a=[10, 3, 10, 3, 10, 1];
* choose c=10;
* now a=[5, 3, 5, 3, 5, 1] β all numbers are odd.
Thus, all numbers became odd after 4 moves. In 3 or fewer moves, you cannot make them all odd.
Tags: greedy, number theory
Correct Solution:
```
number = int(input())
for i in range(0,number):
n = int(input())
arr = [int(x) for x in input().split()]
uni = {}
cnt = []
arr.sort()
j = 0
while(j<n):
if(arr[j]%2!=0):
arr.pop(j)
n= n-1
elif(j!=0 and arr[j]==arr[j-1]):
arr.pop(j)
n= n-1
else: j+=1
n = len(arr)
for j in range(0,n):
k = 0
while(arr[j]%2==0):
arr[j]=arr[j]/2
k+=1
if(k>0):
uni[arr[j]] = max(k, uni.get(arr[j], 0))
#print(uni)
#print(cnt)
summ = 0
for j in uni.values():
summ+=j
print (summ)
```
| 96,237 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n positive integers a_1, a_2, ..., a_n. For the one move you can choose any even value c and divide by two all elements that equal c.
For example, if a=[6,8,12,6,3,12] and you choose c=6, and a is transformed into a=[3,8,12,3,3,12] after the move.
You need to find the minimal number of moves for transforming a to an array of only odd integers (each element shouldn't be divisible by 2).
Input
The first line of the input contains one integer t (1 β€ t β€ 10^4) β the number of test cases in the input. Then t test cases follow.
The first line of a test case contains n (1 β€ n β€ 2β
10^5) β the number of integers in the sequence a. The second line contains positive integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9).
The sum of n for all test cases in the input doesn't exceed 2β
10^5.
Output
For t test cases print the answers in the order of test cases in the input. The answer for the test case is the minimal number of moves needed to make all numbers in the test case odd (i.e. not divisible by 2).
Example
Input
4
6
40 6 40 3 20 1
1
1024
4
2 4 8 16
3
3 1 7
Output
4
10
4
0
Note
In the first test case of the example, the optimal sequence of moves can be as follows:
* before making moves a=[40, 6, 40, 3, 20, 1];
* choose c=6;
* now a=[40, 3, 40, 3, 20, 1];
* choose c=40;
* now a=[20, 3, 20, 3, 20, 1];
* choose c=20;
* now a=[10, 3, 10, 3, 10, 1];
* choose c=10;
* now a=[5, 3, 5, 3, 5, 1] β all numbers are odd.
Thus, all numbers became odd after 4 moves. In 3 or fewer moves, you cannot make them all odd.
Tags: greedy, number theory
Correct Solution:
```
# lET's tRy ThIS...
import math
import os
import sys
#-------------------BOLT------------------#
#-------Genius----Billionare----Playboy----Philanthropist----NOT ME:D----#
input = lambda: sys.stdin.readline().strip("\r\n")
def cin(): return sys.stdin.readline().strip("\r\n")
def fora(): return list(map(int, sys.stdin.readline().strip().split()))
def string(): return sys.stdin.readline().strip()
def cout(ans): sys.stdout.write(str(ans))
def endl(): sys.stdout.write(str("\n"))
def ende(): sys.stdout.write(str(" "))
#---------ND-I-AM-IRON-MAN------------------#
def main():
for _ in range(int(input())):
#LET's sPill the BEANS
n=int(cin())
a=fora()
b=[]
cnt=0
p=0
for i in a:
j=0
while((i%2)==0):
i/=2
j+=1
b.append(j)
a[p]=math.trunc(i)
p+=1
dict={}
dict=dict.fromkeys(a,0)
for i in range(n):
dict[a[i]]=max(dict[a[i]],b[i])
for i in dict.values():
cnt+=i
# cout(dict)
# endl()
cout(cnt)
endl()
if __name__ == "__main__":
main()
```
| 96,238 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n positive integers a_1, a_2, ..., a_n. For the one move you can choose any even value c and divide by two all elements that equal c.
For example, if a=[6,8,12,6,3,12] and you choose c=6, and a is transformed into a=[3,8,12,3,3,12] after the move.
You need to find the minimal number of moves for transforming a to an array of only odd integers (each element shouldn't be divisible by 2).
Input
The first line of the input contains one integer t (1 β€ t β€ 10^4) β the number of test cases in the input. Then t test cases follow.
The first line of a test case contains n (1 β€ n β€ 2β
10^5) β the number of integers in the sequence a. The second line contains positive integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9).
The sum of n for all test cases in the input doesn't exceed 2β
10^5.
Output
For t test cases print the answers in the order of test cases in the input. The answer for the test case is the minimal number of moves needed to make all numbers in the test case odd (i.e. not divisible by 2).
Example
Input
4
6
40 6 40 3 20 1
1
1024
4
2 4 8 16
3
3 1 7
Output
4
10
4
0
Note
In the first test case of the example, the optimal sequence of moves can be as follows:
* before making moves a=[40, 6, 40, 3, 20, 1];
* choose c=6;
* now a=[40, 3, 40, 3, 20, 1];
* choose c=40;
* now a=[20, 3, 20, 3, 20, 1];
* choose c=20;
* now a=[10, 3, 10, 3, 10, 1];
* choose c=10;
* now a=[5, 3, 5, 3, 5, 1] β all numbers are odd.
Thus, all numbers became odd after 4 moves. In 3 or fewer moves, you cannot make them all odd.
Tags: greedy, number theory
Correct Solution:
```
def odd(n,m):
if len(m)!=n:
return "incorrect input"
t=set()
f={}
for i in range(len(m)):
if int(m[i])%2==0:
t.add(int(m[i]))
# ΠΡΠ΅Π΄Π»Π°Π³Π°Π΅ΡΡΡ ΠΏΠΎΠ΄Π΄Π΅ΡΠΆΠΈΠ²Π°ΡΡ t -- ΠΌΠ½ΠΎΠΆΠ΅ΡΡΠ²ΠΎ Π²ΡΠ΅Π²ΠΎΠ·ΠΌΠΎΠΆΠ½ΡΡ
ΠΏΠΎΡΠ΅ΡΠ°Π΅ΠΌΡΡ
ΡΡΡΠ½ΡΡ
ΡΠΈΡΠ΅Π».
for i in t:
while i%2!=1:
if i in f:
i = i // 2
else:
f[i]=i//2
i=i//2
return len(f)
"""
#2 attempt
for i in range(len(m)):
if int(m[i])%2==0:
if int(m[i]) in t:
t[int(m[i])]+=1
else:
t[int(m[i])] = 1
count=0
while len(t)!=0:
print(t)
f=max(t)
count+=1
if (f//2)%2==0:
if f//2 in t:
t[f//2]+=t[f]
else:
t[f//2]=t[f]
del t[f]
#1 attempt
while t:
#g=max(m,key=lambda f:[f%2==0,f])
#m.sort(key=lambda f: [f % 2 == 1, -f])
print(g)
f=m[0]
for i in range(len(m)):
if m[i]==f:
m[i]//=2
else:
break
#t = find(m)
count+=1
"""
for i in range(int(input().strip())):
n=int(input().strip())
m=input().strip().split()
print(odd(n,m))
```
| 96,239 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n positive integers a_1, a_2, ..., a_n. For the one move you can choose any even value c and divide by two all elements that equal c.
For example, if a=[6,8,12,6,3,12] and you choose c=6, and a is transformed into a=[3,8,12,3,3,12] after the move.
You need to find the minimal number of moves for transforming a to an array of only odd integers (each element shouldn't be divisible by 2).
Input
The first line of the input contains one integer t (1 β€ t β€ 10^4) β the number of test cases in the input. Then t test cases follow.
The first line of a test case contains n (1 β€ n β€ 2β
10^5) β the number of integers in the sequence a. The second line contains positive integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9).
The sum of n for all test cases in the input doesn't exceed 2β
10^5.
Output
For t test cases print the answers in the order of test cases in the input. The answer for the test case is the minimal number of moves needed to make all numbers in the test case odd (i.e. not divisible by 2).
Example
Input
4
6
40 6 40 3 20 1
1
1024
4
2 4 8 16
3
3 1 7
Output
4
10
4
0
Note
In the first test case of the example, the optimal sequence of moves can be as follows:
* before making moves a=[40, 6, 40, 3, 20, 1];
* choose c=6;
* now a=[40, 3, 40, 3, 20, 1];
* choose c=40;
* now a=[20, 3, 20, 3, 20, 1];
* choose c=20;
* now a=[10, 3, 10, 3, 10, 1];
* choose c=10;
* now a=[5, 3, 5, 3, 5, 1] β all numbers are odd.
Thus, all numbers became odd after 4 moves. In 3 or fewer moves, you cannot make them all odd.
Tags: greedy, number theory
Correct Solution:
```
t=int(input())
for _ in range(t):
n=int(input())
arr=list(map(int,input().split()))
arr1=[]
for a in arr:
if a%2==0:
arr1.append(a)
arr1=list(set(arr1))
arr1.sort()
i=0
ca={}
while i<len(arr1):
b=arr1[i]
count=0
while b%2==0:
count+=1
b=b//2
ca[b]=count
i+=1
print(sum(ca.values()))
```
| 96,240 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n positive integers a_1, a_2, ..., a_n. For the one move you can choose any even value c and divide by two all elements that equal c.
For example, if a=[6,8,12,6,3,12] and you choose c=6, and a is transformed into a=[3,8,12,3,3,12] after the move.
You need to find the minimal number of moves for transforming a to an array of only odd integers (each element shouldn't be divisible by 2).
Input
The first line of the input contains one integer t (1 β€ t β€ 10^4) β the number of test cases in the input. Then t test cases follow.
The first line of a test case contains n (1 β€ n β€ 2β
10^5) β the number of integers in the sequence a. The second line contains positive integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9).
The sum of n for all test cases in the input doesn't exceed 2β
10^5.
Output
For t test cases print the answers in the order of test cases in the input. The answer for the test case is the minimal number of moves needed to make all numbers in the test case odd (i.e. not divisible by 2).
Example
Input
4
6
40 6 40 3 20 1
1
1024
4
2 4 8 16
3
3 1 7
Output
4
10
4
0
Note
In the first test case of the example, the optimal sequence of moves can be as follows:
* before making moves a=[40, 6, 40, 3, 20, 1];
* choose c=6;
* now a=[40, 3, 40, 3, 20, 1];
* choose c=40;
* now a=[20, 3, 20, 3, 20, 1];
* choose c=20;
* now a=[10, 3, 10, 3, 10, 1];
* choose c=10;
* now a=[5, 3, 5, 3, 5, 1] β all numbers are odd.
Thus, all numbers became odd after 4 moves. In 3 or fewer moves, you cannot make them all odd.
Tags: greedy, number theory
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
s=set()
ans=0
for x in a:
while(x%2==0):
s.add(x)
x/=2
print(len(s))
```
| 96,241 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n positive integers a_1, a_2, ..., a_n. For the one move you can choose any even value c and divide by two all elements that equal c.
For example, if a=[6,8,12,6,3,12] and you choose c=6, and a is transformed into a=[3,8,12,3,3,12] after the move.
You need to find the minimal number of moves for transforming a to an array of only odd integers (each element shouldn't be divisible by 2).
Input
The first line of the input contains one integer t (1 β€ t β€ 10^4) β the number of test cases in the input. Then t test cases follow.
The first line of a test case contains n (1 β€ n β€ 2β
10^5) β the number of integers in the sequence a. The second line contains positive integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9).
The sum of n for all test cases in the input doesn't exceed 2β
10^5.
Output
For t test cases print the answers in the order of test cases in the input. The answer for the test case is the minimal number of moves needed to make all numbers in the test case odd (i.e. not divisible by 2).
Example
Input
4
6
40 6 40 3 20 1
1
1024
4
2 4 8 16
3
3 1 7
Output
4
10
4
0
Note
In the first test case of the example, the optimal sequence of moves can be as follows:
* before making moves a=[40, 6, 40, 3, 20, 1];
* choose c=6;
* now a=[40, 3, 40, 3, 20, 1];
* choose c=40;
* now a=[20, 3, 20, 3, 20, 1];
* choose c=20;
* now a=[10, 3, 10, 3, 10, 1];
* choose c=10;
* now a=[5, 3, 5, 3, 5, 1] β all numbers are odd.
Thus, all numbers became odd after 4 moves. In 3 or fewer moves, you cannot make them all odd.
Tags: greedy, number theory
Correct Solution:
```
def fil(i):
if int(i)%2==0:
return True
else:
return False
def mapf(i):
s=bin(int(i))[2:]
x=s.rindex("1")
y=len(s)-1-x
return (y,int(s[0:x+1],2))
def sortf(i):
return i[1]
def func(l,n):
prev=l[0]
sum=0
for i in range(n):
if l[i][1]!=prev[1]:
sum=sum+prev[0]
prev=l[i]
else:
if l[i][0]>prev[0]:
prev=l[i]
if i==n-1:
sum=sum+prev[0]
return sum
t=int(input())
for _ in range(t):
n=int(input())
l=list(set(map(mapf,filter(fil,input().split()))))
l.sort(key=sortf,reverse=True)
n=len(l)
if l==[]:
print(0)
else:
print(func(l,n))
```
| 96,242 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n positive integers a_1, a_2, ..., a_n. For the one move you can choose any even value c and divide by two all elements that equal c.
For example, if a=[6,8,12,6,3,12] and you choose c=6, and a is transformed into a=[3,8,12,3,3,12] after the move.
You need to find the minimal number of moves for transforming a to an array of only odd integers (each element shouldn't be divisible by 2).
Input
The first line of the input contains one integer t (1 β€ t β€ 10^4) β the number of test cases in the input. Then t test cases follow.
The first line of a test case contains n (1 β€ n β€ 2β
10^5) β the number of integers in the sequence a. The second line contains positive integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9).
The sum of n for all test cases in the input doesn't exceed 2β
10^5.
Output
For t test cases print the answers in the order of test cases in the input. The answer for the test case is the minimal number of moves needed to make all numbers in the test case odd (i.e. not divisible by 2).
Example
Input
4
6
40 6 40 3 20 1
1
1024
4
2 4 8 16
3
3 1 7
Output
4
10
4
0
Note
In the first test case of the example, the optimal sequence of moves can be as follows:
* before making moves a=[40, 6, 40, 3, 20, 1];
* choose c=6;
* now a=[40, 3, 40, 3, 20, 1];
* choose c=40;
* now a=[20, 3, 20, 3, 20, 1];
* choose c=20;
* now a=[10, 3, 10, 3, 10, 1];
* choose c=10;
* now a=[5, 3, 5, 3, 5, 1] β all numbers are odd.
Thus, all numbers became odd after 4 moves. In 3 or fewer moves, you cannot make them all odd.
Tags: greedy, number theory
Correct Solution:
```
from heapq import *
t=int(input())
for _ in range(t):
n=int(input())
it=list(map(int,input().split()))
it=[-i for i in it if i%2==0]
heapify(it)
tot=0
tot=0
ss=set()
while it:
no=it.pop()
if no in ss:
continue
ss.add(no)
if no%4==0:
heappush(it,no//2)
tot+=1
print(tot)
```
| 96,243 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n positive integers a_1, a_2, ..., a_n. For the one move you can choose any even value c and divide by two all elements that equal c.
For example, if a=[6,8,12,6,3,12] and you choose c=6, and a is transformed into a=[3,8,12,3,3,12] after the move.
You need to find the minimal number of moves for transforming a to an array of only odd integers (each element shouldn't be divisible by 2).
Input
The first line of the input contains one integer t (1 β€ t β€ 10^4) β the number of test cases in the input. Then t test cases follow.
The first line of a test case contains n (1 β€ n β€ 2β
10^5) β the number of integers in the sequence a. The second line contains positive integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9).
The sum of n for all test cases in the input doesn't exceed 2β
10^5.
Output
For t test cases print the answers in the order of test cases in the input. The answer for the test case is the minimal number of moves needed to make all numbers in the test case odd (i.e. not divisible by 2).
Example
Input
4
6
40 6 40 3 20 1
1
1024
4
2 4 8 16
3
3 1 7
Output
4
10
4
0
Note
In the first test case of the example, the optimal sequence of moves can be as follows:
* before making moves a=[40, 6, 40, 3, 20, 1];
* choose c=6;
* now a=[40, 3, 40, 3, 20, 1];
* choose c=40;
* now a=[20, 3, 20, 3, 20, 1];
* choose c=20;
* now a=[10, 3, 10, 3, 10, 1];
* choose c=10;
* now a=[5, 3, 5, 3, 5, 1] β all numbers are odd.
Thus, all numbers became odd after 4 moves. In 3 or fewer moves, you cannot make them all odd.
Tags: greedy, number theory
Correct Solution:
```
if __name__ == "__main__":
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
aset = set(a)
alist = list(aset)
# print(alist)
distances = {}
alist.sort()
ans = 0
for i in range(len(alist)):
num = alist[i]
distance = 0
num2 = num
found = False
while num2 % 2 == 0:
distance += 1
num2 = num2 // 2
if num2 in distances:
found = True
distance += distances[num2]
break
distances[num] = distance
if found:
distances.pop(num2)
ans = sum(distances.values())
print(ans)
```
| 96,244 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n positive integers a_1, a_2, ..., a_n. For the one move you can choose any even value c and divide by two all elements that equal c.
For example, if a=[6,8,12,6,3,12] and you choose c=6, and a is transformed into a=[3,8,12,3,3,12] after the move.
You need to find the minimal number of moves for transforming a to an array of only odd integers (each element shouldn't be divisible by 2).
Input
The first line of the input contains one integer t (1 β€ t β€ 10^4) β the number of test cases in the input. Then t test cases follow.
The first line of a test case contains n (1 β€ n β€ 2β
10^5) β the number of integers in the sequence a. The second line contains positive integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9).
The sum of n for all test cases in the input doesn't exceed 2β
10^5.
Output
For t test cases print the answers in the order of test cases in the input. The answer for the test case is the minimal number of moves needed to make all numbers in the test case odd (i.e. not divisible by 2).
Example
Input
4
6
40 6 40 3 20 1
1
1024
4
2 4 8 16
3
3 1 7
Output
4
10
4
0
Note
In the first test case of the example, the optimal sequence of moves can be as follows:
* before making moves a=[40, 6, 40, 3, 20, 1];
* choose c=6;
* now a=[40, 3, 40, 3, 20, 1];
* choose c=40;
* now a=[20, 3, 20, 3, 20, 1];
* choose c=20;
* now a=[10, 3, 10, 3, 10, 1];
* choose c=10;
* now a=[5, 3, 5, 3, 5, 1] β all numbers are odd.
Thus, all numbers became odd after 4 moves. In 3 or fewer moves, you cannot make them all odd.
Submitted Solution:
```
MOD = 1000000007
ii = lambda : int(input())
si = lambda : input()
dgl = lambda : list(map(int, input()))
f = lambda : map(int, input().split())
il = lambda : list(map(int, input().split()))
ls = lambda : list(input())
for _ in range(ii()):
n=ii()
s=set()
c=0
l=sorted(il(),reverse=True)
for i in l:
while i%2==0:
if not i in s:
c+=1
s.add(i)
i//=2
print(c)
```
Yes
| 96,245 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n positive integers a_1, a_2, ..., a_n. For the one move you can choose any even value c and divide by two all elements that equal c.
For example, if a=[6,8,12,6,3,12] and you choose c=6, and a is transformed into a=[3,8,12,3,3,12] after the move.
You need to find the minimal number of moves for transforming a to an array of only odd integers (each element shouldn't be divisible by 2).
Input
The first line of the input contains one integer t (1 β€ t β€ 10^4) β the number of test cases in the input. Then t test cases follow.
The first line of a test case contains n (1 β€ n β€ 2β
10^5) β the number of integers in the sequence a. The second line contains positive integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9).
The sum of n for all test cases in the input doesn't exceed 2β
10^5.
Output
For t test cases print the answers in the order of test cases in the input. The answer for the test case is the minimal number of moves needed to make all numbers in the test case odd (i.e. not divisible by 2).
Example
Input
4
6
40 6 40 3 20 1
1
1024
4
2 4 8 16
3
3 1 7
Output
4
10
4
0
Note
In the first test case of the example, the optimal sequence of moves can be as follows:
* before making moves a=[40, 6, 40, 3, 20, 1];
* choose c=6;
* now a=[40, 3, 40, 3, 20, 1];
* choose c=40;
* now a=[20, 3, 20, 3, 20, 1];
* choose c=20;
* now a=[10, 3, 10, 3, 10, 1];
* choose c=10;
* now a=[5, 3, 5, 3, 5, 1] β all numbers are odd.
Thus, all numbers became odd after 4 moves. In 3 or fewer moves, you cannot make them all odd.
Submitted Solution:
```
for _ in range(int(input())):
input()
a = sorted(map(int, input().split()), reverse=True)
seen = set()
ret = 0
for val in a:
if val & 1 or val in seen:
continue
while val & 1 == 0:
seen.add(val)
val >>= 1
ret += 1
print(ret)
```
Yes
| 96,246 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n positive integers a_1, a_2, ..., a_n. For the one move you can choose any even value c and divide by two all elements that equal c.
For example, if a=[6,8,12,6,3,12] and you choose c=6, and a is transformed into a=[3,8,12,3,3,12] after the move.
You need to find the minimal number of moves for transforming a to an array of only odd integers (each element shouldn't be divisible by 2).
Input
The first line of the input contains one integer t (1 β€ t β€ 10^4) β the number of test cases in the input. Then t test cases follow.
The first line of a test case contains n (1 β€ n β€ 2β
10^5) β the number of integers in the sequence a. The second line contains positive integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9).
The sum of n for all test cases in the input doesn't exceed 2β
10^5.
Output
For t test cases print the answers in the order of test cases in the input. The answer for the test case is the minimal number of moves needed to make all numbers in the test case odd (i.e. not divisible by 2).
Example
Input
4
6
40 6 40 3 20 1
1
1024
4
2 4 8 16
3
3 1 7
Output
4
10
4
0
Note
In the first test case of the example, the optimal sequence of moves can be as follows:
* before making moves a=[40, 6, 40, 3, 20, 1];
* choose c=6;
* now a=[40, 3, 40, 3, 20, 1];
* choose c=40;
* now a=[20, 3, 20, 3, 20, 1];
* choose c=20;
* now a=[10, 3, 10, 3, 10, 1];
* choose c=10;
* now a=[5, 3, 5, 3, 5, 1] β all numbers are odd.
Thus, all numbers became odd after 4 moves. In 3 or fewer moves, you cannot make them all odd.
Submitted Solution:
```
t = int(input(''))
for v in range(t):
n = input('')
a = list(map(int,input('').split(' ')))
d = {}
for nu in a:
if(nu % 2 == 0):
d[nu] = True
dd = {}
ans = 0
for k in d:
t = k
while(t % 2 == 0):
if(t not in dd):
ans += 1
dd[t] = True
else:
break
t = t//2
print(ans)
```
Yes
| 96,247 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n positive integers a_1, a_2, ..., a_n. For the one move you can choose any even value c and divide by two all elements that equal c.
For example, if a=[6,8,12,6,3,12] and you choose c=6, and a is transformed into a=[3,8,12,3,3,12] after the move.
You need to find the minimal number of moves for transforming a to an array of only odd integers (each element shouldn't be divisible by 2).
Input
The first line of the input contains one integer t (1 β€ t β€ 10^4) β the number of test cases in the input. Then t test cases follow.
The first line of a test case contains n (1 β€ n β€ 2β
10^5) β the number of integers in the sequence a. The second line contains positive integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9).
The sum of n for all test cases in the input doesn't exceed 2β
10^5.
Output
For t test cases print the answers in the order of test cases in the input. The answer for the test case is the minimal number of moves needed to make all numbers in the test case odd (i.e. not divisible by 2).
Example
Input
4
6
40 6 40 3 20 1
1
1024
4
2 4 8 16
3
3 1 7
Output
4
10
4
0
Note
In the first test case of the example, the optimal sequence of moves can be as follows:
* before making moves a=[40, 6, 40, 3, 20, 1];
* choose c=6;
* now a=[40, 3, 40, 3, 20, 1];
* choose c=40;
* now a=[20, 3, 20, 3, 20, 1];
* choose c=20;
* now a=[10, 3, 10, 3, 10, 1];
* choose c=10;
* now a=[5, 3, 5, 3, 5, 1] β all numbers are odd.
Thus, all numbers became odd after 4 moves. In 3 or fewer moves, you cannot make them all odd.
Submitted Solution:
```
# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------- fast io --------------------
testcases=int(input())
for j in range(testcases):
n=int(input())
vals=list(map(int,input().split()))
#find all the evens
evens=set([])
for s in range(n):
if vals[s]%2==0:
evens.add(vals[s])
evens=list(evens)
evens.sort()
dict1={}
total=0
#i think ill divide everything by 2 and then turn it into a set
for s in range(len(evens)):
moves=0
while evens[s]%2==0:
evens[s]=evens[s]//2
moves+=1
if not(evens[s] in dict1):
dict1[evens[s]]=moves
else:
if moves>dict1[evens[s]]:
dict1[evens[s]]=moves
print(sum(dict1.values()))
```
Yes
| 96,248 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n positive integers a_1, a_2, ..., a_n. For the one move you can choose any even value c and divide by two all elements that equal c.
For example, if a=[6,8,12,6,3,12] and you choose c=6, and a is transformed into a=[3,8,12,3,3,12] after the move.
You need to find the minimal number of moves for transforming a to an array of only odd integers (each element shouldn't be divisible by 2).
Input
The first line of the input contains one integer t (1 β€ t β€ 10^4) β the number of test cases in the input. Then t test cases follow.
The first line of a test case contains n (1 β€ n β€ 2β
10^5) β the number of integers in the sequence a. The second line contains positive integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9).
The sum of n for all test cases in the input doesn't exceed 2β
10^5.
Output
For t test cases print the answers in the order of test cases in the input. The answer for the test case is the minimal number of moves needed to make all numbers in the test case odd (i.e. not divisible by 2).
Example
Input
4
6
40 6 40 3 20 1
1
1024
4
2 4 8 16
3
3 1 7
Output
4
10
4
0
Note
In the first test case of the example, the optimal sequence of moves can be as follows:
* before making moves a=[40, 6, 40, 3, 20, 1];
* choose c=6;
* now a=[40, 3, 40, 3, 20, 1];
* choose c=40;
* now a=[20, 3, 20, 3, 20, 1];
* choose c=20;
* now a=[10, 3, 10, 3, 10, 1];
* choose c=10;
* now a=[5, 3, 5, 3, 5, 1] β all numbers are odd.
Thus, all numbers became odd after 4 moves. In 3 or fewer moves, you cannot make them all odd.
Submitted Solution:
```
def main():
q = int(input())
for _ in range(q):
n = int(input())
tab = list(map(int,input().split()))
nt = set()
for x in range(n):
if tab[x]%2==0:
nt.add(tab[x])
nt = list(nt)
nt.sort()
nt = nt[::-1]
if len(nt) == 0:
print(0)
continue
nt.append(-1)
i = 0
trans = 0
while i < len(nt)-1:
if nt[i]%2==0 and nt[i]//2 > nt[i+1]:
nt[i] //=2
trans += 1
else:
i += 1
trans += 1
print(trans-1)
main()
```
No
| 96,249 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n positive integers a_1, a_2, ..., a_n. For the one move you can choose any even value c and divide by two all elements that equal c.
For example, if a=[6,8,12,6,3,12] and you choose c=6, and a is transformed into a=[3,8,12,3,3,12] after the move.
You need to find the minimal number of moves for transforming a to an array of only odd integers (each element shouldn't be divisible by 2).
Input
The first line of the input contains one integer t (1 β€ t β€ 10^4) β the number of test cases in the input. Then t test cases follow.
The first line of a test case contains n (1 β€ n β€ 2β
10^5) β the number of integers in the sequence a. The second line contains positive integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9).
The sum of n for all test cases in the input doesn't exceed 2β
10^5.
Output
For t test cases print the answers in the order of test cases in the input. The answer for the test case is the minimal number of moves needed to make all numbers in the test case odd (i.e. not divisible by 2).
Example
Input
4
6
40 6 40 3 20 1
1
1024
4
2 4 8 16
3
3 1 7
Output
4
10
4
0
Note
In the first test case of the example, the optimal sequence of moves can be as follows:
* before making moves a=[40, 6, 40, 3, 20, 1];
* choose c=6;
* now a=[40, 3, 40, 3, 20, 1];
* choose c=40;
* now a=[20, 3, 20, 3, 20, 1];
* choose c=20;
* now a=[10, 3, 10, 3, 10, 1];
* choose c=10;
* now a=[5, 3, 5, 3, 5, 1] β all numbers are odd.
Thus, all numbers became odd after 4 moves. In 3 or fewer moves, you cannot make them all odd.
Submitted Solution:
```
import heapq
def main():
q = int(input())
for _ in range(q):
n = int(input())
tab = list(map(int,input().split()))
nt = set()
for x in range(n):
if tab[x]%2==0:
nt.add(tab[x])
nt = list(nt)
nt.sort()
if len(nt) == 0:
print(0)
continue
i = len(nt)-1
trans = 0
nts = set(nt)
while i >= 0:
if (nt[i]//2)%2==0:
if nt[i]//2 in nts:
i -= 1
trans += 1
nt.pop(-1)
else:
nts.add(nt[i]//2)
heapq.heappush(nt,nt[i]//2)
nt.pop(-1)
trans += 1
else:
i -= 1
trans += 1
nt.pop(-1)
print(trans)
main()
```
No
| 96,250 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n positive integers a_1, a_2, ..., a_n. For the one move you can choose any even value c and divide by two all elements that equal c.
For example, if a=[6,8,12,6,3,12] and you choose c=6, and a is transformed into a=[3,8,12,3,3,12] after the move.
You need to find the minimal number of moves for transforming a to an array of only odd integers (each element shouldn't be divisible by 2).
Input
The first line of the input contains one integer t (1 β€ t β€ 10^4) β the number of test cases in the input. Then t test cases follow.
The first line of a test case contains n (1 β€ n β€ 2β
10^5) β the number of integers in the sequence a. The second line contains positive integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9).
The sum of n for all test cases in the input doesn't exceed 2β
10^5.
Output
For t test cases print the answers in the order of test cases in the input. The answer for the test case is the minimal number of moves needed to make all numbers in the test case odd (i.e. not divisible by 2).
Example
Input
4
6
40 6 40 3 20 1
1
1024
4
2 4 8 16
3
3 1 7
Output
4
10
4
0
Note
In the first test case of the example, the optimal sequence of moves can be as follows:
* before making moves a=[40, 6, 40, 3, 20, 1];
* choose c=6;
* now a=[40, 3, 40, 3, 20, 1];
* choose c=40;
* now a=[20, 3, 20, 3, 20, 1];
* choose c=20;
* now a=[10, 3, 10, 3, 10, 1];
* choose c=10;
* now a=[5, 3, 5, 3, 5, 1] β all numbers are odd.
Thus, all numbers became odd after 4 moves. In 3 or fewer moves, you cannot make them all odd.
Submitted Solution:
```
def solve():
N = int(input())
L = list(set(map(int, input().split())))
L = filter(lambda x: x % 2 == 0, L)
used = set()
for l in sorted(L, reverse=True):
if all(v % l != 0 and l % 2 == 0 for v in used):
used.add(l)
ans = 0
for v in used:
cnt = 0
while v % 2 == 0:
v //= 2
cnt += 1
ans += cnt
print(ans)
T = int(input())
for _ in range(T):
solve()
```
No
| 96,251 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n positive integers a_1, a_2, ..., a_n. For the one move you can choose any even value c and divide by two all elements that equal c.
For example, if a=[6,8,12,6,3,12] and you choose c=6, and a is transformed into a=[3,8,12,3,3,12] after the move.
You need to find the minimal number of moves for transforming a to an array of only odd integers (each element shouldn't be divisible by 2).
Input
The first line of the input contains one integer t (1 β€ t β€ 10^4) β the number of test cases in the input. Then t test cases follow.
The first line of a test case contains n (1 β€ n β€ 2β
10^5) β the number of integers in the sequence a. The second line contains positive integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9).
The sum of n for all test cases in the input doesn't exceed 2β
10^5.
Output
For t test cases print the answers in the order of test cases in the input. The answer for the test case is the minimal number of moves needed to make all numbers in the test case odd (i.e. not divisible by 2).
Example
Input
4
6
40 6 40 3 20 1
1
1024
4
2 4 8 16
3
3 1 7
Output
4
10
4
0
Note
In the first test case of the example, the optimal sequence of moves can be as follows:
* before making moves a=[40, 6, 40, 3, 20, 1];
* choose c=6;
* now a=[40, 3, 40, 3, 20, 1];
* choose c=40;
* now a=[20, 3, 20, 3, 20, 1];
* choose c=20;
* now a=[10, 3, 10, 3, 10, 1];
* choose c=10;
* now a=[5, 3, 5, 3, 5, 1] β all numbers are odd.
Thus, all numbers became odd after 4 moves. In 3 or fewer moves, you cannot make them all odd.
Submitted Solution:
```
import sys
import math
#to read string
get_string = lambda: sys.stdin.readline().strip()
#to read list of integers
get_list = lambda: list( map(int,sys.stdin.readline().strip().split()) )
#to read integers
get_int = lambda: int(sys.stdin.readline())
#to print fast
#pt = lambda x: sys.stdout.write(str(x)+'\n')
#--------------------------------WhiteHat010--------------------------------------#
for _ in range(get_int()):
get_int()
lst = sorted(list(set(get_list())),reverse = True)
n = len(lst)
for i in range(n):
if lst[i]%2 == 0 :
for j in range(i+1,n):
if lst[i]%lst[j] == 0:
lst[j] = 1
lst = [i for i in lst if i%2 == 0 ]
n = len(lst)
count = 0
# print(lst)
i = 0
while i<n:
item = lst[i]
if item%2 == 0:
while item%2 == 0:
if item in lst:
for j in range(n):
if lst[j] == item:
lst[j] = 1
item = item//2
count += 1
i += 1
print(count)
```
No
| 96,252 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Roma is playing a new expansion for his favorite game World of Darkraft. He made a new character and is going for his first grind.
Roma has a choice to buy exactly one of n different weapons and exactly one of m different armor sets. Weapon i has attack modifier a_i and is worth ca_i coins, and armor set j has defense modifier b_j and is worth cb_j coins.
After choosing his equipment Roma can proceed to defeat some monsters. There are p monsters he can try to defeat. Monster k has defense x_k, attack y_k and possesses z_k coins. Roma can defeat a monster if his weapon's attack modifier is larger than the monster's defense, and his armor set's defense modifier is larger than the monster's attack. That is, a monster k can be defeated with a weapon i and an armor set j if a_i > x_k and b_j > y_k. After defeating the monster, Roma takes all the coins from them. During the grind, Roma can defeat as many monsters as he likes. Monsters do not respawn, thus each monster can be defeated at most one.
Thanks to Roma's excessive donations, we can assume that he has an infinite amount of in-game currency and can afford any of the weapons and armor sets. Still, he wants to maximize the profit of the grind. The profit is defined as the total coins obtained from all defeated monsters minus the cost of his equipment. Note that Roma must purchase a weapon and an armor set even if he can not cover their cost with obtained coins.
Help Roma find the maximum profit of the grind.
Input
The first line contains three integers n, m, and p (1 β€ n, m, p β€ 2 β
10^5) β the number of available weapons, armor sets and monsters respectively.
The following n lines describe available weapons. The i-th of these lines contains two integers a_i and ca_i (1 β€ a_i β€ 10^6, 1 β€ ca_i β€ 10^9) β the attack modifier and the cost of the weapon i.
The following m lines describe available armor sets. The j-th of these lines contains two integers b_j and cb_j (1 β€ b_j β€ 10^6, 1 β€ cb_j β€ 10^9) β the defense modifier and the cost of the armor set j.
The following p lines describe monsters. The k-th of these lines contains three integers x_k, y_k, z_k (1 β€ x_k, y_k β€ 10^6, 1 β€ z_k β€ 10^3) β defense, attack and the number of coins of the monster k.
Output
Print a single integer β the maximum profit of the grind.
Example
Input
2 3 3
2 3
4 7
2 4
3 2
5 11
1 2 4
2 1 6
3 4 6
Output
1
Tags: brute force, data structures, sortings
Correct Solution:
```
from sys import stdin, stdout
from bisect import bisect_left, bisect_right
dp = []
add = []
am = []
def build(pos, l,r):
global dp
global am
if l == r - 1:
dp[pos] = -am[l][1]
return
mid = int((l+r)/2)
build (pos*2+1,l,mid)
build (pos*2+2,mid,r)
dp[pos] = max(dp[pos*2+1],dp[pos*2+2])
def push(pos, l,r):
global dp
global add
if add[pos] != 0:
if r- l > 1:
add[pos*2+1] += add[pos]
dp[pos*2+1] += add[pos]
add[pos*2+2] += add[pos]
dp[pos*2+2] += add[pos]
add[pos] = 0
def update(pos, l,r,L,R, val):
global dp
global add
if L>=R:
return
if l == L and r == R:
add[pos] += val
dp[pos] += val
push(pos, l,r)
return
push(pos, l,r)
mid = int((l+r)/2)
update(pos*2+1, l, mid, L, min(R,mid), val)
update(pos*2+2, mid, r, max(L,mid), R, val)
dp[pos] = max(dp[pos*2+1],dp[pos*2+2])
def main():
global dp
global am
global add
n,m,k = list(map(int, stdin.readline().split()))
AMT = 1000002
dp = [0] * (m *4 + 2)
add = [0] * (m *4 + 2)
WP = [-1] * AMT
AM = [-1] * AMT
am2 = []
PWN = [[] for _ in range(AMT)]
max_atk = -1
for _ in range(n):
power, cost = list(map(int, stdin.readline().split()))
if WP[power] == -1 or (WP[power] != -1 and WP[power] > cost):
WP[power] = cost
if power > max_atk:
max_atk = power
for _ in range(m):
power, cost = list(map(int, stdin.readline().split()))
if AM[power] == -1 or (AM[power] != -1 and AM[power] > cost):
AM[power] = cost
for _ in range(k):
atk, df, gold = list(map(int, stdin.readline().split()))
PWN[atk+1].append((df, gold))
for i in range(AMT):
if AM[i] != -1:
am.append((i, AM[i]))
am2.append(i)
m = len(am)
build(0,0,m)
mx = -999999999999
for i in range(AMT):
if i > max_atk:
break
for a,b in PWN[i]:
right = bisect_right(am2, a)
update(0, 0, m, right, m, b)
if WP[i] == -1:
continue
temp = dp[0]
if temp - WP[i] > mx:
mx = temp - WP[i]
stdout.write(str(mx))
main()
```
| 96,253 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Roma is playing a new expansion for his favorite game World of Darkraft. He made a new character and is going for his first grind.
Roma has a choice to buy exactly one of n different weapons and exactly one of m different armor sets. Weapon i has attack modifier a_i and is worth ca_i coins, and armor set j has defense modifier b_j and is worth cb_j coins.
After choosing his equipment Roma can proceed to defeat some monsters. There are p monsters he can try to defeat. Monster k has defense x_k, attack y_k and possesses z_k coins. Roma can defeat a monster if his weapon's attack modifier is larger than the monster's defense, and his armor set's defense modifier is larger than the monster's attack. That is, a monster k can be defeated with a weapon i and an armor set j if a_i > x_k and b_j > y_k. After defeating the monster, Roma takes all the coins from them. During the grind, Roma can defeat as many monsters as he likes. Monsters do not respawn, thus each monster can be defeated at most one.
Thanks to Roma's excessive donations, we can assume that he has an infinite amount of in-game currency and can afford any of the weapons and armor sets. Still, he wants to maximize the profit of the grind. The profit is defined as the total coins obtained from all defeated monsters minus the cost of his equipment. Note that Roma must purchase a weapon and an armor set even if he can not cover their cost with obtained coins.
Help Roma find the maximum profit of the grind.
Input
The first line contains three integers n, m, and p (1 β€ n, m, p β€ 2 β
10^5) β the number of available weapons, armor sets and monsters respectively.
The following n lines describe available weapons. The i-th of these lines contains two integers a_i and ca_i (1 β€ a_i β€ 10^6, 1 β€ ca_i β€ 10^9) β the attack modifier and the cost of the weapon i.
The following m lines describe available armor sets. The j-th of these lines contains two integers b_j and cb_j (1 β€ b_j β€ 10^6, 1 β€ cb_j β€ 10^9) β the defense modifier and the cost of the armor set j.
The following p lines describe monsters. The k-th of these lines contains three integers x_k, y_k, z_k (1 β€ x_k, y_k β€ 10^6, 1 β€ z_k β€ 10^3) β defense, attack and the number of coins of the monster k.
Output
Print a single integer β the maximum profit of the grind.
Example
Input
2 3 3
2 3
4 7
2 4
3 2
5 11
1 2 4
2 1 6
3 4 6
Output
1
Tags: brute force, data structures, sortings
Correct Solution:
```
from bisect import bisect_right
from operator import itemgetter
# quick input by @pajenegod
import io,os
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
class SegmTree:
def __init__(self, size):
N = 1
h = 0
while N < size:
N <<= 1
h += 1
self.N = N
self.h = h
self.t = [0] * (2 * N)
self.d = [0] * N
def apply(self, p, value):
self.t[p] += value
if (p < self.N):
self.d[p] += value
def build(self, p):
t = self.t
d = self.d
while p > 1:
p >>= 1
t[p] = max(t[p<<1], t[p<<1|1]) + d[p]
def rebuild(self):
t = self.t
for p in reversed(range(1, self.N)):
t[p] = max(t[p<<1], t[p<<1|1])
def push(self, p):
d = self.d
for s in range(self.h, 0, -1):
i = p >> s
if d[i] != 0:
self.apply(i<<1, d[i])
self.apply(i<<1|1, d[i])
d[i] = 0
def inc(self, l, r, value):
if l >= r:
return
l += self.N
r += self.N
l0, r0 = l, r
while l < r:
if l & 1:
self.apply(l, value)
l += 1
if r & 1:
r -= 1
self.apply(r, value)
l >>= 1
r >>= 1
self.build(l0)
self.build(r0 - 1)
def query(self, l, r):
if l >= r:
return -float('inf')
t = self.t
l += self.N
r += self.N
self.push(l)
self.push(r - 1)
res = -float('inf')
while l < r:
if l & 1:
res = max(res, t[l])
l += 1
if r & 1:
r -= 1
res = max(t[r], res)
l >>= 1
r >>= 1
return res
n, m, p = map(int, input().split())
weapon = []
for _ in range(n):
a, ca = map(int, input().split())
weapon.append((a, ca))
defense = []
for _ in range(m):
b, cb = map(int, input().split())
defense.append((b, cb))
monster = []
for _ in range(p):
x, y, z = map(int, input().split())
monster.append((x, y, z))
weapon.sort(key=itemgetter(0))
defense.sort(key=itemgetter(0))
monster.sort(key=itemgetter(0))
st = SegmTree(m)
N = st.N
t = st.t
for i, (b, cb) in enumerate(defense):
t[i + N] = -cb
st.rebuild()
i = 0
maxScore = -float('inf')
for a, ca in weapon:
st.inc(0, m, -ca)
while i < p and monster[i][0] < a:
x, y, z = monster[i]
goodDef = bisect_right(defense, (y + 1, 0))
st.inc(goodDef, m, z)
i += 1
currScore = st.query(0, m)
maxScore = max(maxScore, currScore)
st.inc(0, m, ca)
print(maxScore)
```
| 96,254 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Roma is playing a new expansion for his favorite game World of Darkraft. He made a new character and is going for his first grind.
Roma has a choice to buy exactly one of n different weapons and exactly one of m different armor sets. Weapon i has attack modifier a_i and is worth ca_i coins, and armor set j has defense modifier b_j and is worth cb_j coins.
After choosing his equipment Roma can proceed to defeat some monsters. There are p monsters he can try to defeat. Monster k has defense x_k, attack y_k and possesses z_k coins. Roma can defeat a monster if his weapon's attack modifier is larger than the monster's defense, and his armor set's defense modifier is larger than the monster's attack. That is, a monster k can be defeated with a weapon i and an armor set j if a_i > x_k and b_j > y_k. After defeating the monster, Roma takes all the coins from them. During the grind, Roma can defeat as many monsters as he likes. Monsters do not respawn, thus each monster can be defeated at most one.
Thanks to Roma's excessive donations, we can assume that he has an infinite amount of in-game currency and can afford any of the weapons and armor sets. Still, he wants to maximize the profit of the grind. The profit is defined as the total coins obtained from all defeated monsters minus the cost of his equipment. Note that Roma must purchase a weapon and an armor set even if he can not cover their cost with obtained coins.
Help Roma find the maximum profit of the grind.
Input
The first line contains three integers n, m, and p (1 β€ n, m, p β€ 2 β
10^5) β the number of available weapons, armor sets and monsters respectively.
The following n lines describe available weapons. The i-th of these lines contains two integers a_i and ca_i (1 β€ a_i β€ 10^6, 1 β€ ca_i β€ 10^9) β the attack modifier and the cost of the weapon i.
The following m lines describe available armor sets. The j-th of these lines contains two integers b_j and cb_j (1 β€ b_j β€ 10^6, 1 β€ cb_j β€ 10^9) β the defense modifier and the cost of the armor set j.
The following p lines describe monsters. The k-th of these lines contains three integers x_k, y_k, z_k (1 β€ x_k, y_k β€ 10^6, 1 β€ z_k β€ 10^3) β defense, attack and the number of coins of the monster k.
Output
Print a single integer β the maximum profit of the grind.
Example
Input
2 3 3
2 3
4 7
2 4
3 2
5 11
1 2 4
2 1 6
3 4 6
Output
1
Tags: brute force, data structures, sortings
Correct Solution:
```
# quick input by @c1729 and @pajenegod
import io,os
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
from bisect import bisect_right
from operator import itemgetter
class SegmTree:
'''
- increment on interval
- get max on interval
'''
def __init__(self, size):
N = 1
h = 0
while N < size:
N <<= 1
h += 1
self.N = N
self.h = h
self.t = [0] * (2 * N)
self.d = [0] * N
def apply(self, p, value):
self.t[p] += value
if p < self.N:
self.d[p] += value
def build(self, p):
t = self.t
d = self.d
while p > 1:
p >>= 1
t[p] = max(t[p<<1], t[p<<1|1]) + d[p]
def rebuild(self):
t = self.t
for p in reversed(range(1, self.N)):
t[p] = max(t[p<<1], t[p<<1|1])
def push(self, p):
d = self.d
for s in range(self.h, 0, -1):
i = p >> s
if d[i] != 0:
self.apply(i<<1, d[i])
self.apply(i<<1|1, d[i])
d[i] = 0
def inc(self, l, r, value):
if l >= r:
return
l += self.N
r += self.N
l0, r0 = l, r
while l < r:
if l & 1:
self.apply(l, value)
l += 1
if r & 1:
r -= 1
self.apply(r, value)
l >>= 1
r >>= 1
self.build(l0)
self.build(r0 - 1)
def query(self, l, r):
if l >= r:
return -float('inf')
t = self.t
l += self.N
r += self.N
self.push(l)
self.push(r - 1)
res = -float('inf')
while l < r:
if l & 1:
res = max(res, t[l])
l += 1
if r & 1:
r -= 1
res = max(t[r], res)
l >>= 1
r >>= 1
return res
n, m, p = map(int, input().split())
weapon = []
for _ in range(n):
a, ca = map(int, input().split())
# a, ca = n - _, n - _
weapon.append((a, ca))
defense = []
for _ in range(m):
b, cb = map(int, input().split())
# b, cb = m - _, m - _
defense.append((b, cb))
monster = []
for _ in range(p):
x, y, z = map(int, input().split())
# x, y, z = p - _, p - _, p - _
monster.append((x, y, z))
weapon.sort(key=itemgetter(0))
defense.sort(key=itemgetter(0))
monster.sort(key=itemgetter(0))
# store score of each defense item
st = SegmTree(m)
N, t = st.N, st.t
for i, (b, cb) in enumerate(defense):
t[i + N] = -cb
st.rebuild()
i = 0
maxScore = -float('inf')
for a, ca in weapon:
st.inc(0, m, -ca)
while i < p and monster[i][0] < a:
x, y, z = monster[i]
goodDef = bisect_right(defense, (y + 1, 0))
st.inc(goodDef, m, z)
i += 1
currScore = st.query(0, m)
maxScore = max(maxScore, currScore)
st.inc(0, m, ca)
print(maxScore)
```
| 96,255 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Roma is playing a new expansion for his favorite game World of Darkraft. He made a new character and is going for his first grind.
Roma has a choice to buy exactly one of n different weapons and exactly one of m different armor sets. Weapon i has attack modifier a_i and is worth ca_i coins, and armor set j has defense modifier b_j and is worth cb_j coins.
After choosing his equipment Roma can proceed to defeat some monsters. There are p monsters he can try to defeat. Monster k has defense x_k, attack y_k and possesses z_k coins. Roma can defeat a monster if his weapon's attack modifier is larger than the monster's defense, and his armor set's defense modifier is larger than the monster's attack. That is, a monster k can be defeated with a weapon i and an armor set j if a_i > x_k and b_j > y_k. After defeating the monster, Roma takes all the coins from them. During the grind, Roma can defeat as many monsters as he likes. Monsters do not respawn, thus each monster can be defeated at most one.
Thanks to Roma's excessive donations, we can assume that he has an infinite amount of in-game currency and can afford any of the weapons and armor sets. Still, he wants to maximize the profit of the grind. The profit is defined as the total coins obtained from all defeated monsters minus the cost of his equipment. Note that Roma must purchase a weapon and an armor set even if he can not cover their cost with obtained coins.
Help Roma find the maximum profit of the grind.
Input
The first line contains three integers n, m, and p (1 β€ n, m, p β€ 2 β
10^5) β the number of available weapons, armor sets and monsters respectively.
The following n lines describe available weapons. The i-th of these lines contains two integers a_i and ca_i (1 β€ a_i β€ 10^6, 1 β€ ca_i β€ 10^9) β the attack modifier and the cost of the weapon i.
The following m lines describe available armor sets. The j-th of these lines contains two integers b_j and cb_j (1 β€ b_j β€ 10^6, 1 β€ cb_j β€ 10^9) β the defense modifier and the cost of the armor set j.
The following p lines describe monsters. The k-th of these lines contains three integers x_k, y_k, z_k (1 β€ x_k, y_k β€ 10^6, 1 β€ z_k β€ 10^3) β defense, attack and the number of coins of the monster k.
Output
Print a single integer β the maximum profit of the grind.
Example
Input
2 3 3
2 3
4 7
2 4
3 2
5 11
1 2 4
2 1 6
3 4 6
Output
1
Submitted Solution:
```
from bisect import bisect_left as bl
from bisect import bisect_right as br
import heapq
import math
from collections import *
from functools import reduce,cmp_to_key
import sys
input = sys.stdin.readline
M = mod = 10**9 + 7
def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip('\n').split(' ')]
def st():return input().rstrip('\n')
def val():return int(input().rstrip('\n'))
def li2():return [i for i in input().rstrip('\n').split(' ')]
def li3():return [int(i) for i in input().rstrip('\n')]
def makel(l):
helpl = []
for i in l:
while len(helpl) and helpl[-1][-1] >= i[1]:helpl.pop()
helpl.append(i)
return helpl
def do(l3):
ans = [-1]
tot = 0
atmin = -float('inf')
demin = -float('inf')
for i in range(p):
tot += l3[i][-1]
atmin = max(atmin,l3[i][0] + 1)
demin = max(demin,l3[i][1] + 1)
curra = bl(l1,atmin)
if curra == n:break
else:
curra = a[curra]
currb = bl(l2,demin)
if currb == m:break
else:
currb = b[currb]
ans.append(tot - curra - currb)
return max(ans)
n,m,p = li()
l1 = []
l2 = []
a = []
b = []
l3 = []
for i in range(n):
l1.append(li())
l1.sort()
l1 = makel(l1)
n = len(l1)
for i in range(n):a.append(l1[i][1])
l1 = [i[0] for i in l1]
for i in range(m):
l2.append(li())
l2.sort()
l2 = makel(l2)
m = len(l2)
for i in range(m):b.append(l2[i][1])
l2 = [i[0] for i in l2]
ans = -float('inf')
for i in range(p):l3.append(li())
l3.sort()
ans = do(l3)
l3.sort(key = lambda x:(x[1],x[0]))
ans = max(ans,do(l3[:]))
l3.sort(key = lambda x:(x[2],x[0],x[1]))
ans = max(ans,do(l3[:]))
l3.sort(key = lambda x:(x[2],x[1],x[0]))
ans = max(ans,do(l3))
print(ans)
```
No
| 96,256 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Roma is playing a new expansion for his favorite game World of Darkraft. He made a new character and is going for his first grind.
Roma has a choice to buy exactly one of n different weapons and exactly one of m different armor sets. Weapon i has attack modifier a_i and is worth ca_i coins, and armor set j has defense modifier b_j and is worth cb_j coins.
After choosing his equipment Roma can proceed to defeat some monsters. There are p monsters he can try to defeat. Monster k has defense x_k, attack y_k and possesses z_k coins. Roma can defeat a monster if his weapon's attack modifier is larger than the monster's defense, and his armor set's defense modifier is larger than the monster's attack. That is, a monster k can be defeated with a weapon i and an armor set j if a_i > x_k and b_j > y_k. After defeating the monster, Roma takes all the coins from them. During the grind, Roma can defeat as many monsters as he likes. Monsters do not respawn, thus each monster can be defeated at most one.
Thanks to Roma's excessive donations, we can assume that he has an infinite amount of in-game currency and can afford any of the weapons and armor sets. Still, he wants to maximize the profit of the grind. The profit is defined as the total coins obtained from all defeated monsters minus the cost of his equipment. Note that Roma must purchase a weapon and an armor set even if he can not cover their cost with obtained coins.
Help Roma find the maximum profit of the grind.
Input
The first line contains three integers n, m, and p (1 β€ n, m, p β€ 2 β
10^5) β the number of available weapons, armor sets and monsters respectively.
The following n lines describe available weapons. The i-th of these lines contains two integers a_i and ca_i (1 β€ a_i β€ 10^6, 1 β€ ca_i β€ 10^9) β the attack modifier and the cost of the weapon i.
The following m lines describe available armor sets. The j-th of these lines contains two integers b_j and cb_j (1 β€ b_j β€ 10^6, 1 β€ cb_j β€ 10^9) β the defense modifier and the cost of the armor set j.
The following p lines describe monsters. The k-th of these lines contains three integers x_k, y_k, z_k (1 β€ x_k, y_k β€ 10^6, 1 β€ z_k β€ 10^3) β defense, attack and the number of coins of the monster k.
Output
Print a single integer β the maximum profit of the grind.
Example
Input
2 3 3
2 3
4 7
2 4
3 2
5 11
1 2 4
2 1 6
3 4 6
Output
1
Submitted Solution:
```
import sys
readline = sys.stdin.readline
N, M, P = map(int, readline().split())
xcx = [tuple(map(int, readline().split())) for _ in range(N)]
ycy = [tuple(map(int, readline().split())) for _ in range(M)]
enemy = [tuple(map(int, readline().split())) for _ in range(P)]
JM = 10**6+4
table = [-2147483648]*JM
event = [[] for _ in range(JM)]
for x, cx in xcx:
event[x].append((0, cx))
for y, cy in ycy:
table[y] = max(table[y], -cy)
for x, y, z in enemy:
event[x+1].append((y+1, z))
####
N0 = 1<<20
data = [-2147483648]*N0 + table + [-2147483648]*(N0 - JM)
lazy = [0]*(2*N0)
for i in range(N0-1, 0, -1):
data[i] = max(data[2*i], data[2*i+1])
def addN0(l, x):
L = l+N0
R = 2*N0
Li = L//(L & -L)//2
while L < R :
if L & 1:
data[L] += x
lazy[L] += x
L += 1
L >>= 1
R >>= 1
while Li:
data[Li] = max(data[2*Li], data[2*Li+1]) + lazy[Li]
Li >>= 1
####
ans = -2147483648
for i in range(JM):
for y, z in event[i]:
if y:
addN0(y, z)
else:
ans = max(ans, data[1] - z)
print(ans)
```
No
| 96,257 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Roma is playing a new expansion for his favorite game World of Darkraft. He made a new character and is going for his first grind.
Roma has a choice to buy exactly one of n different weapons and exactly one of m different armor sets. Weapon i has attack modifier a_i and is worth ca_i coins, and armor set j has defense modifier b_j and is worth cb_j coins.
After choosing his equipment Roma can proceed to defeat some monsters. There are p monsters he can try to defeat. Monster k has defense x_k, attack y_k and possesses z_k coins. Roma can defeat a monster if his weapon's attack modifier is larger than the monster's defense, and his armor set's defense modifier is larger than the monster's attack. That is, a monster k can be defeated with a weapon i and an armor set j if a_i > x_k and b_j > y_k. After defeating the monster, Roma takes all the coins from them. During the grind, Roma can defeat as many monsters as he likes. Monsters do not respawn, thus each monster can be defeated at most one.
Thanks to Roma's excessive donations, we can assume that he has an infinite amount of in-game currency and can afford any of the weapons and armor sets. Still, he wants to maximize the profit of the grind. The profit is defined as the total coins obtained from all defeated monsters minus the cost of his equipment. Note that Roma must purchase a weapon and an armor set even if he can not cover their cost with obtained coins.
Help Roma find the maximum profit of the grind.
Input
The first line contains three integers n, m, and p (1 β€ n, m, p β€ 2 β
10^5) β the number of available weapons, armor sets and monsters respectively.
The following n lines describe available weapons. The i-th of these lines contains two integers a_i and ca_i (1 β€ a_i β€ 10^6, 1 β€ ca_i β€ 10^9) β the attack modifier and the cost of the weapon i.
The following m lines describe available armor sets. The j-th of these lines contains two integers b_j and cb_j (1 β€ b_j β€ 10^6, 1 β€ cb_j β€ 10^9) β the defense modifier and the cost of the armor set j.
The following p lines describe monsters. The k-th of these lines contains three integers x_k, y_k, z_k (1 β€ x_k, y_k β€ 10^6, 1 β€ z_k β€ 10^3) β defense, attack and the number of coins of the monster k.
Output
Print a single integer β the maximum profit of the grind.
Example
Input
2 3 3
2 3
4 7
2 4
3 2
5 11
1 2 4
2 1 6
3 4 6
Output
1
Submitted Solution:
```
import sys
reader = (s.rstrip() for s in sys.stdin)
input = reader.__next__
if __name__ == '__main__':
n,m,p = [int(x) for x in input().split()]
weapon=[]
weaponprice=[]
armor=[]
armorprice=[]
monster=[]
for i in range(n):
newW,newWP=[int(x) for x in input().split()]
weapon.append(newW)
weaponprice.append(newWP)
for i in range(m):
newA,newAP=[int(x) for x in input().split()]
armor.append(newA)
armorprice.append(newAP)
for i in range(p):
newM=[int(x) for x in input().split()]
monster.append(newM)
currWi,currAi=weaponprice.index(min(weaponprice)),armorprice.index(min(armorprice))
currW,currA=weapon[currWi],armor[currAi]
#print(armorprice,currAi)
currProfit=0
for i in range(p):
#print(i,currAi,currWi)
if monster[i][0]<currW and monster[i][1]<currA:
currProfit+=monster[i][2]
else:
profitGain=monster[i][2]
dmg,defense=False,False
if monster[i][0]>=currW:
if currWi<n-i:
dmg=weapon[currWi+1]
costW=weaponprice[currWi+1]
if monster[i][1]>=currA:
if currAi<n-i:
defense=armor[currAi+1]
costA=armorprice[currAi+1]
#print(dmg,defense,monster[i])
if dmg>monster[i][0] and defense>monster[i][1]:
if costW<=costA and (costW-weaponprice[currWi])<profitGain:
currProfit+=(monster[i][2])
currWi+=1
currW=dmg
elif costW>=costA and (costA-armorprice[currAi])<profitGain:
currProfit+=(monster[i][2])
currAi+=1
currA=defense
else:
continue
elif dmg>monster[i][0]:
if (costW-weaponprice[currWi])<profitGain:
currProfit+=(monster[i][2])
currWi+=1
currW=dmg
else:
continue
elif defense>monster[i][1]:
if (costA-armorprice[currAi])<profitGain:
currProfit+=(monster[i][2])
currAi+=1
currA=defense
else:
continue
else:
continue
print(currProfit,currWi,currAi)
print(currProfit-weaponprice[currWi]-armorprice[currAi])
```
No
| 96,258 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Roma is playing a new expansion for his favorite game World of Darkraft. He made a new character and is going for his first grind.
Roma has a choice to buy exactly one of n different weapons and exactly one of m different armor sets. Weapon i has attack modifier a_i and is worth ca_i coins, and armor set j has defense modifier b_j and is worth cb_j coins.
After choosing his equipment Roma can proceed to defeat some monsters. There are p monsters he can try to defeat. Monster k has defense x_k, attack y_k and possesses z_k coins. Roma can defeat a monster if his weapon's attack modifier is larger than the monster's defense, and his armor set's defense modifier is larger than the monster's attack. That is, a monster k can be defeated with a weapon i and an armor set j if a_i > x_k and b_j > y_k. After defeating the monster, Roma takes all the coins from them. During the grind, Roma can defeat as many monsters as he likes. Monsters do not respawn, thus each monster can be defeated at most one.
Thanks to Roma's excessive donations, we can assume that he has an infinite amount of in-game currency and can afford any of the weapons and armor sets. Still, he wants to maximize the profit of the grind. The profit is defined as the total coins obtained from all defeated monsters minus the cost of his equipment. Note that Roma must purchase a weapon and an armor set even if he can not cover their cost with obtained coins.
Help Roma find the maximum profit of the grind.
Input
The first line contains three integers n, m, and p (1 β€ n, m, p β€ 2 β
10^5) β the number of available weapons, armor sets and monsters respectively.
The following n lines describe available weapons. The i-th of these lines contains two integers a_i and ca_i (1 β€ a_i β€ 10^6, 1 β€ ca_i β€ 10^9) β the attack modifier and the cost of the weapon i.
The following m lines describe available armor sets. The j-th of these lines contains two integers b_j and cb_j (1 β€ b_j β€ 10^6, 1 β€ cb_j β€ 10^9) β the defense modifier and the cost of the armor set j.
The following p lines describe monsters. The k-th of these lines contains three integers x_k, y_k, z_k (1 β€ x_k, y_k β€ 10^6, 1 β€ z_k β€ 10^3) β defense, attack and the number of coins of the monster k.
Output
Print a single integer β the maximum profit of the grind.
Example
Input
2 3 3
2 3
4 7
2 4
3 2
5 11
1 2 4
2 1 6
3 4 6
Output
1
Submitted Solution:
```
import sys
reader = (s.rstrip() for s in sys.stdin)
input = reader.__next__
if __name__ == '__main__':
n,m,p = [int(x) for x in input().split()]
weapon=[]
weaponprice=[]
armor=[]
armorprice=[]
monster=[]
for i in range(n):
newW,newWP=[int(x) for x in input().split()]
weapon.append(newW)
weaponprice.append(newWP)
for i in range(m):
newA,newAP=[int(x) for x in input().split()]
armor.append(newA)
armorprice.append(newAP)
for i in range(p):
newM=[int(x) for x in input().split()]
monster.append(newM)
currWi,currAi=weaponprice.index(min(weaponprice)),armorprice.index(min(armorprice))
currW,currA=weapon[currWi],armor[currAi]
#print(armorprice,currAi)
currProfit=0
for i in range(p):
#print(i,currAi,currWi)
if monster[i][0]<currW and monster[i][1]<currA:
currProfit+=monster[i][2]
else:
profitGain=monster[i][2]
dmg,defense=False,False
if monster[i][0]>=currW:
if currWi<n-i:
dmg=weapon[currWi+1]
costW=weaponprice[currWi+1]
if monster[i][1]>=currA:
if currAi<n-i:
defense=armor[currAi+1]
costA=armorprice[currAi+1]
if dmg and defense:
if costW<=costA and (costW-weaponprice[currWi])<profitGain:
currProfit+=(monster[i][2])
currWi+=1
currW=dmg
elif costW>=costA and (costA-armorprice[currAi])<profitGain:
currProfit+=(monster[i][2])
currAi+=1
currA=defense
else:
break
elif dmg:
if (costW-weaponprice[currWi])<profitGain:
currProfit+=(monster[i][2])
currWi+=1
currW=dmg
else:
break
elif defense:
if (costA-armorprice[currAi])<profitGain:
currProfit+=(monster[i][2])
currAi+=1
currA=defense
else:
break
else:
break
print(currProfit-weaponprice[currWi]-armorprice[currAi])
```
No
| 96,259 |
Provide tags and a correct Python 3 solution for this coding contest problem.
HQ9+ is a joke programming language which has only four one-character instructions:
* "H" prints "Hello, World!",
* "Q" prints the source code of the program itself,
* "9" prints the lyrics of "99 Bottles of Beer" song,
* "+" increments the value stored in the internal accumulator.
Instructions "H" and "Q" are case-sensitive and must be uppercase. The characters of the program which are not instructions are ignored.
You are given a program written in HQ9+. You have to figure out whether executing this program will produce any output.
Input
The input will consist of a single line p which will give a program in HQ9+. String p will contain between 1 and 100 characters, inclusive. ASCII-code of each character of p will be between 33 (exclamation mark) and 126 (tilde), inclusive.
Output
Output "YES", if executing the program will produce any output, and "NO" otherwise.
Examples
Input
Hi!
Output
YES
Input
Codeforces
Output
NO
Note
In the first case the program contains only one instruction β "H", which prints "Hello, World!".
In the second case none of the program characters are language instructions.
Tags: implementation
Correct Solution:
```
command = input()
act = ['H', 'Q', '9']
x = 0
for el in command:
if el in act:
print('YES')
x = 1
break
if x == 0:
print('NO')
```
| 96,260 |
Provide tags and a correct Python 3 solution for this coding contest problem.
HQ9+ is a joke programming language which has only four one-character instructions:
* "H" prints "Hello, World!",
* "Q" prints the source code of the program itself,
* "9" prints the lyrics of "99 Bottles of Beer" song,
* "+" increments the value stored in the internal accumulator.
Instructions "H" and "Q" are case-sensitive and must be uppercase. The characters of the program which are not instructions are ignored.
You are given a program written in HQ9+. You have to figure out whether executing this program will produce any output.
Input
The input will consist of a single line p which will give a program in HQ9+. String p will contain between 1 and 100 characters, inclusive. ASCII-code of each character of p will be between 33 (exclamation mark) and 126 (tilde), inclusive.
Output
Output "YES", if executing the program will produce any output, and "NO" otherwise.
Examples
Input
Hi!
Output
YES
Input
Codeforces
Output
NO
Note
In the first case the program contains only one instruction β "H", which prints "Hello, World!".
In the second case none of the program characters are language instructions.
Tags: implementation
Correct Solution:
```
s = input()
for i in s:
if i in ['H','Q','9']:
print("YES")
break
else:
print("NO")
```
| 96,261 |
Provide tags and a correct Python 3 solution for this coding contest problem.
HQ9+ is a joke programming language which has only four one-character instructions:
* "H" prints "Hello, World!",
* "Q" prints the source code of the program itself,
* "9" prints the lyrics of "99 Bottles of Beer" song,
* "+" increments the value stored in the internal accumulator.
Instructions "H" and "Q" are case-sensitive and must be uppercase. The characters of the program which are not instructions are ignored.
You are given a program written in HQ9+. You have to figure out whether executing this program will produce any output.
Input
The input will consist of a single line p which will give a program in HQ9+. String p will contain between 1 and 100 characters, inclusive. ASCII-code of each character of p will be between 33 (exclamation mark) and 126 (tilde), inclusive.
Output
Output "YES", if executing the program will produce any output, and "NO" otherwise.
Examples
Input
Hi!
Output
YES
Input
Codeforces
Output
NO
Note
In the first case the program contains only one instruction β "H", which prints "Hello, World!".
In the second case none of the program characters are language instructions.
Tags: implementation
Correct Solution:
```
#!/usr/local/bin/python3
string = input()
if ('H' in string) or ('Q' in string) or ('9' in string):
print("YES")
else:
print("NO")
```
| 96,262 |
Provide tags and a correct Python 3 solution for this coding contest problem.
HQ9+ is a joke programming language which has only four one-character instructions:
* "H" prints "Hello, World!",
* "Q" prints the source code of the program itself,
* "9" prints the lyrics of "99 Bottles of Beer" song,
* "+" increments the value stored in the internal accumulator.
Instructions "H" and "Q" are case-sensitive and must be uppercase. The characters of the program which are not instructions are ignored.
You are given a program written in HQ9+. You have to figure out whether executing this program will produce any output.
Input
The input will consist of a single line p which will give a program in HQ9+. String p will contain between 1 and 100 characters, inclusive. ASCII-code of each character of p will be between 33 (exclamation mark) and 126 (tilde), inclusive.
Output
Output "YES", if executing the program will produce any output, and "NO" otherwise.
Examples
Input
Hi!
Output
YES
Input
Codeforces
Output
NO
Note
In the first case the program contains only one instruction β "H", which prints "Hello, World!".
In the second case none of the program characters are language instructions.
Tags: implementation
Correct Solution:
```
s=input()
if any(i in "HQ9" for i in s):
print('YES')
else:
print('NO')
```
| 96,263 |
Provide tags and a correct Python 3 solution for this coding contest problem.
HQ9+ is a joke programming language which has only four one-character instructions:
* "H" prints "Hello, World!",
* "Q" prints the source code of the program itself,
* "9" prints the lyrics of "99 Bottles of Beer" song,
* "+" increments the value stored in the internal accumulator.
Instructions "H" and "Q" are case-sensitive and must be uppercase. The characters of the program which are not instructions are ignored.
You are given a program written in HQ9+. You have to figure out whether executing this program will produce any output.
Input
The input will consist of a single line p which will give a program in HQ9+. String p will contain between 1 and 100 characters, inclusive. ASCII-code of each character of p will be between 33 (exclamation mark) and 126 (tilde), inclusive.
Output
Output "YES", if executing the program will produce any output, and "NO" otherwise.
Examples
Input
Hi!
Output
YES
Input
Codeforces
Output
NO
Note
In the first case the program contains only one instruction β "H", which prints "Hello, World!".
In the second case none of the program characters are language instructions.
Tags: implementation
Correct Solution:
```
n=input()
if '9' in n or 'H' in n or 'Q' in n:
print('YES')
else:
print('NO')
```
| 96,264 |
Provide tags and a correct Python 3 solution for this coding contest problem.
HQ9+ is a joke programming language which has only four one-character instructions:
* "H" prints "Hello, World!",
* "Q" prints the source code of the program itself,
* "9" prints the lyrics of "99 Bottles of Beer" song,
* "+" increments the value stored in the internal accumulator.
Instructions "H" and "Q" are case-sensitive and must be uppercase. The characters of the program which are not instructions are ignored.
You are given a program written in HQ9+. You have to figure out whether executing this program will produce any output.
Input
The input will consist of a single line p which will give a program in HQ9+. String p will contain between 1 and 100 characters, inclusive. ASCII-code of each character of p will be between 33 (exclamation mark) and 126 (tilde), inclusive.
Output
Output "YES", if executing the program will produce any output, and "NO" otherwise.
Examples
Input
Hi!
Output
YES
Input
Codeforces
Output
NO
Note
In the first case the program contains only one instruction β "H", which prints "Hello, World!".
In the second case none of the program characters are language instructions.
Tags: implementation
Correct Solution:
```
str=input("")
c=0
for i in str:
if i=='H' or i=='Q' or i=='9':
c+=1
if c>0:
print("YES")
else:
print("NO")
```
| 96,265 |
Provide tags and a correct Python 3 solution for this coding contest problem.
HQ9+ is a joke programming language which has only four one-character instructions:
* "H" prints "Hello, World!",
* "Q" prints the source code of the program itself,
* "9" prints the lyrics of "99 Bottles of Beer" song,
* "+" increments the value stored in the internal accumulator.
Instructions "H" and "Q" are case-sensitive and must be uppercase. The characters of the program which are not instructions are ignored.
You are given a program written in HQ9+. You have to figure out whether executing this program will produce any output.
Input
The input will consist of a single line p which will give a program in HQ9+. String p will contain between 1 and 100 characters, inclusive. ASCII-code of each character of p will be between 33 (exclamation mark) and 126 (tilde), inclusive.
Output
Output "YES", if executing the program will produce any output, and "NO" otherwise.
Examples
Input
Hi!
Output
YES
Input
Codeforces
Output
NO
Note
In the first case the program contains only one instruction β "H", which prints "Hello, World!".
In the second case none of the program characters are language instructions.
Tags: implementation
Correct Solution:
```
print('YES' if set('HQ9') & set(input()) else 'NO')
```
| 96,266 |
Provide tags and a correct Python 3 solution for this coding contest problem.
HQ9+ is a joke programming language which has only four one-character instructions:
* "H" prints "Hello, World!",
* "Q" prints the source code of the program itself,
* "9" prints the lyrics of "99 Bottles of Beer" song,
* "+" increments the value stored in the internal accumulator.
Instructions "H" and "Q" are case-sensitive and must be uppercase. The characters of the program which are not instructions are ignored.
You are given a program written in HQ9+. You have to figure out whether executing this program will produce any output.
Input
The input will consist of a single line p which will give a program in HQ9+. String p will contain between 1 and 100 characters, inclusive. ASCII-code of each character of p will be between 33 (exclamation mark) and 126 (tilde), inclusive.
Output
Output "YES", if executing the program will produce any output, and "NO" otherwise.
Examples
Input
Hi!
Output
YES
Input
Codeforces
Output
NO
Note
In the first case the program contains only one instruction β "H", which prints "Hello, World!".
In the second case none of the program characters are language instructions.
Tags: implementation
Correct Solution:
```
x= input()
y=True
for i in range (len(x)):
if x[i] in "HQ9":
print("YES")
y= False
break
if y:
print("NO")
```
| 96,267 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
HQ9+ is a joke programming language which has only four one-character instructions:
* "H" prints "Hello, World!",
* "Q" prints the source code of the program itself,
* "9" prints the lyrics of "99 Bottles of Beer" song,
* "+" increments the value stored in the internal accumulator.
Instructions "H" and "Q" are case-sensitive and must be uppercase. The characters of the program which are not instructions are ignored.
You are given a program written in HQ9+. You have to figure out whether executing this program will produce any output.
Input
The input will consist of a single line p which will give a program in HQ9+. String p will contain between 1 and 100 characters, inclusive. ASCII-code of each character of p will be between 33 (exclamation mark) and 126 (tilde), inclusive.
Output
Output "YES", if executing the program will produce any output, and "NO" otherwise.
Examples
Input
Hi!
Output
YES
Input
Codeforces
Output
NO
Note
In the first case the program contains only one instruction β "H", which prints "Hello, World!".
In the second case none of the program characters are language instructions.
Submitted Solution:
```
#Kinda difficile??
j = 1
word = input()
for i in word:
if i == "H":
print("YES")
break
elif i == "Q":
print("YES")
break
elif i == "9":
print("YES")
break
while j == len(word) :
if word[len(word)-1] != "H":
print("NO")
break
elif word[len(word)-1] != "Q":
print("NO")
break
elif word[len(word)-1] != "9":
print("NO")
break
j += 1
```
Yes
| 96,268 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
HQ9+ is a joke programming language which has only four one-character instructions:
* "H" prints "Hello, World!",
* "Q" prints the source code of the program itself,
* "9" prints the lyrics of "99 Bottles of Beer" song,
* "+" increments the value stored in the internal accumulator.
Instructions "H" and "Q" are case-sensitive and must be uppercase. The characters of the program which are not instructions are ignored.
You are given a program written in HQ9+. You have to figure out whether executing this program will produce any output.
Input
The input will consist of a single line p which will give a program in HQ9+. String p will contain between 1 and 100 characters, inclusive. ASCII-code of each character of p will be between 33 (exclamation mark) and 126 (tilde), inclusive.
Output
Output "YES", if executing the program will produce any output, and "NO" otherwise.
Examples
Input
Hi!
Output
YES
Input
Codeforces
Output
NO
Note
In the first case the program contains only one instruction β "H", which prints "Hello, World!".
In the second case none of the program characters are language instructions.
Submitted Solution:
```
x=input()
if ('H' in x) or ('Q' in x) or ('9' in x):
print("YES")
else:
print("NO")
```
Yes
| 96,269 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
HQ9+ is a joke programming language which has only four one-character instructions:
* "H" prints "Hello, World!",
* "Q" prints the source code of the program itself,
* "9" prints the lyrics of "99 Bottles of Beer" song,
* "+" increments the value stored in the internal accumulator.
Instructions "H" and "Q" are case-sensitive and must be uppercase. The characters of the program which are not instructions are ignored.
You are given a program written in HQ9+. You have to figure out whether executing this program will produce any output.
Input
The input will consist of a single line p which will give a program in HQ9+. String p will contain between 1 and 100 characters, inclusive. ASCII-code of each character of p will be between 33 (exclamation mark) and 126 (tilde), inclusive.
Output
Output "YES", if executing the program will produce any output, and "NO" otherwise.
Examples
Input
Hi!
Output
YES
Input
Codeforces
Output
NO
Note
In the first case the program contains only one instruction β "H", which prints "Hello, World!".
In the second case none of the program characters are language instructions.
Submitted Solution:
```
n=input()
l=["H","Q","9"]
count=0
for i in n:
if i in l:
count+=1
if(count>=1):
print("YES")
else:
print("NO")
```
Yes
| 96,270 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
HQ9+ is a joke programming language which has only four one-character instructions:
* "H" prints "Hello, World!",
* "Q" prints the source code of the program itself,
* "9" prints the lyrics of "99 Bottles of Beer" song,
* "+" increments the value stored in the internal accumulator.
Instructions "H" and "Q" are case-sensitive and must be uppercase. The characters of the program which are not instructions are ignored.
You are given a program written in HQ9+. You have to figure out whether executing this program will produce any output.
Input
The input will consist of a single line p which will give a program in HQ9+. String p will contain between 1 and 100 characters, inclusive. ASCII-code of each character of p will be between 33 (exclamation mark) and 126 (tilde), inclusive.
Output
Output "YES", if executing the program will produce any output, and "NO" otherwise.
Examples
Input
Hi!
Output
YES
Input
Codeforces
Output
NO
Note
In the first case the program contains only one instruction β "H", which prints "Hello, World!".
In the second case none of the program characters are language instructions.
Submitted Solution:
```
import sys
input = sys.stdin.readline
def insr():
s = input()
return(list(s[:len(s) - 1]))
source = insr()
if any(x in source for x in ['H','Q','9']):
print("YES")
else:
print("NO")
```
Yes
| 96,271 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
HQ9+ is a joke programming language which has only four one-character instructions:
* "H" prints "Hello, World!",
* "Q" prints the source code of the program itself,
* "9" prints the lyrics of "99 Bottles of Beer" song,
* "+" increments the value stored in the internal accumulator.
Instructions "H" and "Q" are case-sensitive and must be uppercase. The characters of the program which are not instructions are ignored.
You are given a program written in HQ9+. You have to figure out whether executing this program will produce any output.
Input
The input will consist of a single line p which will give a program in HQ9+. String p will contain between 1 and 100 characters, inclusive. ASCII-code of each character of p will be between 33 (exclamation mark) and 126 (tilde), inclusive.
Output
Output "YES", if executing the program will produce any output, and "NO" otherwise.
Examples
Input
Hi!
Output
YES
Input
Codeforces
Output
NO
Note
In the first case the program contains only one instruction β "H", which prints "Hello, World!".
In the second case none of the program characters are language instructions.
Submitted Solution:
```
p = input()
if "H" in p[0] or "Q" in p[0] or "9" in p[0] or "+" in p[0]:
print("YES")
else:
print("NO")
```
No
| 96,272 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
HQ9+ is a joke programming language which has only four one-character instructions:
* "H" prints "Hello, World!",
* "Q" prints the source code of the program itself,
* "9" prints the lyrics of "99 Bottles of Beer" song,
* "+" increments the value stored in the internal accumulator.
Instructions "H" and "Q" are case-sensitive and must be uppercase. The characters of the program which are not instructions are ignored.
You are given a program written in HQ9+. You have to figure out whether executing this program will produce any output.
Input
The input will consist of a single line p which will give a program in HQ9+. String p will contain between 1 and 100 characters, inclusive. ASCII-code of each character of p will be between 33 (exclamation mark) and 126 (tilde), inclusive.
Output
Output "YES", if executing the program will produce any output, and "NO" otherwise.
Examples
Input
Hi!
Output
YES
Input
Codeforces
Output
NO
Note
In the first case the program contains only one instruction β "H", which prints "Hello, World!".
In the second case none of the program characters are language instructions.
Submitted Solution:
```
s =input()
if 'H' in s or 'Q' in s or '9' in s or '+' in s:
print('YES')
else:
print('NO')
```
No
| 96,273 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
HQ9+ is a joke programming language which has only four one-character instructions:
* "H" prints "Hello, World!",
* "Q" prints the source code of the program itself,
* "9" prints the lyrics of "99 Bottles of Beer" song,
* "+" increments the value stored in the internal accumulator.
Instructions "H" and "Q" are case-sensitive and must be uppercase. The characters of the program which are not instructions are ignored.
You are given a program written in HQ9+. You have to figure out whether executing this program will produce any output.
Input
The input will consist of a single line p which will give a program in HQ9+. String p will contain between 1 and 100 characters, inclusive. ASCII-code of each character of p will be between 33 (exclamation mark) and 126 (tilde), inclusive.
Output
Output "YES", if executing the program will produce any output, and "NO" otherwise.
Examples
Input
Hi!
Output
YES
Input
Codeforces
Output
NO
Note
In the first case the program contains only one instruction β "H", which prints "Hello, World!".
In the second case none of the program characters are language instructions.
Submitted Solution:
```
s = input()
def solve(s):
for x in s:
if x in "HQ9":
return "YES"
return "NO"
print(solve(s))
```
No
| 96,274 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
HQ9+ is a joke programming language which has only four one-character instructions:
* "H" prints "Hello, World!",
* "Q" prints the source code of the program itself,
* "9" prints the lyrics of "99 Bottles of Beer" song,
* "+" increments the value stored in the internal accumulator.
Instructions "H" and "Q" are case-sensitive and must be uppercase. The characters of the program which are not instructions are ignored.
You are given a program written in HQ9+. You have to figure out whether executing this program will produce any output.
Input
The input will consist of a single line p which will give a program in HQ9+. String p will contain between 1 and 100 characters, inclusive. ASCII-code of each character of p will be between 33 (exclamation mark) and 126 (tilde), inclusive.
Output
Output "YES", if executing the program will produce any output, and "NO" otherwise.
Examples
Input
Hi!
Output
YES
Input
Codeforces
Output
NO
Note
In the first case the program contains only one instruction β "H", which prints "Hello, World!".
In the second case none of the program characters are language instructions.
Submitted Solution:
```
w=str(input())
a=0
l=['H','Q','9','+']
for i in l:
if w.find(i)!=-1:
a=1
if a==1:
print('YES')
else:
print('NO')
```
No
| 96,275 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Johnny has just found the new, great tutorial: "How to become a grandmaster?". The tutorial tells many strange and unexpected for Johnny things, such as you have to be patient or that very important is solving many harder and harder problems.
The boy has found an online judge with tasks divided by topics they cover. He has picked p^{k_i} problems from i-th category (p is his favorite number). He wants to solve them in two weeks (the patience condition is too hard for Johnny, so for simplicity, he looks only at easy tasks, which can be solved in such a period). Now our future grandmaster has to decide which topics to cover first and which the second week. Help him assign topics in such a way, that workload is balanced.
Formally, given n numbers p^{k_i}, the boy wants to divide them into two disjoint sets, minimizing the absolute difference between sums of numbers in each set. Find the minimal absolute difference. Output the result modulo 10^{9}+7.
Input
Input consists of multiple test cases. The first line contains one integer t (1 β€ t β€ 10^5) β the number of test cases. Each test case is described as follows:
The first line contains two integers n and p (1 β€ n, p β€ 10^6). The second line contains n integers k_i (0 β€ k_i β€ 10^6).
The sum of n over all test cases doesn't exceed 10^6.
Output
Output one integer β the reminder of division the answer by 1 000 000 007.
Example
Input
4
5 2
2 3 4 4 3
3 1
2 10 1000
4 5
0 1 1 100
1 8
89
Output
4
1
146981438
747093407
Note
You have to minimize the difference, not it's remainder. For example, if the minimum difference is equal to 2, but there is also a distribution where the difference is 10^9 + 8, then the answer is 2, not 1.
In the first test case of the example, there're the following numbers: 4, 8, 16, 16, and 8. We can divide them into such two sets: {4, 8, 16} and {8, 16}. Then the difference between the sums of numbers in sets would be 4.
Tags: greedy, implementation, math, sortings
Correct Solution:
```
import sys
input = sys.stdin.readline
for f in range(int(input())):
n,p=map(int,input().split())
k=list(map(int,input().split()))
mod=1000000007
k.sort(reverse=True)
left=-1
right={}
for x in k:
if left==-1:
left=x
else:
if x==left or p==1:
left=-1
right={}
continue
if x in right:
right[x]+=1
else:
right[x]=1
if right[x]==p:
i=x
done=True
while i<left and done:
if right[i]==p:
right[i]=0
if i+1 in right:
right[i+1]+=1
else:
right[i+1]=1
i+=1
else:
done=False
if i==left:
left=-1
right={}
sol=0
if left>=0:
sol=pow(p,left,mod)
for x in right:
sol-=right[x]*pow(p,x,mod)
sol%=mod
print(sol)
```
| 96,276 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Johnny has just found the new, great tutorial: "How to become a grandmaster?". The tutorial tells many strange and unexpected for Johnny things, such as you have to be patient or that very important is solving many harder and harder problems.
The boy has found an online judge with tasks divided by topics they cover. He has picked p^{k_i} problems from i-th category (p is his favorite number). He wants to solve them in two weeks (the patience condition is too hard for Johnny, so for simplicity, he looks only at easy tasks, which can be solved in such a period). Now our future grandmaster has to decide which topics to cover first and which the second week. Help him assign topics in such a way, that workload is balanced.
Formally, given n numbers p^{k_i}, the boy wants to divide them into two disjoint sets, minimizing the absolute difference between sums of numbers in each set. Find the minimal absolute difference. Output the result modulo 10^{9}+7.
Input
Input consists of multiple test cases. The first line contains one integer t (1 β€ t β€ 10^5) β the number of test cases. Each test case is described as follows:
The first line contains two integers n and p (1 β€ n, p β€ 10^6). The second line contains n integers k_i (0 β€ k_i β€ 10^6).
The sum of n over all test cases doesn't exceed 10^6.
Output
Output one integer β the reminder of division the answer by 1 000 000 007.
Example
Input
4
5 2
2 3 4 4 3
3 1
2 10 1000
4 5
0 1 1 100
1 8
89
Output
4
1
146981438
747093407
Note
You have to minimize the difference, not it's remainder. For example, if the minimum difference is equal to 2, but there is also a distribution where the difference is 10^9 + 8, then the answer is 2, not 1.
In the first test case of the example, there're the following numbers: 4, 8, 16, 16, and 8. We can divide them into such two sets: {4, 8, 16} and {8, 16}. Then the difference between the sums of numbers in sets would be 4.
Tags: greedy, implementation, math, sortings
Correct Solution:
```
import sys
readline = sys.stdin.readline
T = int(readline())
Ans = [None]*T
MOD = 10**9+7
mod = 10**9+9
for qu in range(T):
N, P = map(int, readline().split())
A = list(map(int, readline().split()))
if P == 1:
if N&1:
Ans[qu] = 1
else:
Ans[qu] = 0
continue
if N == 1:
Ans[qu] = pow(P, A[0], MOD)
continue
A.sort(reverse = True)
cans = 0
carry = 0
res = 0
ra = 0
for a in A:
if carry == 0:
carry = pow(P, a, mod)
cans = pow(P, a, MOD)
continue
res = (res + pow(P, a, mod))%mod
ra = (ra + pow(P, a, MOD))%MOD
if res == carry and ra == cans:
carry = 0
cans = 0
ra = 0
res = 0
Ans[qu] = (cans-ra)%MOD
print('\n'.join(map(str, Ans)))
```
| 96,277 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Johnny has just found the new, great tutorial: "How to become a grandmaster?". The tutorial tells many strange and unexpected for Johnny things, such as you have to be patient or that very important is solving many harder and harder problems.
The boy has found an online judge with tasks divided by topics they cover. He has picked p^{k_i} problems from i-th category (p is his favorite number). He wants to solve them in two weeks (the patience condition is too hard for Johnny, so for simplicity, he looks only at easy tasks, which can be solved in such a period). Now our future grandmaster has to decide which topics to cover first and which the second week. Help him assign topics in such a way, that workload is balanced.
Formally, given n numbers p^{k_i}, the boy wants to divide them into two disjoint sets, minimizing the absolute difference between sums of numbers in each set. Find the minimal absolute difference. Output the result modulo 10^{9}+7.
Input
Input consists of multiple test cases. The first line contains one integer t (1 β€ t β€ 10^5) β the number of test cases. Each test case is described as follows:
The first line contains two integers n and p (1 β€ n, p β€ 10^6). The second line contains n integers k_i (0 β€ k_i β€ 10^6).
The sum of n over all test cases doesn't exceed 10^6.
Output
Output one integer β the reminder of division the answer by 1 000 000 007.
Example
Input
4
5 2
2 3 4 4 3
3 1
2 10 1000
4 5
0 1 1 100
1 8
89
Output
4
1
146981438
747093407
Note
You have to minimize the difference, not it's remainder. For example, if the minimum difference is equal to 2, but there is also a distribution where the difference is 10^9 + 8, then the answer is 2, not 1.
In the first test case of the example, there're the following numbers: 4, 8, 16, 16, and 8. We can divide them into such two sets: {4, 8, 16} and {8, 16}. Then the difference between the sums of numbers in sets would be 4.
Tags: greedy, implementation, math, sortings
Correct Solution:
```
#!/usr/bin/env python3
import sys
input = sys.stdin.readline
MOD = 10**9 + 7
t = int(input())
for _ in range(t):
n, p = map(int, input().split())
k = [int(item) for item in input().split()]
if p == 1:
print(n % 2)
continue
k.sort(reverse=True)
initialized = False
right = dict()
for val in k:
if not initialized:
left_index = val
initialized = True
continue
if val in right:
right[val] += 1
else:
right[val] = 1
while right[val] == p:
right[val] = 0
if val + 1 in right:
right[val + 1] += 1
else:
right[val + 1] = 1
val += 1
if val == left_index:
initialized = False
left_index = 0
right = dict()
if not initialized:
print(0)
else:
ans = pow(p, left_index, MOD)
for key, val in right.items():
ans -= pow(p, key, MOD) * val
ans %= MOD
print(ans)
```
| 96,278 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Johnny has just found the new, great tutorial: "How to become a grandmaster?". The tutorial tells many strange and unexpected for Johnny things, such as you have to be patient or that very important is solving many harder and harder problems.
The boy has found an online judge with tasks divided by topics they cover. He has picked p^{k_i} problems from i-th category (p is his favorite number). He wants to solve them in two weeks (the patience condition is too hard for Johnny, so for simplicity, he looks only at easy tasks, which can be solved in such a period). Now our future grandmaster has to decide which topics to cover first and which the second week. Help him assign topics in such a way, that workload is balanced.
Formally, given n numbers p^{k_i}, the boy wants to divide them into two disjoint sets, minimizing the absolute difference between sums of numbers in each set. Find the minimal absolute difference. Output the result modulo 10^{9}+7.
Input
Input consists of multiple test cases. The first line contains one integer t (1 β€ t β€ 10^5) β the number of test cases. Each test case is described as follows:
The first line contains two integers n and p (1 β€ n, p β€ 10^6). The second line contains n integers k_i (0 β€ k_i β€ 10^6).
The sum of n over all test cases doesn't exceed 10^6.
Output
Output one integer β the reminder of division the answer by 1 000 000 007.
Example
Input
4
5 2
2 3 4 4 3
3 1
2 10 1000
4 5
0 1 1 100
1 8
89
Output
4
1
146981438
747093407
Note
You have to minimize the difference, not it's remainder. For example, if the minimum difference is equal to 2, but there is also a distribution where the difference is 10^9 + 8, then the answer is 2, not 1.
In the first test case of the example, there're the following numbers: 4, 8, 16, 16, and 8. We can divide them into such two sets: {4, 8, 16} and {8, 16}. Then the difference between the sums of numbers in sets would be 4.
Tags: greedy, implementation, math, sortings
Correct Solution:
```
from sys import stdin, stdout
import math
from collections import defaultdict
def main():
MOD7 = 1000000007
t = int(stdin.readline())
pw = [0] * 21
for w in range(20,-1,-1):
pw[w] = int(math.pow(2,w))
for ks in range(t):
n,p = list(map(int, stdin.readline().split()))
arr = list(map(int, stdin.readline().split()))
if p == 1:
if n % 2 ==0:
stdout.write("0\n")
else:
stdout.write("1\n")
continue
arr.sort(reverse=True)
left = -1
i = 0
val = [0] * 21
tmp = p
val[0] = p
slot = defaultdict(int)
for x in range(1,21):
tmp = (tmp * tmp) % MOD7
val[x] = tmp
while i < n:
x = arr[i]
if left == -1:
left = x
else:
slot[x] += 1
tmp = x
if x == left:
left = -1
slot.pop(x)
else:
while slot[tmp] % p == 0:
slot[tmp+1] += 1
slot.pop(tmp)
tmp += 1
if tmp == left:
left = -1
slot.pop(tmp)
i+=1
if left == -1:
stdout.write("0\n")
continue
res = 1
for w in range(20,-1,-1):
pww = pw[w]
if pww <= left:
left -= pww
res = (res * val[w]) % MOD7
if left == 0:
break
for x,c in slot.items():
tp = 1
for w in range(20,-1,-1):
pww = pw[w]
if pww <= x:
x -= pww
tp = (tp * val[w]) % MOD7
if x == 0:
break
res = (res - tp * c) % MOD7
stdout.write(str(res)+"\n")
main()
```
| 96,279 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Johnny has just found the new, great tutorial: "How to become a grandmaster?". The tutorial tells many strange and unexpected for Johnny things, such as you have to be patient or that very important is solving many harder and harder problems.
The boy has found an online judge with tasks divided by topics they cover. He has picked p^{k_i} problems from i-th category (p is his favorite number). He wants to solve them in two weeks (the patience condition is too hard for Johnny, so for simplicity, he looks only at easy tasks, which can be solved in such a period). Now our future grandmaster has to decide which topics to cover first and which the second week. Help him assign topics in such a way, that workload is balanced.
Formally, given n numbers p^{k_i}, the boy wants to divide them into two disjoint sets, minimizing the absolute difference between sums of numbers in each set. Find the minimal absolute difference. Output the result modulo 10^{9}+7.
Input
Input consists of multiple test cases. The first line contains one integer t (1 β€ t β€ 10^5) β the number of test cases. Each test case is described as follows:
The first line contains two integers n and p (1 β€ n, p β€ 10^6). The second line contains n integers k_i (0 β€ k_i β€ 10^6).
The sum of n over all test cases doesn't exceed 10^6.
Output
Output one integer β the reminder of division the answer by 1 000 000 007.
Example
Input
4
5 2
2 3 4 4 3
3 1
2 10 1000
4 5
0 1 1 100
1 8
89
Output
4
1
146981438
747093407
Note
You have to minimize the difference, not it's remainder. For example, if the minimum difference is equal to 2, but there is also a distribution where the difference is 10^9 + 8, then the answer is 2, not 1.
In the first test case of the example, there're the following numbers: 4, 8, 16, 16, and 8. We can divide them into such two sets: {4, 8, 16} and {8, 16}. Then the difference between the sums of numbers in sets would be 4.
Tags: greedy, implementation, math, sortings
Correct Solution:
```
import os,io
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
def fastExponentiation(a,b,N): # Calculates a^b mod N in log b time
ans = 1
bDiv2 = b
aPow2 = a
while bDiv2 > 0:
if bDiv2 % 2 != 0:
ans *= aPow2
ans %= N
bDiv2 //= 2
aPow2 = aPow2 * aPow2
aPow2 %= N
return ans
MOD = 1000000007
t = int(input())
for _ in range(t):
n,p = map(int,input().split())
k = list(map(int,input().split()))
k.sort(reverse = True)
estDiff = 0
curDiff = 0
curIndex = 0
lastK = -1
currentK = -1
allMinus = False
if p == 1:
print(n % 2)
continue
while True:
if curIndex >= n:
break
currentK = k[curIndex]
if estDiff >= 1 and (lastK - currentK >= 20 or estDiff * pow(p,lastK - currentK) >= n):
allMinus = True
break
else:
multiplier = fastExponentiation(p,lastK - currentK,MOD)
cnt = 0
while curIndex < n and currentK == k[curIndex]:
cnt += 1
curIndex += 1
if cnt >= estDiff * multiplier:
if (cnt + estDiff * multiplier) % 2 == 0:
estDiff = 0
curDiff = 0
else:
estDiff = 1
curDiff = fastExponentiation(p,currentK,MOD)
curDiff %= MOD
else:
estDiff = estDiff * multiplier - cnt
curDiff -= cnt * fastExponentiation(p,currentK,MOD)
curDiff %= MOD
lastK = currentK
if allMinus:
for elem in k[curIndex:]:
curDiff -= fastExponentiation(p,elem,MOD)
curDiff %= MOD
print(curDiff % MOD)
```
| 96,280 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Johnny has just found the new, great tutorial: "How to become a grandmaster?". The tutorial tells many strange and unexpected for Johnny things, such as you have to be patient or that very important is solving many harder and harder problems.
The boy has found an online judge with tasks divided by topics they cover. He has picked p^{k_i} problems from i-th category (p is his favorite number). He wants to solve them in two weeks (the patience condition is too hard for Johnny, so for simplicity, he looks only at easy tasks, which can be solved in such a period). Now our future grandmaster has to decide which topics to cover first and which the second week. Help him assign topics in such a way, that workload is balanced.
Formally, given n numbers p^{k_i}, the boy wants to divide them into two disjoint sets, minimizing the absolute difference between sums of numbers in each set. Find the minimal absolute difference. Output the result modulo 10^{9}+7.
Input
Input consists of multiple test cases. The first line contains one integer t (1 β€ t β€ 10^5) β the number of test cases. Each test case is described as follows:
The first line contains two integers n and p (1 β€ n, p β€ 10^6). The second line contains n integers k_i (0 β€ k_i β€ 10^6).
The sum of n over all test cases doesn't exceed 10^6.
Output
Output one integer β the reminder of division the answer by 1 000 000 007.
Example
Input
4
5 2
2 3 4 4 3
3 1
2 10 1000
4 5
0 1 1 100
1 8
89
Output
4
1
146981438
747093407
Note
You have to minimize the difference, not it's remainder. For example, if the minimum difference is equal to 2, but there is also a distribution where the difference is 10^9 + 8, then the answer is 2, not 1.
In the first test case of the example, there're the following numbers: 4, 8, 16, 16, and 8. We can divide them into such two sets: {4, 8, 16} and {8, 16}. Then the difference between the sums of numbers in sets would be 4.
Tags: greedy, implementation, math, sortings
Correct Solution:
```
import sys
input=sys.stdin.buffer.readline
mod=10**9+7
for _ in range(int(input())):
n,p=map(int,input().split())
limit=0
temp=1
if p==1:
k=list(map(int,input().split()))
print(n%2)
continue
while 10**7>=temp*p:
limit+=1
temp*=p
k=list(map(int,input().split()))
k.sort()
ans=0
for i in range(n):
ans=(ans+pow(p,k[i],mod))%mod
#print(ans)
net=[[k[i],1] for i in range(n)]
for i in range(n-1):
kurai=net[i][0]
kurai2=net[i+1][0]
sa=kurai2-kurai
if limit>=sa:
q=net[i][1]//(p**sa)
net[i][1]%=p**sa
net[i+1][1]+=q
new=[]
while net:
if not new:
new.append(net[-1])
net.pop()
else:
if new[-1][0]==net[-1][0]:
net.pop()
else:
new.append(net[-1])
net.pop()
net=new[::-1]
id=len(net)-1
#print(net)
check=[]
for i in range(n-1,-1,-1):
kurai=k[i]
if kurai!=net[id][0]:
kurai1=net[id-1][0]
kurai2=net[id][0]
sa=kurai2-kurai1
if sa>limit:
net[id-1][1]+=10**7*(net[id][1]!=0)
else:
net[id-1][1]+=min(10**7,net[id][1]*pow(p,sa))
id-=1
#print(kurai,net)
if net[id][1]>1:
ans=(ans-2*pow(p,kurai,mod))%mod
net[id][1]-=2
check.append(kurai)
#print(check)
print(ans)
```
| 96,281 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Johnny has just found the new, great tutorial: "How to become a grandmaster?". The tutorial tells many strange and unexpected for Johnny things, such as you have to be patient or that very important is solving many harder and harder problems.
The boy has found an online judge with tasks divided by topics they cover. He has picked p^{k_i} problems from i-th category (p is his favorite number). He wants to solve them in two weeks (the patience condition is too hard for Johnny, so for simplicity, he looks only at easy tasks, which can be solved in such a period). Now our future grandmaster has to decide which topics to cover first and which the second week. Help him assign topics in such a way, that workload is balanced.
Formally, given n numbers p^{k_i}, the boy wants to divide them into two disjoint sets, minimizing the absolute difference between sums of numbers in each set. Find the minimal absolute difference. Output the result modulo 10^{9}+7.
Input
Input consists of multiple test cases. The first line contains one integer t (1 β€ t β€ 10^5) β the number of test cases. Each test case is described as follows:
The first line contains two integers n and p (1 β€ n, p β€ 10^6). The second line contains n integers k_i (0 β€ k_i β€ 10^6).
The sum of n over all test cases doesn't exceed 10^6.
Output
Output one integer β the reminder of division the answer by 1 000 000 007.
Example
Input
4
5 2
2 3 4 4 3
3 1
2 10 1000
4 5
0 1 1 100
1 8
89
Output
4
1
146981438
747093407
Note
You have to minimize the difference, not it's remainder. For example, if the minimum difference is equal to 2, but there is also a distribution where the difference is 10^9 + 8, then the answer is 2, not 1.
In the first test case of the example, there're the following numbers: 4, 8, 16, 16, and 8. We can divide them into such two sets: {4, 8, 16} and {8, 16}. Then the difference between the sums of numbers in sets would be 4.
Tags: greedy, implementation, math, sortings
Correct Solution:
```
# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------- fast io --------------------
for j in range(int(input())):
n,p=map(int,input().split());vals=list(map(int,input().split()));vals.sort()
if p==1:
if n%2==0:
print(0)
else:
print(1)
else:
mod=10**9 +7;powers={0:1};last=0
for s in range(n):
if not(vals[s] in powers):
powers[vals[s]]=pow(p,vals[s],mod)
a0=0;a1=0;broke=False;ind=0
cv=0;find=0;num=0
for s in range(n-1,-1,-1):
if find==0:
if cv%2==0:
a0+=powers[vals[s]]
else:
a1+=powers[vals[s]]
cv+=1;find=1;num=vals[s]
else:
if vals[s]==num:
find-=1
else:
diff=num-vals[s];num=vals[s]
while diff>0:
find*=p;diff-=1
if find>s:
broke=True;ind=s;break
if broke==True:
break
find-=1
if cv%2==0:
a0+=powers[vals[s]]
else:
a1+=powers[vals[s]]
a0=a0%mod;a1=a1%mod
if broke==True:
if cv%2==0:
for s in range(ind+1):
a0+=powers[vals[s]];a0=a0%mod
print((a1-a0)%mod)
else:
for s in range(ind+1):
a1+=powers[vals[s]];a1=a1%mod
print((a0-a1)%mod)
else:
if cv%2==0:
print((a1-a0)%mod)
else:
print((a0-a1)%mod)
```
| 96,282 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Johnny has just found the new, great tutorial: "How to become a grandmaster?". The tutorial tells many strange and unexpected for Johnny things, such as you have to be patient or that very important is solving many harder and harder problems.
The boy has found an online judge with tasks divided by topics they cover. He has picked p^{k_i} problems from i-th category (p is his favorite number). He wants to solve them in two weeks (the patience condition is too hard for Johnny, so for simplicity, he looks only at easy tasks, which can be solved in such a period). Now our future grandmaster has to decide which topics to cover first and which the second week. Help him assign topics in such a way, that workload is balanced.
Formally, given n numbers p^{k_i}, the boy wants to divide them into two disjoint sets, minimizing the absolute difference between sums of numbers in each set. Find the minimal absolute difference. Output the result modulo 10^{9}+7.
Input
Input consists of multiple test cases. The first line contains one integer t (1 β€ t β€ 10^5) β the number of test cases. Each test case is described as follows:
The first line contains two integers n and p (1 β€ n, p β€ 10^6). The second line contains n integers k_i (0 β€ k_i β€ 10^6).
The sum of n over all test cases doesn't exceed 10^6.
Output
Output one integer β the reminder of division the answer by 1 000 000 007.
Example
Input
4
5 2
2 3 4 4 3
3 1
2 10 1000
4 5
0 1 1 100
1 8
89
Output
4
1
146981438
747093407
Note
You have to minimize the difference, not it's remainder. For example, if the minimum difference is equal to 2, but there is also a distribution where the difference is 10^9 + 8, then the answer is 2, not 1.
In the first test case of the example, there're the following numbers: 4, 8, 16, 16, and 8. We can divide them into such two sets: {4, 8, 16} and {8, 16}. Then the difference between the sums of numbers in sets would be 4.
Tags: greedy, implementation, math, sortings
Correct Solution:
```
import sys
input = sys.stdin.readline
mod=1000000007
t=int(input())
for tests in range(t):
n,p=map(int,input().split())
K=sorted(map(int,input().split()),reverse=True)
if p==1:
if n%2==0:
print(0)
else:
print(1)
continue
ANS=[0,0]
flag=0
for i in range(n):
k=K[i]
if ANS[0]==0:
ANS=[1,k]
elif ANS[1]==k:
ANS[0]-=1
else:
while ANS[1]>k:
ANS[0]*=p
ANS[1]-=1
if ANS[0]>10**7:
flag=1
lastind=i
break
if flag==1:
A=ANS[0]*pow(p,ANS[1],mod)%mod
#print(A,ANS)
for j in range(lastind,n):
A=(A-pow(p,K[j],mod))%mod
print(A)
break
else:
ANS[0]-=1
if flag==0:
print(ANS[0]*pow(p,ANS[1],mod)%mod)
```
| 96,283 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johnny has just found the new, great tutorial: "How to become a grandmaster?". The tutorial tells many strange and unexpected for Johnny things, such as you have to be patient or that very important is solving many harder and harder problems.
The boy has found an online judge with tasks divided by topics they cover. He has picked p^{k_i} problems from i-th category (p is his favorite number). He wants to solve them in two weeks (the patience condition is too hard for Johnny, so for simplicity, he looks only at easy tasks, which can be solved in such a period). Now our future grandmaster has to decide which topics to cover first and which the second week. Help him assign topics in such a way, that workload is balanced.
Formally, given n numbers p^{k_i}, the boy wants to divide them into two disjoint sets, minimizing the absolute difference between sums of numbers in each set. Find the minimal absolute difference. Output the result modulo 10^{9}+7.
Input
Input consists of multiple test cases. The first line contains one integer t (1 β€ t β€ 10^5) β the number of test cases. Each test case is described as follows:
The first line contains two integers n and p (1 β€ n, p β€ 10^6). The second line contains n integers k_i (0 β€ k_i β€ 10^6).
The sum of n over all test cases doesn't exceed 10^6.
Output
Output one integer β the reminder of division the answer by 1 000 000 007.
Example
Input
4
5 2
2 3 4 4 3
3 1
2 10 1000
4 5
0 1 1 100
1 8
89
Output
4
1
146981438
747093407
Note
You have to minimize the difference, not it's remainder. For example, if the minimum difference is equal to 2, but there is also a distribution where the difference is 10^9 + 8, then the answer is 2, not 1.
In the first test case of the example, there're the following numbers: 4, 8, 16, 16, and 8. We can divide them into such two sets: {4, 8, 16} and {8, 16}. Then the difference between the sums of numbers in sets would be 4.
Submitted Solution:
```
import sys
input = lambda: sys.stdin.readline().rstrip()
T = int(input())
P = 10 ** 9 + 7
for _ in range(T):
N, b = map(int, input().split())
A = sorted([int(a) for a in input().split()])
if b == 1:
print(N % 2)
continue
a = A.pop()
pre = a
s = 1
ans = pow(b, a, P)
while A:
a = A.pop()
s *= b ** min(pre - a, 30)
if s >= len(A) + 5:
ans -= pow(b, a, P)
if ans < 0: ans += P
while A:
a = A.pop()
ans -= pow(b, a, P)
if ans < 0: ans += P
print(ans)
break
if s:
s -= 1
ans -= pow(b, a, P)
if ans < 0: ans += P
pre = a
else:
s = 1
ans = -ans
if ans < 0: ans += P
ans += pow(b, a, P)
if ans >= P: ans -= P
pre = a
else:
print(ans)
```
Yes
| 96,284 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johnny has just found the new, great tutorial: "How to become a grandmaster?". The tutorial tells many strange and unexpected for Johnny things, such as you have to be patient or that very important is solving many harder and harder problems.
The boy has found an online judge with tasks divided by topics they cover. He has picked p^{k_i} problems from i-th category (p is his favorite number). He wants to solve them in two weeks (the patience condition is too hard for Johnny, so for simplicity, he looks only at easy tasks, which can be solved in such a period). Now our future grandmaster has to decide which topics to cover first and which the second week. Help him assign topics in such a way, that workload is balanced.
Formally, given n numbers p^{k_i}, the boy wants to divide them into two disjoint sets, minimizing the absolute difference between sums of numbers in each set. Find the minimal absolute difference. Output the result modulo 10^{9}+7.
Input
Input consists of multiple test cases. The first line contains one integer t (1 β€ t β€ 10^5) β the number of test cases. Each test case is described as follows:
The first line contains two integers n and p (1 β€ n, p β€ 10^6). The second line contains n integers k_i (0 β€ k_i β€ 10^6).
The sum of n over all test cases doesn't exceed 10^6.
Output
Output one integer β the reminder of division the answer by 1 000 000 007.
Example
Input
4
5 2
2 3 4 4 3
3 1
2 10 1000
4 5
0 1 1 100
1 8
89
Output
4
1
146981438
747093407
Note
You have to minimize the difference, not it's remainder. For example, if the minimum difference is equal to 2, but there is also a distribution where the difference is 10^9 + 8, then the answer is 2, not 1.
In the first test case of the example, there're the following numbers: 4, 8, 16, 16, and 8. We can divide them into such two sets: {4, 8, 16} and {8, 16}. Then the difference between the sums of numbers in sets would be 4.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------------------
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
def print_list(l):
print(' '.join(map(str,l)))
# sys.setrecursionlimit(300000)
# from heapq import *
# from collections import deque as dq
# from math import ceil,floor,sqrt,pow
# import bisect as bs
# from collections import Counter
# from collections import defaultdict as dc
M = 1000000007
m = 1000000003
for _ in range(N()):
n,p = RL()
a = RLL()
a.sort(reverse=True)
s = 0
ss = 0
for i in a:
t = pow(p,i,M)
tt = pow(p,i,m)
if s==0 and ss==0:
s+=t
ss+=tt
else:
s-=t
ss-=tt
s%=M
ss%=m
print(s)
```
Yes
| 96,285 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johnny has just found the new, great tutorial: "How to become a grandmaster?". The tutorial tells many strange and unexpected for Johnny things, such as you have to be patient or that very important is solving many harder and harder problems.
The boy has found an online judge with tasks divided by topics they cover. He has picked p^{k_i} problems from i-th category (p is his favorite number). He wants to solve them in two weeks (the patience condition is too hard for Johnny, so for simplicity, he looks only at easy tasks, which can be solved in such a period). Now our future grandmaster has to decide which topics to cover first and which the second week. Help him assign topics in such a way, that workload is balanced.
Formally, given n numbers p^{k_i}, the boy wants to divide them into two disjoint sets, minimizing the absolute difference between sums of numbers in each set. Find the minimal absolute difference. Output the result modulo 10^{9}+7.
Input
Input consists of multiple test cases. The first line contains one integer t (1 β€ t β€ 10^5) β the number of test cases. Each test case is described as follows:
The first line contains two integers n and p (1 β€ n, p β€ 10^6). The second line contains n integers k_i (0 β€ k_i β€ 10^6).
The sum of n over all test cases doesn't exceed 10^6.
Output
Output one integer β the reminder of division the answer by 1 000 000 007.
Example
Input
4
5 2
2 3 4 4 3
3 1
2 10 1000
4 5
0 1 1 100
1 8
89
Output
4
1
146981438
747093407
Note
You have to minimize the difference, not it's remainder. For example, if the minimum difference is equal to 2, but there is also a distribution where the difference is 10^9 + 8, then the answer is 2, not 1.
In the first test case of the example, there're the following numbers: 4, 8, 16, 16, and 8. We can divide them into such two sets: {4, 8, 16} and {8, 16}. Then the difference between the sums of numbers in sets would be 4.
Submitted Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
import collections
from itertools import permutations
from collections import defaultdict
from collections import deque
import threading
#sys.setrecursionlimit(300000)
#threading.stack_size(10**8)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------------------------------------------------------------
#mod = 9223372036854775807
class SegmentTree:
def __init__(self, data, default=-10**6, func=lambda a, b: max(a,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
class SegmentTree1:
def __init__(self, data, default=10**6, func=lambda a, b: min(a,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
MOD=10**9+7
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
mod=10**9+7
omod=998244353
#-------------------------------------------------------------------------
prime = [True for i in range(50001)]
pp=[]
def SieveOfEratosthenes(n=50000):
# Create a boolean array "prime[0..n]" and initialize
# all entries it as true. A value in prime[i] will
# finally be false if i is Not a prime, else true.
p = 2
while (p * p <= n):
# If prime[p] is not changed, then it is a prime
if (prime[p] == True):
# Update all multiples of p
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
for i in range(50001):
if prime[i]:
pp.append(i)
#---------------------------------running code------------------------------------------
t=int(input())
for i in range(t):
n,p=map(int,input().split())
power=[int(i) for i in input().split()]
if p==1:
print(n%2)
else:
power.sort(reverse=True)
ans=[0,0]
ok=True
for i in range(n):
k=power[i]
if ans[0]==0:
ans=[1,k]
elif ans[1]==k:
ans[0]-=1
else:
while ans[1]>k:
ans[1]-=1
ans[0]*=p
if ans[0]>=(n-i+1):
#print(ans)
ok=False
ind=i
break
if ok==False:
output=((ans[0]*pow(p,ans[1],mod))%mod)
for j in range(ind,n):
output=((output-pow(p,power[j],mod))%mod)
print(output)
break
else:
ans[0]-=1
if ok:
print((ans[0]*pow(p,ans[1],mod))%mod)
```
Yes
| 96,286 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johnny has just found the new, great tutorial: "How to become a grandmaster?". The tutorial tells many strange and unexpected for Johnny things, such as you have to be patient or that very important is solving many harder and harder problems.
The boy has found an online judge with tasks divided by topics they cover. He has picked p^{k_i} problems from i-th category (p is his favorite number). He wants to solve them in two weeks (the patience condition is too hard for Johnny, so for simplicity, he looks only at easy tasks, which can be solved in such a period). Now our future grandmaster has to decide which topics to cover first and which the second week. Help him assign topics in such a way, that workload is balanced.
Formally, given n numbers p^{k_i}, the boy wants to divide them into two disjoint sets, minimizing the absolute difference between sums of numbers in each set. Find the minimal absolute difference. Output the result modulo 10^{9}+7.
Input
Input consists of multiple test cases. The first line contains one integer t (1 β€ t β€ 10^5) β the number of test cases. Each test case is described as follows:
The first line contains two integers n and p (1 β€ n, p β€ 10^6). The second line contains n integers k_i (0 β€ k_i β€ 10^6).
The sum of n over all test cases doesn't exceed 10^6.
Output
Output one integer β the reminder of division the answer by 1 000 000 007.
Example
Input
4
5 2
2 3 4 4 3
3 1
2 10 1000
4 5
0 1 1 100
1 8
89
Output
4
1
146981438
747093407
Note
You have to minimize the difference, not it's remainder. For example, if the minimum difference is equal to 2, but there is also a distribution where the difference is 10^9 + 8, then the answer is 2, not 1.
In the first test case of the example, there're the following numbers: 4, 8, 16, 16, and 8. We can divide them into such two sets: {4, 8, 16} and {8, 16}. Then the difference between the sums of numbers in sets would be 4.
Submitted Solution:
```
import sys
input = sys.stdin.buffer.readline
def print(val):
sys.stdout.write(str(val) + '\n')
def prog():
for _ in range(int(input())):
n,p = map(int,input().split())
vals = list(map(int,input().split()))
if p == 1:
print(n%2)
else:
vals.sort(reverse = True)
curr_power = 0
last = vals[0]
too_large = False
for a in range(n):
k = vals[a]
if k < last and curr_power > 0:
for i in range(1,last - k+1):
curr_power *= p
if curr_power > n-a:
too_large = True
break
if too_large:
curr_power %= 1000000007
curr_power = curr_power * pow(p, last - i ,1000000007)
mod_diff = curr_power % 1000000007
for b in range(a,n):
k = vals[b]
mod_diff = (mod_diff - pow(p, k ,1000000007)) % 1000000007
print(mod_diff)
break
else:
if curr_power > 0:
curr_power -= 1
else:
curr_power += 1
else:
if curr_power > 0:
curr_power -= 1
else:
curr_power += 1
last = k
if not too_large:
print((curr_power*pow(p,last,1000000007))%1000000007)
prog()
```
Yes
| 96,287 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johnny has just found the new, great tutorial: "How to become a grandmaster?". The tutorial tells many strange and unexpected for Johnny things, such as you have to be patient or that very important is solving many harder and harder problems.
The boy has found an online judge with tasks divided by topics they cover. He has picked p^{k_i} problems from i-th category (p is his favorite number). He wants to solve them in two weeks (the patience condition is too hard for Johnny, so for simplicity, he looks only at easy tasks, which can be solved in such a period). Now our future grandmaster has to decide which topics to cover first and which the second week. Help him assign topics in such a way, that workload is balanced.
Formally, given n numbers p^{k_i}, the boy wants to divide them into two disjoint sets, minimizing the absolute difference between sums of numbers in each set. Find the minimal absolute difference. Output the result modulo 10^{9}+7.
Input
Input consists of multiple test cases. The first line contains one integer t (1 β€ t β€ 10^5) β the number of test cases. Each test case is described as follows:
The first line contains two integers n and p (1 β€ n, p β€ 10^6). The second line contains n integers k_i (0 β€ k_i β€ 10^6).
The sum of n over all test cases doesn't exceed 10^6.
Output
Output one integer β the reminder of division the answer by 1 000 000 007.
Example
Input
4
5 2
2 3 4 4 3
3 1
2 10 1000
4 5
0 1 1 100
1 8
89
Output
4
1
146981438
747093407
Note
You have to minimize the difference, not it's remainder. For example, if the minimum difference is equal to 2, but there is also a distribution where the difference is 10^9 + 8, then the answer is 2, not 1.
In the first test case of the example, there're the following numbers: 4, 8, 16, 16, and 8. We can divide them into such two sets: {4, 8, 16} and {8, 16}. Then the difference between the sums of numbers in sets would be 4.
Submitted Solution:
```
from sys import stdin, stdout
import math
from collections import defaultdict
def main():
MOD7 = 1000000007
t = int(stdin.readline())
pw = [0] * 21
for w in range(20,-1,-1):
pw[w] = int(math.pow(2,w))
for ks in range(t):
n,p = list(map(int, stdin.readline().split()))
arr = list(map(int, stdin.readline().split()))
if p == 1:
if n % 2 ==0:
stdout.write("0\n")
else:
stdout.write("1\n")
continue
arr.sort(reverse=True)
left = 0
i = 0
val = [0] * 21
tmp = p
val[0] = p
slot = defaultdict(int)
for x in range(1,21):
tmp = (tmp * tmp) % MOD7
val[x] = tmp
while i < n:
x = arr[i]
if left == 0:
left = x
else:
slot[x] += 1
if x == left:
left = 0
slot.pop(x)
elif slot[x] % p == 0:
slot[x+1] += 1
slot.pop(x)
if x+1 == left:
left = 0
slot.pop(x+1)
i+=1
if left == 0:
stdout.write("0\n")
continue
res = 1
for w in range(20,-1,-1):
pww = pw[w]
if pww <= left:
left -= pww
res = (res * val[w]) % MOD7
if left == 0:
break
if res == 1:
print(left,n,p)
for x,c in slot.items():
tp = 1
for w in range(20,-1,-1):
pww = pw[w]
if pww <= x:
x -= pww
tp = (tp * val[w]) % MOD7
if x == 0:
break
res = (res - tp * c) % MOD7
stdout.write(str(res)+"\n")
main()
```
No
| 96,288 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johnny has just found the new, great tutorial: "How to become a grandmaster?". The tutorial tells many strange and unexpected for Johnny things, such as you have to be patient or that very important is solving many harder and harder problems.
The boy has found an online judge with tasks divided by topics they cover. He has picked p^{k_i} problems from i-th category (p is his favorite number). He wants to solve them in two weeks (the patience condition is too hard for Johnny, so for simplicity, he looks only at easy tasks, which can be solved in such a period). Now our future grandmaster has to decide which topics to cover first and which the second week. Help him assign topics in such a way, that workload is balanced.
Formally, given n numbers p^{k_i}, the boy wants to divide them into two disjoint sets, minimizing the absolute difference between sums of numbers in each set. Find the minimal absolute difference. Output the result modulo 10^{9}+7.
Input
Input consists of multiple test cases. The first line contains one integer t (1 β€ t β€ 10^5) β the number of test cases. Each test case is described as follows:
The first line contains two integers n and p (1 β€ n, p β€ 10^6). The second line contains n integers k_i (0 β€ k_i β€ 10^6).
The sum of n over all test cases doesn't exceed 10^6.
Output
Output one integer β the reminder of division the answer by 1 000 000 007.
Example
Input
4
5 2
2 3 4 4 3
3 1
2 10 1000
4 5
0 1 1 100
1 8
89
Output
4
1
146981438
747093407
Note
You have to minimize the difference, not it's remainder. For example, if the minimum difference is equal to 2, but there is also a distribution where the difference is 10^9 + 8, then the answer is 2, not 1.
In the first test case of the example, there're the following numbers: 4, 8, 16, 16, and 8. We can divide them into such two sets: {4, 8, 16} and {8, 16}. Then the difference between the sums of numbers in sets would be 4.
Submitted Solution:
```
"""
#If FastIO not needed, use this and don't forget to strip
#import sys, math
#input = sys.stdin.readline
"""
import os, sys, heapq as h, time
from io import BytesIO, IOBase
from types import GeneratorType
from bisect import bisect_left, bisect_right
from collections import defaultdict as dd, deque as dq, Counter as dc
import math, string
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
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 = self.os.read(self._fd, max(self.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 = self.os.read(self._fd, max(self.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:
self.os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#start_time = time.time()
def getInt(): return int(input())
def getStrs(): return input().split()
def getInts(): return list(map(int,input().split()))
def getStr(): return input()
def listStr(): return list(input())
def getMat(n): return [getInts() for _ in range(n)]
def isInt(s): return '0' <= s[0] <= '9'
MOD = 10**9 + 7
"""
[1,1,1,1,4,4]
We want the split to be as even as possible each time
"""
T = getInt()
for _ in range(1,T+1):
N, P = getInts()
A = getInts()
if T == 9999 and _ < 9999: continue
A.sort()
diff = 0
prev = A[-1]
i = N-1
while i >= 0:
diff *= pow(P,prev-A[i],MOD)
j = i
while j > 0 and A[j-1] == A[j]: j -= 1
num = i-j+1
if num <= diff:
diff -= num
else:
num -= diff
if num % 2: diff = 1
else: diff = 0
prev = A[i]
i = j-1
if T == 9999 and _ == 9999:
print(i,j)
print((diff * pow(P,prev,MOD)) % MOD)
```
No
| 96,289 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johnny has just found the new, great tutorial: "How to become a grandmaster?". The tutorial tells many strange and unexpected for Johnny things, such as you have to be patient or that very important is solving many harder and harder problems.
The boy has found an online judge with tasks divided by topics they cover. He has picked p^{k_i} problems from i-th category (p is his favorite number). He wants to solve them in two weeks (the patience condition is too hard for Johnny, so for simplicity, he looks only at easy tasks, which can be solved in such a period). Now our future grandmaster has to decide which topics to cover first and which the second week. Help him assign topics in such a way, that workload is balanced.
Formally, given n numbers p^{k_i}, the boy wants to divide them into two disjoint sets, minimizing the absolute difference between sums of numbers in each set. Find the minimal absolute difference. Output the result modulo 10^{9}+7.
Input
Input consists of multiple test cases. The first line contains one integer t (1 β€ t β€ 10^5) β the number of test cases. Each test case is described as follows:
The first line contains two integers n and p (1 β€ n, p β€ 10^6). The second line contains n integers k_i (0 β€ k_i β€ 10^6).
The sum of n over all test cases doesn't exceed 10^6.
Output
Output one integer β the reminder of division the answer by 1 000 000 007.
Example
Input
4
5 2
2 3 4 4 3
3 1
2 10 1000
4 5
0 1 1 100
1 8
89
Output
4
1
146981438
747093407
Note
You have to minimize the difference, not it's remainder. For example, if the minimum difference is equal to 2, but there is also a distribution where the difference is 10^9 + 8, then the answer is 2, not 1.
In the first test case of the example, there're the following numbers: 4, 8, 16, 16, and 8. We can divide them into such two sets: {4, 8, 16} and {8, 16}. Then the difference between the sums of numbers in sets would be 4.
Submitted Solution:
```
"""
#If FastIO not needed, use this and don't forget to strip
#import sys, math
#input = sys.stdin.readline
"""
import os, sys, heapq as h, time
from io import BytesIO, IOBase
from types import GeneratorType
from bisect import bisect_left, bisect_right
from collections import defaultdict as dd, deque as dq, Counter as dc
import math, string
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
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 = self.os.read(self._fd, max(self.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 = self.os.read(self._fd, max(self.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:
self.os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#start_time = time.time()
def getInt(): return int(input())
def getStrs(): return input().split()
def getInts(): return list(map(int,input().split()))
def getStr(): return input()
def listStr(): return list(input())
def getMat(n): return [getInts() for _ in range(n)]
def isInt(s): return '0' <= s[0] <= '9'
MOD = 10**9 + 7
"""
[1,1,1,1,4,4]
We want the split to be as even as possible each time
"""
def solve():
N, P = getInts()
A = getInts()
A.sort()
diff = 0
prev = A[-1]
for i in range(N-1,-1,-1):
diff *= pow(P,prev-A[i],MOD)
diff %= MOD
if diff:
diff -= 1
else:
diff = 1
prev = A[i]
return diff * pow(P,prev,MOD)
for _ in range(getInt()):
print(solve())
#solve()
#print(time.time()-start_time)
```
No
| 96,290 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johnny has just found the new, great tutorial: "How to become a grandmaster?". The tutorial tells many strange and unexpected for Johnny things, such as you have to be patient or that very important is solving many harder and harder problems.
The boy has found an online judge with tasks divided by topics they cover. He has picked p^{k_i} problems from i-th category (p is his favorite number). He wants to solve them in two weeks (the patience condition is too hard for Johnny, so for simplicity, he looks only at easy tasks, which can be solved in such a period). Now our future grandmaster has to decide which topics to cover first and which the second week. Help him assign topics in such a way, that workload is balanced.
Formally, given n numbers p^{k_i}, the boy wants to divide them into two disjoint sets, minimizing the absolute difference between sums of numbers in each set. Find the minimal absolute difference. Output the result modulo 10^{9}+7.
Input
Input consists of multiple test cases. The first line contains one integer t (1 β€ t β€ 10^5) β the number of test cases. Each test case is described as follows:
The first line contains two integers n and p (1 β€ n, p β€ 10^6). The second line contains n integers k_i (0 β€ k_i β€ 10^6).
The sum of n over all test cases doesn't exceed 10^6.
Output
Output one integer β the reminder of division the answer by 1 000 000 007.
Example
Input
4
5 2
2 3 4 4 3
3 1
2 10 1000
4 5
0 1 1 100
1 8
89
Output
4
1
146981438
747093407
Note
You have to minimize the difference, not it's remainder. For example, if the minimum difference is equal to 2, but there is also a distribution where the difference is 10^9 + 8, then the answer is 2, not 1.
In the first test case of the example, there're the following numbers: 4, 8, 16, 16, and 8. We can divide them into such two sets: {4, 8, 16} and {8, 16}. Then the difference between the sums of numbers in sets would be 4.
Submitted Solution:
```
from sys import stdin, stdout
import math
from collections import defaultdict
def main():
MOD7 = 1000000007
t = int(stdin.readline())
pw = [0] * 21
for w in range(20,-1,-1):
pw[w] = int(math.pow(2,w))
for ks in range(t):
n,p = list(map(int, stdin.readline().split()))
arr = list(map(int, stdin.readline().split()))
if p == 1:
if n % 2 ==0:
stdout.write("0\n")
else:
stdout.write("1\n")
continue
arr.sort(reverse=True)
left = 0
i = 0
val = [0] * 21
tmp = p
val[0] = p
slot = defaultdict(int)
for x in range(1,21):
tmp = (tmp * tmp) % MOD7
val[x] = tmp
while i < n:
x = arr[i]
if left == 0:
left = x
else:
slot[x] += 1
if x == left:
left = 0
slot.pop(x)
elif slot[x] % p == 0:
slot[x+1] += 1
slot.pop(x)
if x+1 == left:
left = 0
slot.pop(x+1)
i+=1
if left == 0:
stdout.write("0\n")
continue
res = 1
for w in range(20,-1,-1):
pww = pw[w]
if pww <= left:
left -= pww
res = (res * val[w]) % MOD7
if left == 0:
break
for x,c in slot.items():
tp = 1
for w in range(20,-1,-1):
pww = pw[w]
if pww <= x:
x -= pww
tp = (tp * val[w]) % MOD7
if x == 0:
break
res = (res - tp * c) % MOD7
stdout.write(str(res)+"\n")
main()
```
No
| 96,291 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In the game of Mastermind, there are two players β Alice and Bob. Alice has a secret code, which Bob tries to guess. Here, a code is defined as a sequence of n colors. There are exactly n+1 colors in the entire universe, numbered from 1 to n+1 inclusive.
When Bob guesses a code, Alice tells him some information about how good of a guess it is, in the form of two integers x and y.
The first integer x is the number of indices where Bob's guess correctly matches Alice's code. The second integer y is the size of the intersection of the two codes as multisets. That is, if Bob were to change the order of the colors in his guess, y is the maximum number of indices he could get correct.
For example, suppose n=5, Alice's code is [3,1,6,1,2], and Bob's guess is [3,1,1,2,5]. At indices 1 and 2 colors are equal, while in the other indices they are not equal. So x=2. And the two codes have the four colors 1,1,2,3 in common, so y=4.
<image> Solid lines denote a matched color for the same index. Dashed lines denote a matched color at a different index. x is the number of solid lines, and y is the total number of lines.
You are given Bob's guess and two values x and y. Can you find one possibility of Alice's code so that the values of x and y are correct?
Input
The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of each test case contains three integers n,x,y (1β€ nβ€ 10^5, 0β€ xβ€ yβ€ n) β the length of the codes, and two values Alice responds with.
The second line of each test case contains n integers b_1,β¦,b_n (1β€ b_iβ€ n+1) β Bob's guess, where b_i is the i-th color of the guess.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, on the first line, output "YES" if there is a solution, or "NO" if there is no possible secret code consistent with the described situation. You can print each character in any case (upper or lower).
If the answer is "YES", on the next line output n integers a_1,β¦,a_n (1β€ a_iβ€ n+1) β Alice's secret code, where a_i is the i-th color of the code.
If there are multiple solutions, output any.
Example
Input
7
5 2 4
3 1 1 2 5
5 3 4
1 1 2 1 2
4 0 4
5 5 3 3
4 1 4
2 3 2 3
6 1 2
3 2 1 1 1 1
6 2 4
3 3 2 1 1 1
6 2 6
1 1 3 2 1 1
Output
YES
3 1 6 1 2
YES
3 1 1 1 2
YES
3 3 5 5
NO
YES
4 4 4 4 3 1
YES
3 1 3 1 7 7
YES
2 3 1 1 1 1
Note
The first test case is described in the statement.
In the second test case, x=3 because the colors are equal at indices 2,4,5. And y=4 because they share the colors 1,1,1,2.
In the third test case, x=0 because there is no index where the colors are the same. But y=4 because they share the colors 3,3,5,5.
In the fourth test case, it can be proved that no solution exists.
Tags: constructive algorithms, graph matchings, greedy, implementation, sortings, two pointers
Correct Solution:
```
from sys import stdin, stdout
from collections import defaultdict
from heapq import heapify, heappop, heappush
def solve():
n, s, y = map(int, stdin.readline().split())
a = stdin.readline().split()
d = defaultdict(list)
for i, x in enumerate(a):
d[x].append(i)
for i in range(1, n + 2):
e = str(i)
if e not in d:
break
q = [(-len(d[x]), x) for x in d.keys()]
heapify(q)
ans = [0] * n
for i in range(s):
l, x = heappop(q)
ans[d[x].pop()] = x
l += 1
if l:
heappush(q, (l, x))
p = []
while q:
l, x = heappop(q)
p.extend(d[x])
if p:
h = (n - s) // 2
y = n - y
q = p[h:] + p[:h]
for x, z in zip(p, q):
if a[x] == a[z]:
if y:
ans[x] = e
y -= 1
else:
stdout.write("NO\n")
return
else:
ans[x] = a[z]
for i in range(n - s):
if y and ans[p[i]] != e:
ans[p[i]] = e
y -= 1
stdout.write("YES\n")
stdout.write(' '.join(ans))
stdout.write("\n")
T = int(stdin.readline())
for t in range(T):
solve()
```
| 96,292 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In the game of Mastermind, there are two players β Alice and Bob. Alice has a secret code, which Bob tries to guess. Here, a code is defined as a sequence of n colors. There are exactly n+1 colors in the entire universe, numbered from 1 to n+1 inclusive.
When Bob guesses a code, Alice tells him some information about how good of a guess it is, in the form of two integers x and y.
The first integer x is the number of indices where Bob's guess correctly matches Alice's code. The second integer y is the size of the intersection of the two codes as multisets. That is, if Bob were to change the order of the colors in his guess, y is the maximum number of indices he could get correct.
For example, suppose n=5, Alice's code is [3,1,6,1,2], and Bob's guess is [3,1,1,2,5]. At indices 1 and 2 colors are equal, while in the other indices they are not equal. So x=2. And the two codes have the four colors 1,1,2,3 in common, so y=4.
<image> Solid lines denote a matched color for the same index. Dashed lines denote a matched color at a different index. x is the number of solid lines, and y is the total number of lines.
You are given Bob's guess and two values x and y. Can you find one possibility of Alice's code so that the values of x and y are correct?
Input
The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of each test case contains three integers n,x,y (1β€ nβ€ 10^5, 0β€ xβ€ yβ€ n) β the length of the codes, and two values Alice responds with.
The second line of each test case contains n integers b_1,β¦,b_n (1β€ b_iβ€ n+1) β Bob's guess, where b_i is the i-th color of the guess.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, on the first line, output "YES" if there is a solution, or "NO" if there is no possible secret code consistent with the described situation. You can print each character in any case (upper or lower).
If the answer is "YES", on the next line output n integers a_1,β¦,a_n (1β€ a_iβ€ n+1) β Alice's secret code, where a_i is the i-th color of the code.
If there are multiple solutions, output any.
Example
Input
7
5 2 4
3 1 1 2 5
5 3 4
1 1 2 1 2
4 0 4
5 5 3 3
4 1 4
2 3 2 3
6 1 2
3 2 1 1 1 1
6 2 4
3 3 2 1 1 1
6 2 6
1 1 3 2 1 1
Output
YES
3 1 6 1 2
YES
3 1 1 1 2
YES
3 3 5 5
NO
YES
4 4 4 4 3 1
YES
3 1 3 1 7 7
YES
2 3 1 1 1 1
Note
The first test case is described in the statement.
In the second test case, x=3 because the colors are equal at indices 2,4,5. And y=4 because they share the colors 1,1,1,2.
In the third test case, x=0 because there is no index where the colors are the same. But y=4 because they share the colors 3,3,5,5.
In the fourth test case, it can be proved that no solution exists.
Tags: constructive algorithms, graph matchings, greedy, implementation, sortings, two pointers
Correct Solution:
```
import sys, math
import io, os
#data = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
from bisect import bisect_left as bl, bisect_right as br, insort
from heapq import heapify, heappush, heappop
from collections import defaultdict as dd, deque, Counter
#from itertools import permutations,combinations
def data(): return sys.stdin.readline().strip()
def mdata(): return list(map(int, data().split()))
def outl(var) : sys.stdout.write(' '.join(map(str, var))+'\n')
def out(var) : sys.stdout.write(str(var)+'\n')
#from decimal import Decimal
from fractions import Fraction
#sys.setrecursionlimit(100000)
INF = float('inf')
mod = int(1e9)+7
for t in range(int(data())):
n,x,y=mdata()
a=mdata()
d=dd(list)
for i in range(n):
d[a[i]].append(i)
q=[]
for i in d.keys():
heappush(q,[-len(d[i]),i])
ind=1
while d[ind]:
ind+=1
ans=[ind]*n
for i in range(x):
j,k=heappop(q)
i=d[k].pop()
ans[i]=a[i]
heappush(q, [-len(d[k]), k])
l1=[]
while q:
j, k = heappop(q)
l1+=d[k]
ind=0
flag=True
y-=x
n-=x
l1=l1[::-1]
while ind<y:
if a[l1[ind]]==a[l1[(n//2+ind)%n]]:
flag=False
break
ans[l1[ind]]=a[l1[(n//2+ind)%n]]
y-=1
if ind>=y:
break
ans[l1[ind+n//2+n%2]]=a[l1[ind]]
ind+=1
if flag==True:
out("YES")
outl(ans)
else:
out("NO")
```
| 96,293 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In the game of Mastermind, there are two players β Alice and Bob. Alice has a secret code, which Bob tries to guess. Here, a code is defined as a sequence of n colors. There are exactly n+1 colors in the entire universe, numbered from 1 to n+1 inclusive.
When Bob guesses a code, Alice tells him some information about how good of a guess it is, in the form of two integers x and y.
The first integer x is the number of indices where Bob's guess correctly matches Alice's code. The second integer y is the size of the intersection of the two codes as multisets. That is, if Bob were to change the order of the colors in his guess, y is the maximum number of indices he could get correct.
For example, suppose n=5, Alice's code is [3,1,6,1,2], and Bob's guess is [3,1,1,2,5]. At indices 1 and 2 colors are equal, while in the other indices they are not equal. So x=2. And the two codes have the four colors 1,1,2,3 in common, so y=4.
<image> Solid lines denote a matched color for the same index. Dashed lines denote a matched color at a different index. x is the number of solid lines, and y is the total number of lines.
You are given Bob's guess and two values x and y. Can you find one possibility of Alice's code so that the values of x and y are correct?
Input
The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of each test case contains three integers n,x,y (1β€ nβ€ 10^5, 0β€ xβ€ yβ€ n) β the length of the codes, and two values Alice responds with.
The second line of each test case contains n integers b_1,β¦,b_n (1β€ b_iβ€ n+1) β Bob's guess, where b_i is the i-th color of the guess.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, on the first line, output "YES" if there is a solution, or "NO" if there is no possible secret code consistent with the described situation. You can print each character in any case (upper or lower).
If the answer is "YES", on the next line output n integers a_1,β¦,a_n (1β€ a_iβ€ n+1) β Alice's secret code, where a_i is the i-th color of the code.
If there are multiple solutions, output any.
Example
Input
7
5 2 4
3 1 1 2 5
5 3 4
1 1 2 1 2
4 0 4
5 5 3 3
4 1 4
2 3 2 3
6 1 2
3 2 1 1 1 1
6 2 4
3 3 2 1 1 1
6 2 6
1 1 3 2 1 1
Output
YES
3 1 6 1 2
YES
3 1 1 1 2
YES
3 3 5 5
NO
YES
4 4 4 4 3 1
YES
3 1 3 1 7 7
YES
2 3 1 1 1 1
Note
The first test case is described in the statement.
In the second test case, x=3 because the colors are equal at indices 2,4,5. And y=4 because they share the colors 1,1,1,2.
In the third test case, x=0 because there is no index where the colors are the same. But y=4 because they share the colors 3,3,5,5.
In the fourth test case, it can be proved that no solution exists.
Tags: constructive algorithms, graph matchings, greedy, implementation, sortings, two pointers
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
import heapq
t = int(stdin.readline())
for _ in range(t):
n, x, y = map(int, stdin.readline().split())
a = list(map(int, stdin.readline().split()))
unused = set(list(range(1, n+2))) - set(a)
r = list(unused)[0]
ans = [r]*n
c = Counter(a)
h = []
ilist = defaultdict(list)
for i, k in enumerate(a):
ilist[k].append(i)
for k in c:
v = c[k]
heapq.heappush(h, (-v, k))
for _ in range(x):
v, k = heapq.heappop(h)
idx = ilist[k].pop()
ans[idx] = k
if v+1 < 0:
heapq.heappush(h, (v+1, k))
todo = y-x
while todo > 0:
v1, k1 = heapq.heappop(h)
idx1 = ilist[k1].pop()
if not h:
if todo == 1:
for i in range(n):
if ans[i] != r and ans[i] != a[i] and ans[i] != k1:
ans[i], ans[idx1] = k1, ans[i]
if ans[i] == a[i] or ans[idx1] == a[idx1]:
ans[i], ans[idx1] = ans[idx1], ans[i]
continue
todo -= 1
break
break
v2, k2 = heapq.heappop(h)
idx2 = ilist[k2].pop()
if todo > 1:
ans[idx1], ans[idx2] = k2, k1
if v1+1 < 0:
heapq.heappush(h, (v1+1, k1))
if v2+1 < 0:
heapq.heappush(h, (v2+1, k2))
todo -= 2
else:
ans[idx1] = k2
todo -= 1
if todo > 0:
print("NO")
continue
print("YES")
print(" ".join(map(str, ans)))
```
| 96,294 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In the game of Mastermind, there are two players β Alice and Bob. Alice has a secret code, which Bob tries to guess. Here, a code is defined as a sequence of n colors. There are exactly n+1 colors in the entire universe, numbered from 1 to n+1 inclusive.
When Bob guesses a code, Alice tells him some information about how good of a guess it is, in the form of two integers x and y.
The first integer x is the number of indices where Bob's guess correctly matches Alice's code. The second integer y is the size of the intersection of the two codes as multisets. That is, if Bob were to change the order of the colors in his guess, y is the maximum number of indices he could get correct.
For example, suppose n=5, Alice's code is [3,1,6,1,2], and Bob's guess is [3,1,1,2,5]. At indices 1 and 2 colors are equal, while in the other indices they are not equal. So x=2. And the two codes have the four colors 1,1,2,3 in common, so y=4.
<image> Solid lines denote a matched color for the same index. Dashed lines denote a matched color at a different index. x is the number of solid lines, and y is the total number of lines.
You are given Bob's guess and two values x and y. Can you find one possibility of Alice's code so that the values of x and y are correct?
Input
The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of each test case contains three integers n,x,y (1β€ nβ€ 10^5, 0β€ xβ€ yβ€ n) β the length of the codes, and two values Alice responds with.
The second line of each test case contains n integers b_1,β¦,b_n (1β€ b_iβ€ n+1) β Bob's guess, where b_i is the i-th color of the guess.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, on the first line, output "YES" if there is a solution, or "NO" if there is no possible secret code consistent with the described situation. You can print each character in any case (upper or lower).
If the answer is "YES", on the next line output n integers a_1,β¦,a_n (1β€ a_iβ€ n+1) β Alice's secret code, where a_i is the i-th color of the code.
If there are multiple solutions, output any.
Example
Input
7
5 2 4
3 1 1 2 5
5 3 4
1 1 2 1 2
4 0 4
5 5 3 3
4 1 4
2 3 2 3
6 1 2
3 2 1 1 1 1
6 2 4
3 3 2 1 1 1
6 2 6
1 1 3 2 1 1
Output
YES
3 1 6 1 2
YES
3 1 1 1 2
YES
3 3 5 5
NO
YES
4 4 4 4 3 1
YES
3 1 3 1 7 7
YES
2 3 1 1 1 1
Note
The first test case is described in the statement.
In the second test case, x=3 because the colors are equal at indices 2,4,5. And y=4 because they share the colors 1,1,1,2.
In the third test case, x=0 because there is no index where the colors are the same. But y=4 because they share the colors 3,3,5,5.
In the fourth test case, it can be proved that no solution exists.
Tags: constructive algorithms, graph matchings, greedy, implementation, sortings, two pointers
Correct Solution:
```
from sys import stdin
from collections import deque
from itertools import chain
input = stdin.readline
def getint(): return int(input())
def getints(): return list(map(int, input().split()))
def getint1(): return list(map(lambda x : int(x) - 1, input().split()))
def getstr(): return input()[:-1]
def solve():
n, x, y = getints()
a = getints()
# table = [[] for _ in range(n + 1)]
ind = [[] for _ in range(n + 2)]
for i, color in enumerate(a):
ind[color].append(i)
not_seen = 0
for i in range(1, n + 2):
if len(ind[i]) == 0:
not_seen = i
break
max_cnt = max(len(g) for g in ind)
# print(max_cnt)
table = [[] for _ in range(max_cnt + 1)]
for i, col in enumerate(ind[1:], 1):
if len(col) > 0:
table[len(col)].append(i)
# for t in table:
# print(t)
ans = [0] * n
for _ in range(x):
assert len(table[max_cnt]) > 0
color = table[max_cnt].pop()
assert len(ind[color]) > 0
i = ind[color].pop()
ans[i] = color
if max_cnt > 1:
table[max_cnt - 1].append(color)
if len(table[max_cnt]) == 0:
max_cnt -= 1
if 2 * max_cnt > 2 * n - x - y:
print("NO")
return
rest = list(chain(*ind))
to_change, offset, m = n - y, (n - x) >> 1, n - x
for i in range(m):
ans[rest[i]] = a[rest[(i + offset) % m]]
if ans[rest[i]] == a[rest[i]]:
ans[rest[i]] = not_seen
to_change -= 1
i = 0
while to_change > 0:
if ans[rest[i]] != not_seen:
ans[rest[i]] = not_seen
to_change -= 1
i += 1
print("YES")
print(*ans)
if __name__ == "__main__":
# solve()
# for t in range(getint()):
# print('Case #', t + 1, ': ', sep='')
# solve()
for _ in range(getint()):
solve()
```
| 96,295 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In the game of Mastermind, there are two players β Alice and Bob. Alice has a secret code, which Bob tries to guess. Here, a code is defined as a sequence of n colors. There are exactly n+1 colors in the entire universe, numbered from 1 to n+1 inclusive.
When Bob guesses a code, Alice tells him some information about how good of a guess it is, in the form of two integers x and y.
The first integer x is the number of indices where Bob's guess correctly matches Alice's code. The second integer y is the size of the intersection of the two codes as multisets. That is, if Bob were to change the order of the colors in his guess, y is the maximum number of indices he could get correct.
For example, suppose n=5, Alice's code is [3,1,6,1,2], and Bob's guess is [3,1,1,2,5]. At indices 1 and 2 colors are equal, while in the other indices they are not equal. So x=2. And the two codes have the four colors 1,1,2,3 in common, so y=4.
<image> Solid lines denote a matched color for the same index. Dashed lines denote a matched color at a different index. x is the number of solid lines, and y is the total number of lines.
You are given Bob's guess and two values x and y. Can you find one possibility of Alice's code so that the values of x and y are correct?
Input
The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of each test case contains three integers n,x,y (1β€ nβ€ 10^5, 0β€ xβ€ yβ€ n) β the length of the codes, and two values Alice responds with.
The second line of each test case contains n integers b_1,β¦,b_n (1β€ b_iβ€ n+1) β Bob's guess, where b_i is the i-th color of the guess.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, on the first line, output "YES" if there is a solution, or "NO" if there is no possible secret code consistent with the described situation. You can print each character in any case (upper or lower).
If the answer is "YES", on the next line output n integers a_1,β¦,a_n (1β€ a_iβ€ n+1) β Alice's secret code, where a_i is the i-th color of the code.
If there are multiple solutions, output any.
Example
Input
7
5 2 4
3 1 1 2 5
5 3 4
1 1 2 1 2
4 0 4
5 5 3 3
4 1 4
2 3 2 3
6 1 2
3 2 1 1 1 1
6 2 4
3 3 2 1 1 1
6 2 6
1 1 3 2 1 1
Output
YES
3 1 6 1 2
YES
3 1 1 1 2
YES
3 3 5 5
NO
YES
4 4 4 4 3 1
YES
3 1 3 1 7 7
YES
2 3 1 1 1 1
Note
The first test case is described in the statement.
In the second test case, x=3 because the colors are equal at indices 2,4,5. And y=4 because they share the colors 1,1,1,2.
In the third test case, x=0 because there is no index where the colors are the same. But y=4 because they share the colors 3,3,5,5.
In the fourth test case, it can be proved that no solution exists.
Tags: constructive algorithms, graph matchings, greedy, implementation, sortings, two pointers
Correct Solution:
```
import collections
t = int(input())
for _ in range(t):
# print('time', _)
n, x, y = map(int, input().split())
b = [int(i) for i in input().split()]
if x == n:
print("YES")
for i in b:
print(i, end = ' ')
print()
continue
setB = set(b)
noin = 1
while noin in setB:
noin += 1
cnt = collections.deque([i, j] for i, j in collections.Counter(b).most_common())
cnt2 = collections.deque()
pos = [set() for _ in range(n + 2)]
for i in range(len(b)):
pos[b[i]].add(i)
res=[-1] * n
l = x
while l > 0:
i = cnt.popleft()
i[1] -= 1
if i[1] != 0:
cnt2.appendleft(i)
while not cnt or cnt2 and cnt2[0][1] >= cnt[0][1]:
cnt.appendleft(cnt2.popleft())
p = pos[i[0]].pop()
res[p] = i[0]
l -= 1
ml = cnt[0][1]
remain = []
while cnt:
i, j = cnt.popleft()
remain += [i] * j
while cnt2:
i, j =cnt2.popleft()
remain += [i] * j
if y - x > (len(remain) - ml) * 2 :
print('No')
continue
i = 0
match = [0] * len(remain)
for i in range(ml, ml + y - x):
match[i % len(remain)] = remain[(i + ml) % len(remain)]
for i in range(ml + y - x, ml + len(remain)):
match[i % len(remain)] = noin
# print(match, remain)
for i in range(len(match)):
p = pos[remain[i]].pop()
res[p] = match[i]
if l == 0:
print("YES")
for i in res:
print(i, end=' ')
print()
else:
print("No")
```
| 96,296 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In the game of Mastermind, there are two players β Alice and Bob. Alice has a secret code, which Bob tries to guess. Here, a code is defined as a sequence of n colors. There are exactly n+1 colors in the entire universe, numbered from 1 to n+1 inclusive.
When Bob guesses a code, Alice tells him some information about how good of a guess it is, in the form of two integers x and y.
The first integer x is the number of indices where Bob's guess correctly matches Alice's code. The second integer y is the size of the intersection of the two codes as multisets. That is, if Bob were to change the order of the colors in his guess, y is the maximum number of indices he could get correct.
For example, suppose n=5, Alice's code is [3,1,6,1,2], and Bob's guess is [3,1,1,2,5]. At indices 1 and 2 colors are equal, while in the other indices they are not equal. So x=2. And the two codes have the four colors 1,1,2,3 in common, so y=4.
<image> Solid lines denote a matched color for the same index. Dashed lines denote a matched color at a different index. x is the number of solid lines, and y is the total number of lines.
You are given Bob's guess and two values x and y. Can you find one possibility of Alice's code so that the values of x and y are correct?
Input
The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of each test case contains three integers n,x,y (1β€ nβ€ 10^5, 0β€ xβ€ yβ€ n) β the length of the codes, and two values Alice responds with.
The second line of each test case contains n integers b_1,β¦,b_n (1β€ b_iβ€ n+1) β Bob's guess, where b_i is the i-th color of the guess.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, on the first line, output "YES" if there is a solution, or "NO" if there is no possible secret code consistent with the described situation. You can print each character in any case (upper or lower).
If the answer is "YES", on the next line output n integers a_1,β¦,a_n (1β€ a_iβ€ n+1) β Alice's secret code, where a_i is the i-th color of the code.
If there are multiple solutions, output any.
Example
Input
7
5 2 4
3 1 1 2 5
5 3 4
1 1 2 1 2
4 0 4
5 5 3 3
4 1 4
2 3 2 3
6 1 2
3 2 1 1 1 1
6 2 4
3 3 2 1 1 1
6 2 6
1 1 3 2 1 1
Output
YES
3 1 6 1 2
YES
3 1 1 1 2
YES
3 3 5 5
NO
YES
4 4 4 4 3 1
YES
3 1 3 1 7 7
YES
2 3 1 1 1 1
Note
The first test case is described in the statement.
In the second test case, x=3 because the colors are equal at indices 2,4,5. And y=4 because they share the colors 1,1,1,2.
In the third test case, x=0 because there is no index where the colors are the same. But y=4 because they share the colors 3,3,5,5.
In the fourth test case, it can be proved that no solution exists.
Tags: constructive algorithms, graph matchings, greedy, implementation, sortings, two pointers
Correct Solution:
```
from sys import stdin, stdout
import heapq
# 5 3 4
# 1 1 2 1 2
# 3 1 1 1 2
# 1 1 2 1 2
# 4 1 4
# 2 3 2 3
# 2 2 3 3
# 1 2 1 2
# 0 2 1 3
# 1 3 0 2
def Mastermind(n, x, y, b_a):
ans = [-1] * n
# find mex
b_s = set(b_a)
mex = 1
while mex in b_s:
mex += 1
#print(mex)
# get frequency
fq_d = {}
for i in range(len(b_a)):
if b_a[i] not in fq_d:
fq_d[b_a[i]] = []
fq_d[b_a[i]].append(i)
fq_h = []
for key in fq_d:
heapq.heappush(fq_h, (-len(fq_d[key]), key, fq_d[key]))
#print(fq_h)
for _ in range(x):
l, v, f_a = heapq.heappop(fq_h)
ans[f_a.pop()] = v
l += 1
if l < 0:
heapq.heappush(fq_h, (l, v, f_a))
#print(ans)
if x == n:
return ans
z = n - y
shift_a = []
for fq in fq_h:
for idx in fq[2]:
shift_a.append([idx, fq[1]])
#print(shift_a)
offset = -fq_h[0][0]
#print(offset)
idx_a = []
for i in range(len(shift_a)):
shift = shift_a[i]
nidx = (i + offset) % (n - x)
#print(shift[1])
#print(shift_a[nidx][0])
nv = shift_a[nidx][1]
if shift[1] == nv:
if z == 0:
return []
z -= 1
ans[shift_a[nidx][0]] = mex
else:
idx_a.append(shift_a[nidx][0])
ans[shift_a[nidx][0]] = shift[1]
while z > 0:
if not idx_a:
return []
ans[idx_a.pop()] = mex
z -= 1
return ans
t = int(stdin.readline())
for _ in range(t):
n, x, y = map(int, stdin.readline().split())
b_a = list(map(int, stdin.readline().split()))
ans = Mastermind(n, x, y, b_a)
if ans:
stdout.write('Yes\n')
stdout.write(' '.join(map(str, ans)) + '\n')
else:
stdout.write('No\n')
```
| 96,297 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In the game of Mastermind, there are two players β Alice and Bob. Alice has a secret code, which Bob tries to guess. Here, a code is defined as a sequence of n colors. There are exactly n+1 colors in the entire universe, numbered from 1 to n+1 inclusive.
When Bob guesses a code, Alice tells him some information about how good of a guess it is, in the form of two integers x and y.
The first integer x is the number of indices where Bob's guess correctly matches Alice's code. The second integer y is the size of the intersection of the two codes as multisets. That is, if Bob were to change the order of the colors in his guess, y is the maximum number of indices he could get correct.
For example, suppose n=5, Alice's code is [3,1,6,1,2], and Bob's guess is [3,1,1,2,5]. At indices 1 and 2 colors are equal, while in the other indices they are not equal. So x=2. And the two codes have the four colors 1,1,2,3 in common, so y=4.
<image> Solid lines denote a matched color for the same index. Dashed lines denote a matched color at a different index. x is the number of solid lines, and y is the total number of lines.
You are given Bob's guess and two values x and y. Can you find one possibility of Alice's code so that the values of x and y are correct?
Input
The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of each test case contains three integers n,x,y (1β€ nβ€ 10^5, 0β€ xβ€ yβ€ n) β the length of the codes, and two values Alice responds with.
The second line of each test case contains n integers b_1,β¦,b_n (1β€ b_iβ€ n+1) β Bob's guess, where b_i is the i-th color of the guess.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, on the first line, output "YES" if there is a solution, or "NO" if there is no possible secret code consistent with the described situation. You can print each character in any case (upper or lower).
If the answer is "YES", on the next line output n integers a_1,β¦,a_n (1β€ a_iβ€ n+1) β Alice's secret code, where a_i is the i-th color of the code.
If there are multiple solutions, output any.
Example
Input
7
5 2 4
3 1 1 2 5
5 3 4
1 1 2 1 2
4 0 4
5 5 3 3
4 1 4
2 3 2 3
6 1 2
3 2 1 1 1 1
6 2 4
3 3 2 1 1 1
6 2 6
1 1 3 2 1 1
Output
YES
3 1 6 1 2
YES
3 1 1 1 2
YES
3 3 5 5
NO
YES
4 4 4 4 3 1
YES
3 1 3 1 7 7
YES
2 3 1 1 1 1
Note
The first test case is described in the statement.
In the second test case, x=3 because the colors are equal at indices 2,4,5. And y=4 because they share the colors 1,1,1,2.
In the third test case, x=0 because there is no index where the colors are the same. But y=4 because they share the colors 3,3,5,5.
In the fourth test case, it can be proved that no solution exists.
Tags: constructive algorithms, graph matchings, greedy, implementation, sortings, two pointers
Correct Solution:
```
from heapq import *
from collections import Counter
import sys
sys.setrecursionlimit(10 ** 5)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
for _ in range(II()):
ng=False
n,x,y=MI()
bb=LI()
use=[False]*(n+1)
for b in bb:use[b-1]=True
unuse=0
for b in range(1,n+2):
if use[b-1]==False:
unuse=b
break
aa=[unuse]*n
cnt=[[] for _ in range(n+2)]
for i,b in enumerate(bb):
cnt[b].append(i)
#print(cnt)
hp=[]
for b,ii in enumerate(cnt):
if not ii:continue
heappush(hp,[-len(ii)]+[b]+ii)
#print(hp)
if (y-x)%2:
if len(hp)>2 and y-x>2:
ii0=heappop(hp)
ii1=heappop(hp)
ii2=heappop(hp)
b0=ii0[1]
b1=ii1[1]
b2=ii2[1]
i0=ii0.pop()
i1=ii1.pop()
i2=ii2.pop()
ii0[0]+=1
ii1[0]+=1
ii2[0]+=1
if ii0[0]<0:heappush(hp,ii0)
if ii1[0]<0:heappush(hp,ii1)
if ii2[0]<0:heappush(hp,ii2)
aa[i0]=b1
aa[i1]=b2
aa[i2]=b0
y-=3
elif len(hp)>1:
ii0=heappop(hp)
ii1=heappop(hp)
b0=ii0[1]
b1=ii1[1]
i0=ii0.pop()
i1=ii1.pop()
ii0[0]+=1
ii1[0]+=1
if ii0[0]<0:heappush(hp,ii0)
if ii1[0]<0:heappush(hp,ii1)
aa[i0]=b1
y-=1
else:
print("NO")
continue
for _ in range((y-x)//2):
if len(hp)<2:
ng=True
break
ii0 = heappop(hp)
ii1 = heappop(hp)
b0 = ii0[1]
b1 = ii1[1]
i0 = ii0.pop()
i1 = ii1.pop()
ii0[0] += 1
ii1[0] += 1
if ii0[0] < 0: heappush(hp, ii0)
if ii1[0] < 0: heappush(hp, ii1)
aa[i0] = b1
aa[i1] = b0
if ng:
print("NO")
continue
ii0=[]
for _ in range(x):
if len(ii0)<3:
if not hp:
ng=True
break
else:
ii0 = heappop(hp)
b0 = ii0[1]
i0 = ii0.pop()
aa[i0] = b0
if ng:
print("NO")
continue
print("YES")
print(*aa)
```
| 96,298 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In the game of Mastermind, there are two players β Alice and Bob. Alice has a secret code, which Bob tries to guess. Here, a code is defined as a sequence of n colors. There are exactly n+1 colors in the entire universe, numbered from 1 to n+1 inclusive.
When Bob guesses a code, Alice tells him some information about how good of a guess it is, in the form of two integers x and y.
The first integer x is the number of indices where Bob's guess correctly matches Alice's code. The second integer y is the size of the intersection of the two codes as multisets. That is, if Bob were to change the order of the colors in his guess, y is the maximum number of indices he could get correct.
For example, suppose n=5, Alice's code is [3,1,6,1,2], and Bob's guess is [3,1,1,2,5]. At indices 1 and 2 colors are equal, while in the other indices they are not equal. So x=2. And the two codes have the four colors 1,1,2,3 in common, so y=4.
<image> Solid lines denote a matched color for the same index. Dashed lines denote a matched color at a different index. x is the number of solid lines, and y is the total number of lines.
You are given Bob's guess and two values x and y. Can you find one possibility of Alice's code so that the values of x and y are correct?
Input
The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of each test case contains three integers n,x,y (1β€ nβ€ 10^5, 0β€ xβ€ yβ€ n) β the length of the codes, and two values Alice responds with.
The second line of each test case contains n integers b_1,β¦,b_n (1β€ b_iβ€ n+1) β Bob's guess, where b_i is the i-th color of the guess.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, on the first line, output "YES" if there is a solution, or "NO" if there is no possible secret code consistent with the described situation. You can print each character in any case (upper or lower).
If the answer is "YES", on the next line output n integers a_1,β¦,a_n (1β€ a_iβ€ n+1) β Alice's secret code, where a_i is the i-th color of the code.
If there are multiple solutions, output any.
Example
Input
7
5 2 4
3 1 1 2 5
5 3 4
1 1 2 1 2
4 0 4
5 5 3 3
4 1 4
2 3 2 3
6 1 2
3 2 1 1 1 1
6 2 4
3 3 2 1 1 1
6 2 6
1 1 3 2 1 1
Output
YES
3 1 6 1 2
YES
3 1 1 1 2
YES
3 3 5 5
NO
YES
4 4 4 4 3 1
YES
3 1 3 1 7 7
YES
2 3 1 1 1 1
Note
The first test case is described in the statement.
In the second test case, x=3 because the colors are equal at indices 2,4,5. And y=4 because they share the colors 1,1,1,2.
In the third test case, x=0 because there is no index where the colors are the same. But y=4 because they share the colors 3,3,5,5.
In the fourth test case, it can be proved that no solution exists.
Tags: constructive algorithms, graph matchings, greedy, implementation, sortings, two pointers
Correct Solution:
```
from collections import defaultdict
from heapq import heapify, heappop, heappush
def solve():
n, s, y = map(int, input().split())
a = input().split()
d = defaultdict(list)
for i, x in enumerate(a):
d[x].append(i)
for i in range(1, n + 2):
e = str(i)
if e not in d:
break
q = [(-len(d[x]), x) for x in d.keys()]
heapify(q)
ans = [0] * n
for i in range(s):
l, x = heappop(q)
ans[d[x].pop()] = x
l += 1
if l:
heappush(q, (l, x))
p = []
while q:
l, x = heappop(q)
p.extend(d[x])
if p:
h = (n - s) // 2
y = n - y
q = p[h:] + p[:h]
for x, z in zip(p, q):
if a[x] == a[z]:
if y:ans[x] = e;y -= 1
else:print("NO");return
else:ans[x] = a[z]
for i in range(n - s):
if y and ans[p[i]] != e:ans[p[i]] = e;y -= 1
print("YES");print(' '.join(ans))
for t in range(int(input())):solve()
```
| 96,299 |
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