message stringlengths 2 48.6k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 318 108k | cluster float64 8 8 | __index_level_0__ int64 636 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Bertown is a city with n buildings in a straight line.
The city's security service discovered that some buildings were mined. A map was compiled, which is a string of length n, where the i-th character is "1" if there is a mine under the building number i and "0" otherwise.
Bertown's best sapper knows how to activate mines so that the buildings above them are not damaged. When a mine under the building numbered x is activated, it explodes and activates two adjacent mines under the buildings numbered x-1 and x+1 (if there were no mines under the building, then nothing happens). Thus, it is enough to activate any one mine on a continuous segment of mines to activate all the mines of this segment. For manual activation of one mine, the sapper takes a coins. He can repeat this operation as many times as you want.
Also, a sapper can place a mine under a building if it wasn't there. For such an operation, he takes b coins. He can also repeat this operation as many times as you want.
The sapper can carry out operations in any order.
You want to blow up all the mines in the city to make it safe. Find the minimum number of coins that the sapper will have to pay so that after his actions there are no mines left in the city.
Input
The first line contains one positive integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then t test cases follow.
Each test case begins with a line containing two integers a and b (1 ≤ a, b ≤ 1000) — the cost of activating and placing one mine, respectively.
The next line contains a map of mines in the city — a string consisting of zeros and ones.
The sum of the string lengths for all test cases does not exceed 10^5.
Output
For each test case, output one integer — the minimum number of coins that the sapper will have to pay.
Example
Input
2
1 1
01000010
5 1
01101110
Output
2
6
Note
In the second test case, if we place a mine under the fourth building and then activate it, then all mines on the field are activated. The cost of such operations is six, b=1 coin for placing a mine and a=5 coins for activating. | instruction | 0 | 100,514 | 8 | 201,028 |
Tags: dp, greedy, math, sortings
Correct Solution:
```
# -*- coding: utf-8 -*-
import sys
from collections import deque, defaultdict, namedtuple
import heapq
from math import sqrt, factorial, gcd, ceil, atan, pi
from itertools import permutations
# def input(): return sys.stdin.readline().strip()
# def input(): return sys.stdin.buffer.readline()[:-1] # warning bytes
# def input(): return sys.stdin.buffer.readline().strip() # warning bytes
def input(): return sys.stdin.buffer.readline().decode('utf-8').strip()
import string
import operator
import random
# string.ascii_lowercase
from bisect import bisect_left, bisect_right
from functools import lru_cache, reduce
MOD = int(1e9)+7
INF = float('inf')
# sys.setrecursionlimit(MOD)
def solve():
a, b = [int(x) for x in input().split()]
s = [int(x) for x in input()]
d = deque(s)
while d and d[0] == 0:
d.popleft()
while d and d[-1] == 0:
d.pop()
if not d:
print(0)
return
s = ''.join([str(x) for x in d])
seg = [len(x) for x in s.split('1') if x]
ans = a * (len(seg) + 1)
for s in seg:
if s * b < a:
ans -= a
ans += s * b
print(ans)
T = 1
T = int(input())
for case in range(1,T+1):
ans = solve()
"""
1 1 1 1 2 2 2 3 3 5 6 6 8 8 11
2 2 2 1 2 2 1
1 1 1 1 1 2 1
"""
``` | output | 1 | 100,514 | 8 | 201,029 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bertown is a city with n buildings in a straight line.
The city's security service discovered that some buildings were mined. A map was compiled, which is a string of length n, where the i-th character is "1" if there is a mine under the building number i and "0" otherwise.
Bertown's best sapper knows how to activate mines so that the buildings above them are not damaged. When a mine under the building numbered x is activated, it explodes and activates two adjacent mines under the buildings numbered x-1 and x+1 (if there were no mines under the building, then nothing happens). Thus, it is enough to activate any one mine on a continuous segment of mines to activate all the mines of this segment. For manual activation of one mine, the sapper takes a coins. He can repeat this operation as many times as you want.
Also, a sapper can place a mine under a building if it wasn't there. For such an operation, he takes b coins. He can also repeat this operation as many times as you want.
The sapper can carry out operations in any order.
You want to blow up all the mines in the city to make it safe. Find the minimum number of coins that the sapper will have to pay so that after his actions there are no mines left in the city.
Input
The first line contains one positive integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then t test cases follow.
Each test case begins with a line containing two integers a and b (1 ≤ a, b ≤ 1000) — the cost of activating and placing one mine, respectively.
The next line contains a map of mines in the city — a string consisting of zeros and ones.
The sum of the string lengths for all test cases does not exceed 10^5.
Output
For each test case, output one integer — the minimum number of coins that the sapper will have to pay.
Example
Input
2
1 1
01000010
5 1
01101110
Output
2
6
Note
In the second test case, if we place a mine under the fourth building and then activate it, then all mines on the field are activated. The cost of such operations is six, b=1 coin for placing a mine and a=5 coins for activating. | instruction | 0 | 100,515 | 8 | 201,030 |
Tags: dp, greedy, math, sortings
Correct Solution:
```
from sys import stdin
input = stdin.readline
t=int(input())
for i in range(t):
a,b=map(int,input().split())
s=input()
n=len(s)-1
i=0
arr=[]
while i<n:
if s[i]=="0":
i=i+1
continue
j=i
while j<n:
if s[j]=="0":
break
j=j+1
arr.append((i,j-1))
i=j
l=len(arr)
res=0
i=0
while i<l:
ans=a
j=i+1
while j<l:
if b*(arr[j][0]-arr[j-1][1]-1)>a:
break
ans=ans+b*(arr[j][0]-arr[j-1][1]-1)
j=j+1
i=j
res=res+ans
print(res)
``` | output | 1 | 100,515 | 8 | 201,031 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bertown is a city with n buildings in a straight line.
The city's security service discovered that some buildings were mined. A map was compiled, which is a string of length n, where the i-th character is "1" if there is a mine under the building number i and "0" otherwise.
Bertown's best sapper knows how to activate mines so that the buildings above them are not damaged. When a mine under the building numbered x is activated, it explodes and activates two adjacent mines under the buildings numbered x-1 and x+1 (if there were no mines under the building, then nothing happens). Thus, it is enough to activate any one mine on a continuous segment of mines to activate all the mines of this segment. For manual activation of one mine, the sapper takes a coins. He can repeat this operation as many times as you want.
Also, a sapper can place a mine under a building if it wasn't there. For such an operation, he takes b coins. He can also repeat this operation as many times as you want.
The sapper can carry out operations in any order.
You want to blow up all the mines in the city to make it safe. Find the minimum number of coins that the sapper will have to pay so that after his actions there are no mines left in the city.
Input
The first line contains one positive integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then t test cases follow.
Each test case begins with a line containing two integers a and b (1 ≤ a, b ≤ 1000) — the cost of activating and placing one mine, respectively.
The next line contains a map of mines in the city — a string consisting of zeros and ones.
The sum of the string lengths for all test cases does not exceed 10^5.
Output
For each test case, output one integer — the minimum number of coins that the sapper will have to pay.
Example
Input
2
1 1
01000010
5 1
01101110
Output
2
6
Note
In the second test case, if we place a mine under the fourth building and then activate it, then all mines on the field are activated. The cost of such operations is six, b=1 coin for placing a mine and a=5 coins for activating. | instruction | 0 | 100,516 | 8 | 201,032 |
Tags: dp, greedy, math, sortings
Correct Solution:
```
t = int(input().strip())
for case in range(t):
a, b = map(int, input().split(' '))
m_line = input().strip()
if len(m_line) == 0:
print(0)
continue
onepart_cnt = 0
pre_m = ''
for m in m_line:
if pre_m == '1' and m == '0':
onepart_cnt = onepart_cnt + 1
pre_m = m
if pre_m == '1':
onepart_cnt = onepart_cnt + 1
min_c = onepart_cnt * a
pre_zero_cnt = 1
flag = False
pre_m = ''
for m in m_line:
if pre_m == '1' and m == '0':
pre_zero_cnt = 0
flag = True
if flag is True and m == '0':
pre_zero_cnt = pre_zero_cnt + 1
if flag is True and m == '1':
flag = False
tmp_c = pre_zero_cnt * b - a
if tmp_c < 0:
min_c = min_c + tmp_c
pre_m = m
print(min_c)
``` | output | 1 | 100,516 | 8 | 201,033 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bertown is a city with n buildings in a straight line.
The city's security service discovered that some buildings were mined. A map was compiled, which is a string of length n, where the i-th character is "1" if there is a mine under the building number i and "0" otherwise.
Bertown's best sapper knows how to activate mines so that the buildings above them are not damaged. When a mine under the building numbered x is activated, it explodes and activates two adjacent mines under the buildings numbered x-1 and x+1 (if there were no mines under the building, then nothing happens). Thus, it is enough to activate any one mine on a continuous segment of mines to activate all the mines of this segment. For manual activation of one mine, the sapper takes a coins. He can repeat this operation as many times as you want.
Also, a sapper can place a mine under a building if it wasn't there. For such an operation, he takes b coins. He can also repeat this operation as many times as you want.
The sapper can carry out operations in any order.
You want to blow up all the mines in the city to make it safe. Find the minimum number of coins that the sapper will have to pay so that after his actions there are no mines left in the city.
Input
The first line contains one positive integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then t test cases follow.
Each test case begins with a line containing two integers a and b (1 ≤ a, b ≤ 1000) — the cost of activating and placing one mine, respectively.
The next line contains a map of mines in the city — a string consisting of zeros and ones.
The sum of the string lengths for all test cases does not exceed 10^5.
Output
For each test case, output one integer — the minimum number of coins that the sapper will have to pay.
Example
Input
2
1 1
01000010
5 1
01101110
Output
2
6
Note
In the second test case, if we place a mine under the fourth building and then activate it, then all mines on the field are activated. The cost of such operations is six, b=1 coin for placing a mine and a=5 coins for activating. | instruction | 0 | 100,517 | 8 | 201,034 |
Tags: dp, greedy, math, sortings
Correct Solution:
```
for _ in range(int(input())):
a,b=map(int,input().split())
c,d=-100000,a
for i in input():
if i=='1':
if c>0: d+=min(a,c*b)
c=0
else: c+=1
print(0 if c<0 else d)
``` | output | 1 | 100,517 | 8 | 201,035 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bertown is a city with n buildings in a straight line.
The city's security service discovered that some buildings were mined. A map was compiled, which is a string of length n, where the i-th character is "1" if there is a mine under the building number i and "0" otherwise.
Bertown's best sapper knows how to activate mines so that the buildings above them are not damaged. When a mine under the building numbered x is activated, it explodes and activates two adjacent mines under the buildings numbered x-1 and x+1 (if there were no mines under the building, then nothing happens). Thus, it is enough to activate any one mine on a continuous segment of mines to activate all the mines of this segment. For manual activation of one mine, the sapper takes a coins. He can repeat this operation as many times as you want.
Also, a sapper can place a mine under a building if it wasn't there. For such an operation, he takes b coins. He can also repeat this operation as many times as you want.
The sapper can carry out operations in any order.
You want to blow up all the mines in the city to make it safe. Find the minimum number of coins that the sapper will have to pay so that after his actions there are no mines left in the city.
Input
The first line contains one positive integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then t test cases follow.
Each test case begins with a line containing two integers a and b (1 ≤ a, b ≤ 1000) — the cost of activating and placing one mine, respectively.
The next line contains a map of mines in the city — a string consisting of zeros and ones.
The sum of the string lengths for all test cases does not exceed 10^5.
Output
For each test case, output one integer — the minimum number of coins that the sapper will have to pay.
Example
Input
2
1 1
01000010
5 1
01101110
Output
2
6
Note
In the second test case, if we place a mine under the fourth building and then activate it, then all mines on the field are activated. The cost of such operations is six, b=1 coin for placing a mine and a=5 coins for activating.
Submitted Solution:
```
t = int(input())
for _ in range(t):
a,b = map(int,input().split())
s = input()
res = []
i = 0
n = len(s)
ans = 0
while(i<n):
if s[i] == '1':
temp = [i]
while(i < n and s[i] == '1'):
i += 1
temp.append(i-1)
if len(res):
if (b * (temp[0]-res[-1][1] -1) + a) <= 2*a:
ans += (b * (temp[0]-res[-1][-1] -1))
res[-1][-1] = temp[1]
else:
res.append(temp)
else:
res.append(temp)
else:
i += 1
print(len(res) * a + ans)
``` | instruction | 0 | 100,518 | 8 | 201,036 |
Yes | output | 1 | 100,518 | 8 | 201,037 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bertown is a city with n buildings in a straight line.
The city's security service discovered that some buildings were mined. A map was compiled, which is a string of length n, where the i-th character is "1" if there is a mine under the building number i and "0" otherwise.
Bertown's best sapper knows how to activate mines so that the buildings above them are not damaged. When a mine under the building numbered x is activated, it explodes and activates two adjacent mines under the buildings numbered x-1 and x+1 (if there were no mines under the building, then nothing happens). Thus, it is enough to activate any one mine on a continuous segment of mines to activate all the mines of this segment. For manual activation of one mine, the sapper takes a coins. He can repeat this operation as many times as you want.
Also, a sapper can place a mine under a building if it wasn't there. For such an operation, he takes b coins. He can also repeat this operation as many times as you want.
The sapper can carry out operations in any order.
You want to blow up all the mines in the city to make it safe. Find the minimum number of coins that the sapper will have to pay so that after his actions there are no mines left in the city.
Input
The first line contains one positive integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then t test cases follow.
Each test case begins with a line containing two integers a and b (1 ≤ a, b ≤ 1000) — the cost of activating and placing one mine, respectively.
The next line contains a map of mines in the city — a string consisting of zeros and ones.
The sum of the string lengths for all test cases does not exceed 10^5.
Output
For each test case, output one integer — the minimum number of coins that the sapper will have to pay.
Example
Input
2
1 1
01000010
5 1
01101110
Output
2
6
Note
In the second test case, if we place a mine under the fourth building and then activate it, then all mines on the field are activated. The cost of such operations is six, b=1 coin for placing a mine and a=5 coins for activating.
Submitted Solution:
```
noOfTestCases = int(input())
for testCase in range(noOfTestCases):
activationCost, placeMine = [int(i) for i in input().split()]
buildings = input()
noOfBuildings = len(buildings)
totalCost = 0
atIndex = 0
previousMine = None
while atIndex<noOfBuildings:
if buildings[atIndex] == '0':
pass
else:
if previousMine is None:
pass
elif (atIndex - previousMine - 1) * placeMine < activationCost:
totalCost += (atIndex - previousMine - 1) * placeMine
else:
totalCost += activationCost
previousMine = atIndex
atIndex += 1
if previousMine is not None:
totalCost += activationCost
print(totalCost)
``` | instruction | 0 | 100,519 | 8 | 201,038 |
Yes | output | 1 | 100,519 | 8 | 201,039 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bertown is a city with n buildings in a straight line.
The city's security service discovered that some buildings were mined. A map was compiled, which is a string of length n, where the i-th character is "1" if there is a mine under the building number i and "0" otherwise.
Bertown's best sapper knows how to activate mines so that the buildings above them are not damaged. When a mine under the building numbered x is activated, it explodes and activates two adjacent mines under the buildings numbered x-1 and x+1 (if there were no mines under the building, then nothing happens). Thus, it is enough to activate any one mine on a continuous segment of mines to activate all the mines of this segment. For manual activation of one mine, the sapper takes a coins. He can repeat this operation as many times as you want.
Also, a sapper can place a mine under a building if it wasn't there. For such an operation, he takes b coins. He can also repeat this operation as many times as you want.
The sapper can carry out operations in any order.
You want to blow up all the mines in the city to make it safe. Find the minimum number of coins that the sapper will have to pay so that after his actions there are no mines left in the city.
Input
The first line contains one positive integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then t test cases follow.
Each test case begins with a line containing two integers a and b (1 ≤ a, b ≤ 1000) — the cost of activating and placing one mine, respectively.
The next line contains a map of mines in the city — a string consisting of zeros and ones.
The sum of the string lengths for all test cases does not exceed 10^5.
Output
For each test case, output one integer — the minimum number of coins that the sapper will have to pay.
Example
Input
2
1 1
01000010
5 1
01101110
Output
2
6
Note
In the second test case, if we place a mine under the fourth building and then activate it, then all mines on the field are activated. The cost of such operations is six, b=1 coin for placing a mine and a=5 coins for activating.
Submitted Solution:
```
from sys import stdin, stdout
from math import factorial as f
from random import randint, shuffle
input = stdin.readline
for _ in range(int(input())):
a, b = map(int, input().split())
s = input()[:-1]
s = s.lstrip('0')
if not s:
print(0)
continue
ans = a
cnt = 0
for i in s:
if i == '0':
cnt += 1
elif cnt:
ans += min(a, cnt * b)
cnt = 0
print(ans)
``` | instruction | 0 | 100,520 | 8 | 201,040 |
Yes | output | 1 | 100,520 | 8 | 201,041 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bertown is a city with n buildings in a straight line.
The city's security service discovered that some buildings were mined. A map was compiled, which is a string of length n, where the i-th character is "1" if there is a mine under the building number i and "0" otherwise.
Bertown's best sapper knows how to activate mines so that the buildings above them are not damaged. When a mine under the building numbered x is activated, it explodes and activates two adjacent mines under the buildings numbered x-1 and x+1 (if there were no mines under the building, then nothing happens). Thus, it is enough to activate any one mine on a continuous segment of mines to activate all the mines of this segment. For manual activation of one mine, the sapper takes a coins. He can repeat this operation as many times as you want.
Also, a sapper can place a mine under a building if it wasn't there. For such an operation, he takes b coins. He can also repeat this operation as many times as you want.
The sapper can carry out operations in any order.
You want to blow up all the mines in the city to make it safe. Find the minimum number of coins that the sapper will have to pay so that after his actions there are no mines left in the city.
Input
The first line contains one positive integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then t test cases follow.
Each test case begins with a line containing two integers a and b (1 ≤ a, b ≤ 1000) — the cost of activating and placing one mine, respectively.
The next line contains a map of mines in the city — a string consisting of zeros and ones.
The sum of the string lengths for all test cases does not exceed 10^5.
Output
For each test case, output one integer — the minimum number of coins that the sapper will have to pay.
Example
Input
2
1 1
01000010
5 1
01101110
Output
2
6
Note
In the second test case, if we place a mine under the fourth building and then activate it, then all mines on the field are activated. The cost of such operations is six, b=1 coin for placing a mine and a=5 coins for activating.
Submitted Solution:
```
# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------- fast io --------------------
from math import ceil,floor,sqrt
from collections import Counter,defaultdict,deque
def int1():
return int(input())
def map1():
return map(int,input().split())
def list1():
return list(map(int,input().split()))
mod=pow(10,9)+7
def solve():
a,b=map1()
l1=list(input().strip("0"))
n=len(l1)
l2=[]
c,flag,sum1=0,0,0
for i in range(n):
if(l1[i]=="0"):
c=c+1
else:
if(flag==0):
v=c*b+a
v1=2*a
flag=1
else:
v=c*b
v1=a
sum1+=min(v,v1)
#print(sum1)
c=0
print(sum1)
for _ in range(int(input())):
solve()
``` | instruction | 0 | 100,521 | 8 | 201,042 |
Yes | output | 1 | 100,521 | 8 | 201,043 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bertown is a city with n buildings in a straight line.
The city's security service discovered that some buildings were mined. A map was compiled, which is a string of length n, where the i-th character is "1" if there is a mine under the building number i and "0" otherwise.
Bertown's best sapper knows how to activate mines so that the buildings above them are not damaged. When a mine under the building numbered x is activated, it explodes and activates two adjacent mines under the buildings numbered x-1 and x+1 (if there were no mines under the building, then nothing happens). Thus, it is enough to activate any one mine on a continuous segment of mines to activate all the mines of this segment. For manual activation of one mine, the sapper takes a coins. He can repeat this operation as many times as you want.
Also, a sapper can place a mine under a building if it wasn't there. For such an operation, he takes b coins. He can also repeat this operation as many times as you want.
The sapper can carry out operations in any order.
You want to blow up all the mines in the city to make it safe. Find the minimum number of coins that the sapper will have to pay so that after his actions there are no mines left in the city.
Input
The first line contains one positive integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then t test cases follow.
Each test case begins with a line containing two integers a and b (1 ≤ a, b ≤ 1000) — the cost of activating and placing one mine, respectively.
The next line contains a map of mines in the city — a string consisting of zeros and ones.
The sum of the string lengths for all test cases does not exceed 10^5.
Output
For each test case, output one integer — the minimum number of coins that the sapper will have to pay.
Example
Input
2
1 1
01000010
5 1
01101110
Output
2
6
Note
In the second test case, if we place a mine under the fourth building and then activate it, then all mines on the field are activated. The cost of such operations is six, b=1 coin for placing a mine and a=5 coins for activating.
Submitted Solution:
```
''' https://codeforces.com/problemset/problem/1443/B '''
from typing import List
def solv(city: List[int], a: int, b: int) -> int:
n_elem = len(city)
if n_elem == 0:
return 0
expl_c = [0] * n_elem
mine_c = [0] * n_elem
if city[0] == 0:
mine_c[0] = b
else:
expl_c[0] = a
for i in range(1, n_elem):
if city[i] == 0:
expl_c[i] = min(expl_c[i-1], mine_c[i-1] + a)
mine_c[i] = mine_c[i-1] + b
else:
expl_c[i] = min(expl_c[i-1] + a, mine_c[i-1] + a)
mine_c[i] = mine_c[i-1]
return expl_c[-1]
TESTS = int(input())
try:
for _ in range(TESTS):
ab = input()
a, b = [int(v) for v in ab.split()]
strarr = input()
arr = [int(v) for v in strarr.strip('0')]
print(solv(arr, a, b))
except Exception as e:
print(strarr)
'''
strarr = "01101110"
arr = [int(v) for v in strarr.strip('0')]
a, b = 5, 1
print(solv(arr, a, b))
'''
``` | instruction | 0 | 100,522 | 8 | 201,044 |
No | output | 1 | 100,522 | 8 | 201,045 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bertown is a city with n buildings in a straight line.
The city's security service discovered that some buildings were mined. A map was compiled, which is a string of length n, where the i-th character is "1" if there is a mine under the building number i and "0" otherwise.
Bertown's best sapper knows how to activate mines so that the buildings above them are not damaged. When a mine under the building numbered x is activated, it explodes and activates two adjacent mines under the buildings numbered x-1 and x+1 (if there were no mines under the building, then nothing happens). Thus, it is enough to activate any one mine on a continuous segment of mines to activate all the mines of this segment. For manual activation of one mine, the sapper takes a coins. He can repeat this operation as many times as you want.
Also, a sapper can place a mine under a building if it wasn't there. For such an operation, he takes b coins. He can also repeat this operation as many times as you want.
The sapper can carry out operations in any order.
You want to blow up all the mines in the city to make it safe. Find the minimum number of coins that the sapper will have to pay so that after his actions there are no mines left in the city.
Input
The first line contains one positive integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then t test cases follow.
Each test case begins with a line containing two integers a and b (1 ≤ a, b ≤ 1000) — the cost of activating and placing one mine, respectively.
The next line contains a map of mines in the city — a string consisting of zeros and ones.
The sum of the string lengths for all test cases does not exceed 10^5.
Output
For each test case, output one integer — the minimum number of coins that the sapper will have to pay.
Example
Input
2
1 1
01000010
5 1
01101110
Output
2
6
Note
In the second test case, if we place a mine under the fourth building and then activate it, then all mines on the field are activated. The cost of such operations is six, b=1 coin for placing a mine and a=5 coins for activating.
Submitted Solution:
```
for _ in range(int(input())):
[x,y] = list(map(int,input().split()))
n = str(input())
output = 0
gear = 0
temp = 0
flag = 0
go =0
for i in range(len(n)):
if n[i]=='1':
go = 1
break
if go == 1:
while n[0]=='0' and len(n)!=0:
n = n[1:]
while n[-1]=='0' and len(n)!=0:
n = n[:(len(n)-1)]
for i in range(len(n)):
if n[i]=='1':
if temp<=x and gear != 0:
output+=temp
temp = 0
flag = 1
if flag == 0:
flag = 1
output+=x
if n[i] == '0':
flag = 0
temp += y
if temp<=x:
gear = 1
else:
output= 0
print(output)
``` | instruction | 0 | 100,523 | 8 | 201,046 |
No | output | 1 | 100,523 | 8 | 201,047 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bertown is a city with n buildings in a straight line.
The city's security service discovered that some buildings were mined. A map was compiled, which is a string of length n, where the i-th character is "1" if there is a mine under the building number i and "0" otherwise.
Bertown's best sapper knows how to activate mines so that the buildings above them are not damaged. When a mine under the building numbered x is activated, it explodes and activates two adjacent mines under the buildings numbered x-1 and x+1 (if there were no mines under the building, then nothing happens). Thus, it is enough to activate any one mine on a continuous segment of mines to activate all the mines of this segment. For manual activation of one mine, the sapper takes a coins. He can repeat this operation as many times as you want.
Also, a sapper can place a mine under a building if it wasn't there. For such an operation, he takes b coins. He can also repeat this operation as many times as you want.
The sapper can carry out operations in any order.
You want to blow up all the mines in the city to make it safe. Find the minimum number of coins that the sapper will have to pay so that after his actions there are no mines left in the city.
Input
The first line contains one positive integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then t test cases follow.
Each test case begins with a line containing two integers a and b (1 ≤ a, b ≤ 1000) — the cost of activating and placing one mine, respectively.
The next line contains a map of mines in the city — a string consisting of zeros and ones.
The sum of the string lengths for all test cases does not exceed 10^5.
Output
For each test case, output one integer — the minimum number of coins that the sapper will have to pay.
Example
Input
2
1 1
01000010
5 1
01101110
Output
2
6
Note
In the second test case, if we place a mine under the fourth building and then activate it, then all mines on the field are activated. The cost of such operations is six, b=1 coin for placing a mine and a=5 coins for activating.
Submitted Solution:
```
for _ in range(int(input())):
a,b=map(int,input().split())
mines=input()
lent=len(mines)
if mines.count('1')==0:
print(0)
continue
for i in range(lent):
if mines[i]=='1':
start=i
break
for i in range(lent-1,-1,-1):
if mines[i]=='1':
end=i
break
mod=mines[start:end+1]
new=''
found=False
for i in range(len(mod)):
if mod[i]=='1' and not found:
new+=mod[i]
found=True
if mod[i]=='0':
new+=mod[i]
found=False
zeros=new.count('0')
if zeros==0:
print(a)
continue
ones=new.count('1')
zeros_cost=(zeros*b)+a
ones_cost=ones*a
print(min(zeros_cost,ones_cost))
``` | instruction | 0 | 100,524 | 8 | 201,048 |
No | output | 1 | 100,524 | 8 | 201,049 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bertown is a city with n buildings in a straight line.
The city's security service discovered that some buildings were mined. A map was compiled, which is a string of length n, where the i-th character is "1" if there is a mine under the building number i and "0" otherwise.
Bertown's best sapper knows how to activate mines so that the buildings above them are not damaged. When a mine under the building numbered x is activated, it explodes and activates two adjacent mines under the buildings numbered x-1 and x+1 (if there were no mines under the building, then nothing happens). Thus, it is enough to activate any one mine on a continuous segment of mines to activate all the mines of this segment. For manual activation of one mine, the sapper takes a coins. He can repeat this operation as many times as you want.
Also, a sapper can place a mine under a building if it wasn't there. For such an operation, he takes b coins. He can also repeat this operation as many times as you want.
The sapper can carry out operations in any order.
You want to blow up all the mines in the city to make it safe. Find the minimum number of coins that the sapper will have to pay so that after his actions there are no mines left in the city.
Input
The first line contains one positive integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then t test cases follow.
Each test case begins with a line containing two integers a and b (1 ≤ a, b ≤ 1000) — the cost of activating and placing one mine, respectively.
The next line contains a map of mines in the city — a string consisting of zeros and ones.
The sum of the string lengths for all test cases does not exceed 10^5.
Output
For each test case, output one integer — the minimum number of coins that the sapper will have to pay.
Example
Input
2
1 1
01000010
5 1
01101110
Output
2
6
Note
In the second test case, if we place a mine under the fourth building and then activate it, then all mines on the field are activated. The cost of such operations is six, b=1 coin for placing a mine and a=5 coins for activating.
Submitted Solution:
```
t = int(input())
for _ in range(t):
a, b = map(int, input().split())
mines = input()
l = -1
mines_s = list()
for i in range(len(mines)):
if l == -1 and mines[i] == "1":
l = i
elif l != -1 and mines[i] == "0":
mines_s.append((l, i - 1))
l = -1
if l != -1:
mines_s.append((l, len(mines) - 1))
s = 0
i = 0
while i < len(mines_s) - 1:
l1, r1 = mines_s.pop(0)
l2, r2 = mines_s.pop(0)
if b * (l2 - r1 - 1) <= a:
s += b * (l2 - r1 - 1)
mines_s = [(l1, r2)] + mines_s
else:
mines_s = [(l1, r1), (l2, r2)] + mines_s
i += 1
s += a * len(mines_s)
print(s)
``` | instruction | 0 | 100,525 | 8 | 201,050 |
No | output | 1 | 100,525 | 8 | 201,051 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a nuts and lots of boxes. The boxes have a wonderful feature: if you put x (x ≥ 0) divisors (the spacial bars that can divide a box) to it, you get a box, divided into x + 1 sections.
You are minimalist. Therefore, on the one hand, you are against dividing some box into more than k sections. On the other hand, you are against putting more than v nuts into some section of the box. What is the minimum number of boxes you have to use if you want to put all the nuts in boxes, and you have b divisors?
Please note that you need to minimize the number of used boxes, not sections. You do not have to minimize the number of used divisors.
Input
The first line contains four space-separated integers k, a, b, v (2 ≤ k ≤ 1000; 1 ≤ a, b, v ≤ 1000) — the maximum number of sections in the box, the number of nuts, the number of divisors and the capacity of each section of the box.
Output
Print a single integer — the answer to the problem.
Examples
Input
3 10 3 3
Output
2
Input
3 10 1 3
Output
3
Input
100 100 1 1000
Output
1
Note
In the first sample you can act like this:
* Put two divisors to the first box. Now the first box has three sections and we can put three nuts into each section. Overall, the first box will have nine nuts.
* Do not put any divisors into the second box. Thus, the second box has one section for the last nut.
In the end we've put all the ten nuts into boxes.
The second sample is different as we have exactly one divisor and we put it to the first box. The next two boxes will have one section each. | instruction | 0 | 100,672 | 8 | 201,344 |
Tags: greedy, math
Correct Solution:
```
def ceiling_div(x, y):
res = x // y
if (x % y != 0):
res += 1
return res
k, a, b, v = map(int, input().split())
sec = ceiling_div(a, v)
res = max(sec - b, ceiling_div(sec, k))
print(res)
``` | output | 1 | 100,672 | 8 | 201,345 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a nuts and lots of boxes. The boxes have a wonderful feature: if you put x (x ≥ 0) divisors (the spacial bars that can divide a box) to it, you get a box, divided into x + 1 sections.
You are minimalist. Therefore, on the one hand, you are against dividing some box into more than k sections. On the other hand, you are against putting more than v nuts into some section of the box. What is the minimum number of boxes you have to use if you want to put all the nuts in boxes, and you have b divisors?
Please note that you need to minimize the number of used boxes, not sections. You do not have to minimize the number of used divisors.
Input
The first line contains four space-separated integers k, a, b, v (2 ≤ k ≤ 1000; 1 ≤ a, b, v ≤ 1000) — the maximum number of sections in the box, the number of nuts, the number of divisors and the capacity of each section of the box.
Output
Print a single integer — the answer to the problem.
Examples
Input
3 10 3 3
Output
2
Input
3 10 1 3
Output
3
Input
100 100 1 1000
Output
1
Note
In the first sample you can act like this:
* Put two divisors to the first box. Now the first box has three sections and we can put three nuts into each section. Overall, the first box will have nine nuts.
* Do not put any divisors into the second box. Thus, the second box has one section for the last nut.
In the end we've put all the ten nuts into boxes.
The second sample is different as we have exactly one divisor and we put it to the first box. The next two boxes will have one section each.
Submitted Solution:
```
k, a, b, v = map(int, input().split())
val = 0
while a > 0:
d = min(b, k - 1)
a -= (d + 1) * v
b -= d
val += 1
print(val)
``` | instruction | 0 | 100,676 | 8 | 201,352 |
Yes | output | 1 | 100,676 | 8 | 201,353 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a nuts and lots of boxes. The boxes have a wonderful feature: if you put x (x ≥ 0) divisors (the spacial bars that can divide a box) to it, you get a box, divided into x + 1 sections.
You are minimalist. Therefore, on the one hand, you are against dividing some box into more than k sections. On the other hand, you are against putting more than v nuts into some section of the box. What is the minimum number of boxes you have to use if you want to put all the nuts in boxes, and you have b divisors?
Please note that you need to minimize the number of used boxes, not sections. You do not have to minimize the number of used divisors.
Input
The first line contains four space-separated integers k, a, b, v (2 ≤ k ≤ 1000; 1 ≤ a, b, v ≤ 1000) — the maximum number of sections in the box, the number of nuts, the number of divisors and the capacity of each section of the box.
Output
Print a single integer — the answer to the problem.
Examples
Input
3 10 3 3
Output
2
Input
3 10 1 3
Output
3
Input
100 100 1 1000
Output
1
Note
In the first sample you can act like this:
* Put two divisors to the first box. Now the first box has three sections and we can put three nuts into each section. Overall, the first box will have nine nuts.
* Do not put any divisors into the second box. Thus, the second box has one section for the last nut.
In the end we've put all the ten nuts into boxes.
The second sample is different as we have exactly one divisor and we put it to the first box. The next two boxes will have one section each.
Submitted Solution:
```
import math
k,a,b,v = map(int,input().split())
ans = 0
while a > 0:
l = min(k-1,b)
b-=l
a-= (l+1)*v
ans+=1
print(ans)
``` | instruction | 0 | 100,677 | 8 | 201,354 |
Yes | output | 1 | 100,677 | 8 | 201,355 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a nuts and lots of boxes. The boxes have a wonderful feature: if you put x (x ≥ 0) divisors (the spacial bars that can divide a box) to it, you get a box, divided into x + 1 sections.
You are minimalist. Therefore, on the one hand, you are against dividing some box into more than k sections. On the other hand, you are against putting more than v nuts into some section of the box. What is the minimum number of boxes you have to use if you want to put all the nuts in boxes, and you have b divisors?
Please note that you need to minimize the number of used boxes, not sections. You do not have to minimize the number of used divisors.
Input
The first line contains four space-separated integers k, a, b, v (2 ≤ k ≤ 1000; 1 ≤ a, b, v ≤ 1000) — the maximum number of sections in the box, the number of nuts, the number of divisors and the capacity of each section of the box.
Output
Print a single integer — the answer to the problem.
Examples
Input
3 10 3 3
Output
2
Input
3 10 1 3
Output
3
Input
100 100 1 1000
Output
1
Note
In the first sample you can act like this:
* Put two divisors to the first box. Now the first box has three sections and we can put three nuts into each section. Overall, the first box will have nine nuts.
* Do not put any divisors into the second box. Thus, the second box has one section for the last nut.
In the end we've put all the ten nuts into boxes.
The second sample is different as we have exactly one divisor and we put it to the first box. The next two boxes will have one section each.
Submitted Solution:
```
import math
k,a,b,v=map(int,input().split())
#3 10 3 3
if(v>a):
print(1)
else:
if b+1<k:
a=a-v*(b+1)
ans=math.ceil(a/v)
print(ans+1)
else:
c=1
a=a-(v*k)#694
b=b-k+1#4
#print("initial a=",a,"b=",b,"k=",k)
while(a>0 and b>0):
if v*(b+1)>a and b<k:
#print("in 1st if")
c+=1
a=0
break
else:
if b>k:
if(v>a):
c+=1
a=0
else:
a=a-(v*(k))#
c+=1#
b=b-(k-1)#
#print("in b>k a=",a,"b=",b,"c=",c)
if(b<=k):
a=a-v*(b+1) #
c+=1#
b=0
#print("in b<k a=",a,"b=",b,"c=",c)
#if(b==k):
if(b==0 and a>0):
c+=math.ceil(a/v)
print(c)
``` | instruction | 0 | 100,678 | 8 | 201,356 |
Yes | output | 1 | 100,678 | 8 | 201,357 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a nuts and lots of boxes. The boxes have a wonderful feature: if you put x (x ≥ 0) divisors (the spacial bars that can divide a box) to it, you get a box, divided into x + 1 sections.
You are minimalist. Therefore, on the one hand, you are against dividing some box into more than k sections. On the other hand, you are against putting more than v nuts into some section of the box. What is the minimum number of boxes you have to use if you want to put all the nuts in boxes, and you have b divisors?
Please note that you need to minimize the number of used boxes, not sections. You do not have to minimize the number of used divisors.
Input
The first line contains four space-separated integers k, a, b, v (2 ≤ k ≤ 1000; 1 ≤ a, b, v ≤ 1000) — the maximum number of sections in the box, the number of nuts, the number of divisors and the capacity of each section of the box.
Output
Print a single integer — the answer to the problem.
Examples
Input
3 10 3 3
Output
2
Input
3 10 1 3
Output
3
Input
100 100 1 1000
Output
1
Note
In the first sample you can act like this:
* Put two divisors to the first box. Now the first box has three sections and we can put three nuts into each section. Overall, the first box will have nine nuts.
* Do not put any divisors into the second box. Thus, the second box has one section for the last nut.
In the end we've put all the ten nuts into boxes.
The second sample is different as we have exactly one divisor and we put it to the first box. The next two boxes will have one section each.
Submitted Solution:
```
import math
boxes = 0
k, a,b, v = map(int, input().split())
#b += 1
if k > (b+1):
k = b + 1
numfb = b // (k - 1)
extrak = b % (k - 1)
while a > 0 and numfb > 0:
boxes += 1
numfb -= 1
a -= k * v
if extrak > 0 and a > 0:
boxes += 1
a -= (extrak + 1) * v
while a > 0:
boxes += 1
a -= v
print(boxes)
``` | instruction | 0 | 100,679 | 8 | 201,358 |
Yes | output | 1 | 100,679 | 8 | 201,359 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a nuts and lots of boxes. The boxes have a wonderful feature: if you put x (x ≥ 0) divisors (the spacial bars that can divide a box) to it, you get a box, divided into x + 1 sections.
You are minimalist. Therefore, on the one hand, you are against dividing some box into more than k sections. On the other hand, you are against putting more than v nuts into some section of the box. What is the minimum number of boxes you have to use if you want to put all the nuts in boxes, and you have b divisors?
Please note that you need to minimize the number of used boxes, not sections. You do not have to minimize the number of used divisors.
Input
The first line contains four space-separated integers k, a, b, v (2 ≤ k ≤ 1000; 1 ≤ a, b, v ≤ 1000) — the maximum number of sections in the box, the number of nuts, the number of divisors and the capacity of each section of the box.
Output
Print a single integer — the answer to the problem.
Examples
Input
3 10 3 3
Output
2
Input
3 10 1 3
Output
3
Input
100 100 1 1000
Output
1
Note
In the first sample you can act like this:
* Put two divisors to the first box. Now the first box has three sections and we can put three nuts into each section. Overall, the first box will have nine nuts.
* Do not put any divisors into the second box. Thus, the second box has one section for the last nut.
In the end we've put all the ten nuts into boxes.
The second sample is different as we have exactly one divisor and we put it to the first box. The next two boxes will have one section each.
Submitted Solution:
```
max_parts, n, divisors, nuts_in_part = map(int, input().split())
parts = (n + nuts_in_part - 1) // (nuts_in_part)
ans = parts - divisors
while ans * (max_parts - 1) < divisors:
ans += 1
print(ans)
``` | instruction | 0 | 100,680 | 8 | 201,360 |
No | output | 1 | 100,680 | 8 | 201,361 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a nuts and lots of boxes. The boxes have a wonderful feature: if you put x (x ≥ 0) divisors (the spacial bars that can divide a box) to it, you get a box, divided into x + 1 sections.
You are minimalist. Therefore, on the one hand, you are against dividing some box into more than k sections. On the other hand, you are against putting more than v nuts into some section of the box. What is the minimum number of boxes you have to use if you want to put all the nuts in boxes, and you have b divisors?
Please note that you need to minimize the number of used boxes, not sections. You do not have to minimize the number of used divisors.
Input
The first line contains four space-separated integers k, a, b, v (2 ≤ k ≤ 1000; 1 ≤ a, b, v ≤ 1000) — the maximum number of sections in the box, the number of nuts, the number of divisors and the capacity of each section of the box.
Output
Print a single integer — the answer to the problem.
Examples
Input
3 10 3 3
Output
2
Input
3 10 1 3
Output
3
Input
100 100 1 1000
Output
1
Note
In the first sample you can act like this:
* Put two divisors to the first box. Now the first box has three sections and we can put three nuts into each section. Overall, the first box will have nine nuts.
* Do not put any divisors into the second box. Thus, the second box has one section for the last nut.
In the end we've put all the ten nuts into boxes.
The second sample is different as we have exactly one divisor and we put it to the first box. The next two boxes will have one section each.
Submitted Solution:
```
def floor(x):
if (x == int(x)):
return x
else:
return int(x) + 1
a = input()
a = list(a.split(' '))
a = list(map(int,a))
ts = floor(a[1] / a[3])
tb = 0
#print(ts)
tb += a[2] // (a[0] - 1)
ts -= tb * a[0]
a[2] -= tb * (a[0] - 1)
#print(ts)
if (ts <= 0):
ts += tb * a[0]
#print(ts)
print(floor(a[1] / (a[0] * a[3])))
else:
tb += 1
ts -= a[2] + 1
if (ts <= 0):
print(int(tb))
else:
tb += ts
print(int(tb))
``` | instruction | 0 | 100,681 | 8 | 201,362 |
No | output | 1 | 100,681 | 8 | 201,363 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a nuts and lots of boxes. The boxes have a wonderful feature: if you put x (x ≥ 0) divisors (the spacial bars that can divide a box) to it, you get a box, divided into x + 1 sections.
You are minimalist. Therefore, on the one hand, you are against dividing some box into more than k sections. On the other hand, you are against putting more than v nuts into some section of the box. What is the minimum number of boxes you have to use if you want to put all the nuts in boxes, and you have b divisors?
Please note that you need to minimize the number of used boxes, not sections. You do not have to minimize the number of used divisors.
Input
The first line contains four space-separated integers k, a, b, v (2 ≤ k ≤ 1000; 1 ≤ a, b, v ≤ 1000) — the maximum number of sections in the box, the number of nuts, the number of divisors and the capacity of each section of the box.
Output
Print a single integer — the answer to the problem.
Examples
Input
3 10 3 3
Output
2
Input
3 10 1 3
Output
3
Input
100 100 1 1000
Output
1
Note
In the first sample you can act like this:
* Put two divisors to the first box. Now the first box has three sections and we can put three nuts into each section. Overall, the first box will have nine nuts.
* Do not put any divisors into the second box. Thus, the second box has one section for the last nut.
In the end we've put all the ten nuts into boxes.
The second sample is different as we have exactly one divisor and we put it to the first box. The next two boxes will have one section each.
Submitted Solution:
```
import math
k,a,b,v = map(int,input().split())
total = (b+1)*v
if a >= total:
g = math.ceil((b+1)/k) + math.ceil((a-total)/v)
else:
re = math.ceil(a/v)
g = math.ceil(re/k)
print(g)
``` | instruction | 0 | 100,682 | 8 | 201,364 |
No | output | 1 | 100,682 | 8 | 201,365 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a nuts and lots of boxes. The boxes have a wonderful feature: if you put x (x ≥ 0) divisors (the spacial bars that can divide a box) to it, you get a box, divided into x + 1 sections.
You are minimalist. Therefore, on the one hand, you are against dividing some box into more than k sections. On the other hand, you are against putting more than v nuts into some section of the box. What is the minimum number of boxes you have to use if you want to put all the nuts in boxes, and you have b divisors?
Please note that you need to minimize the number of used boxes, not sections. You do not have to minimize the number of used divisors.
Input
The first line contains four space-separated integers k, a, b, v (2 ≤ k ≤ 1000; 1 ≤ a, b, v ≤ 1000) — the maximum number of sections in the box, the number of nuts, the number of divisors and the capacity of each section of the box.
Output
Print a single integer — the answer to the problem.
Examples
Input
3 10 3 3
Output
2
Input
3 10 1 3
Output
3
Input
100 100 1 1000
Output
1
Note
In the first sample you can act like this:
* Put two divisors to the first box. Now the first box has three sections and we can put three nuts into each section. Overall, the first box will have nine nuts.
* Do not put any divisors into the second box. Thus, the second box has one section for the last nut.
In the end we've put all the ten nuts into boxes.
The second sample is different as we have exactly one divisor and we put it to the first box. The next two boxes will have one section each.
Submitted Solution:
```
import math
boxes = 0
k, a,b, v = map(int, input().split())
#b += 1
if k > (b+1):
k = b + 1
numfb = b // (k - 1)
extrak = b % (k - 1)
a -= (numfb * k * v)
boxes += numfb
if extrak > 0 and a > 0:
boxes += 1
a -= (extrak + 1) * v
while a > 0:
boxes += 1
a -= v
print(boxes)
``` | instruction | 0 | 100,683 | 8 | 201,366 |
No | output | 1 | 100,683 | 8 | 201,367 |
Provide a correct Python 3 solution for this coding contest problem.
There are N slimes lining up in a row. Initially, the i-th slime from the left has a size of a_i.
Taro is trying to combine all the slimes into a larger slime. He will perform the following operation repeatedly until there is only one slime:
* Choose two adjacent slimes, and combine them into a new slime. The new slime has a size of x + y, where x and y are the sizes of the slimes before combining them. Here, a cost of x + y is incurred. The positional relationship of the slimes does not change while combining slimes.
Find the minimum possible total cost incurred.
Constraints
* All values in input are integers.
* 2 \leq N \leq 400
* 1 \leq a_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 \ldots a_N
Output
Print the minimum possible total cost incurred.
Examples
Input
4
10 20 30 40
Output
190
Input
5
10 10 10 10 10
Output
120
Input
3
1000000000 1000000000 1000000000
Output
5000000000
Input
6
7 6 8 6 1 1
Output
68 | instruction | 0 | 100,984 | 8 | 201,968 |
"Correct Solution:
```
n = int(input())
*A, = map(int, input().split())
S = [0]
for i in range(n):
S.append(S[-1]+A[i])
DP = [[None for j in range(n+1)] for i in range(n)]
def dp(i, j):
if j <= i+1:
return 0
if DP[i][j] != None:
return DP[i][j]
DP[i][j] = S[j]-S[i]+min([dp(i, k)+dp(k, j) for k in range(i+1, j)])
return DP[i][j]
print(dp(0, n))
``` | output | 1 | 100,984 | 8 | 201,969 |
Provide a correct Python 3 solution for this coding contest problem.
There are N slimes lining up in a row. Initially, the i-th slime from the left has a size of a_i.
Taro is trying to combine all the slimes into a larger slime. He will perform the following operation repeatedly until there is only one slime:
* Choose two adjacent slimes, and combine them into a new slime. The new slime has a size of x + y, where x and y are the sizes of the slimes before combining them. Here, a cost of x + y is incurred. The positional relationship of the slimes does not change while combining slimes.
Find the minimum possible total cost incurred.
Constraints
* All values in input are integers.
* 2 \leq N \leq 400
* 1 \leq a_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 \ldots a_N
Output
Print the minimum possible total cost incurred.
Examples
Input
4
10 20 30 40
Output
190
Input
5
10 10 10 10 10
Output
120
Input
3
1000000000 1000000000 1000000000
Output
5000000000
Input
6
7 6 8 6 1 1
Output
68 | instruction | 0 | 100,985 | 8 | 201,970 |
"Correct Solution:
```
from itertools import accumulate
n = int(input())
a = [0] + list(accumulate(map(int, input().split())))
dp = [[0] * (n + 1) for i in range(n + 1)]
for i in range(n)[::-1]:
for j in range(i + 1, n + 1):
if j - i == 1:
continue
dp[i][j] = min([dp[i][k] + dp[k][j] for k in range(i + 1, j)])
dp[i][j] += a[j] - a[i]
print(dp[0][n])
``` | output | 1 | 100,985 | 8 | 201,971 |
Provide a correct Python 3 solution for this coding contest problem.
There are N slimes lining up in a row. Initially, the i-th slime from the left has a size of a_i.
Taro is trying to combine all the slimes into a larger slime. He will perform the following operation repeatedly until there is only one slime:
* Choose two adjacent slimes, and combine them into a new slime. The new slime has a size of x + y, where x and y are the sizes of the slimes before combining them. Here, a cost of x + y is incurred. The positional relationship of the slimes does not change while combining slimes.
Find the minimum possible total cost incurred.
Constraints
* All values in input are integers.
* 2 \leq N \leq 400
* 1 \leq a_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 \ldots a_N
Output
Print the minimum possible total cost incurred.
Examples
Input
4
10 20 30 40
Output
190
Input
5
10 10 10 10 10
Output
120
Input
3
1000000000 1000000000 1000000000
Output
5000000000
Input
6
7 6 8 6 1 1
Output
68 | instruction | 0 | 100,986 | 8 | 201,972 |
"Correct Solution:
```
import sys
INF = 10**14
LI = lambda : [int(x) for x in sys.stdin.readline().split()]
NI = lambda : int(sys.stdin.readline())
N = NI()
a = LI()
sn = [[0 for _ in range(N)] for _ in range(N)]
dp = [[INF for _ in range(N)] for _ in range(N)]
for i in range(N):
x = 0
for j in range(i,N):
x += a[j]
sn[i][j] = x
for i in range(N):
dp[i][i] = 0
for j in range(1,N):
for i in range(j-1,-1,-1):
for k in range(0,j-i):
dp[i][j] = min(dp[i][j],dp[i+k+1][j]+ dp[i][i+k] + sn[i][j])
print(dp[0][-1])
``` | output | 1 | 100,986 | 8 | 201,973 |
Provide a correct Python 3 solution for this coding contest problem.
There are N slimes lining up in a row. Initially, the i-th slime from the left has a size of a_i.
Taro is trying to combine all the slimes into a larger slime. He will perform the following operation repeatedly until there is only one slime:
* Choose two adjacent slimes, and combine them into a new slime. The new slime has a size of x + y, where x and y are the sizes of the slimes before combining them. Here, a cost of x + y is incurred. The positional relationship of the slimes does not change while combining slimes.
Find the minimum possible total cost incurred.
Constraints
* All values in input are integers.
* 2 \leq N \leq 400
* 1 \leq a_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 \ldots a_N
Output
Print the minimum possible total cost incurred.
Examples
Input
4
10 20 30 40
Output
190
Input
5
10 10 10 10 10
Output
120
Input
3
1000000000 1000000000 1000000000
Output
5000000000
Input
6
7 6 8 6 1 1
Output
68 | instruction | 0 | 100,987 | 8 | 201,974 |
"Correct Solution:
```
import sys
from itertools import accumulate
sys.setrecursionlimit(10 ** 9)
N = int(input())
A = list(map(int, input().split()))
acc = [0] + list(accumulate(A))
# dp[i][j] := 区間[i, j]のスライムたちを１匹にまとめるのに必要なコスト
dp = [[None] * N for _ in range(N)]
for i in range(N):
dp[i][i] = 0
def dfs(i, j):
if j - i == 1:
dp[i][j] = A[i] + A[j]
return dp[i][j]
if dp[i][j] is not None:
return dp[i][j]
ret = float('inf')
for x in range(i, j):
ret = min(ret, dfs(i, x) + dfs(x + 1, j) + acc[j + 1] - acc[i])
dp[i][j] = ret
return ret
print(dfs(0, N - 1))
``` | output | 1 | 100,987 | 8 | 201,975 |
Provide a correct Python 3 solution for this coding contest problem.
There are N slimes lining up in a row. Initially, the i-th slime from the left has a size of a_i.
Taro is trying to combine all the slimes into a larger slime. He will perform the following operation repeatedly until there is only one slime:
* Choose two adjacent slimes, and combine them into a new slime. The new slime has a size of x + y, where x and y are the sizes of the slimes before combining them. Here, a cost of x + y is incurred. The positional relationship of the slimes does not change while combining slimes.
Find the minimum possible total cost incurred.
Constraints
* All values in input are integers.
* 2 \leq N \leq 400
* 1 \leq a_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 \ldots a_N
Output
Print the minimum possible total cost incurred.
Examples
Input
4
10 20 30 40
Output
190
Input
5
10 10 10 10 10
Output
120
Input
3
1000000000 1000000000 1000000000
Output
5000000000
Input
6
7 6 8 6 1 1
Output
68 | instruction | 0 | 100,988 | 8 | 201,976 |
"Correct Solution:
```
import sys
input=sys.stdin.readline
sys.setrecursionlimit(100000000)
MOD = 10**9+7
INF = 10 ** 20
from functools import lru_cache
MAXN = 405
a = [0] * MAXN
dp = [[-1] * MAXN for _ in range(MAXN)]
def dfs(l,r):
if dp[l][r] != -1:
return dp[l][r]
if r - l == 1:
dp[l][r] = 0
return 0
ans = INF
for i in range(l + 1,r):
ans = min(ans,dfs(l,i) + dfs(i,r))
if l == 0:
ans += a[r - 1]
else:
ans += a[r - 1] - a[l - 1]
dp[l][r] = ans
return ans
def main():
n = int(input())
b = list(map(int,input().split()))
a[:n] = b
for i in range(1,n):
a[i] += a[i - 1]
ans = dfs(0,n)
print(ans)
if __name__=='__main__':
main()
``` | output | 1 | 100,988 | 8 | 201,977 |
Provide a correct Python 3 solution for this coding contest problem.
There are N slimes lining up in a row. Initially, the i-th slime from the left has a size of a_i.
Taro is trying to combine all the slimes into a larger slime. He will perform the following operation repeatedly until there is only one slime:
* Choose two adjacent slimes, and combine them into a new slime. The new slime has a size of x + y, where x and y are the sizes of the slimes before combining them. Here, a cost of x + y is incurred. The positional relationship of the slimes does not change while combining slimes.
Find the minimum possible total cost incurred.
Constraints
* All values in input are integers.
* 2 \leq N \leq 400
* 1 \leq a_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 \ldots a_N
Output
Print the minimum possible total cost incurred.
Examples
Input
4
10 20 30 40
Output
190
Input
5
10 10 10 10 10
Output
120
Input
3
1000000000 1000000000 1000000000
Output
5000000000
Input
6
7 6 8 6 1 1
Output
68 | instruction | 0 | 100,989 | 8 | 201,978 |
"Correct Solution:
```
N = int(input())
A = list(map(int, input().split()))
dp = [[-1]*N for _ in range(N)]
for i in range(N):
dp[i][i] = 0
def main(l, r):
if dp[l][r] != -1:
return dp[l][r]
ret = 10**15
for i in range(l, r):
ret = min(ret, main(l, i)+main(i+1, r))
ret += sum(A[l:r+1])
dp[l][r] = ret
return ret
print(main(0, N-1))
``` | output | 1 | 100,989 | 8 | 201,979 |
Provide a correct Python 3 solution for this coding contest problem.
There are N slimes lining up in a row. Initially, the i-th slime from the left has a size of a_i.
Taro is trying to combine all the slimes into a larger slime. He will perform the following operation repeatedly until there is only one slime:
* Choose two adjacent slimes, and combine them into a new slime. The new slime has a size of x + y, where x and y are the sizes of the slimes before combining them. Here, a cost of x + y is incurred. The positional relationship of the slimes does not change while combining slimes.
Find the minimum possible total cost incurred.
Constraints
* All values in input are integers.
* 2 \leq N \leq 400
* 1 \leq a_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 \ldots a_N
Output
Print the minimum possible total cost incurred.
Examples
Input
4
10 20 30 40
Output
190
Input
5
10 10 10 10 10
Output
120
Input
3
1000000000 1000000000 1000000000
Output
5000000000
Input
6
7 6 8 6 1 1
Output
68 | instruction | 0 | 100,990 | 8 | 201,980 |
"Correct Solution:
```
# dp[l][r]=min(区間[l,r]を合成するための最小コスト)
n=int(input())
a=list(map(int,input().split()))
INF=10**15
dp=[[0]*n for i in range(n)]
for i in range(n):
dp[i][i]=a[i]
# 合成するときの必要経費を構成
for i in range(n-1):
for j in range(i+1,n):
dp[i][j]=dp[i][j-1]+a[j]
dp[0][-1]=0
# dp[i][j]=min(dp[i][k]+dp[k+1][j])+必要経費 で斜め(0,1),(1,2),(0,2),(1,3),...の順に見ていく
# ループの書き方に手間取った
for i in range(1,n):
for j in range(n-i):
ans=INF
for k in range(j,i+j):
ans=min(ans,dp[j][k]+dp[k+1][i+j])
dp[j][i+j]+=ans
print(dp[0][-1])
``` | output | 1 | 100,990 | 8 | 201,981 |
Provide a correct Python 3 solution for this coding contest problem.
There are N slimes lining up in a row. Initially, the i-th slime from the left has a size of a_i.
Taro is trying to combine all the slimes into a larger slime. He will perform the following operation repeatedly until there is only one slime:
* Choose two adjacent slimes, and combine them into a new slime. The new slime has a size of x + y, where x and y are the sizes of the slimes before combining them. Here, a cost of x + y is incurred. The positional relationship of the slimes does not change while combining slimes.
Find the minimum possible total cost incurred.
Constraints
* All values in input are integers.
* 2 \leq N \leq 400
* 1 \leq a_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 \ldots a_N
Output
Print the minimum possible total cost incurred.
Examples
Input
4
10 20 30 40
Output
190
Input
5
10 10 10 10 10
Output
120
Input
3
1000000000 1000000000 1000000000
Output
5000000000
Input
6
7 6 8 6 1 1
Output
68 | instruction | 0 | 100,991 | 8 | 201,982 |
"Correct Solution:
```
N=int(input())
A=list(map(int,input().split()))
DPLIST=[[None]*N for i in range(N)]
SUM=[0]
for i in range(N):
DPLIST[i][i]=0
SUM.append(SUM[-1]+A[i])
for i in range(1,N):
for j in range(i,N):
ANS=float("inf")
for k in range(j-i,j):
if ANS>DPLIST[j-i][k]+DPLIST[k+1][j]+SUM[j+1]-SUM[j-i]:
ANS=DPLIST[j-i][k]+DPLIST[k+1][j]+SUM[j+1]-SUM[j-i]
DPLIST[j-i][j]=ANS
print(DPLIST[0][N-1])
``` | output | 1 | 100,991 | 8 | 201,983 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N slimes lining up in a row. Initially, the i-th slime from the left has a size of a_i.
Taro is trying to combine all the slimes into a larger slime. He will perform the following operation repeatedly until there is only one slime:
* Choose two adjacent slimes, and combine them into a new slime. The new slime has a size of x + y, where x and y are the sizes of the slimes before combining them. Here, a cost of x + y is incurred. The positional relationship of the slimes does not change while combining slimes.
Find the minimum possible total cost incurred.
Constraints
* All values in input are integers.
* 2 \leq N \leq 400
* 1 \leq a_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 \ldots a_N
Output
Print the minimum possible total cost incurred.
Examples
Input
4
10 20 30 40
Output
190
Input
5
10 10 10 10 10
Output
120
Input
3
1000000000 1000000000 1000000000
Output
5000000000
Input
6
7 6 8 6 1 1
Output
68
Submitted Solution:
```
from itertools import accumulate
N = int(input())
A = list(map(int, input().split()))
dp = [[0 for _ in range(N+1)] for _ in range(N)]
acc = [0] + list(accumulate(A))
for w in range(2, N+1):
for l in range(N-w+1):
r = l + w
dp[l][r] = min([dp[l][m] + dp[m][r] for m in range(l+1, r)]) + acc[r] - acc[l]
print(dp[0][N])
``` | instruction | 0 | 100,992 | 8 | 201,984 |
Yes | output | 1 | 100,992 | 8 | 201,985 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N slimes lining up in a row. Initially, the i-th slime from the left has a size of a_i.
Taro is trying to combine all the slimes into a larger slime. He will perform the following operation repeatedly until there is only one slime:
* Choose two adjacent slimes, and combine them into a new slime. The new slime has a size of x + y, where x and y are the sizes of the slimes before combining them. Here, a cost of x + y is incurred. The positional relationship of the slimes does not change while combining slimes.
Find the minimum possible total cost incurred.
Constraints
* All values in input are integers.
* 2 \leq N \leq 400
* 1 \leq a_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 \ldots a_N
Output
Print the minimum possible total cost incurred.
Examples
Input
4
10 20 30 40
Output
190
Input
5
10 10 10 10 10
Output
120
Input
3
1000000000 1000000000 1000000000
Output
5000000000
Input
6
7 6 8 6 1 1
Output
68
Submitted Solution:
```
import sys
sys.setrecursionlimit(10**8)
N = int(input())
*a, = map(int, input().split())
s = [0]*N
s[0] = a[0]
for i in range(1, N):
s[i] += s[i-1]+a[i]
s = [0]+s
dp = [[0]*(N+1) for _ in range(N+1)]
def solve(l, r):
if l == r:
return 0
if dp[l][r] > 0:
return dp[l][r]
n = float('inf')
m = s[r+1]-s[l]
for i in range(l, r):
n = min([n, solve(l, i)+solve(i+1, r)+m])
dp[l][r] = n
return n
print(solve(0, N-1))
``` | instruction | 0 | 100,993 | 8 | 201,986 |
Yes | output | 1 | 100,993 | 8 | 201,987 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N slimes lining up in a row. Initially, the i-th slime from the left has a size of a_i.
Taro is trying to combine all the slimes into a larger slime. He will perform the following operation repeatedly until there is only one slime:
* Choose two adjacent slimes, and combine them into a new slime. The new slime has a size of x + y, where x and y are the sizes of the slimes before combining them. Here, a cost of x + y is incurred. The positional relationship of the slimes does not change while combining slimes.
Find the minimum possible total cost incurred.
Constraints
* All values in input are integers.
* 2 \leq N \leq 400
* 1 \leq a_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 \ldots a_N
Output
Print the minimum possible total cost incurred.
Examples
Input
4
10 20 30 40
Output
190
Input
5
10 10 10 10 10
Output
120
Input
3
1000000000 1000000000 1000000000
Output
5000000000
Input
6
7 6 8 6 1 1
Output
68
Submitted Solution:
```
import sys
import itertools
input = sys.stdin.readline
n=int(input())
a=list(map(int, input().split()))
b=itertools.accumulate(a)
b=[0]+list(b)
dp=[[0]*(n) for i in range(n)]
x=float('inf')
for i in range(n-1,-1,-1):
for j in range(i+1,n):
if j>=i+2:
for k in range(i,j):
x=min(dp[i][k]+dp[k+1][j],x)
dp[i][j]=x+(b[j+1]-b[i])
x=float('inf')
else:
dp[i][j]=b[j+1]-b[i]
print(dp[0][-1])
``` | instruction | 0 | 100,994 | 8 | 201,988 |
Yes | output | 1 | 100,994 | 8 | 201,989 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N slimes lining up in a row. Initially, the i-th slime from the left has a size of a_i.
Taro is trying to combine all the slimes into a larger slime. He will perform the following operation repeatedly until there is only one slime:
* Choose two adjacent slimes, and combine them into a new slime. The new slime has a size of x + y, where x and y are the sizes of the slimes before combining them. Here, a cost of x + y is incurred. The positional relationship of the slimes does not change while combining slimes.
Find the minimum possible total cost incurred.
Constraints
* All values in input are integers.
* 2 \leq N \leq 400
* 1 \leq a_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 \ldots a_N
Output
Print the minimum possible total cost incurred.
Examples
Input
4
10 20 30 40
Output
190
Input
5
10 10 10 10 10
Output
120
Input
3
1000000000 1000000000 1000000000
Output
5000000000
Input
6
7 6 8 6 1 1
Output
68
Submitted Solution:
```
import sys
sys.setrecursionlimit(10**8)
n = int(input())
a = list(map(int,input().split()))
a_data = [0]
now = 0
for i in a:
now += i
a_data.append(now)
dp = [[10**18 for _ in range(n+1)] for i in range(n)]
#print(a_data)
def solve(i,j):
if dp[i][j] != 10**18:
return dp[i][j]
if i+1 == j:
dp[i][j] = 0
return dp[i][j]
dp[i][j] = a_data[j] - a_data[i]
x = 10**18
for k in range(i+1,j):
s = solve(i,k)
t = solve(k,j)
x = min(x,s+t)
dp[i][j] += x
return dp[i][j]
ans = solve(0,n)
print(ans)
``` | instruction | 0 | 100,995 | 8 | 201,990 |
Yes | output | 1 | 100,995 | 8 | 201,991 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N slimes lining up in a row. Initially, the i-th slime from the left has a size of a_i.
Taro is trying to combine all the slimes into a larger slime. He will perform the following operation repeatedly until there is only one slime:
* Choose two adjacent slimes, and combine them into a new slime. The new slime has a size of x + y, where x and y are the sizes of the slimes before combining them. Here, a cost of x + y is incurred. The positional relationship of the slimes does not change while combining slimes.
Find the minimum possible total cost incurred.
Constraints
* All values in input are integers.
* 2 \leq N \leq 400
* 1 \leq a_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 \ldots a_N
Output
Print the minimum possible total cost incurred.
Examples
Input
4
10 20 30 40
Output
190
Input
5
10 10 10 10 10
Output
120
Input
3
1000000000 1000000000 1000000000
Output
5000000000
Input
6
7 6 8 6 1 1
Output
68
Submitted Solution:
```
n = int(input())
ar = [int(x) for x in input().split()]
INT_MAX = 1000000000
dp = [[INT_MAX for i in range(n + 5)] for j in range(n + 5)]
for i in range(n):
dp[i][i] = 0
prefix = [0, ar[0]]
for i in range(1, n):
prefix.append(prefix[i] + ar[i])
for i in range(n - 1, -1, -1):
for j in range(i + 1, n):
for k in range(0, j - i):
dp[i][j] = min(dp[i][j], (dp[i][i + k] + dp[i + k + 1][j]) + prefix[j + 1] - prefix[i])
print(dp[0][n - 1])
``` | instruction | 0 | 100,996 | 8 | 201,992 |
No | output | 1 | 100,996 | 8 | 201,993 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N slimes lining up in a row. Initially, the i-th slime from the left has a size of a_i.
Taro is trying to combine all the slimes into a larger slime. He will perform the following operation repeatedly until there is only one slime:
* Choose two adjacent slimes, and combine them into a new slime. The new slime has a size of x + y, where x and y are the sizes of the slimes before combining them. Here, a cost of x + y is incurred. The positional relationship of the slimes does not change while combining slimes.
Find the minimum possible total cost incurred.
Constraints
* All values in input are integers.
* 2 \leq N \leq 400
* 1 \leq a_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 \ldots a_N
Output
Print the minimum possible total cost incurred.
Examples
Input
4
10 20 30 40
Output
190
Input
5
10 10 10 10 10
Output
120
Input
3
1000000000 1000000000 1000000000
Output
5000000000
Input
6
7 6 8 6 1 1
Output
68
Submitted Solution:
```
def wczytaj_liste():
wczytana_lista = input()
lista_znakow = wczytana_lista.split()
ostateczna_lista = []
for element in lista_znakow:
ostateczna_lista.append(int(element))
return ostateczna_lista
def potworny_Jan_Kanty():
N = wczytaj_liste()[0]
potwory = wczytaj_liste()
wynik = {}
for dlugosc_przedzialu in range(1,len(potwory)+1):
for poczatek in range(len(potwory)-dlugosc_przedzialu+1):
koniec = dlugosc_przedzialu + poczatek
if dlugosc_przedzialu == 1:
wynik[poczatek,koniec] = 0
else:
odp = 10**18
for i in range(1, dlugosc_przedzialu):
kandydat = wynik[poczatek+i,koniec]+wynik[poczatek, poczatek+i]
if kandydat < odp:
odp = kandydat
wynik[poczatek, koniec] = odp+sum(potwory[poczatek:koniec])
print(wynik[0, N])
potworny_Jan_Kanty()
``` | instruction | 0 | 100,997 | 8 | 201,994 |
No | output | 1 | 100,997 | 8 | 201,995 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N slimes lining up in a row. Initially, the i-th slime from the left has a size of a_i.
Taro is trying to combine all the slimes into a larger slime. He will perform the following operation repeatedly until there is only one slime:
* Choose two adjacent slimes, and combine them into a new slime. The new slime has a size of x + y, where x and y are the sizes of the slimes before combining them. Here, a cost of x + y is incurred. The positional relationship of the slimes does not change while combining slimes.
Find the minimum possible total cost incurred.
Constraints
* All values in input are integers.
* 2 \leq N \leq 400
* 1 \leq a_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 \ldots a_N
Output
Print the minimum possible total cost incurred.
Examples
Input
4
10 20 30 40
Output
190
Input
5
10 10 10 10 10
Output
120
Input
3
1000000000 1000000000 1000000000
Output
5000000000
Input
6
7 6 8 6 1 1
Output
68
Submitted Solution:
```
from itertools import accumulate
import sys
input = sys.stdin.readline
from itertools import chain
dp = []
dp2 = []
aac = []
def fdp(l, r):
global dp, dp2
if dp2[l][r]:
return dp[l][r]
else:
dp2[l][r] = 1
if l == r:
return 0
result_tmp = float('inf')
for kugiri in range(l, r):
result_tmp = min(result_tmp, fdp(l, kugiri) + fdp(kugiri+1, r))
result = result_tmp + (aac[r+1] - aac[l])
dp[l][r] = result
return result
def main():
global dp, dp2, aac
n = int(input())
a = tuple(map(int, input().split()))
aac = list(accumulate(a))
aac = tuple(chain([0], aac))
dp = [[0] * n for _ in range(n)]
dp2 = [[0] * n for _ in range(n)]
fdp(0, n - 1)
print(dp[0][-1])
if __name__ == '__main__':
main()
``` | instruction | 0 | 100,998 | 8 | 201,996 |
No | output | 1 | 100,998 | 8 | 201,997 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N slimes lining up in a row. Initially, the i-th slime from the left has a size of a_i.
Taro is trying to combine all the slimes into a larger slime. He will perform the following operation repeatedly until there is only one slime:
* Choose two adjacent slimes, and combine them into a new slime. The new slime has a size of x + y, where x and y are the sizes of the slimes before combining them. Here, a cost of x + y is incurred. The positional relationship of the slimes does not change while combining slimes.
Find the minimum possible total cost incurred.
Constraints
* All values in input are integers.
* 2 \leq N \leq 400
* 1 \leq a_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 \ldots a_N
Output
Print the minimum possible total cost incurred.
Examples
Input
4
10 20 30 40
Output
190
Input
5
10 10 10 10 10
Output
120
Input
3
1000000000 1000000000 1000000000
Output
5000000000
Input
6
7 6 8 6 1 1
Output
68
Submitted Solution:
```
# coding: utf-8
from functools import lru_cache
from itertools import accumulate as accum
import sys
input = sys.stdin.readline
sys.setrecursionlimit(10**6)
@lru_cache(maxsize = None)
def f(l, r):
if l == r:
return(0)
cost = float('inf')
for x in range(l, r):
cost_ = f(l, x) + f(x + 1, r)
if cost_ < cost:
cost = cost_
return(cost + acsum[r] - acsum[l - 1])
n = int(input()) # 2 <= n <= 400
acsum = list(accum(map(int, input().split()))) + [0]
print(f(0, n - 1))
``` | instruction | 0 | 100,999 | 8 | 201,998 |
No | output | 1 | 100,999 | 8 | 201,999 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice has got addicted to a game called Sirtet recently.
In Sirtet, player is given an n × m grid. Initially a_{i,j} cubes are stacked up in the cell (i,j). Two cells are called adjacent if they share a side. Player can perform the following operations:
* stack up one cube in two adjacent cells;
* stack up two cubes in one cell.
Cubes mentioned above are identical in height.
Here is an illustration of the game. States on the right are obtained by performing one of the above operations on the state on the left, and grey cubes are added due to the operation.
<image>
Player's goal is to make the height of all cells the same (i.e. so that each cell has the same number of cubes in it) using above operations.
Alice, however, has found out that on some starting grids she may never reach the goal no matter what strategy she uses. Thus, she is wondering the number of initial grids such that
* L ≤ a_{i,j} ≤ R for all 1 ≤ i ≤ n, 1 ≤ j ≤ m;
* player can reach the goal using above operations.
Please help Alice with it. Notice that the answer might be large, please output the desired value modulo 998,244,353.
Input
The only line contains four integers n, m, L and R (1≤ n,m,L,R ≤ 10^9, L ≤ R, n ⋅ m ≥ 2).
Output
Output one integer, representing the desired answer modulo 998,244,353.
Examples
Input
2 2 1 1
Output
1
Input
1 2 1 2
Output
2
Note
In the first sample, the only initial grid that satisfies the requirements is a_{1,1}=a_{2,1}=a_{1,2}=a_{2,2}=1. Thus the answer should be 1.
In the second sample, initial grids that satisfy the requirements are a_{1,1}=a_{1,2}=1 and a_{1,1}=a_{1,2}=2. Thus the answer should be 2.
Submitted Solution:
```
# https://www.geeksforgeeks.org/modular-division/
import math
import sys
input = sys.stdin.readline
def modInverse(b,m):
g = math.gcd(b, m)
if (g != 1):
# print("Inverse doesn't exist")
return -1
else:
# If b and m are relatively prime,
# then modulo inverse is b^(m-2) mode m
return pow(b, m - 2, m)
# Function to compute a/b under modulo m
def modDivide(a,b,m):
a = a % m
inv = modInverse(b,m)
if(inv == -1):
return -1
else:
return (inv*a)%m
# print("Result of Division is ",(inv*a) % m)
p=998244353
# p=79
n,m,L,R=map(int,input().split())
x=pow(R-L+1,n*m,p)
# print(x)
# y=modDivide(x,2,p)
if (L-R+1)%2==0:
y=modDivide(x,2,p)
print(y%p)
else:
y=modDivide(x+1,2,p)
print(y%p)
``` | instruction | 0 | 101,361 | 8 | 202,722 |
No | output | 1 | 101,361 | 8 | 202,723 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N logs of lengths A_1,A_2,\cdots A_N.
We can cut these logs at most K times in total. When a log of length L is cut at a point whose distance from an end of the log is t (0<t<L), it becomes two logs of lengths t and L-t.
Find the shortest possible length of the longest log after at most K cuts, and print it after rounding up to an integer.
Constraints
* 1 \leq N \leq 2 \times 10^5
* 0 \leq K \leq 10^9
* 1 \leq A_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 \cdots A_N
Output
Print an integer representing the answer.
Examples
Input
2 3
7 9
Output
4
Input
3 0
3 4 5
Output
5
Input
10 10
158260522 877914575 602436426 24979445 861648772 623690081 433933447 476190629 262703497 211047202
Output
292638192
Submitted Solution:
```
def resolve():
import heapq
n, k = map(int, input().split())
a = list(map(int, input().split()))
q = []
for index, i in enumerate(a):
heapq.heappush(q, (-i, index, 1))
for _ in range(k):
length, index, count = heapq.heappop(q)
heapq.heappush(q, (-((a[index] + count) //
(count + 1)), index, count + 1))
length, index, count = heapq.heappop(q)
print((a[index] + count - 1) // count)
if __name__ == '__main__':
resolve()
``` | instruction | 0 | 102,700 | 8 | 205,400 |
No | output | 1 | 102,700 | 8 | 205,401 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N logs of lengths A_1,A_2,\cdots A_N.
We can cut these logs at most K times in total. When a log of length L is cut at a point whose distance from an end of the log is t (0<t<L), it becomes two logs of lengths t and L-t.
Find the shortest possible length of the longest log after at most K cuts, and print it after rounding up to an integer.
Constraints
* 1 \leq N \leq 2 \times 10^5
* 0 \leq K \leq 10^9
* 1 \leq A_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 \cdots A_N
Output
Print an integer representing the answer.
Examples
Input
2 3
7 9
Output
4
Input
3 0
3 4 5
Output
5
Input
10 10
158260522 877914575 602436426 24979445 861648772 623690081 433933447 476190629 262703497 211047202
Output
292638192
Submitted Solution:
```
import heapq
ceil = lambda a, b: (a + b - 1) // b
N, K = map(int, input().split())
A = tuple(map(int, input().split()))
h = [(-a, a, 1) for a in A]
heapq.heapify(h)
for _ in range(K):
l, a, ct = heapq.heappop(h)
ct += 1
heapq.heappush(h, (-a / ct, a, ct))
_, a, ct = heapq.heappop(h)
print(ceil(a, ct))
``` | instruction | 0 | 102,702 | 8 | 205,404 |
No | output | 1 | 102,702 | 8 | 205,405 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day you wanted to read something, so you went to your bookshelf to grab some book. But when you saw how messy the bookshelf was you decided to clean it up first.
<image>
There are n books standing in a row on the shelf, the i-th book has color a_i.
You'd like to rearrange the books to make the shelf look beautiful. The shelf is considered beautiful if all books of the same color are next to each other.
In one operation you can take one book from any position on the shelf and move it to the right end of the shelf.
What is the minimum number of operations you need to make the shelf beautiful?
Input
The first line contains one integer n (1 ≤ n ≤ 5 ⋅ 10^5) — the number of books.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — the book colors.
Output
Output the minimum number of operations to make the shelf beautiful.
Examples
Input
5
1 2 2 1 3
Output
2
Input
5
1 2 2 1 1
Output
1
Note
In the first example, we have the bookshelf [1, 2, 2, 1, 3] and can, for example:
1. take a book on position 4 and move to the right end: we'll get [1, 2, 2, 3, 1];
2. take a book on position 1 and move to the right end: we'll get [2, 2, 3, 1, 1].
In the second example, we can move the first book to the end of the bookshelf and get [2,2,1,1,1]. | instruction | 0 | 103,108 | 8 | 206,216 |
Tags: data structures, dp, greedy
Correct Solution:
```
n = int(input())
a = [int(t)-1 for t in input().split()]
l, r = [-1]*n, [-1]*n
for i, x in enumerate(a):
if l[x] == -1:
l[x] = i
r[x] = i
dp = [0]*(n+1)
cnt = [0]*n
for i in reversed(range(n)):
x = a[i]
cnt[x] += 1
dp[i] = dp[i+1]
if i == l[x]:
dp[i] = max(dp[i], dp[r[x]+1]+cnt[x])
else:
dp[i] = max(dp[i], cnt[x])
print(n - dp[0])
``` | output | 1 | 103,108 | 8 | 206,217 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day you wanted to read something, so you went to your bookshelf to grab some book. But when you saw how messy the bookshelf was you decided to clean it up first.
<image>
There are n books standing in a row on the shelf, the i-th book has color a_i.
You'd like to rearrange the books to make the shelf look beautiful. The shelf is considered beautiful if all books of the same color are next to each other.
In one operation you can take one book from any position on the shelf and move it to the right end of the shelf.
What is the minimum number of operations you need to make the shelf beautiful?
Input
The first line contains one integer n (1 ≤ n ≤ 5 ⋅ 10^5) — the number of books.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — the book colors.
Output
Output the minimum number of operations to make the shelf beautiful.
Examples
Input
5
1 2 2 1 3
Output
2
Input
5
1 2 2 1 1
Output
1
Note
In the first example, we have the bookshelf [1, 2, 2, 1, 3] and can, for example:
1. take a book on position 4 and move to the right end: we'll get [1, 2, 2, 3, 1];
2. take a book on position 1 and move to the right end: we'll get [2, 2, 3, 1, 1].
In the second example, we can move the first book to the end of the bookshelf and get [2,2,1,1,1]. | instruction | 0 | 103,109 | 8 | 206,218 |
Tags: data structures, dp, greedy
Correct Solution:
```
import bisect
n = int(input())
arr = list(map(int,input().split()))
#arr = arr[::-1]
dp = [-1]*n+[0]
rear = n-1
indexlist = [[] for i in range(n+1)]
for i in range(n):
indexlist[ arr[i] ].append(i)
def getnext(index):
if index>=rear: return 0
if dp[index]>=0: return dp[index]
if index==indexlist[arr[index]][0]:
end = indexlist[arr[index]][-1]
num1 = end-index+1-len(indexlist[arr[index]]) + getnext(end+1)
else:
loc = bisect.bisect_left(indexlist[arr[index]], index )
num1 = n-index-len(indexlist[arr[index]]) + loc
num2 = 1 + getnext(index+1)
dp[index] = min(num1,num2)
return min(num1,num2)
for index in range(n-1,-1,-1):
if index==indexlist[arr[index]][0]:
end = indexlist[arr[index]][-1]
num1 = end-index+1-len(indexlist[arr[index]]) + dp[end+1]
else:
loc = bisect.bisect_left(indexlist[arr[index]], index )
num1 = n-index-len(indexlist[arr[index]]) + loc
num2 = 1 + dp[index+1]
dp[index] = min(num1,num2)
print(dp[0])
``` | output | 1 | 103,109 | 8 | 206,219 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day you wanted to read something, so you went to your bookshelf to grab some book. But when you saw how messy the bookshelf was you decided to clean it up first.
<image>
There are n books standing in a row on the shelf, the i-th book has color a_i.
You'd like to rearrange the books to make the shelf look beautiful. The shelf is considered beautiful if all books of the same color are next to each other.
In one operation you can take one book from any position on the shelf and move it to the right end of the shelf.
What is the minimum number of operations you need to make the shelf beautiful?
Input
The first line contains one integer n (1 ≤ n ≤ 5 ⋅ 10^5) — the number of books.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — the book colors.
Output
Output the minimum number of operations to make the shelf beautiful.
Examples
Input
5
1 2 2 1 3
Output
2
Input
5
1 2 2 1 1
Output
1
Note
In the first example, we have the bookshelf [1, 2, 2, 1, 3] and can, for example:
1. take a book on position 4 and move to the right end: we'll get [1, 2, 2, 3, 1];
2. take a book on position 1 and move to the right end: we'll get [2, 2, 3, 1, 1].
In the second example, we can move the first book to the end of the bookshelf and get [2,2,1,1,1]. | instruction | 0 | 103,110 | 8 | 206,220 |
Tags: data structures, dp, greedy
Correct Solution:
```
"""
#If FastIO not needed, use this and don't forget to strip
#import sys, math
#input = sys.stdin.readline
"""
import os
import sys
from io import BytesIO, IOBase
import heapq as h
from bisect import bisect_left, bisect_right
import time
from types import GeneratorType
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
self.os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
from collections import defaultdict as dd, deque as dq, Counter as dc
import math, string
#start_time = time.time()
def getInts():
return [int(s) for s in input().split()]
def getInt():
return int(input())
def getStrs():
return [s for s in input().split()]
def getStr():
return input()
def listStr():
return list(input())
def getMat(n):
return [getInts() for _ in range(n)]
def isInt(s):
return '0' <= s[0] <= '9'
MOD = 10**9 + 7
"""
"""
def solve():
N = getInt()
A = getInts()
left, right, counts, tmp_counts = [-1]*(N+1), [-1]*(N+1), [0]*(N+1), [0]*(N+1)
for i in range(N-1,-1,-1):
left[A[i]] = i
counts[A[i]] += 1
if counts[A[i]] == 1: right[A[i]] = i
dp = [-1]*(N+1)
dp[N] = 0
for i in range(N-1,-1,-1):
dp[i] = dp[i+1]
tmp_counts[A[i]] += 1
if i == left[A[i]]:
dp[i] = max(dp[i],dp[right[A[i]]+1]+counts[A[i]])
else:
dp[i] = max(dp[i],tmp_counts[A[i]])
return N - dp[0]
#for _ in range(getInt()):
print(solve())
#solve()
#print(time.time()-start_time)
``` | output | 1 | 103,110 | 8 | 206,221 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day you wanted to read something, so you went to your bookshelf to grab some book. But when you saw how messy the bookshelf was you decided to clean it up first.
<image>
There are n books standing in a row on the shelf, the i-th book has color a_i.
You'd like to rearrange the books to make the shelf look beautiful. The shelf is considered beautiful if all books of the same color are next to each other.
In one operation you can take one book from any position on the shelf and move it to the right end of the shelf.
What is the minimum number of operations you need to make the shelf beautiful?
Input
The first line contains one integer n (1 ≤ n ≤ 5 ⋅ 10^5) — the number of books.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — the book colors.
Output
Output the minimum number of operations to make the shelf beautiful.
Examples
Input
5
1 2 2 1 3
Output
2
Input
5
1 2 2 1 1
Output
1
Note
In the first example, we have the bookshelf [1, 2, 2, 1, 3] and can, for example:
1. take a book on position 4 and move to the right end: we'll get [1, 2, 2, 3, 1];
2. take a book on position 1 and move to the right end: we'll get [2, 2, 3, 1, 1].
In the second example, we can move the first book to the end of the bookshelf and get [2,2,1,1,1]. | instruction | 0 | 103,111 | 8 | 206,222 |
Tags: data structures, dp, greedy
Correct Solution:
```
from collections import Counter
n = int(input())
A = list(map(int, input().split()))
left, right = {}, {}
for i, a in enumerate(A):
left.setdefault(a, i)
right[a] = i
dp = [0] * (n + 1)
cnt = Counter()
for i in range(n - 1, -1, -1):
dp[i] = dp[i + 1]
cnt[A[i]] += 1
if i == left[A[i]]:
dp[i] = max(dp[i], dp[right[A[i]] + 1] + cnt[A[i]])
else:
dp[i] = max(dp[i], cnt[A[i]])
print(n - dp[0])
``` | output | 1 | 103,111 | 8 | 206,223 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day you wanted to read something, so you went to your bookshelf to grab some book. But when you saw how messy the bookshelf was you decided to clean it up first.
<image>
There are n books standing in a row on the shelf, the i-th book has color a_i.
You'd like to rearrange the books to make the shelf look beautiful. The shelf is considered beautiful if all books of the same color are next to each other.
In one operation you can take one book from any position on the shelf and move it to the right end of the shelf.
What is the minimum number of operations you need to make the shelf beautiful?
Input
The first line contains one integer n (1 ≤ n ≤ 5 ⋅ 10^5) — the number of books.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — the book colors.
Output
Output the minimum number of operations to make the shelf beautiful.
Examples
Input
5
1 2 2 1 3
Output
2
Input
5
1 2 2 1 1
Output
1
Note
In the first example, we have the bookshelf [1, 2, 2, 1, 3] and can, for example:
1. take a book on position 4 and move to the right end: we'll get [1, 2, 2, 3, 1];
2. take a book on position 1 and move to the right end: we'll get [2, 2, 3, 1, 1].
In the second example, we can move the first book to the end of the bookshelf and get [2,2,1,1,1]. | instruction | 0 | 103,112 | 8 | 206,224 |
Tags: data structures, dp, greedy
Correct Solution:
```
from bisect import *
from collections import *
from math import gcd,ceil,sqrt,floor,inf
from heapq import *
from itertools import *
from operator import add,mul,sub,xor,truediv,floordiv
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())
def A(n):return [0]*n
def AI(n,x): return [x]*n
def G(n): return [[] for i in range(n)]
def GP(it): return [[ch,len(list(g))] for ch,g in groupby(it)]
#------------------------------------------------------------------------
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
mod=10**9+7
farr=[1]
ifa=[]
def fact(x,mod=0):
if mod:
while x>=len(farr):
farr.append(farr[-1]*len(farr)%mod)
else:
while x>=len(farr):
farr.append(farr[-1]*len(farr))
return farr[x]
def ifact(x,mod):
global ifa
fact(x,mod)
ifa.append(pow(farr[-1],mod-2,mod))
for i in range(x,0,-1):
ifa.append(ifa[-1]*i%mod)
ifa.reverse()
def per(i,j,mod=0):
if i<j: return 0
if not mod:
return fact(i)//fact(i-j)
return farr[i]*ifa[i-j]%mod
def com(i,j,mod=0):
if i<j: return 0
if not mod:
return per(i,j)//fact(j)
return per(i,j,mod)*ifa[j]%mod
def catalan(n):
return com(2*n,n)//(n+1)
def isprime(n):
for i in range(2,int(n**0.5)+1):
if n%i==0:
return False
return True
def floorsum(a,b,c,n):#sum((a*i+b)//c for i in range(n+1))
if a==0:return b//c*(n+1)
if a>=c or b>=c: return floorsum(a%c,b%c,c,n)+b//c*(n+1)+a//c*n*(n+1)//2
m=(a*n+b)//c
return n*m-floorsum(c,c-b-1,a,m-1)
def inverse(a,m):
a%=m
if a<=1: return a
return ((1-inverse(m,a)*m)//a)%m
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 ST:
def __init__(self,arr):#n!=0
n=len(arr)
mx=n.bit_length()#取不到
self.st=[[0]*mx for i in range(n)]
for i in range(n):
self.st[i][0]=arr[i]
for j in range(1,mx):
for i in range(n-(1<<j)+1):
self.st[i][j]=max(self.st[i][j-1],self.st[i+(1<<j-1)][j-1])
def query(self,l,r):
if l>r:return -inf
s=(r+1-l).bit_length()-1
return max(self.st[l][s],self.st[r-(1<<s)+1][s])
'''
class DSU:#容量+路径压缩
def __init__(self,n):
self.c=[-1]*n
def same(self,x,y):
return self.find(x)==self.find(y)
def find(self,x):
if self.c[x]<0:
return x
self.c[x]=self.find(self.c[x])
return self.c[x]
def union(self,u,v):
u,v=self.find(u),self.find(v)
if u==v:
return False
if self.c[u]>self.c[v]:
u,v=v,u
self.c[u]+=self.c[v]
self.c[v]=u
return True
def size(self,x): return -self.c[self.find(x)]'''
class UFS:#秩+路径
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 flag
def dij(s,graph):
d={}
d[s]=0
heap=[(0,s)]
seen=set()
while heap:
dis,u=heappop(heap)
if u in seen:
continue
seen.add(u)
for v,w in graph[u]:
if v not in d or d[v]>d[u]+w:
d[v]=d[u]+w
heappush(heap,(d[v],v))
return d
def lcm(a,b): return a*b//gcd(a,b)
def lis(nums):
res=[]
for k in nums:
i=bisect.bisect_left(res,k)
if i==len(res):
res.append(k)
else:
res[i]=k
return len(res)
def RP(nums):#逆序对
n = len(nums)
s=set(nums)
d={}
for i,k in enumerate(sorted(s),1):
d[k]=i
bi=BIT([0]*(len(s)+1))
ans=0
for i in range(n-1,-1,-1):
ans+=bi.query(d[nums[i]]-1)
bi.update(d[nums[i]],1)
return ans
class DLN:
def __init__(self,val):
self.val=val
self.pre=None
self.next=None
def nb(i,j,n,m):
for ni,nj in [[i+1,j],[i-1,j],[i,j-1],[i,j+1]]:
if 0<=ni<n and 0<=nj<m:
yield ni,nj
def topo(n):
q=deque()
res=[]
for i in range(1,n+1):
if ind[i]==0:
q.append(i)
res.append(i)
while q:
u=q.popleft()
for v in g[u]:
ind[v]-=1
if ind[v]==0:
q.append(v)
res.append(v)
return res
@bootstrap
def gdfs(r,p):
if len(g[r])==1 and p!=-1:
yield None
for ch in g[r]:
if ch!=p:
yield gdfs(ch,r)
yield None
t=1
for i in range(t):
n=N()
a=RLL()
l=AI(n+1,n)
r=AI(n+1,0)
for i in range(n):
l[a[i]]=min(i,l[a[i]])
r[a[i]]=max(i,r[a[i]])
dp=A(n+1)
c=A(n+1)
for i in range(n-1,-1,-1):
c[a[i]]+=1
dp[i]=max(dp[i+1],c[a[i]])
if(i==l[a[i]]):
dp[i]=max(dp[i],c[a[i]]+dp[r[a[i]]+1])
ans=n-max(dp)
print(ans)
'''
sys.setrecursionlimit(200000)
import threading
threading.stack_size(10**8)
t=threading.Thr
ead(target=main)
t.start()
t.join()
'''
'''
sys.setrecursionlimit(200000)
import threading
threading.stack_size(10**8)
t=threading.Thread(target=main)
t.start()
t.join()
'''
``` | output | 1 | 103,112 | 8 | 206,225 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day you wanted to read something, so you went to your bookshelf to grab some book. But when you saw how messy the bookshelf was you decided to clean it up first.
<image>
There are n books standing in a row on the shelf, the i-th book has color a_i.
You'd like to rearrange the books to make the shelf look beautiful. The shelf is considered beautiful if all books of the same color are next to each other.
In one operation you can take one book from any position on the shelf and move it to the right end of the shelf.
What is the minimum number of operations you need to make the shelf beautiful?
Input
The first line contains one integer n (1 ≤ n ≤ 5 ⋅ 10^5) — the number of books.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — the book colors.
Output
Output the minimum number of operations to make the shelf beautiful.
Examples
Input
5
1 2 2 1 3
Output
2
Input
5
1 2 2 1 1
Output
1
Note
In the first example, we have the bookshelf [1, 2, 2, 1, 3] and can, for example:
1. take a book on position 4 and move to the right end: we'll get [1, 2, 2, 3, 1];
2. take a book on position 1 and move to the right end: we'll get [2, 2, 3, 1, 1].
In the second example, we can move the first book to the end of the bookshelf and get [2,2,1,1,1]. | instruction | 0 | 103,113 | 8 | 206,226 |
Tags: data structures, dp, greedy
Correct Solution:
```
import sys
input = sys.stdin.readline
from collections import Counter
n = int(input())
A = list(map(int, input().split()))
left, right = {}, {}
for i, a in enumerate(A):
left.setdefault(a, i)
right[a] = i
dp = [0] * (n + 1)
cnt = Counter()
for i in range(n - 1, -1, -1):
dp[i] = dp[i + 1]
cnt[A[i]] += 1
if i == left[A[i]]:
dp[i] = max(dp[i], dp[right[A[i]] + 1] + cnt[A[i]])
else:
dp[i] = max(dp[i], cnt[A[i]])
print(n - dp[0])
``` | output | 1 | 103,113 | 8 | 206,227 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day you wanted to read something, so you went to your bookshelf to grab some book. But when you saw how messy the bookshelf was you decided to clean it up first.
<image>
There are n books standing in a row on the shelf, the i-th book has color a_i.
You'd like to rearrange the books to make the shelf look beautiful. The shelf is considered beautiful if all books of the same color are next to each other.
In one operation you can take one book from any position on the shelf and move it to the right end of the shelf.
What is the minimum number of operations you need to make the shelf beautiful?
Input
The first line contains one integer n (1 ≤ n ≤ 5 ⋅ 10^5) — the number of books.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — the book colors.
Output
Output the minimum number of operations to make the shelf beautiful.
Examples
Input
5
1 2 2 1 3
Output
2
Input
5
1 2 2 1 1
Output
1
Note
In the first example, we have the bookshelf [1, 2, 2, 1, 3] and can, for example:
1. take a book on position 4 and move to the right end: we'll get [1, 2, 2, 3, 1];
2. take a book on position 1 and move to the right end: we'll get [2, 2, 3, 1, 1].
In the second example, we can move the first book to the end of the bookshelf and get [2,2,1,1,1]. | instruction | 0 | 103,114 | 8 | 206,228 |
Tags: data structures, dp, greedy
Correct Solution:
```
from collections import defaultdict
from itertools import accumulate
import sys
input = sys.stdin.readline
'''
for CASES in range(int(input())):
n, m = map(int, input().split())
n = int(input())
A = list(map(int, input().split()))
S = input().strip()
sys.stdout.write(" ".join(map(str,ANS))+"\n")
'''
inf = 100000000000000000 # 1e17
mod = 998244353
n = int(input())
A = list(map(int, input().split()))
# print(*A)
l=[-1]*(n+1)
r=[-1]*(n+1)
F=[0]*(n+1)
cnt=[0]*(n+1)
for num,a in enumerate(A):
if l[a]==-1:
l[a]=num
r[a]=num
for i in range(n-1,-1,-1):
F[i]=F[i+1]
nowcol=A[i]
cnt[nowcol]+=1
if i==l[nowcol]:
F[i]=max(F[i],F[r[nowcol]+1]+cnt[nowcol])
else:
F[i]=max(F[i],cnt[nowcol])
print(n-F[0])
``` | output | 1 | 103,114 | 8 | 206,229 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day you wanted to read something, so you went to your bookshelf to grab some book. But when you saw how messy the bookshelf was you decided to clean it up first.
<image>
There are n books standing in a row on the shelf, the i-th book has color a_i.
You'd like to rearrange the books to make the shelf look beautiful. The shelf is considered beautiful if all books of the same color are next to each other.
In one operation you can take one book from any position on the shelf and move it to the right end of the shelf.
What is the minimum number of operations you need to make the shelf beautiful?
Input
The first line contains one integer n (1 ≤ n ≤ 5 ⋅ 10^5) — the number of books.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — the book colors.
Output
Output the minimum number of operations to make the shelf beautiful.
Examples
Input
5
1 2 2 1 3
Output
2
Input
5
1 2 2 1 1
Output
1
Note
In the first example, we have the bookshelf [1, 2, 2, 1, 3] and can, for example:
1. take a book on position 4 and move to the right end: we'll get [1, 2, 2, 3, 1];
2. take a book on position 1 and move to the right end: we'll get [2, 2, 3, 1, 1].
In the second example, we can move the first book to the end of the bookshelf and get [2,2,1,1,1]. | instruction | 0 | 103,115 | 8 | 206,230 |
Tags: data structures, dp, greedy
Correct Solution:
```
import sys
from sys import stdin
n = int(stdin.readline())
a = list(map(int,stdin.readline().split()))
clis = [ [] for i in range(n+1) ]
for i in range(n):
clis[a[i]].append(i)
dp = [0] * (n+1)
app = [0] * (n+1)
for i in range(n-1,-1,-1):
app[a[i]] += 1
dp[i] = max(dp[i] , dp[i+1], app[a[i]])
if clis[a[i]][0] == i:
dp[i] = max(dp[i] , dp[clis[a[i]][-1] + 1] + len(clis[a[i]]))
print (n - dp[0])
``` | output | 1 | 103,115 | 8 | 206,231 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day you wanted to read something, so you went to your bookshelf to grab some book. But when you saw how messy the bookshelf was you decided to clean it up first.
<image>
There are n books standing in a row on the shelf, the i-th book has color a_i.
You'd like to rearrange the books to make the shelf look beautiful. The shelf is considered beautiful if all books of the same color are next to each other.
In one operation you can take one book from any position on the shelf and move it to the right end of the shelf.
What is the minimum number of operations you need to make the shelf beautiful?
Input
The first line contains one integer n (1 ≤ n ≤ 5 ⋅ 10^5) — the number of books.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — the book colors.
Output
Output the minimum number of operations to make the shelf beautiful.
Examples
Input
5
1 2 2 1 3
Output
2
Input
5
1 2 2 1 1
Output
1
Note
In the first example, we have the bookshelf [1, 2, 2, 1, 3] and can, for example:
1. take a book on position 4 and move to the right end: we'll get [1, 2, 2, 3, 1];
2. take a book on position 1 and move to the right end: we'll get [2, 2, 3, 1, 1].
In the second example, we can move the first book to the end of the bookshelf and get [2,2,1,1,1].
Submitted Solution:
```
# From Tutorial
from collections import defaultdict, Counter
from sys import stdin
input = stdin.readline
n = int(input())
a = list(map(int, input().split()))
fq = Counter()
fl = defaultdict(lambda: [n+1, -1])
for i, el in enumerate(a):
fl[el][0] = min(fl[el][0], i)
fl[el][1] = i
fq[el] += 1
cnt = Counter()
dp = [0] * (n+1)
for i in range(n-1, -1, -1):
el = a[i]
cnt[el] += 1
dp[i] = dp[i+1]
if i == fl[el][0]:
dp[i] = max(dp[i], dp[fl[el][1] + 1] + fq[el])
else:
dp[i] = max(dp[i], cnt[el])
print(max(n - dp[0], 0))
``` | instruction | 0 | 103,116 | 8 | 206,232 |
Yes | output | 1 | 103,116 | 8 | 206,233 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day you wanted to read something, so you went to your bookshelf to grab some book. But when you saw how messy the bookshelf was you decided to clean it up first.
<image>
There are n books standing in a row on the shelf, the i-th book has color a_i.
You'd like to rearrange the books to make the shelf look beautiful. The shelf is considered beautiful if all books of the same color are next to each other.
In one operation you can take one book from any position on the shelf and move it to the right end of the shelf.
What is the minimum number of operations you need to make the shelf beautiful?
Input
The first line contains one integer n (1 ≤ n ≤ 5 ⋅ 10^5) — the number of books.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — the book colors.
Output
Output the minimum number of operations to make the shelf beautiful.
Examples
Input
5
1 2 2 1 3
Output
2
Input
5
1 2 2 1 1
Output
1
Note
In the first example, we have the bookshelf [1, 2, 2, 1, 3] and can, for example:
1. take a book on position 4 and move to the right end: we'll get [1, 2, 2, 3, 1];
2. take a book on position 1 and move to the right end: we'll get [2, 2, 3, 1, 1].
In the second example, we can move the first book to the end of the bookshelf and get [2,2,1,1,1].
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline()
# ------------------------------
def RL(): return map(int, sys.stdin.readline().split())
def RLL(): return list(map(int, sys.stdin.readline().split()))
def N(): return int(input())
def S(): return input().strip()
def print_list(l):
print(' '.join(map(str, l)))
# sys.setrecursionlimit(100000)
# import random
# from functools import reduce
# from functools import lru_cache
# from heapq import *
# from collections import deque as dq
# import math
# import bisect as bs
# from collections import Counter
from collections import defaultdict as dc
n = N()
a = RLL()
count = [0] * (n + 1)
first = {}
last = {}
for i in range(n):
if a[i] not in first:
first[a[i]] = i
last[a[i]] = i
dp = [0] * (n + 1)
for i in range(n - 1, -1, -1):
count[a[i]] += 1
tail = dp[last[a[i]] + 1] if i == first[a[i]] else 0
dp[i] = max(dp[i + 1], count[a[i]] + tail)
print(n - dp[0])
``` | instruction | 0 | 103,117 | 8 | 206,234 |
Yes | output | 1 | 103,117 | 8 | 206,235 |
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