message stringlengths 2 39.6k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 450 109k | cluster float64 2 2 | __index_level_0__ int64 900 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Silver Fox is fighting with N monsters.
The monsters are standing in a row, and we can assume them to be standing on a number line. The i-th monster, standing at the coordinate X_i, has the health of H_i.
Silver Fox can use bombs to attack the monsters. Using a bomb at the coordinate x decreases the healths of all monsters between the coordinates x-D and x+D (inclusive) by A. There is no way other than bombs to decrease the monster's health.
Silver Fox wins when all the monsters' healths become 0 or below.
Find the minimum number of bombs needed to win.
Constraints
* 1 \leq N \leq 2 \times 10^5
* 0 \leq D \leq 10^9
* 1 \leq A \leq 10^9
* 0 \leq X_i \leq 10^9
* 1 \leq H_i \leq 10^9
* X_i are distinct.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N D A
X_1 H_1
:
X_N H_N
Output
Print the minimum number of bombs needed to win.
Examples
Input
3 3 2
1 2
5 4
9 2
Output
2
Input
9 4 1
1 5
2 4
3 3
4 2
5 1
6 2
7 3
8 4
9 5
Output
5
Input
3 0 1
300000000 1000000000
100000000 1000000000
200000000 1000000000
Output
3000000000
Submitted Solution:
```
n, d, a = map(int, input().split())
loans = []
bar = 0
ans = 0
xh = sorted([tuple(map(int, input().split())) for _ in range(n)])
for x, h in xh:
while len(loans) > 0:
xx, dd = loans[0]
if xx < x:
loans.pop(0)
bar -= dd
else:
break
h -= bar
if h > 0:
c = h // a if h % a == 0 else ((h // a) + 1)
ans += c
dd = c * a
bar += dd
loans.append((x + 2 * d, dd))
print(ans)
``` | instruction | 0 | 67,918 | 2 | 135,836 |
Yes | output | 1 | 67,918 | 2 | 135,837 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Silver Fox is fighting with N monsters.
The monsters are standing in a row, and we can assume them to be standing on a number line. The i-th monster, standing at the coordinate X_i, has the health of H_i.
Silver Fox can use bombs to attack the monsters. Using a bomb at the coordinate x decreases the healths of all monsters between the coordinates x-D and x+D (inclusive) by A. There is no way other than bombs to decrease the monster's health.
Silver Fox wins when all the monsters' healths become 0 or below.
Find the minimum number of bombs needed to win.
Constraints
* 1 \leq N \leq 2 \times 10^5
* 0 \leq D \leq 10^9
* 1 \leq A \leq 10^9
* 0 \leq X_i \leq 10^9
* 1 \leq H_i \leq 10^9
* X_i are distinct.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N D A
X_1 H_1
:
X_N H_N
Output
Print the minimum number of bombs needed to win.
Examples
Input
3 3 2
1 2
5 4
9 2
Output
2
Input
9 4 1
1 5
2 4
3 3
4 2
5 1
6 2
7 3
8 4
9 5
Output
5
Input
3 0 1
300000000 1000000000
100000000 1000000000
200000000 1000000000
Output
3000000000
Submitted Solution:
```
from collections import deque
N, D, A = map(int, input().split())
ys = []
hs = []
for i in range(N):
X, H = map(int, input().split())
ys.append((X,-(-H//A)))
ys = sorted(ys, key=lambda x : x[0])
l = ys[0][0] + 2*D
li = dmg = r = 0
hs2 = [0]*N
for i, (y, h) in enumerate(ys):
while l < y:
dmg -= hs2[li]
li += 1
l = ys[li][0] + 2*D
hs2[i] = max(0, h-dmg)
dmg += hs2[i]
r += hs2[i]
print(r)
``` | instruction | 0 | 67,919 | 2 | 135,838 |
Yes | output | 1 | 67,919 | 2 | 135,839 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Silver Fox is fighting with N monsters.
The monsters are standing in a row, and we can assume them to be standing on a number line. The i-th monster, standing at the coordinate X_i, has the health of H_i.
Silver Fox can use bombs to attack the monsters. Using a bomb at the coordinate x decreases the healths of all monsters between the coordinates x-D and x+D (inclusive) by A. There is no way other than bombs to decrease the monster's health.
Silver Fox wins when all the monsters' healths become 0 or below.
Find the minimum number of bombs needed to win.
Constraints
* 1 \leq N \leq 2 \times 10^5
* 0 \leq D \leq 10^9
* 1 \leq A \leq 10^9
* 0 \leq X_i \leq 10^9
* 1 \leq H_i \leq 10^9
* X_i are distinct.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N D A
X_1 H_1
:
X_N H_N
Output
Print the minimum number of bombs needed to win.
Examples
Input
3 3 2
1 2
5 4
9 2
Output
2
Input
9 4 1
1 5
2 4
3 3
4 2
5 1
6 2
7 3
8 4
9 5
Output
5
Input
3 0 1
300000000 1000000000
100000000 1000000000
200000000 1000000000
Output
3000000000
Submitted Solution:
```
from math import *
n, d, a = map(int, input().split())
damage_diff = [0] * n
damage = 0
p = sorted(tuple(map(int, input().split())) for _ in range(n))
j = 0
c = 0
for i in range(n):
damage -= damage_diff[i]
x, h = p[i]
h -= damage
if h <= 0: continue
m = ceil(h/a)
c += m
m = m*a
while j < n and p[j][0] - x <= 2 * d:
j += 1
if j < n:
damage_diff[j] += m
damage += m
print(c)
``` | instruction | 0 | 67,920 | 2 | 135,840 |
Yes | output | 1 | 67,920 | 2 | 135,841 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Silver Fox is fighting with N monsters.
The monsters are standing in a row, and we can assume them to be standing on a number line. The i-th monster, standing at the coordinate X_i, has the health of H_i.
Silver Fox can use bombs to attack the monsters. Using a bomb at the coordinate x decreases the healths of all monsters between the coordinates x-D and x+D (inclusive) by A. There is no way other than bombs to decrease the monster's health.
Silver Fox wins when all the monsters' healths become 0 or below.
Find the minimum number of bombs needed to win.
Constraints
* 1 \leq N \leq 2 \times 10^5
* 0 \leq D \leq 10^9
* 1 \leq A \leq 10^9
* 0 \leq X_i \leq 10^9
* 1 \leq H_i \leq 10^9
* X_i are distinct.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N D A
X_1 H_1
:
X_N H_N
Output
Print the minimum number of bombs needed to win.
Examples
Input
3 3 2
1 2
5 4
9 2
Output
2
Input
9 4 1
1 5
2 4
3 3
4 2
5 1
6 2
7 3
8 4
9 5
Output
5
Input
3 0 1
300000000 1000000000
100000000 1000000000
200000000 1000000000
Output
3000000000
Submitted Solution:
```
from bisect import*
n,d,a,*t=map(int,open(0).read().split())
c=[0]*-~n
z=sorted(zip(*[iter(t)]*2))
s=0
for i,(x,h)in enumerate(z):c[i]+=c[i-1];h-=c[i];t=min(-h//a,0);c[i]-=t*a;c[bisect(z,(x+d+d,9e9))]+=t*a;s-=t
print(s)
``` | instruction | 0 | 67,921 | 2 | 135,842 |
Yes | output | 1 | 67,921 | 2 | 135,843 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Silver Fox is fighting with N monsters.
The monsters are standing in a row, and we can assume them to be standing on a number line. The i-th monster, standing at the coordinate X_i, has the health of H_i.
Silver Fox can use bombs to attack the monsters. Using a bomb at the coordinate x decreases the healths of all monsters between the coordinates x-D and x+D (inclusive) by A. There is no way other than bombs to decrease the monster's health.
Silver Fox wins when all the monsters' healths become 0 or below.
Find the minimum number of bombs needed to win.
Constraints
* 1 \leq N \leq 2 \times 10^5
* 0 \leq D \leq 10^9
* 1 \leq A \leq 10^9
* 0 \leq X_i \leq 10^9
* 1 \leq H_i \leq 10^9
* X_i are distinct.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N D A
X_1 H_1
:
X_N H_N
Output
Print the minimum number of bombs needed to win.
Examples
Input
3 3 2
1 2
5 4
9 2
Output
2
Input
9 4 1
1 5
2 4
3 3
4 2
5 1
6 2
7 3
8 4
9 5
Output
5
Input
3 0 1
300000000 1000000000
100000000 1000000000
200000000 1000000000
Output
3000000000
Submitted Solution:
```
from collections import deque
n, d, a = map(int, input().split())
XH = [list(map(int, input().split())) for _ in range(n)]
XH.sort()
cnt = 0
bomb = 0
current_bomb = 0
owari = deque([])
for i in range(n):
x, h = XH[i]
temp = (h + (a - 1)) // a
if owari:
if owari[0][0] < x:
p, q = owari.popleft()
current_bomb -= q
if temp <= current_bomb:
continue
delta = temp - current_bomb
current_bomb += delta
bomb += delta
owari.append(((x + 2 * d), delta))
print(bomb)
``` | instruction | 0 | 67,922 | 2 | 135,844 |
No | output | 1 | 67,922 | 2 | 135,845 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Silver Fox is fighting with N monsters.
The monsters are standing in a row, and we can assume them to be standing on a number line. The i-th monster, standing at the coordinate X_i, has the health of H_i.
Silver Fox can use bombs to attack the monsters. Using a bomb at the coordinate x decreases the healths of all monsters between the coordinates x-D and x+D (inclusive) by A. There is no way other than bombs to decrease the monster's health.
Silver Fox wins when all the monsters' healths become 0 or below.
Find the minimum number of bombs needed to win.
Constraints
* 1 \leq N \leq 2 \times 10^5
* 0 \leq D \leq 10^9
* 1 \leq A \leq 10^9
* 0 \leq X_i \leq 10^9
* 1 \leq H_i \leq 10^9
* X_i are distinct.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N D A
X_1 H_1
:
X_N H_N
Output
Print the minimum number of bombs needed to win.
Examples
Input
3 3 2
1 2
5 4
9 2
Output
2
Input
9 4 1
1 5
2 4
3 3
4 2
5 1
6 2
7 3
8 4
9 5
Output
5
Input
3 0 1
300000000 1000000000
100000000 1000000000
200000000 1000000000
Output
3000000000
Submitted Solution:
```
from collections import deque
import math
n, d, a = map(int, input().split())
data = []
for i in range(n):
data.append(list(map(int, input().split())))
data.sort()
dam = 0
ans = 0
q = deque()
for i in range(n):
x,h = data[i]
while q:
k = q.popleft()
if k[1] >= x:
q.appendleft(k)
break
else:
dam -= k[0]
h -= dam
count = math.ceil(h / a)
dam += count * a
q.append([count * a, x + 2*d])
ans += count
print(ans)
``` | instruction | 0 | 67,923 | 2 | 135,846 |
No | output | 1 | 67,923 | 2 | 135,847 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Silver Fox is fighting with N monsters.
The monsters are standing in a row, and we can assume them to be standing on a number line. The i-th monster, standing at the coordinate X_i, has the health of H_i.
Silver Fox can use bombs to attack the monsters. Using a bomb at the coordinate x decreases the healths of all monsters between the coordinates x-D and x+D (inclusive) by A. There is no way other than bombs to decrease the monster's health.
Silver Fox wins when all the monsters' healths become 0 or below.
Find the minimum number of bombs needed to win.
Constraints
* 1 \leq N \leq 2 \times 10^5
* 0 \leq D \leq 10^9
* 1 \leq A \leq 10^9
* 0 \leq X_i \leq 10^9
* 1 \leq H_i \leq 10^9
* X_i are distinct.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N D A
X_1 H_1
:
X_N H_N
Output
Print the minimum number of bombs needed to win.
Examples
Input
3 3 2
1 2
5 4
9 2
Output
2
Input
9 4 1
1 5
2 4
3 3
4 2
5 1
6 2
7 3
8 4
9 5
Output
5
Input
3 0 1
300000000 1000000000
100000000 1000000000
200000000 1000000000
Output
3000000000
Submitted Solution:
```
# -*- coding: utf-8 -*-
N, D, A = map(int, input().split(' '))
X_H_pairs = [list(map(int, input().split(' '))) for _ in range(N)]
X_H_pairs.sort(key=lambda x: x[0])
import math
i, ans = 0, 0
while i < N:
x, h = X_H_pairs[i]
if h <= 0:
i += 1
continue
else:
times = int(math.ceil(h / A))
ans += times
dh = times * A
dx = x + 2*D
while i < N and X_H_pairs[i][0] <= dx:
X_H_pairs[i][1] -= dh
i += 1
print(ans)
``` | instruction | 0 | 67,924 | 2 | 135,848 |
No | output | 1 | 67,924 | 2 | 135,849 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Silver Fox is fighting with N monsters.
The monsters are standing in a row, and we can assume them to be standing on a number line. The i-th monster, standing at the coordinate X_i, has the health of H_i.
Silver Fox can use bombs to attack the monsters. Using a bomb at the coordinate x decreases the healths of all monsters between the coordinates x-D and x+D (inclusive) by A. There is no way other than bombs to decrease the monster's health.
Silver Fox wins when all the monsters' healths become 0 or below.
Find the minimum number of bombs needed to win.
Constraints
* 1 \leq N \leq 2 \times 10^5
* 0 \leq D \leq 10^9
* 1 \leq A \leq 10^9
* 0 \leq X_i \leq 10^9
* 1 \leq H_i \leq 10^9
* X_i are distinct.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N D A
X_1 H_1
:
X_N H_N
Output
Print the minimum number of bombs needed to win.
Examples
Input
3 3 2
1 2
5 4
9 2
Output
2
Input
9 4 1
1 5
2 4
3 3
4 2
5 1
6 2
7 3
8 4
9 5
Output
5
Input
3 0 1
300000000 1000000000
100000000 1000000000
200000000 1000000000
Output
3000000000
Submitted Solution:
```
import numpy as np
import math
N, D, A = map(int, input().split())
xh = [list(map(int, input().split())) for _ in range(N)]
x = np.array([i[0] for i in xh])
h = np.array([i[1] for i in xh])
h_int = np.ceil(h/A)
cnt=0
for i in range(len(x)):
if h_int[i]==0: continue
cnt+=h_int[i]
h_int = np.where(x<x[i]+2*D, h_int-h_int[i], h_int)
print(int(cnt))
``` | instruction | 0 | 67,925 | 2 | 135,850 |
No | output | 1 | 67,925 | 2 | 135,851 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
While roaming the mystic areas of Stonefalls, in order to drop legendary loot, an adventurer was given a quest as follows. He was given an array A = {a_1,a_2,...,a_N } of length N, and a number K.
Define array B as B(q, A) = { q-a_1, q-a_2, ..., q-a_N }. Define function F as F(B,K) being sum of products of all K-tuples of elements in array B. For example, if the array B is [2,3,4,5], and with K=3, sum of products of all 3-tuples is $$$F(B, 3) = 2*3*4+2*3*5+3*4*5+2*4*5$$$
He was then given a number Q, number of queries of two types:
* Type 1: Given q, i, and d calculate F(B(q, A), K) where we make change to initial array as A[i] = d.
* Type 2: Given q, L, R, and d calculate F(B(q, A), K) where we make change to initial array as A[i] = A[i] + d for all i in range [L, R] inclusive.
All changes are temporarily made to initial array, and don't propagate to following queries. Help the adventurer calculate the answer to a quest, and finally get that loot!
Input
In the first two lines, numbers N (1 ≤ N ≤ 2*10^4) and K (1 ≤ K ≤ N), the length of initial array A, and tuple size, followed by a_1,a_2,a_3,…,a_N (0 ≤ a_i ≤ 10^9) , elements of array A, in the next line. Then follows number Q (Q ≤ 10), number of queries. In the next Q lines come queries of the form:
* 1 q i d, for type 1,
* 2 q L R d, for type 2,
as explained above (0 ≤ q, d ≤ 10^9, 1 ≤ i,L,R ≤ N)
Output
Print Q lines, the answers to queries, modulo 998244353.
Example
Input
5
2
1 2 3 4 5
3
1 6 1 1
1 6 5 2
2 6 2 3 1
Output
85
127
63
Note
In the first query array A = [1, 2, 3, 4, 5], B = [5, 4, 3, 2, 1], sum of products of 2-tuples = 85.
In second query array A = [1, 2, 3, 4, 2], B = [5, 4, 3, 2, 4], sum of products of 2-tuples = 127
In third query array A = [1, 3, 4, 4, 5], B = [5, 3, 2, 2, 1], sum of products of 2-tuples = 63
Submitted Solution:
```
from math import gcd
lcm = 1
for i in range(1, 10):
lcm = lcm * i // gcd(lcm, i)
print(lcm)
``` | instruction | 0 | 68,241 | 2 | 136,482 |
No | output | 1 | 68,241 | 2 | 136,483 |
Provide tags and a correct Python 3 solution for this coding contest problem.
"Night gathers, and now my watch begins. It shall not end until my death. I shall take no wife, hold no lands, father no children. I shall wear no crowns and win no glory. I shall live and die at my post. I am the sword in the darkness. I am the watcher on the walls. I am the shield that guards the realms of men. I pledge my life and honor to the Night's Watch, for this night and all the nights to come." — The Night's Watch oath.
With that begins the watch of Jon Snow. He is assigned the task to support the stewards.
This time he has n stewards with him whom he has to provide support. Each steward has his own strength. Jon Snow likes to support a steward only if there exists at least one steward who has strength strictly less than him and at least one steward who has strength strictly greater than him.
Can you find how many stewards will Jon support?
Input
First line consists of a single integer n (1 ≤ n ≤ 105) — the number of stewards with Jon Snow.
Second line consists of n space separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) representing the values assigned to the stewards.
Output
Output a single integer representing the number of stewards which Jon will feed.
Examples
Input
2
1 5
Output
0
Input
3
1 2 5
Output
1
Note
In the first sample, Jon Snow cannot support steward with strength 1 because there is no steward with strength less than 1 and he cannot support steward with strength 5 because there is no steward with strength greater than 5.
In the second sample, Jon Snow can support steward with strength 2 because there are stewards with strength less than 2 and greater than 2. | instruction | 0 | 68,595 | 2 | 137,190 |
Tags: constructive algorithms, sortings
Correct Solution:
```
n = int(input())
arr = list(map(int, input().split()))
maximum = arr[0]
minimum = arr[0]
for value in arr:
if maximum < value:
maximum = value
if minimum > value:
minimum = value
print(len([value for value in arr if value != minimum and value != maximum]))
``` | output | 1 | 68,595 | 2 | 137,191 |
Provide tags and a correct Python 3 solution for this coding contest problem.
"Night gathers, and now my watch begins. It shall not end until my death. I shall take no wife, hold no lands, father no children. I shall wear no crowns and win no glory. I shall live and die at my post. I am the sword in the darkness. I am the watcher on the walls. I am the shield that guards the realms of men. I pledge my life and honor to the Night's Watch, for this night and all the nights to come." — The Night's Watch oath.
With that begins the watch of Jon Snow. He is assigned the task to support the stewards.
This time he has n stewards with him whom he has to provide support. Each steward has his own strength. Jon Snow likes to support a steward only if there exists at least one steward who has strength strictly less than him and at least one steward who has strength strictly greater than him.
Can you find how many stewards will Jon support?
Input
First line consists of a single integer n (1 ≤ n ≤ 105) — the number of stewards with Jon Snow.
Second line consists of n space separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) representing the values assigned to the stewards.
Output
Output a single integer representing the number of stewards which Jon will feed.
Examples
Input
2
1 5
Output
0
Input
3
1 2 5
Output
1
Note
In the first sample, Jon Snow cannot support steward with strength 1 because there is no steward with strength less than 1 and he cannot support steward with strength 5 because there is no steward with strength greater than 5.
In the second sample, Jon Snow can support steward with strength 2 because there are stewards with strength less than 2 and greater than 2. | instruction | 0 | 68,596 | 2 | 137,192 |
Tags: constructive algorithms, sortings
Correct Solution:
```
t=int(input())
a=list(map(int,input().strip().split(" ")))
a.sort()
c=1;d=1
if t==1:
print(0)
else:
for i in range(t-1):
if a[i]==a[i+1]:
c+=1
else:
break
for i in range(t-1,0,-1):
if a[i]==a[i-1]:
d+=1
else:
break
if len(list(set(a)))==1:
print(len(a)-c)
else:
print(len(a)-c-d)
``` | output | 1 | 68,596 | 2 | 137,193 |
Provide tags and a correct Python 3 solution for this coding contest problem.
"Night gathers, and now my watch begins. It shall not end until my death. I shall take no wife, hold no lands, father no children. I shall wear no crowns and win no glory. I shall live and die at my post. I am the sword in the darkness. I am the watcher on the walls. I am the shield that guards the realms of men. I pledge my life and honor to the Night's Watch, for this night and all the nights to come." — The Night's Watch oath.
With that begins the watch of Jon Snow. He is assigned the task to support the stewards.
This time he has n stewards with him whom he has to provide support. Each steward has his own strength. Jon Snow likes to support a steward only if there exists at least one steward who has strength strictly less than him and at least one steward who has strength strictly greater than him.
Can you find how many stewards will Jon support?
Input
First line consists of a single integer n (1 ≤ n ≤ 105) — the number of stewards with Jon Snow.
Second line consists of n space separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) representing the values assigned to the stewards.
Output
Output a single integer representing the number of stewards which Jon will feed.
Examples
Input
2
1 5
Output
0
Input
3
1 2 5
Output
1
Note
In the first sample, Jon Snow cannot support steward with strength 1 because there is no steward with strength less than 1 and he cannot support steward with strength 5 because there is no steward with strength greater than 5.
In the second sample, Jon Snow can support steward with strength 2 because there are stewards with strength less than 2 and greater than 2. | instruction | 0 | 68,597 | 2 | 137,194 |
Tags: constructive algorithms, sortings
Correct Solution:
```
n = int(input())
stuards = [int(x) for x in input().split()]
ans = 0
mn, mx = min(stuards), max(stuards)
for i in stuards:
if i == mn or i == mx:
continue
else:
ans += 1
print(ans)
``` | output | 1 | 68,597 | 2 | 137,195 |
Provide tags and a correct Python 3 solution for this coding contest problem.
"Night gathers, and now my watch begins. It shall not end until my death. I shall take no wife, hold no lands, father no children. I shall wear no crowns and win no glory. I shall live and die at my post. I am the sword in the darkness. I am the watcher on the walls. I am the shield that guards the realms of men. I pledge my life and honor to the Night's Watch, for this night and all the nights to come." — The Night's Watch oath.
With that begins the watch of Jon Snow. He is assigned the task to support the stewards.
This time he has n stewards with him whom he has to provide support. Each steward has his own strength. Jon Snow likes to support a steward only if there exists at least one steward who has strength strictly less than him and at least one steward who has strength strictly greater than him.
Can you find how many stewards will Jon support?
Input
First line consists of a single integer n (1 ≤ n ≤ 105) — the number of stewards with Jon Snow.
Second line consists of n space separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) representing the values assigned to the stewards.
Output
Output a single integer representing the number of stewards which Jon will feed.
Examples
Input
2
1 5
Output
0
Input
3
1 2 5
Output
1
Note
In the first sample, Jon Snow cannot support steward with strength 1 because there is no steward with strength less than 1 and he cannot support steward with strength 5 because there is no steward with strength greater than 5.
In the second sample, Jon Snow can support steward with strength 2 because there are stewards with strength less than 2 and greater than 2. | instruction | 0 | 68,598 | 2 | 137,196 |
Tags: constructive algorithms, sortings
Correct Solution:
```
n = int(input())
v = [int(x) for x in input().split()]
mn = min(v)
mm = max(v)
ans = 0
for i in range(len(v)):
if v[i] > mn and v[i] < mm:
ans += 1
print(ans)
``` | output | 1 | 68,598 | 2 | 137,197 |
Provide tags and a correct Python 3 solution for this coding contest problem.
"Night gathers, and now my watch begins. It shall not end until my death. I shall take no wife, hold no lands, father no children. I shall wear no crowns and win no glory. I shall live and die at my post. I am the sword in the darkness. I am the watcher on the walls. I am the shield that guards the realms of men. I pledge my life and honor to the Night's Watch, for this night and all the nights to come." — The Night's Watch oath.
With that begins the watch of Jon Snow. He is assigned the task to support the stewards.
This time he has n stewards with him whom he has to provide support. Each steward has his own strength. Jon Snow likes to support a steward only if there exists at least one steward who has strength strictly less than him and at least one steward who has strength strictly greater than him.
Can you find how many stewards will Jon support?
Input
First line consists of a single integer n (1 ≤ n ≤ 105) — the number of stewards with Jon Snow.
Second line consists of n space separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) representing the values assigned to the stewards.
Output
Output a single integer representing the number of stewards which Jon will feed.
Examples
Input
2
1 5
Output
0
Input
3
1 2 5
Output
1
Note
In the first sample, Jon Snow cannot support steward with strength 1 because there is no steward with strength less than 1 and he cannot support steward with strength 5 because there is no steward with strength greater than 5.
In the second sample, Jon Snow can support steward with strength 2 because there are stewards with strength less than 2 and greater than 2. | instruction | 0 | 68,599 | 2 | 137,198 |
Tags: constructive algorithms, sortings
Correct Solution:
```
number = int(input())
strenght = list(map(int, input().split()))
strenght.sort()
if strenght[0] != strenght[-1]:
borders = strenght.count(strenght[0]) + strenght.count(strenght[-1])
else:
borders = strenght.count(strenght[0])
print(number-borders)
``` | output | 1 | 68,599 | 2 | 137,199 |
Provide tags and a correct Python 3 solution for this coding contest problem.
"Night gathers, and now my watch begins. It shall not end until my death. I shall take no wife, hold no lands, father no children. I shall wear no crowns and win no glory. I shall live and die at my post. I am the sword in the darkness. I am the watcher on the walls. I am the shield that guards the realms of men. I pledge my life and honor to the Night's Watch, for this night and all the nights to come." — The Night's Watch oath.
With that begins the watch of Jon Snow. He is assigned the task to support the stewards.
This time he has n stewards with him whom he has to provide support. Each steward has his own strength. Jon Snow likes to support a steward only if there exists at least one steward who has strength strictly less than him and at least one steward who has strength strictly greater than him.
Can you find how many stewards will Jon support?
Input
First line consists of a single integer n (1 ≤ n ≤ 105) — the number of stewards with Jon Snow.
Second line consists of n space separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) representing the values assigned to the stewards.
Output
Output a single integer representing the number of stewards which Jon will feed.
Examples
Input
2
1 5
Output
0
Input
3
1 2 5
Output
1
Note
In the first sample, Jon Snow cannot support steward with strength 1 because there is no steward with strength less than 1 and he cannot support steward with strength 5 because there is no steward with strength greater than 5.
In the second sample, Jon Snow can support steward with strength 2 because there are stewards with strength less than 2 and greater than 2. | instruction | 0 | 68,600 | 2 | 137,200 |
Tags: constructive algorithms, sortings
Correct Solution:
```
n = int(input())
l = [int(x) for x in input().split()]
if len(l) <= 2:
print(0)
else:
mx, mn = max(l), min(l)
if mx == mn:
print(0)
else:
cnt_feed = l.count(mx)+l.count(mn)
print(n-cnt_feed)
``` | output | 1 | 68,600 | 2 | 137,201 |
Provide tags and a correct Python 3 solution for this coding contest problem.
"Night gathers, and now my watch begins. It shall not end until my death. I shall take no wife, hold no lands, father no children. I shall wear no crowns and win no glory. I shall live and die at my post. I am the sword in the darkness. I am the watcher on the walls. I am the shield that guards the realms of men. I pledge my life and honor to the Night's Watch, for this night and all the nights to come." — The Night's Watch oath.
With that begins the watch of Jon Snow. He is assigned the task to support the stewards.
This time he has n stewards with him whom he has to provide support. Each steward has his own strength. Jon Snow likes to support a steward only if there exists at least one steward who has strength strictly less than him and at least one steward who has strength strictly greater than him.
Can you find how many stewards will Jon support?
Input
First line consists of a single integer n (1 ≤ n ≤ 105) — the number of stewards with Jon Snow.
Second line consists of n space separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) representing the values assigned to the stewards.
Output
Output a single integer representing the number of stewards which Jon will feed.
Examples
Input
2
1 5
Output
0
Input
3
1 2 5
Output
1
Note
In the first sample, Jon Snow cannot support steward with strength 1 because there is no steward with strength less than 1 and he cannot support steward with strength 5 because there is no steward with strength greater than 5.
In the second sample, Jon Snow can support steward with strength 2 because there are stewards with strength less than 2 and greater than 2. | instruction | 0 | 68,601 | 2 | 137,202 |
Tags: constructive algorithms, sortings
Correct Solution:
```
n = int(input())
a = list(map(int,input().split()))
a.sort()
b = [a[0]] + [a[-1]]
for i in range(1,n):
if a[i] in b: b.append(a[i])
else: break
for i in range(n-2,-1,-1):
if a[i] in b: b.append(a[i])
else: break
print(max(len(a)-len(b),0))
``` | output | 1 | 68,601 | 2 | 137,203 |
Provide tags and a correct Python 3 solution for this coding contest problem.
"Night gathers, and now my watch begins. It shall not end until my death. I shall take no wife, hold no lands, father no children. I shall wear no crowns and win no glory. I shall live and die at my post. I am the sword in the darkness. I am the watcher on the walls. I am the shield that guards the realms of men. I pledge my life and honor to the Night's Watch, for this night and all the nights to come." — The Night's Watch oath.
With that begins the watch of Jon Snow. He is assigned the task to support the stewards.
This time he has n stewards with him whom he has to provide support. Each steward has his own strength. Jon Snow likes to support a steward only if there exists at least one steward who has strength strictly less than him and at least one steward who has strength strictly greater than him.
Can you find how many stewards will Jon support?
Input
First line consists of a single integer n (1 ≤ n ≤ 105) — the number of stewards with Jon Snow.
Second line consists of n space separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) representing the values assigned to the stewards.
Output
Output a single integer representing the number of stewards which Jon will feed.
Examples
Input
2
1 5
Output
0
Input
3
1 2 5
Output
1
Note
In the first sample, Jon Snow cannot support steward with strength 1 because there is no steward with strength less than 1 and he cannot support steward with strength 5 because there is no steward with strength greater than 5.
In the second sample, Jon Snow can support steward with strength 2 because there are stewards with strength less than 2 and greater than 2. | instruction | 0 | 68,602 | 2 | 137,204 |
Tags: constructive algorithms, sortings
Correct Solution:
```
n = int(input())
t = [int(i) for i in input().split()]
s, mn, mx = 0, min(t), max(t)
for x in t:
if mn < x < mx:
s += 1
print(s)
``` | output | 1 | 68,602 | 2 | 137,205 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"Night gathers, and now my watch begins. It shall not end until my death. I shall take no wife, hold no lands, father no children. I shall wear no crowns and win no glory. I shall live and die at my post. I am the sword in the darkness. I am the watcher on the walls. I am the shield that guards the realms of men. I pledge my life and honor to the Night's Watch, for this night and all the nights to come." — The Night's Watch oath.
With that begins the watch of Jon Snow. He is assigned the task to support the stewards.
This time he has n stewards with him whom he has to provide support. Each steward has his own strength. Jon Snow likes to support a steward only if there exists at least one steward who has strength strictly less than him and at least one steward who has strength strictly greater than him.
Can you find how many stewards will Jon support?
Input
First line consists of a single integer n (1 ≤ n ≤ 105) — the number of stewards with Jon Snow.
Second line consists of n space separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) representing the values assigned to the stewards.
Output
Output a single integer representing the number of stewards which Jon will feed.
Examples
Input
2
1 5
Output
0
Input
3
1 2 5
Output
1
Note
In the first sample, Jon Snow cannot support steward with strength 1 because there is no steward with strength less than 1 and he cannot support steward with strength 5 because there is no steward with strength greater than 5.
In the second sample, Jon Snow can support steward with strength 2 because there are stewards with strength less than 2 and greater than 2.
Submitted Solution:
```
n = int(input())
l = list(map(int,input().split()))
Min = min(l)
Max = max(l)
count=0
for i in l:
if (i > Min) and (i < Max) :
count+=1
print(count)
``` | instruction | 0 | 68,603 | 2 | 137,206 |
Yes | output | 1 | 68,603 | 2 | 137,207 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"Night gathers, and now my watch begins. It shall not end until my death. I shall take no wife, hold no lands, father no children. I shall wear no crowns and win no glory. I shall live and die at my post. I am the sword in the darkness. I am the watcher on the walls. I am the shield that guards the realms of men. I pledge my life and honor to the Night's Watch, for this night and all the nights to come." — The Night's Watch oath.
With that begins the watch of Jon Snow. He is assigned the task to support the stewards.
This time he has n stewards with him whom he has to provide support. Each steward has his own strength. Jon Snow likes to support a steward only if there exists at least one steward who has strength strictly less than him and at least one steward who has strength strictly greater than him.
Can you find how many stewards will Jon support?
Input
First line consists of a single integer n (1 ≤ n ≤ 105) — the number of stewards with Jon Snow.
Second line consists of n space separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) representing the values assigned to the stewards.
Output
Output a single integer representing the number of stewards which Jon will feed.
Examples
Input
2
1 5
Output
0
Input
3
1 2 5
Output
1
Note
In the first sample, Jon Snow cannot support steward with strength 1 because there is no steward with strength less than 1 and he cannot support steward with strength 5 because there is no steward with strength greater than 5.
In the second sample, Jon Snow can support steward with strength 2 because there are stewards with strength less than 2 and greater than 2.
Submitted Solution:
```
# Hey, there Stalker!!!
# This Code was written by:
# ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
# ▒▒╔╗╔═══╦═══╗▒▒▒╔╗▒▒▒
# ▒╔╝║║╔═╗║╔═╗╠╗▒▒║║▒▒▒
# ▒╚╗║║║║║║║║║╠╬══╣║╔╗▒
# ▒▒║║║║║║║║║║╠╣║═╣╚╝╝▒
# ▒╔╝╚╣╚═╝║╚═╝║║║═╣╔╗╗▒
# ▒╚══╩═══╩═══╣╠══╩╝╚╝▒
# ▒▒▒▒▒▒▒▒▒▒▒╔╝║▒▒▒▒▒▒▒
# ▒▒▒▒▒▒▒▒▒▒▒╚═╝▒▒▒▒▒▒▒
# ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
#from __future__ import division, print_function
#mod=int(1e9+7)
#import resource
#resource.setrlimit(resource.RLIMIT_STACK, [0x100000000, resource.RLIM_INFINITY])
#import threading
#threading.stack_size(2**26)
#fact=[1]
#for i in range(1,100001):
# fact.append((fact[-1]*i)%mod)
#ifact=[0]*100001
#from collections import deque, defaultdict, Counter, OrderedDict
#from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians, gcd
#from heapq import heappush, heappop, heapify, nlargest, nsmallest
#ifact[100000]=pow(fact[100000],mod-2,mod)
#for i in range(100000,0,-1):
# ifact[i-1]=(i*ifact[i])%mod
# sys.setrecursionlimit(10**6)
from sys import stdin, stdout
import bisect #c++ upperbound
from bisect import bisect_left as bl #c++ lowerbound bl(array,element)
from bisect import bisect_right as br #c++ upperbound
import itertools
from collections import Counter
import collections
import math
import heapq
import re
def modinv(n,p):
return pow(n,p-2,p)
def cin():
return map(int,sin().split())
def ain(): #takes array as input
return list(map(int,sin().split()))
def sin():
return input()
def inin():
return int(input())
def Divisors(n) :
l = []
for i in range(1, int(math.sqrt(n) + 1)) :
if (n % i == 0) :
if (n // i == i) :
l.append(i)
else :
l.append(i)
l.append(n//i)
return l
def most_frequent(list):
return max(set(list), key = list.count)
def GCD(x,y):
while(y):
x, y = y, x % y
return x
def ncr(n,r,p):
t=((fact[n])*((ifact[r]*ifact[n-r])%p))%p
return t
def Convert(string):
li = list(string.split(""))
return li
def SieveOfEratosthenes(n):
global prime
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
f=[]
for p in range(2, n):
if prime[p]:
f.append(p)
return f
prime=[]
q=[]
def dfs(n,d,v,c):
global q
v[n]=1
x=d[n]
q.append(n)
j=c
for i in x:
if i not in v:
f=dfs(i,d,v,c+1)
j=max(j,f)
# print(f)
return j
#Implement heapq
#grades = [110, 25, 38, 49, 20, 95, 33, 87, 80, 90]
#print(heapq.nlargest(3, grades)) #top 3 largest
#print(heapq.nsmallest(4, grades))
#Always make a variable of predefined function for ex- fn=len
#n,k=map(int,input().split())
"""*******************************************************"""
def main():
#Write Your Code Here
n=inin()
a=ain()
if len(set(a))<=2:
print(0)
else:
a.sort()
length = len(a)
temp1=a[0]
temp2=a[-1]
i=0 #list length
if a[0]!=a[-1]:
while(i<length):
if(a[i]==temp1):
a.remove (a[i])
# as an element is removed
# so decrease the length by 1
length = length -1
# run loop again to check element
# at same index, when item removed
# next item will shift to the left
continue
elif a[i]==temp2:
a.remove(a[i])
length-=1
continue
i = i+1
print(length)
else:
print(0)
######## Python 2 and 3 footer by Pajenegod and c1729
py2 = round(0.5)
if py2:
from future_builtins import ascii, filter, hex, map, oct, zip
range = xrange
import os, sys
from io import IOBase, BytesIO
BUFSIZE = 8192
class FastIO(BytesIO):
newlines = 0
def __init__(self, file):
self._file = file
self._fd = file.fileno()
self.writable = "x" in file.mode or "w" in file.mode
self.write = super(FastIO, self).write if self.writable else None
def _fill(self):
s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
return s
def read(self):
while self._fill(): pass
return super(FastIO,self).read()
def readline(self):
while self.newlines == 0:
s = self._fill(); self.newlines = s.count(b"\n") + (not s)
self.newlines -= 1
return super(FastIO, self).readline()
def flush(self):
if self.writable:
os.write(self._fd, self.getvalue())
self.truncate(0), self.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
if py2:
self.write = self.buffer.write
self.read = self.buffer.read
self.readline = self.buffer.readline
else:
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')
if __name__== "__main__":
main()
#threading.Thread(target=main).start()
``` | instruction | 0 | 68,604 | 2 | 137,208 |
Yes | output | 1 | 68,604 | 2 | 137,209 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"Night gathers, and now my watch begins. It shall not end until my death. I shall take no wife, hold no lands, father no children. I shall wear no crowns and win no glory. I shall live and die at my post. I am the sword in the darkness. I am the watcher on the walls. I am the shield that guards the realms of men. I pledge my life and honor to the Night's Watch, for this night and all the nights to come." — The Night's Watch oath.
With that begins the watch of Jon Snow. He is assigned the task to support the stewards.
This time he has n stewards with him whom he has to provide support. Each steward has his own strength. Jon Snow likes to support a steward only if there exists at least one steward who has strength strictly less than him and at least one steward who has strength strictly greater than him.
Can you find how many stewards will Jon support?
Input
First line consists of a single integer n (1 ≤ n ≤ 105) — the number of stewards with Jon Snow.
Second line consists of n space separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) representing the values assigned to the stewards.
Output
Output a single integer representing the number of stewards which Jon will feed.
Examples
Input
2
1 5
Output
0
Input
3
1 2 5
Output
1
Note
In the first sample, Jon Snow cannot support steward with strength 1 because there is no steward with strength less than 1 and he cannot support steward with strength 5 because there is no steward with strength greater than 5.
In the second sample, Jon Snow can support steward with strength 2 because there are stewards with strength less than 2 and greater than 2.
Submitted Solution:
```
n = int(input())
strength = input()
strength = strength.split(' ')
strengthlist = []
for s in strength:
strengthlist.append(int(s))
ssorted = strengthlist[:]
ssorted.sort()
support = 0
if (n <=2):
support = 0
else:
for i in range(0,n):
sup1 = 0
sup2 = 0
if (ssorted[0]!=strengthlist[i]):
sup1 = 1
if (ssorted[-1]!=strengthlist[i]):
sup2 = 1
if (sup1+sup2==2):
support = support+1
print(support)
``` | instruction | 0 | 68,605 | 2 | 137,210 |
Yes | output | 1 | 68,605 | 2 | 137,211 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"Night gathers, and now my watch begins. It shall not end until my death. I shall take no wife, hold no lands, father no children. I shall wear no crowns and win no glory. I shall live and die at my post. I am the sword in the darkness. I am the watcher on the walls. I am the shield that guards the realms of men. I pledge my life and honor to the Night's Watch, for this night and all the nights to come." — The Night's Watch oath.
With that begins the watch of Jon Snow. He is assigned the task to support the stewards.
This time he has n stewards with him whom he has to provide support. Each steward has his own strength. Jon Snow likes to support a steward only if there exists at least one steward who has strength strictly less than him and at least one steward who has strength strictly greater than him.
Can you find how many stewards will Jon support?
Input
First line consists of a single integer n (1 ≤ n ≤ 105) — the number of stewards with Jon Snow.
Second line consists of n space separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) representing the values assigned to the stewards.
Output
Output a single integer representing the number of stewards which Jon will feed.
Examples
Input
2
1 5
Output
0
Input
3
1 2 5
Output
1
Note
In the first sample, Jon Snow cannot support steward with strength 1 because there is no steward with strength less than 1 and he cannot support steward with strength 5 because there is no steward with strength greater than 5.
In the second sample, Jon Snow can support steward with strength 2 because there are stewards with strength less than 2 and greater than 2.
Submitted Solution:
```
n = int(input())
a = list(map(int,input().split()))
Min, Max = min(a) , max(a)
c1,c2 = a.count(Min),a.count(Max)
n = n-c1-c2
print(n if n > 0 else 0)
'''
ans = 0
for i in range(n):
if a[i] is not Min and a[i] is not Max:
ans+=1
print(ans)
'''
``` | instruction | 0 | 68,606 | 2 | 137,212 |
Yes | output | 1 | 68,606 | 2 | 137,213 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"Night gathers, and now my watch begins. It shall not end until my death. I shall take no wife, hold no lands, father no children. I shall wear no crowns and win no glory. I shall live and die at my post. I am the sword in the darkness. I am the watcher on the walls. I am the shield that guards the realms of men. I pledge my life and honor to the Night's Watch, for this night and all the nights to come." — The Night's Watch oath.
With that begins the watch of Jon Snow. He is assigned the task to support the stewards.
This time he has n stewards with him whom he has to provide support. Each steward has his own strength. Jon Snow likes to support a steward only if there exists at least one steward who has strength strictly less than him and at least one steward who has strength strictly greater than him.
Can you find how many stewards will Jon support?
Input
First line consists of a single integer n (1 ≤ n ≤ 105) — the number of stewards with Jon Snow.
Second line consists of n space separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) representing the values assigned to the stewards.
Output
Output a single integer representing the number of stewards which Jon will feed.
Examples
Input
2
1 5
Output
0
Input
3
1 2 5
Output
1
Note
In the first sample, Jon Snow cannot support steward with strength 1 because there is no steward with strength less than 1 and he cannot support steward with strength 5 because there is no steward with strength greater than 5.
In the second sample, Jon Snow can support steward with strength 2 because there are stewards with strength less than 2 and greater than 2.
Submitted Solution:
```
n = int(input())
b= list(map(int,input().split()))
count=0
if len(b) <= 2:
print(count)
else:
b.sort()
for i in range(1,n-1):
if b[i] > b[i-1] and b[i+1] > b[i]:
count +=1
print(count)
``` | instruction | 0 | 68,607 | 2 | 137,214 |
No | output | 1 | 68,607 | 2 | 137,215 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"Night gathers, and now my watch begins. It shall not end until my death. I shall take no wife, hold no lands, father no children. I shall wear no crowns and win no glory. I shall live and die at my post. I am the sword in the darkness. I am the watcher on the walls. I am the shield that guards the realms of men. I pledge my life and honor to the Night's Watch, for this night and all the nights to come." — The Night's Watch oath.
With that begins the watch of Jon Snow. He is assigned the task to support the stewards.
This time he has n stewards with him whom he has to provide support. Each steward has his own strength. Jon Snow likes to support a steward only if there exists at least one steward who has strength strictly less than him and at least one steward who has strength strictly greater than him.
Can you find how many stewards will Jon support?
Input
First line consists of a single integer n (1 ≤ n ≤ 105) — the number of stewards with Jon Snow.
Second line consists of n space separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) representing the values assigned to the stewards.
Output
Output a single integer representing the number of stewards which Jon will feed.
Examples
Input
2
1 5
Output
0
Input
3
1 2 5
Output
1
Note
In the first sample, Jon Snow cannot support steward with strength 1 because there is no steward with strength less than 1 and he cannot support steward with strength 5 because there is no steward with strength greater than 5.
In the second sample, Jon Snow can support steward with strength 2 because there are stewards with strength less than 2 and greater than 2.
Submitted Solution:
```
__author__ = "Ryabchun Vladimir"
n = int(input())
stuards = list(map(int, input().split()))
if n == 1: print(0)
elif n == 2: print(0)
else:
print(len(set(stuards)) - 2)
``` | instruction | 0 | 68,608 | 2 | 137,216 |
No | output | 1 | 68,608 | 2 | 137,217 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"Night gathers, and now my watch begins. It shall not end until my death. I shall take no wife, hold no lands, father no children. I shall wear no crowns and win no glory. I shall live and die at my post. I am the sword in the darkness. I am the watcher on the walls. I am the shield that guards the realms of men. I pledge my life and honor to the Night's Watch, for this night and all the nights to come." — The Night's Watch oath.
With that begins the watch of Jon Snow. He is assigned the task to support the stewards.
This time he has n stewards with him whom he has to provide support. Each steward has his own strength. Jon Snow likes to support a steward only if there exists at least one steward who has strength strictly less than him and at least one steward who has strength strictly greater than him.
Can you find how many stewards will Jon support?
Input
First line consists of a single integer n (1 ≤ n ≤ 105) — the number of stewards with Jon Snow.
Second line consists of n space separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) representing the values assigned to the stewards.
Output
Output a single integer representing the number of stewards which Jon will feed.
Examples
Input
2
1 5
Output
0
Input
3
1 2 5
Output
1
Note
In the first sample, Jon Snow cannot support steward with strength 1 because there is no steward with strength less than 1 and he cannot support steward with strength 5 because there is no steward with strength greater than 5.
In the second sample, Jon Snow can support steward with strength 2 because there are stewards with strength less than 2 and greater than 2.
Submitted Solution:
```
n=int(input())
a=[str(i) for i in input().split()]
a=sorted(a)
b="".join(a)
b=b.strip(b[0])
b=b.strip(b[len(b)-1])
print(len(b))
``` | instruction | 0 | 68,609 | 2 | 137,218 |
No | output | 1 | 68,609 | 2 | 137,219 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"Night gathers, and now my watch begins. It shall not end until my death. I shall take no wife, hold no lands, father no children. I shall wear no crowns and win no glory. I shall live and die at my post. I am the sword in the darkness. I am the watcher on the walls. I am the shield that guards the realms of men. I pledge my life and honor to the Night's Watch, for this night and all the nights to come." — The Night's Watch oath.
With that begins the watch of Jon Snow. He is assigned the task to support the stewards.
This time he has n stewards with him whom he has to provide support. Each steward has his own strength. Jon Snow likes to support a steward only if there exists at least one steward who has strength strictly less than him and at least one steward who has strength strictly greater than him.
Can you find how many stewards will Jon support?
Input
First line consists of a single integer n (1 ≤ n ≤ 105) — the number of stewards with Jon Snow.
Second line consists of n space separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) representing the values assigned to the stewards.
Output
Output a single integer representing the number of stewards which Jon will feed.
Examples
Input
2
1 5
Output
0
Input
3
1 2 5
Output
1
Note
In the first sample, Jon Snow cannot support steward with strength 1 because there is no steward with strength less than 1 and he cannot support steward with strength 5 because there is no steward with strength greater than 5.
In the second sample, Jon Snow can support steward with strength 2 because there are stewards with strength less than 2 and greater than 2.
Submitted Solution:
```
n = input()
k = input()
stewards = k.split()
stewards = sorted(stewards)
supported = 0
if int(n) > 2:
for i in range(1, int(n) - 1):
if int(stewards[i + 1]) > int(stewards[i]) > int(stewards[0]):
supported += 1
else:
continue
else:
supported = 0
print(supported)
``` | instruction | 0 | 68,610 | 2 | 137,220 |
No | output | 1 | 68,610 | 2 | 137,221 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n knights sitting at the Round Table at an equal distance from each other. Each of them is either in a good or in a bad mood.
Merlin, the wizard predicted to King Arthur that the next month will turn out to be particularly fortunate if the regular polygon can be found. On all vertices of the polygon knights in a good mood should be located. Otherwise, the next month will bring misfortunes.
A convex polygon is regular if all its sides have same length and all his angles are equal. In this problem we consider only regular polygons with at least 3 vertices, i. e. only nondegenerated.
On a picture below some examples of such polygons are present. Green points mean knights in a good mood. Red points mean ones in a bad mood.
<image>
King Arthur knows the knights' moods. Help him find out if the next month will be fortunate or not.
Input
The first line contains number n, which is the number of knights at the round table (3 ≤ n ≤ 105). The second line contains space-separated moods of all the n knights in the order of passing them around the table. "1" means that the knight is in a good mood an "0" means that he is in a bad mood.
Output
Print "YES" without the quotes if the following month will turn out to be lucky. Otherwise, print "NO".
Examples
Input
3
1 1 1
Output
YES
Input
6
1 0 1 1 1 0
Output
YES
Input
6
1 0 0 1 0 1
Output
NO | instruction | 0 | 69,461 | 2 | 138,922 |
Tags: dp, math, number theory
Correct Solution:
```
def divisors(n):
i = 2
res = set()
tmp = n
while i * i <= n:
while tmp % i == 0:
tmp //= i
res.add(i)
i += 1
if tmp != 1:
res.add(tmp)
if n % 4 == 0:
res.add(4)
res.discard(2)
return res
def main():
n = int(input())
knights = [int(c) for c in input().split()]
for p in divisors(n):
for i in range(n // p):
if all(knights[i % n] for i in range(i, i + n, n // p)):
print('YES')
return
print('NO')
if __name__ == "__main__":
main()
``` | output | 1 | 69,461 | 2 | 138,923 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n knights sitting at the Round Table at an equal distance from each other. Each of them is either in a good or in a bad mood.
Merlin, the wizard predicted to King Arthur that the next month will turn out to be particularly fortunate if the regular polygon can be found. On all vertices of the polygon knights in a good mood should be located. Otherwise, the next month will bring misfortunes.
A convex polygon is regular if all its sides have same length and all his angles are equal. In this problem we consider only regular polygons with at least 3 vertices, i. e. only nondegenerated.
On a picture below some examples of such polygons are present. Green points mean knights in a good mood. Red points mean ones in a bad mood.
<image>
King Arthur knows the knights' moods. Help him find out if the next month will be fortunate or not.
Input
The first line contains number n, which is the number of knights at the round table (3 ≤ n ≤ 105). The second line contains space-separated moods of all the n knights in the order of passing them around the table. "1" means that the knight is in a good mood an "0" means that he is in a bad mood.
Output
Print "YES" without the quotes if the following month will turn out to be lucky. Otherwise, print "NO".
Examples
Input
3
1 1 1
Output
YES
Input
6
1 0 1 1 1 0
Output
YES
Input
6
1 0 0 1 0 1
Output
NO | instruction | 0 | 69,462 | 2 | 138,924 |
Tags: dp, math, number theory
Correct Solution:
```
def gen_prime(n):
prime = []
prime.append(2)
prime.append(3)
prime.append(4)
for i in range(5, n+1):
div = False
for j in prime:
if not i%j:
div = True
break
if not div:
prime.append(i)
return prime
n = int(input())
prime = gen_prime(n)
prime = prime[1:]
prime.append(n)
a = [int(i) for i in input().split()]
possible = False
for i in prime:
if not n%i:
found = False
l = n//i
for k in range(0, l):
mood = True
for j in range(k, n, l):
if not a[j]:
mood = False
break
if mood:
found = True
break
if found:
possible = True
break
if possible:
print("YES")
else:
print("NO")
``` | output | 1 | 69,462 | 2 | 138,925 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n knights sitting at the Round Table at an equal distance from each other. Each of them is either in a good or in a bad mood.
Merlin, the wizard predicted to King Arthur that the next month will turn out to be particularly fortunate if the regular polygon can be found. On all vertices of the polygon knights in a good mood should be located. Otherwise, the next month will bring misfortunes.
A convex polygon is regular if all its sides have same length and all his angles are equal. In this problem we consider only regular polygons with at least 3 vertices, i. e. only nondegenerated.
On a picture below some examples of such polygons are present. Green points mean knights in a good mood. Red points mean ones in a bad mood.
<image>
King Arthur knows the knights' moods. Help him find out if the next month will be fortunate or not.
Input
The first line contains number n, which is the number of knights at the round table (3 ≤ n ≤ 105). The second line contains space-separated moods of all the n knights in the order of passing them around the table. "1" means that the knight is in a good mood an "0" means that he is in a bad mood.
Output
Print "YES" without the quotes if the following month will turn out to be lucky. Otherwise, print "NO".
Examples
Input
3
1 1 1
Output
YES
Input
6
1 0 1 1 1 0
Output
YES
Input
6
1 0 0 1 0 1
Output
NO | instruction | 0 | 69,463 | 2 | 138,926 |
Tags: dp, math, number theory
Correct Solution:
```
n=int(input())
b=list(map(int,input().split()))
p=int(n**0.5)+1
j=1
r=n
c=[]
while(j<p):
if n%j==0:
k=j
r=n//j
if k>=3:
c.append(k)
if r>=3 and r!=k:
c.append(r)
j+=1
c=sorted(c)
f=0
for r in c:
k = n // r - 1
dp = [1] * (k + 1)
j = 0
p = 0
while (j < n):
p = (j // (k + 1))
if b[j] == 0:
dp[j - p * (k + 1)] = 0
j += 1
for j in range(k + 1):
if dp[j] == 1:
f = 1
break
if f == 1:
break
if f==1:
print("YES")
else:
print("NO")
``` | output | 1 | 69,463 | 2 | 138,927 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently you've discovered a new shooter. They say it has realistic game mechanics.
Your character has a gun with magazine size equal to k and should exterminate n waves of monsters. The i-th wave consists of a_i monsters and happens from the l_i-th moment of time up to the r_i-th moments of time. All a_i monsters spawn at moment l_i and you have to exterminate all of them before the moment r_i ends (you can kill monsters right at moment r_i). For every two consecutive waves, the second wave starts not earlier than the first wave ends (though the second wave can start at the same moment when the first wave ends) — formally, the condition r_i ≤ l_{i + 1} holds. Take a look at the notes for the examples to understand the process better.
You are confident in yours and your character's skills so you can assume that aiming and shooting are instant and you need exactly one bullet to kill one monster. But reloading takes exactly 1 unit of time.
One of the realistic mechanics is a mechanic of reloading: when you reload you throw away the old magazine with all remaining bullets in it. That's why constant reloads may cost you excessive amounts of spent bullets.
You've taken a liking to this mechanic so now you are wondering: what is the minimum possible number of bullets you need to spend (both used and thrown) to exterminate all waves.
Note that you don't throw the remaining bullets away after eradicating all monsters, and you start with a full magazine.
Input
The first line contains two integers n and k (1 ≤ n ≤ 2000; 1 ≤ k ≤ 10^9) — the number of waves and magazine size.
The next n lines contain descriptions of waves. The i-th line contains three integers l_i, r_i and a_i (1 ≤ l_i ≤ r_i ≤ 10^9; 1 ≤ a_i ≤ 10^9) — the period of time when the i-th wave happens and the number of monsters in it.
It's guaranteed that waves don't overlap (but may touch) and are given in the order they occur, i. e. r_i ≤ l_{i + 1}.
Output
If there is no way to clear all waves, print -1. Otherwise, print the minimum possible number of bullets you need to spend (both used and thrown) to clear all waves.
Examples
Input
2 3
2 3 6
3 4 3
Output
9
Input
2 5
3 7 11
10 12 15
Output
30
Input
5 42
42 42 42
42 43 42
43 44 42
44 45 42
45 45 1
Output
-1
Input
1 10
100 111 1
Output
1
Note
In the first example:
* At the moment 2, the first wave occurs and 6 monsters spawn. You kill 3 monsters and start reloading.
* At the moment 3, the second wave occurs and 3 more monsters spawn. You kill remaining 3 monsters from the first wave and start reloading.
* At the moment 4, you kill remaining 3 monsters from the second wave.
In total, you'll spend 9 bullets.
In the second example:
* At moment 3, the first wave occurs and 11 monsters spawn. You kill 5 monsters and start reloading.
* At moment 4, you kill 5 more monsters and start reloading.
* At moment 5, you kill the last monster and start reloading throwing away old magazine with 4 bullets.
* At moment 10, the second wave occurs and 15 monsters spawn. You kill 5 monsters and start reloading.
* At moment 11, you kill 5 more monsters and start reloading.
* At moment 12, you kill last 5 monsters.
In total, you'll spend 30 bullets. | instruction | 0 | 70,046 | 2 | 140,092 |
Tags: dp, greedy
Correct Solution:
```
n, k = list(map(int, input().split()))
L = []
R = []
A = []
for i in range (0, n):
x = list(map(int, input().split()))
L.append(x[0])
R.append(x[1])
A.append(x[2])
L.append(R[-1])
i = n-1
x = 0
y = 0
ans = 0
v = True
N = [0 for i in range (0, n)]
while i >= 0:
if R[i] == L[i+1]:
x = max(x + A[i] - k * (R[i] - L[i]), 0)
N[i] = x
else:
x = max(A[i] - k * (R[i] - L[i]), 0)
N[i] = x
if N[i] > k:
v = False
i = i - 1
m = k
N.append(0)
i = 0
while i < n and v == True:
if m < N[i]:
ans = ans + m
m = k
m = m - A[i]
ans = ans + A[i]
while m < 0:
m = m + k
i = i + 1
if v == True:
print(ans)
else:
print(-1)
``` | output | 1 | 70,046 | 2 | 140,093 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently you've discovered a new shooter. They say it has realistic game mechanics.
Your character has a gun with magazine size equal to k and should exterminate n waves of monsters. The i-th wave consists of a_i monsters and happens from the l_i-th moment of time up to the r_i-th moments of time. All a_i monsters spawn at moment l_i and you have to exterminate all of them before the moment r_i ends (you can kill monsters right at moment r_i). For every two consecutive waves, the second wave starts not earlier than the first wave ends (though the second wave can start at the same moment when the first wave ends) — formally, the condition r_i ≤ l_{i + 1} holds. Take a look at the notes for the examples to understand the process better.
You are confident in yours and your character's skills so you can assume that aiming and shooting are instant and you need exactly one bullet to kill one monster. But reloading takes exactly 1 unit of time.
One of the realistic mechanics is a mechanic of reloading: when you reload you throw away the old magazine with all remaining bullets in it. That's why constant reloads may cost you excessive amounts of spent bullets.
You've taken a liking to this mechanic so now you are wondering: what is the minimum possible number of bullets you need to spend (both used and thrown) to exterminate all waves.
Note that you don't throw the remaining bullets away after eradicating all monsters, and you start with a full magazine.
Input
The first line contains two integers n and k (1 ≤ n ≤ 2000; 1 ≤ k ≤ 10^9) — the number of waves and magazine size.
The next n lines contain descriptions of waves. The i-th line contains three integers l_i, r_i and a_i (1 ≤ l_i ≤ r_i ≤ 10^9; 1 ≤ a_i ≤ 10^9) — the period of time when the i-th wave happens and the number of monsters in it.
It's guaranteed that waves don't overlap (but may touch) and are given in the order they occur, i. e. r_i ≤ l_{i + 1}.
Output
If there is no way to clear all waves, print -1. Otherwise, print the minimum possible number of bullets you need to spend (both used and thrown) to clear all waves.
Examples
Input
2 3
2 3 6
3 4 3
Output
9
Input
2 5
3 7 11
10 12 15
Output
30
Input
5 42
42 42 42
42 43 42
43 44 42
44 45 42
45 45 1
Output
-1
Input
1 10
100 111 1
Output
1
Note
In the first example:
* At the moment 2, the first wave occurs and 6 monsters spawn. You kill 3 monsters and start reloading.
* At the moment 3, the second wave occurs and 3 more monsters spawn. You kill remaining 3 monsters from the first wave and start reloading.
* At the moment 4, you kill remaining 3 monsters from the second wave.
In total, you'll spend 9 bullets.
In the second example:
* At moment 3, the first wave occurs and 11 monsters spawn. You kill 5 monsters and start reloading.
* At moment 4, you kill 5 more monsters and start reloading.
* At moment 5, you kill the last monster and start reloading throwing away old magazine with 4 bullets.
* At moment 10, the second wave occurs and 15 monsters spawn. You kill 5 monsters and start reloading.
* At moment 11, you kill 5 more monsters and start reloading.
* At moment 12, you kill last 5 monsters.
In total, you'll spend 30 bullets. | instruction | 0 | 70,047 | 2 | 140,094 |
Tags: dp, greedy
Correct Solution:
```
#!/usr/bin/env python3
# from typing import *
import sys
import io
import math
import collections
import decimal
import itertools
import bisect
import heapq
def input():
return sys.stdin.readline()[:-1]
# sys.setrecursionlimit(1000000)
# _INPUT = """10 89
# 1 2 82
# 2 2 31
# 3 4 63
# 6 7 18
# 9 9 44
# 10 11 95
# 13 13 52
# 13 15 39
# 15 16 70
# 17 18 54
# """
# sys.stdin = io.StringIO(_INPUT)
INF = 10**15
def solve(N, K, Waves):
dp = [list() for _ in range(N+1)]
dp[0].append((K, 0))
for i in range(N):
L, R, A = Waves[i]
dic = collections.defaultdict(int)
for remaining, reload in dp[i]:
# リロードしないで撃ち始める
reload_this_time = math.ceil((A-remaining)/K)
if reload_this_time <= R-L:
remaining1 = remaining + K* max(reload_this_time, 0) - A
reload1 = reload + reload_this_time
dic[reload1] = max(dic[reload1], remaining1)
if reload_this_time <= R-L-1:
dic[reload1+1] = max(dic[reload1+1], K)
# 1ターン目の前にリロードしておく（前のウェーブから1日ある場合のみ）
if i > 0 and Waves[i-1][1] < L and remaining < K:
remaining = K
reload_this_time = math.ceil((A-remaining)/K)
if reload_this_time <= R-L:
remaining1 = remaining + K* max(reload_this_time, 0) - A
reload1 = reload + reload_this_time + 1
dic[reload1] = max(dic[reload1], remaining1)
if reload_this_time <= R-L-1:
dic[reload1+1] = max(dic[reload1+1], K)
# 1ターン目にリロード（前のウェーブから1日も開けずにこのウェーブが始まる場合のみ）
if i > 0 and Waves[i-1][1] == L and remaining < K:
remaining = K
reload_this_time = math.ceil((A-remaining)/K)
if reload_this_time <= R-L-1:
remaining1 = remaining + K* max(reload_this_time, 0) - A
reload1 = reload + reload_this_time + 1
dic[reload1] = max(dic[reload1], remaining1)
if reload_this_time <= R-L-2:
dic[reload1+1] = max(dic[reload1+1], K)
dp[i+1] = [(dic[reload], reload) for reload in dic]
if dp[N]:
ans = min((reload+1)*K-remaining for remaining, reload in dp[N])
return ans
else:
return -1
N, K = map(int, input().split())
Waves = []
for _ in range(N):
L, R, A = map(int, input().split())
Waves.append((L, R, A))
Waves.sort()
print(solve(N, K, Waves))
``` | output | 1 | 70,047 | 2 | 140,095 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently you've discovered a new shooter. They say it has realistic game mechanics.
Your character has a gun with magazine size equal to k and should exterminate n waves of monsters. The i-th wave consists of a_i monsters and happens from the l_i-th moment of time up to the r_i-th moments of time. All a_i monsters spawn at moment l_i and you have to exterminate all of them before the moment r_i ends (you can kill monsters right at moment r_i). For every two consecutive waves, the second wave starts not earlier than the first wave ends (though the second wave can start at the same moment when the first wave ends) — formally, the condition r_i ≤ l_{i + 1} holds. Take a look at the notes for the examples to understand the process better.
You are confident in yours and your character's skills so you can assume that aiming and shooting are instant and you need exactly one bullet to kill one monster. But reloading takes exactly 1 unit of time.
One of the realistic mechanics is a mechanic of reloading: when you reload you throw away the old magazine with all remaining bullets in it. That's why constant reloads may cost you excessive amounts of spent bullets.
You've taken a liking to this mechanic so now you are wondering: what is the minimum possible number of bullets you need to spend (both used and thrown) to exterminate all waves.
Note that you don't throw the remaining bullets away after eradicating all monsters, and you start with a full magazine.
Input
The first line contains two integers n and k (1 ≤ n ≤ 2000; 1 ≤ k ≤ 10^9) — the number of waves and magazine size.
The next n lines contain descriptions of waves. The i-th line contains three integers l_i, r_i and a_i (1 ≤ l_i ≤ r_i ≤ 10^9; 1 ≤ a_i ≤ 10^9) — the period of time when the i-th wave happens and the number of monsters in it.
It's guaranteed that waves don't overlap (but may touch) and are given in the order they occur, i. e. r_i ≤ l_{i + 1}.
Output
If there is no way to clear all waves, print -1. Otherwise, print the minimum possible number of bullets you need to spend (both used and thrown) to clear all waves.
Examples
Input
2 3
2 3 6
3 4 3
Output
9
Input
2 5
3 7 11
10 12 15
Output
30
Input
5 42
42 42 42
42 43 42
43 44 42
44 45 42
45 45 1
Output
-1
Input
1 10
100 111 1
Output
1
Note
In the first example:
* At the moment 2, the first wave occurs and 6 monsters spawn. You kill 3 monsters and start reloading.
* At the moment 3, the second wave occurs and 3 more monsters spawn. You kill remaining 3 monsters from the first wave and start reloading.
* At the moment 4, you kill remaining 3 monsters from the second wave.
In total, you'll spend 9 bullets.
In the second example:
* At moment 3, the first wave occurs and 11 monsters spawn. You kill 5 monsters and start reloading.
* At moment 4, you kill 5 more monsters and start reloading.
* At moment 5, you kill the last monster and start reloading throwing away old magazine with 4 bullets.
* At moment 10, the second wave occurs and 15 monsters spawn. You kill 5 monsters and start reloading.
* At moment 11, you kill 5 more monsters and start reloading.
* At moment 12, you kill last 5 monsters.
In total, you'll spend 30 bullets. | instruction | 0 | 70,048 | 2 | 140,096 |
Tags: dp, greedy
Correct Solution:
```
import sys
n, k = map(int, input().split())
L, R, A = [0] * n, [0] * n, [0] * n
for i in range(n):
L[i], R[i], A[i] = map(int, input().split())
dp = [0] * (n + 1)
for i in range(n - 1, -1, -1):
need = A[i]
if i < n - 1 and R[i] == L[i + 1]:
need += dp[i + 1]
if (R[i] - L[i] + 1) * k < need:
print(-1)
sys.exit()
dp[i] = max(0, need - (R[i] - L[i]) * k)
rem, ans = k, 0
for i in range(n):
if rem < dp[i]:
ans += rem
rem = k
ans += A[i]
rem = (rem - A[i] + k) % k
print(ans)
``` | output | 1 | 70,048 | 2 | 140,097 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently you've discovered a new shooter. They say it has realistic game mechanics.
Your character has a gun with magazine size equal to k and should exterminate n waves of monsters. The i-th wave consists of a_i monsters and happens from the l_i-th moment of time up to the r_i-th moments of time. All a_i monsters spawn at moment l_i and you have to exterminate all of them before the moment r_i ends (you can kill monsters right at moment r_i). For every two consecutive waves, the second wave starts not earlier than the first wave ends (though the second wave can start at the same moment when the first wave ends) — formally, the condition r_i ≤ l_{i + 1} holds. Take a look at the notes for the examples to understand the process better.
You are confident in yours and your character's skills so you can assume that aiming and shooting are instant and you need exactly one bullet to kill one monster. But reloading takes exactly 1 unit of time.
One of the realistic mechanics is a mechanic of reloading: when you reload you throw away the old magazine with all remaining bullets in it. That's why constant reloads may cost you excessive amounts of spent bullets.
You've taken a liking to this mechanic so now you are wondering: what is the minimum possible number of bullets you need to spend (both used and thrown) to exterminate all waves.
Note that you don't throw the remaining bullets away after eradicating all monsters, and you start with a full magazine.
Input
The first line contains two integers n and k (1 ≤ n ≤ 2000; 1 ≤ k ≤ 10^9) — the number of waves and magazine size.
The next n lines contain descriptions of waves. The i-th line contains three integers l_i, r_i and a_i (1 ≤ l_i ≤ r_i ≤ 10^9; 1 ≤ a_i ≤ 10^9) — the period of time when the i-th wave happens and the number of monsters in it.
It's guaranteed that waves don't overlap (but may touch) and are given in the order they occur, i. e. r_i ≤ l_{i + 1}.
Output
If there is no way to clear all waves, print -1. Otherwise, print the minimum possible number of bullets you need to spend (both used and thrown) to clear all waves.
Examples
Input
2 3
2 3 6
3 4 3
Output
9
Input
2 5
3 7 11
10 12 15
Output
30
Input
5 42
42 42 42
42 43 42
43 44 42
44 45 42
45 45 1
Output
-1
Input
1 10
100 111 1
Output
1
Note
In the first example:
* At the moment 2, the first wave occurs and 6 monsters spawn. You kill 3 monsters and start reloading.
* At the moment 3, the second wave occurs and 3 more monsters spawn. You kill remaining 3 monsters from the first wave and start reloading.
* At the moment 4, you kill remaining 3 monsters from the second wave.
In total, you'll spend 9 bullets.
In the second example:
* At moment 3, the first wave occurs and 11 monsters spawn. You kill 5 monsters and start reloading.
* At moment 4, you kill 5 more monsters and start reloading.
* At moment 5, you kill the last monster and start reloading throwing away old magazine with 4 bullets.
* At moment 10, the second wave occurs and 15 monsters spawn. You kill 5 monsters and start reloading.
* At moment 11, you kill 5 more monsters and start reloading.
* At moment 12, you kill last 5 monsters.
In total, you'll spend 30 bullets. | instruction | 0 | 70,049 | 2 | 140,098 |
Tags: dp, greedy
Correct Solution:
```
from bisect import *
from collections import *
from math import *
from heapq import *
from typing import List
from itertools import *
from operator import *
from functools import *
#------------------------------------------------------------------------
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())
#------------------------------------------------------------------------
@lru_cache(None)
def fact(x):
if x<2:
return 1
return fact(x-1)*x
@lru_cache(None)
def per(i,j):
return fact(i)//fact(i-j)
@lru_cache(None)
def com(i,j):
return per(i,j)//fact(j)
def linc(f,t,l,r):
while l<r:
mid=(l+r)//2
if t>f(mid):
l=mid+1
else:
r=mid
return l
def rinc(f,t,l,r):
while l<r:
mid=(l+r+1)//2
if t<f(mid):
r=mid-1
else:
l=mid
return l
def ldec(f,t,l,r):
while l<r:
mid=(l+r)//2
if t<f(mid):
l=mid+1
else:
r=mid
return l
def rdec(f,t,l,r):
while l<r:
mid=(l+r+1)//2
if t>f(mid):
r=mid-1
else:
l=mid
return l
def isprime(n):
for i in range(2,int(n**0.5)+1):
if n%i==0:
return False
return True
def binfun(x):
c=1
res=0
for w in arr:
res+=w
if res>x:
res=w
c+=1
return c
def lowbit(n):
return n&-n
class BIT:
def __init__(self,arr):
self.arr=arr
self.n=len(arr)-1
def update(self,x,v):
while x<=self.n:
self.arr[x]+=v
x+=x&-x
def query(self,x):
ans=0
while x:
ans+=self.arr[x]
x&=x-1
return ans
class smt:
def __init__(self,l,r,arr):
self.l=l
self.r=r
self.value=(1<<31)-1 if l<r else arr[l]
mid=(l+r)//2
if(l<r):
self.left=smt(l,mid,arr)
self.right=smt(mid+1,r,arr)
self.value&=self.left.value&self.right.value
#print(l,r,self.value)
def setvalue(self,x,val):
if(self.l==self.r):
self.value=val
return
mid=(self.l+self.r)//2
if(x<=mid):
self.left.setvalue(x,val)
else:
self.right.setvalue(x,val)
self.value=self.left.value&self.right.value
def ask(self,l,r):
if(l<=self.l and r>=self.r):
return self.value
val=(1<<31)-1
mid=(self.l+self.r)//2
if(l<=mid):
val&=self.left.ask(l,r)
if(r>mid):
val&=self.right.ask(l,r)
return val
class UFS:
def __init__(self,n):
self.parent=[i for i in range(n)]
self.ranks=[0]*n
def find(self,x):
if x!=self.parent[x]:
self.parent[x]=self.find(self.parent[x])
return self.parent[x]
def union(self,u,v):
pu,pv=self.find(u),self.find(v)
if pu==pv:
return False
if self.ranks[pu]>=self.ranks[pv]:
self.parent[pv]=pu
if self.ranks[pv]==self.ranks[pu]:
self.ranks[pu]+=1
else:
self.parent[pu]=pv
def Prime(n):
c=0
prime=[]
flag=[0]*(n+1)
for i in range(2,n+1):
if not flag[i]:
prime.append(i)
c+=1
for j in range(c):
if i*prime[j]>n: break
flag[i*prime[j]]=prime[j]
if i%prime[j]==0: break
return prime
def dij(s,graph):
d={}
d[s]=0
heap=[(0,s)]
seen=set()
while heap:
dis,u=heappop(heap)
if u in seen:
continue
for v in graph[u]:
if v not in d or d[v]>d[u]+graph[u][v]:
d[v]=d[u]+graph[u][v]
heappush(heap,(d[v],v))
return d
'''
import time
s=time.time()
for i in range(2000):
print(0)
e=time.time()
print(e-s)
'''
n,k=RL()
arr=[]
for i in range(n):
arr.append(RLL())
dp=[inf]*n
dp[0]=0
ans=inf
for i in range(n):
rem=k
total=dp[i]
for j in range(i,n):
c=ceil(max(0,arr[j][2]-rem)/k)
if c>arr[j][1]-arr[j][0]:
break
newrem=rem+c*k-arr[j][2]
total+=arr[j][2]
if j+1<n:
if arr[j+1][0]>c+arr[j][0]:
dp[j+1]=min(dp[j+1],total+newrem)
rem=newrem
else:
ans=min(ans,total)
#print(i,dp,ans)
print(ans if ans!=inf else -1)
'''
t=N()
for i in range(t):
n=N()
s=input()
arr=[]
for ch,g in groupby(s):
arr.append(len(list(g)))
n=len(arr)
if len(arr)==1:
ans=1
else:
ans=0
i=j=0
while i<n:
j=max(i,j)
while j<n and arr[j]<=1:
j+=1
if j==n:
i+=2
ans+=1
if i>=n:
break
else:
arr[j]-=1
i+=1
ans+=1
print(ans)
'''
``` | output | 1 | 70,049 | 2 | 140,099 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently you've discovered a new shooter. They say it has realistic game mechanics.
Your character has a gun with magazine size equal to k and should exterminate n waves of monsters. The i-th wave consists of a_i monsters and happens from the l_i-th moment of time up to the r_i-th moments of time. All a_i monsters spawn at moment l_i and you have to exterminate all of them before the moment r_i ends (you can kill monsters right at moment r_i). For every two consecutive waves, the second wave starts not earlier than the first wave ends (though the second wave can start at the same moment when the first wave ends) — formally, the condition r_i ≤ l_{i + 1} holds. Take a look at the notes for the examples to understand the process better.
You are confident in yours and your character's skills so you can assume that aiming and shooting are instant and you need exactly one bullet to kill one monster. But reloading takes exactly 1 unit of time.
One of the realistic mechanics is a mechanic of reloading: when you reload you throw away the old magazine with all remaining bullets in it. That's why constant reloads may cost you excessive amounts of spent bullets.
You've taken a liking to this mechanic so now you are wondering: what is the minimum possible number of bullets you need to spend (both used and thrown) to exterminate all waves.
Note that you don't throw the remaining bullets away after eradicating all monsters, and you start with a full magazine.
Input
The first line contains two integers n and k (1 ≤ n ≤ 2000; 1 ≤ k ≤ 10^9) — the number of waves and magazine size.
The next n lines contain descriptions of waves. The i-th line contains three integers l_i, r_i and a_i (1 ≤ l_i ≤ r_i ≤ 10^9; 1 ≤ a_i ≤ 10^9) — the period of time when the i-th wave happens and the number of monsters in it.
It's guaranteed that waves don't overlap (but may touch) and are given in the order they occur, i. e. r_i ≤ l_{i + 1}.
Output
If there is no way to clear all waves, print -1. Otherwise, print the minimum possible number of bullets you need to spend (both used and thrown) to clear all waves.
Examples
Input
2 3
2 3 6
3 4 3
Output
9
Input
2 5
3 7 11
10 12 15
Output
30
Input
5 42
42 42 42
42 43 42
43 44 42
44 45 42
45 45 1
Output
-1
Input
1 10
100 111 1
Output
1
Note
In the first example:
* At the moment 2, the first wave occurs and 6 monsters spawn. You kill 3 monsters and start reloading.
* At the moment 3, the second wave occurs and 3 more monsters spawn. You kill remaining 3 monsters from the first wave and start reloading.
* At the moment 4, you kill remaining 3 monsters from the second wave.
In total, you'll spend 9 bullets.
In the second example:
* At moment 3, the first wave occurs and 11 monsters spawn. You kill 5 monsters and start reloading.
* At moment 4, you kill 5 more monsters and start reloading.
* At moment 5, you kill the last monster and start reloading throwing away old magazine with 4 bullets.
* At moment 10, the second wave occurs and 15 monsters spawn. You kill 5 monsters and start reloading.
* At moment 11, you kill 5 more monsters and start reloading.
* At moment 12, you kill last 5 monsters.
In total, you'll spend 30 bullets. | instruction | 0 | 70,050 | 2 | 140,100 |
Tags: dp, greedy
Correct Solution:
```
import sys
input = sys.stdin.buffer.readline
n,k = map(int,input().split())
tank = [(k,0)]
new = []
L = [0]*n
R = [0]*n
A = [0]*n
for i in range(n):
L[i],R[i],A[i] = map(int,input().split())
for i in range(n):
new = []
l,r,a = L[i],R[i],A[i]
val = -1
for p,q in tank:
if p + (r-l)*k < a:
continue
ct,v = 0,0
if p >= a:
new.append((p-a,q+a))
v = p-a
else:
tmp = a-p
if tmp%k == 0:
new.append((0,q+a))
ct = tmp//k
else:
amari = (tmp//k + 1)*k
new.append((p+amari-a,q+a))
ct = tmp//k + 1
v = p+amari-a
if r-l > ct or (i < n-1 and r < L[i+1]):
if val == -1 or val > q+a+v:
val = q+a+v
if val != -1:
new.append((k,val))
tank = new
#print(tank)
res = -1
for p,q in tank:
if res == -1:
res = q
elif res > q:
res = q
print(res)
``` | output | 1 | 70,050 | 2 | 140,101 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently you've discovered a new shooter. They say it has realistic game mechanics.
Your character has a gun with magazine size equal to k and should exterminate n waves of monsters. The i-th wave consists of a_i monsters and happens from the l_i-th moment of time up to the r_i-th moments of time. All a_i monsters spawn at moment l_i and you have to exterminate all of them before the moment r_i ends (you can kill monsters right at moment r_i). For every two consecutive waves, the second wave starts not earlier than the first wave ends (though the second wave can start at the same moment when the first wave ends) — formally, the condition r_i ≤ l_{i + 1} holds. Take a look at the notes for the examples to understand the process better.
You are confident in yours and your character's skills so you can assume that aiming and shooting are instant and you need exactly one bullet to kill one monster. But reloading takes exactly 1 unit of time.
One of the realistic mechanics is a mechanic of reloading: when you reload you throw away the old magazine with all remaining bullets in it. That's why constant reloads may cost you excessive amounts of spent bullets.
You've taken a liking to this mechanic so now you are wondering: what is the minimum possible number of bullets you need to spend (both used and thrown) to exterminate all waves.
Note that you don't throw the remaining bullets away after eradicating all monsters, and you start with a full magazine.
Input
The first line contains two integers n and k (1 ≤ n ≤ 2000; 1 ≤ k ≤ 10^9) — the number of waves and magazine size.
The next n lines contain descriptions of waves. The i-th line contains three integers l_i, r_i and a_i (1 ≤ l_i ≤ r_i ≤ 10^9; 1 ≤ a_i ≤ 10^9) — the period of time when the i-th wave happens and the number of monsters in it.
It's guaranteed that waves don't overlap (but may touch) and are given in the order they occur, i. e. r_i ≤ l_{i + 1}.
Output
If there is no way to clear all waves, print -1. Otherwise, print the minimum possible number of bullets you need to spend (both used and thrown) to clear all waves.
Examples
Input
2 3
2 3 6
3 4 3
Output
9
Input
2 5
3 7 11
10 12 15
Output
30
Input
5 42
42 42 42
42 43 42
43 44 42
44 45 42
45 45 1
Output
-1
Input
1 10
100 111 1
Output
1
Note
In the first example:
* At the moment 2, the first wave occurs and 6 monsters spawn. You kill 3 monsters and start reloading.
* At the moment 3, the second wave occurs and 3 more monsters spawn. You kill remaining 3 monsters from the first wave and start reloading.
* At the moment 4, you kill remaining 3 monsters from the second wave.
In total, you'll spend 9 bullets.
In the second example:
* At moment 3, the first wave occurs and 11 monsters spawn. You kill 5 monsters and start reloading.
* At moment 4, you kill 5 more monsters and start reloading.
* At moment 5, you kill the last monster and start reloading throwing away old magazine with 4 bullets.
* At moment 10, the second wave occurs and 15 monsters spawn. You kill 5 monsters and start reloading.
* At moment 11, you kill 5 more monsters and start reloading.
* At moment 12, you kill last 5 monsters.
In total, you'll spend 30 bullets. | instruction | 0 | 70,051 | 2 | 140,102 |
Tags: dp, greedy
Correct Solution:
```
import sys
from array import array # noqa: F401
import typing as Tp # noqa: F401
def input():
return sys.stdin.buffer.readline().decode('utf-8')
def output(*args):
sys.stdout.buffer.write(
('\n'.join(map(str, args)) + '\n').encode('utf-8')
)
def main():
n, k = map(int, input().split())
wave = [tuple(map(int, input().split())) for _ in range(n)]
ans = inf = 10**18
dp = [0] + [inf] * (n - 1)
for i in range(n):
if dp[i] == inf:
continue
time = wave[i][0]
bullet = k
cost = dp[i]
for j in range(i, n):
time = max(time, wave[j][0])
if bullet >= wave[j][2]:
bullet -= wave[j][2]
else:
ub = wave[j][1] - time
reload = (wave[j][2] - bullet - 1) // k + 1
if ub < reload:
break
time += reload
bullet = (-wave[j][2] + bullet) % k
cost += wave[j][2]
if j < n - 1 and time < wave[j + 1][0]:
dp[j + 1] = min(dp[j + 1], cost + bullet)
else:
ans = min(ans, cost)
print(ans if ans < inf else -1)
if __name__ == '__main__':
main()
``` | output | 1 | 70,051 | 2 | 140,103 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently you've discovered a new shooter. They say it has realistic game mechanics.
Your character has a gun with magazine size equal to k and should exterminate n waves of monsters. The i-th wave consists of a_i monsters and happens from the l_i-th moment of time up to the r_i-th moments of time. All a_i monsters spawn at moment l_i and you have to exterminate all of them before the moment r_i ends (you can kill monsters right at moment r_i). For every two consecutive waves, the second wave starts not earlier than the first wave ends (though the second wave can start at the same moment when the first wave ends) — formally, the condition r_i ≤ l_{i + 1} holds. Take a look at the notes for the examples to understand the process better.
You are confident in yours and your character's skills so you can assume that aiming and shooting are instant and you need exactly one bullet to kill one monster. But reloading takes exactly 1 unit of time.
One of the realistic mechanics is a mechanic of reloading: when you reload you throw away the old magazine with all remaining bullets in it. That's why constant reloads may cost you excessive amounts of spent bullets.
You've taken a liking to this mechanic so now you are wondering: what is the minimum possible number of bullets you need to spend (both used and thrown) to exterminate all waves.
Note that you don't throw the remaining bullets away after eradicating all monsters, and you start with a full magazine.
Input
The first line contains two integers n and k (1 ≤ n ≤ 2000; 1 ≤ k ≤ 10^9) — the number of waves and magazine size.
The next n lines contain descriptions of waves. The i-th line contains three integers l_i, r_i and a_i (1 ≤ l_i ≤ r_i ≤ 10^9; 1 ≤ a_i ≤ 10^9) — the period of time when the i-th wave happens and the number of monsters in it.
It's guaranteed that waves don't overlap (but may touch) and are given in the order they occur, i. e. r_i ≤ l_{i + 1}.
Output
If there is no way to clear all waves, print -1. Otherwise, print the minimum possible number of bullets you need to spend (both used and thrown) to clear all waves.
Examples
Input
2 3
2 3 6
3 4 3
Output
9
Input
2 5
3 7 11
10 12 15
Output
30
Input
5 42
42 42 42
42 43 42
43 44 42
44 45 42
45 45 1
Output
-1
Input
1 10
100 111 1
Output
1
Note
In the first example:
* At the moment 2, the first wave occurs and 6 monsters spawn. You kill 3 monsters and start reloading.
* At the moment 3, the second wave occurs and 3 more monsters spawn. You kill remaining 3 monsters from the first wave and start reloading.
* At the moment 4, you kill remaining 3 monsters from the second wave.
In total, you'll spend 9 bullets.
In the second example:
* At moment 3, the first wave occurs and 11 monsters spawn. You kill 5 monsters and start reloading.
* At moment 4, you kill 5 more monsters and start reloading.
* At moment 5, you kill the last monster and start reloading throwing away old magazine with 4 bullets.
* At moment 10, the second wave occurs and 15 monsters spawn. You kill 5 monsters and start reloading.
* At moment 11, you kill 5 more monsters and start reloading.
* At moment 12, you kill last 5 monsters.
In total, you'll spend 30 bullets. | instruction | 0 | 70,052 | 2 | 140,104 |
Tags: dp, greedy
Correct Solution:
```
#!/usr/bin/env python
from __future__ import division, print_function
import os
import sys
from io import BytesIO, IOBase
from types import ModuleType
class _TailCall(Exception):
def __init__(self, f, args, uid):
self.func, self.args, self.uid, self.follow = f.func, args, uid, id(f)
def _tailCallback(f, uid):
"""
This is the "callable" version of the continuation, which sould only
be accessible from the inside of the function to be continued. An
attribute called "C" can be used in order to get back the public
version of the continuation (for passing the continuation to another
function).
"""
def t(*args):
raise _TailCall(f, args, uid)
t.C = f
return t
class _TailCallWrapper():
"""
Wrapper for tail-called optimized functions embedding their
continuations. Such functions are ready to be evaluated with
their arguments.
This is a private class and should never be accessed directly.
Functions should be created by using the C() class first.
"""
def __init__(self, func, k):
self.func = func( _tailCallback(self, id(self)),
*map( lambda c: _tailCallback(c, id(self)), k) )
def __call__(self, *args):
f, expect = self.func, id(self)
while True:
try:
return f(*args)
except _TailCall as e:
if e.uid == expect:
f, args, expect = e.func, e.args, e.follow
else:
raise e
class C():
"""
Main wrapper for tail-call optimized functions.
"""
def __init__(self, func):
self.func = func
def __call__(self, *k):
return _TailCallWrapper(self.func, k)
def with_continuations(**c):
"""
A decorator for defining tail-call optimized functions.
Example
-------
@with_continuations()
def factorial(n, k, self=None):
return self(n-1, k*n) if n > 1 else k
@with_continuations()
def identity(x, self=None):
return x
@with_continuations(out=identity)
def factorial2(n, k, self=None, out=None):
return self(n-1, k*n) if n > 1 else out(k)
print(factorial(7,1))
print(factorial2(7,1))
"""
if len(c): keys, k = zip(*c.items())
else: keys, k = tuple([]), tuple([])
def d(f):
return C(
lambda kself, *conts:
lambda *args:
f(*args, self=kself, **dict(zip(keys, conts)))) (*k)
return d
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
from collections import defaultdict
____dm = None
ONLINE_JUDGE = __debug__
def inputline():
global inp
inp = input().split()
inp.reverse()
def readint(): return int(inp.pop())
def readfloat(): return float(inp.pop())
def nextline(): return inputline()
def ____void(*arg): return None
displayln = print
if ONLINE_JUDGE:
dbgln = ____void
else:
dbgln = print
def main():
def asc(k, d):
try:
return d.__getitem__(k)
except KeyError:
return False
def dictmap(f, d):
rt = []
for (k, v) in d.items():
rt.append(f((k, v)))
return rt
n = readint()
k = readint()
____dm = nextline()
ret = None
pr = None
@with_continuations()
def loop(ans = {0: k}, n = n, lastr = -1, self = None):
if ans == {} or n == 0:
return ans
ll = readint()
rr = readint()
a = readint()
____dm = nextline()
nans = {}
rl = rr - ll
def add(a, r):
cs = asc(a, nans)
cs0 = asc(a - 1, nans)
if cs0 == False or cs0 < r:
if cs == False or nans[a] < r:
nans[a] = r
def f(xx):
ax = xx[0]
dx = xx[1]
aa = a - dx
q = aa // k
r = aa % k
if aa <= 0:
add(ax, dx - a)
if rl > 0:
add(ax + 1, k)
else:
nax = ax + q + 1
nrl = rl - q
rst = k - r
if nrl > 1:
add(nax, rst)
add(nax + 1, k)
elif nrl == 1:
add(nax, rst)
elif nrl == 0 and r == 0:
add(nax - 1, 0)
def g(xx):
ax = xx[0]
dx = xx[1]
if dx < k:
add(ax + 1, k)
if lastr < ll:
nans = ans.copy()
dictmap(g, ans)
ans = nans
nans = {}
dictmap(f, ans)
dbgln(nans)
return self(nans, n - 1, rr)
def h(xx):
ax = xx[0]
dx = xx[1]
return (ax + 1) * k - dx
ret = loop()
dbgln(ret)
ret = dictmap(h, ret)
if ret == []:
pr = -1
else:
pr = min(ret)
displayln(pr)
# 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")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
inputline()
main()
``` | output | 1 | 70,052 | 2 | 140,105 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently you've discovered a new shooter. They say it has realistic game mechanics.
Your character has a gun with magazine size equal to k and should exterminate n waves of monsters. The i-th wave consists of a_i monsters and happens from the l_i-th moment of time up to the r_i-th moments of time. All a_i monsters spawn at moment l_i and you have to exterminate all of them before the moment r_i ends (you can kill monsters right at moment r_i). For every two consecutive waves, the second wave starts not earlier than the first wave ends (though the second wave can start at the same moment when the first wave ends) — formally, the condition r_i ≤ l_{i + 1} holds. Take a look at the notes for the examples to understand the process better.
You are confident in yours and your character's skills so you can assume that aiming and shooting are instant and you need exactly one bullet to kill one monster. But reloading takes exactly 1 unit of time.
One of the realistic mechanics is a mechanic of reloading: when you reload you throw away the old magazine with all remaining bullets in it. That's why constant reloads may cost you excessive amounts of spent bullets.
You've taken a liking to this mechanic so now you are wondering: what is the minimum possible number of bullets you need to spend (both used and thrown) to exterminate all waves.
Note that you don't throw the remaining bullets away after eradicating all monsters, and you start with a full magazine.
Input
The first line contains two integers n and k (1 ≤ n ≤ 2000; 1 ≤ k ≤ 10^9) — the number of waves and magazine size.
The next n lines contain descriptions of waves. The i-th line contains three integers l_i, r_i and a_i (1 ≤ l_i ≤ r_i ≤ 10^9; 1 ≤ a_i ≤ 10^9) — the period of time when the i-th wave happens and the number of monsters in it.
It's guaranteed that waves don't overlap (but may touch) and are given in the order they occur, i. e. r_i ≤ l_{i + 1}.
Output
If there is no way to clear all waves, print -1. Otherwise, print the minimum possible number of bullets you need to spend (both used and thrown) to clear all waves.
Examples
Input
2 3
2 3 6
3 4 3
Output
9
Input
2 5
3 7 11
10 12 15
Output
30
Input
5 42
42 42 42
42 43 42
43 44 42
44 45 42
45 45 1
Output
-1
Input
1 10
100 111 1
Output
1
Note
In the first example:
* At the moment 2, the first wave occurs and 6 monsters spawn. You kill 3 monsters and start reloading.
* At the moment 3, the second wave occurs and 3 more monsters spawn. You kill remaining 3 monsters from the first wave and start reloading.
* At the moment 4, you kill remaining 3 monsters from the second wave.
In total, you'll spend 9 bullets.
In the second example:
* At moment 3, the first wave occurs and 11 monsters spawn. You kill 5 monsters and start reloading.
* At moment 4, you kill 5 more monsters and start reloading.
* At moment 5, you kill the last monster and start reloading throwing away old magazine with 4 bullets.
* At moment 10, the second wave occurs and 15 monsters spawn. You kill 5 monsters and start reloading.
* At moment 11, you kill 5 more monsters and start reloading.
* At moment 12, you kill last 5 monsters.
In total, you'll spend 30 bullets. | instruction | 0 | 70,053 | 2 | 140,106 |
Tags: dp, greedy
Correct Solution:
```
import sys
n, k = map(int, input().split())
L, R, A = [0] * n, [0] * n, [0] * n
for i in range(n):
L[i], R[i], A[i] = map(int, input().split())
dp = [0] * (n + 1)
for i in range(n - 1, -1, -1):
need = A[i]
if i < n - 1 and R[i] == L[i + 1]:
need += dp[i + 1]
if (R[i] - L[i] + 1) * k < need:
print(-1)
sys.exit()
dp[i] = max(0, need - (R[i] - L[i]) * k)
rem, ans = k, 0
for i in range(n):
if rem < dp[i]:
ans += rem
rem = k
ans += A[i]
rem = (rem - A[i]) % k
print(ans)
``` | output | 1 | 70,053 | 2 | 140,107 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently you've discovered a new shooter. They say it has realistic game mechanics.
Your character has a gun with magazine size equal to k and should exterminate n waves of monsters. The i-th wave consists of a_i monsters and happens from the l_i-th moment of time up to the r_i-th moments of time. All a_i monsters spawn at moment l_i and you have to exterminate all of them before the moment r_i ends (you can kill monsters right at moment r_i). For every two consecutive waves, the second wave starts not earlier than the first wave ends (though the second wave can start at the same moment when the first wave ends) — formally, the condition r_i ≤ l_{i + 1} holds. Take a look at the notes for the examples to understand the process better.
You are confident in yours and your character's skills so you can assume that aiming and shooting are instant and you need exactly one bullet to kill one monster. But reloading takes exactly 1 unit of time.
One of the realistic mechanics is a mechanic of reloading: when you reload you throw away the old magazine with all remaining bullets in it. That's why constant reloads may cost you excessive amounts of spent bullets.
You've taken a liking to this mechanic so now you are wondering: what is the minimum possible number of bullets you need to spend (both used and thrown) to exterminate all waves.
Note that you don't throw the remaining bullets away after eradicating all monsters, and you start with a full magazine.
Input
The first line contains two integers n and k (1 ≤ n ≤ 2000; 1 ≤ k ≤ 10^9) — the number of waves and magazine size.
The next n lines contain descriptions of waves. The i-th line contains three integers l_i, r_i and a_i (1 ≤ l_i ≤ r_i ≤ 10^9; 1 ≤ a_i ≤ 10^9) — the period of time when the i-th wave happens and the number of monsters in it.
It's guaranteed that waves don't overlap (but may touch) and are given in the order they occur, i. e. r_i ≤ l_{i + 1}.
Output
If there is no way to clear all waves, print -1. Otherwise, print the minimum possible number of bullets you need to spend (both used and thrown) to clear all waves.
Examples
Input
2 3
2 3 6
3 4 3
Output
9
Input
2 5
3 7 11
10 12 15
Output
30
Input
5 42
42 42 42
42 43 42
43 44 42
44 45 42
45 45 1
Output
-1
Input
1 10
100 111 1
Output
1
Note
In the first example:
* At the moment 2, the first wave occurs and 6 monsters spawn. You kill 3 monsters and start reloading.
* At the moment 3, the second wave occurs and 3 more monsters spawn. You kill remaining 3 monsters from the first wave and start reloading.
* At the moment 4, you kill remaining 3 monsters from the second wave.
In total, you'll spend 9 bullets.
In the second example:
* At moment 3, the first wave occurs and 11 monsters spawn. You kill 5 monsters and start reloading.
* At moment 4, you kill 5 more monsters and start reloading.
* At moment 5, you kill the last monster and start reloading throwing away old magazine with 4 bullets.
* At moment 10, the second wave occurs and 15 monsters spawn. You kill 5 monsters and start reloading.
* At moment 11, you kill 5 more monsters and start reloading.
* At moment 12, you kill last 5 monsters.
In total, you'll spend 30 bullets.
Submitted Solution:
```
import sys
n, k = list(map(int, sys.stdin.readline().strip().split()))
L = []
R = []
A = []
for i in range (0, n):
x = list(map(int, sys.stdin.readline().strip().split()))
L.append(x[0])
R.append(x[1])
A.append(x[2])
L.append(R[-1])
i = n-1
x = 0
y = 0
ans = 0
v = True
N = [0 for i in range (0, n)]
while i >= 0:
if R[i] == L[i+1]:
x = max(x + A[i] - k * (R[i] - L[i]), 0)
N[i] = x
else:
x = max(A[i] - k * (R[i] - L[i]), 0)
N[i] = x
if N[i] > k:
v = False
i = i - 1
m = k
N.append(0)
i = 0
while i < n and v == True:
if m < N[i]:
ans = ans + m
m = k
m = m - A[i]
ans = ans + A[i]
while m < 0:
m = m + k
i = i + 1
if v == True:
print(ans)
else:
print(-1)
``` | instruction | 0 | 70,054 | 2 | 140,108 |
Yes | output | 1 | 70,054 | 2 | 140,109 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently you've discovered a new shooter. They say it has realistic game mechanics.
Your character has a gun with magazine size equal to k and should exterminate n waves of monsters. The i-th wave consists of a_i monsters and happens from the l_i-th moment of time up to the r_i-th moments of time. All a_i monsters spawn at moment l_i and you have to exterminate all of them before the moment r_i ends (you can kill monsters right at moment r_i). For every two consecutive waves, the second wave starts not earlier than the first wave ends (though the second wave can start at the same moment when the first wave ends) — formally, the condition r_i ≤ l_{i + 1} holds. Take a look at the notes for the examples to understand the process better.
You are confident in yours and your character's skills so you can assume that aiming and shooting are instant and you need exactly one bullet to kill one monster. But reloading takes exactly 1 unit of time.
One of the realistic mechanics is a mechanic of reloading: when you reload you throw away the old magazine with all remaining bullets in it. That's why constant reloads may cost you excessive amounts of spent bullets.
You've taken a liking to this mechanic so now you are wondering: what is the minimum possible number of bullets you need to spend (both used and thrown) to exterminate all waves.
Note that you don't throw the remaining bullets away after eradicating all monsters, and you start with a full magazine.
Input
The first line contains two integers n and k (1 ≤ n ≤ 2000; 1 ≤ k ≤ 10^9) — the number of waves and magazine size.
The next n lines contain descriptions of waves. The i-th line contains three integers l_i, r_i and a_i (1 ≤ l_i ≤ r_i ≤ 10^9; 1 ≤ a_i ≤ 10^9) — the period of time when the i-th wave happens and the number of monsters in it.
It's guaranteed that waves don't overlap (but may touch) and are given in the order they occur, i. e. r_i ≤ l_{i + 1}.
Output
If there is no way to clear all waves, print -1. Otherwise, print the minimum possible number of bullets you need to spend (both used and thrown) to clear all waves.
Examples
Input
2 3
2 3 6
3 4 3
Output
9
Input
2 5
3 7 11
10 12 15
Output
30
Input
5 42
42 42 42
42 43 42
43 44 42
44 45 42
45 45 1
Output
-1
Input
1 10
100 111 1
Output
1
Note
In the first example:
* At the moment 2, the first wave occurs and 6 monsters spawn. You kill 3 monsters and start reloading.
* At the moment 3, the second wave occurs and 3 more monsters spawn. You kill remaining 3 monsters from the first wave and start reloading.
* At the moment 4, you kill remaining 3 monsters from the second wave.
In total, you'll spend 9 bullets.
In the second example:
* At moment 3, the first wave occurs and 11 monsters spawn. You kill 5 monsters and start reloading.
* At moment 4, you kill 5 more monsters and start reloading.
* At moment 5, you kill the last monster and start reloading throwing away old magazine with 4 bullets.
* At moment 10, the second wave occurs and 15 monsters spawn. You kill 5 monsters and start reloading.
* At moment 11, you kill 5 more monsters and start reloading.
* At moment 12, you kill last 5 monsters.
In total, you'll spend 30 bullets.
Submitted Solution:
```
from sys import stdin
n, k = tuple(int(x) for x in stdin.readline().split())
data = []
for _ in range(n):
data.append(tuple(int(x) for x in stdin.readline().split()))
data.append((float('inf'), float('inf'), 0))
mini = []
ans = 0
for i in range(n):
l, r, a = data[i]
ans += a
remain = max(0, a-(r-l)*k)
if remain > k:
print(-1)
break
mini.append(remain)
else:
last = k
remain = k
for i in range(n):
l, r, a = data[i]
#print(remain, mini[i], last)
if remain < mini[i]:
ans += remain
remain += (k-last)
remain_cpy = remain
if remain >= mini[i]:
remain = (remain - a) % k
if (a-remain_cpy) <= k * (data[i+1][0] - data[i][0] - 1):
last = remain
else:
print(-1)
break
else:
print(ans)
``` | instruction | 0 | 70,055 | 2 | 140,110 |
No | output | 1 | 70,055 | 2 | 140,111 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently you've discovered a new shooter. They say it has realistic game mechanics.
Your character has a gun with magazine size equal to k and should exterminate n waves of monsters. The i-th wave consists of a_i monsters and happens from the l_i-th moment of time up to the r_i-th moments of time. All a_i monsters spawn at moment l_i and you have to exterminate all of them before the moment r_i ends (you can kill monsters right at moment r_i). For every two consecutive waves, the second wave starts not earlier than the first wave ends (though the second wave can start at the same moment when the first wave ends) — formally, the condition r_i ≤ l_{i + 1} holds. Take a look at the notes for the examples to understand the process better.
You are confident in yours and your character's skills so you can assume that aiming and shooting are instant and you need exactly one bullet to kill one monster. But reloading takes exactly 1 unit of time.
One of the realistic mechanics is a mechanic of reloading: when you reload you throw away the old magazine with all remaining bullets in it. That's why constant reloads may cost you excessive amounts of spent bullets.
You've taken a liking to this mechanic so now you are wondering: what is the minimum possible number of bullets you need to spend (both used and thrown) to exterminate all waves.
Note that you don't throw the remaining bullets away after eradicating all monsters, and you start with a full magazine.
Input
The first line contains two integers n and k (1 ≤ n ≤ 2000; 1 ≤ k ≤ 10^9) — the number of waves and magazine size.
The next n lines contain descriptions of waves. The i-th line contains three integers l_i, r_i and a_i (1 ≤ l_i ≤ r_i ≤ 10^9; 1 ≤ a_i ≤ 10^9) — the period of time when the i-th wave happens and the number of monsters in it.
It's guaranteed that waves don't overlap (but may touch) and are given in the order they occur, i. e. r_i ≤ l_{i + 1}.
Output
If there is no way to clear all waves, print -1. Otherwise, print the minimum possible number of bullets you need to spend (both used and thrown) to clear all waves.
Examples
Input
2 3
2 3 6
3 4 3
Output
9
Input
2 5
3 7 11
10 12 15
Output
30
Input
5 42
42 42 42
42 43 42
43 44 42
44 45 42
45 45 1
Output
-1
Input
1 10
100 111 1
Output
1
Note
In the first example:
* At the moment 2, the first wave occurs and 6 monsters spawn. You kill 3 monsters and start reloading.
* At the moment 3, the second wave occurs and 3 more monsters spawn. You kill remaining 3 monsters from the first wave and start reloading.
* At the moment 4, you kill remaining 3 monsters from the second wave.
In total, you'll spend 9 bullets.
In the second example:
* At moment 3, the first wave occurs and 11 monsters spawn. You kill 5 monsters and start reloading.
* At moment 4, you kill 5 more monsters and start reloading.
* At moment 5, you kill the last monster and start reloading throwing away old magazine with 4 bullets.
* At moment 10, the second wave occurs and 15 monsters spawn. You kill 5 monsters and start reloading.
* At moment 11, you kill 5 more monsters and start reloading.
* At moment 12, you kill last 5 monsters.
In total, you'll spend 30 bullets.
Submitted Solution:
```
def min_bullets(k, n, l, r, a):
initial = k
used_bullets = 0
bonus = (False, None)
for i in range(n):
necessary_reload = max((a[i] - initial), 0) // k + 1 if max((a[i] - initial), 0) % k else max((a[i] - initial), 0) // k
if necessary_reload > r[i] - l[i]:
if not bonus[0] or necessary_reload > r[i] - l[i] + 1 or a[i] > k*(r[i] - l[i] + 1):
return -1
used_bullets += bonus[1]
bonus = (False, None)
initial = k
used_bullets += a[i]
if initial >= a[i]:
initial -= a[i]
else:
initial = k - (a[i] - initial)%k if (a[i] - initial)%k else 0
bonus = (True, initial) if (necessary_reload < r[i] - l[i]) or (i<n-1 and r[i]<l[i+1]) else bonus
return used_bullets
if __name__ == "__main__":
n, k = [int(x) for x in input().split()]
l, r, a = [], [], []
for _ in range(n):
l_element, r_element, a_element = input().split()
l.append(int(l_element))
r.append(int(r_element))
a.append(int(a_element))
print(min_bullets(k, n, l, r, a))
``` | instruction | 0 | 70,056 | 2 | 140,112 |
No | output | 1 | 70,056 | 2 | 140,113 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently you've discovered a new shooter. They say it has realistic game mechanics.
Your character has a gun with magazine size equal to k and should exterminate n waves of monsters. The i-th wave consists of a_i monsters and happens from the l_i-th moment of time up to the r_i-th moments of time. All a_i monsters spawn at moment l_i and you have to exterminate all of them before the moment r_i ends (you can kill monsters right at moment r_i). For every two consecutive waves, the second wave starts not earlier than the first wave ends (though the second wave can start at the same moment when the first wave ends) — formally, the condition r_i ≤ l_{i + 1} holds. Take a look at the notes for the examples to understand the process better.
You are confident in yours and your character's skills so you can assume that aiming and shooting are instant and you need exactly one bullet to kill one monster. But reloading takes exactly 1 unit of time.
One of the realistic mechanics is a mechanic of reloading: when you reload you throw away the old magazine with all remaining bullets in it. That's why constant reloads may cost you excessive amounts of spent bullets.
You've taken a liking to this mechanic so now you are wondering: what is the minimum possible number of bullets you need to spend (both used and thrown) to exterminate all waves.
Note that you don't throw the remaining bullets away after eradicating all monsters, and you start with a full magazine.
Input
The first line contains two integers n and k (1 ≤ n ≤ 2000; 1 ≤ k ≤ 10^9) — the number of waves and magazine size.
The next n lines contain descriptions of waves. The i-th line contains three integers l_i, r_i and a_i (1 ≤ l_i ≤ r_i ≤ 10^9; 1 ≤ a_i ≤ 10^9) — the period of time when the i-th wave happens and the number of monsters in it.
It's guaranteed that waves don't overlap (but may touch) and are given in the order they occur, i. e. r_i ≤ l_{i + 1}.
Output
If there is no way to clear all waves, print -1. Otherwise, print the minimum possible number of bullets you need to spend (both used and thrown) to clear all waves.
Examples
Input
2 3
2 3 6
3 4 3
Output
9
Input
2 5
3 7 11
10 12 15
Output
30
Input
5 42
42 42 42
42 43 42
43 44 42
44 45 42
45 45 1
Output
-1
Input
1 10
100 111 1
Output
1
Note
In the first example:
* At the moment 2, the first wave occurs and 6 monsters spawn. You kill 3 monsters and start reloading.
* At the moment 3, the second wave occurs and 3 more monsters spawn. You kill remaining 3 monsters from the first wave and start reloading.
* At the moment 4, you kill remaining 3 monsters from the second wave.
In total, you'll spend 9 bullets.
In the second example:
* At moment 3, the first wave occurs and 11 monsters spawn. You kill 5 monsters and start reloading.
* At moment 4, you kill 5 more monsters and start reloading.
* At moment 5, you kill the last monster and start reloading throwing away old magazine with 4 bullets.
* At moment 10, the second wave occurs and 15 monsters spawn. You kill 5 monsters and start reloading.
* At moment 11, you kill 5 more monsters and start reloading.
* At moment 12, you kill last 5 monsters.
In total, you'll spend 30 bullets.
Submitted Solution:
```
mod = 1000000007
eps = 10**-9
def main():
import sys
input = sys.stdin.buffer.readline
N, K = map(int, input().split())
r_prev = -1
mag = K
ans = 0
for _ in range(N):
l, r, a = map(int, input().split())
if l != r_prev:
if (r - l) * K + mag >= a:
ans += a
mag = (mag - a) % K
elif (r - l + 1) * K < a:
print(-1)
exit()
else:
ans += mag
mag = (-a) % K
ans += a
else:
if mag + K * (r - l) >= a:
ans += a
mag = (mag - a) % K
else:
print(-1)
exit()
r_prev = r
print(ans)
if __name__ == '__main__':
main()
``` | instruction | 0 | 70,057 | 2 | 140,114 |
No | output | 1 | 70,057 | 2 | 140,115 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently you've discovered a new shooter. They say it has realistic game mechanics.
Your character has a gun with magazine size equal to k and should exterminate n waves of monsters. The i-th wave consists of a_i monsters and happens from the l_i-th moment of time up to the r_i-th moments of time. All a_i monsters spawn at moment l_i and you have to exterminate all of them before the moment r_i ends (you can kill monsters right at moment r_i). For every two consecutive waves, the second wave starts not earlier than the first wave ends (though the second wave can start at the same moment when the first wave ends) — formally, the condition r_i ≤ l_{i + 1} holds. Take a look at the notes for the examples to understand the process better.
You are confident in yours and your character's skills so you can assume that aiming and shooting are instant and you need exactly one bullet to kill one monster. But reloading takes exactly 1 unit of time.
One of the realistic mechanics is a mechanic of reloading: when you reload you throw away the old magazine with all remaining bullets in it. That's why constant reloads may cost you excessive amounts of spent bullets.
You've taken a liking to this mechanic so now you are wondering: what is the minimum possible number of bullets you need to spend (both used and thrown) to exterminate all waves.
Note that you don't throw the remaining bullets away after eradicating all monsters, and you start with a full magazine.
Input
The first line contains two integers n and k (1 ≤ n ≤ 2000; 1 ≤ k ≤ 10^9) — the number of waves and magazine size.
The next n lines contain descriptions of waves. The i-th line contains three integers l_i, r_i and a_i (1 ≤ l_i ≤ r_i ≤ 10^9; 1 ≤ a_i ≤ 10^9) — the period of time when the i-th wave happens and the number of monsters in it.
It's guaranteed that waves don't overlap (but may touch) and are given in the order they occur, i. e. r_i ≤ l_{i + 1}.
Output
If there is no way to clear all waves, print -1. Otherwise, print the minimum possible number of bullets you need to spend (both used and thrown) to clear all waves.
Examples
Input
2 3
2 3 6
3 4 3
Output
9
Input
2 5
3 7 11
10 12 15
Output
30
Input
5 42
42 42 42
42 43 42
43 44 42
44 45 42
45 45 1
Output
-1
Input
1 10
100 111 1
Output
1
Note
In the first example:
* At the moment 2, the first wave occurs and 6 monsters spawn. You kill 3 monsters and start reloading.
* At the moment 3, the second wave occurs and 3 more monsters spawn. You kill remaining 3 monsters from the first wave and start reloading.
* At the moment 4, you kill remaining 3 monsters from the second wave.
In total, you'll spend 9 bullets.
In the second example:
* At moment 3, the first wave occurs and 11 monsters spawn. You kill 5 monsters and start reloading.
* At moment 4, you kill 5 more monsters and start reloading.
* At moment 5, you kill the last monster and start reloading throwing away old magazine with 4 bullets.
* At moment 10, the second wave occurs and 15 monsters spawn. You kill 5 monsters and start reloading.
* At moment 11, you kill 5 more monsters and start reloading.
* At moment 12, you kill last 5 monsters.
In total, you'll spend 30 bullets.
Submitted Solution:
```
n,k = map(int,input().split())
l = []
tt = 0
bu = [0]*(n+1)
ans = 0
for i in range(n):
l.append(list(map(int,input().split())))
for i in range(n):
a = l[i][0]
b = l[i][1]
c = l[i][2]
if i == 0:
if i == 0:
if (b-a+1)*k >= c:
z = (k - c%k)%k
d = (b-a+1)*k - c
ans+=c
else:
tt = 1
break
else:
if a == l[i-1][1]:
if d>=k:
if z+(b-a)*k>=c:
ans+=c
if z>=c:
z = z-c
else:
z = (k-(c-z)%k)%k
d = z+(b-a)*k - c
elif (b-a+1)*k>=c:
ans+=(z+c)
z = (k-c%k)%k
d = (b-a+1)*k - c
else:
tt = 1
break
else:
if z+(b-a)*k>=c:
ans+=c
if z>=c:
z = z-c
else:
z = (k-(c-z)%k)%k
d = z+(b-a)*k - c
else:
tt = 1
break
else:
if z+(b-a)*k>=c:
ans+=c
if z>=c:
z = z-c
else:
z = (k-(c-z)%k)%k
d = z+(b-a)*k - c
elif (b-a+1)*k>=c:
ans+=(z+c)
z = (k-c%k)%k
d = (b-a+1)*k - c
else:
tt = 1
break
if tt == 1:
print(-1)
else:
print(ans)
``` | instruction | 0 | 70,058 | 2 | 140,116 |
No | output | 1 | 70,058 | 2 | 140,117 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Harry Potter has a difficult homework. Given a rectangular table, consisting of n × m cells. Each cell of the table contains the integer. Harry knows how to use two spells: the first spell change the sign of the integers in the selected row, the second — in the selected column. Harry's task is to make non-negative the sum of the numbers in each row and each column using these spells.
Alone, the boy can not cope. Help the young magician!
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100) — the number of rows and the number of columns.
Next n lines follow, each contains m integers: j-th integer in the i-th line is ai, j (|ai, j| ≤ 100), the number in the i-th row and j-th column of the table.
The rows of the table numbered from 1 to n. The columns of the table numbered from 1 to m.
Output
In the first line print the number a — the number of required applications of the first spell. Next print a space-separated integers — the row numbers, you want to apply a spell. These row numbers must be distinct!
In the second line print the number b — the number of required applications of the second spell. Next print b space-separated integers — the column numbers, you want to apply a spell. These column numbers must be distinct!
If there are several solutions are allowed to print any of them.
Examples
Input
4 1
-1
-1
-1
-1
Output
4 1 2 3 4
0
Input
2 4
-1 -1 -1 2
1 1 1 1
Output
1 1
1 4 | instruction | 0 | 70,113 | 2 | 140,226 |
Tags: constructive algorithms, greedy
Correct Solution:
```
import math
import sys
input = sys.stdin.readline
n, m = map(int, input().split())
a = [[] for _ in range(n)]
for i in range(n):
a[i] = [int(_) for _ in input().split()]
row = []
col = []
rowStat = [1] * n
colStat = [1] * m
while True:
bad = False
for i in range(n):
total = 0
for j in range(m):
total += a[i][j]
if total < 0:
bad = True
rowStat[i] *= -1
for j in range(m):
a[i][j] *= -1
for i in range(m):
total = 0
for j in range(n):
total += a[j][i]
if total < 0:
bad = True
colStat[i] *= -1
for j in range(n):
a[j][i] *= -1
if not bad:
break
for i in range(n):
if rowStat[i] == -1:
row.append(i + 1)
for i in range(m):
if colStat[i] == -1:
col.append(i + 1)
print(len(row), ' '.join(map(str, row)))
print(len(col), ' '.join(map(str, col)))
``` | output | 1 | 70,113 | 2 | 140,227 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Harry Potter has a difficult homework. Given a rectangular table, consisting of n × m cells. Each cell of the table contains the integer. Harry knows how to use two spells: the first spell change the sign of the integers in the selected row, the second — in the selected column. Harry's task is to make non-negative the sum of the numbers in each row and each column using these spells.
Alone, the boy can not cope. Help the young magician!
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100) — the number of rows and the number of columns.
Next n lines follow, each contains m integers: j-th integer in the i-th line is ai, j (|ai, j| ≤ 100), the number in the i-th row and j-th column of the table.
The rows of the table numbered from 1 to n. The columns of the table numbered from 1 to m.
Output
In the first line print the number a — the number of required applications of the first spell. Next print a space-separated integers — the row numbers, you want to apply a spell. These row numbers must be distinct!
In the second line print the number b — the number of required applications of the second spell. Next print b space-separated integers — the column numbers, you want to apply a spell. These column numbers must be distinct!
If there are several solutions are allowed to print any of them.
Examples
Input
4 1
-1
-1
-1
-1
Output
4 1 2 3 4
0
Input
2 4
-1 -1 -1 2
1 1 1 1
Output
1 1
1 4
Submitted Solution:
```
import math
import sys
input = sys.stdin.readline
n, m = map(int, input().split())
a = [[] for _ in range(n)]
for i in range(n):
a[i] = [int(_) for _ in input().split()]
row = []
col = []
while True:
bad = False
for i in range(n):
total = 0
for j in range(m):
total += a[i][j]
if total < 0:
bad = True
row.append(i + 1)
for j in range(m):
a[i][j] *= -1
for i in range(m):
total = 0
for j in range(n):
total += a[j][i]
if total < 0:
bad = True
col.append(i + 1)
for j in range(n):
a[j][i] *= -1
if not bad:
break
print(len(row), ' '.join(map(str, row)))
print(len(col), ' '.join(map(str, col)))
``` | instruction | 0 | 70,114 | 2 | 140,228 |
No | output | 1 | 70,114 | 2 | 140,229 |
Provide tags and a correct Python 2 solution for this coding contest problem.
Jon Snow now has to fight with White Walkers. He has n rangers, each of which has his own strength. Also Jon Snow has his favourite number x. Each ranger can fight with a white walker only if the strength of the white walker equals his strength. He however thinks that his rangers are weak and need to improve. Jon now thinks that if he takes the bitwise XOR of strengths of some of rangers with his favourite number x, he might get soldiers of high strength. So, he decided to do the following operation k times:
1. Arrange all the rangers in a straight line in the order of increasing strengths.
2. Take the bitwise XOR (is written as <image>) of the strength of each alternate ranger with x and update it's strength.
Suppose, Jon has 5 rangers with strengths [9, 7, 11, 15, 5] and he performs the operation 1 time with x = 2. He first arranges them in the order of their strengths, [5, 7, 9, 11, 15]. Then he does the following:
1. The strength of first ranger is updated to <image>, i.e. 7.
2. The strength of second ranger remains the same, i.e. 7.
3. The strength of third ranger is updated to <image>, i.e. 11.
4. The strength of fourth ranger remains the same, i.e. 11.
5. The strength of fifth ranger is updated to <image>, i.e. 13.
The new strengths of the 5 rangers are [7, 7, 11, 11, 13]
Now, Jon wants to know the maximum and minimum strength of the rangers after performing the above operations k times. He wants your help for this task. Can you help him?
Input
First line consists of three integers n, k, x (1 ≤ n ≤ 105, 0 ≤ k ≤ 105, 0 ≤ x ≤ 103) — number of rangers Jon has, the number of times Jon will carry out the operation and Jon's favourite number respectively.
Second line consists of n integers representing the strengths of the rangers a1, a2, ..., an (0 ≤ ai ≤ 103).
Output
Output two integers, the maximum and the minimum strength of the rangers after performing the operation k times.
Examples
Input
5 1 2
9 7 11 15 5
Output
13 7
Input
2 100000 569
605 986
Output
986 605 | instruction | 0 | 70,360 | 2 | 140,720 |
Tags: brute force, dp, implementation, sortings
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
def in_num():
return int(raw_input())
def in_arr():
return map(int,raw_input().split())
def pr_num(n):
stdout.write(str(n)+'\n')
def pr_arr(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return map(int,stdin.read().split())
range = xrange # not for python 3.0+
n,k,x=in_arr()
l=in_arr()
for i in range(min(k,8+(k%4))):
l.sort()
for i in range(n):
if i%2==0:
l[i]^=x
l.sort()
pr_arr([l[n-1],l[0]])
``` | output | 1 | 70,360 | 2 | 140,721 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jon Snow now has to fight with White Walkers. He has n rangers, each of which has his own strength. Also Jon Snow has his favourite number x. Each ranger can fight with a white walker only if the strength of the white walker equals his strength. He however thinks that his rangers are weak and need to improve. Jon now thinks that if he takes the bitwise XOR of strengths of some of rangers with his favourite number x, he might get soldiers of high strength. So, he decided to do the following operation k times:
1. Arrange all the rangers in a straight line in the order of increasing strengths.
2. Take the bitwise XOR (is written as <image>) of the strength of each alternate ranger with x and update it's strength.
Suppose, Jon has 5 rangers with strengths [9, 7, 11, 15, 5] and he performs the operation 1 time with x = 2. He first arranges them in the order of their strengths, [5, 7, 9, 11, 15]. Then he does the following:
1. The strength of first ranger is updated to <image>, i.e. 7.
2. The strength of second ranger remains the same, i.e. 7.
3. The strength of third ranger is updated to <image>, i.e. 11.
4. The strength of fourth ranger remains the same, i.e. 11.
5. The strength of fifth ranger is updated to <image>, i.e. 13.
The new strengths of the 5 rangers are [7, 7, 11, 11, 13]
Now, Jon wants to know the maximum and minimum strength of the rangers after performing the above operations k times. He wants your help for this task. Can you help him?
Input
First line consists of three integers n, k, x (1 ≤ n ≤ 105, 0 ≤ k ≤ 105, 0 ≤ x ≤ 103) — number of rangers Jon has, the number of times Jon will carry out the operation and Jon's favourite number respectively.
Second line consists of n integers representing the strengths of the rangers a1, a2, ..., an (0 ≤ ai ≤ 103).
Output
Output two integers, the maximum and the minimum strength of the rangers after performing the operation k times.
Examples
Input
5 1 2
9 7 11 15 5
Output
13 7
Input
2 100000 569
605 986
Output
986 605 | instruction | 0 | 70,361 | 2 | 140,722 |
Tags: brute force, dp, implementation, sortings
Correct Solution:
```
n, k, x = map(int, input().split())
arr = list(map(int, input().split()))
dp = [[0 for i in range(1024)], [0 for i in range(1024)]]
for i in range(n):
dp[0][arr[i]] += 1
for i in range(k):
for j in range(1024):
dp[~i&1][j] = 0
ctr = 0
for j in range(1024):
if ctr == 0:
dp[~i&1][j^x] += dp[i&1][j] + 1 >> 1
dp[~i&1][j] += dp[i&1][j] >> 1
else:
dp[~i&1][j] += dp[i&1][j] + 1 >> 1
dp[~i&1][j^x] += dp[i&1][j] >> 1
ctr ^= dp[i&1][j] & 1
minv, maxv = 1024, 0
for i in range(1024):
if dp[k&1][i] == 0:
continue
minv = min(minv, i)
maxv = max(maxv, i)
print(maxv, minv)
``` | output | 1 | 70,361 | 2 | 140,723 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jon Snow now has to fight with White Walkers. He has n rangers, each of which has his own strength. Also Jon Snow has his favourite number x. Each ranger can fight with a white walker only if the strength of the white walker equals his strength. He however thinks that his rangers are weak and need to improve. Jon now thinks that if he takes the bitwise XOR of strengths of some of rangers with his favourite number x, he might get soldiers of high strength. So, he decided to do the following operation k times:
1. Arrange all the rangers in a straight line in the order of increasing strengths.
2. Take the bitwise XOR (is written as <image>) of the strength of each alternate ranger with x and update it's strength.
Suppose, Jon has 5 rangers with strengths [9, 7, 11, 15, 5] and he performs the operation 1 time with x = 2. He first arranges them in the order of their strengths, [5, 7, 9, 11, 15]. Then he does the following:
1. The strength of first ranger is updated to <image>, i.e. 7.
2. The strength of second ranger remains the same, i.e. 7.
3. The strength of third ranger is updated to <image>, i.e. 11.
4. The strength of fourth ranger remains the same, i.e. 11.
5. The strength of fifth ranger is updated to <image>, i.e. 13.
The new strengths of the 5 rangers are [7, 7, 11, 11, 13]
Now, Jon wants to know the maximum and minimum strength of the rangers after performing the above operations k times. He wants your help for this task. Can you help him?
Input
First line consists of three integers n, k, x (1 ≤ n ≤ 105, 0 ≤ k ≤ 105, 0 ≤ x ≤ 103) — number of rangers Jon has, the number of times Jon will carry out the operation and Jon's favourite number respectively.
Second line consists of n integers representing the strengths of the rangers a1, a2, ..., an (0 ≤ ai ≤ 103).
Output
Output two integers, the maximum and the minimum strength of the rangers after performing the operation k times.
Examples
Input
5 1 2
9 7 11 15 5
Output
13 7
Input
2 100000 569
605 986
Output
986 605 | instruction | 0 | 70,362 | 2 | 140,724 |
Tags: brute force, dp, implementation, sortings
Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Tue Aug 25 02:16:02 2020
@author: Dark Soul
"""
[n,k,x]=list(map(int, input().split()))
arr=list(map(int, input().split()))
if k%2:
y=1
else:
y=2
if n*k<1000000:
y=k
if k==0:
print(max(arr),min(arr))
else:
for j in range(y):
arr.sort()
for i in range(0,n,2):
arr[i]=arr[i]^x
print(max(arr),min(arr))
``` | output | 1 | 70,362 | 2 | 140,725 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jon Snow now has to fight with White Walkers. He has n rangers, each of which has his own strength. Also Jon Snow has his favourite number x. Each ranger can fight with a white walker only if the strength of the white walker equals his strength. He however thinks that his rangers are weak and need to improve. Jon now thinks that if he takes the bitwise XOR of strengths of some of rangers with his favourite number x, he might get soldiers of high strength. So, he decided to do the following operation k times:
1. Arrange all the rangers in a straight line in the order of increasing strengths.
2. Take the bitwise XOR (is written as <image>) of the strength of each alternate ranger with x and update it's strength.
Suppose, Jon has 5 rangers with strengths [9, 7, 11, 15, 5] and he performs the operation 1 time with x = 2. He first arranges them in the order of their strengths, [5, 7, 9, 11, 15]. Then he does the following:
1. The strength of first ranger is updated to <image>, i.e. 7.
2. The strength of second ranger remains the same, i.e. 7.
3. The strength of third ranger is updated to <image>, i.e. 11.
4. The strength of fourth ranger remains the same, i.e. 11.
5. The strength of fifth ranger is updated to <image>, i.e. 13.
The new strengths of the 5 rangers are [7, 7, 11, 11, 13]
Now, Jon wants to know the maximum and minimum strength of the rangers after performing the above operations k times. He wants your help for this task. Can you help him?
Input
First line consists of three integers n, k, x (1 ≤ n ≤ 105, 0 ≤ k ≤ 105, 0 ≤ x ≤ 103) — number of rangers Jon has, the number of times Jon will carry out the operation and Jon's favourite number respectively.
Second line consists of n integers representing the strengths of the rangers a1, a2, ..., an (0 ≤ ai ≤ 103).
Output
Output two integers, the maximum and the minimum strength of the rangers after performing the operation k times.
Examples
Input
5 1 2
9 7 11 15 5
Output
13 7
Input
2 100000 569
605 986
Output
986 605 | instruction | 0 | 70,363 | 2 | 140,726 |
Tags: brute force, dp, implementation, sortings
Correct Solution:
```
n,k,x=map(int,input().split())
r=list(map(int,input().split()))
k%=64
while k:
k-=1
r.sort()
for i in range(0,n,2):#alternate
r[i]^=x
print(max(r),min(r))
``` | output | 1 | 70,363 | 2 | 140,727 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jon Snow now has to fight with White Walkers. He has n rangers, each of which has his own strength. Also Jon Snow has his favourite number x. Each ranger can fight with a white walker only if the strength of the white walker equals his strength. He however thinks that his rangers are weak and need to improve. Jon now thinks that if he takes the bitwise XOR of strengths of some of rangers with his favourite number x, he might get soldiers of high strength. So, he decided to do the following operation k times:
1. Arrange all the rangers in a straight line in the order of increasing strengths.
2. Take the bitwise XOR (is written as <image>) of the strength of each alternate ranger with x and update it's strength.
Suppose, Jon has 5 rangers with strengths [9, 7, 11, 15, 5] and he performs the operation 1 time with x = 2. He first arranges them in the order of their strengths, [5, 7, 9, 11, 15]. Then he does the following:
1. The strength of first ranger is updated to <image>, i.e. 7.
2. The strength of second ranger remains the same, i.e. 7.
3. The strength of third ranger is updated to <image>, i.e. 11.
4. The strength of fourth ranger remains the same, i.e. 11.
5. The strength of fifth ranger is updated to <image>, i.e. 13.
The new strengths of the 5 rangers are [7, 7, 11, 11, 13]
Now, Jon wants to know the maximum and minimum strength of the rangers after performing the above operations k times. He wants your help for this task. Can you help him?
Input
First line consists of three integers n, k, x (1 ≤ n ≤ 105, 0 ≤ k ≤ 105, 0 ≤ x ≤ 103) — number of rangers Jon has, the number of times Jon will carry out the operation and Jon's favourite number respectively.
Second line consists of n integers representing the strengths of the rangers a1, a2, ..., an (0 ≤ ai ≤ 103).
Output
Output two integers, the maximum and the minimum strength of the rangers after performing the operation k times.
Examples
Input
5 1 2
9 7 11 15 5
Output
13 7
Input
2 100000 569
605 986
Output
986 605 | instruction | 0 | 70,364 | 2 | 140,728 |
Tags: brute force, dp, implementation, sortings
Correct Solution:
```
N, K, X = map( int, input().split() )
A = list( map( int, input().split() ) )
freq = [ [ 0 for i in range( 1024 ) ], [ 0 for i in range( 1024 ) ] ]
for i in range( N ):
freq[ 0 ][ A[ i ] ] += 1
for i in range( K ):
for j in range( 1024 ):
freq[ ~ i & 1 ][ j ] = 0
ctr = 0
for j in range( 1024 ):
if ctr == 0:
freq[ ~ i & 1 ][ j ^ X ] += freq[ i & 1 ][ j ] + 1 >> 1
freq[ ~ i & 1 ][ j ] += freq[ i & 1 ][ j ] >> 1
else:
freq[ ~ i & 1 ][ j ] += freq[ i & 1 ][ j ] + 1 >> 1
freq[ ~ i & 1 ][ j ^ X ] += freq[ i & 1 ][ j ] >> 1
ctr ^= freq[ i & 1 ][ j ] & 1
minv, maxv = 1023, 0
for i in range( 1024 ):
if freq[ K & 1 ][ i ] == 0: continue
minv = min( minv, i )
maxv = max( maxv, i )
print( maxv, minv )
``` | output | 1 | 70,364 | 2 | 140,729 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jon Snow now has to fight with White Walkers. He has n rangers, each of which has his own strength. Also Jon Snow has his favourite number x. Each ranger can fight with a white walker only if the strength of the white walker equals his strength. He however thinks that his rangers are weak and need to improve. Jon now thinks that if he takes the bitwise XOR of strengths of some of rangers with his favourite number x, he might get soldiers of high strength. So, he decided to do the following operation k times:
1. Arrange all the rangers in a straight line in the order of increasing strengths.
2. Take the bitwise XOR (is written as <image>) of the strength of each alternate ranger with x and update it's strength.
Suppose, Jon has 5 rangers with strengths [9, 7, 11, 15, 5] and he performs the operation 1 time with x = 2. He first arranges them in the order of their strengths, [5, 7, 9, 11, 15]. Then he does the following:
1. The strength of first ranger is updated to <image>, i.e. 7.
2. The strength of second ranger remains the same, i.e. 7.
3. The strength of third ranger is updated to <image>, i.e. 11.
4. The strength of fourth ranger remains the same, i.e. 11.
5. The strength of fifth ranger is updated to <image>, i.e. 13.
The new strengths of the 5 rangers are [7, 7, 11, 11, 13]
Now, Jon wants to know the maximum and minimum strength of the rangers after performing the above operations k times. He wants your help for this task. Can you help him?
Input
First line consists of three integers n, k, x (1 ≤ n ≤ 105, 0 ≤ k ≤ 105, 0 ≤ x ≤ 103) — number of rangers Jon has, the number of times Jon will carry out the operation and Jon's favourite number respectively.
Second line consists of n integers representing the strengths of the rangers a1, a2, ..., an (0 ≤ ai ≤ 103).
Output
Output two integers, the maximum and the minimum strength of the rangers after performing the operation k times.
Examples
Input
5 1 2
9 7 11 15 5
Output
13 7
Input
2 100000 569
605 986
Output
986 605 | instruction | 0 | 70,365 | 2 | 140,730 |
Tags: brute force, dp, implementation, sortings
Correct Solution:
```
from itertools import accumulate
R = lambda: map(int, input().split())
n, k, num = R()
cnts = [0] * 1025
for x in R():
cnts[x] += 1
for _ in range(k):
nxt_cnts = [0] * 1025
acc = list(accumulate(cnts)) + [0]
for i in range(1025):
if cnts[i]:
nxt = num ^ i
if cnts[i] % 2 == 0:
nxt_cnts[nxt] += cnts[i] // 2
nxt_cnts[i] += cnts[i] // 2
elif acc[i - 1] % 2 == 0:
nxt_cnts[nxt] += cnts[i] // 2 + 1
nxt_cnts[i] += cnts[i] // 2
else:
nxt_cnts[nxt] += cnts[i] // 2
nxt_cnts[i] += cnts[i] // 2 + 1
cnts = nxt_cnts
print(max(i for i, x in enumerate(cnts) if x), min(i for i, x in enumerate(cnts) if x), sep=' ')
``` | output | 1 | 70,365 | 2 | 140,731 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jon Snow now has to fight with White Walkers. He has n rangers, each of which has his own strength. Also Jon Snow has his favourite number x. Each ranger can fight with a white walker only if the strength of the white walker equals his strength. He however thinks that his rangers are weak and need to improve. Jon now thinks that if he takes the bitwise XOR of strengths of some of rangers with his favourite number x, he might get soldiers of high strength. So, he decided to do the following operation k times:
1. Arrange all the rangers in a straight line in the order of increasing strengths.
2. Take the bitwise XOR (is written as <image>) of the strength of each alternate ranger with x and update it's strength.
Suppose, Jon has 5 rangers with strengths [9, 7, 11, 15, 5] and he performs the operation 1 time with x = 2. He first arranges them in the order of their strengths, [5, 7, 9, 11, 15]. Then he does the following:
1. The strength of first ranger is updated to <image>, i.e. 7.
2. The strength of second ranger remains the same, i.e. 7.
3. The strength of third ranger is updated to <image>, i.e. 11.
4. The strength of fourth ranger remains the same, i.e. 11.
5. The strength of fifth ranger is updated to <image>, i.e. 13.
The new strengths of the 5 rangers are [7, 7, 11, 11, 13]
Now, Jon wants to know the maximum and minimum strength of the rangers after performing the above operations k times. He wants your help for this task. Can you help him?
Input
First line consists of three integers n, k, x (1 ≤ n ≤ 105, 0 ≤ k ≤ 105, 0 ≤ x ≤ 103) — number of rangers Jon has, the number of times Jon will carry out the operation and Jon's favourite number respectively.
Second line consists of n integers representing the strengths of the rangers a1, a2, ..., an (0 ≤ ai ≤ 103).
Output
Output two integers, the maximum and the minimum strength of the rangers after performing the operation k times.
Examples
Input
5 1 2
9 7 11 15 5
Output
13 7
Input
2 100000 569
605 986
Output
986 605 | instruction | 0 | 70,366 | 2 | 140,732 |
Tags: brute force, dp, implementation, sortings
Correct Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
import sys
from io import BytesIO, IOBase
def main():
n, k, x = map(int, input().split())
a = list(map(int, input().split()))
ma = max(max(a),max(i^x for i in a))
b = [0] * (ma + 1)
for i in a:
b[i] += 1
for _ in range(k):
c, su = [0] * (ma + 1), 1
for i in range(ma + 1):
if b[i]:
z = b[i] // 2 + ((b[i] & 1) &(su & 1))
c[i ^ x] += z
c[i] += b[i] - z
su += b[i]
b = c
maa,mi=0,ma
for i in range(ma, -1, -1):
if b[i]:
maa,mi=max(maa,i),min(mi,i)
print(maa,mi)
# 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")
if __name__ == "__main__":
main()
``` | output | 1 | 70,366 | 2 | 140,733 |
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